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an Surveying, Higher Surveying, Mine Surveying, Hydrographic
urv ylng, Topographic Surveying, Astronomy, Simple Curves, Compound Curves,
plral Curves, Reversed Curves, Parabolic Curves, Sight Distance, Earthworks,
ass Diagram, Highway Engineering, Transportation Engineering, Traffic Engineering.
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.>
DETI~
for CIVIL and
LieeQSure Exam
Copyn~;t 1987
Venancio I. Besavilla, Jr.
(BSCE, MSME, AS, F. (PICE)
-CIT (2nd Place) - August, 1969
Civil Engineer
. Geodetic Engineer - CIT (7th Place) - July, 1966
Former Instructor: Cebu Institute of Technology
Former Instructor: University of the Visayas
Former Chairman: CE Dept. University of the Visayas
Dean: College ofEngineering and Architecture, University of the Visayas
Awardee: As an Outstanding Educator from the Phil.
Veterans Legion on May 1984
Awardee: As Outstanding Alumnus in the Field of Education
from CIT Alumni Association, Inc., Marcil 1990
Awardee: As Outstanding Engineering Educator from the
CIT High School Alumni Association, December 1991
Member: Geo-Institute of the American Society of Civil Engineers (ASCE)
Member: Structural Engineering Institute (ASCE)
Member: American Concrete Institute (ACI)
(Membership No. 104553)
Member: American Society of Civil Engineers (ASCE)
(Membership No. 346960)
Member: PICE Delegation to the ASCE (Minneapolis, Minnesota, USA) (Oct. 1997)
Head PICE Delegation to VFCEA, Hanoi, Vietnam (April 2009)
Head PICE Delegation to JSCE, Fokuka Japan (September 2009)
Head PICE Delegation to A SCE.~ Kansas City, USA (October 2009)
Director: PICE National Board (1997-2008)
Director: PICE Cebu Chapter 1991- 2008
~lce President: PICE Cebu Chapter 2006, 2008
President: PICE Cebu Chapter 2009
Vice President: PICE National Board 2009
President: PICE National Board 2009
Chairman International Committee (PICE) Busan, Korea (February 2010)
President: Cebu Institute ofTechnology Alumni Association (20OJ-up to the present)
ISBN 971- 8510-11-7
Available at:
BISAVlllA
Engineering Review Center
CEBU
DAVAO
2nd Floor, Pilar-Bldg.
4th Floor,
Cor. Osmena Blvd.
Porras Bldg.
& Sanciangko Sts.
Magallanes St.,
Cebu City
Davao City
Tel. No. (032) 255-5153 Tel. No. (082) 222-3305
BAGUIO
Lujean Chalet Bldg.
Lourdes Grotto
Dominican Road
No. 68 San Roque SI. Baguio City
Tel. Nos. (074) 445-5918
CAGAYAN DE ORO
3rd Floor,
Ecology Bank Bldg.
Tiano Bros. St.
Cagayan De oro City
Tel. (08822) 723-167(Samsung)
TACLOBAN
Door 11.-303, F. Mendoza
Commercial Complex
141 Sto. Nino Street
Tacloban City
Tel. No. (053) 325-3706
MANILA
2nd Floor, Concepcion
Villaroman Bldg., P. Campa
St., Sampaloc Melro Manila
Tel. (02) 736-0966
GENERAL SANTOS
RD. Rivera Bldg.
Constar Lodge, Pioneer Ave.
General Santos City
Tel. No. (083) 301-0987
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SURVEYING
for
CIVIL and GEODETIC
Licensure Exam
Copyright 1984 by Venancio I. Besavilla, Jr. All Rights
Reserved. No part of this publication may be
reproduced, stored in a retrieval system, or transmitted,
in any form or by any means, electronic, mechanical,
photocopying, recording, or otherwise, without the prior
written permission of the publisher.
ISBN 971· 8510·11.7 •
•
~,.. ~;:,_;_~_,m.,
Punta Princesa, Cebu City. Tel. 272-2813
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"~..'III.. I~ •• I~~ ~ '•• ~"I~I~XTS ~
@
X~'''~~"''{i:'t'¥m"",
0%l'0fu~!;·*kW
~" ~W<fm_~" """it"
DESCRIPTION OF TOPICS
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
31.
32.
0«
W::~ill.Th,.
~~"
PAGE NO.
Tape Correction ------------------------------ S- 1 - S- 12
Errors and Mistakes -------------------------- S- 13 - S- 129
Leve Ii ng ------------------------.- -,- ------------- S- 30 - S- 54
Compass SUNeying ------------------------- S- 55 - S- 66
Errors in Transit Work -------------------------- S~ 67 - S- 83
Triangulation -----------------:------------------ S- 84 - S-108
Spherical Excess ------------~------------------ S-1 08 - S-111
Areq of Closed Traverse -----------------"- S-112 - S-130
Missing Data ------------------------------------ S-130 - S-144
Subdivision -------------------------------------- S-144 - S-172
Straightening of Irregular Boundaries -- S-172 - S-175
Areas of Irregular Boundaries ------------ S-176 - S-178
Plane Table ------------------------------------- S-179 - S-182
Topographic SUNey ------------------------- S-183 - S-187
Route SUNeying ------------------------------S-188
Stadia SUNeying ------------------------------ S-189 ,. S-194
Hydrographic SUNeying -----"-------------- S-195 - S-208
Three Point Problem ------------------------- S-209 - S-213 .
Mine SUNeying ------------------------------.- S-214 - S-222
Practical Astronomy ------------------------- S-223 - S-251
Simple CUNes ---------------------------------- S-252 - S-292
Compound CUNes --------------------------. S-292 - S-318
Reversed CUNes - '--------.----------------- S-318 - S-333
Parabolic CUNes --- -------------- ------------ S-333 ~ S-361
Sight Distance ---------------------------------- S-361 - S-376
Head Lamp Sight Distance ---------------- S-376 - S-386
Sight Distance ----- ---------------- --- -- ------- S-387 - S-388
Reversed Vertical Parabolic CUNe ---- S-389 - S-390
Spiral CuNe----"-------------------------------- S-391 - S-404
Earthworks -- -- --- ---- ------ ---------------------- S-405 - S-430
Transportation Engineering -------.-------- S-431 - S-523
Miscellaneous ----- ------ ---------- ---- ---- ---- S-524 - S-542
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TAPE CORRECTION
I-~~-~~----'
--~~-
i
2.
Tape not standard length
Imperfect alinement of tape
Tape not horizontal
Tape notstretch straight
Imperfection of observation
Variations in temperature
Variations in tension
1.
2.
3.
4.
5.
6.
7.
Pull Correction:
(To be added or sUbtracted)
P2 = actual pull dUring measurement
P1 = applied pull when the length of
tape is L1
A = Cross-sectional area of tape
E = Modulus of elasticity of tape
I~~------------'-~-
3.
I
1.
2.
3.
4.
Adding or dropping a full tape length.
Adding a em., usually in measuring the
fractional part of tape length at the end
of the line.
Recording numbers incorrectly,
example 78 is read as 87.
Reading wrong meter mark.
w= weight of tape in pit. or kg.m.
L = unsupported length of tape
p = actual pUll or tension applied
4.
1.
Sag Correction:
(To be subtracted only)
Slope Correction:
(To be subtracted only)
Temperature Correction: (To be
added or subtracted)
K =000000645 ft. per degree F.
K =0.0000116 m. per degree C.
T1 =temp. when the length of tape is L1
T2 =temp. dUring measurement
H=S-Cs
H = horizontal distance or corrected
distance
S = inclined distance
h = difference in elevation at the end of
the tape
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5-2
TAPE CORRECTION
5.
Sea Level COffection:
Reduction factor=
1.~
The 3:4:5 Method:
1.
To erect a perpendicular to the line
AB, from a given point C a point a on line
AB is assumed to be on the
perpendicular and a pin is set at a. With
sides a multiple of 3, 4 and 5 m. such as
24.32 and 40 m., a right triangle abc is
constructed as follows: A pin is set on
line AB at b, 32 m. from a. The zero end
of (he tape is fixed with a pin at a, and
the 100 m. end at b. The zero end of the
tape is fixed with a pin at a, and the 100
m. end at b. The head chainman moves
to c and holds the 24 m. and the 60 m.
marks of the tape in one hand, with the
tape between these marks laid out so as
to avoid kinking. He then sets a pin at c.
The rear chairman moves from a to b as
necessary to check the position of the
tape at these points as c is established.
He then sights along ae to C' beside C,
usually Cto the a(::' is measu"red, and the
foot a of the perpendicular aC' is moved
along the line AB by an equal amount, to
the point a'. If the trial perpendicular ae'
fails to include the point C by several
feet, the process is repeated for a', the
new point, otherwise the location of a'
may be assumed as correct.
B = horizontal distance Gorrected for
temperature, sag and pUll.
B' = sealevel distance
h = average altitude or observation
R = Radius.of curvature
6.
Normal Tension:
It is the tension which is applied to
a t.ape supported over two supports
which balances the cofrec!ion due to
pull and due to sag. The application of
the tensile force increases the length of
the tape whereas the sag decreases its
length, the normal tension neutralizes
both corrections, therefore no correction
is necessary.
P = applied 'normal tension
P1 = tension at which the tape is
standardized
W = total weight of tape
A = cross-sectional area of tape
E = modulus of elasticity of tape
2.
The Chord Bisection Method:
a) Add correction when measuring distances
b) Subtract correction when laying out
distances
o
a) Subtract correction when measuring
distances
b) Add correction when laying out distances
To erect a perpendicular to the line
AB; from a given point C, the position of
the perpendicular is estir,lated, and a pin
IS set at d on this estimated
perpendicular, somewhat less than one
tape length from the line AB. With d as
center and the length of tape as radius,
the head chainman describe the arc ED
of a circle, setting pins at the
intersections band c of the arc with the
line AB. The rear chainman stationed at
A or B determines the location of the
intersections band c on line. The point a
is established midway between band c.
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TAPE CORRECTION
The line ad is prolonged to C' besides C,
and the point a is moved if necessary as
described for the 3:4:5 method.
2.
ParaJle/lines:
A'i ..·
··
~B'
c c'
r: y!
i *f,
ii
A
'~~~~,
A.---.o-o-~-...."
a'a
b
3:4:5 Method
B
Chord Bisection Method
1----"""'8
0-'" - - - -
If the necessary distance from the
line AS is short, perpendicular AA' = BB'
are erected by either using 3:4:5 method
of the chord bi-section method to clear
the obstacle. The line A'B' is then
chained, and its length is taken as that of
AB.
3.
Similar Triang/es:
C
A /
1.
..
~.<:~:>~~'" •
Let C be a point from A and Bare
visible. AC and BC are measured. CD
and CE bears to CB: that is CD/CA =
CEICB. It will generally be convenient to
make this a simple ratio such as 1/2 of
1/3. The triangles ACB and DCE are
similar. DE is measured and AS is
computed.
Swing offsets:
At'/
To find the distance AS by the sWing
offset method, the head chainman
attached the end of the tape to one end
of the line as at B and describes an arc
with center S and radius 100 m. The
rear chainman stationed at A lines in the
endiof the tape with some distant object
as 0 and directs the setting of pins a1
points a and b where the end of the tape
crossed line AO. A point C midway
between a b lies on the perpendicular
CB. A pin is set at C, and the distances
BC and CA are measured to obtain the
necessary data for computing the length
of AS,
.~i.tl'"iI
rll~anteTperature<()t~5·9·,.a@us$djl)l)f
stan9a~l~ngth.1lt20·QunderaP4IIAf§~g,
Cr9$csezliQnClI"are<l.of,l~pe.,is • • Q.Q~~q;pm;,.
CQefficfe!1t()ltheyrMex:e~n$iM>I~
Q.009011~WC,.M~dtllu$ • ofela$tic~Y.(jf,tllP~ls,
2x1mkg1Crl1 2.
' , ",
[)eterllline.the•• errot•• of .the.,tapedu#.to
changeintempera1ure. ,. ' ",' '.
'
® DetElfrninelheerrprduetotension,
••'.','
® Oetenninethe correetedlengthofthe line.
G)
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8-4
TAPE CORRECTION
Solution:
G)
Temp. correction:
Ct =KL (T - Ts)
C, =0.0000116 (2395.25}(35 - 20)
Ct = + 0.4168 m.
@
Tension correction:
(P-Ps ) L
Cp=---xE
@
c - (4 - 5)(2395.25)
.A·••$Q.·.rIl·•• • m~EH • • f~p~WaW.$t@9a.rdlzed • ·and
0.03 (2) 106
Cp =• 0.0399 m.
·1QlJJldJ9b{l;.Q.O()~Q5nl. long~r~t·an .• pb~ery~d
Corrected length:
L =2395.25 + 0.4168 - 0.0399
L = 2395.6269 m.
Wfl$f~~'nd • f~b$ .• 66Z.702.% • at • an•• ~v~mge
·~~Beralljr~ • • 8f•• 24·B·9.@ingJh~ • • s~me· • pull,
sj~ppporteq1:ttrougholJttts
p-
@
True length of the line:
Corrected hor. distance =673.92 + 0.1348
Corrected hor. distance =674.055 m.
•
• • wbole •. lergth.·.~nd
t~mp~rliIW~(ff~1.~·C~Qd~pulh~flON~~,
Thl$taReW<1l.SlJ$edlomea~ure a• liqe•• v.'hjCh
•
Y$e.·C()~fflci$nt.pf.e~Bansion.·pf.0.()Ooo116·m·
pl:ifgegreeCE!ntigiage~<
(j)..¢oIl1IM~lhesl!lnd?~.terilp· • • • • • ·
~· • ··G9IllP~~me.t()!al • telTlp>Wq'El(;tiOn-
®
A$QrmtAA~W~~$t~rm@l4eg.MdW!l$f@hd.
l$bep:Q<l42m:IQO)?n~>tf1~nJhe$t<1ln~~rd
.lE!gg1:tt~t~r:Jop~ry@.t£!rnP~r~fw~6fR~~Q·{:l[l9
CQmPlJf~the.{icllTE!CflenglhJ:iflh~·IiI1~;·
Solution:
CD Standard temperature:
• ~~~t~~~~~iri~r~~~~~aW~~~e~td·
Cr=K(Tz- Ts)L
• 19~~7~.~4.m···long.M.~n.(l~S7!Ye<J.·~~mp:.Of
+0.00205 =0.0000116(31.8- Ts)(50)
.~~~[t·:~~~~~lIfs°6;~.~~~j~··idi~ji:~~I .• o.f
31.8 - Ts = 3.53
qj\P~I~!QElme$!Md~Jem@@tt!rEf;
Ts = 28.27'C (standard temp.)
··~.··~~{~I~~m:I'~I'~ihe.lioe ..
.
@
Cr =K(T2 - Ts ) L
Cr =0.0000116(24.6 - 28.27)(50)
Cr =;_0.00213 (too short)
Solution:
G)
@
Total correction:
Standard temperature:
Cr = K (TZ - T1) L1
+ 0.0042 =0..0000116 (58 - T1}(50)
+ 0.0042 =0.03364 - 0.00058 T1
T1 = 50.76'C (standard temp.)
· 0.00213(662.702)
7iataI correctIOn =
50
Total correction =0.02823 m.
Total correction:
Cr=K(T- T1)L1
@ Corrected length of line:
Cr =0.00(J01Q (68' 50.76)(50)
Corrected horizontal distance
Cr =0.01 (tape is too long)
b.;;; G 1J
=662.702 - 0.02823
Ttl
t·
673.92 (0.01)
,oa correc Ion =
50
~D
=662.67377 m.
"" G .-Total correction = 0.1348 m.
\...0
::co,O I
;:r~,"1'2.\
~
I') \
\-:1,t.\&llv Db
-)
.
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TAPE CORRECTION
@
Tota/error:
Too shortby =30 - 29.992
Too short by =0.008 m
- 472.90 (0.008)
7iot'aJ error30
Total error = 0.126 m. (to be subtraded)
@
True horizontal distance:
Corrected inclined distance
=472.90 •0.126
=472.774m
.e
Solution:
h
= 472.774
h =472.774 (0.03)
h= 14.183m
. for
(14.183f
Correction slope = 2 (472.774)
Sin
CD Actual length:
Cr = K (T2 - T1) L1
Cr =0.0000116(5·20)(30.005)
Cr=· 0.00522 mm
(P2· P1) L1
CpAE
C - (75· 5OX3O.005)
3(200 x 103)
Cp = tQ.00125 m
p-
Correction for slope =0.213 m
Corrected horizontal distance
=472.774·0.213
= 472.561 m.
w2L3
CS=24P2
0.65 x9.81 \'
w=.30":' .
C «(0.65 x 9.8115(30)3
S - . (3O)t(24)(75t
Cs =- 0.00904 m
Total correction =·0.00522 + 0.00125
·0.00904
- Tata/ Cf)ffection=- 0.009G4Total correction =·0.00522 + 0.00125
,
• 4IJling
mAA~~rerrEl9t. • '> • • • i • • • • • • • • • • • • • • • •>•• • • • •>./.
® Wl'!litlsttletfuEl·atilll' .••.'••'. •. •••••.. '".< • • < . . .• .
-0.00904
Total correction =·0.013 m
.(!)... (;q~PIJtEl~.~¢@Illeng{h9f.tcl~
Actual length of tape during measurement
@. ·Whab$lh$¢~&ln~rea.ln$q.m.<········<·······
=30.005·0.013
·=29.992m.
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S-6
TAPE CORRECTION
Solution:
CD Actual length:
Cr = K(T2 - Tl) Ll
Cr =0.‫סס‬OO16(30 - 2OX30·002)
Cr =0.00348 m (too long)
Actual length of tape during measurement
=30.002 + 0.00348
=30.00548 m.
® True area:
Therefore the tape is 0.00548 mtoo long
Forthe 144.95mside:
=0.26 m.
7iotal error =144.95(0.00548)
30
True length = 144.95 + 0.026
True length = 144.976 m
For the 113 m side:
Total error
=113(0~) =0.021
True length =113 + 0.021
True length = 113.021 m
True area =(144.976X113.021)
True area = 16,385.33 ~
® Error in area:
Erroneous area = (144.95)(113)
Erroneous area::: 16,379.35 m2
Error in area = 16,385.33 -16,379.35
Error in area = 5.982 ~
Solution:
CD Cross-sectional area:
A(1OOX100X7.86) = 267
1000
.
A = 0.034 sq.m.
W1"f1:
® Total correction:
Cr=K(T2' Tl)L
jt1
Cr= 7 x 10-7 (20 • 15X1,l:1J
Cr =0.00035 m.
Pull correction:
. (P2- P1)L
Cp= AE
C - (16 ·10)(100)
p - 0.034 (2) 106
Cp= 0.009 m
Correction =0.00035 + 0.009
Correction = 0.00935 m
No. of tape lengths =~~6 =4.306
Total correction =4.306 (0.00935)
Total correction =0.0403 m
® True length of baseline:
True length of baseline =430.60 + 0.0403
True length ofbaseline =430.6403 m
A~pm.~tEleIJap~W~jghl~gA·f$~}$@f
~tMdatQI~rglll.Ulld~r~ • pUII.~f?@~UPP9rt~·
forfull.. . I~l"\gth, • • .• • TMta~ • • ..v~$U95~ . . . I~
m~asyHrt9.Allne~~~·§§m.IMg.l?~~·smWm
f~~elgW~n~~flder~$te~f1~ . P\,ltIQf19t<~;
AsslJming.I:.=.2•• l(.·.1()~ . k9tcmZ•• ~@ll1e • uriil
'Il¢ighl.m.$I~I.t()be7.9X1(t3.kglbrn3.
®•• ·.Oalermina.·me·crosssticllonal.atea·Ofthe
®~8IuleI16@~~@n,
®Compuletl'lEl1ruelElng!tiW1beb(lseli®•....
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TAPE CORRECTION
Solution:
<D Cross-sectional area of tape:
w=AL Ys
@
@
1.45 = A(3000)(7.9) x 10.3
A = 0.061 cml
Pull Correction:
(P- Psi L
Cp=-xE
_ (10-5X30)
Cp - 0.061 (2 x 106)
Cp =+ 0.00123
Total Correction = 0.00123 (938.55)
30
Total Correction = + 0.038 m.
Correct length ofline:
Corrected length = 938.55 + 0.038
Corrected length = 938.588 m.
Ill'll'• •••
@ •••. ·Per~rm(oe.·lIl$·hOrtzOO~I.dlStatt¢.,.
Solution:
~~~~jbg~~ •.•j~611~~~~.~h~/i~i~
<D Total correction per tape length:
Gr= K (T2· T1) L1
Cr= 0.0000116 (15 ·10)(30)
Gr= + 0.00174 rn
jsmown•• 19be.5(JnV.19@@?9·¢'Tl1~t~p~
wasusedtome<l$lJre<.~ljffll'Wtti¢l't\¥~s@YQg
l[).pe.532@.meter~l@g~~·tl1~Jelll~Wt~
W<ls.35·c, • • ~t€lITl1If\~.~.@IQ\Vil'l9:·>i.· .• · ·.••·•·•·•· · ..
0!•• • ternper~tureC9ttecti6rll¢nap~·l®$tti<> ••••••••
®TelJ1~raflJreCllrre¢ticmJprfh~fTlffll!l~rE:!9
WilfuL··································
@)1\C~edlength()ff&ljne...•
Solution:
<D Temperature correction per tape length:
Cr=K(T- Ts)L
Cr = 0.0000116 (35 -20)(50)
Cr =0.0087 m. too long
® Temperature correction for the measured
line:
ft· :-1···
,.
- 532.28 (0.0087) . r'\.!J)
loa correCdon 50
C !, -.- Total correction =+ 0.0926 m.
\. ""0
@
Corrected length ofline:
COrrected length =532.28 + 0.0926
Corrected length =532.3726 m.
.
PUll correction:
(P2 ,P1)L1
Cp
AE
_ (75 • 50)(30)
Cp- 6.50 (200 x 103)
Cp =+ 0.00058 rn
Sag Correction:
. .l-L 2
Cs= 24 p2
r
_ (0.075 x 9.81)2 (30)3
vs-
24 (75)2
Cs =0.10827 m
Total correction per tape length:
C =0.00174 + 0.00058 ·0.10827
C=· 0.10595 m
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S-8
TAPE CORRECTION
® Correction for slope:
.
459.20 (·0.10595)
Tolaf correction =
30
Total correction =- 1.622
® Total correction:
Temp. corrections:
Cr=KL(T- T1)
Cr =0.‫סס‬OO116 (624.95~32 -20)
Cr =.;. 0.087 (too long)
Correct slope distance =459.20 ·1.622
Correct slope distance =457.578 m
Pull correction:
(P2' P1) L
Cp = AE
_ (15 -10) (624.95)
Cp- 0.06 (2) 106
Cp := +0.026 (too long)
fl2
CS=2S
(1.25f
Cs =2 (457.578)
Cs = 0.002
@
Horizontal distance:
Corrected horizontal distance
=457.578· 0.002
= 457.576m
Total correction = + 0.0l7 • 1.781 +0.026
Total correction =• 1.668 m.
@
Il,.'.
Solution:
CD Sag correction:
w2 L3
- CS1 = 24 p2
CS1 =
(0.04j2(100)3
24 ~15)2 . =0.296 m.
_ (0.04) (24.95)3 =0 005
CS2'"
24 (15)2 . .
Total sag correction =6(0.296) + 0.005
Total sag correction = 1.781 (too short)
Corrected length of the liIle:
Corrected length =624.95 .1.668
Corrected length = 623.182 m.
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5-9
Pdfbooksforum.com
TAPE CORRECTION.
Actual length of tape at 40.6'C Tension
10 kg
99.986 - 0.00232 =99.98368
14 kg
99.992 - 0.00232 = 99.98968
100.003 -0.00232 =100.00068 20 kg
Solution:
CD Tension applied at 32'.
'4
99.992
0.008
{
100.00
0.011 x
{
'4
99.98968
6
0.01032
{
0.011 x
100.0000
100.003
~_9·008
{
6
100.00068
6 - 0.011
x=4.36 kg.
Tension applied =14 +~4.36'~·~·'~··
Tension applied = 18.36 kg.
x 0.01032
6=llO11
,) ~ .', ~:.,
x= 5.63 kg
',k
® Tension applied at 40.6'.
Temperature correction:
Cr=K(T2- T,}L
Tension applied =14 + 5.63
Tension applied = 19.63 kg
Cr= 0.0000116 (40.6 - 32) 100
Cr=O.00998
Actual length of tape at 40.6'C Tension
99.986 + 0.00998 = 99.99598
10 kg
99.992 + 0.00998 = 100.00198
14 kg
100.003 + 0.00998 = 100.01298 20k~
0.006
r·~98}
0.00402
100.0000
{'
100.00198
~_ 0.00402
4 - 0.006
x= 2.68 kg
Tension to be applied =10 + 2.68
Tension to be applied = 12.68 kg
4
14
GJ Compute the correctiartdue to the applied
pulfofS kg. ,'"
.'
@ Compute the cortecUOn due to weight.ot
tape.
@
@
Tension applied at 30'C:
Temperature coffection:
Cr=K(T2- T,}L
Cr =0.0000116 (30 - 32) 100
Cr=-0.00232
'
Compute the true length of the measured
lineAB ,due to the combined effects of
tension, sag arid temperature.
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S-10
TAPE CORRECTION
Solution:
@
CD Pull correction:
C - (P,- PM,
PAE
C - (8 -5.5X458.650) _ +
P - 0.04 (2.10 x 106) - 0.014 m.
@
No. ofpaces =893.25':.-:;.
® Distance of the new line:
.Distance of new line =893.25 (0.691)
Distance of new line =617.236 m.
Correction due to weight oftape:
W2L3
Cs = 24 rJ2
Number ofpaces for the new line:
AI
f
_ 893 =893.5 + 891 + 895.5
IVO. 0 paces 5
'I.
C = (o.05f (50j3 (9)
24 (8f
(0.05)2 (8.65)3
+
24(8f
Cs =·1.832 m. (always negative)
s
.@'.
1'hl$Sk!es•• of.a.M9a(elat••havIM•• ®J@a
·····>··Clf•• ?~!)pe9~rAAWfl~m¢-~~qt~.9$j.J19 • •~·
·······~~~~ih:.~~ • i~i~J~I.·m.~~~@W
® True length ofmeasured line AB:
Cr =K(T2 -T1)L
Cr =0.0000116 (18 - 2OX458.65)
Cr =-0.011 m.
··ttiE!•• (:Qrr1'!¢t.dl~ta6@ • ~~!lh@~ • P9.@•• ~··
'. · · • • ~P.4$.m' • • l@ng.~ • jQQm;mjll;l.lMl#,,#!
@
m,@~ltl~!~\th~l¢hSll1l~~~I~J(j96t/'l~
••
·••• ·• ·· ··>.Woyl1~.®()ut~.~.2@Q*:Itlm;
·.•
." . . . . . . .tl$ffi~
•. /••. . •. . . .
Total correction =0.014 -1.832 - 0.011
Total correction =-1.829 m.
True length AB =458.65 - 1.829
True length AB =456.821 m.
Ii'~i:$.·of.:x·?··.····.·····
@•• • T\lij.~jstlil'lCElfrOm.P.Il)ei@fJi~@~d;.j$·.
• • • • • • • • 1§$.2.rl1.•••• lf.tM.$Q.m;t#p~~eaj~Q~Q1 • m;.
tQP$lyjrt;Wh~!i$~#grf~W(ll$tM¢¢m
rie?
..
... ... .... ...
Solution:
~~~I~~~!~~n~~~~i~I~~at~jl~_\\~~
.~~5f:o~~~~~~~Ii~~~~"'1~11;1~~l~'~.·
..' .. .
.
CD Error in area:
(99.962 _ (1002)
A - 2.25
A=2.2482 hectares
Enur in area =2.25 - 2.2482~.l\
Errorin area =0.0018 hectares,
Error in area = 0.0018 x10000 ~\
Error in area = 18 sq.m.
6~3,893,5dI9talj(18~S;&/<'"
@-p~lerl11IMltlE!pa¢¢taClor+<
~·< •. [)~t~rmjlie· • I'l~mbefgf.p~c~s.·fCll'~~¢-,,·new
•
une,>
~.·• •. Oetennineme·pjfltanc:eMlbe.fleWljne.
Note: 1 hectare =1000 sq.m.
@
Value of x:
~~5 x =220.45 - 220.406
Solution:
CD Pace factor:
142 + 145 + 145.5 + 146
5
k,t'::'l:1,
No. of paces = 144.625" " '
100
1:.; L -:.
Pace factor =144.625 = 0.691
•
x =0.02 m.
N0.0 f paces =
® Corrected distance:
/.
Correct distance = 165.2 _1:.2 (0.01)
Correct distance
=165.167 m.
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TAPE CORRECTION
@
\i)•• • 8()rnpqte.the.O()rrnlJl•• t~~$IQI1.wh~hwin • b~
·. ·.• •.
....•. ~ppU®t~~t~®SlJPPA.~~9pV~r.twq
. . ·Sl.IPP()rtsillord~tt()fl1aketr~tape~~alt£>
·.it$•. n()[llill~I.I~ngtb •.VjI'l~ll~IlPP()rtE!d.ptJly'~t
•.
~~~~.~f~~~ ~~4$t~I • _1~ ~.~6~~~
Unit weight of tape:
_ 0.204 w...fAE
PNPN- Ps
-J
.•
•.•. t@)~9hourit$ • leo91h.und~r.~.ll\at\!j~rd·PW
· • •pt$·§.·kg•• wilh.•!J'tEl•. rnod~N~.9felastiglyfS
.• ·~ • x.1~kgf(@fW"ldaieaOf.Q,06(:mf • • • • • •.• • ·•• •·
16 =0.204 w'./'--0.-05-(2-)1-06'
w=
=0.026 kglm
® Cross sectional area:
_ 0.204 w...fAE
PN-J PN-Ps
·®•• • Aste~tapei$30:lt1 .•• ~ogu@era • ~nd~@.·
•.• • • • • plJu • • ~f.9 • • • k~, • • • With•• • ~ • • • %J~~I~nt • • ett)$lh.
·(:.lilli'~_'1
18 =
O.204[~O.OO25)(40i...fAE
~
me¢M·P9IIlt~~ • tl)9~ls.W~.~ff®!·.()f$a~Wlll·
W=279.02
•.••.••.. ·.M.~I@i~W~pYlffil~l()l'lg~~9f;l§flIW¥lP~.
AE =77854.67
A = 77854.67 = 0039 rrf
2 x 106
.
c
d~Mo.tI'lE!M~lj¢atl@Qft~I$IMdi~¢M~1
••.t6.t6.M....(feterlnio~lhe9~il·W~i9htcitt~e.
••
iaPE!•• j~
~*lQ~Mtcnt.>
taM·.MQ(I~ltl~ ()f•• elast@fy.Qf••
®Urid$ra$tandardpullQf~~g,Jhel@el
tapej~4QlTl"()(lg,An9rm~IJ~n~i9Q:qf
1&.k9rll~~~s •
~I()ng~~qiJ9f.~h~ • • l~Il$··
··()ff$e;t.t~~en~pt.pf.·~~·.· • • lfm~t~w(#Stm··
0.:
~
w=0.784 kg
m¢••
llqZ#·kgfril,•• •arld.E • =••z•)(•• j06•• • kglciril%)
Qeletrlline .• • it~ • • cr(Jss • • $ecll()nlll.··ar~.· • . in.
s:(P::Il'l{
Solution:
CD Normal tension:
PN = 0.204
{Ai!
..,j Pw Pt"
_ 0.204(0.84) ..,j'-O.-OO-(2-)1-06'
PN-
..,j Pw P1
_ 59.3608
PN-
..,j PN - 5.6
By trial and error:
PN= 17.33 kg
_ 59.3608
PN-
..,j 17.33 - 5.6
PN= 17.33 kg
.0) ..• Det~rminelli~ IEltlgll1 of the fine in meters if
... .there were 3 tallies'S phis aildthe last pin
\ was9>·Iil,Jromtheend of tile Hne. The
...•..• tapeMectwas so. m.IOhg. .• .
.
lID A line was measured wilh a50 m. tape
and fo~nd .to beJOO m. long. It was
. <Jis(:oileredlhaltneflrst pin was stUck
.... 306m. tathe left oftheUne.andthesecond
. . pin 30cm. to .the right. Fino the error In the
measurement in em?
® A line was measured with a 50 m. tape
and recorded ·100 m. long. While
measuring lhe first pin was stuck 20 em to .
the right of the line and the second pin
40 em. 10 the left. Find the correct length
oflhe line.
Solution:
(j)
Length of the line:
L = 3(10)(50) +8(50) + 9
L =1.909 m
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S-12
TAPE CORRECTION
@
Total distance = 8 (16.5) t 6 (16.5)
Error in the measurement:
If
45(33)
Error=2s
+ 12
Total distance =354.75 ft.
Total distance = 108.16 m.
E
(0.30f (0.60)2
-rror = 2(50) + 2(50)
Error =0.0045 m. =0.45 em.
@
Correct length of the line:
E
- (0.20f (0.60)2
rror - 2(50) + 2(50)
Error = 0.004 m.
Correct length ofline = 100 - 0.004
Correct length of/ine = 99.996 m.
@
.·A.. 1M • • rrl· • • !~pe • • is•• • 1~.·mrT!·.wide.<lhd
O.80mm·.th~ •.• lf.the·tapl.j$·90rr~unq~r·
~ • PlJIl()f.94·.N;.Wll'tputEl.tIJ~~rJ'9r.ll'Iadl'by
··uSlngapuII.Qf.eaN,••·.e.::;:.. 4®,OOQ.MP!i.· . .
@·.1'lWIElr1Qlh9f••~• • ~~ries • of.Une5.iSf()@gll)
. • • bE!34.2f·Q2mdDWef9~rq.directlOn.~nd
·•342f.~·rn·.ln·1~e.reverseddire(;t~n, • •. VV/'lal
@A Ilnewasmeasure<l tC)ha~5 tallies; 6
• marking pins lind ~3.5Ijnks. How long is
thelme In ft.?' .... .
@
.
A line wasmeasiJred witha 50 m. lape.
There were 2 .tallles. 8 pirys, and the.
distance from the last pin iothe elid of the
line was 2.25 nf. Find the length of the
lirie in meters?
.
A distance was Measured and was
recorded to.. tlave a value equiYalentio
8 perch,6 rods and 45 Yarn. Compute the
.•• tolal distance inmeters. .
..
@
i~lh~ratJopftlj¢l:lrror?···
@.·.. A~ytJsterlS~ • p~r • il)m®nt~~<lt • ~ • • ~rtair
.... ·9i$~n~frpmjlj~jnstru~~t~lldthean~l~
... ·~~~t~ndeq.l:!Yff@.bal'.i$p·$".9:JrnP~t~t~e
. . .•. ~r\~8n!~! • . d~tM~~.·.frpmtr.~ • j~strlJrr~nl
.. sl1ltl(')flIQ OOlloAA«"nmthesl,lblen$f:l par,
Solution:
CD Error made by using a pull of 68 N:
_ (P2 - P1) L_ (68- 54) 100
Cp AE
-12 (0.8)(200000)
Cp =0.0007
Error = 0.0007 x 100
100
Error = 0.0007%
Solution:
CD Distance ofline:
Note: 1 tally = 10 pins
1 link =1 ft
1 pin = 100 links
@
Length of the line:
Note: 1 tally =10 pins
1 pin = 1 chain
2 (10)(50) + 8 (50) + 2.25 =1402.25 m.
(3)
Total distance:
Note: 1 perch =1 rod =16.5 feet
1 vara =33 inches
Ratio ofthe error:
A
I gth 3427.fJ2 + 3427.84
verage en =
2
Average length = 3427.73
3427.84 - 3427.62
. f
Ratlo a error=
3427.73
L = 5 (10)(160) + 6 (100) + 63.5
L =5663.5 ft
@
.
0.22
1
· f
Rat10 a error =3427.73 =15581
@
Horizontal distance:
1
tan 0.2' =/1
H = 1,718.87 m.
Note: Subtense bar is standard
to be 2 m. long
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ERRORS AND MISTAKES
Gli2D
2.
Probable Error of the Mean:
is defined as the difference between the
true value and the measured value of a
quantity.
E
Em=-
MISTAKES
are inaccuracies in measurements
which occur because some aspect of a
surveying operation is performed by the
Geodetic Engineer with carelessness,
poor judgment and improper execution.
1.
2.
Systematic Error
Accidental Error
1.
2.
3.
Instrumental Error
Natural Error
Personal Error
-r;,
1"1"
3.
Standard deviation:
4.
Standard error:
1.
The weights are inversely proportional
to the square of the corresponding
probabl errors.
2.
The weights are also proportional to the
number of observations.
3.
Errors are directly proportional to the
square roots of distances.
~
is define as the number of times
something will probably occur over the
range of possible occurrences.
1.
Probable Error a
single observation:
Where E =probable error
,£V2 = sum of the squares of the
residuals
n =number of observation
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S-14
ERRORS AND MISTAKES
® Most probable value of diff. in elevation:
Route
1
2
3
4
®Whaflsttleweightnfr(lqte2a$$~mI6g
. • •. . .• • ·.•. ~ig~t·.¢f·.·.l'O.Ut~ • 1•.• I~.~ql¥l.I§ •.1.•·..•. • . . ·.. .•.•. . . . . .·.•• . ·
.• ~• • Detem1ine1hemos1~b'eValueof.diff.
jnel~'ffltl$l'I.<
• 'W•. • !fW~.·~I£tlfatio/' • 9fa~h • ls.~@,4f.m;'M\at
••..• ·.isth~elev~TIWL(lf~M2asslJrnlllgiti~
. ··fjigheftMhaMW .
.
Weight
1
0.25
0.1111
0.0625
Sum = 1.4236
Weighted Observation
340.22 (1)
= 340.220
~.30 (0.25)
= 85.075
340.26 (0.1111) = 37.803
340.32 (0.0625) = 21.270
Sum
= 484.368
484.368 i'/";
Most Probable Value = 1.4236- .• ',: .
Most Probable Value =340.242
Solution:
G)
Diff. in Elev.
340.22
340.30
340.26
340.32
.
Weight of route 2:
The weights are invserseley proportional
to the square of the corresponding probable
errors.
® Elev. ofBMi
Bev. = 650.42 + 340.242
Elev. =990.662 m.
4 W1 = 16 Wz = 36 W3 = 64 W4
W1 = 4 Wz
=9 W3 = 16 W4
Assume W1
=1
1
WZ=4"
Wz
=0.25
1
W3=g
W3=0.111
1
=16
W4 =0.0625
W4
\D Compute lfle probable weight oftrtal 3..
® Determine the most probable diff. in
elevatiOn.
.
.
® Compute the elevation of B if .elevation of
AIs 1000 with Bhigher than A,
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ERRORS AND MISTAKES
Solution:
Solution:
ill Weight of trial 3:
CD Probable error:
The weights are also proportional to the
number of observation.
Mean value
··120.68 + 120.84 + 120.76 + 120.64
~ =
4
Weight of trial 3 = 6.
@
Mean value =120.73
Probable diff. in elevation:
Distance
520.14
520.20
520.18
520.24
Weight
1
3
6
8
Sum =18
Residual V
V2
120.68 -120.73 =- 0.05
120.84 -120.73 =+0.11
120.76 -120.73 =+0.03
120.64 -120.73 =- 0.09
0.0025
0.0121
0.0009
0.0081
'LV2 =0.0236
Weighted Values
520.14(1) = 520.14
520.20 (3) = 1560.60
520.18 (6) = 3121.08
520.21 (8) = 4161.92
Sum
= 9363.74
· I
Probable value 0 fdiff.. In eev.
_ f"iV2t:.(
Probable erro~.= 0.6745 -" n(rJ~1r'- 'l
Probable error =0.6745
@
Probable error =± 0.0299
9363.74
=-18-
Probable value ofdiff. in elev. = 520.208
~0.0236.
4(3)
@
Standard deviation:
.. - fiV2
Elevation of B:
Elev. ofB =1000 '10520.208
Elev. of B = 1520.208 m.
Standard deViation =-" ~
. = ~0.0236
Standard deviation
-3Standard deviation = ± 0.0887
@
y\..
Standrad error:
Standard deviation
Standard error =
{;;
±0.0887
Standard error =
{4
Standard error =± 0.0443
["11,\
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5-16
ERRORS AND MISTAKES
1--------------------- •
Iftiglf!tma2Z;- At ;i)i'<1
Three iridepen<lentJilieof levelsarerunJrom
.BM1t~BM~.R<JuteAjS6 kll'l, 1~,rOt1tee 1$
4 knt long andrpute Pis 8 km 8y muff! A,
"M()bserveO•• afi91~s • 9faW~nfll~ar~ • • as
follOWS: .. A"'34'20'36~<B"'49~t6'34·
... '..'
' .
¢7®'?2'41~
.:~JS~~·;~~ej~~~a~~~Y:y:~'.:~.:;'
SHOm. shove-BM,. ···TM eJevatloo{jfBM1 Is
6M2>'"
. ....
. ...
Using the weighted mean valUes,WtiaLis
the weight~froufeB.
'
.
00 Whatls the PrPfutblevalue 9ftheVil!lgtited.
.0)
niEian.
.•.
.
@ . WhatJstheelevalion ofB~;
••. . .
. .
Soiution:
G)
Solution:
G)
Weight of route A:
ROUTE
A
B
C.
DISTANCE
6
4
8
1 1 1
LCD =24
6 4 8
Sum of all angles = 180'
DIFF. IN ELEV.
82.27
82.40
82.10
34'20'36" +49'16'34" +96'22'41"
=179'59'51"
=
=
Error 180' ·179'59'51" 09" (too small)
F"
.
9
CorrectlOn =3
L)
Weight computations
24
Correction =3"
24
B W2=4"=6
Probable value of angle C = 96'22'41" +3"
24
C W3=a=3
Probable value of angle C = 96'22'44"
A W1 =6=4
Weight of 8 = 6
® Probable value of weighted mean:
82.27(4) =329.08
82.40(6) =494.40
82.10(3) =246.30
1069.78
Probable value of the weighted mean
1009.78
=-13-
=82.29
@
Probable value of angle C:
Elevation of8M2.'
8M2 =82.46+82.29
8M2 = 168.71 m.
® Probable value of angle A:
Probable value of angle A = 34'20'36" + 3"
Probable value of angle A = 34'20'39"
@
Probable value of angle B:
Probable value of angle B = 49'16'34" +3"
Probable value of angle B = 49'16'37"
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ERRORS AND MISTAKES
Sum = (n - 2) 180 =(5 - 2) 180 =540'
Enor = 540' - 539'59'40" 20"
=
G)
Adjusted value of angle 0:
Adjusted value of angle 0 = 167'02'07.11"
® Adjusted value of angle B:
Adjusted value ofangle B = 134'44'41,31"
® Adjusted value ofangle E:
Adjusted value ofangle E =76'08'53.16"
Solution:
ANGLE
OBSERVED
VALUE
A
86'15.20"
1
6= 0.167
8
134'44'35"
1
2=0.SO
C
75'48'SO"
1
2=0.SO
0
167'02.05"
1
6= 0.167
E
76'08'SO"
1
4=0.25
Sum = 539'59'40"
1.584
CORRECTION
WEJGHT
ADJUSTED
ANGLES
O.~~~O) = 2.11"
86'15'22.11"
O.~~O) = 6.31"
134'44'41.31"
0.~.~O)=6.31"
75'48'56.31"
O.~~~O) = 2.11"
167'02'07.11"
O.~~O) = 3.16"
76'08'53.16"
Sum -20"
Solution:
G)
Probable value of angle A:
A+8+C=41 +77+63=181'
Error= 181' -180' =01'
Error= 60 mins.
LCD of 5, 6and 2 is 30
Sta.
540'00'00"
Weight
I
c~v(
Correction
6
A ~. 4.~ = 6 ~:;..;
Z6 (50) = 13.84'
8
16 (60) =11.54'
C
30 = 5
6
30 = 15
2 26
15 (60) = 34.62'
26
50'
Corrected value ofA =41' - 13.84'
Corrected value of A=40'46.16'
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5-18
ERRORS AND MISTAKES
® Probable value of angle B:
Corrected value of B = 77' - 11.54'
Corrected value of B = 76'48.46'
@
® Probable value of angle B:
Corrected value ofB = 65' + 13.85'
Corrected value of B = 65'13.85'
Probable value of angle C:
Corrected value of C = 63' - 34.62'
Corrected value of C = 62'25.38'
@
Solution:
ill Average mean value:
Average value (mean)
200.58 + 200.40 + 200.38 + 200.46
=
4
Average value (mean) = 200.455
Solution:
ill Probable value ofangle A:
A + B + C= 39 + 65 + 75 = 179'
Error =180' - 179' =01'
Error =60 mins.
LCD of3, 4 and 2 is 12
Sta.
A
B
C
Weight
12
-=4
3
12 =3
4
12 =6
2
Correction
4
Probable value of angle C:
Corrected value of C = 75' +27.69'
Corrected value of C=75'27.69'
/
13 (60)= 18.46
3
13 (50) = 13.85
6
13 (50) =27.69
13
Corrected value of A = 39' + 18.46'
Corrected value of A = 39'18.46'
60
® Probable error of the mean:
Length
200.58
200.40
200.38
200.46
V
200.58 - 200.455 = +0.125
200.40 - 200.455 = - 0.055
200.38 - 200.455 = - 0.075
200.46 - 200.455 = +0.005
V2
0.015625
0.003025
0.005625
0.000025
,,£V2 =0.0243
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ERRORS AND MISTAKES
)',,,·v.
fd.
-V
.-.~.
2
XV-1)
Probable error of mean =0.6745 ..n(n
P.E. =0.6745
-v
0.0243
4(3}
PE =±0.03
Probable error= 0.6745
- r""iV2
-\I
n(n:1)
Probable error =0.6745
~0.0236
4{3}
Probable error =± 0.0299
® Precision of the measurements:
0.03
p "
Precision = 200.455
i ,,,,
1
·Y.
Precision =6681.83
® Standard deviation:
""IG2
~
. t·Ion = Standard devta
Standard deviation =
..
1
PreCISion =6682
..y0.0236
-3-
Standard deviation =± 0.0887
® Standard error.
Standard error =
Standard error =
Standard deviation
-{;
±O.OBB?
{4
=± 0.0443
Solution:
CD Probable error.
Mean value
120.68 + 120.84 + 120.76 + 120.64
=
4
Mean value = 120.73
Residual V
120.68 • 120.73 =·0.05
120.84 -120.73 =+0.11
120.76 -120.73 =+0.03
120.64 -120.73 =- 0.09
V2
0.0025
0.0121
0.0009
0.OOB1
LV2 =0.0236
CD What is thewelghfofroute 3as$uming the
weightof route f equal tD 1.
® What is the .sum of the weighted
obserVation.
.
@ What i$ the most probable value of the
elevatlon.
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5-20
ERRORS AND MISTAIES
Solution:
CD Weight ofroute 3:
The weights are inversely proportional to
the square of the corresponding probable
errors.
" K
1 J/1 = (2)2 .
K 'J
W3= (6)2
4W1 = 16W2 = 36W3 = 64W4
W1 = 4W2 = 9W3 = 16W4
IfW1 = 1
1
W3 =9=0.1111
1
W2 =4'=0.25
1
W4 = 16 =0.0625
Therefore the weight ofroute 3 = 0.1111.
@
Sum of the weighted obseNation:
DIFFERENCE IN WEIGHT'
t·
ELEVATION
1
340.22
0.25
340.30
'I
0.1111
(,
340.26
0.0625
340.32
I'.
Sum=
1.4236
ROUTE
1
2
3
4
F"'.
Solution:
CD Probable value under each set:
Most probable value using the Invartape in
measurements:
571.185 + 571.186 + 571.179 + 571.180 + 571.183
5
:; =571.183
Most probable value using the Steel tape
in measurements:
571.193 + 571.190 + 571.185 + 571.189 + 571.182
5
WEIGHTED
OBSERVATION
240.22 (1)
= 240.220
340.30 (0.25)
= 85.075
340.36(0.1111)
= 37.803
340.32 (0.0625) = 21.270
Sum
= 484.368 :'5.'
The sum of weighted obseNation
~) = 571.188
@
Probable Errors under each set:
Probable error using Invar tape:
~H
=484.368
@
Most probable value of the elevation:
484.368
Most Probable Value = 1.4236
Most Probable Value = 340:424
~.
~.,
"~,
V2
0.000004
0.000009
0.000016
0.000009
0.000000
'Dfl =0.000038
Invar tape
Residual (V)
571.185 - 571.183 = 0.002
571.186 - 571.183 = 0.003
571.179 - 571.183 = - 0.004
571.180·571.183 = ·0.003
571.183·571.183 =
0
PE = 0.6745
PE =
-GY2
'\J
nTtJ=1)
~ 0.000038
= 00009'1
5(4)
± .
v
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ERRORS AND MISTAKES
Probable erro~us;ng Steel tape:
WE2 = W1 E12
WE2 = W2 E22
1;.
Steel tape ::I.. Residual (V)
V2
571.193 - 571.1~
0.005
0.000025
571.190-571.188
0.002
0.000004
571.185 - 571,188
- 0.003
0.000009
571.189 - 57-V,188
t{).OO1
0.000001
571.182-57t188
-0.006
0.000036
{
}:.V2 =0.000075
. PE = 0.6745
~=
2.98
E= ±O.DOO76 (probable errorofmean)
-GT
'\J
~
~=
'2 =
K
/
K
Ei ')
k r.
W1 Ei 2 = W2 Ei
W1 (0,00093Y = W2 (0.00131)2
Ass. W2 = 1
Wi (0,OOO93Y =1 (0,OO131j,2
Wi = 1.98
" ' c:~F:'~~;11
.~
Weight/
.. poC
t','
W1 = 1,98
W2 = 1.00
Sum= W=.
298"'.'.'1
.-
I·'
Wt. x value
1130.94
571.188
. .,"
1702.18.:'//
Most probable value ofthe two sets = 571.184
Probable error ofthe general mean:
K
W=E2
K
W1=-2
E1
2
E2 _ W1 E1
- W
E2_ W2E
l
- W
W
2.98
E=tO.OOO76
..••-
fd!:··~;W·:J:~1
i l ! !f~jl:l!lll~~;l.
li'_liiil~ii:!~1
1702.128
Most probable value ofthe two sets = ~
@
..-
W2 E,2
£:2 = 1.00(0.00131)2
® Most probable value ofthe two sets:
Probable value
Probable error
.-;. 571.183
Ei =0.00093
"1571.188
E2=0,oo131
.
1
=E2
W
E2 = 1.98(0.00093)2
PE = 0 6'745'" J0.000075
' \ { 5(4)
PE =to.OO131
W1
W1 E12
Solution:
G)
Probable error:
40'31' + 40'34' +40'36'
Mean value =
3
Mean value =40'33.7'
?
Residual
v
40'31' - 40'33.7' = 2.7
40'34' -40'33.7' = t{).3
40'36' -40'33.7 = +2.3
V2
7.29
0.09
5.29
}:. V2 =12.67
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5-22
ERRORS AND MISTAKES
-G:V2
G)
Probable error:
Probatie error =0.6745 -,,~
-52
Probable error =0.6745 ." ~
_f12.67
Probable error =0.6745 -" 3(3=1)
Probable error = ± 0.98
Probable efTDr= 0.6745
® Standard deviation:
@
-52
Standard deviation = -" ;;'::1
-fiV2
="_112.67
Standard deviation =
Standard error =
® Standard error:
Standard error =
Standard error.
Standard deviation =-" ~
2-2Standard devia,tion= ± 2.52
Standard deviation
~ = ± 0.039
2.52
13
~ =0.10
Standard deviation
v;,
Standard error = ~ = ± 0.577
Standard error = ± 1.453
@
Precision:
0.039
Precision = 141.70
1
Precision = 3633
ii;Atili1:~' 11'1~;
Solution:
Average value (mean)
141.60 + 141.80 + 141.70
=
(j)Flrldthepr°bable¥~I»~\)faogleA.
FiMthepto~~~I~Y~19~f>1~r~I~B,<
3
@
Average value (mean) = 141.70
V
v2
141.60 ·141.70 =·0.10
141.80 -141.70 = +0.10
141.70-141.70= 0
+0.01
+0.01
o
@tIMt!'leproMPi¢VaI4¢~f~OglE~%
Solution:
Error= 180· (39' +65' +75')
.Error = 01'
Error = 50' (too smalQ
,',
.,
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S·23
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ERRORS AND MISTAKES
. Weight
Solution:
CD Corrected angle A:
39'
A
12 =6
2
B
65'
12=4
3
C
75'
12 = 3
4
13
Correction
Corrected Angles
13 (50) =27.69'
39' 27'41"
1~ (50) = 18.46'
65'18' 28"
3
13 (50) =13.85'
75' 13'51"
6
50'
Angles Value
A
92'
B
C
o
88'
71'
110'
Weight
12/6 =6
12/4 =3
12/3 =4
12/6 = 2
15
Corrections
6/15 (60) =24'
3/15 (60) =12'
4/15 (60) =16'
2/15 (50) =..!
00'
Error =(92 + 88 +71 +110) - 360
Error =01' = 60' (too big)
Corrected angle A = 92' ·24'
Corrected angle A = 91'36'
® Corrected angle B:
Correded angle B =88' • 12'
Corrected angle 8 = 87'48'
® Corrected angle C:
Correctedangle C= 71' • 16'
Correctedangle C = 70' 44'
CD Probable value of angle A:
Probable value of angle A =39' 27' 41"
® Probable value of angle B:
Probable value of angle B = 65'18' 28"
® Probable value of angle C:
Probable value of angle C= 75'13' 51"
@•• • qO/tlflM~m~#Jffep(#1.val®(Jt.ittg~.!} •.•
.~ • • GOI1)P~t~lhe(;Orr¢¢t#lv~lpe9f.~leEl.·i·<.··
·®••··CornplJ@ftle90rrec!ed·vBll.le.ofangle.Q...
Solution:
CD Probable value of angle A:
Sum ofinterior angles = (n· 2)180
Sum ofinferior angles = (5· 2)(180)
Sum ofinterior angles = 540'
Sum = 110' +98' + 108' +120' +105'
Sum=541'
Error =0l' or 50' (to be subtracted)
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S-24
ERRORS AND MISTAIES
Sta.
Angles
A
110'
B
98'
C
108'
0
120'
E
105'
COITection
6
18 (50) =20'
Weight
12 -6
2-
Solution:
28'34' + 61'15' =89'49'
Error = 40"
12=4
3
12 =3
4
12 =2
6
12 =3
4
18
1~ (50) =10'
101' 50'00"
2
18 (60) =6.67'
119' 53'19.8"
1~ (50) =10'
104' 50' 00"
50'
540'00'00"
Probable value ofangle A :: 109'40' 00"
@
Probable value ofangle C= 107'50' 00"
@
Probable value ofangle 0 = 119'53'19.8"
BAC
1
2=0.5
BAD
1
4=0.25
CAD
2=0.5
Correction
40 (9.5) = 16"
1.25
40 (0.25) =.8"
1.25
40 (0.5) = 16"
1.25
109' 40' 00"
91' 46' 40.2"
Weight
1
1.25
Corrected Angles
1~ (60) =13.33'
Angle
Corrected Values
28'34'16"
89'49'32"
61'15'16"
<D Probable value ofangle BAC = 28'34'16"
@
Probable value ofangle BAD = 89'49'32"
® Probable value of angle CAD = 61'15'16"
(1).. f)eleWIM1heWeight.(lfr@W.@lTIber.g.
® ·Oelertrilne fhemO$fpt()Pabledlffe$/1¢eJo
elevatioll,
..
.
® P~t~it1e.tnelnosfprOb@IEleIWalion·tJfQ
lnmetars.· .
.
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ERRORS AND MISTAKES
Solution:
CD Weight ofroute no. 2.:
W1D1 = W2D2 = waDa = W4 D4
Assume:
W1 =6
6(2)= W2 (6)
W2=2
2(6) = wa (4)
wa =3
3(4) =w4 (8)
w4 = 1.5
Weight ofroute 2 = 2
@
Probable difference in elev:
Route Weight WI. x Diff in elevation
1
6
6(0.86) =5.16
2
2
2(0.69) =1.38
3
3
3(0.75) =2.25
4
1.5
1.5(1.02) =1.53
-12.5
10.32
Solution:
Determine first the weight of each route
111
10 16 40
To find the weight, divide the L.C.D. by its
distance.
CD Probable weight of route B:
WEIGHT
ROUTE
LENGTH
A
10
160 =16
10
B
16
160 = 10
16
C
40
160 =4
40
Sum=3O
Probable diff. in elev. = 1102~52
Probable diff. in elev. = 6.826
Weight ofroute B= 10
@
Probable elevation of C:
Probable elevation ofC= 825 + 0.82
Probable elevation of C = 825.82
@
Mast probable d·ff.·
. 18984.86
I.melev.=~
lines of levels are run from BM1 to BM2 over
three different routes. Ifthe elevation of BM1 is
100 m. above the sea level.
Route
Length
A
B
C
10
16
40
(Diff. In Elev.)
Between Bm1 & Bm2
.632.81
632.67
633.30
Probable difference in elevation:
16(632.81) = 10124.96
10(632.67) = 6326.78
4(633.30) = 2533.28
18984.86
Most probable diff. in elev. = 632.83
@
Probable elevation of 8M2:
Probable elevation of 8M2
= 100 + 632.83
= 732.83 m. above sea level
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ERRORS AND MISTAIES
·.I.•
[dlrr2;~lull~lfll_·.
i I• • •;'
·~~~.··.O•·.••·n. ~.$ · ., 4v
.~ t >.>
~• .• •.• • •.·•.~~I~i
6i• • • • • • • • •
•.••.•. '.•. •.•.•.•·.•.•. '.·.•L.• ..•·.•.•.1••
. ·• .E.•.·••.• •.•.•.•.•.•.•.• . ·•. ••. •. I.I.
•.••. : ..•
1.•11..• •.k1.•·. .•.• .•.•.L.•.•.• . E
.•
l.·•.
U.
~~~:~»<i~
Solution:
® Probable weighted mean:
Solution:
CD Probable error of the resulting computation:
PE =..J (b Eh)2 + (h Eb)2
b =314.60
h=92.60
Eb =+0.16
Eh =0.14
.-----~---
PE =..J[314.6(0.14)]2 + [92.60(0.16)~
PE =+46.47
@
Diff. in Elev.
1
2
3
41.16
41.20
41.12
WI. x Diff. in Elev.
V
WV2
6(41.16)=246.96
4(41.20) =164.80
3(41.12)=123.36
535.12
0
+0.04
-0.04
0.0064
0.0048
0.0112
535.12
® Standard deviation:
Standard deviation =
Relative precision:
0.043
.
..
ReIatlve precIsion = 860
Relative precision = 2~O
o
Weighted mean = ~ = 41.16
PE =..J (PE1)2 + (PE2)2 + (PE3)2 + (PE4)2
@
Weight
6
4
3
13
V1 =41.16-41.16=0
V2 =41.20 - 41.16 = +0.04
V3=41.12 ~ 41.16 =- 0.04
Probable error of the sum of the sides: .
PE =..J (0.04)2 +(0.08)2 +(0.04~ + (0.08~
PE =+0.126
Line
. .
Standard deViation =
~~21)
0.30112
13(3 _1)
Standard deviation =±O.021
@
Elevation 8Mi
Elevation 8M2 =212.40 - 41.16
Elevation 8M2 = 171.24
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ERRORS AND MlSTAIES
8M,
· hted d·ff.·
I
'1436.36
Wieig
I .meev.=~
Weighted diff. in elev. = 143.636
@
Bevation of 8M3:
Elev. of 8M3 = 143.636 + 30.162
Bev. of 8M3 = 173.798 m.
@
Adjusted elevation of8M2:
Total Correction for route 1
=143.70 ·143.636
=O.064m.
3
Correction for 8M2 = 10 (0.064)
Solution:
0)
Weighted difference in elevation between
BM1 & 8M3:
Diff. in elev.
Route Distance
1
10 km. 68.258 + 75.442 = 143.70
143.62
2
6km.
143.58
3
15 km.
w1 01 =w2 02 =w3 03
wd10) =6w2 =15w3
Ass: W2=5
w1 (10) =6(5)
w1 =3
3(10) =15 w3
w3=2
Ditt. in
elev.
Weight
143.70
143.62
143.58
3
5
2
10
Correction for 8M2 =0.0192
Correction for diff. in elev. of BM1 and 8M2
= 68.258·0.0192
= 68.2388 m.
Adj. Elev. of8M2 =30.162 + 68.2388
Adj. Elev. of 8M2 =98.4008 m.
From starting .P:OiN A,eleyatipn340;6S
theelevat~nof asecondpotnt6is f6Und,
m·,
lile
rottle, distance<liidefevationof 8 being
respeClivelyas f6HoWs: Route J ~ 4 km, 3&h64
WI. x Ditt.
in Elev.
3(143.7) =431.10
5(143.62) = 718.10
2(143.58) = 287.16
1436.36
m., Roule2· 2,5 krri., 364.2.0 m. Route 3-3
km.,365.01 m.,R()ute4- 6 km., 364.31m,
Midway alOng route 1~8M1 is located with an
elevation of 35U9. Along route 4, 2.5 km.
from AIQwards 6,6M2 is located, with an
elevation of 349;86 m..
~.,
.
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S-28
ERRORS AND MISTAKES
il.ipillelB
Solution:
Distance
4 km.
2.5 km.
3 km.
6 km.
4
@
Adjusted elevation of 8M2 using route 4:
Correction for 8M2 =2 5 (023)
Correction for 8M2 =0.096 m.
Corrected £Iev. of 8M2 = 349.86 + 0.96
Corrected Bev. of 8M2 =349.956
A
1
2
3
Error in elevation of 8 using route 4:
Error in route 4 =364.60 - 364.37
Error in route 4 = 0.23 m
6
CD Weighted elevation of 8:
Route
@
Elevation of 8
364.84 m
364.20 m.
365.01 m.
364.37 m.
k
k
w2 =02
k
k
WJ=w4=03
04
w1 01= w2 ~ = w3 03 = w4 04
w1
=01
Ass: w1 = 1.0
(1)(4) =2.5(W2)
w2 = 1.6
w1 01 = w3 0 3
(1 )(4) = w3 (3)
w3= 1.33
w1 01 = w4 04
(1)(4) = w4 (6)
w4 =0.67
l
Assume w2 = 1
_ (1.6)2
WI. x Elev.
WI.
1.0
1.6
1.33
0.67
4.60
Solution:
CD Probable weight of A:
k
w1=-2
E1
w1 E1 2 =W2 E
w1 (3.2)2 =w2 (1.6)2
1(364.84)
1.6(364.20)
1.33(365.01)
0.67(364.37)
= 364.84
= 582.72
= 485.13
= 244.13
1677.15
Weighted elev. of 8:
1677.15
Elev. 8 =4:60 = 364.60
w1- (3.2f
w1 = 0.25
Observer
A
8
Angle
42'16'25"
42'16'20"
Error
±3.2"
±1.6"
Weight
WI. x Angle
0.25
1.00
1.25
25(0.25) = 6.25
20(1) = 20.0
26.25
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ERRORS AND MISTAKES
@
Sum of the weight of A and B:
Sum of the weight of Aand B = 1 + 0.25
Sum of the weight of Aand B =1.25
@
Most probable value of the angle:
Actual ground area = (0.999775612)2
Best value of the angle =42'16' + 21~;;
Actual ground area = 25436.41 sq,m.
@
Actual ground area:
Combined factor = 0.9998756 (0.9999)
Combined factor = 0.999775612
25425
Best value of the angle =42'16'21"
(IJ
CD A line tneasures6846.34 in. at elevation
.. 993.9 m. The average radiusofcurvature
in the area is 6400 km~Corilpute thesEla
. leYeldistance.
®. The ground distance as corrected for
temp., sag and puUcorrection Is 10000 m.
ff the sea level reduction factor is
. 0.9998756 and the flrid scale factor is
0.9999000,corripute the grid distance of
the same1ine,
®The' geodetic length qf a line on theearlh's
surface is found 10 be 5280 m. and Its grid
dl$tanceis equal tClS279.67 m.· Compute
the scate factor used. . . ........ ...
@
@
The grid area of a parcel of landis 25425
sq.m. If the sea level re:ductionfactor is
0.9998756 and the griel scale factor is
0.9999, determine the actual grOUnd area.
CD Difference in elevation:
CD Sealevel distance:
Vertical angle =90' - 83'14'20"
Vertical angle =6'45'40"
Diff. in elevation = 1486.72 Sin 6'45'40"
Diff. in elevation =175.03 m.
Reduction factor =1 - ~
I
1
993.9
ReductJon lactor = - 6400000
Reduction faqtor= 0.99984
Sea level distance =6846.34 (0.99984)
Sea level distance =6845,24 m.
@
The corrected field distance on the surface
of the earth was found to be 3296.43 m.·· If
the elevation factor isl).9999()4zand a
scale factor ofO.9999424,compufe the grid
distance.
..
Solution:
Solution:
·
The dlfference of elevation between two
points\vas deterlllil1eq by tng~nometrjc
l~!Veling, . The slope . distance was
measured electronically andw~s .found to
be 148e.72m.. andtheieh~h dis!<lncewas
83'14'20". Calcul~te fhediffetence In
elevation between the two points.. .•... .....
@
5279.67
Scale factor = 5280
Grid distance:
Combination factor
=0.9998756(0.9999000)
=0.9997756
Grid distance =10000(0.9997756)
Grid distance =9997.756 m.
Scale factor:
Grid distance =Geodetic length
x scale factor
Scale factor =0.9999375
@
Grid factor'
Grid factor = elevation factor x scale factor
Grid factor = 0.9999642(0.9999424)
Grid factor =0.9999066
Grid distance =3296.43(0.9999066)
Grid distance =3296.12 m.
5-30
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LEVEliNG
1.
1.
Dumpy Level
2.
Wye Level
1.
In the dumpy level, the bubble tube is
attached to the level bar while in the wye
level it is attached to the telescope.
2.
In the dumpy level, the bubble tube can
be adjusted in a vertical plane only,
while in the wye level it may be adjusted
vertically and laterally.
3.
In the dumpy leVel, the telescope is
rigidly fastened to the level bar and can
not be removed there from, while in the
wye level, the telescope rests in Yshaped supports which permit it to be
removed and reversed end for end or
revolved abolit the axis of collars.
4.
The dumpy is more rigidly constructed
that the wye that it has fewer
adjustments. However, to adjust a
dumpy level would require at least two
men whereas a wye level may be
adjusted by only one man.
5.
In the dumpy the supports are not
adjustable, while in the wye, one end of
the level bar may be adjusted vertically.
Adjustment of Level Tube:
To make the axis of the level tube
perpendicular to the vertical axis.
2.
Adjustment of Horizontal Cross~Hair:
To make the horizontal cross-hair lie in a
plane perpendicular to the vertical axis.
3.
Adjustment of the Line of Sight:
To make the line of sight parallel to the
axis of level tube.
1.
Adjustments of Level Tube:
To make the axis of level tUbe lie in the
same plane with the axis of the wyes.
2.
Adjustment of Level Tube:
To make the axis of the level tube
parallel to the axis of wye.
3.
Adjustment of Horizontal Cross-Hair:
To make the horizontal cross-hair lie in a
plane perpendicular to the vertical axis.
4.
Adjustment of Line of Sight:
To make the line of sight coincide with
the axis of the wyes.
5
Adjustment of Level Tube:
To make the axis of level tube
perpendicular to the vertical axis.
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LEVELING
1.
Imperfect adjustment of instrument:
This could be eliminated by adjusting the
instrument or by balancing the sum of
foresight and backsight distances.
2.
Rod not of standard length:
This could be eliminated by
standardiZing the rod and apply
corrections same as for tape.
3.
Parallax:
This could be eliminated by focusing
carefully.
4.
Bubble not centered at instant of sighting:
This could be eiiminated by checking the
bubble before making each sight.
5.
Rod not held plumb:
This could be eliminated by waving the
rod or using rod level.
6.
Faulty of reading the rod:
This could be eliminated by checking
each rod reading before recording.
7.
Faulty turning point:
This could be eliminated by choosing
definite and stable points.
8.
Variation of temperature:
This could be eliminated by protecting
the level from the sun while making
observations.
9.
Earth's curvature:
This could be eliminated by balancing
each backsight and foresight distance, or
. apply the computed correction.
10. Atmospheric refraction:
This could be eliminated by balancing
each backsight and foresight distance,
also take short sights well above ground
and take backsight and foresight
readings qUick succession.
11. Settlement of tripod or turning points:
This could be eliminated by choosing
stable locations, and taking backsight
and foresight readings in qUick
succession.
1.
2.
3.
4.
5.
6.
Imperfect adjustment of instrument
Parallax
Earth's curvature
Atmospheric refraction
Variation in temperature
Rod not standard length
7. Expansion or contraction of rod
8. Rod not held plumb
9. Faulty turning points
10. Settlement oftripod ortuming points
11. Bubble not exactly centered at the
instant of sighting
12. Inability of observer to read the rod
exactly.
1.
2.
3.
4.
5.
6
Confusion of numbers in reading and
recording.
Recording B.S. on the F.S. column and
vice-versa.
Faulty additions and subtractions.
Rod not held on the same point for both
B.S. and F.S.
Wrong reading of the vernier when the
target rod is used.
Not having target set properly when the
long rod is used.
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S-31-A
lEVELING
R =radius of earth
R::6400km.
K2
h =2R
Horizonrul Lint!
h = K2 (1000)
2(6400)
h =0.078 K2
hr--~
7
h = 1 (0.078 K2)
Horizontal Line = a straight line tangent to a
level surface.
Level Surface =a curved surface every
element of which is normal to the plumb
line.
r
7
hr =0.011 K2
her = h - hr
her = 0.078 K2 - 0.011 K2
Level line = a line in a level surface.
From the figure shown, an object actually
at C would appear to be at B, due to
atmospheric refraction, wherein the rays of
light transmitted along the surface of the earth
is bent downward slightly. The value of h
represents the effect of earths curvature and
atmospheric refraction and has the following
values.
her = in meters
K =in thousand of meters
Derivation:
DERIVATION OF
C.URVATURE and
~
REFRAqTION CORRECTION
Conditions:
K2 + R2 :: (R + h)2
K2 + R2 =F{2 + 2 Rh + h2
'Since h is so small, h2 is negligible
h =height in m. of the line of sight, at the
intervening hill C, above sea level.
h1 =height in m. of the station occupied A,
above sea level.
h2 :: height in m. ofthe station observed B,
above sea level.
0 1 :: distance in miles of the intervening
hill C from A.
O2 :: distance in miles of the intervening
hill C from B.
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lEVELING
Since h1, h, and h2 are vertical heights, and considering the effects of
curvature and refraction at A and S, as reckoned from a tangent (horizontal) line at
sea level vertically below C, the figure can be reconstructed in its plane sense.
Hoeizontal
Line
In triangle ABE, by proportion:
(h 1 - 0.067012) - (h2 - 0.0670,2) _h- (h2 - o.067Dll
0 1 +~
~
hl - h2 - 0.067 (0 12 - Ol) _ h- h2 + 0.067Q2
01+~
-
~
h- h2 + 0.0670l =: ~ [h 1 - h2 - 0.067(0, + O2)(01 • O2)]
h =: h2 - 0·067Ol + 0
h =h2 - 0·067Oi + 0
~D-t (hl • h2) - 0.067~ (01- D-tl
1
~2O2 (h l - h2) - 0.0670 102 + 0.067D-t2
1
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5-32
LEVEUNG
@
Diff. in elevation ofstation 7 and station 4:
Diff. in elevation of station 7 and 4
=400.78 - 389.01
=11.77
@
Elevation of station 3 = 392.61
Inm~.pla~ • beIOW$Ho~s.adlffer~ntial.levelln9
frorlt~rct1.ITla.rt<.to~nomer.~nph • rnark,al?(lg'
ea®llqerepr~$~n~~~~rShUnthe~etualr()d
r~~din~· • • • Tbe • dj~~~n • • of•• lbtl.Jjeh:lwork•• is
indica~~,pythe.nlJrnber.oft1,Jllllng.polnts·
ill, 90rnPlItet/lE!el~YliijortpfTP2'
@
9ptnplJleth~ele¥~ti@Qf,~~.w
® '. ()o@i.JteJneelevau®mJPs·
Solution:
Note: H.I. = Elev. + B.S.
Elev. = H.I. - F.S.
BM,
Sta.
1
2
3
4
5
6
7
B.S.
5.87
7.03
3.48
7.25
10.19
9.29
H.I.
398.12
398.86
396.09
396.26
400.88
405.72
IBS = 43.11
F.S.
6.29
6.25
7.08
5.57
4.45
4.94
Elev.
392.25
391.83
392.61
389.01
390.69
396.43
400.78
IFS =34.58
Arithmetic check:
IBS - IFS "43.11 - 34.58 = 8.53
400.78 - 392.25 = 8.53
CD Diff. in elevation of station 7 and station 5:
Diff. in elevation of station 7 and 5
= 400.78 - 390.69
=10.09m.
El,33.971
Solution:
Sta.
B.S.
BM,
2.565
TP 1
10.875
TP 2
7.035
BM2
3.560
TP3
7.186
BM 1
36.536
41.59
46.679
44.498
41.948
Sta.
BM 1
TP,
TP 2
BM 2
TP 3
BM 1
Remarks
Bench Mark No.1
Turning point
Turning point
Bench Mark NO.2
Turning point
Bench M;:jrk NO.1
Elev.
33.971
30.715
39.644
40.938
34.762
33.971
H.I.
F.S.
5.821
1.946
5.741
9.736
7.977
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lEVELING
Arithmetical check:
Solution:
LF.S. = 5.821 + 1.946 + 5.741
+ 9.736 + 7.977
LF.S. =31.221
LB.S. =2.565 + 10.875 + 7.035
+ 3.560 + 7.186
LB.S. =31.221
LF.S.. B.S. =31.221 ·31.221 =0
33.971 - 33.971 =0
STA
8M,
4
TP1
5
6
7
Elevation of TP 2 = 39.644 m.
Elevation of 8M2:
Elevation of 8M2 = 40.938 m.
@
Elevation of TP3:
Elevation of TP3 = 34.762 m.
TP~
1.7
2.2
1.2
0.9
2.77 330.36 3.43
2.2
3.7
1.6
2.22 329.52 3.06
8
9
10
11
2.8
3.6
2.0
1.1
7.31
%hefj@r~M4w~ • i:t•• $RMmajjc.~$l®¢mElnt • !)f
~• • pro~I~ •. I~y~I • fl'llt~frgrn·aM1 •.lir9••
\I~W~~lhdiC~ledrE!pre$~NPMK$lgrt,.
a.Mi··.·.TM.
.f()(eSi9nt, • ~rld • •jl'ltl3fm~(jiate • f()J'.esigh1•• re~~i.~.~·
l1:ikEln.·on.~tMOl'l~.aI9ng.the. rolJle.·.Elevation.• 9f
IFS
2.32 331.02
BM~
13M,£W!.70m,· .
FS
HI
1
2
3
CD Elevation of TP2:
@
8S
2.45
8.94
ELEV
328.70
329.32
328.82
329.82
330.12
327.59
328.16
326.66
328.76
327.30
326.72
325.92
327.52
328.42
327.07
Arithmetic check:
8.94 - 7.31 = 1.63
328.70 - 327.07 = 1.63
.
CD Difference in elevation between stations
Sand 9:
= 328.16 - 325.92
=2.24m.
® Elevation of TP 2:
Elevation of TP2 =327.30 m.
® Elevation of 8M2:
Elevation of 8M 2 =327.07 m.
CD Find the difference in elevation between
stations 5 and 9.
@ Find theelevatiQn of TP2.
@ Find the elevation ofBM2.
From the given prUflleleveling notes:
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S-34
lEVEliNG
I
BS· . FS
BM. 1 0.95
-
225.50
}
.....
~I
::c:
3.13
~
0.64
Arithmetic check:
9.03 . 8.41 = 0.62
226.12 . 225.50 =0.62
CD Difference in elevation between station
Sand 2:
I
=225.8 ·224.2
= 1.6m.
® Bevation ofTPi
Elevation of TP2 = 227.66 m.
® Elevation ofBMi
Elevation ofBM2 = 224.88 m.
..
0.62
2.37
·3.50
,
..
~···1.24
BM.
(J)·>What•. . . j~ . •'.t@•• • diff¢ri3nce•• · in .··elevallon·
betW~~~$ta1k>o5~6d2, • •.• ' \
@jyomputelh&l'll&Y<ltioopfTF'ti .
@ c:ornput~theeleva~Cll1of8M:(
Solution:
STA
BM,
1
2
TP,
3
4
5
6
7
TPo
8
9
10
11
TP~
TPA
12
BMo
BS
0.95
l~vellS$l;!t4PBll~~r~dlngcrlg.~~5l'rl.I$
tilken•• on•• a.tlench•• mal'l<..llie.~levam)l'l.6fWhjch
iS12·13qrn·Atth~tlegihrying()f~e@et6tl~
prBfl'ed'••l®•• WdJeadjn~,lsg·Q?5~.·.~~m·fl'9hj.
ml:ll:>egin~jl1g,)tis1·~1TI'l1·aJepm·,ifi$
HI
226.45
FS
IFS
3.0
2.3
3.13 228.94
0.64
2.7
2.8
3.1
0.5
0.8
2.16 229.82
1.28
0.9
1.2
1.7
2.8
0.82 228.27 2.37
1.35 226.12 3.50
3.0
8.41
M~09&.tMJ()ll9Wlng.(l~$crlRfl®.lothemtfu.()f
.prMIEl.~¥l'll.hQt~~C9mp~I~· • tPel~Y@()r1, • • • A·
1.24
9.03
ELEV
225.50
223.50
224.2
225.81
226.2
226.1
225.8
228.4
228.1
227.66
228.9
228.6
228.1
227.0
227.45
224.88
223.1
224.88
{l;702m.atpem.and~1m.,thet()~~at!lng$
~te • l.2~1 • nl;~@.O.7§2.lil. • r~spe¢tj@M·.·.()(ja
rocklh~tistlQtoPlll1e,tMtPdrl;!~qiMl~
0.p55•• lTl.•• Jh~·.le\l&liS,me{l·remo\lM.ahead,$et
upand•• aroqf dingqt.1.95Z.m· i QbseNed,
*.
ea
the}odsljllbelngh~ld()nlherock,Ttte
readjllgs • al()h~.th~.PtOfil#.~re.thenrll~lJrn¢d;
90m.• ftoll'llhebegir,"jl1gofth~ljn~/therod
reading.ls.1AS$.m.! ·'ZO.rn.• frorn••IHe.beginniog
of•• tile ·linl;!.tM.·readil19.is • 1AM.I'Il.,•• ~IlCllly • 1(iO
rn·frornt@p~9InningQfthemnethe@j
readfngfs2J9611l.
.
(i). COrnPlJtethEle~vationatJhepQlnt60rn.
fr°rn.lhe.t¥9inning•• oflhe.line.
® CompUI~.the.eIeVatiOn.l)f.the.tumjn9.pol{lt
@ CornPufe lhe difference in &/evation ala
P<Jlflt.·1~0I11;·.ar1d • 81•• lTl..• fTorn.the·\:ieginning
oftheHne.
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lEVELING
Solution:
HI
BS
2.995 15.130
STA
BM
0
30
60
66
81
TP
90
120
150
cD Find thedlff. in elevation belweEm TPl and
FS
IFS
2.625
1.617
0.702
1.281
0.762
1.952 16.527
0.55~
1.159
1.434
2.196
4.947
ELEV
12.135
12.505
13.513
14.428
13.849
14.368
14.575
15.368
15.093
14.331
0.555
Arithmetic check:
4.947 - 0.555 =4.392
16.527 - 12.135 =4.392
CD Elevation at point 60 m. from the beginning
of the line:
=14.428 m.
TP3'
.
'.'
@ .. Find the elevation of~.
@ What is the difference
Solution:
STA
BM 1
BM 1
TP -L
TP 1 -R
TP?-L
TP?-H
TP1 -L
TP,·H
BM?
8M?
BS
9.08
9.08
12.24
10.10
11.04
9.92
1.75
0.55
HI
758.14
758.14
766.65
766.64
775.48
775.48
767.39
767.41
63.76
® Elevation of the tuming point = 14.575 m.
Arithmetic check:
GD Difference in elevation at point 150 m. and
81 m. from the beginning of the line:
= 14.368 -14.331
14~/4 =7.07
=O.037m.
in elevation
between BM 1 and TP2'
FS
ELEV
749.06
3.73
1.60
2.21
1.08
9.84
8.62
11.27
11.27
49.62
754.41
756.54
764.44
765.56
765.64
766.86
756.12
756.14
63.76 - 49.62 =14.14
Ave. elev. of BM2
756.12 + 756.14
2
=756.13
756: 13 - 749.06 =7.07
.Data shown is obtained from a double rodded
line of levels of a certain cross-seclion of the
proposed ManUa-Bataan Road.
IA.
I.
TP1 -L
RS.
9.08
9.08
12.24
F.S
749.06
3.73
1
TPrL
11.04
TP~-l
1.75
ELEV.
1.08
9.84
11.27
11.27
Elev. of TP I =755.475
- 765.64 + 766.86
EIev.of TP3 2
Elev. of TP3 = 766.250
Diff. in elevation =766.250 - 755.475
Diff. in elevation =10.775 m.
~n
K62
CD Diff. in elevation between TP I and TP3:
:: 754.41 + 756.54
EIev.o f TP I
2
® Elevation of BMi
I
756.12 + 756.14
2
=756.13 m.
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S-36
LEVELING
® Difference in elevation betwwen 8M1 and
TP2:
'
f TP - 764.44 + 765.56
E,ev.o
22
Elev. of TP2 :: 765.00
Diff. in elevation:: 765 - 749.06
Diff. in elevation:: 15.94 m.
'l'b~ • fCtI19Wing • • @()w,s•• • fl•• • t<lbtllate(j•• • d~ta • • ()f
~VE!lil'iS·.r()t~~.#Slng.ri~~.~nd.fall.rnettiod
..
STA.
BM 1
1
2
3
4
BM,
Fall
Rise
+0.860
+1.153
+0.059
-1.046
+0.672
2.744
Reduce Level
346.75
347.61
348.763
348.822
347.776
348.448
1.046
Rise:: 3.755-2.895
Rise:: 0.860 m.
Rise:: 2.895 -1.742
Rise:: 1.15Jm.
Rise:: 1.742 -1.683
Ris~ :: 0.059 m.
Fall:: 1.683 - 2.729
Fall :: 1.046 m.
Rise:: 2.729- 2.057
Rise:: 0.672 m.
Rise at station 2:: 1.153 m.
® Reduced elevation at station 3:
Reduced elevation at station 1
:: 346.75 +0.86
:: 347.61 m.
Reduced elevation at station 2
=347.61 +1.153
=348.763m.
Reduced elevation at station 3
= 348.763 + 0.059
=348.822 m.
Solution:
cD Rise or fall at station 2:
STA.
BM1
1
2
3
4
BM,
ROD READINGS
B.S.
I.F.S.
3.755
2.895
1.742
1.683
2.729
3.755
F.S.
2.057
2.057
® Reduced Level of BMi .
Reduced Level of station 4
=348.822 - 1.046
= 347.776 m.
Reduced Level of 8M2
= 347.776 + 0.672
:: 348.448 m.
Arithmetic check:
IBS· LFS:: 3.755-2.057:: 1.698 (check)
IRise:: 0.860 + 1.153 + 0059 + 0.672
IRise:: 2.744
IFall:: 1.046
IRise· IFall:: 2.744 - 1.046 :: 1.698 (check)
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lEVELING
8 :: 4.478; 4.476:: 4.477
m
A•• ffi9ipro~I • I~'J~IiDfl • • ·ls?~~~~gacmss • • a·
~ide.• • W~er<'lQ~ • • ·ll'l~.·.t~C!pr9¢aLI~¥~I • r~~m~9:S
W~r~taken[)etWEl~nP9jl'lts~aod~>as
Diff. in elevation between Aand B
:: 3.143 •4.477
=·1.334 m.
reRI!lrClC<llleYtllre~~ing()~·thT()Pposlt~~I~()f
True difference in elevation Aand B
_ (-1.336) + (·1.334)
fgll?W$,.·•• Wllhlnstrtkoeo{setup.nearA,.the.rod.
rE!a~mgs()nA~f~~.28~~nti?·285rn,Ih~
lhtlJiVerCltp9iml3ar~~.61a,~·919,~·9?1~M
-
3.622:m··•• • ·\Nilh.th~jllsWlTlerrtsetupnear • ~;
lh$rad.reaqing$.(jIl••e• <'ire .4A78@d.4·47tl!n···
an9•• trye.rod•• r@~iM$ • ()n • lhe9P1lO$it~ • si~9f
Elevation of B:
B= 300 -1.335
B= 298.665 m.
@
~river·llctpaitlt·A,.lhe.@lrtladln9$·<'irEl~.143,·
M4Qi3.14u~@~.1.:t4ffl.<
2
=1.335m.
. . .... . ......
CD Compute the difference In elevation
. betweeliA and B with the. instrument set
. .• up near A ...
•. .
..• . . ..../ ..• .
® What· is the true difference in· elevation
between A and B. . . •• .•. . . •.. . . . :
.~.. If the elevation at A i$~OOm:, whatis.the
.·elevationofB.
.
/
.
IO.lev~llngacro$$ • ~.Wi.d~rivE!l'6n·eall'lPa"~a.a
WPiPf9cal.lev~lt~~dings • W;E!~I'~~~ .• ·b~W~l'n
p(')iht$§anqR~~$~()Y"nlnth~~~~I~~~6<
Solution:
CD Difference in elevation between A and B
with instrument set up near A:
With instrument near A:
"'1ean rod reading on A.
~ :: 2.283 + 2.285 = 2 284
i m
2
. m.
2.284
2.286
Mean rod reading on B:
B :: 3.618 + 3.619 + 3.621 + 3.622
m
4
Bm :: 3.62 m.
Diff. in elevation between A and B
=2.284 - 3.62
::·1.336m
® True dirt. in elevation between A and B:
With instrument near B:
Mean rod reading on A:
A =3143 +3.140+ 3.146 + 3.144
m
4
Am:: 3.143
Mean rod reading on B:
• 2.283··
.
2~48Q
2;476
·2.478
I
I
.
1.674
3.140 . .
1.677
3.145
:1.6743.142 I .
.
1.6773:1431
1.678
I
3.146
Compute the difference in elevation
between Band C with instrument set up
nearS.
® Compute the true difference in elevation
between 6 and C.
@ If the elevation of 8 is 346.50 m., compute
fhe elevation of C.
(j)
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5-38
LEVEliNG
Solution:
CD Difference in elevation between 8 and C
with instrument set up near B:
Mean rod reading on 8:
8 =2.283 +2.284 +2.286 +2.283
m
4
8m =2.284 m.
••
•
tomput~~ • lR••.tl¢•• H.·~~·m·\ It•• wa$•• f()ung9ut
tJQ',Y~Y~r • ttJat•• tIle•• ~Yel.settIEls.5.rl11'!l.b~~n.
Mean rod reading on C:
C =1.675 + 1.674 + 1.677 + 1.674 + 1.677 + 1.678
m
~.llne.9ftev~IS.1(lkCll· • • §ng.WE\ft4n()vet~ff.
9~Otlll~ • • <§Ia,rting•• f~~m ~1 • Wlf~.·.·~I~Y~~An
.:!f·§<I11Elt~r~·.· • • • • rhe·•• $le~li()n • • Qf•• ~M?·.\y~s
6
fbe•• in~lanf.Af.ey~rt • ba,d($ighf•• r~a.ding,.me.Wd
mrn••
~tlttl~~ •. 2••
if•• th~ • • ~c;k~ight.<1nd • • fo(esjghl
d:islanl;ElheveanE\verage100m.< FindJhe
(x)rte~teJevati6l'ldfaM2·<·
.. .
.
Cm = 1.676 m.
Find.l~errO(duetO.$MllementW~~I .• •
.~w • •Oet~rrnjne
• the.elT9r•• duet()•• s~tt~Il1El.~tOf
•.•1..
Diff. in elevation between 8 and C with
instrument set up near 8
=2.284 • 1.676
=+O.60Bm.
f()(j.\
WCpfJ1Pulilthecprrecled elevaticiflpfBMi·<
Solution:
@
True diff. in elevation between 8 and C:
Mean rod reading on C:
C - 2.478 +2.480 +2.476 +2.478
m
4
CD Error due to settlement oflevel:
10000
No. of set ups = 100 + 100
No. ofsetups = 50
Cm =2.478 m.
Error due to settlement oflevel
=50(0.005)
=O.25m.
Mean rod reading on 8:
8 = 3.143 + 3.140 + 3.145 + 3.142 + 3.143 + 3.146
m
6
8m =3.143 m.
@
Error due to settlement ofrod:
N
0.0
Diff. in elevation between Band C
=3.143·2.478
=0.665 m.
True difference in elevation
0.608+ 0.665
=
No. of turning points = 49
Error due to settlement of rod
=49(0.002)
= O.098m.
2
=O.6365m.
® Elevation of C:
Elevation of C= 346.50 +0.6365
Elevation of C =347.1365 m.
ft'
.t
10000 1
ummgpoms= 100 + 100·
@
Corrected elevation of8Mi
Total error =0.25 + 0.098
Total error =0.348 m.
Corrected elevation of8M2
= 17.25 - 0.348
= 16.902m.
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lEVEliNG
® Rod reading on A with iilsflUment near B:
x+ e = 0.549
e = 0.549 - 0.53
e = 0.019
.
.'
'.
I','
.....
..
Instrument . lristrortlenl
. .Sl:}t unearAsetuDnearB
on8'" I
Rod reading on A=0.938 - 0.019
Rod reading on A = 0.919 m.
@
Error in line ofsight:
= 0.019m.
-.,7
Rod readlng·1 .'
.
.,':
. .•...
¥
;."
<D Whaf is the. difference Jnelevatlon
. . between Aand at '. .
.
® If the line of sight is not in adjustment,
determire the correct rod reading on AwUh
thernsttument still setup at B.
.
® Deterrnlne fhe error in the line of sight
In.a.lWoipeglesfOSing.rriqdel\'Vjldf>.lPi2liuI11PY.·
level,the.f9"Q\'!'mgobsepj§tioMw~r~t~~~~i.·.
Solution:
CD Diff. in elevation between Aand B:
l.ine ofsif<hr
-rll----...::lII--'
0 C
~- -
- -
"- - - - - - - --I --
{/{)ri:oftwl {iue
x
CD What is the true difference in elevation
betWeen A and6? ..•.. ..... ...• .... .....'. .... .
® With tl)elevel inthe..samepqsiijonat D; to
what rod reading on B shouldJhe Une of
sight be adjusted. • '.' . i ..... '...'
.
(3). Whatls the corresponding rod reading on A
for a hotizontalline of sight Withlnsltument
still at D?
.
Solution:
CD True diff. in elevation between Aand B:
1.505 + x = 2.054 - e
x+ e:: 0.549·
x + 0.938 - e =1.449
x- e =0.511
x+ e:: 0.549
2x =1.06
x = 0.53 m. (diff. in elevation)
e
B
O.99lm
Ik~"7'I1 x
1.103 + e :: 0.991 + e + x
x=O.112m.
5-4Q
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lEVElIN~
® Rod reading on B with level at D:
<D•• C()(nplJW.·.tM•• • lfiffet¢nte • • • jry•• • ~leya. I. io.•.•.n
pelwi¥rrAalldEL/
~ • 'Nhat~~ould • ~~.tbecoIT@tJ()(tf'l<:l9Jngon
A.to•• g~tl • • ~ • • leM~I • • line•• of.si9htwth.~e
1I'1$II'\1ltlflllt$~llsiHupat~1
® • Whllt.~hPul~ • h~lYe9~~ • lhtl••rellpln9••pn§
..........ilhJhein$t!'ul'MntatA16 giYe a leVelline
QfsJghtT
.
Solution:
CD Diff. in elevation between A and B:
0.568 + e1 =e2 + 0.289 + 0.112
0.568 + e1 =~ +0.401
e2 - e1 = 0.167
Hori:.onwl finf'
..-11--"'1:--.... = 0
~-~
d -- -}--- --
12 -72
e2 = 6e1
LlTll'
6e1 • e1 =0.167
5e1 =0.167
e1 = 0.0334 m.
~ = 6(0.0334)
e2 =0.2004
_-_~:~_-
e
of :nghl
__ \_O~_­
line Of5ight
Rod reading on B = 0.289 + 0.2004
Rod reading on B = 0.4P94 m.
@
1.623 + X =C + 2.875
Rod reading on A to have a horizontal line
ofsight with instrument still at D:
Rod reading on A =0.568 + e1
Rod reading on A =0.568 + 0.0334
Rod reading on A = 0.6014 m.
e + 0.362 + x= 1.622
x- e =1.252
x+ e = 1.26
2x =2.512
X =1.256 (diff. in elev. bet. A and B)
® Rod reading on A to give
@~.t~~ra~hip$Ui\le.YLJM~~kent>Y.Kawa~
··sorv·.~rp .• b~te • ~nY • leyeljng•• I$¢QmjUcted.
Ifye • Mgllleers··H~u.~lIy • • ch!ck•• wh~lhEm •. lhe
engil'le~t~.mY~Hs.lry.PeJ'feCt.~djUSlment. • • AtW()
peg .f@t·is•• use<i • to.j;;ryeck.""hetherfhe.llne•• of
$i9hlls.lIlPerfect.~djU$trnentiln(j • t®.fOIlQViing
rqd.W'¥iings.<lre.tilken, ",itfi•• instrument.set.9P
n~llrA,I:l~¢f<si9hloIlAjst~2~rn;allcl
f()l'esI9ht•• readlllg • on•• B•• l$•• 2.~7S • • m.• • ""llh•• the
instrumentseluPHear.8,backsighIOI'l~is
1.622··m.• and.afo~ght.onA.jsO,3Q2.m.··
a level line of
sight with instrument at B:
RA =e+0.362
x+e=1.26
1.256 + e =1.26
e=0.004
RA =0.004 + 0.362
RA =O.366m.
@
Rod reading on B with the instrument at A
to give a level line of sight:
RB =2.875 +e
RB = 2.875 + 0.004
RB =2.879m.
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lEVEliNG
x + 1.563 + e1 =ez + 2.140
0.614 + 1.563 + e1 =2.140 + ez
0.037 + e1 =ez
~-~
2.5 -79.27
e1 = 0.0315 ez
0.037 + 0.0315 ez = ez
ez = 0.038 m.
@
kOint••M.l$••~~ldiS~~I.1r6fu.~lh.·.A.1nd.~f.Whil~.
Pj$f.50%a:w~y JrgfJlAl:lI()ngthe~~flSi()n
Qfllt1~AElaod1~~27m;fr9ma.
. ...
Rod reading on B for a horizontal line of
sight with level at P:
RB =2.140 + ez
RB = 2.140 + 0.038
RB = 2.178m.
•~ • • De!ermjn.~.th!3.true • difference••irlElI:Vatfon
b~WJeen~~ndB><.
·.@.f.)eterlllit1~me~rt't:lt.,n,.~.·r9(j.J'B.a9illgatB
WilhmelJjsfCJJmemsti~alg.< . , '.' .•
~ • • O~in~tpe.~~dlllg·9I).rQ(i~.f9r.a •
·••··• ··sfIUatP.
• • • no[iiM@lirleof•
• ~lght.With.the·.im~!ttiment
.
. ..
Soluticm:
G)
Difference in elevation between A and B:
Attigo®~m4 • 1WElliils~q@«¥I.W.J~re~~
§urVey.lng.pgmp~1I~y,.ttie:tW9·P91t1t$A.aMB.pt
Cl¢e~inr0ti9h·tert'''ill~ree"chl:lil>tanc:e • 2900
1TI·•• • frqrtj •.• a••·.miTlt••PQint.Pi • • fr°ITl .• ·.\Yhic:~ •.• th~·
.ll)~$~r~d • veijl~!.~~S!G)Ai$.t.~ .•~9'.~nq19.
Bj~f1·~'·ftevM9nClfCjs~t1()'-mlobe
342.pqm.abQ\I~~~lev~kPollltCjsm
b~~rjA~l'ldEl;<"
P
G)
0.296
2.5'-+---f-----~79.27-----I
x + 0.296 + e =0.910 + e
x= 0.910 . 0.296
x= 0.614m.
@
Error in rod reading at B with instrument
still at P:
tL----------
...' .. ...
.
.CqmpriJ¢•• • th~ • • (l!ffer~OClr • •ln•• .• ~I~y~ti9r
·• • • ····~~;6·~cjlw~d~fet. •.
effect.Of
~··qWllPl.lt~thw.pjffer~llp~ in Jllevation
.beweM13 and.q.•• • • • • • • • • • • • • • • •·.·•·•·
@ Compuletheel~vatlonmk'"
Solution:
CD Diff. in elevation between Aand B:
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S-42
LEVEliNG
hCr1 = 0.067 x2
hCr2 =0.067 (2)2
hCr2 = 0.268 m.
h1 =x tan 18'30'
h1 =0.3346x km.
h1 =334.6x m.
~ =2000 tan 8'15'
f7 =289.99 m.
h1 + hCr, = 111.356 + h2 t hCr2
334.6x + 0.067x2
= 111.356+289.99+0.268
x2 + 4994x - 5994.24 =0
x= 1.2 km.
x= 1200m.
tan 3'30' =J!L
2000
h2 = 122.33 m.
h0 =0.067 (2)2
h0 =0.268 m.
h 1 =2000 tan 1'30'
h1 =52.37
hc, = 0.067 (2)2
hc, =0.268 m.
H + h, + hC1 = ~ + h0
H +52.37 +0.268 = 122.33 +0.268
H= 69.96m.
@
@
Difference in elevation between Band C:
=h1+ hC1
=52.37 +0.268
= 52.638m.
@
Elevation of B:
Elev. of B = Elev. A + h1 t hCr,
Elav. of B = 200 + 334.6('1.200)
+0.067(1.2')2
Elev. of B = 601.62 m.
@
Elevation of C:
Elev. C = Elev. B - 111.:356
Elev. C = 601.62 - 111.2,56
Elev. C = 490.264 m.
Elev.ofA:
Elev. A = 342.60 + 122.33 +0.268
Elev. A = 465.20 m.
l@n$lq¢rlhgt@ • • ~ff~(;~$ • •§f•• Bury~ture • • al1d
ff}fr~pl!og,Jh~ •. q1ff~r~~¢El.inele¥ati9n • of.p()ints
• ~o~id~Ci~$.~~@U~IOa~~ • •~.~~Tf~~hill~O~
··eleVation.of.eal'\d.q.are.t8'SO'.!'f1spl';!(ltively.
(j)•• • lfq.ls2000.rl1·•• fr()I1l.~,ho . .lfari~.~.'fr()m.A?
@ . • If.th(:).~~.Vali9r·()ffJ..i$.~qUal·t(j.2OQm.,.ffod
. . .•.• JMtllEtvatiQnpfB.)
@ • ··Plndal$O·ihe.elevatiOOof.C.
Solution:
CD Distance of B from A:
. ..
.
Ai$i~pointh~vih~@ijl~vatl®l:rr1~.4fil}@
~qove9atMl11lal)dB.:lm:iC~t~pojl'lI~(:)f
unkno\\,I).~leva~i()ni .• aJ$•• jn·•.l1e.tWElen6.a~d¢.
BY.l11eansof.an•• jl1~trlJm~nt • $~t·1;22lli,.a~O\le
B,V~rl~l~nQle$.~re.9bserVed,.ttl~ttoA.b~irlQ
14'15'andJMtt9Cpeltlg+·~r~2',The
hOrizontaldlstanceABi$5~;r.20anqthe
horizol1tal•• djstance • aqi~ • 923,2;5 • nV•• M?king
urviitureiil'ld
almosphericrefrl@lon· ... .
due<lllQv.'ancefore~l'th'sc
®.·•• •Cqmpute•• • • lhe • • difference•• • • jn•• • elevation.
@
betw~nAandB:.
Oetermine • • th~ . •·djffl:lrence .• ·.in•• .• elevallon
betweenB?ndC· .•
® OeterminelheelevatiOnofG.
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LEVELING
Solution:
CD Oiff. in elevation ot A and 8:
A man's eyes t 75 m. above sea level can
barely see the lop of a lighthouse which is at a
certain distance away from a man.
What. is .th~ elevation of the lop of the
lighthouse·· above sea level if the
. lighthouse Is 20 km. away from the man~
® How far is the lighthouse from the nian in
meters if the top of the lighthouse 15
14.86 m.above sea level.
@ What is the height of the loWer at a
distance 20km, away from the man thaI
will just be visible without the Hne of sight
approaching nearer than US m. to the
water,
.
(j)
54~~20
tan 14'45' =
h1 = 144.07 m.
hCl 0.067 K12
hc,
=
=0.067(0.5472)2
hC1 =0.02
H= 144.07 -0.02
H= 144.05 m.
Solution:
CD Elevation of the top of the lighthouse:
Oiff. in elevation of A and 8
= 144.05 - 1.22
=142.83m.
1.75m
® Ditt. in elevation between 8 and C:
~~
(c
1.75 = 0.067 K,2
K1 = 5.11 km.
h=0.067 Kl
K2 =20 - 5.11
K2 =14.89 km.
h =0.067 (14.89)2
h =14.86 m. above sea level
® Distance from lighthouse from the man:
h =923.25 tan 8'32'
h =138.53
h~ = 0.067(0.92325)2
h~ =0.057
H =138.53 + 0.057 + 1.12
H= 139.81
@
Elev.otC:
Elev. ofC =130.48 + 142.83 + 139.81
Elev. of C =413.12 m.
h1 = 0.067 K12
1.75 =0.067 K12
K1 =5.11
h2 =0.067
14.86 =0.067 Ki
K2 =14.89 km.
0= K1 +K2
0=5.11+14.89
0= 20km.
Kl
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S-44
LEVELING
® Height of tower at a distance of 20 km.
away from the man:
h = h + fb (h 1 - h2) _0 067 0 0
2
0 1 + D-;,
.
1 2
h = 800 + 10 (600 - 800) _0 067 (12)(10)
12 +10
.
h =701.05
----.-----------71
h1
~
1 75m
Obstruction = 705 - 701.05
Obstruction = 3.95 m.
175m
h1 =0.067 (20)2
h 1 =26.8 m.
H =h1 + 1.75
H =26.8 + 1.75
H = 28.55 meters
® Height of tower at C so that it could be
visible from A with a 2 m. clearance above
hill 8:
Two hUls A and C have elevalkJns of 600 m.
and 800 m, respectively. In between A andC
is another hill 8 which has an elevation of
705 m. and is located at 12 km. from A and
10 km. from C.
.
CD Determine the clearance or obstruction of
the line of sight at hill 8 if the observer is
at A so that Cwill be visible from A.
@ If C is not visible from A, what height of
tower must be constructed at C so that it
could be visible from A with the line of
~l~ht having a clearance of 2 m. above hill
h =h + D-;, (h 1 - h2) _0 067 D 0
2
0 1 + D-;,
.
1 2
707 = (800 + x) +10 [6~2'1~~ +x)]
- 0.067 (12)(10)
707 =800 + x - 90.91 - 0.4545x - 8.04
x =10.91 m.
@
Height of equal towers at A anti Cso that it
will be intervisible:
® Whal height of equal towers at A and C
must be constructed in order that A, Band
C will be inlervisible,
Solution:
CD Obstruction of the line of sight at hill B:
h = h + O2 (h 1 - h2) - 0 067 0 0
2
0 1 + D-;,
.
1 2
) 10 [(600 + x) - (800 + x)]
+x +
12 + 10
- 0,067 (12)(10)
705 =800 + x- 90.91·8.04
x=3.95m.
705 = (800
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lEVEliNG
@
Equal height of towers at Alpha and
Charlie:
\
------"-----n--------- x
:
X
h = h. + 0 (h, - hz) • 0067 0 n.
"L
0, + ~
.
1......L
649 = (620 +x) + 15 [(6801+2~ ~~620 + x)]
¢PQrrPtltEl • tH.~ • e@,atiOtt9ftl1elj~Qf.~1ght •
~ts®io~.Bt~Vo.¥llththe.instrtJme.lt.plaqed
• • 1.lt•. $t;;lIIOt"\NPb~·sUchltJ~t •. $t@~(lCnaMj~
·0.067 (12)(15)
649 =620 +x +0.556(60) -12.06
x= 7.7m.
. 'k0uld•• • M•• VlsiblE!.<f~orri·· • •~t~tiClni • ~lpha_
{;9n$id~rlrg • Jh~· • ~ff#8 • • 91·.99rv~tute)and
tef!'a¢lklncortl3Ctl0n.}
®•• • ASS9lnirygttH~t.$fJ.llloriat~y9Wi(k9~~tl'\lpt
.. .··t~E;.IiQEl(lf~j~htJtCJnl$lati9~.AIPhil,·~ile
@
Height of tower at station Charlie:
A
<)b$~lhg$ta!l/)nCharlieaM~4m.J()Yw
B
.ls••con~!tu9t~d • • ~m • ·t9pQf.~tatI9m~rayCJ'
. . . . . P(lmp~t~ . f!m.·.b~i9hl·9t~qll~!t9VJE!rs.·.Elt
.•·.·• • • • §la~AA • AlpM•• an~ • ~Wioryqh~m~ln • 9@er••
·mat.bQ!tl.• three.$ti!111l0sas·Ob$rvedfrom··
• • •. ~a~on • ~phawlll s~n.l¥lihtr&l~tJlE! .• • • • • .• • • •.
.@..•WilMgllt•. (;;9IJstrtJc~ngElnyt<lWl!lr.Elt •. ~t~U9!1
•
..•••••~r~\j()~ •••• WhElfIW9ht9ftOW~r@ij$~b~
.···@nsttu(:t~~ • ~·.$t.aI16~·.ynar1i~SCl·thal.f)()tb.
. . .. .~ti9m§r~¥(lam:fQMl'lt~W99Id~~\li~tJl~
fromstatl6l1AJpM:
Solution:
G)
Elevation of line of sight at station Bravo:
.
h = h. + 0 (h, - hU _0 067 0 n.
"L
01 +~
.
l"'"L
15
645 = (620 + x) + 12 +15 [(680) - (620 + x)]
•0.067 (12)(15)
645 =620 + x +0.556(60 - x)· 12.06
37.06 = x + 33.36 . 0.556x
x= 8.33 m.
Thr~(h!!I~%a,jMPhM~IWM9Mqf
600••
6~nl,,6@5.'W • ~~4 •
m>respe¢ijMelY.•·•• ~• •~
in•• betWeeilAEll'ld··C~nd.·is·10~rit·fJ;9Il1Aa!l<l·
h =0. + 01~~ (h1· hi)·0.067 01~
12krn·®rn(k·
15
h=620 +12 + 15 (680-620)
(1).<Con~ideriJ@ • •
effect•• of.curVamrea.riiI·
.·.·.refraction.cotrEl%iorl,.whatisth~·creat~nce .
·0.067 (12)(15)
h= 641.27m.
tne.• •
%obstrycti91l.of•. • th.~ .• line • • of•• siQht•• al•• B
consldenngtt\alCisvlslblefromA . .
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S-46
lEVELING
@)•• • ff.~ • $m,.t@ffltW~.~.~.~tAAm~i.WM~.
.
. . •.•.•.•. ·Wl:)\:ll~J;lEi.~~~ • M19W¢~~il,tI~@.tlf! •
• .• • • !•• ~f~~~I1~.1~.~.~at.~.~.~~.~
@
Height of tower at C:
A
B
••~.• ,• •.er~et~·.at.CsQll'l~tl3aii'l
vm~ • $®61~.Miti~ij~9ID~~~ft9.~·
• •tWlllbe:
••jht~M$lb~ftClm.A· • • • ·i.···.·>······ ············::···.···••·•·•·•• ·•• ·>·U •
Solution:
~=600+X
CD Clearance orobstrucfion of the line:
h=625
h, =660
h = h. + 0, (h , - ~ _0 067 D n..
"l
D, + ~
.
' .....L
625
,
h = h.. + 0, (h • 0) . 0 067 0 n..
D, +~
"L
.
+x+
12[(660) • (600 + xll
10+12
- 0.067 (10)912)
,.....L
25 = x+ 12 (60 - x) _8.04
22
33.04 = x + 32.73·0.545 x
h = 600 12 (660.600).0067 (10)(12)
+ 10 + 12
h = 624.69 < 625
=600
.
x= 0.68 m.
Obstruction = 625 -624.69
Obstruction = 0.31 m.
® Equal height of towers at A and C:
fQuthillsA,I3.¢arldq~f$lfisl~i~htline.
tMele\'~tiClnsareA;(~1~m,B::;23~ro}
G.:;'•• ~H • rn.~rld.P • "'.3~~.l1J;.·te~~etlv~Iy. • • Th~
di$lanc~s()fB)Catldqfr6111~~re12k1tl,
4$l@anq§Ol<J"tl·fElsp~lively.¢onsidering
tMElff®f.Of•• (jlJrv!lltirEl • ~M • • tE!ft'a(jtioll•• ()f•• ll1e
h=625+5
h=630
h, = 660 +x
h2 =600+x
h = h + 0, (h , • 0) _0 067 D n..
2
0 , +~
.
,.....L
630
. 12 [(660 + x) - (600 + xl)
+x +
(10 + 12)
-0.067 (10)(12)
= 600
30 =x+
12~0). 0.067 (10)(12)
x= 5.31 m.
MM.
(j)
.p()ll1pute•• th$·t$I~ht • PfM4al•• tPWElf$•• ()n·A
$nd{)Josighf()v€r~ilrld8~j~~31T1.
>c!elll'llnce'......:: .'. ...••.•.• •••••..•.•.••••
@•• • ¢dmpute•• th&•• ¢I¢va@Q8t.th~ • lin~ • ot.siSht
13tWWiltlth$inMallaliofl9ttheElQlJal
tie@11$9ft9Wet~tA<1l@O;
~. .C()rnplJt~ • the • heighl•• qf.lo~r
' ...
• $t.A·.WlltJ•• a
ch.ara[1¢eoC3ri'lCaFP$omafPWlIlbe
Visibfe.jJ'°lTi.A••
2m.
. lf.tM.hei9ptgfIWWat.p.i$
..
. .
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LEVELING
Solution:
CD Equal heights of tower at Aand 0:
@
h1 = 247 + x
Considering hi/ss A, C and 0
h1 =247+x
hz =396+x
h =314 + 3
h = 317
h = h + Dz (h] +
Z
hz =396 +2
hz =398
h= 314 + 3
h= 317
h =h. + P2 (h 1 + hz) _0067 0 n.
f?L 0.067 0 1....n_L
0 1 +~
"L
15 [(247 + xl· (396 +x)]
+x+
45 + 15
- 0.067 (45)(15)
317 = 396 + x- 37.25 - 45.225
x= 3.475m.
317
@
Height oftower at A:
01 +~
.
1V Z
317 = 398 15 [247 +x- 398]
+
45 + 15
·0.067 (45)(15)
317 =378 + 0.25x - 37.75 - 45.225
x= 7.9m.
= 396
Elevation of line of sight at B:
qonsidElrir.g¢[/tVatureand·re~Oli()(j@ctiol1
aftnE!earth@~.·
0)"
..'
·Tt@.f,$,rn~~lng • on.·.•ffi~.l'pd~t • P6lht••$• j$:.
1;86rtt••·•• J:h¢~rteClit)hJ9r¢~rv~~~9Ply
iF • 9·Q48.• nk·.I(HJ·•• : : •. g98-,17.Jll.ar~the
r,orrepl~~I~vatj0l'l8fa.~.·f~6,~~m·;o/bat.
i~ .• tl1e.G<1[El~t\(inf()rreft~ct!9n()f\1Y?:<>·· .• • • •
h1 =247 + 3.475
h1 =250.475
hz= 396 + 3.475
hz = 399.475
h = he + ~ (h 1 +
L
0 1 + Dz
hzl. 0.067 0 1....L
n_
~ = 399 475 + 48 (250.475 - 399.475)
12+48
- 0.067 (12)(48)
h = 241.633 m. > 236 m.
1/.
.. .. '.'
® AIPtlll)tBI.ihE!•• F,§.•• readIQgi$Z.~nt,th£',
correctEld • e'~V{lti6n.of8.· • isi114·~ • ·m"
Cons19f!l'ing•• • reffaClIoJl•• and.p~rvaW%.····lf
HJ. ",117.Q§3m. atldlhe.C(jrrepti 011 fof
refraBtlon.is?005••• ~at • j%reCFrreFiol1
f!JrcllrvatU~?/
@
Considerins.curvalure.aryd • rSifacfion.the
corret;ledelevatian•• ofpoil1t.CiS$lh8$m·
Thef·S.reading.ontherodaIC.is.2J6.m.
TheC9ffecUon·for•• curvature••lsQ.046while
thatforrefi'actiOnis 0.004.· Determine HJ
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S-48
LEVE1I1G
Solution:
CD Correction for CUNature only:
Corrected F.S. = 238.17 - 236.35
Corrected F.S. = 1.82 m.
Error in F.S. reading = 1.86 - 1.82
Error in F.S. reading = 0.04
CUNature and refraction correction =0.04
CUNature and refraction = curvature - refraction
Solution:
CD Oiff. in elevation between Band C:
L~~·}.A.{8:1·5'
0.04 = 0.048 • x
x = 0.008 refraction correction
.!--
,
® CUNature correction:
F.S. =147.063 - 144.86
F.S. =2.203
CUNatureand refraction correction
= 2.23 • 2.203
=0.027
CUNature and refraction = Curvature - refraction
correction
0.027 = CUNature· 0.005
Curvature = 0.032 m.
@
Corrected F.S. = 2.16 - 0.042
Corrected F.S. = 2.118 m.
H.I. =Elev. + Corrected F.S.
H.I. =311.85+2.118
H.I. =313.968 m.
,--'"
ElfN.219.42m
hc1 = 0.067 (1.2)2
hC1 = 0.09648 m.
hc2 = 0.067 (2)2
hC2 =0.268
h1 =1200 tan 18'30'
h1 =401.51 m.
h2 =2000 tan 8'15'
~
=289.99 m.
H= (h l + hCl)· (h2• h0.)
H = (401.51 + 0.09648)· (289.99 + 0.268)
H= 111.348m.
ValueofH.J.
Curvature and refraction correction
=0.046 - 0.004
=0.042
1
h e & . ""':2600"'-..;
® Oiff. in elev. between A and C:
Oiff. in elev. = h2 + hC2
Oiff. in elev. = 289.99 + 0.268
Diff. in elev. =290.258 m.
@
Elevation of B:
E1ev. B =' 219.42 + 401.51 ... 0.09648
Elev. B = 621.026 m.
,Frornp(Jit1f.A.ill•• bEltween••Efan(JC,•• the.an~le~.·
ofeleyallon9tBaMqafe18'3Q'and8'15~
r~sPectiMelY .••• A8irtPi~20()Orn·fr9rn.A.al1dl?
is••12,OO.m·•• Jr'°mA, • • e.I#\laflgn•• of.A.is•. 219AZ.Ill.
abo\lt3se<lJilVill·
.
cD Compute the difference in elevation
betWeen B and C, considering the effect of
fheeartl1s curvature and refraction.
@) CQmpute the difference in elevatiori
between Aand C.
@ Gompute the elevation ofB.
A transit is sef up at point Bwhich is between
Aand B. The Yertical angle obsef\led towards
Ais known to be . 20' and thai of C is +12'.
The honzontaldlstance between A and B is
64~.80m. and that of Band C is 1032.4Q m.
The height of Instrument is 1.5 m.abOY8 8
with A having an elevation of 146.32 m.
Considering the effect of curvature and
refraction correction.
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5-49
lEVELING
®Gornpul~>th~<qff'fElre6Qe
bEltvl~enA,gr9~,>
.• •. . • .
in ~HNaliQn
i:ID.Pmnput¢•• tf\~.)differEll#¢e '>lnelElYaI160
pel>veenl\~nd8··)
@'.Comput!3c.ftIe·.ElIEl\l()ti()n.of.~i •.
Solution:
G)
Dift. in elevation between A and B;
Miradorhilhvfth.an. elev9tionofp26rn,iSoll.a
lifle•• ~Elwee8.A~ror() • ~ill • ~hlh$~· .• el~YWipry·.i~
660.·m·.9nd•• Q()thedr()lhillna...i'19.<3f1.~IW<:l~!Ofl
Of.60Qtii.·•• • Di~I~n~.()H.-1ir~~orhiUfml'll)\tirQl'll
hHljsl0kfl1argdistaryCe8fMir~dorniRfrPm
cathetlral.hi1l.i~ • 14·.km·•• • 9aO$i9~ringP~rY~tQte·
aodrefractioncorrectlOn.
.
.....
..
<D CompUte.theQbS!tuciIQtlqMMllM9lsltM
at.• Miragor.mll•• VItlElr·.o~~eiyjq~Q.1ftlEldqll
hlll.fr()Ill~WOl'a.tlill, • • • • • •.• • • • • • • • • • .• / • . ).<...<
.
® \foIhaf.would.betheDejghlof.ElqY~lm~rs·
tRe.~reQled.9t.i\uror~.hill.affij·cmhe9ral~IH
$o•• • that.Cal~eqr~I • • hill,lwror~Nll.lln~
Miradorhilj.wil'•• beirtervisi~l~wilh?·.4rfi~
t<>wererected.ft.the.~QP.of.MiW~%blll? •.• •.•
>••
® Ifou·. tow~r • wi!I .• b~.eract~d .• atAur()r<3·.hill
hC1
hC1
=0.067 (0.6428W
=0.028
' ~
tan 20. = 642.80
h1 =233.96 m.
H1 = 233.96 m. - 0.028
H1 = 233.932
artd•• Mirador••hitf,•• ~>twould~e.tbehe~ht
()f.tower•• tQbEl.e~cIElq •. (ltGlllheQfilllJilI~Q
Ihat.·.Mlrador•• and•• • Sathedral•• hill WiU•• be
lntervisiblefromA\Jrqrahill,
'.' .
Solution:
CD Obstruction of line of sight at Mirador hill:
Dift. in elevation between A and B
=233.932 - 1.55
= 232.382m.
(?)
Diff. in elevation between Aand C:
h~ = 0.067 (1.0324)2
hC2
=0.071
h2 =1032.4 tan 12'
h2 = 219.44 m.
H2 =219.44 + 0.071
H2 = 219.511 m.
Dift. in elevation between A and C
=H1 +H2
= 233.932 + 219.511
=453.443 m.
. ® Elev. of B:
Elev. of B =146.32 + 233.932 - 1.55
Elev. of B = 378.702 m.
(~) (h 1 - hz)- 0.067 Dl~
h -?2+
0 +~ /
1
12(660 - 600) - 0.067 (10)(12)
h=600+
h = 624.69 m.
10+12
Obstruction =626 - 624.69
Obstruction = 1.31 m.
5-50
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LEVEliNG
@
Equal heights of towers at Aurora and
Cathedral hills:
A line of levels is run from 8M, to 8M z which
is 12 km long. Elevation of 8M 1was found out
to be 100 m. and that of 8M 2 is 125.382 m.
Backsight and foresight distances were 150 m.
and 100m. respectively.
G)
h =h + QzJ.h1 - hz)· 0.067 °lfh
z
0 1 + fh
630=600+x
+ 12 [(660 +x) - (600 + x))· 0.067 (10)(12)
10 + 12
x= 5.31 m.
@
(2)
@
Height oftower at Cathedral hill:
Determine the corrected elevation of 8M 2
considering the effect of curvature and
refraction correction.
If during the leveling process the line of
sight is inclined downward by 0.004 m. in
a distance of 10 m, what would be the
corrected elevation of 8M 2?
If the average backsight reading is 3.4 m.
and every time it is taken, the rod is
inclined to the side from the vertical by 4',
what should be the corrected elevation of
8M2?
Solution:
G)
Error per set up = 0.00151 - 0.00067
Error per set up = 0.00084
12000
No. of set ups =150 + 100 =48
h =h + Oz (h, - hz)- 0.067 O,D:?
z
0, + Dz
626 = 600 + x
+ 12 [660· (600 =x)]- 0.067(10)(1~
10 + 12
12 (60-x)
3404
. -x+
22
748.88 = 22x + 720 -12x
10x= 28.88
x= 2.89 m.
Corrected elevation of 8M 2 considering
curvature and refraction correction.
hC1 =0.067 (0.15)z
hc, =0.00151
hC2 = 0.067 (0.100)2
hcz =0.00067
Total error =48(0.00084)
Total error = 0.04032
Corrected elevation of 8M2
. =125.382 - 0.04032
= 125.34168 m.
@
Corrected elevation of 8M 2 if the line of
sight is inclined downward by 0004 m.
every 10m:
!!L _0.004
150 - 10
h1 = 0.06
J1L _0.004
100 - 10
hi = 004
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S-51
LEVEliNG
Errorper set up =0.06 - 0.04
Error per set up =0.02
Total error = 0.02(48)
Total error = 0.96 m.
Corrected elev. of 8M2
= 125.382 + 0.96
=126.342 m.
Solution:
CD Corrected elevation of 8M 2 due to
curvature and refraction correction:
hI =0.067 (0.11W
h1 = 0.0008107
h2 =U.067 (0.070)2
h2 = 00003283
Error per set up = 0.0008107 - 0.0003283
Error per set up = 0.0004824 m.
9360
No. of set ups = 110 + 70 =52
® Corrected elev. of 8M2 if the rod is inclined
by 4' from the vertical:
Error in reading per set up
Total error = 52 (0.0004824)
T(}tal error = 0.0251 m.
Corrected elevation of8M2
= 31.388 - 0.0251
=3.4 - 3.4 Cos 4'
=0.0083m.
Total error =48(0.0083)
Total error = 0.3984 m.
Corrected elev. of 8M2
= 125.382 - 0.3984
=124.9836 m.
= 31.3629 m.
'.?1 Corrected elev. of 8M 2 due to line of sight
inclined upward by 0.003 m. every 25 m:
Diff. in distance per set up = 110 -.70
Diff. in distance per set up =40 m.
x 0.003
,40=20
x = 0.006 m.
9360
No. of set ups =1Tclt7O =52
A line of levels 9.36 km is run to check the
elevation of 8M 2 which has been found to be
31.388 meters, with 8M 1 of elevation at sea
level (reference datum). backsight and
foresight distances are consistently 110 m.
and 70 m. respectively.
cD Detemline the corrected elevation of 8M2
@
'j
considering the effect of curvature and
refraction correction.
if the level used is out of adjustment so
that when the bubble was centered the line
of sight was inclined 0.003 m. upward in a
distance of 20 m. Determine the corrected
elevation of 8M 2, ,
If at every turning points the rod settles
about 0.004 m., determine the corrected
elevation of BM 2.
Total error =52(0.006)
Total error = 0.312 m.
Corrected elevation of 8M2
= 31.388 - 0.312
=31.076 m.
@
Corrected elevation of 8M2 due to rod
settlement·
9360 1 51
.
.
No. af turnmg
pOints = 110 + 70 - =
Total error =0.004 (51)
Total error = 0.204 m.
Corrected elevation of 8M2
= 31.388 - 0.204
= 31.184m.
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S-52
LEVEliNG
P#i~g.M.$llgineers • lfw~l,tMreilqiM.on.arod
8qfu·~WW~S9b$IW"edt(l~~.2·~1t:n·Th~
.\)\;jbble·was••lf)l(el@·.ttlry•. ~·ll·p~~~Qll.·lbe • leVElI
·t\.lbE!aM.thE!rOd•• @Mingihdr$$~Wz.a74rrl.
ii) • D~te®lr@.t® .• ~ngletl1at.tl1~.blJbbl~.()fl.lh13 •
•·.·.·.···lul)¢WM•• qeYiMi!d•• d(je.tO.M.inM~aseln
M~f()(jljl#<ljol!lbYfu9\@9fD~I$I~C()Il~
ijpW~rdin$ec()fld~¢f*c.«·
•.. . •.••
·~ • ·.·P13~wjn~.m~.all~@M"~1~9r~11~Sp~Gl3.
i:iflhelubeirlsecondSofar&>
.·&>•.•.• D~teri1ijOO··t@r@iU$Qf • @&afOr$•• 6ftHe
:i~yelt~pejf@@~P;~c~@th~hibei$
<MOmmlong. .
.
Solution:
CD Angle the bubble on the tube was deviated
due to an increase in the rod reading:
~:F=
R\ iR
~8~
S =2.874 - 2.81 =0.064
0.064
tan8=-80
8" (0.000005) = 0-:4
8"
@.·.Gprn@~.mll.COlTec~Oll.fbbl:l.~pplied.tq.tb~.
~~vatl()nOfBM2-
®.•• ~OIl1P9tethecorrecte9.elElvftiOnofBM2.··.
&:i•• Cornpme.tl1E!.correc!ed.elev<ltionofBM3·····
Solution:
CD Correction to be applied to BMi
,
Station
BM 1
BM?
BMo
BM 1
Distance
Ikm)
0
4
6
10
-Observed
Elevation
1oo.00m.
121.42 m.
131.64m.
100.15 m.
Error of closure = 100.15 - 100
Error of closure =0.15 m.
-fL_.!
0.15 -10
C1 =0.06 Correction to be applied to 8M2
= 160"
® Angular value of one space:
160
0=-=32"
5
® Radius of cUNature:
o S
R-L
0=0.6(5)
0=3 mm =0.003 m.
0.003 _ 0.064
R - 80
R=3.75m.
® Corrected elevation of BMi
Corrected elevation = 121.42 -0.06
Corrected elevation =121.36 m.
® Corrected elevation of BM3:
Correction 6
0.15
10
Correction = 0.09
Corrected elev. = 13164 - 0.09
Corrected elev. =131.55 m.
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5-53
LEVELING
@
The .elevation of the base at station A is
1584m. above sea level. The barome1ric
r~;ldi(ig at station A was 65,53 cm, ofHg.at
thelnst..ml when the barometric reading at
statlon i B Which Is higher than A was
M,39 cm~ of Hg. The temp. at the time of
observation at A W3S g'C While 1hat I'll B was
22'C. rime of observation on both slatlorl
10:20 AM.
is
Adjusted elevation of B:
Corrected diff. in elev. =475.31 + 12.03
Corrected diff. in e/ev. =487.34 m.
Adjusted elev. of B =1584 +487.34
Adjusted elev. of B =2071.34 m.
Give~below.are.th$ • porfe$p6rKling.b~ro~triS
reildir9~.i1lI~o~lVerlplil~ .•••.• T~re¥Vi1ltionAf
•
·rv1o~mM~wpn ·is.1~9Q/Il'l· • • above.• ~a • le,,~1.
Motlm.~YQn •. is.loWeJ"tI1~n.Mpu~t.J\pIJ .• • • • • • • >•·•·•
CD .Compute the barometric reading at MoUnt
Apo at10,30 A.M.
••... ....• .
. ...
@Compule IhediffereMe in elevation
between Mount Apo and Mount Mayan.·
® What is the elevation of Mount Apo above
sealevel.
.
Solution:
Solution:
CD Uncorrected difference in elevation
belYieen A and B:
76
CD Barometric reading at Mount Apo at
10:30 AM:
76
z =19122 log h~- 19122 log h
2
76
76
z =19122 log 65.53 -1912210g 69.39
z =1230.95 - 755,64
z =475.31 m.
@
Station
Time
Mt. Arm
Mt. Mayan
Mt.Apo
9:25
10:30
10:57
Correction for the difference in elevation:
8 + 22
Mean temp. =- 2Mean temp. =15'C
Correction for diff. in elevation
=0.0253 (475.31)
= 12.03 m.
92 mn
f] ~
10:57
Barometric
Reading
.(cmofHq)
74.73
68.96
74.57
Air
Temp.
8.3'C
He
6.rC
m;, 7473} ' } 0.16
74.57
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5-54
lEVEliNG
Oiff. in time = 10:57·9:25
Diff. in time = 1:32 hrs. = 92 min.
Oiff. in time = 10:30·9:25
Oiff. in time =1:05 hrs. =65 min.
Diff. in reading = 74.73·74.57
Diff. in reading =0.16 em.
TM~re\latiCln°tlM~~perba~eAis37~1l1.
Whll~t~*\l.l)tth6 • lpwerb~~at~,m~~~vallQr
Is•• 1Sq.rn.• • • At.~giM~rt • • jnsta~tt~f¢ealtirn~~er •
re.adirg$ lndiCali~ that lhe i(jiffete.~.~ ••. ·W .
By ratio and proportion
¢l~aliCln • of~n •. lr~~rined~t~p()jll~·.qftpmth~ .
65_2...-
qpp¥P~$~AI~~~m·~I1d;th~dl~~rt~Bem
el#if9tj9Jifrom@!l.·19Yler.b~$tl§ . tOPOl!1t•• q IS.•
2?.. rrl.~in~thElttlJ~~I¢va~9!lpfpolrtCirl
92 - 0.16
x=0.11 em.
Barometric reading of Mount Apo at 10:30
=74.73·0.11
= 74.62cm.
b~tWeeni\~OdEl·
.. .
. ...
Solution:
® Difference in elevation between Mount Apo
and Mount Mayan:
H = 18336.6 (log h1 • log h2) [1 + T~oT2]
h 1 = 74.62 em. (barometric reading of
Mount Apoat 10:30)
h2 = 68.96 em. (barometric reading of
Mount Mayan at 10:30)
T1 = temp. at Mount Apo at 10:30
T2 = temp. at Mount Mayan at 10:30
{3'C
9:25 }
92
10:30
65
I
y
x=true difference in elevation between
AandC
2.2
{
10:57
X
6.1·C
92 _2.2
By ratio and proportion:
x 219
209 = 234
x = 195.60
65-x
x= 1.6'C
T1 = 8.3 • 1.6 = 6.7'C
T2 = 1.1'C
H = 18336.6 (log 74.62 • log 68.96)
[1 + (6.7500+ 1.1)]
H= 637.97m.
® Elevation of Mount Apo:
Elev. = 1200 + 637.97
Elev. =1837.97 m.
Elevation of C= 375 • 195.60
Elevation of C = 179.40 m.
y=219-195.60
y= 23.40 m.
Elevation of C= 156 + 23.40
Elevation of C = 179.40 m.
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8·55
Pdfbooksforum.com
COMPASS SURVEYING
Surveyor's .Compass - an instrument for
determining the. horizontal direction of a
line with reference to the direction of the
magnetic needle.
1. Needle bent· if the needle is not perfectly
straight, a constant error is introduced in all
observed bearings. The needle can be
corrected by using pliers.
1. Compass box = with a circle graduated
2. Pivot bent • if the point of the pivot
supporting the needle is not at the center of
the graduated circle, there is introduced a
variable systematic error, the magnitude of
which depends on the direction in which
the compass is sighted. The instrument
can be corrected by bending the pivot until
the end readings of the needle are 180'
apart for any direction of pointing.
from O· to 90' in both directions from the N.
and S. points and usually having the E and
W points interchanged.
3. Plane of sight not vertical or graduated
circle not horizontal.
2. Sight Vanes - which defines the line of
sight in the direction of the SN points of the
compass box.
3. Magnetic needle - has the property of
pointing a fixed direction namely, the
magnetic meridian.
1. Pocket compass - which is generally
held in the hand when bearings are
observed; used on reconnaissance or other
rough surveys.
2. Surveyor's compass - which is mounted
usually on a light tripod, or s.ometimes on a
Jacob's staff (a point stick about 1.5 m.
long).
3. Transit compass - a compass box
similar to the surveyor's compass,
mounted on the upper or vernier plate of the
engineer's transit.
4. Sluggish
5. Reading the needle
6. Magnetic variations
1. Compass is light and portable and it
requires less time for setting up, sighting
and reading.
2. An error in the direction of one line does
not necessarily affect other lines of the
survey.
3. The compass is especially adopted to
running straight lines through woods and
other places where obstacles are likely to
interfere with the line of sight.
8-56
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COMPASS SURVEYING
Why is the East and West points of a compass
interchanged?
1. The compass reading is not very accurate.
2. The needle is unreliable especially with
the presence of local attractions, such as
electric wires, metals, magnets that may
render it practically useless.
From the figure shows a compass having
a NS and EW calibration. In using a compass,
always sight the object with. the north end of
the compass and the compass needle when
pivoted and brought to rest gives the magnetic
bearing.
Magnerit.Norrh
Magnetic declination· the angle that a
magnetic meridian makes with the true
meridian.
Magnetic dip. the vertical angle which the
magnetic needle makes with the horizontal
due to uneven magnetic attraction from the
magnetic poles.
Isogonic lines - an imaginary lines passing
through places having the same magnetic
declination:
Isoclinic lines - an imaginary line passing
through points having the same magnetic
dip.
Let us sayan object on the right side is
observed, sight this object with the north end
of the compass. The needle atthis instant will
point steadily on the magnetic north, so a
reading could now be obtained as shown as
NE.
Agonic lines· imaginary line passing through
places having a zero declination.
(J)1'hei ol:i~rVedcomp~5sbearlng of a line in
,1981 W8S$, 37'3:0~S. ~nd the magnetic
When the compass traverse forms a
closed figure, the interior angle at each station
is computed from the observed bearings at that
particUlar point, the computed value which is
free from local attraction. The sum of the
interior angles of a closed polygon must be
equal to (n • 2) 180' in which nis the number of
sides of the polygon. Since the error of
observing a bearing is accidental, it is
assumed to be distributed equally at each
interior angle. The bearings are then adjusted
from a line whose observed bearing is to be
correct using the adjusted values of each
interior angle.
',' "detli~tioriQftheplade then W8S lmown to
be 3'10'W. ,'If'has also discovered that
"dUring' the6bserv~ijon local attraction of
" '. the place at that mom~t of 5'Eexisled,
'Fifldthe trueazirnuth aHhe line.
@
'The ~aring of aline from A to B was
measured as S. 16'3Q'W. It was found
that there-was local attraction at both A
'and
and therefore a forward and a
backward bearing w,ere taken between A
'and apoin! Cat which there was no local
attraction. If the bearing of AC was
8,30'10' E. and that of CA was N. 28'20'
W., What is the corrected bearing of AS?
a
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COMPASS SURVEYING
@
··m•• • •a•• • pattic~IN • • •y¢llr,.ihll·.·lTJa9flellc
d~l;lin~~()l'I·.~.~.·.1··1W • l:~n~I~~m~9neti9
wanngQflill~.peWas~,la'3{)'.W .• ••lfthe
~~cHI<!r'larja,tl(mp~y~aris>3·E.,
cl~tE!fl1'\ire .• ·~ • ~g~ti¢~rtI'lgQftlneOE
!5yearslarer?/
... ..... . . .
@
Magnetic bearing of line DE:
Solution:
<D True azimuth of the line:
True bearing DE =16'30' ·1'10'
True bearing DE =N 15'20' W
True bearing =S 37'30' E-1'50'
True bearing = S 35'40' E
Magnetic bearing of DE
= 15'20' + 1'25'
True azimuth = 324'20'
=N16'45'W
® Corrected bearing of AB:
Angle at A=16'30' +30'10'
Angle at A = 46'40'
Bearing of AB =46'40' - 28'20'
Bearing of AB = S 18'20' W
A field ish; the form ofa tegularpentagol1:
The direction oflha bounding sides were
surveyed wffh an aSSttmed meridian S' to the
right of the true. nOrth and south meridian. As
$urveyed .With an assumed meridian, the
beating Mone side AS ls N. $3'20' W.
<D Compute the true bealingof tine BC.
@
@
Compute the true azimuth of line CO.
Compute the true bearing of line AE.
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5-58
COMPASS SURVEYING
Solution:
Sum of all interior angles of
polygon.
(n- 2) 180 (5 - 2) 180 540'
=
a closed
=
Value of each inferior angle = ~O
Value of each inferior angle = 108
TN
MN
N. 73"00' W.
. . S. 72'15 E. ..
\DP®lput~fl'le~@l'ing9fljMIilQ.
®¢PmPlJleth~~eanjjg(jfII~¢p.
@ ()QfflPute(!lEl6ij<IDngb1Hh~.DI: .• • •
Solution:
LINES
AB
Be
CD
DE
AE
AB
BEARING
N.28'20'W
N.43'40· E
S. 64'20' E
S.7'40'W
S.79'4Q'W
N.28'20'W
AZIMUTH
151'40'
223'40'
295'40'
7'40'
79'40'
151'40'
CD True bearing of line BC =N. 43'40' E
® True azimuth of line CD = 295'40'
@
True bearing of line AE
=S. 79'40' W
A
E~
\~7'Jo,3
Aw
'30
B
60'
7~B
c~
..........
D
A
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COMPASS SURVEYING
Point
A
B
C
D
E
Interior angles
59'00' + 37"30'
180' • (37'30' +43'15')
73'00' +44'15'
180' • (12'45' +72'15')
120' + 13'15'
LINES
AS
=96'30'
= 99'15'
= 117'15'
=95'00'
= 133'15'
541'15'
Sum of interior angles
= (n - 2) 180 =(5· 2X180) =540'
Error = 541'15' - 540'00'
Error = 1'15' (toa big)
BC
CO
. = 5"
75' = 15'
Error per station
POINTS
A
B
f--C
0
E
DE
CORRECTED
96'15'
99'00'
117'00' .
94'45'
133'00'
EA
AE
Bearing of line CO = N, 73'30' W
® Bearing of line DE =N. 11'45' E
INTERIOR ANGLE
BEARING
S. 37'30' E
S.43'30'W
N.73'30'W
N.11'45' E
S.3T30'E
403'30'
- 360'00'
43'30'
+ 53'00'
106'30'
+ 85'15'
191'45'
45'00'
238'45'
83'45'
322'30'
CD Bearing ofline BC = S. 43'30' W
@
LINES
AB
BC
CO
DE
AB
AZIMUTH
322'30'
+ 81'00'
403'30'
(since there is no azimuth
greater than 360', subtract 360')
AZIMUTH
322'30'
43'30'
106'30'
191'45'
322'30'
CD Compute the deflection angte at C.
CV Compute the bearing of line DE.
@ Compute the bearing of line AE.
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5-60
COMPASS SURVEYING
Solution:
CD Deflection angle at C:
N
Solution:
Ar--__
C =180' - 142'54'
C=37'06'R
~~
Station
A
B
C
0
E
F
G
Interior Anales
180- L
180+R
180- L
180-L
180- L
180- L
180 -L
® Bearing of line DE:
AB =180' - 96'32'
AB =N 83'28' E
CD =142'54' - 83'28'
CD = S 59'26' E
DE::: 180' - 59'26'
DE= 120'34'
DE= 132'18'-120'3'l'
DE =S 11'44' E
® Bearing ofline AE:
EA =11'44' + 50'46'
,EA =N 62'30' W
Sum ofinterior angles
=1080 - LL + 180 + 'LR
=1260-2,L + LR
Sum ofinterior angles = (7 - 2) 180
Sum of interior angles = 900
900 =1260· LL + LR
360=LL-LL
Therefore the difference of the sum of
deflection angles is always 360'
2,L =50'20 +83'32 +63'27
+34'18' +72'56' + 30'45'
2,L =370'18' L
LR= 10'11' R
2,L·LR=360
370'18' ·10'11' =360'07'
CD Total error of the deflection angle:
Error = 360W . 360'
Error = 07' too big
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S-61
COMPASS SURVEYING
® Bearing of line DE:
Points
A
8
C
0
E
F
G
Corrected
85'20'L
10'32'R
83'32' L
63'2,l
34'18' L
72'56' L
30'45' L
Reflection
-01
+01
- 01
-01
-01
- 01
-01
Anale
85'19'l
10'12'R
83'31'L
63'26'L
34'1TL
72'55'L
30'44'L
Beanng ofline DE =N. 3"15' E
Check:
@
Bearing of line GA = S, 45'19' W
LL =85'10' +93'31' +63'26'
+ 34"1, + 72'55' + 30'44'
L.L = 10'12'
LL -'LR = 370'12' - 10'12' = 360 (check)
An engineers notebook gives the observed
•magnetic bealingsQf tile fOlklWlng traverSe, .
LINES
AB
BC
CD
DE
EF
FG
GA
AB
LINES
AS
Be
CD
DE
EF
FG
GA
AB
AZIMUTH
320'
320' -10'12'
=330'12'
330'12' - 83'31' =246'41'
246'41' - 63'26' = 183'15'
183'15' - 34'1, = 148'58'
148'58' " 72'55' = 76'03'
76'03' - 30'44'
= 45'19'
45'19' + 180' +(180 - 85'19')
= 320'00'
BEARING
S.40'E
S. 29'48' E
N. 66'41' E
N. 3'15' E
N,3"02'W
S.76'03'W
S. 45'19' W
S,40' E
AZIMUTH
320'00'
330'12'
246'41'
183'15'
148'58'
76'03'
45'19'
320'00'
CD Compute the local attraction at A
Compute the local attraction atB,
@ Compute thetoeali'lttractlon at C,
@
Solution:
B
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S-62
COMPASS SURVEYING
Solution:
TN
D
B
'MN
,
,,
,
0'32'
B
LINES
AB
BC
OJ
DA
CORRECTED
BEARING
S.68'19'E
N. 39'41' E
N.S1'OO'W
S. 63'30' N
LOCAL
ATTRACTION
A =0'41' E
B=1'19'W
C=09'W
D=O
Check:
48'11' + 108' + 83'19' + 120'30' =360'00'
A
c
CD True beanng of AB:
Tn/e bearing ofAB =48'45' + 0'52'
True bearing of AB = N49'37' E
® Length of AD:
Since the triangle is equilateral, it is also
eqUiangular.
B
CD Leal attraction at A:
Lcal attraction at A= 69' - 68'19'
Leal attraction at A = 0'41' E
® Local attraction at B:
Local attraction at B = 68'19' - 61'
Local attraction at B = 1"19' W
@ Local attraction at C:
[Dcal attraction at C= 39'50' - 39'41'
Local attraction at C = 09' W
A
c
AB=BC= CA
Area = (AB)(AC)
2 S'In 60'
The side A~ of an equilafenilfleld' ABC with'
an ~rea of 692:80 ~q",ri>hasamag1leti:C
bearing of N 48:45' Fin>1930 wheil the
magn~lic declination WB$O'52' E. A$sumeB
and C is on. the north eastsidji,
. . .. .
CD FInd the true bearing of A6.·••••. '.
.
® .Find the length of AD with pOint Don the
line Be and makillg the area of thetl'iangle
ABD one third of the Whole area.
.
@ Compute the bearing of line AD.
69280 _ (AB)2 Sin 60'
2
AB=40m.
1
A1 = 3" (692.80)
A1 =230.93
A 40 (x) Sin 60'
1
2
x= 13.3 m.
(AD)2 = (40)2 + (13.3)2.2(40)(13.3) Cos 60'
(AD)2 = 1245 .
AD= 36.3m.
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COMPASS SURVEYING
@
Bearing of line AD:
Sin f1I Sin 50'
13.3'=36:3
f1I = 20'08'
Bearing of AD =49'37' +20'08'
Bearing of AD = N 69'45' E
A triangUlar lot has for \Jne of its boundaries a.
nne 1500 rn, long Which runs due East «pm AThe eastern boundary is 900 m. long and the
western boundClry 1200 m. long.. Astraight li~e
@
Bearing of line 8E:
=90' • 53'0748"
=N 36'52'12" W
@
Bearing of line DA:
= 79'41'44" - 26'33'56"
= S 53'07'48" W
In the defiectionaogie trClverse wilh atraO$il
survey data below. Assume deflection T, T2
T3 and bearing Tl T2 is correct.
.:.
cuts the western· bqundaiyal the iniddle point
and meets Ihe easterly boundary E, 6{)O tit
from the Sf. comerR·
.
o
CD Find the bearing af line ED.
@ Find the bearing of line BE
@ Find the bearing of line DA.
Solution:
CD Bearing of line ED:
T2 • Ta,
Find the beClring of line T3• T4.
Find the bearing of line T4 • Tl .
(1) Find the bearing af line
Angle ACB is a right angle having the ratio
of its side as 3:4:5.
.
900
Sin A = 1500
A = 36'52'12"
B =90' - 36'52'12" = 53'OT48"
300
CotE= 600
E = 63'26'04"
0= 90' - 63'26'04" = 26'33'56"
DE Sin 63'26'04 = 600
DE=670.83
Bearing of line ED
=180' - (36'52'12" + 63'26'04")
=S 79'41'44" W
@
@
Solution:
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S-64
COMP'SS SURVEYING
I Def. < S to the right
R 96'42'
R 176'33'
R
IDef. <S to the left
IL = 69'16'
156'00'
429'15'
LR=429'15'
LL = 69'16'
359'59' Error 01' too small
Correction is applied only at T-4 =156'01'
156'01 '
Solution:
Line Azimuth
Back
Azimuth
Bearing
Distances
1- 2
2-3
3-4
4-1
1-2
42'25'
218'58'
149'42'
305'43'
42'25'
N.42'25' E
S. 38'58' IA
S. 30'18' E
N. 54'17' ....
N.42'25' E
118.38
83.22
83.44
85.26
225'25'
38'58'
329'42'
125'43'
222'25'
CD Bearing of line T2 - T3:
=S.38'58'W
~
Bearing of line T3 = S. 30'18'E
h
:;v Bearing of line T4 - T{
=N,54'17'w
Interior Ls:
LB =55'45' +58'40'
LC =180 + 14'30' - 58'30'
LD =180 -14'00' - 77'10'
LE =40'20' + 77'10'
LA =180 -40'15' -55'30'
= 114'25'
= 136'08'
=
88'50'
= 111'30'
= 84'15'
541'00'
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COMPASS SURVEYING
CD Error ofmisclosure:
Error ofmisclosure =541' - 540'
Errorofmisclosure =1'00'
® Adjusted interior angle at station C:
Correction per interior angle
= roo' =12'
5
Corrected interior Ls:
LB =114'25' - 0'12'
= 114'13'
LC = 136'08' - 0'12'
= 135'56'
LD = 88'50' - 0'12'
=
LE = 117'30' - 0'12'
=117'18'
LA =84'15' - 0'12'
= 84'03'
88'38'
Angle at station C = 135'56'
Solution:
@
Adjusted forward bearing ofline CD:
a = 180 - 77'10' - 88'38'
a= 14'12'
Bearing ofline CD = S 14'12' E
CD Error of deflection angle:
IR =55'30' + 99'30' + 44'00' + 92'00' +68'55'
LR=359'55'
LL=O
LR- IL =359'55'
LR- IL=360'Error = OS' (to be added)
Distribute the error equally at each station
=01'
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5-66
COMPASS SURVnlNG
CORRECTED
55'31' R
99'31' R
44'01' R
92'01'R
68'56' R
STATION
1
2
3
4
5
Arithmetical sum ofdepartures
= 1096 + 1088.84
= 2185.67
Correction in latitude:
64:13
15.97 -1870.97
~_
C1
~
~
C4
Cs
= 0.00854 (640.13) =
= 0.00854 (299.05) =
= 0.00854(281.98) =
= 0.00854 (362.44). =
= 0.00854 (287.37) =
5.47
2.55
2.41
3.10
2.44
, 15.97
Correction in departure:
--.fL =J 12.87
7.99 2185.67
LINES
1·2
2-3
3-4
4-5
5-1
LINES
1-2
2-3
3-4
4-5
5-1
@
BEARING
N 10' E
S70'20' E
S26'28' E
S65'32'W
N45'30'W
Distance
650
895
315
872
410
AZIMUTH
190'00'
289'31'
333'32'
65'32'
134'30'
LAT
+640.13
- 299.05
- 281.98
- 362.44
+287.37
+927.50
- 943.47
• 15.97
Linear enor of closure:
DEP
+112.87
+843.56
+140.40
·796.41
- 292.43
+1096.83
-1088.84
+ 7.99
_
Linear error of closure =.yr-(1--'5.'-97-)2-+-(7-.9-9i
Linear error of closure = 17,86
@
Area by DMD method:
Balance the traverse using transit rule.
Arithmetical sum of latitudes
=972.50 +343.47
;, 1870.97
C1
~
~
C4
Cs
= 0.00366 (112.87)
= 0.00366 (843.56)
= 0.00366 (140.40)
= 0.00366 (796.41)
= 0.00854 (292.43)
=
=
=
=
=
0.41
3.09
0.51
2.91
1.07
7.99
(uncorrected)
Lines
LIn
5-1
+5.47
+640.13
- 2.55
- 239.05
- 2.41
- 281.98
- 3.10
- 362.44
+2.44
+287.37
LINE
LAT
1- 2
2-3
3-4
4-5
5-1
+645.60
- 296.50
- 279.57
-359.34
+289.81
'1-2
2-3
3-4
4-5
DEP
- 0.41
+112.87
- 3.09
+843.56
- 0.51
+140.40
+2.91
- 786.41
+1.07
- 292.43
DMD
(corrected)
LAT
DEP
+645.60 +112.46
- 296.50 +840.47
- 279.57 +139.89
- 359.34 - 799.32
+289.81 - 293.50
DOUBLE
AREA
+7260418
+112.46
+1065.39
- 315888.14
+2054:75
- 57193033
+1386.32
-49816023
+293.50
+8505924
2A -- 1228315.28
A =614157.64 m2
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ERRORS IN TRANSIT WORK
2. Lower plate:
a. Outer plate
b. Lower clamp
c. Outer spindle
Transit - it is an instrument of designed
3. Leveling plate group:
. a.
b.
c.
d.
primarily for measuring horizontal and
vertical angle.
Lower clamp and tangent screw
Leveling screws
Leveling head
Foot plate
Line of collimation - a line segment joining
the intersection of the cross hairs and the
optical center of the objective~ens when in
proper adjustment.
1. Engineer's transit - a transit provided
with vertical circle and a long level tube on
its telescope.
Line of sight- the line joining the intersection
of the cross hairs and the optical center of
the objective lens, regardless of whether it
is in adjustment or not. When in
adjustment, the line of sight and the line of
collimation can be termed either of the
other.
2. Plain transit: a transit without a vertical
circle and telescope level.
3. City transit - a transit without a compass
and having a U-shaped one piece
standard.
4. Mining transit - a transit provided with an
Focusing - consists in the adjustment of the
eyepiece and the objective so that the
cross hairs and the image can be seen
clearly at the same time.
auxiliary telescope, a reflector for
illuminating the cross hairs and a diagonal
prismatic eyepiece for upward sighting, 60'
above the horizon.
5.
Theodolite - a transit designed for
surveying of high precision.
6. Geodimeter - a transit which can measure
distances using the principles of the speed
of light.
Three principal subsidivions of a
transit and parts under each
subd ivision:
1. Upper plate:
a.
b.
c.
d.
e.
f.
Telescope and telescope level
Telescope standard
Telescope clamp and tangent screw
Vertical circle and vertical vernier
Plate levels, compass box, upper
tangent screw
Vernier and inner spindle
1.
2.
3.
4.
5.
6.
The adjustment of the plate bubble
The adjustment of the vertical cross h'air
The adjustment of the line of sight
The adjustment of the standards
The adjustment of the telescope bubble
The adjustment of the vertical vernier
Four adjustments of the transit
which is not ordinarii performed:
7. To make the line of sight as defined by the
horizontal hair coincide with the optical
axis.
8. To make the axis of the objective slide
perpendicular to the. horizontal axis.
9 .To center the eyepiece slide.
10. To make the axis of the striding level
parallel to the horizontal axis.
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ERRORS IN TRANSIT WORK
1. Adjustment of the Plate Bubble:
Object:
To make the axis of the plate level
lie in a plane perpendicular to the
vertical axis.
Test:
Rotate the instrument about the
vertical axis until each level tube is
parallel to a pair of opposite
leveling screws.
Center the
bubbles fly means of the leveling
screws. Rotate the transit end for
end about the vertical axis. If the
bubble remains on the center, then
the axis of the plate level tube is
perpendicular to the vertical axis.
Correction: If the point appears to depart from
the cross hair, loosen the two
adjacent capstan screws and rotate
the cross hair ring in the telescope
tube until the point traverses the
enUre length of the hair. Tighten the
same screws.
Correction: I.f the bubbles become displaced,
bnng them halfway back by means
of the adjusting screws. Level the
instrument again and repeat the test
to verify the results.
Point sighted 1sf position
3. Adjustment of the line of sight:
2. Adjustment of the vertical cross hair:
Object:
To make the vertical cross hair in a
plane perpendicular to the
horizontal axis.
Test:
Slight the vertical cross hair on a
well defined point not less than 60
m. away. With both horizontal
mo~ions of the instrument clamped,
sWing the telescope through a
small vertical angle, so that the
point traverses the length of the
vertical cross hair. If the point
appears to move continuously on
the hair, then the cross hair is in
adjustment
Object:
To make the line of sight
perpendicular to the horizontal axis.
Test:
Level the instrument. Sight on the
point A about 150 m. away, with the
telescope on the normal position.
With both horizontal motions of the
instrument clamped, plunge the
telescope and set another point B
on the line of sight and about the
same distance away on the
opposite side of the transit.
Unclamp the upper motion, rotate
the instrument about the vertical
axis, and again sight at A with the
telescope inverted, Clamp the
upper motion. Plunge the telescope
as before, if B is on the line of sight,
the desired relation exist.
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ERRORS IN TRANSIT WORK
Correction: If the line of sight does not fall on
B set a point C on the line of sight
beside B. Marked a point D, 1/4 of
the distance from Cto B, and adjust
the cross hair ring by means of the
two opposite horizontal screws
until the line of sight passes
through D. The point sighted
should be at the same elevation as
the station occupied by the transit.
Correction: If the line of sight does not fall on
B, set a point C, on the line of sight
beside B. A point D, halfway
between Band C, will lie in the
same vertical plane with the height
point A. Sight on D, elevate the
telescope until the line of sight is
beside A, loosen the crews of the
bearing cap, and raise or lower the
adjustable end of the horizontal
axis until the line of sight is in the
same vertical plane with A.
4. Adjustment of standards:
5. Adjustment of the telescope bubble:
Object:
To make the horizontal axis
perpendicular to t he vertical axis.
Object:
Test:
Set up the transit near a building or
other object on which is some welldefined point A at a certain vertical
angle. Level the instrument very
carefully thus making the vertical
axis truly vertical. Sight at the high
point A and with the horizontal
motions clamped depress the
telescope and set a point B on the
ground below A. Plunge the
telescope, rotate the instrument end
for end about the vertical axis, and
again sight on A. Depress the
telescope as before, it the line of
sight falls on B, then the desired
relation exist.
r
d
To make the axis of the telescope
level parallel to the line of sight.
--------.----,~_-Ih_
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S-70
DRDRS IN TRANSIT WORK
Test & Correction: Use the two-Peg Test
method. Select two point A and B
say, 60 m. apart. Set up the transit
close to A so that when the rod is
held upon it, the eyepiece will be
about a quarter of an inch from the
rod. Look through the telescope
with the wrong end - to at the rod
and find the rod reading at the cross
hair if visible. If not take the
reading by means of a pencil point
opposite the center of field of view.
Tum the telescope toward Band
take a rod reading on it. Subtract
one reading from the other to secure
the apparent difference in elevation
betWeen the two pegs. The transit
is then taken to B and the operation
is repeated. The mean of the two
apparent difference in elevation is
the true difference in elevation
between the two pegs. The rod
reading on A with the instrument
still at B, is then computed. With
the computed value for the rod
reading at A known, the end of the
telescope bubble tube is raised or
lowered by means of the adjusting
screws until the telescope bubble
is centered.
Test:
Level the instrument first by means
of the plate levels and then by
means of the telescope bubble,
center the telescope bubble
carefully and observe if the vernier
reads zero. If not proceed as
follows.
Correction: Slightly loosen the capstan
screws holding the vernier and shift
the vernier lightly by tapping lightly
with a pencil until the zeros
coincide.
1. Non-adjustment, eccentricity of circle, and
errors of graduation.
2. Changes due to temperature and wind.
3. Uneven setting of tripod
4. Poor focusing (parallax) ,
5. Inaccurate setting over a point
6. Irregular refraction of atmosphere
6. Adjustment of the vertical circle and
vernier.
Object:
To make the vernier read zero when
the telescope bubble is centered.
1. Reading in the wrong direction from the
index in a double vemier.
W:'mier
2. Reading the vernier opposite the one
which was set.
3. Reading the circle wrongly that is reading
59' to 60'.
Vernier
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ERRORS IN TRANSIT WORI
2. Line of sight deflected to the left of
collimation.
clockwise.
Angle measurement
A) Error of line of sight: Line of sight not
perpendicular to the horizontal axis.
x Cos h Sin E = X Sin e
Sin E = Sin e
Cosh
Sin E = Sin 2 Sec h
For small angle, Sin E = E
Sin e =e
E =e Sec h
oJ"
/'"
Line ofcollimation
..
Line of sight ' .....
M- E2 =T - E1
T:: M- E2 +E1
T:: M- (E2 - E1)
T = M- E (sec h2- sec h1)
T=M-E'
1) When h1 = h2,there is no error.
1. Line of sight deflected to the right of line of
collimation. (clockwise)
. . . <."
Line of collimation
2) When one angle is depression and the
other is angle of elevation having
numerically equal values, there is no error.
B) Error of traverse axis of the telescope is
not horizontal or horizontal axis not
perpendicular to the vertical axis.
Line ofsight
Line ofcolIimorion .
...
/...
Line ofsighl
T = true horizontal angle
T- E2 =M- E1
T=M +Er E1
T = M + (e sec h2- e sec h1)
T = M + E (sec h2 - sec h1)
T= M + E'
whereE' = e (sec h2 - sec h1)
:clan h
~
e
:clan
Une of sight
y
Tan E = X tan htan e
Tan E = tan h tan e
E=etanh
For small angles,
tan E =E
tane=e
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5-72
ERRORS IN TRANSIT WORK
1) left end of transverese axis higher. Angle
measurement clocwise.
1
Line of sighl
Line ofcoUimatiofl
Line of sjg~t
T- E2 = M- E1
T= M + Er E1
T = M+ (e sec hr tan hI)
Tan = M + E'
E' = E2 - E1
2) Right end of transverse axis higher.
\
\
\
\
\
I
\. .t.--- Line oj collimufion
To measure an angle by repetition means
to measure it several times, allowing the
vernier to remain clamped at each time at the
previous reading instead of setting it back at
zero when sighting at the backsight.
The first measurement is made in exactly
the same manner as that described for a single
angle. Then, do not touch the upper clamp or
upper tangent screw, but loosen the lower
clamp turn the telescope back to the first
object and set exactly on it by means of the
lower clamp and tangent screw. The circle
now reads, not 0', but the first single angle.
Next loosen the upper clamp, turn the
telescope to the second object and set exactly
on it by the use of the upper clamp and its
tangent screw. The index of the vernier now
points to the double angle on the horizontal
circle. Half the angle now read is the improved
value of the required angle. If the process is
repeated and a third angle is mechanically
added to the last reading, the circle reading is
divided by three and still more exact values of
the angle is obtained. Six readings are
usually the greatest number of times taken
with the telescope in one position.
Laying Off An Angle By Repetition:
Line ofcollimation
To layoff an angle of 18'30'20" with a
transit to the nearest min. first layoff an angle
of 18'30' by a single setting and establish a
temporary stake. Measure this angle that has
just been laid off by repetition. Assume that
repetition determines 18'29'40" as the value of
the angle to the temporary stake. A new stake
must then be set a short perpendicular
distance called an offset from the temporary
stake, by 40".
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ERRORS IN TRUSIT WORK
Consequently if the temporary stake is
600 m. from the transit it would be necessary
to set the final stake.
600(0.0003)(40) - 012
f
h
60
-.
m. rom t e
temporary stake.
ANGLES BY REPETITION
The transit is set up at B, with the
telescope in normal position and a backsight
at A was taken. Assume that the true position
of the line of sight (line of collimation) is
deflected by an amount "e" as shown. When
the telescope was plunged and point C was
sighted, the line of sight is now deflected by
an amount equal to 2e from the prolongation of
line AB. The telescope is then rotated at 180'
about its vertical axis and point A is again
sighted but this time the telescope is in
inverted position.
Tnu position
0/
f{OfPronrg/ Axis
A
-'=~~.
(As applied to the Adjustment of Bubble Tube)
Let us say that there is an error of the axis
of the bubble tube fro its position by an amount
"e". If the telescope is rotated at 180', the
position of the axis of the bubble tube is now
doubled as shown in the figure, with reference
to its original, position in order to adjust the
bubble just move it at half this value.
Total error from first to second position is
2e. Therefore to place the axis of the bubble
tube to its true position. move by an amount
"e",'
The telescope is again plunged and point
D is established on the ground. Point 0 is
erroneous by an amount 2e from the
prolongation of line AB. The line of sight is
adjusted' by an amount e. backwards that is
determine first the location of E, that is
DE = 1/4 CD.
A
Liue of coli/marion
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5-74
ERRORS IN TRANSIT WORK
t@$ln!~letllr@t.vefui~rbftnElhbfifOIl~I<:irA~
A vernier is a device for measuring the
fractional part of one of the smallest divisions
of a graduated scale more accurately than can
be estimated by eye. The amount by which.
the smallest division on the vemier differs from
the smallest division on the vemier differs from
the smallest division on graduated scale
determines the least count of the vemier.
Least Count:
S
L=N
where L = least count
S = smallest division on scale
N = Number of divisions on the vemier
:~ • ~~~ffa~~~~I~~~d~~t~.am.~ • •19·4q'·
® ·.Wh~t~lbesJJl~II¢§~diYi$jPn.qfth~Ci~!e? •.
®Ho~ • • rnanY • diVi$iprl$ • • iltElthere.·.bn!b~··
YEirl1iE!t?
2. Retrograde vernier - the division in the
vernier is longer than the division of the
scale.
3. Folded vernier· is a direct vernier' it is
used where a double vemier would
long as to make it impracticable.
b~ too
..
Solution:
CD Number ofdivisions on the vernier:
Nv = number of divisions on the vernier
Ns =number of divisions' of the scale
Lv =least count of vemier
Ls = least reading of circle
L = length of vemier
For retrograde vernier,
Nv= Ns-1
For direct vernier,
Nv=Ns+1
L =19'40' =1180'
1
LV=20"=3
1. Direct vernier • is one in which the
smallest division on the vernier is shorter
than the smallest division on the scale.
.
Equation CD
L=NsLs
1180 = Ns Ls
Equation ®
Ls
Lv=Nv
Nv
Ls=3
From Equation ®
Ls=Ns+1
3
From Equation CD
Ls = 1180
Ns
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ERRORS IN TRANSIT WORI
Solving Equations CD &®
Ns+1_1180
3 - Ns
Ns2 + Ns = 3540
Ns- -1 ±...j'(1-?-.4-(-.3540-)
-
2
Ns = 59 divisions
Nv= Ns + 1
Nv=59+1
Nv = 60 divisions on the vernier
® Least reading of circle:
Nv
Ls-- 3
lia,.
Solution:
VERNIER
10
9
3
7
6
5
4
3
2
1
0'
IIIIIIIIII~III'IIJ~
I I Ii 1
1ST
Co<nad""",
60
LS="3
156'
ISS'
CIRCLE
Ls = 20' smallest division of the circle
L=30"
5=60=20'
3
S
L=N
pesigr.a1q~i!ernjei.bi • ~~ • ~mWilft~.~@t
readinQofaO'(m.th~sCl:lI~ .••.. lll~§t®El~·~lrlg··
of1Q()'3Z30";
'..'. ....
..
Solution:
r-15Spaces,
'I'
I.. 5
0
5
10
Itlll~ Itl tltI~' tltltltI" ,)tltI
II
C""ad",,,
fj i i '
95,1 ,
100
10
96
30=20 (60)
N
N = 40 divisions on the vernier which is
equivalent to 39 divisions on the
scale.
Since there are 7 spaces on the vernier,
only 6 spaces on the circle will give us the
coincide reading. Coincide reading on the
scale:: 155'40'
6 (20):: 157'40'.
LI4Spaces
5
L=N
30 =20(60)
N
N =40 spaces in the vernier
N- 1 =39 spaces in the scale
14 spaces = 14 (20) = 280'
=4'40'
O~n.a~rogradevemi~rf(ll'li • VernWrh~v(OQ
ale$tteadiOS.Qf.• gQ.~%,aM.~ •. I~ast • ~~(flryg • in
the .Circle.of • 3Q•• inln•••• lfldicale.a.·re~9hl$ • 9f
150'34'20... (3lVelher~lngofm~·C()irCi~.jr1
the stelle. > ..'
Solution:
VERNIER
Reading on scale coincide is
10<Y20' -4'40' = 95'40'
Folded vernier with a reading of 100'32'30"
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S-76
ERRORS IN TRANSrr WORK
S
L=-N
20 = 30(60)
N
N = 90 spaces on the vemier which is
equivalent to 91 spaces on the scale.
There are 13 spaces on the vernier,
therefore 14 spaoes on the scale must be
laid out to determine the coincide.
·.tJe~igQa·f()@ed.v~rni~r.fClr~d~~Wlth.~ • I~~
r$adln9pf2{)'~q • U'l£!••~~ .•• ·.1ll~~tr~t~.?.r~Mi~9
flf1()()'~Zillcl()t:kWi$edjr~Glltln.<>
. ....
Solution:
VERNIER
Therefore the reading on scale
= 150'30' - l' = 143'30'
, 14 spaces = T.
; 1 spaces =30 min.
~c.io'
99" 98-
9r 9~'·
l.----14 spaccs~
95'
SCALE
IJifiiJlli1i
Solution:
30
15
VERNIER
0
15
30
S
L=-N
_l' (60)
L-
12
L=5'
Number of divisions on the SCALE
Scale or circle is 12 - 1 = 11 division
Vernier coincidence
= ~ = 9 division marks on the vemier
Start counting from A to S, then proceed
from C, the coincidence is at 0 which is 3
divisions from C. Scale coincidence is
112"· 2(1) = 110',
L=§.
N
30 =20 (60)
N
N = 40 division in the vemier
N- 1 = 39 spaces in the scale
Oiff. in reading =100'32'30" • 100'20"
Oiff. in reading = 12'30".
12'30"·25 spaces in the vemier,
Start counting from A to S, then from C to
D. From A to 0 in the vernier there are 15
spaces, therefore it is equivalent to 14
spaces in the scale.
Therefore the reading in the scale for the
location of the coincidence
= 100'20'·4'40' =95'40'.
14 spaces on the scale =4'40',
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ERRORS IN TRANSrr WORK
Divide the corrected 3rd reading by 6
142'01'30" - 23'40'15"
6
Sta.
Sta. Tel. Repetitions Vernier
W
O~, O~.
-
Mean
~
BAD
C D
0
1
0-00 180'01' 00-00-30"
83'40'
C
R
6
142'02' 3;12'03'
A
R
6
0'01' 180'02' 00'01'30"
Since the 2nd reading is 83'40' add
mUltiple of 60', 120', 180',240'
142'02'30'
True horizontal angle = 60' +23'40'15"
True horizontal angle =83'40'15"
Determine the true value of angle ABC.
Solution:
:'::«::::/:::}?/<~:,:<::·:·}"<:\::i·.:)U.?·)])::-:::C) ?J.::::::
Mean values
.
.
First readmg =
00-00 + 00-01
2
First reading =00'30"
Third reading = 142'02'30"
FOl1lth reading = 00-01'-30"
Take the mean of the first and fourth
reading:
OO..()()..30
00-01-30'
OO-OZ-OO
Mean = 00-01' (too big)
Correction of third reading
= 142'02'30" - 00'01'00"
= 142'01'30"
Check on the adjustment of the Instrument
reveals the following errors. The line of sight
with the telescope on the normal position is
deflected 30" to the .left of its correct position
and the horizonlaLaxiS (fight end lower) maKeS
an angle of 15' with the true horizontal. ..
Compute the correction due to line of sight
not perpendicular to the hanzontalaxlS~ .
® Compute the correction due to the
horizontal axis no! perpendiCUlar to the
(j)
vertical aXis.
@
..
Compute the corrected horizontal angle
between Aand B.
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S-78
ERRORS III TRANSIT WORK
Solution:
CD Correction due to line ofsight:
ActyiF~ng@~ard$W.~!todElt~rmltleth$
lltimlJ1~ • (;)fl~Ae;.··Witl1.~ • fransit.at.st<lti()!'lA,.
LUte 0/
collimation
Une ofcollimation
E = e (sec ~ - sec h1)
E =30 (sec 60' - sec 45')
E =20 (2 -1.414)
E =30 (0.586)
E'= 17.58"
@
@
Correction due to the horizontal axis:
E = e (tan h2 -tan hl )
E: =15 (tan 60' -tan 45')
E' =15 (1.932 -1.0)
E' =10.98"
Corrected horizontal angle:
Line 0/ sight
~e.sjQht~ • pqlnt•• ¢.Wfll1:tl;i~.on.the • leff.pO~itfpll.
ofpPlnteandnl~$Yt~dayem9alatlgl$atg
to·.·~ • • 45'.•••·••·He•• ·ttiE#l•• tum$ • th~ • • jnslru@l~t • In.
c'o"q~$e • qlta#lQqarl9$lghlat•• pgjmtv·.··lm=;
rn~iJlj~redlW~~PnW~Q~I~.·RAf:li~~~·~O'<lIl~ •
·the.V~rt~I • .a.nglr·.·!l@di~~·.at.·.a • l'.la;$·•.90.·.•·.··The.
li~~9f.$jghl.\*Jlt,ti • mel~e~cope9nlhe.nOrm~l.
PO$tti()fljsdetfficlE!<tQ3\lothElrigl'ltQfjt~
CCirr~tpO~ll/<
".
....
CD ·R6inpqlalt1~~@r.qu~ • IO••.li;lle•• of~19ht.Mt
p~rpen~iptJt4rt9th~hBtiz9~lalal(is .•. • ;.; •.•. •. . •. •.•
® Cl)ruPllte.th~·.C#11'~ed .• h6riiont'd.angle
••
@. ···.~e~~El~~~j~~b~ • line • ~c.iS ~~.o.·$6,1·;
comR\-ffi:lth~aZltUuthPflloeAB.·
...'.....
SOlution:
CD Error due to line of sight not perpendicular
to the horizontal axis:
Line of
collimation
E =e (sec h2' sec h1)
T- El = M- E2
T= M- (E2 + El )
T= M- (Er El )
T=M-E'
T- E2 =M- El
T= M+ Er E1
T=M+E
Correctedhorizontal angle
=80'10'10' -17.58" + 10.98"
.=80'10'3.4"
E = 03' (sec 60' - sec 45')
E= 1.758'
® Corrected horizontal angle:
T=M+E
T=43'30' + 1.758'
T=43'31.76'
T =43'31'46"
® Azimuth of AB:
Azimuth of AB = 210'30' + 43'31'46"
Azimuth of AB =254'01'46"
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5-79
ERRORS IN TRANSIT WORK
® Angular error in segment CD:
8=30+30
Maladjustment of the. transit is such that the
Hne. of sight with the t~lescope In normal
position. is deflected '~ensec,onds to the leftaf.
its correct position ofnol perpendicular to the
horizontal axis•. This causesanerrorof8}9~'
in the measured horizontal angle when the
vertical angle to the firSt point Is 45' and thatpf
the second point is W....
of
8 =60"
8 =01'
® Offset distance:
Offset distance =X1 + x2
Offset distance = 50 Sin 30" + 50 Sin 60"
Offset distance =0.218 m.
(j) What is the value "e" in seconds,
..
® If this transit is used
layout a straight
to
line by prolonging a line. AS by setting up·
the transit at siJcceedingpolnts A BandC
and plunging the telescope.lftl'le
procedure were such that each. backsighl
were taken with .tha telescope at normal
position, what would be the angular error in
the segment CD,
...
....
@ What is the offset diStance from the
prolongation of lirl!! AS. from point () .• if
AB=BC",CD= 50m. .
.
'Alhat.~ltorwould.~.intro<luce~nthe.m~s~r~~
Mri4ontal.~ggle.lfthr<lllgl1non,adjustm~m ••• th~
hl)r.izgnl~taxlswer~h'rClinedO~rWlmtM:
hblitClhlal.
true
Solution:
A
A
Solution:
B
CD Error with one sight at the same elevation:
E =e (tan h:1- tan h1)
E=0.05 (tan 45' - tan 0')
E= 05'
B
® Error with both sights are 45',
E=e (tan h2 - tan h1)
E=05 (tan 45' - tan 45')
E=O
D
@
CD Value of "e":
E =e (sec h2 - sec h1)
8.79" = e (sec 60' - sec 45')
e" == 15"
Error with one sight is +45' and the other
is-45':
E =05 [tan 45' - tan (-45))
E= 05(1 + 1)
E= 10'
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S-80
ERRORS IN TRANSIT WORK
.~~1$~f:~zd~r~\~~I~.~~~;~~~~.~ffi~P:~~~.
of.elevatiol1.pfll1e~rst.poim·js.4f·~.0'\Yll'[~fh~t.
·~~~i:!t~1o~$e~Wd1~'6fWo~i~~~~~~dw~i.·
In••
Pr(ll<lngil1ga•• straigntnl1~lhetraO$ills$et.at
~, • a•~ck$jgm.~.tat<etl.a~A.··~Mt~et~lesg~w,·
.i$'plun~~d.t9 • P,•• 3O(l·I1'l;ir.·~ctv!lr.Cl:I.of.8 . ••. 1fW~
·V~iB?'a.Xhr.W~fejl1cll~~d.·p·Withth~.tr:Ue
yert~!.jn.a vElrticalplan~ • makirlg9W•• WiW.1he
•
prbba[)l~ln911n~flOn9n~tr~n$VE!r$~~ls;Jt
·~jrectien.of.fhe
WhiChls•• defleqted • tQ•• t~~.~gtrt.9fthe • • llne.pf
Cplnrnation••by•. an·~mQ9~t .• I3q~<ll • ftl.15·,<.lnt~t;l.
ill
Wa$fouM9Yt~~erm~AAuremel1tl'wme
instl'l.lrnent•• h~s .• an • Elrr()fjnltle • •line••. ()f.• ~i~ht.·
·.l?~er .c~se, • sttidingleY$I • v&~.H~.~ • • t(') • • C~~k.
~~~~dcJ~~~~6~Wg~eJ@r~~~~~1oa;j~
.fine•• what.y.'0ullL@•• the•• lihe?f
l!rrQl'inthelocaledposltionofQ. . ...
Wb,•• and • Bareat•• th~ • same•• ~evafion, • but
thf:l.vel1~al.anglefr°rTl •.~• ~Q9js+1AW . • • .·•• ·•·
®.. ffA,·.B.~ndC • are.alli'l.ttb~$<lll1e.eleVatl<lnf···
.~ •.• ·li.lhe••Vertlcal.an91.~*9fIl·B.t9·.A..~M.frorn6
toCis+1S'?
. ..
~n~ottfJ~ltiln$Yer$~~XI$jnJerm$?f
Solution:
• ~~idt;jons' .• • TMMQ[Jl<lrY~IIJ~o/rm~~i"i§19ri
c
®•• Cornpute•• lhe.·err()f~lIe •.~ • •line • ofsl9ht
.•.•. deflecled.t<lmeOgtit•• ><••.• • • • .• • • • .•.• • • •.• • • • • <•• • • • • • • •.
®C()rTlPlJle the~rrp@1.!~!w~AJnsv~r~a)(ls
Wllh.left~ndhighep«.· • • • • • >i<•• • • i•• • •. • • • • .• •. • .• •. •. .
® CornPule.the • • .s9(f~¢f • • hoH~om.~laJl~le
~el.YJeenlhelW<lp()lr\t$;
..
.
E
.
Solution:
CD Error due to line ofsight deflected to the
right:
E1 = e (sec h2 - sec h1)
E1 =15" (sec 63'58' - sec 42'30')
E, = 13.83"
@
@
Eror due to transverse axis with left end
higher:
E2 =e (tan h2 -tan h,)
e =2 (10)
e =20"
E2 =20 (tan 63'58' - tan 42'30')
E2 =22.62"
COlTected horizontal angle:
T=M+ E1 +E2
T= 150'20'20" + 13.83" + 22.62"
T'" 150'20'56.45"
B
CD Linear error if A and B are at the same
elevation:
E= etan h
E=01tan15'
E= 0.268'
x
tan E= 300
x = 300 tan 0.268'
Note: tan l' = 0.0003
x = 0.024 m. linear offset
® Linear error if A, Band C are all at the
same elevation:
= There is no error
@
Linear error if the vertical angle from
and from B to cis +15':
The error is doubled
E = 0.268 (2) = 0.536'
x = 300 tan 0.536'
x = 300 (0.0003)(0.536)
x= 0.048 m
e to A
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ERRORS IN TRANSn WORK
T~~ • • boritontal/~ngle/.peiw~~~twqpQifl~.
mea$wedCIO~KWISfij$17$'20'2o",Jl)~~ngl~
Of.i'lI~ttoIl9ff~e~n·tP9iflt~.42'Aq'whili'ltMt
·.of.• the.$ecqI'l9•• i$6a·~a'· • • TfuJ.II"l~tmment·w~$
th~rt • t$$tedWrerr0r5.ptqollirn<ltjon•• atidf()b111e
pr9bable.lnplill~llon.of •.~ • tran.sver%~axl$ .•••••!g
® Error in the transverse axis:
E2 =e (tan h2 - tan h,)
E2 =20" (tan 63'58'· tan 42'30')
E2 = 22,62" (is added if the left end is
higher than the right end)
@
lpe•• fgnn~r • ~ase • • ttWmsplacem~ht •.• ()t•• t~e
~eSOIl9.B6inf.es!abli~he<l.on. thefnre¥l~@ • ~ide
ofthetr!J~$itl$3cm·totnerigbtotth~mrsl
Horizontal angle:
H =179'20'20" + E1 + E2
H = 179'20'20" + 14.27" +22.62"
H = 179'20'56,89"
polllkT~e5epqjntsare1{)qm,fr()fl1lhetrary~t
sta~on-ln}h~latter@$!:!~slrldlngleYeIW~s
\)$eci•• t9.p~eCl<·theJnQlim~ti()n.ot
• thetran$¥etse·
a~i~~1l9v.'~sfClul'ldJClWJrthanll1ete~~of
.lhetrMsYer$e}lXl$•• • in .• Jerm$•• Qf••·f.·~IYISlQn$ .•
Thi'l•
seC()l'• • ~flg~ICit
ldS. . • • V~lue..• • 9f•. • 9~e •..•tllXiSi(ln•. • • j§ ••. 10·
CD C(jmputethE!err6f()fc~1Ilfl1atl® .•·
®. C9mpUle.th~.el'tQrm.l/1elranSV~r$e.;1l.Xi~, . .•.•.
@. CPrrlPlltEl.·thl:l•• hortzClntal.ansle•• !lEl.tvI~n • Jh~
fWOPOil'l!S.
..
Solution:
CD Error of collimation:
c
A
B
TM@jCl~In9.fuea$~tement • • were • • taker • ·tQ
9heCktl1epl~rP!tldiclllafity.ofa.t~p~ril19faBt°lY
chil1'ln~Y • • Qf • cirOUI~rCrOS$cseOIi9n, • • pplm•• ~·
Vias~~ta!)1j~h~~()r1We • • Qf()llnd.. ~9(ll.lt·4R·.·rn;
frQll1W~.b~s~6fth~l:hilllll~Y • ~lldth~~Mrf~~t
dlrnensIQ~tq • Itle··~se.me~$wep.carety11y~n~
fQ~l1<l •. tQ.be.~5·996rrl. • • 'fhe.WsfWrnentVl<lsset
up~t.~ • ~ryd • $ight~dso~~tQ.bi~eClthet(ip9f
fh~Fhimney;.m~ • tele:>C()pe.wa5lhenloW~~ed
and~ • l1Oiot.8~lon • .ll;ie.;¢irC!lmfereOMalthe
!)aSe,'Nhlchwa$JW~JWnhlhetel~%?lm~
ThejnstrorT)entVia:s.r~yersed •.• ~ndth~si~l1tina
repel.lted,ahdll1ejnSlrj,lm~l'It~~if;f~ln
adjU$lrn~n~ •••• tfJ~ • •lil'lfl • ()f~i~ht .•. a~ain • WIl~t.~·
With • lh~ • (ral1~lt.atNI~!:! • anQlel't'as.rnea$lJr~q
from.~to.aIIO~ • t?n~(lnftO • the.·right••~.~e.()f.th~
chitllney.~tth~b~s~.~ndfouhdbYrep~ijti~nfp
be.2'1~2Q·.$jmil~rty;~nglewa5measUrEld
0.03/4
tane=1OQ
e =15.47"
f ro m·.A8fo.lhee)(trt!ll1El·left5ide offheChil'Tln~y
~ndfoundtobe2·$8'40".
..
(j) • • lNhatls.theradlgsofthechjfl1ney? . .•. . . .
®
Error of collimation:
E1 = e (sec h2 - sec h1)
E1 = 15.47 (sec 63'58' - sec 42'30')
E1 = 14,27" (is added if the fine of sight is
to the right of the line of collimation)
.
@
···l'iow•• m\JC~.iStfl;l8hlmney.out.()(pl~trIb~t
ttle.top.inadlrectiQ~.atrightangles.~.~? •
HowfarjSthe qenter of chimney ITomthe
pointofQbse~tion.
.
.
.
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5-82
ERRORS IN TRANsn WORK
Solution:
.(1)
Radius of the chimney:
Mean value offhe angle
2'58'40" +2'12'20"
=
2
=2'35'30"
Angle AOS' = 2'35'30" • 2'12'20"
Angle AOB' = 0'23'10"
Sin 2'35'30" = 45.0~ + R
Solution:
R
0.04522 = 45.096+ R
45.096 (0.04522)+ (0.04522) R = R
0.95478 R =45.096 (0.04522)
R= 2.136
® Direction at rightangles to AB:
OB'= (45.096 +2.136) Sin 0'23'10"
OB' =0.318 m.
@
Distance of the center of chimney from the
point of observation:
AO =46.096 + R
AO = 46.096 +2.136
AO =48.232 m.
(1)
Elevation of the H.I. of the instrument atA:
=261.60 +5
=266.60
@
Elevation of the H.I. of the instrument at C:
=261.60 +5+2.8
=269.4
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ERRORS IN TRANSrr WORK
@
Elevation of 8:
tan 24'25' =2.8
x
x=6.17m.
200 - x =193.83 m.
DB
193.83
Sin 155'35' =Sin 7'56'
DB =580.52 fl.
h = 580.52 Stin 16'29'
h= 164.71 m.
Solution:
<D Emor in horizontal angle:
E = e (tan h2 -Ian h1)
E =04' [tan 50' • tan (-3D')]
E= 7'4,6"
Angular error ofline:
@
c
Elevation of B = 261.60 t 5 t 164.71
Elevation of B = 431,21 m.
<!51
B
A
I
I
I
I
D
1
x=4(0.145)
x= 0.03625
CD The horizontal axis of a· transit was
inclined at 4'wilhthehOlMihilildueto
nOIl"adjuslmeill.. The first5lgh.t.bada
vertical angle Of 50', lhenext had;;' 30';
Oetermine the error in themeasur~d
horizontal angle.
.
@
250
e =30"
Angular error =2(30'}
Angular error = 60"
Atransitis set upatB ;:Inda backsightat
A. By daublaraversal twopoiills Cand I)
~.t a distanceequill to 0.145hi,were
established. IfBe =250 m. and BD = 150
m. (app.), how much is the aiigtilarerrorof
the line of sight from true position: . .. .
@
Sin e = tan e =0.03625
In testing fofiM magnifyln9powet ota
level telescope, a transit is sst up and the
angle between two very far points which
are very near each other has been found to
be 5'15'. The level telescope whose,
magnifying power is desired is placed in
front of the transit telescope with its
objective close 10 the objective end of the
transit telescope. Again the same angle is
measured thru the two telescope a.nd found
to be 09', What is the magnifying power of
the level telescope?
@
Magnifying power:
5'15'
M.P. = 09'
MP = 315
.. 9
MP =35 diameters
5-84
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TRIANGUIiTiON
Triangulation - a method for extending
horizontal control for topographic and
similar surveys which require observations
of triangular figures whose angles are
measured and whose sides are determined
by trigonometric computations.
Four common geometric figures used in
triangulation:
1. Chain of single and independent triangles .
2. Chain of quadrilaterals formed with
overlapping triangles.
3. Chain of polygons or central-point figures.
4. Chain of polygons each with an extra
diagonal.
Solution:
CD Angle CAD:
A
Approximate method of adjusting the angles
and sides of triangulation systems.
c
1. Station Adjustment
2. Figure Adjustment
Two methods of adjustment of Quadrilateral
D
Angle CAD = 180' - 49'30' - 33' - 34'30'
Angle CAD = 63'
@
Angle BAD:
Consider triangle BCD:
. 1. Angle Condition Equations
2. Side Condition Equations
C~...l-_------'-..hD·
1. Sum of angles about a station = 360'
2. Sum of three angles in each triangle = 180'
Assume CD = 1
BC
1
Sin 72' = Sin 75'
BC= 0.985
BD
1
Sin 33' =Sin 75'
BD= 0.564
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TRIANGUlaTION
Consider triangle AOC:
@
Angle ABC:
A
C'b-.l.--------L..-"'oD
AC __
1_
Sin 34'30' - Sin 63'
AC =0.636
0.636
0.749
Sin B =Sin 49'30'
B=40'13'
Angle ABC = 40'13'
Consider triangle ABC:
A
Using Cosine Law:
(AB)2 = (0.636)2 + (0.985)2
- 2 (0.636)(0.985) Cos 49'30'
AB =0.749
Consider triangle ABO:
A
Solution:
CD Distance BC:
D
D
Using Sine Law:
0.564
0.749
Sin A =Sin 37'30'
A = 27'17'
Angle BAD =27'17'
84'30'
A ~...J...----:;::::-----'-""""'B
BC
500
Sin 79'30' Sin 47'30'
BC =666.81 m.
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5-86
RIANGUlATION
@
Distance BD:
AD
1
Sin 80' =Sin 40'
BD
500
Sin 28'30' =Sin 67'
Distance CD:
AD = 1.532
AB
1
Sin 50' =Sin 50'
(CDf =(8<12 + (BDf
AB= 1.0
·BD = 259.18m.
@
•2(BC)(BD) Cos 31'30'
(CO)2 =(666.81'f +(259.18)2
(BD)2 =(AB)2 + (AD)2 • 2(AB)(AD) Cos 20'
·2(666.81 )(259.18) Cos 31'30'
CD = 465.94 m.
(BD)2 = (1)2 + (1.532)2·2(1)(1.532) Cos 20'
BD=0.684
AD
BD
Sin (50 + 0) =Sin 20'
1.532 _ 0.684
Sin (50 + 8) - Sin 20'
50 + 8= 130'
8 = 80' (angle CBD)
@
Angle BOA.'
A
11_lllllil~'llli!jii;
Solution:
BOA = 180 - (20 + 130)
CD Angle CBD:
BOA =30'
A
@
Assume AC = 1.0
_1__ ~
Sin 50' - Sin 80'
BC= 1.2856
CD
1
Sin 60' = Sin 40'
CD =1.3473
Ang/eBDC:
C~~---,---...JD
BDC = 180· (30 +80)
BDC= 70'
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TRIANGlllATlON
<D Distance BC:
Two stations A and l:lare 540 in. apart. From
the following triangulation staflansCand Don
opposite sicies of AB, lhe fol19Wing angles
were observet!. "
'
. .. '.' .
., '.
.','
',.,'._
,',.,
,',
..,.
AngleACD=54'12' .'
M9le
~ 4fZ4'
ace
A
'.
~
540m
B
a ~ 180' • 54'12'·49'18'
An9leADg=49'18' ...' .
a~
Angle BOC:: 47'12'
~~
76'30'
180' ·41'24'·47'12'
~ ~91'24'
0~54'12'+41'24'
o ~95'36'
Using Cosine Law
Considering triangle ABC:
(AB)2 :: (BC}2 + (AC)2 •2(BC)(AC) Cos 95'36'
(540)2 = (BCf + (1.062 BC)2
• 2(BCX1.062 BC) Cos 95'36'
BC = 353.38 m.
Solution:
@
Distance CD:
Using Sine Law
Considering triangle AOC:
CD _ AC
Sin 76'30' - Sin 49'18'
CO= 1283AC
c c
Considering triangle COB:
CD _ BC
sin 91'24' - Sin 47'12'
CD = 1.362 BC
CO = 1.362 (353.38)
CD = 481.30 m.
@
Distance AC:
1.283 AC =1.362 BC
AC= 1.062 BC
AC = 1.062 (353.38)
AC = 375.38 m.
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8-88
TRIANGUlaTION
® AngleDAC:
Angle DAC + 42' + 30' + 78' : 180
Angle DAC : 30'
lIi&il;.~~i~1
® Angle DAB:
Angle DAB: 42' + 30'
Angle DAB: 72'
Allies
OM.. >
cao>
• .~• • • PqmMt~~M~l~.~; • • •i• • • • • • • • • • ·•• • • .• > .
T~~.q~~~ljn~ • • AI? • ()f.a.lri@941~I!M • $Y$IMtis
~~lijl_~6~1IJ··<
egY~to4()Q • nl·.~~g· • • • §taliptlS.G~r~ • R·~r~
Qthllr.p()inf~.pfth~triMgglation • $Y5@JJ.• • • T~e
al1gles•• ob§erV~dfr<lrtiA.~nd • •~ • ~r~.~sfollAW~.
Angle.[)/l$.*2r3Q\~ngleGA13·#.78'30·,.an9r~
Solution:
.
CBA=:$2'~H~~qgl~1:$C:#~:~Q'.>·
CD Angle BOA:
c
Solution:.
CD Distance CB:
A
B
Assume BC: 1.0
~_-1:L
Sin 79' - Sin 42'
AC: 1.467
CD
1
Sin 49' : Sin 52'
CD: 0.958
Using Cosine Law:
(AD)2 : (0.958)2 + (1.467)2
'. (0.958)(1.467) Cos 20'
AD:0.655m.
A
B
78'30'
Using Sine Law:
AC
_~
Sin (52 +8) - Sin 20'
1.467
0.655
Sin (52 +6) : Sin 20'
52 + e: 50' or 130'
52 +8: 130'
8: 78'
Angle BOA: 78'
400
CB
Sin 49'30' =Sin 78'30'
CB: 515.47 m.
® Distance DB:
DB
400
Sin2T30' =Sin 70'
DB: 196.55 m.
82'30'
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TRIANGULAnON
@
Distance DC:
(DCP = (515.47)2 + (196.SSj2
- 2(515.47X196.55) Cos 30'30'
DC = 360,21 m.
<D Distance AD:
Consider triangleCDA:
AD
180
Sin SO' =Sin 41'
AD =210.18 m.
A:·,I.d.B.are.tw(}.polnt$·IOcaledQll~acn~t1k
ofa.riy~r·13lld.n~rtl1~abut1nElntsllf(ipmp~d
brldg~· • Wd~tflrrnirflltsdi$l?nC~,~~~$~lirfl
CO 180.m.•• 10llQ.W<lSE:lS@bli$l)ed·onp0E:l9~®-.
of\herjVet13r@tl'leJrl3nsltwa$~~II.lPat
statjl)qs•• C.··iln~·Oandtti~.aZiJ11Utb • w¢rff·.t~kElii··
asfollOws:
..
..
Solution:
N
c
. .... .
@
Distance BD:
Considering triangle COB:
Using Sine Law
180
BD
Sin 25' =Sin 80'
BD = 419.45 m,
@
Distance AB:
Consider triangle ABO:
Using Cosine Law
(ABf =(210.18j2 + (419.45)2
- 2 (210.18)(419.45) Cos 14'
AB =221.43m.
tn a topographic survey, three triangulatiOn
stations A, Band Care sighted from a point P.
The distance between the stations are
AS =500 m., Be =3SO m. and CA =450m.AI
P, the angle sublending AC is 4S'while for Be
is 30'. AC is due North.
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5-90
TRIANGUlaTION
Taking triangle ABC:
a=3S0 b=450 c=500
Solution:
CD Distance CD:
c
Using Cosine Law:
(3S0}2 = (45W + (SOO}2 - 2(450)(SOO} Cos A
A=42.8'
From triangle ACD:
B =42.8' - 3D'
From the triangle ABO:
~-~.
Sin 45' - Sin 105'
AD = 500 Sin 45'
Sin 105'
AD'=366 m.
B = 12.8'
Using Cosine Law:
(CD}2 =(450}2 + (366f
- 2(450)(366} Cos 12.8'
CD = 123.65 m.
@
BD
500
Sin 30' =Sin 75'
BD =500 Sin 30'
Sin 75'
BD=2S9 m.
Distance PA:
c
A,&-..I...---~B
c=500
123.65_ 366
Sin 12.8' - Sin a
. _366 Sin 12.8'
SIn a - 123.65
a=41'
c
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TRIANGUlaTION
450
PA
Solution:
Sin 45' =Sin 41'
PA =450 Sin 41'
Sin 45'
PA =417.55 mm
@
CD Distance CO:
D
~
30'
Azimuth of PA:
A
N
B
C
A
u_~--p
c
~-----~p
Azimuth of AP = 274'
Azimuth ofPA =274' -180'
Azimuth of PA = 94'
.
.
>-•.
.'.
.
."
,
~.~
.
'
....
.,.
.
.
Considering triangle ACB:
Using Cosine Law
(400)2 = (600f +(300)2·2(600)(300) Cos "
360000 COS" =(600)2 +(300)2 - (400)2
36.34'
" =36'20'
c
c
Construct a circle passing through A, B
andP.
Considering triangle ABO:
Using Sine Law
AD
300
Sin.15' = Sin 135'
AD = 109.81 m.
,,=
AL-.__.----,7
,,
,,
,,
,
,,
,, ,,,
\lY.30
e
'.....,...,'
" ,
\~I
P
,
B
Aftt!:.~~;:--7~
,,
,
I
I
\
\
I
,
,
,,
,
\
"
- p _.'
."
.I"
Considering triangle ACD:
Using Cosine Law:
(CO)2 = (600)2 + (109.81)2
- 2 (600)(109.81) Cos 6.34'
CO = 491.01 m.
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5-92
TRIANGUlAnOI
® Distance AP:
c
Using Sine Law
Sin f!, Sin 6.34'
109.81 = 491.01
II = 1.42'
II :: 1'25'
Considering triangle ACP:
Using Sine Law
~_
AP
Sine 15' - Sine 1'25'
AP= 57,45m,
@
Azimuth of BP:
Solution:
CD Angle of intersection FEC:
Angle FEC = 77'10'
p
Considering triangle ABP:
Sin B Sin 45'
57.45 =300
B=T47
A = 180' - 45' ·7'47
A = 127'13'
BP
300
Sin.e 127'13' =Sin 45'
BP = 337.85 m.
® Distance BC:
(BC)2 = (376)2 + (417)2
- 2(376)(417) Cos 138'29'
(BC)2 = 137500 + 16900 + 236000
BC= 737 m.
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5-93
TRIANGUlATION
@
Distance EC:
Solution:
CD Distance BD:
Be
~
~
737m
Sin" _ sin 138'29'
.-;1] 737
B
C
A
. - 417 Sin 41'31'
SIn" 737
,,= 21'36'
a
+" + 138'29' =180;
a
=19'55'
BE
737
Sin 53'22' = Sin 102'50'
BE = 737 Sin 35'22'
Sin 77'10'
BE=435 m.
CE _
737
Sin 41'48' - Sin 102'50'
CE
= 737 Sin 41'48'
Sin 77'10'
CE=505m.
A
c
Consfden'ng tdangle ADC:
Using Sine Law
Sin A Sin 140'
1000 = 2355.45
A= 15'50'
C = 180' - 140' - 15'50'
C= 24'10'
Considering ABC:
Using Sine Law
AB
2355.45
Sin 30' = Sin 125'
AS = 1437.74 m.
Be 2355.45
Sin 25' = Sin 125'
BC:;; 1215.23m.
Considering triangle ABO:
" = 25' -15'50'
9'10'
Using Cosine Law
(BD)2 = (150W + (1437.74)2
- 2(1500)(1437.74) Cos 9'10'
BO:;; 242.90 m.
,,=
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8-94
TRIANGUlAnON
@
Distance AP:
Consider triangle APC:
Sin ACP Sin 55'
2829.86 =2355.45
Angle ACP= 79'47
Angle CAP = 180' - 55' • 79'47'
Angle CAP = 45'43'
A
Using Sine Law:
PC
2355.45
Sin 45'43' = Sin 55'
PC= 2058.66m.
B
D
11_.it
A
lIilllliill:
p
I:E4i."lli~l,
Using Sine Law
Sin B Sin 9'10'
1500 = 242.90
B =70'47'
Using Cosine Law
(ACf =(1500f + (1ooW
- 2(1500)(1000) Cos 140'
AC =2355.45 m.
·@·.·PPttiP@:i~M~~g~······"'······
Solution:
CD Angle ACB:
B
Consider triangle ABP:
Using Sine Law
AP
1437.74
Sin 79'47' =Sin 30'
AP = 2829.86 m.
@
Distance PC:
c
p
c
Angle CPB = 360' - 71'30' • 10'30'
Angle CPB = 178'
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TRIANGULATION
Using Sine Law:
7.46
17560
Sin 0 = Sin 110'30'
0=01'22"
Solution:
CD Distance AB:
B
7.46
24614
Sin a. = Sin 178'
a = 00' 2.18"
{I, = 180'
- 110'30' - 01 '22"
= 69'31' 22"
0= 180' -178' -0'2.18"
o = 1'59' 57.82"
Angle ACB =69'31'22" + 1'59' 57.82"
Angle ACB = 71'31' 19,82"
216'43'20"
{I,
B
A
58'12'30"
@
@
Distance AP:
AP
17560
Sin 69'31' 22" = Sin 110'30'
AP = 17562,61 m.
Distance PB:
PB
24614
Sin 1'59' 57.82" = Sin 178'
PB = 24606.55 m.
Ecc.A
Using Cosine Law:
(AB)2 = (4.50)2 +(18642)2
.- 2(4.50)(18642)Cos 58'12' 30"
AB = 18639.63 m,
@
Angle BA Ecc. A:
B
InatriansUlati06~Il~@ttl§SfatlOJ'\(~cc;Al
IsoCCllpi¢~jl1st~~~()fll)l!tM~~tatj96A;
ObserVationsar~ thenmade·lo • tffie·.staliori.A
amt lostatiooB, T@otJsaNatl<>llareas
f\:flloWs:i>
. . ..
AZIMQTHOlSIAN.GS
158~3mS(l"
.4.50fu,<
216'43'20" 18642,OOm;
<D Find the distance AB.
...
~ Find thEl angle BA Ecc. A .
@ Compute the aZimuth of AB.
Ecc.A
Using Sine Law:
18642
18639.63
Sin e = Sin 158'12' 30"
e =121'46'47.6"
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S-96
TRIAIGUlAnOI
® Azimuth AB:
Using Cosine Law:
(AC)2 = (1017)2 + (800.63)2
- 2(1017.22)(800.63) Cos 74'44' 50"
AC = 1116.80 m.
B
Using Sine Law:
800.63
1116.80.
Sin 8 = Sin 74'44' 50"
e =43'45' 47.25"
Ecc.A
Azimuth AB = 180' + 36'44' 2.32"
Azjmuth AB = 216·44' 2.32"
A
B
c
Angle BCA = 180' - 43'45' 37.5" - 74'44' 50"
Angle BAC =61'29'32.5"
x + 118'25' 40" + 158'13' 50" + Y
+43'45' 37.5" = 360
x + y =39'34' 52.5"
AP
1017.22
Sin x = Sin 118'25' 40"
p.p = 1156.70 Sin x
AP
116.80
Sin y = Sin 158'13' 50"
AP =3011.28 Sin y
Solution:
CD Angle PCA:
c
1156.70 Sin x =3011.28 Sin y
Sin x = 2.603 Sin y
x = 39'34' 52.5" - Y
Sin (39'34' 52.5" - y)= 2.603 Sin y
Sin 39'34' 52.5" Cos Y- Sin y Cos 39'34' 52.5"
= 2.603 Sin y
0.63717 Cos Y- 0.77072 Sin y =2.603 Sin y
0.63717 Cos Y= 3.37372 Sin y
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TRIANGULATION
0.63717
tan y =3.37372
y= 10'41'42.23"
Angle peA = 10'41' 42.23"
Solution:
CD Angle ABP:
B
® Azimuth of BP:
p
e + (J + 257'15' +- 26'35' +44'15' =360'
e + (J = 31'55'
x + y =39'34' 52.5"
x = 39'34' 52.5" ·10'41' 42.23"
x =28'53'10.27"
Using Sine Law:
AP _ 6600
Sin e - Sin 26'35'
Angle BAP = 180'·28'53' 10.27" ·118'25' 40"
Angle BAP = 32'41' 9.73"
Azimuth ofBP = 225'20' 10" + 105'15' 10"
+45'51' 39.73"
Azimuth ofBP = 376'26' 59.7"
Azimuth of BP = 16'26' 59.7"
@
Distance BP:
'BP
Sin 32'41' 9.73"
BP =624.66 m.
AP= 1474.64 Sin e
Sin AP
6800
Sin r.. =Sin 44'15'
AP =9745.05 Sin t:\
1017.22
.1474.64 Sin e = 9745.05 Sin 8
= Sin 118'25' 40"
Sin e =0.6607 Sin 8
e = 31'55' • 8
Sin (31'55' • 8) = 0.6607 Sin (J
Sin 31'55' Cos 8· Cos 31'55' Sin (J
= 0.6607 Sin 8
1.50952 Sin (l, = 0.52869 Cos 8
(J=19'18'
e = 31'55' ·19'18'
e =12'37'
Angle ABP =12'37'
@
Angle ACP:
Angle ACP = 8
Angle ACP =19'18'
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S-98
TRIANGULATION
Angle BCA = 180' • 80'27' 35.8" - 54'14' 37.8"
Angle BCA =45'17' 46.4"
® Distance BP:
B
Using Sine Law:
895.86
_
Be
Sin 45'17' 46.4" Sin 80'27' 35.8"
BC = 1243.01 m.
® Distance AC:
TN
p
Angle B4P= 180' ·12'37·26'35'
Angle BAP = 140'48'
BP
_ 6600
Sin 140'48' - Sin 26'35'
BP = 9321.57 m.
TN
~
["05'54.2"
.~~~W.$I~~6~S.~Wl~~0~.· • ~~~i6le.~~~~~~~tr~.1
cO(:@nMt~$9t¢QttWA • M~OQtlP.N9rthlOg$
~M4QP~Q·§Mlil1Q~,]Q~#~Ii#1AA~@
•
c
~flWJtb<t@l'iQ¢tthQf·lhe.·.IJ~eAtQ·a.are.
l$~;$§ • m,.<l#d?$$'?Q~ • ~~~ • t~@~¢lw~M"~$·
m~~$ijt~~hM~6m~I~MI~$.~f~ah9lW
..
••
U~#~r:I·'~]i~D'ji ~?·$" • ,~r9i ~?Q.I~ .
mq®lP~t~IM~i#lal1¢ij~¢)
®Pl:ll'li,@~~~~~i$l(l~A(;·>
··~ • • • ~r~~'~~~j"cp6rd't@~~Qt¢QrMt • 9••bY··
Solution:
CD Distance BC:
Using Sine Law:
AC
895.86
Sin 54'14' 37.8" =Sin 45'17' 46.4"
AC =1022.86 m.
® Coordinates of comer C:
STA.
A
LINE
AB
B
BC
BEARING
DISTANCE
N. 55'20'32" E
895.86
S. 1'05'54.2" W.
1243.01
TN
STA.
A
LAT
20000.00
~
~
B
20736.90
20509.45
- 1242.7~
19266.67
~
C
c
(20000. 200(0)
20713.07
DEP
20000.00
Coordinates ofC =20713.07 N., 19266.67 E.
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TRIANGUlaTION
AB Sin L2 CD Sin LS
Sin L7
Sin L8
CD =.:...A=B-=S.::.,:in-=L:..:.1-=S.::.,:in-=L::.:3
Sin L4 Sin L6
AB Sin L2 AB Sin L1 Sin L3 Sin L5
A. Angle condition equations.
Sin L7
Sin L4 Sin L8 Sin L6
Sin L1 Sin L3 Sin LS Sin L7
=1
Sin L2 Sin L4 Sin L6 Sin L8
A"-..I.::.---------.:..L-:.~D
Strength of Figure:
1. L1 + L2 + L3 + L4 = 180'
2. L3 + L4+ L5 +L6= 180'
3. L1 +L2+L7+L8=180'
4. L5 + L6 + L7 + L8 = 180'
5. L1 +L2+L3+L4
+L5 + L6 + L7 + L8 = 180'
6. L1 + L2 = L5 + L6
7. L3 + L4 =L7 + L8
In a triangulation system,. to be able to
adopt the best shaped triangulation network it
is necessary to apply a criterion of strength to
the different figures that maybe formed such an
index or criterion is known as the strength of
figure and is given in the equation:
B. Side Consition equations.
where
R = relative strength of figure
o= number of directions observed (forward
and backward) not including the fixed
or known side of agiven figure.
C =number of geometric conditions to be
satisfied in a given figure.
F =a factor for computing the strength of
D-C
figure and is equal to 0
Sin L1 Sin L3 Sin L5 Sin L7
Sin L2 Sin L4 Sin L6 Sin L8
~=~
Sin L4 Sin L1
ABSin L1
BC-
SinL4
(1J
-BC
---
Sin L6 Sin L3
BC
CO Sin L6
SinL3
CO Sin L6
Sin L4 - Sin L3
AD-AB
--Sin L7 Sin L2
AB Sin L2
AB Sin L1
Ao----
Sin L7
AD
- - - -(1JSin LS Sin.L8
CO Sin L5
AD
SinL8
D-C
R=O I(di +dA~B+di)
D-C
F=o
=1
=tabular difference for 1 second,
expressed in units of the 6th decimal
place corresponding to the distance
angles A and B of a triangle.
I (dA 2 + !:lA ~B + ~B2) =summation of
values for particular chain of triangles
through which computation is carried
from the known line to the line
required.
C =(n' - s' + 1) + (n - 2s + 3)
n' = number of lines observed in both
directions, including the known side of
the given figure.
s' =number of occupied stations
n = total number of lines in the figure
including known line.
s =total number of stations
dA , dB
S-IOO
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TRIANGULATION
Station B:
Angle 2 = 59'10'05"
Angle 4 = 60'29'10"
Angle 11 = 240'21'00"
360'00'15"
15"
Error=- =05"
3
Adjusted angle 2 = 59'10'00"
Adjusted angle 4 = 60'29'05"
Adjusted angle 11 = 240'20'55"
360'00'00"
Station C:
Angle 3 = 62'25'10"
Angle 5 = 59'25'10"
Angle 8 = 63'10'08"
Angle 14 =174'59'24"
359'59'52"
o
Error = OS"
08'
Correction =- = 02"
4
Adjusted angle 3 = 62'25'12"
Adjusted angle 5 = 59'25'12"
Adjusted angle 8 = 63'10'10"
Adjusted angle 14 =174'59'26"
360'00'00"
11111lIIi181;';';
Solution:
CD Corrected value of angle 3:
Station Adjustment:
Station A:
58'25'15" + 301'34'49" =360'00'04"
Error =04'
04'
Correction ="2 =02'
Adjusted angle:
Angle 1 = 58'25'15" - 02" = 58'25'13"
Angle 10 = 301'34'49" - 02" = 301'34'47"
360'00'00"
Station 0:
Angle 6 = 60'05'10"
Angle 7 = 71'40'20"
Angle 12 =22S'14'52"
360'QO'03"
Error = 03"
03"
Correction =3 =01"
Adjusted angle 6 = 60'05'09"
Adjusted angle 7 = 71'40'01"
Adjusted angle 12 =22S'14'50"
360'00'00"
Station E:
Angle 9 = 45'10'20"
Angle 13 =314'49'42"
360'00'02"
Error = 02"
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TRIANGUlATION
02"
Correction =- =01 "
2
Adjusted angle 9= 45'10'19"
Adjusted angle 13 = 314'49'41"
360'00'00"
FrOl1lthegivencjl1Clgril~teral.~f!$mTh99s~~re
OC()UPie~• i:l@.. .all.•. Ii~~~ .• ~f~.ql:j~~t¥~~i",·.~~m
qi~tk>ij~t: . , ·
":-:<::::;:</:::;:;::.::',:::.>\
Figure Adjustment
Considering triangle ABC
Angle 1 = 58'25'13"
Angle 2 = 59'10'00"
Angle 3 = 62'25'12"
180'00'25"
Error =25"
Adjusted Angle 1 =58'25'05" - 08" =58'25'05"
Adjusted Angle 2 =59'10'00" - 08" =59'09'52"
Adjusted Angle 3 =62'25'12" - 09" = ~
180'QO'00"
Corrected value ofangle 3 = 62'25'03"
® Corrected value of angle 6:
Considering triangle BCD
Angle 4 = 60'29'05"
Angle 5 = 59'25'12"
Angle 6 = 60'05'09"
179'59'26"
Error = 34"
Adjusted angle 4 = 60'29'05" + 12" = 60'29'17"
Adjusted angle 5 =59'25'12" + 11" = 59'25'23"
Adjusted angle 6 =60'05'09" + 11" = ~
180'00'00"
Corrected angle 6 =60'05'20"
@
Corrected value of angle 9:
Considering triangle COE:
Angle 7 = 71'40'01"
Angle 8 = 63'10'10"
Angle 9 = 45'10'19"
180'00'30"
Error =30"
. =3
30" =10"
CorrectIOn
Adjusted angle 7 = 71'39'51"
Adjusted angle 8 = 63'10'00"
Adjusted angle 9 = 45'10'09"
180'00'00"
Corrected angle 9 =45'10'09"
AI::::..-...i:...------:-::-....l~
Baseline = 1420 m
111.ti'(lli'~~
Solution:
CD Constant F:
O-C
Constant F = 0
o = 10 (no. of directions observed forward
and backward not including Ab)
C= (n' - s' + 1) + (n - 25 + 3)
n' = no. of lines observed in both
directions.
n'=6
s' = no. of occupied stations
s'=4
n = total no. of lines in the figure including
known lines
n=6
s = total no. of stations
5=4
C= (6 -4 + 1) +[6 - 2(4) +3)
C=3+1
C=4
D-C
F=O
F= 10-4
10
F= 0.60
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S-102
TRIUGUIlTION
@
Strength of figure which gives the
strongest route:
Consider triangle ABC dand ACD with AC
as common side.
-AL_~
R =0.60(31.979)
R = 19.19
AD
AB
Sin 90' = Sin 53'
=AB Sin 90'
AD Sin 53'
CD
Distance angles are 42' and 60' for triangle
ABC
9.825510895
9.825513234
2339
AA =2.339
log Sine 60'00'00"
log Sine 60'00'01"
hc) L (Ai + AA I.'>.a + I.'>.l)
Consider triangle ABO and ACD: with AD
as common sides.
Sin 42' - Sin 60'
AC - AB Sin 42'
- Sin 60'
CD
AC
Sin 41' =Sin 35'
CD=ACSin41'
S;n35'
CD = AS Sin 42' Sin 41'
Sin 60' Sin 35'
log Sine 42'00'00"
log Sine 42'00'00"
R = (0
9.937530632
9,937531847
1215
AD
Sin 40' = Sin 104'
CD =AD Sin 40'
Sin 104'
= AB Sin 90' Sin 40'
CD
Sin 53' Sin 104'
Distance angles are 53' and 90' for ABO
log Sin 53'00'00" = 9.902348617
log Sin 53'00'01" = 9.902350203
1586
AA =1.586
Aa = 1.215
log Sin 90'00'00" = 0
log Sin 90'00'01" =0
Aa=O
(Ai + AA Aa + Art)
=(2.339f + 2.339(1.215) + (1.215)2
(6} + AA Aa Art) =9.789
(Ai + AA Aa + Ai) = (1.586)2 + 0 + 0
(I.'>.i + AA Aa + Ai) = 2.51
+
Distance angles for triangle ACD are 41'
and 35'
log Sin 41'00'00" =
log Sin 41'00'01" =
9.816942917
~~
2422
log Sin 35'00'00" = 9.758591301
log Sin 35'00'01" = 9.758594308
3007
AA = 2.422
Aa = 3.007
(Ai + AA Aa + Ai)
=(2.422)2 + (2.422)(3.007) + (3.007)2
(Ai + AA Aa + Ai) =22.19
L (Ai + AA Aa + Ai) =9.789 + 22.19
L (~i + AA Aa + Ai) :a: 31.979
Distance angles are 41' and 104' for
triangle ACD
log Sin 41'00'00" = 9.816942917
log Sin 41'00'01" = 9.816945339
2422
AA = 2.422
log Sin 104'00'00" = 9.986904119
log Sin 104'00'01" = 9.986903594
525
I.'>.a = 0.525
(Ai + I.'>.A I.'>.a + I.'>.il
=(2.422)2 + 2.422(0.525) + (0.525)2
(Ai + AA I.'>.a + Ail = 7.41
L (Ai + I.'>.A Aa + I.'>.i) =2.51 + 7.41
L (Ai + AA I.'>.a + Ai) = 9.92
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TRIANGULATION
D-C)
2
2
R= ( C
L(~A +~A~B+~B)
R = (0 ~.~ L (Ill + ~A t1B+ Ili)
R = 0.60(9.92)
R= 5.952
R =0.60(5.96)
R=3.58
Considering triangle ABC and BCD with BC
as common side:
Jin 78' - Sin 60'
= AB Sin 78'
BC
Sin 60'
CD
BC
Sin 48' = Sin 88'
Consider triangles ABO and BCD with BD
as common side.
BD
AB
Sin 37' = Sin 53'
BD =AB Sin 37'
Sin 53'
CD
BD
Sin 48' = Sin 44'
- BC Sin 48'
CD - Sin 88'
= AB Sin 78' Sin 48'
CD
Sin 60' Sin 88'
CD _ Sin 48'
- Sin 44'
=AB Sin 37' Sin 48'
CD
Sin 53' Sin 54'
The distance angles are 60' and 78' for
triangle ABC and 48' and 88' (or BCD
Distance angles of triangle ABO are 37'
and 53' and for triangle BCD are 44' and
48'
~-~
log Sin 60'00'00" =
log Sin60'OO'01" =
~A
9.937530632
9.937531847
1215
= 1.215
log Sin 78'00'00" =
log Sin 78'00'01" =
9.990404394
9.990404842
448
Il B = 0.448
(Ili + ~A ~B + Ili)
= (1.215)2 +(1.215)(0.448) + (O.44W
(Ili + IlA IlB + Ili) = 2.22
log Sin 48'00'00" =
log Sin 48'00'01" =
9.871073458
9;871075354
1896
IlA = 1.896
log Sin 88'00'00" =
log Sin 88'00'01" =
9.999735359
9.999735432
073
Il B = 0.073
(Ill + IlA IlB + Ili)
=(t.896)2 + (1.896)(0.073) + (0.073)2
(t1l + IlA Il B+Ili) =3.74
L (Ill + IlA IlB + Ili) = 2.22 + 3.74
L (Il/ + IlA IlB +Ili) = 5.96
eo
log Sin 37'00'00" =
log Sin 37'00'01" =
9.779463025
9.779465819
2794
IlA = 2.794
log Sin 53'00'00" =
log Sin 53'00'01" =
9.902348617
9.902350203
1586
~B=1.586
(Ill +IlA ~B + Ili)
=(2.794)2 + (2.794)(1.586) + (1.586)2
(Ill + IlA IlB + ~i) = 14.75
log Sin 44'00'00":::
log Sin 44'00'01" =
9.841771273
9.841773454
2181
IlA =2.181
log Sin 48'00'00· =
log Sin 48'00'01" =
Il B=:.896
9.871073458
9.871075354·
1896
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5-104
TRIANGUlaTION
Solution:
(di + dA dB + di)
=(2.181f + (2.181)(1.896) + (1.896)2
(di + dA dB + di) = 12.49
L (di + dA dB + di) =14.75 + 12.49
L (di + dA dB + di) = 27.24
CD Adjusted value of angle 4:
ADJUST A
ADJUST 8
L 1= 23'44' 38"
23'44' 37"
c0- C) L(di+dAdB+di)
R= (
R =0.60(27.24)
R= 16.34
Relative strength of the quadrilateral R
= 3,58 (smallest value)
@
Length of check base CD:
CO =AS Sin 78' Sin 48'
Sin 60' sin SS'
CO = 1420 Sin 78' Sin 48'
Sin 60' Sin 88'
CO = 1192.61 m,
ADJUSTC
23'44'35"
L 2 = 42'19' 09"
42'19' 08"
42'19' 06"
Z. 3 = 44'52' 01"
44 '52' 00"
44'52'00"
L 4 = 69'04' 21"
69'04' 20"
69'04' 19"
L 5 = 39'37' 48"
39'37' 47"
39'37' 49"
L 6 = 26'25' 51"
26'25' 50"
26'25' 52"
L 7 = 75'12' 14"
75'12' 13"
75'12' 13"
L 8 = 38'44' 06"
38'44' 05"
Sum = 360'00' 08"
360'00' 00"
3.8'44' 06"
360'00'00
Error =8"
Correction =§.
8
Correction = 1" (sub)
L1+L2=L5+L6
23'44' 37"
42'19' 08"
66'03' 45"
• rfit~~ • foIIOWJdg"•• qM~drllatfjtal • ·.Wlth
¢or~~l;ljng·ClnSle~.·.~®lated.~h()wn.
a
39'37' 47"
26'25'50"
66'03' 37"
Error = 45·37
Error = 8"
c
Correction =§.
4
Correction = 2" (subtract from L1 and L2
and add to L5 and L6)
L3+ L4 = L7 + L8
A O'-u....-
.-.L.~B
d) Compute the adjusted value of angle 4 by
appIyihg the artgleconditioll only.
® Compute the adjusted value of angle 7 by
. applying the angleconditlon only.
@
Compute the strength of figure factor.
44'52' 00"
69'04' 20"
113'56' 20"
75'12' 13"
38'44' 05"
113'56' 18"
Error = 20 - 18 =2"
Correction = Add 1" to La and subtract 1"
from L4
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TRlAliGUlADON
Check:
L1 23'44' 35"
L8 38'44' 06"
L2 42'19' 06"
L7 75'12'13"
180'00'00"
L1
L2
L3
L.4
23'44' 35"
42'19' 06'
44 '52' 00"
69'04'19"
180'00' 00"
L8
L.7
L.6
L.8 .
L.3
L4
L.5
L.6
38'44' 06"
75'12' 13"
26'25' 22"
39'37' 49"
180'00'00"
44'52' 00"
69'04'19"
39'37' 49"
26'25; 52"
180'00'00"
Angle 4 = 69'04'19"
Gtv~nthe~uCidrilateral.shownYJfjichh~~·qeerl
adjU$t~d • u~lng·.IJrgle.C9ndi~()~, • • • ltl$rMUire~
1(l.a<lj\l!lI.tl)e•• l3rtglElslJ$ing·tb¢$j~ClQMitil>O,
ZD99mPutelheadju$te~~~glCl-~' ...i
®i.9omp\lte.ttmadjU!;ted.arl~Ie-.~ .• • • •. ·.· · · · .
@PQmp\.ltetf)eadjllsted.al1gl~6.
L.1
.
=39'3749"
L.2 = 26'25' 52"
L3 = 75'12' 13"
L4 = 38'44' 06"
@
@
Angle 7 = 75'12'13"
L5 =23'44' 35"
Strength of figure factor.
L6 = 42'19' 06"
D=10
L7 = 44'52' 00"
n'=6
n=6
L8 =69'04' 19"
Sum = 360'00' 00"
s=4
s'=4
C =(n' - s +4) +(n - 2s +3)
C = (6 - 4 + 4) + (6 - 8 + 3)
C=4
F=D-C
o
F= 10 -4
10
F =0,60 (Strength offigure factor)
Aif'-o"........................-
..........- -.........:.J..:.~B
Solution:
CD Adjusted angle 3:
.
Sin L.2 Sin L4 Sin L6 Sin L8
1
~~-'-------=
Sin L1 Sin L3 Sin L5 Sin L.7
log Sin 26'25' 52" = 9.64847855
log Sin 38'44' 06" = 9.796379535
log Sin 42'19' 06" = 9.82817581
log Sin 69'04'19" = 9.970360677
9.243394570
.
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TRIANGUlATION
log Sin 26'25' 52"
log Sin 26'25' 53"
log Sin 38'44' 06"
log Sin 38'44' 07"
log Sin 42'19' 06"
fog Sin 42'19' 07"
log Sin 69'04'19"
log Sin 69'04' 20"
log Sin 39'37' 49"
log Sin 75'12' 13"
log Sin 23'44' 35"
log Sin 44'52' 00"
log Sin 39'37' 49"
log Sin 39'37' 50"
log Sin 75'12' 13"
log Sin 75'12' 14"
log Sin 23'44' 35"
log Sin 23'44' 36"
log Sin 44'52' 00"
log Sin 44'52' 01"
= 9.64847855
= 9.648482785
= 9.796379535
= 9.79638216
= 9.82817581
= 9.828178122
= 9.970360677
= 9.970361483
=
=
=
=
=
=
=
=
=
=
=
=
Diff. in 1"
4.2
2.62
Add 2" to all angles in the numerator:
(smaller)
Subtract 2" to all angles in the denominator:
(bigger)
2.31
- L1 = 39'37' 49"
+L2
26'25' 52"
=
.75'12'13"
- L3
+L4 = 38'44'06i,
- L5 = 23'44' 35"
+L6 = 42'19' 06"
- L7 = 44'52' 00"
+L8 = 69'04' 19"
0.81
__
9.94
9.804705675
9.985354379
9.604912331
9.848471997
9.243444381
Diff. in 1"
9.804705675 2.54
9.804708217
9.985354379 0.56
9,985354935
9.604912331 4.79
9.604917117
9.848471997 2.12
9.848474112 _ _
10.01
• 02"
+02"
• 02"
+ 02"
- 02"
+ 02"
- 02"
+ 02"
39'37' 47"
26'25'54"
75'12' 11"
38'44'08"
23'44'33"
42'19'08"
44'51' 58"
69'04' 21"
360'00' - 00"
Adjusted angle 3 =75'12' 11'~
@
Adjusted angle 5 = 23'44' 33"
@
Adjusted angle 8 =69'04' 21"
-----/
J
,c
••••
"II1thE!figwe•• ~hpw§ • .~,.qy~drla~~r~IWl!htheit
~9q~$Pon(jin9,
• • • • t*ngUItl.r•
'
deSignated.
"....,'
... .. ., ."•.• ','•, m~#$l)r~m~nls
.'..'. ,', . . ,
Subtract: 9.243394570 - smaller
9.243444381' bigger
0.000049811
Add: 9.94 + 10.01 = 19.95
Difference =49.81
49.81 '
0=-80=6.23
(J = 19.95
8
(J = 2.49
.
6.23
Correcllon =2,49
Correction = 2.5" say 2"
(j)
WhiCh. of InemilOWing equation dOes not
®ltsfy the figure shown.
L2 + L3 =L7 + L6
b) .L1 + L8 =L4 + L5
c) L1 + L2 + L3 + L4 = 180'
d) L1 + L8 + L6 + L7 = 180'
a)
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5·107
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TRIANGUlATION
~
~
c}
d} ,
Sin L1 Sin L3 Sin L5 Sin L.7
Sin L8 Sin L2 Sin L4 Sin L6
Sin L2 Sin L4 Sin L6 Sin L8
=0
=1
Sin L3 Sin L5 Sin L7 Sin L1
Sin L1 Sin L3 Sin L5 Sin L1
Sin L2 Sin L4
Sin L2 Sin L4
Sin L6 Sin L8
@
n = total number of lines in figure, including
the known side
n=6
Sin L6 Sin L8
Sin L1 Sin L3
= ---=--....:..:...--=.
Sin L5 Sin L7
.·WMt.Will • ~th~#~~Qr..• (.....F)l".~lvihg
• me
..... . .
.
C----~::---~~~D
slt~ngtI'l9fflgur~.<
a} 0.80
b} 0.60
Values ofn
c) 0.90
d} 0040
n' =no. of lines observed in both directions
including the known side
n'=6
F=D-C
o
o= no. of direcfions observed (forward and
backwards) not including the fixed or
unknown side of a given figure.
C = no. of geometric conditions to be
satisfied in a given figure.
C= (n' - s' +1) + (n - 2s +3)
n' = no. of lines observed in both
directions, including the fixed or
known side of a given figure.
n = total number of lines in the figure
including fixed or known line.
s = total number of stations
Values of n'
$ = total no. of stations
$=4
C=(n'-s!+1} +'(n-2s+3)
~ =(6 - 4 + 1) + ,[6 - 2(4) + 3]
C=3+1
C=4
No. of directions observed (forward and
backwards) not including the known
side of
0=10
. luded in computarion of D
normc,
B
F=D-C
.
0
F=10-4
10
F= 0.60
Answer:
G) c
Values of D
@ b
@ b
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S-108
TRIANGUlATION
@
Fraction F:
F=O-C
o
o =no. of directions observed(forward and
backward) not including the known
side.
0=24
24- 9
F0-C- - 0 - 24
F= 0.625
D
E
@
Strength of figure R:
R= F(!::>} + ~A ~B + ~i)
R = 0.625(5.02)
R= 3.14
G
Solution:
CD Value of C:
n' =no. of lines observed in both directions
including known side
n'= 13
n =total no. of lines in figure unciuding
known side
n =13
s' =no. of occupied stations
s' =7
s =total no. of stations
s=7
C =(n' - s' + 1) + (n - 2s + 3)
C=(13-7+1) + [13-2(7)+3]
C=7+2
C=9
CD Compute the adjU$tedvalue of angle Aby
@
@
diWibuting lhespherical excess and the
remalning error equally.
.
Compute the adjusted value of angle B by
distributing the spherical excess and the
remaining error equally.
.
Compute the adjusted value of angle Cby
distributing the spherical excess and the
remaining error equally.
Solution:
A
e=R2 Sin01"
A =be Sin A
2
b
Sin 56'10'30"
b =33814.89
35965.47
= Sin 62'04'11"
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SPHERICAl EXCESS
A = be Sin A
2
33814.89 (35965.47) Sin 61'45'20"
A=
2
A = 53568365b.2 m2
"
A
The interior angles in triangle ABC are
A "'. 57'30' 29", B ::: 65'17'27" • and
C =57'12' 16". The distance from A to B is
equal'to 180,420 m, ASsuming fh~ average
radius Qf curvature is 6400 km.
',
e = R2 Sin 01"
.'
535683650.2
e"= (6372000)2 Sin 01"
[."=2.72"
CD Compute the area of fhe triangl~;
@ Compute the second term of the spherical
excess.
61'45'20"
56'10'30"
62'04'11"
,
'
@ Compute Ihe total spherical eXcess,
Solution:
180'-00'01"
180'-00'02.72"
G)
Area of triangle:
B
Error =1.72"
1.72
1st Carr. = -31st Corr. =0.573 (added)
2.72
2nd Corr. =-3-
A
2nd Corr. =0.907 (to be subtracted)
.... c
~--~
Using Sine Law:
61'45'20" + 0.573" - 0.907 = 61'45' 19.676"
56'10'30" t 0.573" ·0.907 = 56'10' 29.676"
62'04'11" t 0.573" - 0.907 = 62'04' 10.676"
180'00' 00"
B
b
180420
Sin 65'17' 27" = Sin 57'12' 16"
b =194978.94 m.
A
bcSin A
rea=-ZA
rea
194978.94 (180420) Sin 57'30' 29"
2
6
Area = 14836 x 10 rrf
® Second term of the spherical excess:
A
......... c
~-_
When the sides ofthe triangle are over 100
miles (160,000 m.) use the accurate
formula for spherical excess.
Area
[
a2t b2t
e" =R2 Sin 01" 1 t
24 R2
el) Adjusted value of angle A = 61'45'19.676"
=56'10'29.676"
c;"
Adjusted value of angle B
(;'t
Adjusted value of angle C =62'04'10.676"
The second term is
2 2
Area
(a + b t
R2 Sin 01"
24 R2
c2)
c2]
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S-J10
SPIIIBICIl EXCESS
a
180420
=
Sin 57'30' 29" Sin 57'12' 16"
a =181033.49 m.
Second tent .
(j)()omPOle • the.adjlJ~tedvalue • ofMgle.A.PY
91~WputingmemmericaL~X(\{Jssandthe
~i:lrnlng:e@tequ~:"IY,>
·®·.Cp!l'lp~t~ • • the.9djl.lste9Wlue•• ofan~le~ • W•
. . . .•··.·.4iS1riblJti/lg • lhe.spherj"al•• til)(ee$$~l'ld • • m~·
••.•..•..• @l)airyins~tI"QreqUEln}"
a2 =32773 x 1tl6
'.
.
• ~• • • Gompute•• t/)e.adJustedv~J~~Qf.~r~J~Cby .
••.•.•.•. •. . g~trll?~ljlls • . th~ • $P~~ig~I.·~~e;;san~ • the
tElll'lafl'ling error eqU(lny. '. .. ..... .'
b2 =38017 x 106
c?- =32551 x 106
W- =40960000 x 106
Area = 14836 x 106
Solution:
B
(a2
Area
+ ~ + c?-)
2nd term = R2 Sin 01" .24W-
14836 x 106
2nd term =40960000x1 06 Sin 01"
A
(32773+38017+32551)106]
[
24(4096‫סס‬OO)106
2nd tenn
,
= 74.7106"
(
103341 )
24(40900000)
2nd term = 0.00785"
® Total spherical excess:
Area
[
a2+~+c2]
e" =W- Sin 01" 1 + 24 We" =74.7106 + 0.OD7e5
d' =74.71845"
·~-_
....... c
C
5260
=
Sin 52'03'17" Sin 88'33'05"
C = 4149.3$ m.
log m = 1.40658 ·10
m = 2.55023 x10-9
e"=mbc Sin A
e' = 2.55023 x 10,9 (5260)(4149.35) Sin 39'23'40
e"= 0.035"
A = 39'23'40"
B = 88'33'05"
C = 52'03'17"
180'· 00'02"
180',00,00.035"
Error= 1.965
First Correction: 1,~5 ::; 0.655"
Second Correction:
0,~35 =0,012
39'23'40" • 0.655" • 0,012" = 39'23' 39.333"
88'33'05"·0.655"·0.012" = 88'33' 4.333'.'
52'03'17"·0.655·0,012" = 52'03' 16,333"
180'00' 00"
CD Adjusted value of angle A ::; 39'23'39,333"
@
@
=88'33' 4,333"
Adjusted value of angle C =52'03'16.333"
Adjusted value of angle B
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SPHERICAl EXCESS
m = 2.536 x 10.9
log m =1.40415·10
_b
c_
Sin B - Sin C
b
3012
Sin 63'44'59" = Sin 79'59'57"
b = 2743.05
o
@
m.
e" = 2 A
.I'fR'IS known.
R Sin 01"
e" = m bc Sin A if no radius is given
1
m- 2 R NArc 1"
CD Compute the adjusted value of aogle Eby
.' ," 'distnbutingthe spherical excess and the
remaining error equally.
,,
@ ,Compute ,the adjusted value of angle Nby
, distributing the spherical excess and the
remaining ~rror equally. , '
,'. ,
® Cbmputeth,e adjusted value of angle L by
distributing the spherical excess and the
remaining error equally.
"
1t
Arc 1 = 180(3600)
N = 6376032 m.
At a given latitude
e"=mbcSinA
e" =2.536 x 10.9 (2743.05)(3012) Sin 36'15'07"
e"= 0.012" (spherical excess)
79'59'27"
63'44'59"
36'15'07"
180',00'- 03"
180'- 00'· 0 012"
Error = 2.988" (error of spherical triangle)
Solution:
N
Correction for each angle
2.988
=-3-
= 0.966" (First Correction)
L
Second Correction
0.012
=-3-
~--""""-E
A
e"= R2 Sin 01"
= 0.004" (spherical excess,subtracted
from each angle)
Arc 1" =180 ~600)
79'59'57" ' 0.996" • 0.004" =79'59'56"
63'44'59",0.996" - 0.004" =63'44'58"
36'15'07"·0.995",0.004" =~
180'00' 00"
bcSin A
,
A = - 2 - (area oftnangle)
e =m bcSin A
1
m= 2RNArc1"
m=
CD Adjusted value of angle E = 79'59'56"
1
1t
2(6378160)(6376032) 180(3600)
® Adjusted value ofangle N = 63'44'58"
@
Adjusted value of angle L
=36'15'06"
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5-112
AREA Of ClOSED TIIAIERSE
AREA OF CLOSED
TRAVERSE
In any closed traversed, there is always an
error. No survey is geometrically perfect, until
proper adjustment are made. For a closed
traversed, the sum of the north and south
latitudes should always be zero.
BALANCING A SURVEY
Latitude of any line - is the projection on a
north and south lines. It may be called as
north or positive latitude and south or
negative latitude.
Departure of any line - is the projection on
the east and west line. West departure is
sometimes called negative departure and
East departure is sometimes called
positive departure.
C
DEPARTURE
B
1. Compass rule - the correction to be
applied to the latitude or departure of any
course is to the total correction in latitude
or departure as the length of the course is
to the length of the traverse.
2. Transit rule - the correction to be applied
to the latitude or departure of any course is
to the total correction in latitude or
departure as the latitude or departure of that
coUrse is to the arithmetical sum of all the
latitudes or departures in the traverse
without regards to sign.
Error of closure = ."j LL2 + L02
RIf
Error of closure
eaIve error = Perimeter of all courses
LL =error in latitude
LO =error in departure
A
Line AB has its latitude AC and departure BC.
The angle 0 is the bearing of the line AB.
BC =AB.Sin 0
.
1 Area by Triangle Method
Departure = Distance x Sin Bearing
AC =AB Cos 0
Latitude = Oistance x Cos Bearing
Dist = Latitud~ .
Cos Bearrng
O' t _ Departure
IS - Sin Bearing
0
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AREA OF ClOSED TRAVERSE
or
A=A1+A2+A3
d1d2 Sin a
A1 =
2
·A - d3d4Sin {!,
22
- dsd s Sin l2I
A3 2
4 Area by Double Meridian Distance
2 Area by Rule of Thumb Method
2A = [Y1 (X1 - X2) + Y2 (X1 - X3) + Y3 (X2 - ~)
+ Y4 (X3 - Xs) + Ys(~ - X1)]
Double Meridian Distance of line BC is the
sum of meridian distances of the two
extremeties.
3 Area by coordinates
Area ABCDE =Area BCEF + Area CDIF
- Area EDIH • Area AEGH •Area ABGE
c
~ .'...x j....A~·
D.M.D. of BC = EB + FC
Latitude of BC = EF
H······:':$·f,····
, !""I('»,.
Area of BCFE:
_ (EB + FC) EF
A2
- (X2 + x 3) (Y2 - Y3) + (X3 + X4) (Y3 - Y4)
A2
2
(X4 + xs) (Ys • Y4) (xs + X1) (Y1 - Ys)
2
-
2
(1'1 + X2) (Y2 • Y1)
2A = (EB + FC) EF
2A = (D.M.D.)tatitude
Double Area = Double Meridian Distance
x Latitude
2
Simplifying 'this relation:
CD 2A = • [Y1 (xs - X2) + Y2(X1 - X3)
+Y3 (X2 • X4) + Y4 (X4 - Xs) + Ys (~ • X1)]
or
@
2A = Yt X1+ Y3Y2 + X4Y3 + XSY4 + X1YS
• XtY2 • X2Y3 - Y3Y4 . X4YS· XSY1
D.M.D. of the first course is equal to
the departure of that course.
2. D.MD. of any other course is equal to
the DMD of the preceding course, plus
the departure of the preceding course
plus the departure of the course itself:
3. D.M.D. of the last course IS
numerically equal to the departure of
the last course but opposite in sign.
1.
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5-114
AREA OF ClOSED TRAVERSE
Computing Area by D.M.D. Method:
1. Compute the latitudes and departures
of all courses.
2. Compute the error of closure in
latitudes and departures.
3. Balance the latitudes and departures
by applying either transit rule or
compass rule.
4. Compute for the D.M.D. of all courses.
5. Compute the double areas by
mUltiplying each D.M.D. by the
corresponding latitude.
6. Determine the algebraic sum of the
double areas.
7. Divide the algebraic sum of the double
area to obtain the area of the whole
tract.
Double Area =D.M.D. x Latitude
5
Area by: Double Parallel Distance
1. D.P.D. of the first course is equal to
the latitude of that course.
2. D.P.D. of any other course is equal to
the D.P.D. of the preceding course,
plus the latitude of the preceding
course, pius the latitude of the course
itself.
3. D.P.D. of the last course is
numerically equal to the latitude of the
last course but opposite in sign.
Double Area =Double Parallel Distance
x Departure
Example:
Area by Double Meridian Distance
Lines LAT. DEP.
DMD
-30
-30
-40
+40
+SO
1-2
2-3
3-4
4-1
+60
-20
-80
+40
+20
+60
-SO
Double
Area
·30160\ =-1800
·401- 20) =+800
+40(. 80\ =-3200
+SOI4O\- +2000
2A =- 2200
A=-1100m2
Area by Double Parallel Distance
Lines LAT. DEP.
1-2
2-3
3-4
4-1
+60
-20
- 80
+40
- 30
+20
+60
-SO
Double
Area
+60 601- 30\ =- 1800
+100 100120\ = +2000
d(6Q) = 0
0
-40 -401- SO\ = +2000
2A 2200
A=1100m 2
DPD
=-
•..••••..<•• • 4OQ·,OQitJ.·.·>···
.·llOll,OOffi)
700.00fft·· ..
600.00trl.
CD Compute the correction <i1latitudeotlline
CO using transit rule, ...• ..
...
@ Compute the linear error of closure.
® Compute the relative error or precision. .
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AREA OF CLOSED TRAVERSE
Solution:
Solution:
Lines Bearina Distances LAT
DEP
AB Due North 400.00m !+400.
0
N45' E 800.00m +565. +565.69
BC
860' E 700.00m 3SO. +606.22
CD
DE S2Q'W 600.00m 563.8 -205.21
EA S86'S9'W 966.34m - SO.86 -965.00
Penmeter -- 3466.34 +1.01
+1.7
400 + 565.69 +3S0 + 563.82 + SO.86 = 1930.37
CD Error of closure:
Lines
AB
BC
CD
DA
Bearinll
N.53'3TE.
S.66'54'E.
S.29'Oaw.
N.S2'OOW.
Distances
59.82m
70.38 m
76.62m.
9S.75m
302.57
ill Correction of latitude on line CD using
transit rule:
3SO.00
1.01 - 1930.37
Ceo =0.18
LAT
+35.62
- 27.61
-66.93
+58.95
+94.S7
- 94.54
+0.03
DEP
+48.06 '
+64.74
-37.30
- 7S.45
+112.aO
- 112.75
+O.OS
fm_
Error ofclosure = -V (0.03)2 +(0.05)2
error of closure = 0.0583
Linear error ofclosure:
@
LEG = -V (1.01~ +(1.7)2
LEG = 1.97740
@
•
® Precesion of linear measurement:
..
0.0583
PrecIsion =302.S7
Relative error or precision:
1.97740
Relative error =3466.34
1
Precision =5190
.
1
RelatIVe error = 1753
Precision = 1:5190
@
•~~i~I.~~;t$h:~d~~~~~~~ijjjti~.lf~~~.
>.•• • • • • •.• • .•.
!ljid$s.$reshOW!l: '.
... . . . •. . . . •••.. •. ••••••
~
Area in acres:
LAT
DEP
DIv1D
+3S.61
- 27.61
- 66.94
+58.94
+48.0S
+64.73
- 37.31
- 75.47
+48.0S
+160.83
+188.25
+75,47
o
ill
~:~S~~h~~:err9r.·Of• • cIO$~re • • • fOt•• • ~e.
the" pre<:e~Jon cif linear
measurement of this tl'averse. '. .
.
What is the lotal area'locliJdedwlthlJi the
traverse inacfes.
® What
is
0
Note: 1acre = 4047 sq.m.
, . 5441.36
.
Area = 4047
Area = 1.34 acres
DOUBLE
AREA
+1711.06
-4440.46
-12601.46
+4448.20
2A --10882.72
A =5441.36 m2
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S-1I6
AREA OF ClOSED TRaVERSE
Corrected Latitude:
..f-_
'~liiiii~1
249.40
0.68 -1868.94
C=O.09
Corrected lat =249.40 •0.09
Corrected lat = 249.31
• H•HW?~;~r:W~M~P<!~~&§;~pg;p~@]jjn~
• ·••·•• ·.t~~.¢~tt~~J~m~p~ll.h~9~rmrt~@9f~.
·.·.bYAAmM~~rn@>····
@AgW@ij#~yijr~$1lj~~llj~@I~W4M<
.• • •·•• • •i$t.lI~W>·.·i·...···.//·>.·./ •. H:
• · •·•·• • • •.•.• • .•. . . .
;!lll:t~'Jl'
•·• • • • •~~'~il~l~eEl~.~ll~0te
Solution: \'\-1"1,
CD Correction of latitude and departure AB:
Corrected Latitude:
_C__ 483.52
(+) 0.44 - 2915.80
C=+0.07
Corrected lat =326.87 +0.07
Corrected lat =326.94
Corrected Departure:
C 483.53
0.37 =2915.80
C=O.06
Corrected dep = 356.30 +0.06
Corrected dep = 356.36
® Correction of departure and latitude BC:
Corrected Departure:
C
364.20
- 0.42 =1842.64
C=-O.08
Corrected dep =364.20·0.08
90rrected dep =364.12
Solution:
CD Corrected latitude of DE by comp~ss rule:
-.L._ 518.40
.... .. . .
/0.56 - 2628.5
E = +0.11
"",
ll-?t
Corrected latitude of DE =259.2 + 0.11
Corrected latitude of DE =259.31
® Correctedlatitude of DE by transit rule:
E
259.2
.. ;KJ.56 = 1726.8
E=+0.08
Corrected latitude of DE =259.2 + 0.08.
Corrected latitude of DE =259.28
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S-l17
AREA OF ClOSED TRAVERSE
@
Corrected departure of DE by compass
rule:
(j)
Relative error:
- 0.34 - 2628.5
E=- 0.07
Error of closure =--J (1.01)2 + (1.7)2
Error of closure =1.977
.
1.977
RelatIVe error =3466.34
Corrected departure =448.9 - 0.07
Corrected departure =448.83
Relative error =
~_518.4
175~.33
® Adjusted distance of EA using
'f: Transit Rule:
For A- B: (latitude)
F"9t1)fhe1ieldnOtesOfa\cl~sed{trav~r~e<
sh~mtleIoW,<ldjlJl3~.m~.tr<av~rs~.lIl3il'lg
....
Q!¢()fuP~W·l!J~r~~pv~~rr8rOf.pl()~re.
• • /\•• •·••.•·
..fL_~
. t1.01 - 1930.37
C1 = 0.00052 (400)
C1 =0.21 ." " .
"-f\1
®.····Cw.nput~ • th~··~4jll~t~~Ai~t~I'l~()t .• litl~eA
lJ~it'9Trcllisit~~lEl,
Total distance =3466.34
@qomputethei~dJlJ§ted~a.ringOfliIW)CO
using Compa$$Rllle:< ... .
STA.
OCC.
A
B
C
0
E
STA.
OBS.
B
C
D
E
A
. ..
Bearings
Distances
Due North
N45' E
S60' E
S20'W
S86'59'W
400.00m.
800.00m.
lOO.OOm.
600.00m.
9OO.34m.
Solution:
Line Bearinas
A-S Due North
S-C N45'E
CoD 860' E
D-E 820'W
E-A S86'59'W
965.69
~
LINES
A-B
B-C
CoD
D-E
E-A
CORRECTION FOR LATITUDE
C1 = 0.000523 (400)
=+0.21
~ = 0.000523 (565.69) =W·30
0.000523 (350.00) =\0.18
CJ
C4 = 0.000523 (563.82) =>\{).29
C5 = 0.000523 (50.86) =J{).Q3
1.01
Correction for Departure:
For line A - B:
..fL __0_
Distance
400.00
800.00
700.00
600.00
966.34
1171.91
LAT
+400.00
+565.69
·350.00
- 563.82
-50.86
+965.69
DEP
0
+565.69
+606.22
- 205.21
-965.00
+1171.91
11lQ2.1 ~
:..11ZQ21
1930.37 2342.12 + 1.01
;~(:t~~
Total distance = 3466.34
Error in latitude = 1.01
Error in departure = 1.70
+ 1.70
1.70 - 2342.12
C1 = 0.0007258 (0)
C1 =0 (., \.
•...•....
LINES
A-S
S-C
CoD
D-E
E-A
J
CORRECTION FOR DEPARTURE
C1
0.0007258 (0)
= 0
~ = 0.0007258 (565.69) = 0.41
~ = 0.0007258 (606.22) = 0.44
C4
0.0007258 (205.21) = 0.15
C5 = 0.0007258 (965)
= 0.70
1.70
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S-118
AREA OF ClOSED TUVERSE
ADJUSTED LATITUDES AND DEPARTURE
(uncorrected)
(corrected)
Lines
~·B
A-B
B-C
CoD
D-E
E-A
LAT
DEP
GO.21
'1'400 I ·
+400.00
0
-0.30
- 0.41
+565.69 +565.69
-0.18
-0.44
- 350.00 +606.22
-0.29
+0.15
- 563.82 - 205.21
-0.03
+0.70
~ 50.86 -965.00
LAT
Correction for departure:
LineA -B:
.ft_~
DEP
1.7 - 3466.34
C1 =0.00049043 (400)
C1 =0.20
..,
+399.79
0
LINES CORRECTION FOR DEPARTURE
A- B C1 = 0.00049043 (400)
= 0.20
B-C C:1 = 0.00049043 (800) = 0.40
C-D ~ = 0.00049043 (700)
= 0.34
D- E C4 = 0.00049043 (600)
= 0.29
E-A Cs = 0.00049043 (966.34) = 0.47
+565.39 +565.28
- 350.18 +605.78
- 564.11 -205.36
-50.89 -965.70
o
o
1.70
ADJUSTED LATITUDES AND DEPARTURE
(uncorrected)
(corrected)
Lines
A-B
Adjusted bearing ofEA:
q;;
tan bearing =
)
·
- 965.70
tan beanng
= _50.89
B-C
Bearing =S. 86'59' W.
CoD
Adjusted distance of EA:
965.70
·t
DIS
ance =Sin 86'59'
Distance =967,04 m.
D-E
E-A
® Adjusted bearing of CD using
Compass Rule:
Correction for latitude:
LIneA -B:
f
LAT
DEP
-0.20
- 0.12
+400.00
0
-0.24
-0.40
+565.69 +565.69
-0.20
-0.34
- 350.00 +606.22
- -1'0.17
+0.29
- 563.82 - 205.21
- lO.28 +0.47
-50.86 -965.00
LAT
DEP
+399.88
-0.20
+565.45 +565.29
-350.20 +605.88
-563.99 - 205.50
- 51.14 - 965.47
o
0
.
605.88
tan beanng = 350.20
Bearing = S. 59'58' E
~-~
1.01 - 3466.34
C1 =0.000291374 (400)
C1 =0.12
LINES CORRECTION FOR LATITUDE
A- B C1 = 0.000291374 (400)
= 0.12
B - C C:1 = 0.000291374 (800)
= 0.24
C- D ~ = 0.000291374 (700)
= 0.20
D-E C4 = 0.000291374(600)
= 0.17
E-A Cs = 0.000291374(966.34) =~
1.01
(j)
@
compute the be~ of line 2- 3... •
Compute the dls~nce ofline 4 , 1.
® Determine the area of the lot using DMD.
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5-119
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AREA OF CLOSED TRAVERSE
Solution:
For line 3-4:
.
800
tan Beanng :: 800
Since the property line intersect each other,
then we could only apply DMD, if we divide it
into areas where no property lines will
intersect.
Line~
LAT
-1200
+200
+800
+200
1- 2
2-3
3-4
4·1
Bearing:: N4S' E
Distance :: S~S'
Distance = 1131.37
Bearing
DEP
+400
-400
+800
-800
Distance
1265.02
N63'2f>W 447.23
N4S'E
1131.37
N7S'58'W 824.63
S 18'26' E
For line 4 -1:
.
Bearing:: N7S'58' N
·t
800
DIS
ance - Sin 7S'S8
Distance:: 824,03
CD Bearing ofline 2 • 3:
@
Farline 1- 2:
·
400
(,
tan Beanng:: 1200
Distance:: 1265.02
For line 2- 3:
400
.
tan Beanng:: 200
Bearing:: N 63'26' W
. ';
Distance of line 4- 1....
D'
IS
tance:: Sin400
63'26~
Distance:: 447.23
r'
Area by DMD:'''<i
Using Sine Law:~ ,: ,.,
X
824.63
Sin S7'32' = Sin 63'26'
X= 777.88
Y
824.63
Sin 59'02':: Sin 63'26'
Y=790.56
.f:.
a= 1265.02 - 790.56
a=474.46
b=1131.37-777.88
~
b::: 353.49
'Y)
("7"
J/
Far Lot A:
Bearing:: S 18'26' E
400 '
Distance:: Sin 18'26'·
@
800
tan Beanng :: 200
Lines
1-4
4-S
5-1
Bearina
S7S'58' E
S4S'W
N18'66'W
LINES
1· 4
4-5
5·1
Distance LAT
824.63 -200
777.88 -S5O
790.56 +750
DMD
+800
+1050
+250
DEP
+800
-550
-2S0
DOUBLE AREA
-160000
·577500
+187500
2A:: 550000
A = 275000 m2
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5-120
ARU OF ClOSED TRAVERSE
For Lot B:
To compute the DMD:
Linel Bearina
5-2 S18'26' E
2-3 N63'26'W
3-5 N45'E
LINES
5-2
2-3
3-5
Distance
474.46
447.23
353.49
DMD
+150
-100
-250
LAT
-450
+200
-250
DEP
+150
-400'
+250
DOUBLE AREA
·67500
-20000
62500
2A= 150000
A=75000 m2
Lines
DMD
Double
Area
1-2
-11.77 +219.275
2-3 -11.77-11.77-5.96= -29.50 -236.885
3-4 - 29.50 - 5.96 -1.36 = -36.82 -1n.104
4-1 -36.82-1.36-19.09= +19.09 -110.531
To compute for double area =DMD xlatitude
Lines
1-2
2-3
3-4
4-1
Total area = 275000 +75000
Total area = 350000 m2
-11.77 (-18.63)
- 29.50 (8.03)
- 36.82 (4.81)
-19.09 (5.79)
Double Area
= 219.275
= -236.885
= -1n.104
= -110.531
Negative double areas
=236.885 +1n.104 + 110.531
= 524.520 sq.m.
Positive double area = 219.275
524.5b - 219.275
Area=
2
Area = 152.622 sq.m.
@
G)
flndthea~ea()Nhelqtby.DMt>rnf!lhod,······
@
Finq.thearea.ofk:llbyDpo·rnethgd.•
(!)
Area by DMD method:
Departure = distance x sin bearing
Latitude = distance x cos bearing
@)findtMPAt>9ffi®3,A. iii.·
.
Solution:
u
Line
1- 2
2-3
3-4
4·1
BearinQs
S32'17W
N36'25W
N15'47W
N73'07'E
';,
DPD offine 3 - 4:
Line's
LAT
1-2
2-3
3-4
4-1
-18.63
8.03
4.81
5.79
DEP
DPD
-11.77 -18.63
-5.96 -29.23
-1.36 -16.39
+19.09 -5.79
DPD offine 3 - 4 =-16.39
<'~
. .",
Dist
LAT DEP DMD
22.04 -18.63 -11.77 -17.77
10.00 +8.03 -5.96 -29.50
5.00 +4.81 -1.36 -36.82
19.95 +5.79 -19.09 +19.0E
@
Area by DPD method:
Double Area =219.275 + 174.211
+ 22.290 - 110.531
Double area = 305.245
Area - 305.245
-
2
Area = 152.622 sq.m.
Double
Area
+219.275
+174.211
+22.290
+110.531
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IREA Of ClOSED TRAVERSE
Line BC:
Latitude
.£L_ 30.98
20.5 -1357.44
C2 = 0.47
From the given technical description of a lot.
LINES
AB
BC
CD
DE
EA
BEARINGS
N.48'20'E.
N.87"OO'E.
S.7'59'E.
S.80'OO'W.
N.48'12'W.
DISTANCES
529.60 m.
592.00 m.
563.60 m.
753.40 m.
428.20 m.
Line CD:
Latitude
~_ 558.14
20.5 -1357.44
CD Find thE! corrected bearing of line BC using
transit rule.
@ Find the corrected bearing of line DE using
transit rule.
@ Find the corrected distance of line EA
using transit rule.
C3 =8.42
Line DE:
Latitude
~_ 130.83
20.5 -1357.44
C4 = 1.98
Solution:
CD Corrected bearing of line BC using transit
rule:
Lines
AB
BC
CD
DE
EA
Bearina
N.48'20'E.
N.8TOO'E.
S.7'59'E.
S.80'OO'W
N.48'12'W.
Distance
529.60
592.00
563.60
753.40
428.20
LAT
+352.08
+30.98
-558.14
-130.83
+285.41
+668.47
-688.97
Line EA:
Latitude
~_ 285.41
20.5 -1357.44
DEP
+395.62
+591.19
+78.28
-741.95
-319.21
+1065.09
-1061.16
Error = ('20.5
Cs = 4.31
+.3.93
i-f" 668.47 1065.09;\,<·
688.97 .1Q2.1.12
/I"",
LINES
AS
BC
CD
DE
EA
LineAB:
Latitude
-fL _ 352.08
Departure
~_
20.5 -1357.44
395.62
3.93 - 2126.25
C1 = 5.32
C1 =0,73
3.93 - 2126.25
C2 = 1.09
Departure
78.28
3.93 - 2126.25
C3 =0.15
~_
Departure
74.95
3.93 - 2126.25
C4 = 1.37
~_
Departure
319.21
3.93 - 2126.25
~_
Cs = 0.59
CORRECTED LATITUDES
\ 352.08 + 5.32 = + 357.40
r 30.98 + 0.47 = + 31.45
.- 558.1#-8.42 =- 549.72
130.83+1.98 = -128.85
T 285.41 + 4.31 = + 289,72
o
1357.44 2126.25
Corrections using transit rule:
Departure
.£L_ 591.19
LINES
AB
BC
CD
DE
EA
CORRECTED DEPARTURES
I. 395.62 - 0.73 = + 394.89
.j., 591.19 -1.09 = + 590.10
78.28 - 0.15 = + 78.13
-741.95~1.37 =- 743.32
.. 319.21 f 0.59 = - 319.80
o
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5-120-B
AREI OF CLOSED lUVERSE
Corrected bearing of line BC:
tan bearing = Dep.
Lat.
Using the given data in the traverse shown:
. BC 590.10
ta n beanng
= 31.45
Corrected Bearing BC = N. 86'57' E
POINTS
A
B
C
@
0
E
F
Corrected bearing of line DE:
tan bearing = Dep.
Lat.
t be' DE - 743.32
an anng
=_128.85
Corrected bearing DE = S. 80'10' W.
@
Corrected distance ofline EA:
·
Dep.
ta n beanng =
Lat.
t be' EA - 319.80
an anng
=+ 289.72
Corrected bearing EA= N. 4T50'W.
D' t
Dep.
IS ance - Sin bearing
CD Compute the bearing of line BC.
@
@
Compute the distance of line FA.
Compute the area enclosed by the straight
line bounded by the points ABCDEFA.
Solution:
CD Bearing ofline BC:
LINES
AB
BC
CD
DE
EF
FA
r'\"
319.80
ulstance = Sin 47'50'
LATITUDES
425 - 75 = +350
675 - 425 = +250
675 - 675 =0
425 - 675 =-250
175 - 425 =-250
75-175=-100
DEPARTURE
150 - 250 =-100
450 -150 = +300
675 - 450 = +225
700 - 675 = +25
550 - 700 =-150
250 - 550 =-300
Bearing of line BC:
Distance =431.52 m.
·
Den.
tan beanng =-=
Lat.
.
+300
tan beanng =+250
Check:
Distance =..J (Latf+
EASTINGS
. 250 m.
150 m.
450m.
675 m.
700 m.
550m.
NORTHINGS
75m.
425m.
675m.
675m.
425m.
175m.
(Dep~
Bearing BC = N. 50'12' E.
Distance =..J (289.72)2 + (319.80~
Distance = 431.52 m.
@
Distance offine FA:
FA = ..J (Dep)2 + (Lat)2
FA=..J (- 30W + (-10W
FA =316.23 m.
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AREA OF ClOSED 'IIVERSE
@
Area bounded the straight lines:
)(
Lines
AB
BC
CD
DE
EF
FA
LAT.
+350
+250
0
-250
-250
-100
DEP
-100
+300
+225
+25
-150
-300
.n
f' _<.f,
107.28
\,\
D~D
-100
+100
+625
+875
+750
+300
~ It~
,
Double Area
-100(350) = -35000
100250 = + 25000
625 0) = O.
8751-250 = -21875(
7501 -250 = -18750(
300 -100) =-30000
2A= -446250
A=223125.r
JM.5.3.
211.81
131.70
lli..ZQ
273.40
Corrections using transit rule:
LineAB:
Latitude
Departure
~_
36.13
2.75 - 211.81
~_
C1 = 0.47
C1 =0.94
25.77
10 - 273.40
Line BC:
In the traverse table below shows the
Latitudes and Departures of the closed
traverse.
LINES
AB
BC
CD
DE
EA
LAT.
- 36.13
+ 74.56
+ 12.82
+ 19.90
- 68.40
DEP.
-25.77
-115.93
+0.39
+61.74
+69.57
CD Compute the corrected bearing of line BC
using transit rule.
@ Comp.ute the corrected distance of line EA
using transit rule.
.
@ Compute the area of the traverse by
balancing the traverse by transit rule.
Solution:
CD Corrected bearing of line Be using transit
rule:
LINES
AB
BC
CD
DE
EA
LAT.
-36.13 .
+ 74.56
+ 12.82
+ 19.90
- 68.40
+ 107.28
~
+ 2.75
DEP.
- 25.77
-115.93
+ 0.39
+61.74
+69.57
+ 131.70
-141.70
-10.00
Latitude
Departure
~_
74.56
2.75 - 211.81
f2. _115.93
C2 = 0.97
C2 =4.24
10 - 273.40
Line CD:
LatitUde
Departure
12.82
2.75 - 211.81
f.a. _ 0.39
C3 = 0.16
C3 = 0.01
~_
10 - 273.40
Line DE:
Latitude
Departure
~_ 19.90
2.75 - 211.81
~_
C4 =0.26
C4 = 2.26
61.74
10 - 273.40
Line EA:
Latitude
Departure
~_
68.40
2.75 - 211.81
~_
Cs = 0.89
Cs =2.55
69.57
10 - 273.40
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S-120-D
.IEI OF ClOSED TIAVElSE
LINES
AB
.BC
CD
DE
. EA
CORRECTED LATITUDES
- 36.13 + 0.47 =- 36.6
+ 74.56 - 0.97 = + 73.59
+ 12.82 - 0.16= + 12.66
+ 19.90 - 0.26 = +.19,.64'
.- 68.40 + 0.89 =- 69.29
CORRECTEO DEPARTURES
- 25.77 - 0.94 =- 24.83
-115.~3 - 4.24 =-111.69
+0.39 +0.01 = + 0.40
+61.74 +2.26 = +64.00
+ 69.57 +'2.55 = + 72.12
c
Corrected beanng of BC:
.
.Qep.
tan be anng = Lat.
. BC - 111.69
tan beanng
= + 73.59
..
'-'
~
Bearing BC = NS6'37' W
,
@ . Corrected distance Qf line
. AE'=
EA:
4(Dep)2 + (Latf
AE =...j (72.12)2 + (- 69.29)2
AE = 100.01 m.
@
Area of the traverse:
LINES
AS
BC
CD
DE
EA
LATITUDE
- 36.6
+ 73.59
+ 12.66
+ 19.64
- 69.29
DEPARTURE
- 24.83
- 111.69
+ 0.40
+ 64.00
+ 72.12
DMD
- 24.83
- 161.35
- 272.64
- 208.24
- 72.12
DOUSLEAREA
- 24.83(- 36.6) = + 908.78
-161.35(73.59) = -11873.75
- 272.64(12.66) =- 3451.62
- 208.24(19.64) =- 4089.83
- 72.12(- 69.29) = + 4997.19
2A =-13509.23
A= 6754.62 m2
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AREA OF ClOSED TRAVERSE
Correction for Departure:
C1
368.76
13.35 ;.; 2075
13.35
C1 = 2075 (368.76) = 2.37
From the given data of aclosed traverse
DISTANCE
368.76 m.
645.38 m.
467.86 m.
593.00 m.
LINES
AB
BC
CD
DA
BEARING
N.15'18'E.
S.85'46'E.
S.18'30W
N.7T35W
13.35
C2 = 2075 (645.38) =4.15
13.35
C3 = 2075 (467.86) = 3.01
13.35
C4 = 2075 (593)
Using compass rule of balancing atraverse.
Determine the corrected bearing of BC.
Determine the corrected bearing of CD.
Determine the adjusted distance of BC.
(j)
@
@
LINES
Solution:
LINE
BEARING
DISTANCE
AB
N:15·18'E.
368.76 m
BC
S.85·46'E.
645.38 m.
CD
S.18·30'W
467.86 m.
DA
N.7T35'W
59300 m.
207500
LAT.
+355.69
-47.64
-443.68
+127.51
+483.20
-491.32
Error = -8.12
Total distance = 2075.00 m.
DEP.
+97.31
+643.62
-148.45
-579.13
+740.93
-727.58
+13.35
1-2.37]
+355.69
= +357.13
+97,31
= +94,94
1-2.531
1-4.151
BC
-47.64
=-45.11
+643.62
=+639.47
1-1.831
-443.68
=-441.85
1+3.011
CD
DA
368.76
8.12 =2575
@
812
C2 = 2675 (645.38) = 2.53
8.12
C4 = 2075 (593)
1+2.321
1+3.821
+127.51
= +129,83
-579.13
=-582.95
=0
=0
Bearing
812
C3 = 2075 (467.86) = 1.83
= 232
8.12
-148.45
=-151.46
Corrected bearing of line BC:
·
Dep. + 63947
tan beanng = La!. =--=45.1T'
.Correction for Latitude:
C1 = ~O~~ (368.76) = 1.44
Corrected Dep.
I +1.441
Using compass rule of balancing:
C1
Corrected La!.
AB
Corrected bearing of BC:
(j)
= 3.82
13.35
=S.SS'SS'E.
Corrected bearing of CD:
t b'
Dep,
-151.46
an eanng = La!. = - 441.85
Bearing = S.1S'SSW
@
Adjusted distance of BC:
S' b ' - Dep.
In eanng - Dis!.
0' t -
Dep,
639.47
Sin bearing = Sin 8558'
Dist =641.06 m.
IS -
S~120-F
AREA OF CLOSED TRAVERSE
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tAT
A closed ·traverse has the following data:
LINES
AB
BC
CD
DE
EA
DISTANCE
895
315
875
410
650
BEARING
S. 70'29' E.
S. 26'28' E.
S. 65'33' W.
N.45'31' W.
N. 10'00' E.
<D Find the· corrected bearing of line BC by
using Transit Rule.
® Find the corrected bearing of line CD by
using Transit Rule.
~=~
15.73 1870.57
AB =299(0.0084092)
AB =-2:51 (to be subtracted
BC =281.99(0.0084092) .
BC =-2.37 (subtracted)
CD =362.16(0.0084092)
CD =-3.05 (subtracted)
DE =287.29(0.0084092)
DE =+2.42 (added)
EA =640.13(0.0084092)
EA =+5.38 (added)
15.73
DEP
AB
843.58
-=-7.79 2184.89
AB =843.58(0.0035654)
AB =3.01 (to be subtracted)
BC = 140.39(0.0035654)
BC =0.50 (subtracted)
CD =796.53(0.0035654)
CD =2.84 (added)
DE =292.52(0.0035654)
DE = 1.04 (added)
EA:: 112.87(0.0035654)
EA =0.40 (subtracted)
7.79
@ Find the corrected bearing of line EA by
using Transit Rule.
Corrected bearing for line BC:
Solution:
<D Corrected bearing of line BC using Transit
Rule:
LINES
AB
BC
CD
DE
EA
BEARING
S. 70'29' E.
S. 26'28' E.
S. 65'33' W.
N. 45'31' W.
N.10'OO' E.
DISTANCE
895
315
875
410
650
tan bearing = dep
lat
·
+ 139.89
tan beanng= - - - 279.62
Bearing = S. 26' 34' 42" Eo
® Corrected bearing for line CD:
ta be . - - 799.37
n
anng - _359.11
Bearing = S. 65' 48' 30" W.
Lines
AB
BC
CD
DE
EA
LAT
- 299
-281.99
-362.16
+287.29
+640.13
-943.15
+927.42
1870.57
: 15.73
DEP
+843.58
+140.39
-796.53
-292.52
+112.87
-1089.05
+1096.84
2184.89
+7.79
CORRECTED
LAT
DEP
-296.49 +840.57
-279.62 +13989
-799.37
-359.11
-293.56
+289.71
+645.51 +112.47
Sum of lat & dep.
Error
@ Corrected bearing for line EA:
ta be'
+112.47
n anng = + 645.51
Bearing = N. 9' 53' 01" E
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S-12]
ABU OF CLOSED TRAVERSE
@
Area enclosed by the traverse:
3
i$Cpmputelhe b~arinsqfline4.t.
·~· •. ··.ColTlPu~ .•thE!.dlstal'l~9f •~l®.4 ••• 1.••••
tID9Wl1PW
t$vetse.e ·the51t~~WWI(l~~9l)}'
... . ... . fhe
B
Solution:
The sketch shows that the traverse lines 1 - 2
and 3 - 4 crossed each other, hence we could
not adopt the DMD method of determining its
area.
Lines Bearina Distance
1-2 N48'30'W 81.00
2-3 N77'00' E 66.00
3-4 S55'OO'W 94.00
4-1
(j)
Bearing ofline 4 - 1:
.~
tan beanng = lat
.
73.40
tan beanng = 14.65
Bearing = S 78'42' E
@
Distance of side 4 - 1:
.
_ Departure
Distance - Sin bearing
.
73.40
Distance = Sin 78'42'
Distance (4 -1) = 74.85 m.
LAT
+53.70
+14.85
- 53.90
+14.65
DEP
-60.70
+64.30
-77.00
-73.40
A'
From Plane Trigonometry: 74.85
Area of triangle ABC =~ a c sin B
Using the Law of Sine:
a_
_c
Sin C-SinA
a Sin C
c= Sin A
~ (a) (a) Sin CSin B
Area =
Area -
Sin A
a2 Sin BSin C
2 Sin A
Considering triangle 230:
A _(66)2 Sin 22' Sin 54'30'
12 Sin 103'20'
Al = 683 sq.m.
Considering triangle 014:
A - (74.85)2 Sin 46'18' Sin 30'12'
22 Sin 103'30'
A2 =1046 sq.m.
Total area =Al +A2
A=683 + 1046
A = 1729 sq.m,
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S-122
AREA OF CLOSED TRAVERSE
G
.
r.
~
Ind 5
LINES
1-2
2-3
3-4
4-5
5-1
.
2
".
B
BEARING
S 10'00' E
N 56'00' E
N63'OOW
........
S33'OO'W
DISTANCES
485.00
780.00
975.00
........
890.00
c
.4
. (
)
tan Beanng 4 - 5
622.59
=345.24
Bearing (4 - 5) =N 60'59' E
Solution:
@
Distance of line 4 - 5:
.
.
622.59
D/stance (4 - 5) =Sin 60'59'
Distance (4 - 5) =711,90 m.
@
Area enclosed by the line 3 - 4, 4 - 5 and
5" 1:
A = bc Sin A
2
2
CD Bearing of line 4 - 5:
Lines
5-1
1- 2
2-3
3-4
Bearina Distances LAT
DEP
S33'W
890
- 746.42 - 484.73
S 10' E
485
- 477.63 +84.22
+436.17 +646.65
N56' E
780
N63'W
+442.64 - 868.73
575
+345.84 +622.59
+878.81 +730.87
- 1224.05 - 1353.46
- 345.24 - 622.59
c
b
Sin C~ Sin B
bSin C
c= Sin B
A _ b2 Sin A Sin C
2 Sin B
Area of shaded section
A
=(711.90}2 Sin 56'01' Sin 27' 59
2 Sin 96'
A =99169.28 m2
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AREA OF CLOSED TRAVERSE
A Civil Engineer, in his haste, forgot to record
the data of the closing line of his traverse, the
field noles of which reflects the following
record.
@
Distance of line 5 - 1:
53.51
Distance (5 - 1) Sin 54'20'
Distance (5 -1):: 65,86 m,
@
Area enclosed by the traverse:
A _ (65.86)2 Sin 16' Sin 80:40'
12Sin 83'20'
A1 :: 593.91 sq.m.
Using Sine Law:
(j)
@
@
5
Compute the bearing of line 5- 1.
Compute the distance oHine 5 -1. .
Compute the area enclosed by the
traverse.
Solution:
Sketch the traverse and nnd out if the lines do
not intersect each other, if so, then application
of DMD in determining the area will not suffice.
5
5
4
3
I
(j)
Bearing of line 5 - 1:
Lines
1- 2
2-3
3-4
4-5
5-1
Bearinq
S 30'20' E
S51'57' W
N49'10'W
N45'00' E
Distance
LAT
DEP
140.25 -110.02 +86.99
77.52
-47.78 - 61.04
65.10
+42.57 -~.26
108.64
+76.82 +76.82
+38.41 53.51
53.51
tan bearing (5 -1):: 38.41
Bearing (5 -1):: N 54'20' W
x _ 65.86
Sin 80'40' - Sin 83'20'
x:: 65.43
y
65.86
Sin 16' :: Sin 83'20'
Y:: 18.28 m.
Distance 4 to 0:: 108.64 - 18.28
Distance 4to 0:: 90.36
Distance 2 to 0:: 140.25 - 65.43
Distance 2 to 0:: 74.82
Ar
- 90.36 (74.82) Sin 83'20'
2
A2 :: 3357.49 sq.m.
- 65.10(77.52)Sio 101'07'
A32
A3 =2475.90 sq.m.
Total A:: A1 + A2 + A3
A:: 593.91 +3357.49 + 2475.90
A :: 6427.30 sq.m.
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8-124
AREA OF ClOSED TRAVERSE
Solution:
cD Location of the point of intersection of the
overlapping areas from corner 4 of lot
PSU-171211:
AC
33.86
Sin 87'18' =Sin 53'30'
AC=42.08
The point of intersection from comer 4
= 56.65 ·42.08
= 14.57
® •Location of the point of intersection of the
overlapping areas from corner 5 of lot
PStJ-187773:
BC
33.86
Sin 39'12':;: Sin 53'30'
BC= 26.62
Point of intersection from comer 5
= 37.74·26.62
= 11.12
® Area of the overlapping portion:
A =33.86 (42.08) Sin 39'12'
2
A =450.20 sq.m.
4
li.lilill1ll
~CteS.····
B
('
»..
1
.
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AREA OF CLOSED TRf"VERSE
Solution:
CD DMD afline 3 - 4:
LINES
1-2
2-3
3-4
4-5
DMD
DEP
- 40.12
- 36.82
+ 50.42
+ 30.36
- 52.34
+48.50
5-6
6-1
- 40.12
-117.06
-103.46
- 22.68
- 44.66
·48.50
DMD afline 3 - 4 = • 103.46
@
DPDafline4- 5:
LINES
1-2
2-3
3-4
4-5
5-6
6-1
DPO
+80.16
+ 120.19
+ 150.24
+ 190.28
+220.34
+140.27
LAT
+ 80.16
- 40.13
+ 70.18
- 30.14
+60.20
-140.27
DPD ofline 4 - 5 = + 1190.28
@
Area of Jot:
LINES
LAT
OM)
DOUBLE AREA
(LATxDMD)
1-2
2·3'
3-4
4-5
5-6
6-1
+80.16
-40.13
+70.18
-30.14
+60.20
-140.27
- 40.12
-117.06
-103.46
- 22.68
- 44.66
-48.50
- 3216.02
+4697.62
- 7260.82
+683.58
Solution:
CD BeiJring ofline FA:
Lines Bearina Distance LAT.
DEP.
AB N. 20' E. 17.42 +16.37 +5.96
Be N.68' E. 18.46
+6.92 +17.12
CD S. 22' E. 22.40 -20.n +8.39
DE S.40·W. 12.60
- 9.65 -8.10
EF S.62'W. 10.20
-4:79 - 9.01
FA
+11.92 -14.36
tan bearing =
·
14.36
tan beanng = 11.92
- 2688.53
+6803.10
- 13165.37
+ 12184.30
2A =-981.07
A= 490.54
A= 490.54
4047
A= 0.121 acres
f!i
Bearing FA = NSO'18' W
@
Distance FA:
O'
t
IS ance
-
Dep
- Sin bearing
'D' t
14.36
Isance::: Sin 50'18'
Distance::: 18.66 m.
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S-126
AREA OF ClOSED TRAVERSE
® Area of closed traverse:
Lines
LAT.
DEP.
AB
BC
CD
DE
+16.37
+6.92
-20.n
-9.65
-4.79
+11.92
+5.96
+17.12
+8.39
-8.10
- 9.01
-14.36
EF
FA
Double
Area
+5.96 +97.51
+29.04 +200.96
+54.55 -1133.00
+54.84 - 529.21
+37.73 -180.73
+14.36 +171.17
2A 373.24
A = 186.62
A = 186.62
4047
A=0.046 acres
=-
Solution:
CD
DA
tangent bearing = S. 11'24' W
® Distance DA:
.t
10.79
=Sin 11'24'
Distance =54.55 m.
DIS ance
@
Area:
_ 3745.01 m2
Area - 4047
Area = 0.925 acres
I!rll.ill~
Solution:
CD Bearing DA:
LINES
AB
BC
.
10.79
tangent beanng =53.47
D\1D
BEARING
S.8'S1'W.
N.1S'S1'W.
N. 32'27' E.
DISTANCES
126.90 m.
90.20 m.
110.80 m.
-
-
LAT
DEP
Line
AB -125.3£ -19.52
BC +8S.36 - 29.14
CD +93.50 - 59.45
DE - 53.47 -10.79
lJv1D
2A
-19.52 +2447.61
- 68.18 - 5819.84
- 37.87 - 3540.85
+10.79 - 576.94
2A = 7490.02
A = 3745.01 m2
CD Bearing of line 4. - 1:
Line:
LAT
DEP
1-2
2-3
3-4
4 -1
+104.1(
+ 18.75
- 74.97
-47.88
+60.10
+88.23
+46.84
-195.17
Double
Area
+60.10 3760102.4
+208.43 +3908.06
+343.5 -25752.20
+195.17 - 9344.74
2A = 3728913.53
O'v1D
'
195.17
tangentbeanng = 47.88
tangent bearing =S. 76'13' W
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AREA OF ClOSED TRAVERSE
@
Distance 4 - 1:
195.17
=
Sin 76'13'
= 200,96
@
DMD ofline 3-4 =+ 3434,5
@
Area oflot in acres:
Area = 4800 m2 .
4800
Area =4047
Area = 1.186 acres
® Area = 1,864,456.77
Area = 186.45 hectares
fr9mth~~Wem~~#9fl~ry9<h~ymSIM
fplloWil'lgr.lB1a ,C()fJ1Pule'fhejfq1IriWing(>
..
tiNES l. tAl1WP/l'·
DI1J'Af\r~
...
CD
,
<90<· ······>+50
o.;;}\
09Ublemeridtandist~~{)fllheC8'
.
.
® j)oup~par<llfeldist~%~.oflillElGP' .
@ Ateaoftracl()flan¢il'la¢res. ... ..
Solution:
CD Bearing of CD:
Double area = Lat x DMD
- 8108.71 = Y(189.50)
Solution:
CD DMD ofline CD:
LAT
+40
+80
- 30
-90
DEP
DMD
2A
-80
-40
- 80
-200
-170
- 3200
-16000
+ 5100
+ 4500
2A = - 9600
A= 4800ml
+70
+50
-50
y=-42.79
Lat. CD =- 42.79
Dep. CD = +13.36
f!i
tan bearing =
t b
.
13.36
an eanng = 42.79
Bearing = S 17'20' E
DMD of/ine CD =. 170
@
DPD ofline CD:
LAT
+40
+80
- 30
- 90
DEP
DPD
- 80
+40
+160
+210
+90
-40
+70
+50
OPO ofline CD = + 210
® OMD of/ine DE:
+ 16.03
+72.04
+ 13.36
+101.43
- 48.18
- 53.25
Dep. ofDE =. 53.25
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5-128
AREA OF ClOSED TRAVERSE
® Area of the 5 sided lot in acres.
Lines
LAT
DEP.
AB
Be
+57.81
- 9.63
- 42.79
- 18.75
+ 13.36
+ 16.03
+72.04
+ 13.36
- 53.25
- 48.18
()\1()
Solution:
Double
Area
OJ
DE
EA
+ 16.03 +926.69
+104.10 -1002.48
+189.50 - 8108.71
+149.61 - 2805.19
+48.18 +643.68
2A = 10346.01
A =5173.005 m2
tan bearing =~
·
tan beanng
=_1430
Bearing = S 25'01' E
® DMD ofline 4-5:
I+LAT
I-LAT
+57.81
+ 13.36
+ 71.17
-42.79
-18.75
-61.54
x = 71.17 - 61.54
x=-9.63
A = 5173.005 m2
A = 5173.005
4047
A =1.278 acres
CD Bearing ofline 3 - 4:
LAT(DMD) = Double area
Lat (186) =- 5580
Lat=- 30
Course
1- 2
2-3
3-4
4-5
5-1
La!
OeD
()\1()
+60
+16
+70
+14
-54
-46
+16
+102
+186
+146
+46
-14
- 30
-28
+12
Latof2-3:
60 +12-30-28= 14
Depof4-5:
16 + 70 + 14 - 46 =54
DMD ofline 4 -5 =+146
® Area oflot:
Lines
LAT
DEP
DMD
1- 2
2-3
3-4
4-5
5-1
+60
-14
-30
-28
+12
+16
+70
+14
-54
-46
+16
+102
+186
+146
+46
4792
Area =4047
Area = 1.184 acres
Double
Area
+960
-1428
-5580
-4088
+552
2A =9584
A =4792 m2
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5-129
Pdfbooksforum.com
.AREA OF CLOSED TRAVERSE
® Area of whole lot:
LAT
AB
Be
- 310.95
- 640.21
+28.80
+576.94
+345.42
We.fOIlO\'ii.,~ • • cl<ita9f.~.relq~tionsurveyof
QOn•• Marian().E.sClJtIElr().i~Jo~·.rec()nstl\lot~;
IUne:
~ ~'.
I
I
....
.
DMD
LINES
CD
Double
.hea
DE
EA
AS .31().95+469J~4·. <.:.:.: I··· <
•.....
BG-':':;' • • ·+112,87 1"+':·l!)7~1p.93
··a)"·~·'··109M2 +es.a0+1895;04
DOUBLE
AREA
+469.84 -146096.75
+1052.55 -673853.04
+1895.04
+65.8
-1055.68 - 609064.02
- 538.77 -186101.93
2A =1613220.70
A = 806610.35 m2
DE +576.94 >\-_.:
.
EN +345.42+538>77>
An englnee; sets upa lTaJiSil at apoint inside a
triangular lot and observes' the bearinfjS and
distances of the corners A, Band C ofthelbt
as follows:
Solution:
CD Bearing of line CD:
. CORNERS
Double area = Lat xDMD
BEARING
1895.04 = Lat (65.80)
Lat= +28.80
tan bearing =~
ill Compute the area of the triangUlar lot.
.Compute the perimeter of the lot.
® • If the bearing from C to thepolnl instdethe
. . trian.gular lot is due north. compute lhe
beanng of CB.
...
@)
-1099.62
tan beanng = +28.80
.
bearing = N. 88'30' W
Solution:
® DMD of line DE:
LINES
AB
Be
CD
DE
EA
CD Area of the triangular lot:
DEP
DMD
+469.84
+112.87
-1099.62
- 21.86
+538.77
+469.84
+1052.55
+65.8
-1055.68
- 538.77
A
B
DMD ofline DE =• 1055.68
c
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5-130
AREA OF CLOSED TRAVERSE
- 17(22) Sin 105'
A12
A,
=180.63
- 22(32) Sin 110'
A22
A2=330.77
Ar
- 32(17) Sin 145'
2
A3 =156.01
A=A, +A 2 +A 3
A =667.41 m2
CV Perimeter of the lot:
(AB)2 = (17)2 +(22)2 - 2(17)(22) Cos 105'
AB = 31.09 m.
(BCf =(22)2 + (32)2 - 2(22)(32) Cos 110'
BC=44.60m.
(AC)2 =(17)2 + (32)2 - 2(17)(32) Cos 145'
AC=46.95 m.
CD¢omPOI~tI'\ebMtltl9()tlioeDE.
@QQflJ@l~tt1El~@MOfllne
EA·
@CarnpLJiEnft$ar~()fthel()t
Solution:
Line~
BearinQ Distance LAT
DEP
AB S 35'30'W 44.37 - 36.12 -25.77
BC N5T15'W 137.84 +74.57 -115.93
CD N 1'45' E 12.83 +1282 +0.39
.. ----- - 51.27 +14131
DE
~
Perimeter = 31.09 + 44.60 +46.95
E
Perimeter =122.64 m.
@
Bearing of CB:
22
44.60
Sin 8 =Sin 110'
8 =27'37'
c
B
Bearing of CB = N. 27'37' E.
D
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5-131
MISSING DATA
CD Bearing of line DE:
Bearing and distance of line DA:
.
141.31
tan beanng = 51.27
Bearing (DA) = S 70'04' E.
141.31
Distance (DA) =Sin 70'04' =150.31 m.
Considering triangle DCA:
Using Cosine Law:
(64.86)2 = (106.72f + (150.31)2
·2(150.31)(106.72) Cos .,
., =21'52"
Using Sine Law:
Sin B Sin 21'52
106.72 - 64.86
B=37'48'
Sin 21'52' _ Sin a
64.86 - 150.31
Ct =120'20'
Solution:
Line~
Bearina
Distance LAT DEP
OE S 8'51' W
126.9
125.39 -19.52
EA N 18'51' W
90.2
+85.42 - 28.96
AB N 32'21E
110.8 +93.50 +59.45
BO
- 53.53 -10.97
?earing ofline DE = N, 72'08' E.
® Bearing of line EA:
Bearing ofline EA =70'04' - 21'52'
Bearing of line EA =S. 48'12' E.
@
Note: OE is equal and parallel to CD, likewise
CO is equal and parallel to line DE.
Area of the lot:
Line~
Bearina
AB
BC
CD
DE
EA
S 35'30'W
N 57'15'W
N 1'45' E
N 72'08' E
S. 48'12' E
Lines
AB
BC
CD
DE
EA
<D Compute the bearing of Hne BC.
® Compute the distance of line DE..
® Compute the area of the lot.
.
A
Distance
44.36
137.84
12.83
64.86
106.72
DIv1D
- 25.77
-167.47
- 283.01
-220.88
·79.57
LAT
- 36.13
+74.56
+12.82
+19.90
- 71.15
DI;:P
-25.77
115.93
+0.39
+61.74
+79.57
2A
+ 931.07
- 12486.56
- 3628.19
·4395.51
+ 5661.41
2A = 13917.78
A = 6958.89
E
Bearing and distance of line (BO)
. 10.97
tan beanng =53.53
Bearing of (BO) =S 11 '35' W
(BO)
10.97
·
DIstance
= Sin 1-1' 35'
Distance (BO) =54.65 m.
B
Area of the lot = 6958.89
o~----:;;L.~C
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5-132
MISSING DATA
Consider the triangle BOC
Angle BOC = 73'31' - 11'35'
Angle BOC = 61'56'
Sin '" Sin 61'56'
54.65 = 83.6
'" = 35'14'
CD Bearing ofline BC:
B
<D
G®JpqtElth~fJil§singsi(jEl~P­
~QomPllt<ltherllis$l@sideQA
.. ",., ,,'•• ',.,
~ • • C®@t~.tfIe.~r~a.of.tfIe • (()~itlMre$ .•,
Solution:
B =180 - 61'56' - 35'14'
B =180 - 90'10'
B=82'5O'
CD SideBC:
B
Bearing of line BC = S 71'15' E
~
DIstance of line DE:
OC
_ 83.6
Sin 62'50' - Sin 61'56'
OC= 94.00 m.
OC=DE
Distance of line DE =94,00 m.
@
Area of the lot:
Line~
Bearinq
N32'2TE
S 71'15' E
CD S8'51'W
DE S 73'31'W
EA N 18'44'W
AB
BC
Lines
AB
BC
CD
DE
EA
LAT
+93.50
- 26.87
-125.39
- 26.66
+85.42
Distance
110.8
83.60
126.90
94.00
90.20
LAT DEP
+93.50 +59.45
·26.87 +79.16
125.39 -19.52
-26.66 - 90.13
+85.42 - 28.96
Drv'ID
+59.45
+198.06
+257.70
+148.05
+28.96
2A
+5558.58
- 5321.87
-32313.00
- 3947.01
+2473.76
2A = 33549.54
A = 16774.77
Area of the lot =16774.77 sq.m.
BC
73.2
Sin 60' = Sin 45
BC= 89.65m.
® SideCA:
CA
73.2
Sin 75' = Sin 45'
CA = 99.99m.
@
Area ofABC:
A = (73.2)(99.99) Sin 60'
2
A = 3169.34 m2
A = 3169.34
4047
A = 0.783 acres
Note: 4047 m2 =1 acre
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MISSING DATA
Using Cosine Law:
(64.86)2 = (150;32)2 +(107.72)2
- 2(150.32)(1()7.72) Cos e
e = 22' 09'
The technical description of a closed traverse
is as follows.
LINE
1-2
2-3
3-4
4-5
5- 1
DISTANCE(m)
64.86
107.72
44.37
137.84
12.83
BEARING
?
107.72 = 64.86
Sin B Sin 22' 09'
B = 38' 47' .
AzilTl.uth of line 1 to 3 = 360' - 70' 03'
Azimuth of line 1 to 3 = 289' 57'
?
S. 35' 30'W.
N. 57' 15' W.
N. l' 45' E.
Azimuth of line 1- 2 = 289'57' - 38'47'
Azimuth of line 1- 2 = 251' 10'
Bearing of line 1- 2 =N. 71'10' E.
Find the bearing of line 1- 2.
® Find the bearing of line 2- 3.
@ Find the area of the closed traverse.
(j)
® Bearing of line 2 - 3:
Solution:
(j)
Azimuth of line 2 - 3 = 289' 57' - 22' 09'
Azimuth of line 2 - 3 = 312' 06'
Bearing of line 2 - 3 =S 47~ 54' E.
Bearing of line 1 - 2:
@
Area of closed traverse:
Line
1-2
2-3
3-4
4-5
5-1
4
Lines
Bearin!l
Distance
LAT
DEP
3-4
4·5
5 -1
1-5
S.35·30'W
N.57'15'W
N."45'E
44.37
137.84
12.83
-36.12
+74.57
+12.82
-51.27
-25.77
-115.93
+0.39
+141.31
Bearing afline 1-3:
·
141.31
ta n beanng=--
51.27
Bearing of line 1-3 =S. 70' 03' E.
Distance =
141.31
Sin 70' 03'
Distance =150.32 m.
Dist
64.86
107.72
44.37
137.84
12.83
Bearina
N.71'10'E
S.47'54'E
S.35·30'W
N.5T15'W
N.1'45'E
Line
DEP
DMD
1-2
2-3
3-4
4-5
5-1
+61.39
+79.92
- 25.77
-115.93
+ 0.39
+141.70
- 141.70
+61.39
+202.70
+256.85
+115.15
-0.39
Area = 7023.57 m2
LAT
+20.94
- 72.21
- 36.12
+74.57
+12.82
+108.33
-108.33
Double
Area
+1285.51
-14636.97
- 9277.42
+8586.74
-'5.00
2A=-14047.14
A=-7023.57
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S-l32-B
MISSING DATA
A closed traversed shows tabulated values of
latitudes and departures.
LINES
1- 2
2-3
3-4
4-5
5-6
6 -1
LATITUDE
+ 84.60
+ 95.32
+ 62.66
·48.16
-43.04
-
Given the following descriptions of a four sided
lot.
DEPARTURE
---
- 57.52
- 31.40
+ 59.70
+47.63
--
<D Compute the DMD ofline 3 - 4.
@ Compute the length of line 6 to 1.
@ Compute the bearing of line 6 to 1.
BEARING
N 30'30' E.
N75'30'W
S45'30'W
LINE
AB
BC
CD
DA
c 56.11
...
DISTANCE
56.5m.
46.5m.
87.5 m.
---
<D What is the length of line DA?
What is the bearing of line DA?
@ Compute the area of the enclosed
traverse.
@
Solution:
<D DMD of line 3 - 4.
Latitude of (6 - 1)
~(Iatitude) = 0
84.60 + 95.32 + 62.66 ·48.16 - 43.04 - Y= 0
y= -151.38
Departure of (1 - 2)
L(departure) = 0
x- 56.11- 57.52 - 31.40+59.70+47.63 =0
x = 37.70
LINES
LATITUDE
DEPARTURE
DMD
1- 2
2-3
+84.60
+95.32
+62.66
- 48.16
- 43.04
-151.38
+37.70
-56.11
- 57.52
- 31.40
+59.70
+47.63
37.70
19.29
- 94.34
-183.26
-154.96
- 47.63
3-4
4-5
5-6
6-1
Solution:
<D Length of line DA:
LINE
LAT
AB
+48.68
BC
+ 11.64
CD .
-61.33
DA
@
Length of line 6 to 1.
(Dislance)2 =(latitude)2 + (departuref
(0 6 _1)2 = (- 151.38)2 + (47.63)2 1.
0 6 _1 =158.70 m.
w- - I
Bearing of line 6 10 1.
43.63
tan e =151.38
8 = 11' 29'
:. Bearing is S 17" 29' E
·151.38
@
Bearing of line DA:
78.82
tan 11 = "'1.ii1
11 = 78'02'
Bearing ofDA =N 78'02' E.
@
Area of the enclosed traverse:
LINE tAT
OEP
OMO
POA
AB + 48.68 + 28.68 +28.68 +1396.14
BC + 11.64 - 45.09 + 12.27 + 142.82
CD - 61 .33 - 62.41 - 95.23 + 5840.46
DA
s
+ 78.82
Distance DA =..J(+ 1.01)2 + (78.82)2
Distance DA = 78.83 m.
DMD of line 3 - 4 = • 94.34
@
+ 1.01
DEP
+ 28.68
- 45.09
- 62.41
+ 1.01
+ 78.82
2A= 7299.81
A=3,649.91 m2
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MISSING DATA
Using Sine Law:
30.17
24.22
Sin 95'13' - Sin f3
f3 =53'05'
Bearing CA =69'11' - 53'05'
Bearing CA =N 16'06' W
Using Sine Law:
0) FindlhedisfahooDAinmeters,
@ • • FilldthedistandeCtlinr1lefers:
@FiodtM<ilreM6sQ,m,
Solution:
CD Distance DA in meters,'
180- (15'36' +69'11')
95'13'
,,=
,,=
c
Using Cosine Law,'
(AC)2 = (2422)2 + (15.92)2
- 2(24.22)(15.92) Cos 95'13'
AC =30.17
DA
_ 30,17
Sin 74'04' - Sin 22'45'
DA= 75.02m.
® Distance CD in meters:
30.17
CD
Sin 22'45' =Sin 83'11'
CD= 77.47
@
Area in sq.m:
LINES
AB
BC
CD
DA
A
BEARING
S. 15'36' W.
S. 69'11' E.
N. 57'58' E.
S. 80'43' W.
Lines LAT
DEP
AB - 23.33 -6.51
- 5.66 +14.88
BC
CD +41.09 +65.67
DA -12.10 -7404
c
DISTANCES
24.22 m.
15.92 m.
77.47 m.
75.02 m.
DMD
- 6.51
+1.86
+82.41
+74.04
2A
+151.88
- 10.53
+3386.23
- 895.88
2A - 2631.7
A =1315.85 m2
S-134
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MISSING DATA
0). Q()/1'\plJt~ffl¢loW~09!tll)tttiet~~;
@ .QptlJp~te:thei)~n1~thQflitj~.9A' • • • • "•• • '• ".
@~llte:tl'w!!lreai:lftl'w!klOOllcre$.
Solution:
Solution:
<D Total length of traverse:
N
(AC)2 = (1200)2 + (1400)2
- 2(1200)(1400) Cos 59'
AC = 1292.08 m.
<D Distance AB:
AB
129.86
Sin 58' = Sin 60'
AB = 127,16 m,
® Distance CD:
x
129.86
Sin 62' =Sin 60'
x = 132.40rn.
CD= 132AOm.
@
Area of lot:
A=
t? - b,2
2 (Cot e + Cot ~)
A = (380.48)2 - (250.62)2
2 (Cot 62' + Cot 58')
A = 34084.72
A = 34084.12
4047
A= 8,42 acres
Total length of traverse
= 1200 + 1400 +1292.08
=3892.08 m.
® Azimuth of line CA:
Sin 8 _ Sin 59'
1200 -1292.08
e = 52'45'
Bearing =N. 66'45' W.
Azimuth of CA =113'15'
@
Area of lot:
A 1200 (1400) Sin 59'
2
A = 720020.53 m3
A =720020.53
4047
A =177.91 acres
.
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MISSING DATA
(j) • • • COlTll?ute.~.~te~.(lf.~k)tirlacre$·
~q()mput~th~mi$Sing~l$~np~9:tM~1.g.
@GqnlPu!$JhemlssiQ99~f1jnpe(1f1jnl'l4<t
®ql>l'fJPlJlE!fh~affl~Qf:~MA9tj~SQQ~r~
¢Bi'le.IDe.djstahce.6flin~.~
.®..
• .• 3,•• • • • • • • ·•
@ . COll1putE!!tlE!~i$~o®mll~~3"4·
Solution:
CD Area of the lot:
Solution:
CD Area of lot in square meters:
4
216.60(116.40) Sin 34'
A
rea =
2
216.60(174.40) Sin 24'
+
2
Area = 14731.50 m2
A - 14731.50
rea - 4047
Area = 3.64 acres
4
- 142(260) Sin 36' 260(240) Sin 62'
Area 2
+
2
Area =38,398.48 sq.m.
".
® Distance 1 - 2:
(1 - 2)2 =(216.60)2 + (116.40)2
·2(216.60)(116.40) Cos 34'
Distance 1- 2 =136.60 m,
® Distance 2 - 3:
(2 - 3)2 =(142)2 + (26W - 2(142)(260) Cos 36'
Une (2 - 3) =167.41 m.
® Distance 4 - 1:
(4-'1)2 =(174.40)2 +(216.60)2
- 2(174.40)(216.60) Cos 24'
Distance 4 - 1 =91,17 m.
® Distance 3 - 4.
(3 - 4)2 = (260)2 +(24W - 2(260)(240) Cos 62'
Une (3 - 4) =258.09 m.
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S-136
MISSING DATA
~~'eI!ijt~I~I~~tlS~b~!hiClbS0~
·1·.··.~I~:I·~~.a;I~~8.10t
.• • •. . .
Solution:
CD Area of closed traversed:
Solution:
CD Area of lot:
N
- (120)(220) Sin 34' 220 (180) Sin 24'
Area2
+
2
Area =15434.73
15434.73
Area=400Area =3.81 acres
® Distance AB:
(AB)2 =(120)2 + (220)2 - 2(120)(220) Cos 34'
AB = 137.94 m.
@
Distance DA:
(DA)2 = (180}2 + (220f - 2(180)(220) Cos 24'
DA =91.91 m.
Area = 1400(1800) Sin 57'
2
Area = 1056724.92 m2
- 1056724.92
Area - 4047
Area =261.11 acres
® Total perimeter of lot:
.;. = (1400)2 + (1800)2.2(1400)(1800) Cos 57'
x = 1566.85 m.
Totalperimeter= 1400 + 1800 + 1566.85
Total perimeter =4766.85 m.
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MISSING DATA
® Side AD:
® Bearing of 3-1:
1800 1566.85
Sin e = Sin 57'
B
c
e = 74' 28'
11 = 180·45' • 74'28'
11 = 60'32'
Bearing of 3-1 is N 60'32' W
A
D
AD=BE
BE
100
Sin 77' =Sin 41'
BE = 148.52 m.
® Areaoflot:
_ (25Of· (150)2
A - 2 (cot 77' + cot 62')
A =26,226.84 sq.m.
.(!).. . yM}put~fM~l$$m~$l~~~p .• • • • • • • • • • •·•· .
•1••6~~ill1.i~~rl~~ldt6t~~·.···············
FfQlllfh~giYMJfl~h~'~~I~~~pt~~.Clf~@t.
I..JNES> ·BSARIN~S<DISTANCE;S
.N4a'20·/E·.·.····. .·•·•·••·.·.S29,6QIrL..
. 592mOrtl/
Solution:
S63.60m.
CD Side BC:
Q)
®
GClmpllte thebeal"ihg (jfline DE..
ComputetIJ$beanr19oflineBC.
® ComputelhearaaofthelOt
Solution:
Draw a line BO parallel to CD at B:
B
A
D
BC
100
Sin 62' =Sin 41'
BC = 134.58 m.
E
Closing Line
S-138
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MISSING DATA
CD Bearing ofline DE:
Line Bearinas Distances LAT
DEP
EA N48'12'W 428.20 +285.41 - 319.21
AB N48'20'E 529.60 +352.10 +395.60
BO S70'59' E 563.60 - 558.14 +78.24
- 79.37 -154.63
. (DE) 154.63
tan Beanng.
= 79.37
Bearing (DE) =S 62'50' W
154.63
0 IS, tance (DE) = Sin-62'80'
OE= 173.81 m.
Area of/of:
@
LINES
AB
BC
a>
DE
EA
Lines
AB
BC
CD
DE
AE
BEARING
N.48'20' E.
N. 87'33' E.
S. 7'59' E.
S.82'01'W.
N.48'12'W.
LAT
+352.01
+25.31
- 558.1'
-104.64
+285.41
Consider triangle ODE:
DEP
+395.60
+591.44
+78.28
- 746.12
- 319.20
DISTANCES
529.60
592.00
563.60
753.40
428.20
2A
DMD
+395.60 +139278.89
+1382.64 +34994.62
+2052.36 - 1145524.73
+1384.52 -144876.17
+319.20 +91102.87
2A =- 1025024.52
A= 512,512,26 m2
Using Cosine Law:
(592)2 = (193.81~ + (753.40~
- 2(173.81)(753.40) Cos '"
0=19'11'
Bearing (ED) =62'SO' +19'11'
Bearing (ED) =N82'01' E
Bearing ofDE =S 82'01' W
cD
Pl)rop@~tM~j:lian®6(;
@
CotnputelheareabYOMOmelhod.
Solution:
® Bearing of/ine BC:
Sin a _ Sin 19'11'
173.81 - 592.00
a =5'32'
Bearing DO =82'01' +5'32'
Bearing DO =S87'33' W
OO=BC
Bearing BC = N 87'33' E
.•.•. .•.
® C()lttplJt¢tbedi~~A13.<
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MISSING DATA
CD Distance BC:
Using Cosine Law:
(AC)2 = (75~ + (77.45~ - 2(75)(77.45) Cos 22'45'
AC=30.16 m.
In.!he•• surveY•• ofl3..()~$~d§t • willl·.flve.@dll$,
~M • f~lIE)wi~g • ~~til~r~.gliJe9.Wher~ • I(l•• all·•• m~
~~ril1~.llf'ldAlstM~ciflillsjq~8"X'cllJ:l1th8"
Sin" Sin 22'45'
T = 30.16
I~Q~~$()f~~~s;4f§~n95+.t~.mnittecl·
" =74'05'
74'05' - 57'58' = 16'Or
Bearing (AC)= S 16'07' E
Angle B4C = 16'07' + 15'36'
Angle BAC =31'43'
Angle BCA =69'11' -16'07'
Angle BCA =53'04'
l..lNSS> . ElEARlNG<
1)2 ···S73'21'E<
•·· · • ·•• S4l))1Q\E·.·•. • • ·
· · • S2ef4ZW)·
• ·•· • • ·N14'20~W)··
ill ¢0mPute.tI'l~.qi$¥tn®pfll~.e.4."
• 1,
® qolTIpQle.tI1~di~t~Il(;eQHllle4 • • ~.
@" ()ornplit~·tfu!·l.li~~~otline$.·
Using Sine Law,
Considering triangle ABC:
30.16
BC
Sin 95'13' =Sin 31'43'
BC= 15,92m.
Distance AB:
30.16
AB
Sin 95'13' = Sin 53'04'
@
• • • • •·
• 1•.
4
Solution:
AB= 24.21 m.
CD Distance of line 4 . 1:
® Area by DMD method:
Line Bearinas
AB S 15'36'W
BC S69'11' E
CD N57'58' E
DA S80'43'W
Distances LAT
24.21 m -23.32
15.92 m ·5.66
75.45 m -41.07
75.00m -12.00
Line~
DEP
AB
Be
CD
DA
LAT
-23.32 -6.51
- 5.66 +14.88
+41.07 +65.66
-12.09 -74.03
DEP
- 6.51
+14.88
+65.66
- 74.03
Linel Bearina
1-2 S 73'21' E
2-3 S40'10' E
3-4 S26'42' W
Distance
DEP
LAT
247.20
-70.83 +236.83
154.30 -117.91 +99.53
611.90 -546.65 -274.94
o/J
tan bearing =
. 61.42
tan beanng= 735.39
DMD
Double
Alea
Bearing (4 -1) =N4'47' W
-6.51
-1.86
+82:40
+74.03
+151.81
-10.53
+3384.17
-895.02
2A =630.43
Distance (4 -1) = Si~~47'
A=1315.22 rTf
61.42
·t
DIS ance =Sin 4'47'
Distance =746.53 m.
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5-140
MISSING DATA
@
Distance of line 4• 5:
Consider triangle 1- 4 - 5:
r~~
[
100
--"
,,= 14'20' ·4'47
,,=9'33'
a. = 12'20' +4'47
0.= 17'07
f!, =180·9'33' • 17'07
f!, =153'20'
AC==:;;;==::jD
Using Sine Law:
736.54 _ (4 - 5)
Sin 153'20' - Sin 17'07
Solution:·
CD Distance of CD:
Distance (4 - 5) = 483.02 m.
@
CD =--J (100)2 + (60)2
CD = 116.62 m.
Distance of line 5- 1:
. 736.54 _ J§.:1L
Sin 153'20' - Sin 9'33'
@
Distance (5 -1) = 272.88 m.
Bearing of CD:
50'
tan 8 = 100
8=30'58'
Bearing CD = S. 30'58' E.
@
Floor area of the builidng:
Atrapei4jd~lll*tabc~~astrefbH(lwiM
.~niqElld~p~fln~ho~~.be'9V1· • • A6~tBrey
cqncrele.~4I1diry9j$tq~~¢pn~tru!!Ie4Prl
• fhe
·.slladedporli()f1.a$ • • shO/,ni•• • \¥h~rei~ •••••• th~
.9(Jlller$tQne··f".@n•• be'p~~~(j.~y • ~JI$Unng
•. 45.nt.ftpl'l1Clllo~gCI) • t~en.3p·m . .·frol1l..C[).
[hebliildlng?lIMHKfallsaJonglhe
$UPdlVl$io?lIn~.m~l.mVtd~$.Wfflr~p~oldal
• tPt
.• into.tWb.• equal.afl!as..•• ·GKis·par~llel.toC[}.and
is5fu.#omit. .... . ....
Lf----_~
A
- 8 0 - - - - -..- - 0
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MISSING DATA
LM=~mbi+nb12
m+n
LM=
1(80)2+1(20)2
1+1
LM=58.3Om.
x= 80- 58.30
x=21.70
x 60
y= 100
_ 100 (21.70)
y60
y=36.17
.
Triangle FGI is similar to MCD
K
Area ofshaded section:
A=(45.32 + 29.15) (26.96)
2
A =1003.86 sq.m.
Total floor area ofbuilding
=6(1003.86)
= 6023.16 sq.m.
25 _36.17
a -21.70
a=15m.
fi2 = (25f + (15)2
b = 29.15 m.
Triangle COP is similar to MED
5 36.17
e= 21.70
e=3m.
f 36.17
42 =42.18
f= 36.87 m.
b = 100·36.17
b =63.83 m.
FH =63.83 - 36.07
FH =26.96 m.
26.96 ~ 36.17
d - 21.70
d = 16.17 m.
HK=45.32 m.
<J) • • qompule.the.• dl$lan~()tlihe • ~@ • •
\IDqoIl1JluleWe<lreaofWE!lotin~qres'<
®pQmputethel~ngthoflhetliVkJim;rHM
W'hich • • js•• p<lr<l"el•• IQ • lir~ • • t\~ • • ~llCh.ttl<lt • • it
cliVides.th~ • lot•• jnt°tvm~ls • havi,w.a.rali6
Of .t;2,.theblgger.lot.adjacent.lo.line.CD,
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S-142
MISSING DATA
Solution:
G)
Length of dividing line:
@
Distance BC;
c
--V
--V
xx-
mp 12 +nbl
m+n
2{4(0)2 + (1)(200f
2+1
x=346.41 m.
B
DE=BC
BC
200
Sin 56' =Sin 60'
BC = 191.46 m,
@ Area:
Area =(PI + ~) h
2
h =191.46 Sin 64'
h =172.08 m.
A= (200 + 4ooX172.08)
2
A= 51625 sq.m.
A =51625
4047
A= 12.76 acres
(j) ·Find.o~ijp®~ib@l¢fI9t1l.pt.~
@)' Andtlnepq~$lblEl)bearing.ofDS
.•••
.•
~•.' 8@.,:)notnEl(P()~$ibl~.bllatlrtg • QfD,E.
Solution:
CD .One possible length of EA:
Lines
AB
Bearing
Distance
Be
S30'W
S 5'04' E
CD
Due west
500
720
592
.. ---
DA
~
.... -
LAT
DEP
-433.01 '-250
- 717.19 +63.59
- 592.00
+1150.2 +778.41
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MISSING DATA
A
First Possible
Position
.B
.,.::..__--,~-i',-...,c
,,
,,
,
"
""
""
'\
"
\
"
',t'
-fl.'
'b\, ! ;'
\
:
, ' ,
/
,If
\~/ Second Possible
,E Position
,,
I
.
778.41
tan beanng DA = 1150.2
tan bearing DA = N34'05' E
778.41
.
Distance DA = Sin 34'05'
Distance DA = 1389.03
Considering triangle AED:
Sin" Sin 14'05'
1389.03 = 800
" =25'
a = 180 - 14'05 - 25'
a= 140'55'
AE
800
Sin 140'55 =Sin 14'05
AE =2072.72 m.
® One possible bearing of DE:
2072.72
800
~ = Sin 14'05'
f!,= 39'05'
Bearing DE =34 '05' +39'05'
Bearing. DE =N. 73'10' E.
® Second possible bearing of DE
= S. 5' E.
Comers 1 and 4 can be divided on ·tflElgr6und;
The engineer is to reset comers ~ alld3where
they were originally and determine the titie
bearings of all the courses. Dalenf survey
unknown. Upon runnhiS araJidom line, the
random line missed the !iue comer by 1.5 m.
The bearing from the end of the random line to
comer 4 was S62'30' E.
Compute the bearing of line. 4 - 1.
Compute the distance ofline 4 • 1.
@ What was the magnetic declination at the
time of the original survey?
(j)
@
5-144
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MISSING DATA
Solution:
SUBDMSION OF AREAS
a) To cut off an area from a given point.
let us say, that the entire area of the
lot is 10,000 sq.m. It is reqUired to divide
the lot into two equal parts such that the
division line shall pass thru comer one of
the lot concern. Bearings and distances of
all courses are known.
2
CD Bearing ofline 4 - 1:
Lines BearinQ Distance
lAT
DEP
1- 2 N 16'30' E 105.30 +100.96 +29.91
2-3 N 15'15' E 33.50
+32.32 +8.80
3-4 NT10'W 15.20
+15.08 -1.90
...........
....... 4-1
-148.36 - 36.82
36.82
tan bearing (4 - ~) = 148.36
Bearing (4 -1) =S 13'56' W
@
Distance of line 4· 1:
36.82
Distance (4 -1) =Sin 13'56
Distance (4 -1) =152.91 m.
@
Magnetic declination:
f12 =(1.5)2 +(152.91 'f
- 2(1.5)(152.91) Cos 103'34'
h = 153.27 m.
Sin 103'34' _ Sin '"
153.27 - 1.5
'" =0'33'
Since the random line is supposed to be the
true position of 1 - 4 based on true bearing,
then the magnetic declination during the
survey is 0'33' E.
Division line
Since it is difficult to approximate the
actual position of the subdivision line, it is
therefore advisable to solve for the bearing
and distance of line 3 to 1. Knowing the
bearing of the line 3-1 and 3-4, '" could be
computed. let us say A1 =2000 sq.m.
only, so we still have 3000 sq.m. more to
be added in order to obtain the required
area. A2 therefore would be equal to
5000 - 2000 =3000 sq.m. Knowing the
distance "a" and the angle "', we could
compute the distance "b" from the relation.
abSin '"
A2 =-2-
A, + A2 =5000 sq.m.
(the required area to be cut off!
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5-145
SDlomSION
)
/
b) To cut off an area by a line whose
direction is given.
This requires a longer computation in
the sense that it would be difficult to
assume the position of the subdivision
line. Let us sayan area of 12,000 sq.m. is
to be segregated from the whole area, the
direction of hte dividing line known (see
sketch below). The total area of the lot is
40,000 sq.m. and the area to be segregated
is adjacent to the line 2-3.
2
3
~
:: h1 - d cot '" - d cot a
~
:: h1 - d (cot", + cot a)
Given values:
A1, h1, '" and a
_(h1 +~)d
A1-
2
r
A1 =[h 1 +h1 - d (cot", + cot a)]
6
5
Thru the comer that seems likely to be
the nearest line cUlling the required area, a
trial. line AD is drawn whose direction is
given. Distances AD and A2 could be
computed by using trigonometric
principles. Considering lines A2, 2-3, 3-4
and 4-A to be closed polygon, its area
could be determined by using D.M.D.
method. Let us say the area of lines A 234
A is 20,000 sq.m., so there is an excess
area A1 :: 20,000 - 12,000 :: 8,000 sq.m.
The values of '" and a could be
determined cause the bearings of lines A2,
3-4 and 4-A is known, length of line 4-A is
known.
We could solve for the value of "d"
d=ABSin '"
d
4c=-Sin a
d
AB=-.Sin '"
2B=2A-AB
3C=3-4-4C
Points Band C is now the tru\! position of
the dividing line.
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5-146
SUBDIVISION
@
Bearing of dividing line:
c.
•
~lfh~.~ra~l~a~lj~Q~.~j $esr~~iea •
(j).·..·.•m~~I~iA~e~t~~aio~th~h~@1¢~···
fu~~~~~q.>
®....••~tj• t~0 ~~~jS ~~j ~lM'dl:~IT· • li~~.·
•
• Gf•• •
@
·PomI'MElth~t~f\9tfjPfthEl~jViqifflJlil'l~W··············
(1)
Distance CD:
541.71 714.68
Sin e = Sin 60'
e =41'02'
Bearing of diving line BD = S. 78'58' E
Solution:
c
,4.let~b@lIge~b~3~ttfMsht$m~sri~rn~y,
Aa,'N;4S·~,169m,IQng,~q~mt9~M9Qm.
il$!;lilli.
APEisIObe2l~oMhelotal~rfta9flh~~r
Thet6titf!3r~of@l/(Jfi$.11,~~,&~nW.··· ...
$·~~¥ml~~lhe~i~~rGElffqmg@A.Y
@q9ImW~lt\~beactr~pO!O~AP'
~¢Ol'YlPllt&thedjs1anceOE.
A
A=
Solution:
CD Distance OA:
81g x Sin.60'
190000 180 ~n 60'
. . ..
.
x
x= 541.71 m.
® Length of dividing line:
(BO)2 =(541.71)2 +(81of
- 2 (541.71)(810) Cos 60'
BO = 741.68 m.
c
• •.•·•
..
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S-147
Pdfbooksforum.com
SUBDIVISION
100 (DA) Sin e _2 (160)(190) Sin e
2
-5
2
DA= 121.60
b 2_b}
A = ~---_._-2 (cot e + cot (3)
_ (3aW· (200)2
A - 2 (cot 70' + cot 58)
A == 25,282.16 sq.m.
® Bearing of/ine AD:
A=
160 (190) Sin e
2
® Length of dividing line:
A =11,643.88
e = SO'
x=
Bearing ofAD = S. 85' E.
Distance DE:
(DE)2 = (10W +(121.6)2 - 2(100)(121.6) Cos SO'
DE= 95.68 m.
.L~~CJQi
x=
1 +2
x= 254.95m.
@
cD•• Fihdthe area of the lot in m2 .
i])FincUhe length ofthedivldlng line (EF)
that is parallel to line OA which will divide
thelotinto two equal areas,
..
@i Oetermine the location of arieendof the
dividingllne Efrom comer A along line AB.
Solution:
CD Area of/of:
+0d.
mb 12
m+n
® Distance AE:
a'" 300 - 254.95
a=45.05
~~_ 45.05_
Sin 58' - Sin 52'
AE==48A8 m.
A lot is bounded by 3 straight sides A, 8, C.
AS is N. 45' E. 95 m.long and AC is due East,
88 m. long. From point D, 43 m. from A on
side AB, a dividing line runs to E whiclt is on
side CA. The area ADE is to be 1/7 of the
total area of the lot.
(j) Determine the distance DE.
i]) Determine the bearing of side BC.
@ Determine the distance AE.
Solution:
(j)
Distance DE:
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S-148
SUBDIVISION
Solution:
(AE)(43) Sin 45' = .!(95)(88) Sin 45'
272
AE = 27.77 m.
cD Area of AFE:
-7-_...."
A _--
Using Cosine law:
(DE)2 = (43)2+(27.77)2 - 2(43)(27.77) Cos45'
DE= 30,52m.
® Bearing of line BC:
(BC)2 = (95)2 + (88)2 - 2(95)(88) Cos 45'
Be = 70.33 m.
G
K
1
Using Sine Law:
AC =~
~ = 70.33
AC=42.43m.
Sin ()
1
Area of AGH= '2 (60)(60) = A1
Sin 45'
e = 72'46'
. A1
a':: 90' - 72' 46'
-
Bearing of BC =S, 17'14' E,
2
A2 = 1200trf
®
Distance AE:
AE= 27,77 m.
=1800 m2
Area of AEF - 60(60) - 1200
a= 17'14'
@
'J'2 (60)
Width of the road:
&_(AC)2
A2 - (AB)2
1800 _ (42.43? .
1200 - (AB)2
AB=34.64
BC =42.43 • 34.64
BC= 7.79m.
BO=2(7.79)
BO = 15.58 m. (width of road)
A_-_----.;.....:.:.;.;,:..."
.The centerline of apfoposed rOadbaving' a.
bearing eIN. 45' E.· passesttm.lug'h the
diagonal of a square lot AHIO having sIdes of'
60 rnx 60 m. !flhe area oCcupied by the
propoSed road Is equal to 1200 sq.l'li·
.
CD Compute the area of section AFE, where
Eft IS parallel 10 the dragonal of Ihe square
lot.
.
..
® . Compute the widthaf the road. .
@ Qompute Ihe total. perimeter of the
proposed road inside the lot.
k:..:1.~,,:,L--------II
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S-149
Pdfbooksforum.com
SUBDIVISION
@
Total perimeter:
E,c= GH- 7.79(2)
x (960.22) Sin 40' = 210000
2
x= 680.47
EF =60 -{2 -15.58
EF=69.27m.
® Length of dividing line:
(y)2 = (680.47? '+ (960.22)2
- 2(680.47)(960.22) Cos 40'
y= 619.67m.
p= 69.27 + 11.02 + 11.02 + 69.27
+ 11.02 + 11.02
p= 182.62m.
® Bearing of dividing line:
680.47 _ 619.67
Sin 8 - Sin 40'
8=44'54'
Bearing = S. 84'54' W.
Azimuth = 84'54'
'
.
G?tl1ElIO(i$t6@divideasuchlhaftM
M¢~.ottH~$()uthem®~lohW:Rul~be
•. • •· • • •·41q,QPQ.lll{.•·•• GQl'tlPlJtflJf1~P9~~i<l~.9fth~·
i')tb~r~tl~()N~l'ldIJtldit\gliri~jfth~Jine
·iltaHs~ICgm¢rapft@lol·t1<~r&$$the
·•~• • ~:8¥tl.~medf1he • div~,og •
·q}J<Pomp~~.tbl'l.~i1numQfthedividing.iine·
Solution:
CD Location ofx from corner 1:
2
'O\lM®rth
ljrie? • • .• •
ill
QOmpytettle•• ril@;ing•• dlstance•• @C•• is•• the
ar~.Qf.tl1¢IQtls·43560$gm.
@ CQIllPEJtMMtli$@'l~qfCD.
®
;ig:..::1'l~w
@rnptJteihebeiilirigCO.
Solution:
CD Distance Be:
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S-150
SUBDIVISION
200
=300
lil = 33.69'
Ll = 45' . 33.69'
Ll = 11.31'
tan
lil
.S4bdi\lide•• the.r()t•• ha'liri~ • • thEl9ivllrttecMiC<l1
de~MPtjon>l/)t() . • MO·.~qW!I • • llrt1JM.·.'oY • ~•. ·lir~·
paralleltQthe sideA6'
.... '.' ....''.' .
Sin 3369'
. =200
x
x= 360.56 m.
_200(300}
A1-
2
A1 = 30,000
A2 = 43560 • 30000
A2 =13,560 m2
.(1)
A2 = x (BC) Sin f3
CO/llPu~the.area9fthe~hQleWtirl~C~$?
~ • C()l'llPtJtTtn~.leJ'l9!hpftbe ~lyi~#rig~h~ . • • • /.·.
2
13560 = 360.56 (Be) Sin 11.31'
2
BC = 383,53 m.
®CoiJJPl.Jtelhem~jng$ideBCL«>
Solution:
@
Distance of CD:
(CD)2 =(360.56)2 + (383.53)2
- 2(360.56}(383.53) Cos 11.31'
CD= 76.80m
@
Beating of CD:
383.53. _ 76.80
Sina -Sin 11.31'
(j)
Area of whole lot:
u = 78'21'
Bearing BD =45' + 11'19'
Bearing BD =56'19'
Bearing BD =S.56'19' W.
Bearing DB = N. 56'19' E.
A~
~
-.
--",B
A=
b-}.b 2
1
2 (cot e + cot ~)
_ (200)2 - (100)2
A - 2 (cot 62' + cot 70')
A = 16747.06 m2
c
Bearing CD =N. 45'20' W.
4047
A=4.14 acres
.
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5-151
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SUBDIVISION
® Length of dividing line:
Solution:
CD Location of the dividing line from comer 2 if
the dividing line starts from comer 1:
10·
_ (1000)2 Sin 70' Sin 80'
A2 Sin 30'
A= 925416.58 m2
--V nb mb-}
--V 1(200f1 11(100f
x-
x-
2
1 +
m+n
2
Al ='3 (925416.58)
+
+
Al = 616944.39 m2
- 1000 (y) Sin 80'
x= 158.11 m.
@
A1 -
SideBC:
BC
100
Sin 62' = Sin 48'
BC = 118.81 m.
2
\
616944.39 = Sin 80'J1000) y
y= 1252.92m.
® Length of the dividing line:
.;. = (1000)2 + (1252.92)2
·2 (1000)(1252.92) Cos 80'
x =1461.05 m.
@
Bearing of the dividing line from comer 1:
-L=_x_
Sin II Sin 80'
1252.92 1461.05
Sfn II = Sin 80'
\D qomputetH~16catj§n
• • ()ttM•• ~iVidWIUn~
·fr()lTIcorner2.jfthe~ividingline~tarls.fr0f11
comer 1.
@
@
>
ColtlPute.th~.leryslh()fthedjyjdlpgljne ••••••••••••
CQl'l'lPtlt~.·We • b~rlng • Of•• the•• diViding•. nn¢
frorncotrlei't.
..
.
II
=57'37
Bearing of dividing line: (ll + 30')
=N 87'37' E from corner 1
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5-152
SUBDIVISION
Using Sine Law:
4-1
150
Sin 64' = Sin 48'
4 -1 = 181.42 m.
® Area of the Jot:
2
A=
b-/-b1
2 (cot e + cot 11)
DISfllNOl;)
_
· • •.·OOOrlk»···
(300f - (15Of
A - 2 (Cot 64' + Cot 68')
A = 37846.56 m2
A =37846.56
4047
A = 9,35 acres
@
Length of dividing line:
Solution:
CD Side 4 -1:
_..y nb
x-
2
1 + mt>i
m+n
--V
3(300}2 + 2(150)2
x3+2
x= 251 m.
Anequilater~tl:1'iangufarlrack Of land has a
length of one '(jfits sides equal to 3600 m.
Iori9' ·111$ requited to dIVide the lot into 3 equal
.sharesoy aline parallel to one of Ihe 3sides.
CD COmpulethe whole area of Ihe lot jfl-£lCf8S.
® Compute the iength {If lhedivlding line on
the portion near the vertex.
.
@ Compute the length of the dividing line on
the se<:ond lot.
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SUBDIVISION
Solution:
G)
Solution:
Area of whole Jof:
G)
Distance of dividing line from comer B:
N
B
3600
A :;: (3600)2 Sin 60'
2
A:;: 5611844.62
4047
A:;: 1386.67
G)
c
Length of DE:
A :;: ~ (5611844.62)
000 (xl Sin SO' :;: 200000
2
x:;: 530.18 m.
A:;: 1870614.87
2
1840614.87 :;: x s~n 60'
® Length of dividing line:
(AD)2 :;: (900j2 +(580.18)2
·2(900)(580.18) Cos SO'
x:;: 2078.46 m.
® Length of dividing line y:
AD :;: 672. 70 m.
A :;: ~ (5611844.62)
® Bearing of dividing line:
~ (5611844.62) :;: t s~n 60'
N
y:;: 2939.39 m.
An area of 200()OO rnZ ls to be $egregatedfr~m
the northern portion of triangUlar 101 ABC. fr~m
camerA bearing and distancieot AS is
N. 50' E., 900 m., Be is due South and CA Is
N.42'E.
.
ill Compute Ihe distance of the dividing line
from comer 8 along~ne BC.
..
Compute the length of the dMdingljne~
® Compute the bearing of the dividing line
from comer A.
@
¢-+----~D
580.18 672.70
Sin e :;: Sin SO'
e:;: 41'21'
SO' + 41'21':;: 91'21'
Bearing:;: S, 88'39' E.
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S-154
SUBDIVISION
;.. <:>:::,:":":::::::::>::::::
UNks{ ··AZ1MUlH ..• . D,S1'ANCE
. . 1 · 2 ' 187'00'27.90 m.
W __
2·~
••.•. .• .. 268'4T<34;12m,
.·>.358'33'
21.72m.
. 4; 1
aa~57'
. 38:.21m.
~j...----4
.... 3~4
G>ComPlltetl1~ .Ienglh: Oflhe dividing'line;
is
® HoW far the lntefSe<;ti6opolntof the·
. diViding line on the northempart of the lot
from
2?
. .
® How far is the inlersectidnpoint of the
dividing line on the southern part of the lot
. from comer 1 of the boundary.
comer
800 =(27.90 +0.003 h) h + 27.90 h
800 =55.80 h +0.003 rf
rf + 18600'h - 266666.67 =0
h = -18600 + 18628.65
2
h= 14.33 m.
b = 27.90 + 0.003 (14.33)
b = 27.94 m. (length of diViding line)
Solution:
CD Length of dividing line:
A
"1
=bh + 27.90h
2
2
400 = bh + 27.90 h
2 . 2
800 = bh + 27.90 h
tan 81'4T =!.!
y
y= 0.1444 h
tan 8'03' =~
h
x=0.1414h
b=27.90-x+y
b = 27.90·0.1414 h + 0.1444 h
b := 27.90 + 0.003 h
@
Distance of Afrom comer 2:
Sin 81'47' = ~
A-2
14.33
A- 2 = Sin 81'47'
A- 2 = 14.48 m.
® Distance of B from comer 1:
Cos 8'03' = _h_
B-1
14.33
B-1 = Cos 8'03'
B-1 = 14.47 m.
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5-155
SUBDIVISION
@
A parcel of land has a technical description as
shown in the tabulated data.
5.01'27£,··
S. BB+S7'W.
27.72m.
38.21m·····
3·4
N.OrOO'E.
27.90m..
4-1
A, = 236.45 m2
Bf:ARING . . DISTANCE.
LINES ..
1-2
2-3 .
. N.88'4TE.
Total ama =A1 + A2
Total ama = 236.45 + 236,07
Total ama =472.52 m2
34;12m.
more
The area of the lotis
or lessHlOO sq.m;
If !he lot Is to be subdivided into two parts
such thaUhedivldltig li/remuSlslM atthemld
point of line 4- 1 and must be parallel to line to
1·2 oftheboundary. ...
.
Ama on the eastem part:
- 32.61 (27.67) Sin 31'33'
A2 2
A2 = 236.07 m2
- 17.06 (27,72) Sin 90'14'
A,2
@
Distance of other end of dividing line from
comer 2:
Using Sine Law:
a
_ 32.61
Sin 31'33' - Sin 90'24'
a= 17,06m,
CD What is the diS1ance of the subdividing
line? .
.
....
@ What is the area of the lot subdivided on
the eastern part?
.
® What is the dIstance of the other end of the
dividing line from comer 2 of the lot? .
Solution:
CD Distance of dividing line:
Using Cosine Law:
=(17.06)2 + (27.72)2 • 2(17.06)(27.72)
Cos 90'14'
y= 32.61 m.
y2
Using Sine Law:
17.06
32.61
Sin e - Sin 90'14'
e = 31'33'
(J. =89'36' • 31'33'
a =58'03'
fJ = 180' - 89'36'
fJ =90'24'
a = 180' - 90'24' - 58'03'
ct =31'33'
Using Sine Law:
3261
AB
Sin 90'24' =Sin 58'03'
AB =27.67 m.
89'36'
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5-156
SUBDIVISION
---------------------_....
.li.~ill~ii~8~~4,e~
~,;;~
Length of dividing line
Using Cosine Law:
(abj2 =(32.61}2 + (26.28)2
- 2(32.61 }(26.28) Cos 58'03'23"
ab= 29,11 m.
W<¢rw@~~et~~I~gmpttl}¢~q~l~~IM@~J
[~~.II'J
® Bearing of dividing line from mid· point of
line 2 - 3:
3
Solution:
CD Length of dividing line:
89'36'
4
89'36'
4
32.61 _ 29.11
Sin a - Sin 58'03'23"
a= 71'55'
- 17.06 (27.72) Sin 90'14'
A1-
2
AZimuth of ba:: 268'57'
Az :: 600 - 236.45
A2 :: 363.55 m2
AZimuth of ba:: 19T02'
A1 :: 236.45 m2
Using Cosine Law:
(4· a}2 :: (17.06)2 + (27.72)2
• 2(17.06)(27.72) Cos 90'14'
, 4 - a :: 32.61 m.
Using Sine Law:
17.06 _ 32.61
Sin e - Sin 90'14'
~
Bearing ab :: S 7'02' W
@
Bearing and distance from T - 1 to comer
"b":
LINES
1
DISTANCE
27,89
N88'47' E
34.12
S 1'27' E
27.72
S88'57'W
38.22
2
e == 31'32'37"
2
- 32.61 (x) Sin 58'03'23"
A2 2
3
363.55 :: x (32.61) s~n 58'03'23"
4
x== 26.28 m.
BEARING
NTOO'E
3
4
1
,
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S-157
SUBDIVISION
LINES
1
NorthinCls
21412.63
2
21440.32
Eastinas
17424.86
:!:-.MQ
19428.26
±...Jill
.±....M..1.1
3
21441.04
19462.37
.:t....2L2a
2
=-.2Z.Z1.
±-.Q.lQ
1
21413.33
.:.-QJQ
21412.63
19463.07
~
19424.86
LINES
T -1
BEARING
N66'27E
DISTANCE
18.60
LINES
T -1
NorthinCls
21433.61
3
21441.04
EastinQS
19445.32
~
19462.37
3
4
4
3
LIM
itilA.Vi
SU~~IVii~6~I~ilnj~~(2)~U~l~~rts;
pmvtel~d!.flj~t• ~.$qpdiYjding.·!l9~·.rnu$t.Staff·<lt
th~cMWlj/'l~9n!@~M()1Mu®~rYljn~;
Coordinates ofb:
21413.33
4
~
21412.85
b
19463.07
~
19436.79
Solution:
CD Distance of the subdividing line:
T -1
b
21433.65
21412.85
- 20.76
19445.32
~
- 8.53
.
8.53
tan beanng = 20.76
Bearing (T - 1to b) = S. 22'20' W.
Distance =Sin82;~20'
=22.45 m.
Bearing and distance from T· 1 to dividing
point on the southern portion is
S. 22'20'
22.45 m.
w..
e = 88'4.7 + 1"27'
e =90'14'
f3 =180' - 88'47' + T
f3 =98'13'
.....
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8-158
SUBDIVISION
Distance of subdividing line AB:
Cor. 3
BEARING
S. 01'27' E.
DISTANCE
13.86
Northinas
42935.27
Eastinas
34584.29
~
34584.64
A
Cor. 3
~
A
42921.41
Bearing and distance from 2 to A:
- 34.12 (13.86) Sin 90'14'
A12
Cor. 2
A
A, = 236.45 m2 .
A2 = 500·236.45
A2 =263.55 m2
BBM#1
BEARING
S.37'33'W.
34550.18
~
+ 34.46
.
34.46
tan beanng= 13.14
DISTANCE
237.32
Bearing =S. 69'08' E.
.t
DIS ance
34.46
= Sin 69'08'
Distance =36.88 m.
1
1
42934.55
42921.41
• 13.14
N. 07'00' E.
27.89
N. 88'4r E.
34.12
S. 01'27' E.
27.72
S.88'57'W.
38.22
Northinas
43095.02
Eastinas
34691.42
- 144.64
34546.78
2
2
3
3
4
4
1
BBM#1
1
34550.18
IJ =69'08' +7'
IJ =76'08'
- (x)(36.88) Sin 76'Q8'
Ar
2
.:t......D..ZZ
.t...M..11
263.55 = x (36.88);in 76'08'
42935.27
34584.29
~
2
42906.87
:t....2L.§.8.
42934.55
3
1
2
3
4
4
:JJ..J..Q
.t.-QlQ
42907.57
34584.99
- 38.21
34546.78
=--.Q1Q
1
LMQ
42906.87
x= 14.72 m.
Using Cosine Law:
(AB}2 = (14. 72f + (36.88)2
- 2(14.72)(36.88) Cos 76'08'
AB= 36.28 m.
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5-159
SUBDIVISION
@
Solution:
CD Distance ofsubdividrrg line:
Bearing of subdividing line:
Using Sine Law:
36.28 _36.88
Sin 76'08' - Sin a
3 S1 '2TE
N88'47' E -y..l'l
2
a =80'43'
S07'OO~
or
Bearing of subdividing line
=80'43' +7'
=N.8T43'E
e -;;,
.'"
--.._------@ . ""
36.88------;;,
<X
B
N 88'47' E
~
4
I
® Distance to be laid out from comer 2 of the
boundary to subdividing line:
= 14.72m.
e = 88'47' + 1'27'
e= 90'14'
BBM#20
BEARING
S. 37'33' W.
DISTANCE
237.32
N, 07'00' E.
27.89
N. 88'47' E.
34.12
S. 01'27' E.
27.72
S,88'57' W.
38.22
Northinos
43095,02
~
42906.87
42906.87
Eastinqs
34691.42
1
•
···l~~~~I.o~~.I#I~~i.~~~P\iM m.lqt
1
2
2
···a~IN~·.··.·.·
3
3
'N:Q7'QQ\~,>
4
·······N:ea~4nlS'··
4
"'S,Q1T2ne:
1
.1.· .
j.:
BBM#20
1
1
2
2
3
3
4
4
1
.:!:-.2ZM
42934.55
:t......Q.ll
42935.27
~
34546.78
34546.78
±.-..MQ
34550.18
:!:...M..11
34584.29
:....11J!1
2:..-JUQ
42907.57
34584.99
=-...QlQ
::..-Ja21
42906.87
34546.78
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5-160
SUBDIVISION
Cor. 3
BEARING
S. 01'27' E.
DISTANCE
13.86
Northings
42935.27
EastinQs
® A1aa oflot to be subdivided on the northem
part.
_(34.12)(13.86) Sin 90'14'
A1·2
A
Cor. 3
~
A
42921.41
A1 =2:36.45~
=(36.88)(36.18) Sin 22'05'
2
~
34584.29
~=250.82m2
.~
34584.64
Total area = 236.45 + 250.82
Total area = 487. 27 rrI
Bearing and distance from 2to A
@
Cor. 2
42934.55
A
~
-
.
13.14
34550.18
34584. 64
+ 34.46
Coordinates of the subdiViding point on the
Eastern part:
Coordinates ofA = 42921.41 N" 34584,64 E.
34.46
tan beanng =13.14
Bearing = S. 69'08' E.
·t
34.46
DI§
ance =Sin 69'08
Distance = 36.88 m.
~a.!·n~.: I·.:'~.ti.•·f~•..~ t.\.:.'.I.;.~.ilil
•.
.. .• .
I.• ,.l.•·.:.·.&.·.•.•.••.
p•..!.i:
r.•.. o o . ·
.l}'iA,iI.
fJ =69'OS' + 7'
fJ = 76'08'
::(>:::::::::::::::::
.
1~I!JtBII'I11
:@•• • P~rmiil~:~l'lEl • at~~ • ()f.t!J~ .• @ilpiry9••n~~t.w.
m~$i~l:!qP@lt!$~lijg.¢iWPff~YtM
@0(jinglil'l@<·········································
.
LINES
BEARINGS
DISTANCES
AS
Due north
N61'E
Due south
Due west
20.00 m.
114.30m.
75.30 m.
98.00 m.
Be
Q)
DA
AB
36.88
Sin 76'08' =Sin 81'47'
AB= 36.18m.
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S-161
SUBDmSIOI
(20 + 55.09) (98 • Yi _(55.09 + 75.3) Y
2
2
Solution:
CD Length of this dividing line:
130.39y = 75.09 (98) • 75.09y
205.48y =7358.82
y= 35.81 m.
h=58·y
h= 58·35.81
h= 22.19
Area ofbuilding cut offnear the line AB
= 10(22.19)
A, =A2
n=m
n=1
m=1
= 221.90m2
_.... /nbl + mb12
x- 'I
® Area ofbUilding cut offnear CD:
A = 10 (33·22.19)
m+n
_
1(75.3)2 + 1(20)2
1+ 1
\ X=55.09m.
x-
@
A = 10B.10m2
Area ofbuilding cut offnear the line AB
98-]--1----,
·piVenbe,()wl$.fb~tecnni()aLde$¢@fi~nofa
m•
IBh·.~aVi~g • •~l'l· • • an~a • • ·6~Q·~?$9m< • J li$
f~q4ir~tQ~p~qiVipethl~1?tiht()lVm~qH~
1
are~s.siJcbthat·.th~Y·?"ill~~veElql!~lf'B~I~gEl
20
l. -4-0--+\-3....
3 ---f.........-4~
~1(l1l~lMUr~q,P'@lph~jQ~$~~~~{.<
!-I
IJNES·· .. l:lEARINGi .
• • N7~·2~'g,·········
·$39'$1'1:'
·S4$T4e'W/
I
I
I
• • • ·N3~r52W
~;'j:l;]ll~ I}o
I '
I
I
I
I
~-hl
I
58-------001
- (20 + x)(98 - y)
A,-.
2
Ar
- (x + 75.3) Y
A1 =A2
2
N1&~50'W····
(j)
Compute the distance of the ofher end(jf
the dividing line from COtner B..
® Compute the distance of the <lMdll1g line.
® Compute the bearing of the dividing Hne.
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5-162
SUBDMSION
Solution:
<D Distance of the end of dividing line from B:
® Distance of dividing line:
Using Cosine Law:
(FG)2 = (25.36~ + (22.63)2
·2(25.36)(22.63) Cos 53'2T
FG= 21.72m.
@
Bearing of diViding line:
Using Sine Law:
JJJ"V·fN7J·V'X)
E
E
- 19.625 (9.21) Sin 00'43'
A12
A1 =89.75 m2
A2 = ~.56 • 89.75
22.63
21.72
Sin a =Sin 53'2T
A2 = 230.53 m3
a= 56'49'
(BG)2 = (9.625}2 + (9.21}2
BG =22.63 m.
Azimuth of FG = 253'23' + 56'49'
Azimuth of FG = 310'12'
Using Sine Law:
Bearing of FG = S. 49'48' E.
·2(19.625)(9.21) Cos 00'43'
19.625
22.63
Sin e =Sin 96'43'
8 =59'2T
/!, = 112'54'·59'27
/!, =53'2T
A - {FB) (22.63) Sin 53'2T
2-
2
230.53 = (FB) (22.63J Sin 53'2T
FB= 25,36m.
<D Compute the area at thetdt. ", ',.
In, the same lot; a dividing line is drawn
from comer 5 to the midpOint of line 2 • 3.
Rnd lheazitnuthof the divk:iing line.
@ Find the distance of the diViding line.
@
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5-163
SUBDIVISION
Solution:·
CD Area oflot:
® Azimuth of dividing line:
For/ot 1
Comer LAT
DEP
Northines
19955.95
- 43.20
19912.75
2-3
-43.20
+23.23
1
Eastinas
20081.7
+23.2
20104.9
49.045
Line
1-1
1-2
2-3
3-4
4-5
5-1
Bearine
S52'14 E
S4T49'W
N28'16'W
N79'21'E
S63'26' E
S6'44'W
LAT
Distance
DEP
198.66 -121.67 +157.04
-20.47 - 22.59
30.48
111.37
+98.09 - 52.73
+5.38 +28.59
29.09
-26.69 +53.38
56.68
- 56.31 -6.65
56.70
Comers LAT
DEP
COORDINATES
BLLM 1 121.67 +157~ 20000.00 20000.00
-121.67 +157.04
1
19878.33 20157.04
- 20.47 - 22.59 -20.47
- 22.59
19857.86 20134.45
2
+98.09 - 52.73 +98.09
- 52.73
19955.95 20081.72
3
+5.38 +23.59
+5.38
+28.59
19961.33 20110.31
4
-26.69 +53.38 - 26.69 +53.38
19934.64 20163.69
5
+56.31 -6.65
·56.31
·6.65
19878.33 20157.04
1
LINE
1- 2
2-3
3-4
4-5
5-1
LAT
DMD
BLLM 1
1
2
3
4
1
Comer
LAT
DMO
1-2
2-3
3-4
4-1
+43.20
+5.38
- 26.69
-21.89
-23.23
-17.87
+64.10
+58.74
CORNERS
1-2 .
2-3
3-4
4-1
@
BEARING
N28'16'W
N79'21' E
S63'26' E
S69'34'W
Distance of dividing line:
D· t
58.74
69'34'
Distance =62.69 m.
IS ance = Sin
20000.00
20104.95
20081.72
20110.31
20163.69
20104.95
DEP
DOUBLE
AREA
- 23.23 -1003.54
-28.59
- 96.14
+53.38 -1710.83
- 58.74 -1285.82
2A =4096.33
A=2048.17m2
Line 4 - 1 (Dividing line)
58.74
.
tan beanng= 21.89
bearing = S69'34' W
azimuth = 69'34'
DEP
DOUBLE
AREA
-20.47 - 22.59 - 22.59 +462.42
+98.09 - 97.91 - 52.73 - 9603.99
+5.38 -122.05 +28.59 - 656.63
- 26.69 - 40.08 +53.38 +1069.74
- 56.31 +6.65
-6.65
- 374.46
2A -9102.92
A =4551.46 m2
20000.00
19912.75
19955.95
19961.33
19934.64
19912.75
DISTANCE
49.05 m.
29.09 m.
59.68 m.
62.69 m.
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S-164
SUBDIVISION
Lines
LAT
DEP
AB
BC
+57.81
-9.63
- 42.79
-18.75
+13.36
+16.03
+72.04
+13.36
- 53.25
-48.18
m
DE
EA
DOUBLE
AREA
+926.09
+16.03
-104.10 -1002.48
-189.50 - 8108.71
+149.61
·2805.19
+643.68
+48.18
2A = 10346.01
A =5173.005 m2
DMD
A _ 5173.005
c.~.
JBC7
2
6O:00fu.N1S'$'E .
A = 2586.50 sq.m.
>··.·.···.. ·72~tl9@<·.····
••·.<.·Sa2~23r·E • • • · · ·
..... 44.$3m: ... ·······S1TW·e.··
@
. nJ;
·•• • S(tOOfu,>.·.··.··· • • • • • • N.V4tatrW·•• ·• •·
Distance of dividing line:
Considering triangle ABF:
D
A
mfJn911'l~C1l~9f~chl<lh
. ~..•..Fln~tIl~ •. ~j~I~~pfm~d!@llnQ.ljn~,·······
@. ·Fi~~@!ltl~~ijg.tJfthEl.djyiding.line.·
Solution:
G)
Area ofeach lot:
A - 60(25)
12
A, = 750 sq.m.
A2 =2586.50 - 750
A2 =1836.50 sq.m.
60
tan 0=25
0=67'23'
LINES
AS
BC
m
DE
EA
BEARING
N 15'30' E
S82'23' E
S 1T20'E
S70'36'W
N74'30'W
DISTANCES
60.00
72.69
44.83
56.45
50.00
AF= FB cos 0
25 .
FB = cos 67'23'
FB = 65.01 m.
Considering triangle BFG:
Bearing of FB: NTOT W
- 65.01 (BG) Sin 75'16'
A2 2
__.1l!?36.50)
BG - 65.01 Sin 75'16'
BG =58.42 m.
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5-165
SUBDIVISION
Using Cosine Law:
(FG't =(65.01}2 + (58.42)2
- 2(65.01)(58.42) Cos 75'16'
FG = 75,55 m. (distance of dividing line)
® Bearing of dividing line:
Using Sine Law:
Sin a. Sin 75'16'
BG= FG
.S'
- 58.42 Sin 75'16'
In a. 75.55
ill Compute the Iengthofthe centerline of the
road that traverses along the lot.
.
How far is the other end of the center Une
"" :::~irJgtheproperty boundary from
@
\Wlf the property is located at·Barangay Puflla.
Princesa;whete .cost of land is P2000:QO
per square Meter. eSUmate thecostoHhe '
property to be expropriated for the service
road,
.
Soltition:
a. =48'24'
CD Length of the center line:
Bearing ofdividing line FG = N 41"17' E
A proposed 10 m.~ervice road crosses the
property. Of JFN Holdings whose technical
descriptions are as follows.
lll'ilESA21MUTH
.··QISTANCI;
••••••••••. •· • • • •··zn:r
•>
Using Cosine Law:
>? = (9.55)2 + (14.02}2
- 2(9.55)(14.02) Cos 71'40'
x= 14.27 m.
The centerline
of the' proposed service roM
Using Sine Law:
14.02
14.27
Sin", =Sin 71'40'
0=68'5"
f3 = 180' - 71'40'·68'51'
f!,=39'29'
AB
14.27
Sin 50'31' = Sin 88'06'
AB::: 11.02m.
cro~es at 9.55 m. from corner 4 along the line
3·4 and runs in adirection afN3'45' E.
1
@
S7J'Ji'W
Distance of other end of center line of
proposed road from comer 1;
AB
B-1
Sin 50'31' = Sin 41'23'
B-1 =11.02 Sin 41'23'
Sin 50'31'
B-1=9.44m.
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S-166
SUBDIVISIOI
@
Cost ofproperly to be expropriated:
tan 88'00' =~
X2
X2
=0.17
X1
5
= tan 88'06'
= 0.17
X1
tan 20'14' =~
X4 = 1.84 m.
Xs = 1.84 m.
CD =AB- x2 + X4
CD = 11.02 - 0.17 + 1.84
CD=12.69
EF = 11.02 -1.84 +0.17
EF=9.35
A =A 1 +A2
A1 = (12.69 +211.02) 5 = 59.275 m2
Line!
1- 2
2-3
3·4
4-5
5-1
Bearinas Distance
N61'57' E 74.18
........
--_ ..
S9'03'W 54.13
N68'21'W 55.43
N13'56'W 58.85
LAT
+34.88
.......
DEP
+65.47
-
·53.46 -8.51
+20.45 - 51.52
+57.12 -14.17
+112.45 +65.47
::.lli2 :.l!2.Q
A = (11.02 ; 9.35) 5 = 50.925
+58.99
2
A = 59.275 + 50925
A = 110.20 m2
........
- 8.73
CD Length of (he boundary of the proposed
road on the sou/hem portion of the lot:
Fortine 2- 3:
Total Cost =110.2 (2000)
Total Cost = P220400,OO
tangent bearing =~
.
8.73
tangent beanng =58.99
bearing = S 8'25' E
h,!•• Vi~W • •9t•• lh~~@ffi • ·%.tll~9PY~mm@lJp
relleyepet~~riial~ht¢ulararyqp~e~W~I1·
·.ttiif1ipcRmi~~ti(1~pf:a.city,a.dead.endrqligl~·
.to•• b¢•• eXf~nded • arid·.coQ~tfUeted • .\Vl.th.~rroM
r1ghtqfw~YClf:2qgL • whi9h.iOter~s • ~.eYe~.1
Pl'jvat~IY • ()Wf\~dprpP~rtlM ..•• Qoe.• of·.ltieSe··lot
D'
t
dep
IS ance = Sin bearing
.
8.73
Distance = Sin 8'25'
Distance = 59.64 m.
ha$tMfQlIOWI09f1e.I~Mttl~k
.
.:-:.:-:-:-:'.-:'"-:.:.', .• '.-.-.-.:-:.'.:.»:.'.",'.','.'.' .....:.:.:.:-,.:-:",-- ....
LINES··
SEARING
/. N6flm.s
.·····blStANCS···
lA
Complete Field Notes
LINES
1- 2
2-3
3-4
4-5
5-1
BEARING
N61'57' E
SS'25' E
S9'03'W
N68'2fW
N13'56'W
DISTANCE
74.18 m.
59.64 m.
54.13 m.
55.43 m.
58.85 m.
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S-167
SUBDIVISION
X
68.50
Sin 0'45' = Sin 104'07'
X=0.92m.
y
68.50
Sin 75'08' =Sin 104'Or
Y=68.27
h1 =0.92 Sin 75'53'
h1 =0.89m.
~ =10- 0.89
~=9.11 m.
2
tan 52'54' =!!.l
a
0.89
a=tan 52'54'
a=0.67m.
4
Bearinqs Distance
3-4 S9'03'W 54.13
4-5 N68'21' W 55.43
........
. -...
5··1
Line~
LAT
-53.46
+20.45
DE?
- 8.51
- 51.52
.... - ..
.. ......
- 33.01
- 60.03
tan 52'54' = &
d
d
9.11
=tan 52'54'
d=6.89m.
tan 75'53' = 10
e
e= 2.51 m.
tan 70'2Z = 10
f
f= 3.57 m.
Distance ofline F - 5:
2
3
5
'
60.03
tangentb eanng =33.01
Bearing =N61'1 Z E
60.03
.
.
Distance =Sin 61'12'
Distance = 68.50 m.
4
F - 5 =68.27· 0.67 +0.22
F- 5 =67.82 m.
Distance DE:
DE =67.82 - 7.86·6.89
DE=53.07
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5-168
SUBDIVISION
@
Lenght ofthe boundary ofhte proposal road
on the northern portion of the lot:
Distance AB:
arotlljjrsABlldBare~Q'lhef:jtare.$idjjl1tiali()t
Whl~6·~h~II• • ~• dl~ed~llaIlY.lh~area • • ao~
•
.m~ lEl~liI\l'l.·~tl~,.Jt9~l~~.9butti99··~ • Nl1lipn<:ll
R()~ll;~~~QWl:l •• ·.·.AS.·Plilt·AAIl!~gQtd@:lIlCl!'·Il.·.
Wi9¢Ji.ing·ClfrP~6JdghlQt~~yqf6 •. meter§
A~_--n~
·~M~9rl··.P9lb·.$i(l~\If~I~~ • i$.·.fElq~ir~~, • • ·l.J!liog·
DMP~(·(·············
.
68.27
sCl------,g
5Z054'
"L.----jE
D
AB =68.27 - 2.51 + 3.57
AB= 69.33m.
@
Area of the proposed road to be
expropriated from the property:
A - (69.33 + 68.2D (10)
12
A1 =688 sq.m.
- (68.27 +67.82) (0.89)
Ar
2
A2 =60.56 sq.m.
- (67.82 + 53.07) (9.11)
A32
A3 =550.65 sq.m.
Proposed Road Right of ffily
ill • • PQlllP#t~m~ • beadog • a@dl$tatlCept.lir~
~G'(})
•.
each Ofthebrothef.
@·.CQrtipijt~.~ • q~rillga@di~l8.rlge()f.~tl~
·¢Qmri:i6$idJ~sqM6thlotS{·
... .....'.' . .
@ .·G9mMtelffi.at~Mll.ff~rr9M.WldE!lli@}fdr
Solution:
A=A1 +A2 +A 3
A=688 +60.56 + 550.65
A =1299.21 sq.m.
A
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5·169
SUBDIVISION
CD Bearing and distance ofEC:
LINES
AB
BC
BEARING
N 30E.
.........
----
CD
Due South
Due West
12m.
36m.
DA
Lines
LAT
DEP
AB
BC
+41.57
- 29.57
-12.00
0
+24
+12
0
-36
CD
DA
DISTANCE
48m.
D\1D
+24
+60
+72
+36
Double
Area
+997.68
-1174.20
-864.00
. 0
2A -1640.52
A = 820.26
ForlineBC:
. -~
tan beanng - Latitude
12
tan bearing =29.57
Bearing = S22'05' E.
12
Distance =Sin 22'05'
Distance = 31.91 m.
Bearing and distance ofEC:
Bearing = S 22'05' E.
Distance =31,91 m,
@
Areas for each brother after widening:
18
tan l!l = 12
l!l
=56'19'
820.26
Area required =-2- =410.13 sq.m.
- 18 (12)
A1- 2
Aj =108 sq.m.
A2 =410.13 -108.00
A2 =302.13 sq.m.
.y
EC = (18)2 + (12)2
EC=21.63 m.
_(FC) (Ee) Sin 101'36'
Ar
2
(Fe) (21.63) Sin 101'36'
302.13 =
2
FC=28.52m.
BF =31.91 - 28.52
BF=3.39 m.
. (FE'f =(28.52)2 +(21.63)2
- 2(28.52)(21.63) Cos 101'36'
FE=39.11 m.
Sin a. Sin 101'36'
21.63 = 39.11
a. = 32'48'
Bearing ofFE =32'48' - 22'05'
. BearingofFE=S 10'43'W.
x = 6.93 Cos 60'
x= 3.47 m.
y= 6.11 Cos 79'17'
y= 1.14m.
ME= 18 - 3.47
ME= 14.53 m.
GH =14.53 + 1.14
GH= 15.67 m.
HL =18 +14.53 -15.67
HL = 16.86m.
BG=48-6.93
BG=41.07m.
FH =39.11 - 6.11
FH=33m.
CL=6m.
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, 5-170
SUBOIVISIOI
LOTA:
LINES
HG
GB
EF
FH
DISTANCE
15.67
41.07
3.39
33.00
BEARING
DueWesl
N30' E.
S 22'05' E.
S 10'43'W.
Lines
LAT
HG
GB
EF
FH
0
+35.56
- 3.14
- 32.42
DEP
DMO
Double
Plea
-15.67 -15.67
+20.54 -10.80
+1.27 +11.01
-6.14' +6.14
0
-384.05
- 34.57
-199.06
2A = 617.68
A =308.84m2
.frolll•• • t~e • giveht~c/1nisal •.• de$cription·.ltis
·re91lIt¢<l • • l().• de~~tmin~ the•• • l09~ti91'101 • • tt)e
~iviqil'lg!if1~that'/fUllnt~rseytJh~line$r§q
And,pA·~Wha~yt~atJ~ • WiUe~%~.1htP~h •
p()irlt()d~~jifetMk)k~hichi~9~e~~tot
comerAandadistance'.Pf.703,80ro·i.··.·.Tnl$
.•~jYidlng.H~·.~n' • .·dj\lidefhe·.Wl1()!¢•• '(it.·jnto.. tW.o••
~QJ¥II~~~,/",
'. .'.
."
lINES> ..... 13l:ARING
···········1.492A71'tl(.···
S9-Mtltl..
LOTB:
DISTANCE
16.86
33.00
28.52
6.00
BEARING
Due West
N 10'43' E.
S 22'05 E.
Due South
LINES
LH
HF
FC
CL
Line
LAT
DEP
DMD
LH
0
32.42
- 26.42
-6.00
-16.86
+6.14
+10.72
0
-16.86
-27.58
-10.72
0
Double
Plea
HF
FC
a.
0
- 894.14
+283.22
0
2A = 610.92
A =305.46 m2
Areas for each brother after Widening:
Al
= 308.84 m2
A2 = 305.46 m2
@
Bearing and distance of common sides of
both lots (line FE):
Bearing =S 10'43' W.
Distance =33.00 m.
D
Compute the area of the whole lot.
® Compute the location of diViding line from'
C()rtlerA alOOg the line AD. . .... . '.
® Compute the location of lhedividinglihe
from coiner Balong line Be.
CD
Solution:
ill Area of whole lot:
Lines
LAT
DEP
DMD
Double
Area
A-B +699.4~ +1068.26 +1068.26 +747237.19
B-C - 1428.1 -433.45 +1703.07 -2432222.3f
CoD - 164.7t - 634.81 +634.81 ·104591.30
D-A +893.41
0
0
0
2A =1789596.50
A =894788.25
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5-171
SUBDIVISION
@
Location of dividing line from comer A:
Using Sine Law:
AG
1276.90
Sin 39'54' = Sin 106'53'
AG = 855.96
OG = 855.96 - 703.90
OG = 152.16
1
Area offriangle ABG:
- 1276.90 Y
Ar
2
- 1276.00 (468.9)
A22
A2 = 299369.21
Area of triangle OFG:
- 152.16 x
Area ofABFE ="2 (894788.25)
A3-
Area ofABFE = 447394.125
x = FG Sin 73'07'
FG 152.16
Sin", = Sin 13
IJ = 180 - ('" + 106'531
IJ =73'07· '"
FG _ 152.16
Sin", - Sin (73'07' - "')
FG =
152.16 Sin '"
Sin 73'07' COS" - Sin", Cos 73'07'
152.16 Sin", Sin 73'07'
x =Sin 73'07' COS" - Sin", Cos 73'07'
152.16 Sin 73'07'
x = Sin 73'07' Cot ~ - Cos 73'07'
145.60
x =0.96 cot ~. 0.29
A _152.16 (145.60)
3 - 2 (0.96 col", - 0.29)
11077.25
As = 0.96 cot '" - 0.29
Area oftriangJe AOE:
2
A =A 1 + A2 -As
447394.125 =247667.22 tan" +299369.21
11077.25
0.96 col '" - 0.29
148024.92 =247676.22 tan '"
11077.25
0.96 cot", - 0.29
142103.92 cot '" - 42927.23
= 237760.53· 71823.49 tan" -11077.25
142103.92 cot" + 71823.49 Ian
269610.51
1.98 cot", + tan" - 3.75 = 0
tan2 '" - 3.75 Ian
1.98 = 0
,,=
703.80 (h)
A, =
2
h
tan ~ =703.80
703.80 (703.80) Ian '"
A, =
2
A, =247667.22 tan '"
y= 855.96 Sin 33'13'
y=468.90
"+
tan" = 3.75 + ~ (3.75)2.4 (1. 982
2
t
3.75 + 2.48
an", =
2
'" = 32'25'
h =703.80 tan 32'25'
h= 446.93 m.
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5-172
SUBDIVISION
® Location of dividing line from B:
STRAIGHTENING ABOUNDARY
The figure below shows an irregular
boundary ABeD and is to be replaced by a
single line AE. The general procedure is to
solve for the choosing line AD and compute for
the values of AI and A2. When AI is greater
than A2 move the new property line towards
the bigger area at A1 but if A2 is greater than
A1, move the property line towards A2.
A
o
FG
152.16
Sin 32'25' ::: Sin 40'41
FG::: 125.09 m.
BG
855.96
Sin 33'13'::: Sin 39'54'
BG::: 731.00
BF =731 -125.09
BF= 605.91
The dividing line is 446.93 m. from corner
A along the line AD and 605.91 m. from
corner B along the line Be.
p
D
Since A2 is greater than A1 move the new·
property line to A2.
A
A
I
E 1
~
Z
D
A:::Ar A1
A::: ab Sin 0
2
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STRAlGHnllNG OF IRREGUlAR BOUNDARIES
When Al is greater than A2• move towards AI'
A
A
@ . . Pjndlhe•• dislan(;~.~I1(~long·Web()\JM;irY
....fF)•• such•• tb<ltth~ • • $!r<'iljg~t • • line.t\I+.Vtin
• pmYjMtfle$am.e.~ipt~~propert~ •.?f·
><MLPetez anqMr. $i#I~za~~st~case
• • • • ••• V!ljelllh~.~9~ndary.~·%'rve·
®•·•• qomtM~ttjEl~~glhmfll1iJ.AH··.
@
COll'lPlItelhElMaringoflineAH.
G)
Distance BH:
Al =a + c+e
A1 = 3040 + 2384 +68
Al =5492 sq.m.
A2 = b+d
A2 = 926 + 1592
A2 = 2518 sq.m.
Solution:
c
D
a
A =A I -A2
A =5492 - 2518
A = 2974 sq.m.
Therefore, the dividing line should be
moved towards the property of Mr. Perez..
F
Mr. Parezano Suarez own pieces of land
Whlchareadjacentlo each other. The curve
·Iirie in the figure fE':!pi'esentsastreamforming a·
•boundary between the two· pieces of propertY.·
It is proposed to place this stream in a storm
dtaiilandto straigtiteri the boundary. Starling
at poihl A, iti'andonfline AS was run and the
area "a", "~"/'d" and '!e" were detelTllinedfrwn
offset measurements using SImpson's RUletd
be 3040, 926, 2384, 15B2 and 68 sq.m,
resPl¥tively. .. . . . . .
.
AS =<375 m: and has a bearing of S. 77'4fY E.
CO:::EF
.
Bearing of CO '" N.15'30' W.
D
. ...
F
c
" = 77'40' - 15'30'
,,= 62'10'
A _ 375 (x) Sin 62'10'
2
_2(2974)
X - 375 Sin 62'10'
X= 17.93m.
® Length of line AH:
Sin 62'10' = 1~~3
H/= 15.86 m.
c
S 77"40' E
E
B/ = 17.93 Cos 62'10'
B/= 8.37 m.
E
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5-174
STRAIGHTENING OF IRREGUlAR BOUNDARIES
N
Solution:
G)
B
Distance EB:
Solve for the line EB:
LAT
100.27 Cos 13'10' = +97.86
91.26 Cos 0'11' = +91.26
112.48 Cos 27'39' = +97.59
LINES
BC
CD
DE
+28.71
.AI = 375 - 8.37
AI=366,63
15.86
tan a = 366.63
0.=2'29'
366.63
AH = Cos2'2fJ
AH = 366,97 m,
@
Bearing of line AH:
Bearing AH =77'40' +2'2fJ
Bearing AH= S 80'09' E
BI= 8.37 m.
DEP
LINES
BC
CD
100,27 Sin 13'10' = - 22.84
91.26 Sin 0'11' = -0.25
112,48 Sin 27'39' = +52.20
DE
25.07
F
D
c
T~~ • foll()wm~I$.~ • • $ef9f.n()t~$l}fahl%®9~(
l>9Ql1dary.of.apl~·.9f.IaM ..·.··lt•• ls~e$~ • lQ••
str<ti9ht~n • • jhi$ • crBglMI•• • bp~nd~ry>II~~ • • ~Y •
. .lihe/:F.
~~~~~Mlqsa.$tralght.llr)~l"\jnhil19ft@l~·.t9tfl~
.
.•.
A
E
n<f}
<D Find the distanceEB.
.
® Find the distance along EF frompo!ntE to
the point where the new Une cuts EF, .. .
@ Fin9 the bearing of the new boundary line
BX
.
ti>,
B
B
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STRAIGHTENING OF IRREGUlAR BOUNDARIES
.
29.07
tan beanng (EB) = 286.71
Bearing (EB) =S 5'55' W
@
Bearing ofBX:
Using Cosine Law:
(BX)2 = (6322)2 + (288.25)2
• 2(63.22)(288.25) Cos 102'
286.71
Distance (EB) = Cos 5'55'
Distance (EB) = 288.25 m.
(BX? = 3996.77 + 83088.06 + 7577.56
(BX)2 = 94662.39
BX=307.67
Using Sine Law:
® Distance along EF from point E to the point
where the new line cuts EF:
From Plane Trigonometry:
A - c?- Sin A Sin B
- 2 Sin C
- (100.2D 2 Sin 7'15' Sin 166'39'
A1 2 Sin 6'06'
B
A, = 1378.42 sq.m.
A2
(112.48)2 Sin 21'44' Sin 152'10'
2 Sin 6'06'
A2= 10291.43 s~.m.
A=A 2 -A 1·
307.67 _63.22
Sin 102' -Sin Il
. - 63.22 Sin 78'
SIn Il 307.67
Il
= 11'36'
A= 1D291.43 - 1378.42
A = 8913.01 sq.m.
Bearing ofBX = 11'36'·5'55' =5'41'
A = (EX) 288.25 Sin 102'
Bearing of BX = N5'41' W
2
EX = 17826.02 esc 102'
288.25
EX= 63.22m.
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5-176
ARUS OF IRREGUlAR BOUNDARIES
b) Simpson's One Third Rule: (Applicable
only to even inteNals or odd offsets)
hs
Methods of computlnq Areas of IrreqUli1r
Boundaries at Re ular Intervals
A
a} Trapezoidal Rule
d
d
B
d
d
C
d = common interval
h1 =first offset
hn = last offset
d
d
d
d
A =A, +A2 +A3 + ~
d
. A = 2[(h 1 +h2) + (h2 +h3) + (h3 +h4) + (h4 +h,J]
d
A = 2(h, +2h 2 +2h3 +2h4 + hnl
]
h1 + hn
A = d [ 2 + h2 +h3 +h4
A=d
[h hn +I,h]
1
;
I,h=h2 +h3 +h 4
I,h = sum of intermediate offsets.
For the next two inteNal
d
A2 =3[h 3 +hs +4h 41
hs = last offset
h1 = first offset
h2 and h4 = even offset
h3 = odd offset
d
A = 3 [(h , +hn) +2I,h odd +4 I,h even]
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AREAS Of IRREGUW BOUNDARIES
@
Trapezoidal Rule:
A =d [h 1 ; hn + h2 + h3 + h4 + hs + h6]
A series of perpendicular offsets were taken
.ftom a lralls~ ijne 10 a curved bOUllldaryline.
These offsets were taken 9 meters apart and
were taken .In the folQ.Wlng order: 2m., 3;2 m.,
4 m., 15 m.,S in.4.5m~, 6 m., 7m,Deleml.ine
the area included between theJransilline and
the curved uSIng: ... .
.
..
d=9m .
h1 =2m.
hn = 7 m.
A = 9 [(2 ; 7) + 3.2 + 4 + 3.5 + 5 + 4.5 + 6]
A =9(30.7)
A = 276.3 sq.m.
@
Difference between Simpson's One Third
Rule and Trapezoidal Rule:
=276.3 - 270.9
= 504m 2
Solution:
CD Simpson's One Third Rule:
.d
A1 ="3 [h 1 + hn + 2 LfJcxJd + 4 LfJevenl
d=9m.
h1 =2m.
hn =6 m.
LfJOdd = 4 + 5
LfJodd= 9 m.
"Lheven =3.2 + 3.5 + 4.5
LfJeven = 11.2 m.
Shown 1n the accompanying sketch are the
measured offsets from a traverse line AB 10 an
irregular boundary and the spacing .between
the offsets. Determine the area boUnded by the
traverse line, the irregular boundary and the
end offsets using:
.
A1 = ~ [2 + 6 + 2(9) + 4(11.2)]
A1 "3(70.8)
A1 "212.40 sq.m.
A2 = (6; 7)9
A2 = 58.8 sq.m.
Total area':: 212.40 + 58.5
Area =270,90 sq.m.
CD Trapezoidal RUle.
Simpson's One· Third RUle.
Compute the difference between
Trapezoidal Rule and Simpson's One·
Third Rule.
S-178
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AREAS OF IRREGUlAR BOUNDARIES
Solution:
<D Trapezoidal Rule:
Solution:
<D Trapezoidal Rule:
2 Lh]
A=d [h 2 + h +
d=20
h1 = 12.22
hn = 10.35
Lh =11.32 + 8.82 +6.52 + 16.38
Lh=43.04
d
A = 2" (h 1 + hn + 2 LhinlJ
@
6
A= 2" [5.60 + 2.70 + 2 (6.40 + 7.90 + 6.20
A= 20
+ 7.50 + 9.50 + 12.30 + 10.80)1
A = 388.50m2
A = 1086.50 sq.m.
Simpson's One-Third Rule:
d
A = 3" (h 1 +hn + 2 Lhodd +4 Lheven)
6
A =3" [5.60 + 2.70 + 2(7.90 + 7.50 + 12.30)
+ 4(6.40 + 6.20 + 9.50 + 10.80)]
A = 390.60m2
® Difference between Trapezoidal Rule and
Simpson's One-Third Rule:
Difference in area = 390.60· 388.50
Difference in area = 2.1 m2
@
[C 2.22 ; 10.35) + 43.04]
Simpson's One Third Rule:
(Treat the last area as trapezoid)
d
A1 = 3" [(h1 + hn) +2 Lhodd +4 Lhevenl
h1 = 12.22
hn = 16.38 .
d=20
Lhood=8.82
Lheven =11.32 + 6.52 =17.84
20
A1 =3" [12.22 + 16.38 + 2(8.82) + 4(17.84)] .
A1 = 784 sq.m.
(16.38 + 10.35) (20)
Ar
. 2
A2 = 267.30 sq.m.
A =A 1 +A 2
A =784 + 267.30
A = 1051.30 sq.m.
® Difference in area:
Difference in area = 1086.50 • 1051.30
Difference in area = 35.20 m2
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PLANE TABLE
Five Methods of Orienting the Plane Table:
1. By the lise of magnetic compass.
2. By backsighting.
3. By solving the three-point problem
a) Trial Method (Lehman's Method)
b) Bessel's Method
4. By solving the two-point problem
5. By using the Baldwin Solar Chart
A. TRIAL METHOD (LEHMAN'S METHOD)
The 3 points A. Band C plotted on the
tracing paper with any convenient scale.
The tracing paper is then placed on top of
the plane table and the plane table is set
up over the station whose position is to be
determined and is oriented either by
compass or by estimation. Resection
lines from the three stations A, Band Care
drawn through the corresponding plotted
points "a", "b" and "c". These lines will
not intersect at a common point unless the
trial orientation happens to be correct.
Usually, a small triangle called· the
"triangle or error" is formed by the three
lines. Let us say, the vertices of this
triangle is ab, bc and ac as shown and
point P is called .the point sought, the
position of which is to be determined as
follows:
a) Draw a circle passing through points
"a'\ "b" and "ab".
b) Draw a circle passing through points
"b", lie" and "bc".
c) Draw another circle passing through
points "a", "c" and "ac".
d) The three circles will intersect at point
"P", the point sought.
B. BESSEL'S METHOD
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S-180
PlANE TABLE
Points a, band c are the plotted points
of A, Band C on the ground. With the
straight edge of the alidade placed along
line "ab", tum the table until a backsight at
A is taken as shown in the figure 1, with
point "a" towards point "A" on the ground.
Clamp the table, then take a foresight to
point C and draw a line passing through
"b". Reset the straight edge of the alidade
at line "ab", tum the table and backsight at
point "S' with "b" towards point "B" as
shown on figure 2. Clamp the table and
take a foresight towards point C and draw
a line passing through point "a". The two
lines drawn intersects at point "e". Set the
straight edge along the line "ec" and take a
backsight at C. Clamped the table. Then
draw resection lines through "a" and "b",
these two lines will intersect each other at
point "P", then point sought.
A
8
~\------7
'\
/'
\
\"
/
,/
The location of points A and B are plotted
on the plane table sheet at "a" and "b" as
shown in the figure. These points must be
plotted using convenient scale. The plane
table is then set up over point C on the ground
which points A and Sare visible. The board is
then oriented by either compass or by
estimation. Point "c" corresponding to C on
the ground is plotted by. estimation on the
plane table sheet. Point D is also established
on the ground with the distance C to D
estimated. With the table at "c", foresights are
taken on A, Sand Dand lines are drawn on the
sheet. The corresponding position of D is
plotted on the sheet as "d". The table is then
transformed to station D and is oriented
tentatively by backsighting at C. Foresights
are taken on points A and B and lines are
drawn intersecting the previous lines drawn
before, say at points e' and f. The line joining
e' and f is parallel to the AB. With the straight
edge of the alidade placed along line ef, a
point Eat some distance from the table is set
on the line of sight. The alidade is then moved
to the line ab and the board is turned until the
same point E is sighted. The plane table is
now properly oriented. Sy section through a
and b, the correct position of the plane table is
plotted at P.
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PlANE TABLE
SOURCES OF ERRORS IN PLANE
TABLE WORK:
1. Setting over a point.
2. Drawing rays.
3. Instability of the table.
E
RADIA.TION WITH PlANE TABLE
o
,"' .....
-- A,
-- '
\\ "#..<~~~><:~}~;~~:,!
....
[ZJ
,'
e
~
a-----'
iNTERSEcnON
a
wrrn PLANE TABLE
1.!?~j)'\~./
.... ..-
E
GRAPI-DCAl TRJAN(iIJl..ATlON
1. Relatively few points need be located
because the map is drawn as the survey
proceeds.
2. Contours and irregular objects can be
presented accurately because the terrain is
in view as the outlines are plotted.
3. As numerical values of angles are not
observed, the consequent errors and
mistakes in reading and recording are
avoided.
4. As plotting is done in the field, omissions
in the field data are avoided.
5. The useful principles of intersection and
resection are made convenient.
6. Checks on the location of plotted points
are obtained readily.
7. The amount of office work is relatively
small.
1. Plane table is very cumbersome and
several accessories must be carried.
2. Considerable time is required for the
topographer to gain J::roficiency.
3. The time required in the field is relatively
large.
4. The usefulness of the method is limited to
relatively open country.
5-182
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PlaNE TABlE
ADJUSTMENTS OF THE PLANE
TABLE ALiDADE:
1. To make the axis of each Plate Level
Parallel to the Plate:
Center the bubble of the plate level
when manipulating the board. On the
plane table sheet, mark a gUide line alone
one edge of the straightedge. Tum the
alidade end for end, and again plane the
straight edge along the guide line. If the
bubble is off center, bring it back halfway
by means of the adjusting screws. Again
center the bubble by manipulation the
board and repeat the test.
4. (For alidade on Tube-in Sleeve Type).
To make the axis of the Striding Level
parallel to the axis of the telescope
Sleeve, and parallel to the Line of
Sight.
Place the striding level on the
telescope and center the bubble. Remove
the level, tum it end for end, and replace it
on the telestope tube. If the bubble if off
center, bring it back halfway by means of
the adjustment screw at one end of the
level tube. Again center the bubble and
repeat the test.
5. (For alidade of Fixed tube Type). To
make the axis of the telescope level
parallel to the Line of Sight.
2. To make the Vertical Cross-hair lie in a
Plane Perpendicular to the Horizontal
Axis.
This adjustment is the same as the
two-peg adjustment of the dumpy level.
Sight the vertical cross-hair on a well
defined point about 100 m. away and
swing the telescope through a small angle
(vertical). If the point appears to depart
from the vertical cross-hair loosen two
adjacent screws of the cross-hair ring, and
rotate the ring in the telescope tube until by
further trial the point sighted traverses the
entire length of the hair.
6. (For Alidade having a Fixed Vertical
Vernier). .To make the vertical vernier
read zero when the Line of Sight is
horizontal.
3. (For alidade of Tube in Sleeve Type).
To make the line of sight coincide with
the axis of the Telescope Sleeve.
Sight the intersection of the crosshairs on some well define point. Rotate
the telescope in the sleeve through 180'. If
the cross-hairs have apparently moved
away from the point bring each hair
halfway back to its origin position by
means of the capstan screws, holding the
. cross-hair ring. The adjustment is made
by manipulating opposite screws, bringing
first one cross-hair and then the other to its
estimated correct position. Again sight on
the point and repeat the test.
With the board level, center the bubble
of the telescope level. If the vertical
vernier does not read zero, loosen it and
move it until it will read zero.
7. (For Alidade hav~ng a movable Vertical
Vernier with Control Level). To make
the axis of the Vernier Control Level
parallel to the axis of the Telescope
when the Vernier reads zero.
Center the bubble of the telescope
level, and move the vernier by means of its
tangent screw until it reads zero, if it is off
center move it to the center by means of
the capstan screws at the end of the
control level tube.
8. (For Alidade having Tangent
movement to Vertical Vernier Arm).
This type of vernier needs no
adjustment.
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TOPOGRAPHIC SURVEY
Contour interval - on a given map,
successive contour lines represents
elevations which differs by a fixed vertical
distance called contour interval.
Topographic Survey· is a survey made in
order to secure important data from which a
topographic map could be made.
Hachures . artificial shade lines drawn in the
direct of steepest slope for the purpose of
representing a relief.
Saddle - a dip at the junction of two ridges.
Scheme of work of a Topo raphic Survey:
1. Establishment of a horizontal control by
measuring angular and linear
measurements of a center point.
2. Establishment of the vertical control by
determining the elevation of control points
by leveling or using plane table.
3. Determining the elevations and location of
some important features as many deem
necessary for the preparation of the
topographic map.
4. Computations of elevations, distances and
angles as obtained from the previ0us field
work undertaken.
5. Preparation of the topographic map, which
is actually a representation of the terrestrial
relief.
Thalwegs - are the lines where the two sides
of a valley meet.
CHARACTERISTICS OR PROPERTIES
OF CONTOURS
1. All points on the same contour have the
same elevation.
Refief - configuration of earth's surface.
Methods of representing relief:
1. Relief models
2. Shading
3. Hachure lines
. 4. Form lines
5. Contour lines
Contour· an imaginary line of constant
elevation on the ground surface.
Contour fine - a line on the map joining points
of the same elevation.
2. Every contour closes upon itself either
within or outside the limits of the map.
5-184
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TOPOGRAPHIC SURVEY
3. On unifonn slopes, the contour lines are
spaced unifonnly.
4. A single contour can not lie between two
contour lines of higher or lower elevation.
5. Along plane surfaces (such as those of
railroad cuts and fills) the contour lines are
straight and parallel to one another.
6. As contour lines represents level lines,
they are perpendicular to the line of
steepest slope. They are perpendicular to
ridge and valley lines where they cross
such lines.
7. On steep slopes contours are closely
spaced and on gentle slopes, contours are
spaced far apart.
8. As contour lines represent contours of
different elevation on the ground, they not
merge or cross one another on the map,
except in cases where there is an
overhanging cliff or cave, or bridge
abultments.
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TOPOGUPHIC SURln
9. A closed contour indicate either a summit
or depression. A hachured, dosed contour
line indicates a depression.
Hachure Lines
FIVE MOST COMMON TYPES OF
GROUND FORMATION
1. Depression
. 10. A contour never splits.
3. End of a Ridge
11. No two contours can run into one.
4. End of a Valley
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S-186
TOPOGRAPHIC SURVEY
5. Saddle
1. Cross-sections and profiles from contour
maps.
2. Earthwork for grading areas.
3. Earthwork for roadway.
4. Reservoir areas and volume.
5. Route location.
Four Systems of Ground Points
for locating Contours:
1.
The ground points form an irregular
system along ridge and valley lines and at
other critical features of the terrain. The
ground points are located in plan by
radiation or intersection with transit or
plane table and their elevations determined
by trigonometric leveling or sometimes by
direct leveling.
2.
Trace Contour System:
in this system, the contours are traced
out on the ground. The various contour
points occupied by the rod are located by
radiation using a transit or aplane table.
4.
1. Transit and Level Method
2. Stadia Method
3. Plane Table Method
Cross Profile System:
The ground points are on relatively
short lines transverse to the main traverse.
The distances from traverse to ground
points are measured with the tape and the
elevation of the ground points are
detenmined by direct leveling.
3.
Three General Methods Employed in
Undertaking a Topographic Survey:
Control point system:
Checker Board System:
This is used in areas whose
topography is smooth. The tract is then
divided into squared or rectangles with
stakes set at the comers. The elevation of
the ground is determined at these comers
and at intermediate critical points where
changes in slope occurs, usually by direct
leveling.
A level rectangular field of 1800 sq.m. has for
its sides in the ratio of 2: 1. Three poles of
equal heights are located at three consecutive'
corners. To measure the heights of the poles,
a CMI Engineer set a transit (H.I. =15 m.) at a
point within the lot and took the angles of
elevation of the top of the poles. The angles of
elevation of the top of the three poles taken are
25', 25' and 30'.
CD Compute the height of each pole.
Compute the distance of the transit from
the nearest comer. .
@ Compute the distance of the transit from
the farthest comer.
@
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TOPOGRAPHIC SURVEY
Solution:
0.2171 t 48.91 = 0.571 - 69.36y + 2210.98
G) Height of each pole:
0.3531· 69.26y+ 2162.07 = 0
1 -196.5 t 6124.8 = 0
r - - - - 3 0 - -.....
Y = 196.5 ~ (196.5j2· 4(6124.8)
2
y=
196.5 ± 118.8
2
y= 38.85 m.
~:: 0.2171 t 48.91
~
~xI2-LxI2-
~
~
=0.217(38.85)2 t 48.91
h= 19.40 m.
Height of each pole =19.41 + 1.50
25"
Height of each pole = 20,90 m.
h cot 25'
.
@
30'
Distance of the transit from the nearest
comer:
Distance of transit from nearest comer
h cot 30'
= 19.40 cot 30'
= 33.60m.
3x2 = 1800
x2 =900
x =30
@
Distance of the transit from the farthest
comer:
Distance of transit from farthest comer
(1)
1 + (15)2
(h cot 25')2 =
4.6h2 =
2)
1 + 225
(h cot 30') = (60· y)2 t (15)2
1.73~ = 3600· 120y t I t 225
CD ~ = 0.2171 + 48.91
(2)
h2 = 0.57; - 69.36y + 2210.98
= 19.40 cot 25'
= 41.60 m,
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5-188
ROUTE SURVEYING
3. Location;survey
a. Five survey teams go out in the field.
Route Surveys - are surveys made for the
p~rpose of locating any buildings,
highways. canals, power transmission
lines, pipe lines, and other utilities which
are constructed for purposes of
transportation or communications.
1. Transit party - stakes the location
of circular curves with proper
stationing.
2. Level party - checks the selected
bench mark and executes the
profile work.
3. Cross-section party - slope stakes
are set on the ground.
4. Land line party - property lines and
other important details are
indicated on the plan.
1. Reconnaissance:
a. General routes are selected and
horizontal and vertical controls are
established.
5. Special team - takes care of the
special surveys for structure.
b. All surv~y works are consolidated
with the following prepared:
1. Location map
b. Reconnaisance report is made
accompanying a reconnaissance map.
2. Location profile
3. Cross-sections
2. Preliminary survey:
4. Earthwork estimates
5. Right of way maps
a. There are survey parties that execute
this phase of work.
6. Structure maps and plans
1. Transit parly - runs the traverse.
2. Level party - sets bench marks
and determines the profile.
3. Topographic party - runs the crosssectioning work.
b. A preliminary map is prepared for
determination of possible cost of the
project.
4. Construction survey:
a. Slope stakes for construction works
are staked, spiral are laid and lines
and grades for tract or pavement are
defined.
b. Final plans' are prepared, profile
sections, as revised during
construction.
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STADIA SURVEYING
By ratio and proportion:
d
f= SCos 0
f
f
a) Horizontal Sights:
d=~SCOS0
I
H =(f + c +d) COS"
f
H =~S Cos2" + (f+ c}Cos"
I
V= (d +f+ c) Sin"
V=
•
f Sin 2 I2J
V =i S -2-'- + (f + c) Sm I2J
F = principal focus
f = focal length
o = optical center
i = distance between stadia hairs
c = distance from optical center to center of
instrument
By ratio and proportion:
! =s!
i s
d =~S
[f S COS" + (H c)l Sin"
1. Stadia interval factor not that assumed.
Rod not of standard length
Incorrect stadia interval
Rod not held plumb
Unequal refraction
2.
3.
4.
5.
I
D=d+f+c
f
D =i S + (f + c)
f= stadia interval factor
f + c = stadia constant
S = stadia interval or intercept
b) Inclined Sights:
tan m = 0.006
m = 17' (too small and is negligible)
1. The telescope be of excellent quality, with
good illumination.
2. The magnifying power should be about 25
to 30.
3. The stadia hairs should be fixed and
f
should be set accurately so that ~ = 100.
I
4. The transit should have a good compass
needle.
5. The transit should have a complete
vertical circle.
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5-190
STADIA SURVEYING
D, =K8, +R
fJ< =KS2 +R
fJ< . D1 =(KS2 + R) • (K81+ R)
K(~,S1}=fJ<-D1
n.. - D.
82 .81
K=~
200 - 60
2.001 - 0.600
K = 99.93 (stadia inteNal factor)
K=
® Horizontal distances DE and DF:
Assume elevation of D = 100 m.
f
H=:SCOS 2 0 +(f+c}cos 0
I
H= 99.93(2.12) cos 2 4'22' + (0.30) cos 4'22'
H=210.92m.
Horizontal distance DE = 210.92 m,
f
H=: S cos 2 0 + (f + c) cos
I
0
H=99.93 (3.56) cos2 3'1T + (0.30) cos 3-,17'
H= 354.88 m.
Horizontal distance DF = 354,88 m,
@
Differences in elevation between points D
and E and points D and F:
sin 20
.
V=/f S-2-+
(f+ c) Sin 0
V=99.93(2.12} sin ~'44' +(0.3) sin 4'22'
Solution:
CD Stadia inteNal factor:
--- ----I-=D:...;1:..-=..:...60=--_D=200_ _- - I
f sin 20
.
V= /S-2-+(f+c}SIn
0
V= S9.93(3.56}Sin ~'34' +(0.30) sin 3'17'
fS
D =--:- + (f+ c)
I
f=stadia inteNal factor
K=
=
V=16.11m.
Elev. of E = 100 + HI + 16.11 - HI
Elev. of E = 116.11
Difference in elevation between D and E
= 16.11 m.
=
R (f + c) stadia constant
(f+ c) = 0.30
V= 20.36 m.
Elevation of F = 100 +HI- 20.36 - HI .
Elevation of F = 79.64 m.
Difference in elevation between Dand F
=20.36m.
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STADIA SURVEYING
@
Horizontal distance between Band D:
H=tSCos2
(j)
@
@
I
A transit with a stadia constant equal to
0.30 is used to determine the horizontal
distance be1ween points Band C, with a
stadia intercept reading of
1.85 m. the
distance BC Is equal to 182.87 m.
Compute the stadia interval faclor of the
instrument
.
Using the same instrument, it was use~ to
determine the difference in elevation
between Band 0 having a stadia intercept
reading of 2.42 m. at 0 at a vertical angle
of + 6'30'. Compute the difference in
elevation of Band D.
+ (f+ c) Cos 0
H =98.69(2.42) CoS2 6.5' + 0.30 Cos 6.5'
H= 236.07m.
A Civil Engineer proceedeq to do the stadia
survey work to determine thEdopagraphy o~ a
certain area. The transit was set up at a poInt
A, with the line of sighthorizontal, took rod
readings from the ((}ds placed at B a~d C
whiCh is 2QO m. and 60 m,from Arespectively.
Compute also the horizontal distance
between 8 and O.
Stadia Intercept
2.001 m.
O.ro<lm.
Rod at B
RCldatC
Solution:
(i)
0
Stadia interval factor:
fI
(j)
D = S + (f+ c)
.
f
182.87 =j(1.85) +0.30
t/ = 98.69
® Ditt. in elevation between Band D:
@
Compute the stadia interval factor.
.
Using the same instrument this was .used
for determining the elevation of pomt D
With a stadia intercept of 2.12 m. and a
vertical angle of +4'22'. If the elevation of
the point where the instrument was set ~p
is 10Q m., compute the elevation of pOInt
0, Stadia constant is 0,30 m.
Compute the horizonlaldistance from the
point where the instrument was set up to
paintD.
Solution:
CD Stadia interval factor:
s
D =t + (f + c)
/
tI
200 = (2.001) + (f + c)
Sin 20
.
V= i S-2- + (f+ c) Sin 0
f
Sin 13' +.
030 S·In.
65'
V= 98.69(2.42) -2V= 26.90m.
f
60 =~ (0.600) + f + C
I
140 =(1.401)
!/ =99.93
f
(stadia interval factor)
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5-192
STADIA SURVEYING
@
Elev.ofD:
Solution:
f
:=99.5
f Sin2fil
.
V=iS-.-2-+(f+'C)Slnfil
I
f + c = 0 (interior focusing)
V= 99.93(2.12) Sin ~'44' + (0.30) Sin 4'22' .
V=16.11
Elev. D= 100 + H/+ 16.11- HI
Elev. D = 116.11
CD Inclined stadia distance:
f
D=:SCOSfil+(f+c)
I
@
Horizontal distance:
f
H=: Scos2 fil + (f + e) cos fil
D=99.5(2.50) cos 23'34' + 0
D= 228m,
I
H =99.93(2.12) cos2 4'21 + (0.30) cos 4'22'
H=210.92m.
® Difference in elevation between the two
stations:
f Sin 2fil
.
V=jS-2-+ (f+ e) Sin fil
··@ij••~PM@MI9WM~I~~@trffi~gi6g~.9@
.
~;~IIII~~II~~~'i~~'\rtJI~
... at~~h~lt.Wi'h • •~lJ • lqt~t@lfP¢ijSlj@.t~I$¢Qp~.
@g~#Ylllg~~mg~@~W~lf@19fgf~~($,
• 'M
~r~~~o1t6')t~~$~~I6~.r~~~~:~~.~g.*~:0~
th~y~ijl¢~IM91~9g11~ty~#i$h?~'~M,
.i1¢tettnj~¢tMJ(!II6Wl~~C··
V= 91.16 m.
DEAB =2.25 + 91.16 - 1.45
•.
0Yf@l%ed$lMiaglslariCe,>
®••.·.Oifferenceil'l•• elevatioil·belWeen.··the·two
~~9l)l)fl.)
V=99.5(2.50) ~ sin (2 x 23'34') + 0
··1··••••~~~~l~.1$1~~.~~~~eril1~~~~~.·
. DEAB = 91.96 m.
@
Elevation of station B:
Elev. at B = 155.54 - 91.96
Elev. at B = 63.58 m..
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STADIA SURVOING
@
Difference in elevation between points A
andB:
V=fS~sin
20
+ (f+c) sin 0
VA = 100.8 (0.31) ~ sin (2 x 15'35')
+ 0.381 sin 15'35'
VA =8.19 m.
i
Va = 100.8 (0.236) sin (2 x8'08') + 0.381 Sin 8l O8'
Va =3.39 m.
Diff. in elev. between A and B
= 1.854 + 3.39 + 8.19 -1.175
= 12.259m.
: =::~~;Km
@
@
beM."poinl' A
Find the' horizontaHdtstance from the
lraHsit to !herod held at S, .'.."
Horizontal distance frem the transit to the
rod held at B:
Ha = 100.8 (0.236) cos 2 8'08'
+ 0.381 cos 8'08'
Ha= 23.69m.
Solution:
1330
1m 1
y.
o~ ~.~.~:
" ' • •~
~'~~~:r'w~:~',)
...
1 7361 8~4
B
SA = 1.330 -1.020
SA = 0.31
Sa = 1.972 -1.736
Sa = 0.236
CD Length of line AB:
f
H = ~ S cos 2 13 + (f + e) cos '"
-I
A survey party proceeded to dothelrsmdill
.sLirvey work as fdlows. The Iransit was s~t
up at A and wtlh the line of sighftlotiibntal.
took rod readings at poinls13 and C whiCh is
;300 m. and 80rruespectively, . ...> . . . . . •
. Wtlh· rod' at ·the·stadlaloterveJw.a$
recorded 10 be 3.001 m. and withth~rod afC
the stadia interval wastecOrded 10 be. o.aOorll.
the distance from the inslrumen'ttothe
principal ·foco.s was recorded to be 0;30 m,
Then they went to survey other points With
some of the data recorded as follows witnthe
transit at point Of the two paints E aridf were
sighted.
."
..
a
Rod al E
I
HA = 100.8 (1.330· 1.020) cos 2 15'35'
+ 0.381 cos 15'35'
HA = 29.36 m.
Ha = 100.8 (0.236) cos 2 8'08'
+ 0.381 cos 8'08'
Ha = 23.69 m.
HAa = 29.36 + 23.69
HAa = 53.05
Rod at F
®•
Stadia interval
Vertical angle
Stadia interval
Vertical angle
:::: 225 m.
= +4'30'
= 3,56 m.
= - 3'30' .
~ . Compute the stadia jnlerval~ctot. •.
@
Compute the horizontal distance DE. .. .
Compute the difference in' elevatiOn
between E and F assuming elevation of
0= 350.42 m. above sea level.
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STADIA SURVEYING
Solution:
f Sin 20
.
V =, S -2- + (f + c) Sin I2l
CD Stadia inteNal factor.
f
S = ~ S + (f+ c)
V =99.95(2.25) Si; 9' +0.30 Sin 4'30'
I
fI
V= 17.61 m.
300 = (3.001) + (f + c)
80
=,f (0.80)+ (f+ c)
Elev. E = 350.42 + HI + 17.61 - HI
f
Elev. E = 368.03 m.
220 = (2.201)
fI = 99.95 (stadia inteNal
factor)
.
.-7J30-· ------1
~~~llmnm--"""'~'_"" ~
® Horizontal Distance DE:
H.I.
J
f
Sin 20
.
V =, S --2- + (f + c) Sin"
V = 99.95(3.56) Si;E +(0.30) Sin 3'30'
f
.
{-., =~ S Cos2 0 + (f + c) Cos 0
I
H= 99.95(2.25) Cos2 4'30' + (0.30) Cos 4'30'
H= 223.80m,
@
V=21.70m.
Elev. F = 350.42 + HI- V- HI
Elev. F = 350.42 - 21.70
Elev. F = 328.72 m.
Diff. in elevation between E and F:
Ditt. in elev. between E and F
;: 368.03 - 328.72
=39.31 m.
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S-195
I
HYDROGRAPHIC SURIEfI.NG
2. By means of range line and an angle from
the shore.
Purpose of Hydrographic Surveying:
1. To determine shore lines of harbors, lakes
and rivers from which to draw an outline
map of the body of water.
3. By means of range line and an angle from
the boat.
2. To determine by means of soundings, the
submerged relief of ocean bottoms.
3. To observe tidal conditions for the
establishedof standard datum.
4. To obtain data, in case of rivers, related to
the studies of flood control, power
development, water supply and storage..
4. Two angles from the shore.
5. To locate channel depths and obstruction
to navigators.
6. TO determine quantities of underwater
excavations.
7. To measure areas subject to scour or
silting.
8. To indicate preferred locations of certain
engineering works by stream discharge
measurement.
5, Two angles taken simultaneously at the
boat by using a sextant, and three stations
on the shore,
Methods of locating soundings:
1. By means of a boat towed at uniform
speed along a known range line at equal
intervals of time.
\Rallge Lille
:
~----~----~---t
I
I
~.
I
I
,"
I
I
I
Baal
::',' ,;'\-t-\-l----~
I
I
I
1
\>'----o-J---.---.
:
\
I
I
\
I
I
II
~----~----~----~ ~
-
~--
-~----~
----~----~
6. By transit and stadia.
'~l
6
II .
.stadia
all boat
...-"'--'''''.-'-.
'~
- .•~=:r---', .--0
I~
. ---- .--- ,&"
I
~
5-196
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7. By intersection of fixed ranges.
Methods of Plotting Soundings:
1. By using the Two Polar Contractor
2. By using Two Tangent Protractors
I
3. By the Tracin,!1 Cloth Method
4. By using the Three Arm Protractor
8. By a wire stretched along a river at known
distances.
5. By the use ofPlolting Charts
Methods of Measuring Velocity
in a Vertical Line:
Hydrographic maps - is similar to the
ordinary topographic map but it has its own
particular symbols. The amount and kind
of informations shown on the hydrographic
map varies with the use of the map.
A hydrographic map contains the
following informations:
1. Data used for elevation.
2. High and low water lines.
3. Soundings usually in feet and tenths, with
a decimal point occupying the exact
plolted location of the point.
4. Lines of equal depths, interpolated from
soundings. On navigation charts the
interval of line of equal depth is equal to
one fanthom or six feet.
5. Conventional signs for land features as in
topographic maps.
6. Light houses, navigation lights, bouys,
etc., either shown by conventional signs or
leIters on the map.
1. Vertical-velocity-culVe method:
Measurements of horizontal velocity
are made at 0.5 beneath the surface and at
each tenth of the depth from the surface to
as near the bed of stream as the meter will
operate. If the stream is relatively shallow,
measurements are taken at each one fifth
of the depth. These measured velocities
are plotted as abscissas and the
respective'depths as ordinates. A smooth
curve drawn through the plotted points
defines the velocity at each point in the
vertical. The are under this curve is equal
to the product of the mean velocity and the
total depth in that vertical line. This area
may be computed by using aplanimeter or
by Simpson's One Third Rule. The
vertical velocity curve method gives us the
most precise method of determining mean
velocity but requires only too much time.
2. Two-tenths and Eight-tenths Method:
The current meter is lowered
downward at 0.2 and 0.8 of the total depth
where observations are made. The mean
of this two velocities is taken as the mean
horizontal velocity in that vertical.
3 Six-tenths Method:
Only one observation is made at a
distance below the water surface equal to
0.6 the total depth of the stream. The
velocity obtained at that particular depth is
considered to be the mean velocity of
vertical.
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4. Integration Method:
The cllrrent meter is lowere at a
uniform rate down to the bed of the. tream
and is raised also at the same rate up to
the surface. The total time and the m mber
of revolution during this interval con itute
a measurement.
~
This method is based upon the th ry
that all horizontal velocities in the ve 'cal
have acted equally upon the meter w el
thereby giving the average as the mea~ 01'
all the velocity reading.
I
5. Subsurface Method:
In this particular method. the current
meter is held at just sufficient depth below
the surface usually 150 10m to 200 10m to
avoid surface disturbance. The mean
horizontal velocity is obtained by
multiplying the sub-surface velocity by a
coefficient. This coefficient varies with
the depth and velocity of stream. This
coefficient varies from 0.85 to 0.95.
Float Method of Measuring Stream Velocity
From the figure shown, a base line AB is
well selected and is established near the' bank
of a river where no obstruction will interfere the
line of sight during the observation period.
Points Cand 0 are established on the opposite
side of the river such that the sections AC and
BO are' perpendicular to the line AB, hence
they are parallel to each other. One transit is
set up at A and the other at B. The transitman
at B with vernier at zero, follows the float
where it is being released at point E, at a
distance of 15 m. above section BO. As the
float approaches section BO, the transitman at
A keeps the line of sight pointing at the float
until the transitman at B shouts "shot" a the
float passes section AB.
The transitman at A then clamps the lower
plate, turns the line of sight to the signal
station C and reads the angle 0. The
transitman at Balso follows the float, until the
transitman at A gives the "get ready' signal
and by means of the upper tangent screw
angle B is measured the moment the float
passes the section AC. The time that the float.
passes the section BO and AC is also
recorded.
I
The base line AB is then measured
accurately and the position C and 0 is then
plotted. The path of the float is either scaled or
computed using trigonometric principles. The
distance divided by the time gives the mean
velocity of the float.
Three Distinct Methods of
Determining the Flow Channels or in
. Open Channels or Stream:
1. Velocity-Area Method:
,
The velocities at any vertical line is
observed by using a current meter based
on the five different method of velocity
measurement using current meters. The
area of a certain section is obtained by
sounding, or by stretching a wire across
the stream and marking the points where
observations were made referred from an
initial zero point. The depths at this
particular points are also measured. The
area of the section could then be computed
by dividing the section into triangles and
trapezoids. The product of the area and
the mean velocity gives us the discharge
of flow of a certain section. The sum of all
the discharges at all sections gives us the
total discharge or flow.
2. Slope Method:
The 'slope method involves a
detennination of the following:
aj Slope of water surface.
bj Mean area of channel cross-section
cj Mean hydraulic radius
d) Character of stream bed and the proper
selection of roughness coefficient
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I
HYDROGRAPHIC SURVEYING
Mean Velocity is computed by
applying the Chezy Formula:
V=c{RS
3..)"'_00,
~il
A weir method is an obstruction place
in a channel, over which water must flow.
D' scharged of a stream using this method
i valves the necessary information.
where
V = mean velocity
C = coefficient of roughness of stream
bed
R = hydraulic radius
R=pA
c)
A :: cross-sectional area of stream
P = wetted perimeter of stream
d)
e)
~
Computing values of C by
Kutter's Formula:
g)
h)
Depth of water flowing over the crest
of weir, H.
Length of crest, L for rectangular or
trapezoidal weir.
Angle of side slopes if weir is
triangular or trapezoidal.
Whether flat or sharp crested.
Height of crest above bottom of
approach channel, P.
Width and depth of approach channel
Velocity of approach
Nature of end contractions
a) English:
C=
41.65+ 0.00281 + 1.811
s
n
1+
(41.65 + 0.0~281).
..JR
C = coefficient of roughness of stream
bed
n =retardation factor of the stream bed
R = hydraulic radius
s = slope of water surface
End Contracted Weir
Q = 1.84 (L - 0.2H)
b) Metric:
H312
23 +0.00155 +1
c=
s
n
1 + ~ (23 + 0.00155)
{R
s
Computi~ values of C by
Manning's Formula:
Rl/6
C=n
Discharge = Area x Velocity
Suppressed Weir
Q::: CLH3/2
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Three common types of floats used
in measuring stream velocity:
1. Surface floats - it is designed to measure
surface velocities and should be made
light in weight and of such a shape as to
offer less resistance to floating debris,
wind, eddy currents and other extraneous
forces. The use of surface float ;s the
quickest and the most economical method
of measuring stream velocity.
Triangular Weir
Q = 1.4 H2.5
2. Sub-surface floats - this is sometimes
called a double floats. It consists of a
small surface float from which is
suspended a second float slightly heavier
than water. The submerged float. is a
hollow cylinder, thus offering the same
resistance in all directions and the
minimum vertical resistance to rising
currents.
Cipolletti Weir
1.86 LH3I2
o 1
when tan-::;:Q::;:
3. Rod float - the rod float is usually' a
2 4
cylindrical tube of thin, copper or brass
25 mm to 50 mm in diameter. The tube is
sealed at the bottom and in weighted with
shot until it will float in an upright position
with 50 mm to 150 mm, projecting above
the surface of the water.
Q ::;: 1.84 LH3/2 (Francis Formula Neglecting
Velocity of Approach)
Q::;: 1.84 L [(H + hv)3/2 • hv 3/2 ] (Considering
Velocity of Approach)
Instruments used for measuring
difference in level of water:
Discharge measurements are made for the
following purposes:
1. To determine a particular flow without
regard to stage of stream.
2. To determine flows for several definite
gage readings throughout the range
stage, in order to plot a rating curve for the
station. From this curve the discharge for
any subsequent period is computed from
the curVe of water stage developed in the
recording gage.
3. To obtain a formula or coefficient of dams,
or rating flumes.
1.
2.
3.
4.
5.
6.
7.
Hook gauge
8taff gauge
Wire-Weight gauge
Float gauges
Automatic gauges
Piezometers
Plumb bob
of
Instruments used for measuring
the velocit of flow:
1. Floats
a) surface float
b) sub-surface float
c) rod float
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2. Current meters
a) Those which the revolving element is
cup-shaped, or of the anemometer type
and acts under differential pressure.
Types of current meter:
1.
2.
3.
4.
5.
Price meter
Ellis meter
Haskell meter
Fteley meter
Ott meter
A. Measurement of dredged materials:
Measurement in place:
Soundings of fixed section are
taken both before and after dredging
and the change in the cross-sectional
area is obtained by calculation or by
using a planimeter. The volume of the
material removed is computed by
using the borrow pit method or by the
end-area method.
2. Scow measurement:
Each scow is numbered and the
capacity of each is carefully
determined. When the scow is filled
to the capacity the inspector records
the full measurements. Materials is
scow is sometimes measured by the
amount displaced in loading.
B. Measurements of Surlace Current:
Certain engineering problems require
important information about the direction
and velocity of currer)ts at all tidal stages.
This. is done by locating the path and
computing the velocity of floats from points
whose locations are known and can be
determined. Floats should be designed to
give minimum wave resistance and to
extend underwater to a sufficient depth to
measure the current in question. The
direction of the current may be determined
by sextant angles from the boat between
. known signals and the floats.
C. Wire drag or Sweep:
This method is used in harbor or a bay
Where corral reefs and pinnacle rocks are
likely to occur. This consists of a wire of
any length up to 120 m. which may be set
at any desired depth. Depths are
maintained by means of bouys placed at
the wires and whose length can be
adjusted. The drag is pulled through the
water by means of a power launches,
steering diverging forces to keep the drag
taut. When an obstruction is met, the
bouys are shown with the position of two
straight lines intersecting at the
obstruction. These intersection is located
by sextant observations to reference
points on the shore. Soundings are taken
for the minimum depth.
D. Determination of stream slope:
To determine surface slope, a gauge
is installed on each side of the. stream at
the end of the section. The zero's of the
gauge are connected to permanent bench
marcks on the shore. The gauges are read
simultenaously every ten to fifteen minutes
for six to eight hours. The mean of these
elevation at that point of the stream. The
difference in elevation between the ends of
the section divided by the distance is the
slope.
Capacity of Existing Lakes
or Reserviors:
•
1. Contour Method:
A traverse is run from a shore line and
the desired shore topography are located
by stadia. Take sufficient number of
soundings by any method suited 'for the
particular job and plot the sub-ageous
contour. The area inclosed between
contours are determined by planimeter.
The average area of two consecutive
contours multiplied by the contour interval
gives the partial volume. The summation
of the partial volumes gives the total
volume.
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2. Cross-Section Method:
The outline of the water line is
obtained as in the contour method. The
water line is then plotted and divided into
approximate trapezoids and tri-angles.
Soundings are taken along the boundary
lines between each station and are plotted
on cross section paper. A perpendicular
distances between sections are then
obtained by the end area method. The
summation of these partial volumes gives
the total volume.
. Two General Methods of Determining
the Capacity of a Lake or Reservoir:
1. Contour Method:
a) End-area method
b) Prismoidal formula
The' areas A1, A2, etc. are determined
by using a planimeter and h represents t~e
contours interval. Area below A5 IS
neglected.
b) prismoid,1 Formula:
L
V =6(A +4A", +A2)
In this case the midqle area Am is the Area
A2 and ~ while L is equivalent to 2h.
2. Parallel Cross-Section Method:
a) End-area method
b) Prismoidal formula
a) End-area method:
a) End-area method:
Parallel ranges are laid out across the lake and
soundings are then taken along the ranges.
From the observed sounding the corresponding
cross-sections could be plotted and its
corresponding areas would then be computed.
5-202
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F'rOll) • • the.culTent~~Wtlo~s • take~8r • th~
R<:tsjgBIYtr,m~m~i6BpU~toftheh~9Un~
·~lll<,e, • • AlI.~~~lJr~m~I'l~l!r~ll'lm~~~, • • · ·
Total Volume = VI + V2 +V3 + V4
+Vs +Vs
b) Pismoidal Formula:
The problem arises here in the
determination of Am. since the distances
between parallel sections are not equal, it
is therefore necessary to evaluate or
interpolate the values of Am.
Solution:
CD Total discharge:
V1 =!!J.
6 (A l +4Am + A2)
V2 =b2
6 (A2 +4Am + A3)
V3 =&
6 (A3 +4Am + At)
V4 =·~(At +4Am +As)
Vs =b5.
6 (As +4Am + Ae,)
h
Vs =-W (~+4Am +A7)
v = (0.32 + 0.22)
a
2
Va =0.27 m/sec
l / _ (0.40 +0.24)
Vb 2
Vb =0.32 m/sec
Vc = 0.21 m/sec
Velocity:
- (0 +0.27)
V12
V1 = 0.135
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Vr
- (0.27 + 0.32)
2
V2 = 0.295
_(0.32 + 0.21)
V3-
2
V3 = 0.265
_ 0 + 0.21
\I
2
V4 -
V4 = 0.105
.0;739
0.720···
Discharge: Q =AV
Q1 = 15 (0.135) = 2.03
~ = 43.5 (0.295) = 12.83
OJ = 39.75 (0.265) = 10.53
C4 = 10 (0.105) = 1.05
Total Q = 2.03 + 12.83 + 10.53 + 1.05
Total Q= 26.44 cu.m.lsec.
1 cU.m. = 1000 liters
Total Q = 26440 liters/sec
1243
0.852
0,524
.0.473
0.469
(j)
Compute the veloCity at distance of 30 m.
from l.P.
@
Total area:
- 12(2.5)
A1A1 = 15.00
- (2.5 +3:3)(15)
Ar
2
.
® . Compute the discharge in ~tersfsec.·
@ .•
Solution:
(j)
Velocity at distance of 30 m. from IP.
A2 =43.50
A (3.3 + 2)(15)
3-
2
A3 = 39.75
A4=~¥
A4 = 10.00
V= 0.739 + 0.720
2
V= 0.7295
Total area:
A = A1 + A2 + A3 + A4
A = 15 +43.5 +39.75 + 10
A = 108.25 sq.m.
@
Discharge in liters/sec:
Va =045
Vb =0.739; 0.720 =0.7295
® Mean velocity:
Q
V=f1
v=}644
108.25
V =O. 244 mlsec.
.
Compute lliemean veloCity in section. . ..
V =1.251 ; 1.243 = 1.2465
c
Vd =0.852
Va =0.524
V - 0.473-+ 0.469 0.471
f-
2
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V1 = 0 + ~.45 =0.225
V2 = 0.45 +20.7295 = 0.5898
V
3
!A~@~lllflo\l{rllM~Qfem~~tWe~Cp®99~.M
0.7295;1.2465 = 0.988
~l'iYMB$ll'lg~¢(lI'l'~tm~WWlmth¢
ll(lrr~(j119ioo.\f~IV@(lf.Ptl.~sWN~
• ~~m.·il~~ .•":pfl.•.•
Th~fpll(lWl~9@·W.tb~.·9b$~~dJW~.·ta~M··
V4 = 1.2465; 0.852 = 1.049
d@Mtl'tEl.@l~$9~M~l'lt$· • • P~IYQ·gl'@l@d
W~$U$l¥lirl;ob$~N<ltl()rlL • • >}.·•• ·•·• •·• • ·• · • · ·•· ·
.
V = 0.852 ; 0.524 = 0.688
s
Vs = 0.524 ; 0.471 = 0.4975
V7 = 0.47~ + 0 =0.2355
A1 = 10.5~1.25}=6.5625
A = (1.25)
2
+p.7} 10 -14.75
cD • • • GOrrlpu~.th¢VeIOMy~la.~J$@H;~Qf···· • •16·
rn·frQ(l1\ME:'/
A = (1.7 +;.3) 10 = 20
3
@ Deternljn~the8iS@.~tge9~tffl~~et.
@ .. O~term1nEl.!hel'llMIl·VEllocJtYl)lltlllll'i'l~
.•••••.•.
A4 = (2.3 +;85) 10 =25.75
As = (2.85 +;.55) 10 = 22
Solution:
CD Velocity at distance of 16 m. from WE
As = (1.55 +20.9) 10 = 12.25
A7 = 8.35~0.90} =3.7575
Q=AV
Area
A1 =6.5625
Ai = 14.75
A3 =20
~ = 25.75
As=22
As = 12.25
6I~
A = 105.07 m2
@
Velocity.
Discharge
0 1 =1.477
O2 = 8.700
0 3 = 19.760
<It = 27.012
0 5 =15.136
Os = 6.094
V, =0.225
V2 =0.5898
V3 =0.988
V4 =1.049
Vs =0.688
Va =0.4975
V7 =0.2355
Mean velociy:
Q=AV
79.064 = (105.07) V
V= 0.752m1s
.Qz~
0= 79.064 m3/sec
= 79064 liters/sec
o
V=aN+ b
V=0.232 (~~) + 0.022
V= 0.1782 m/s
@
Discharge of a certain river:
Va=O
V =aN +b (Straight line equation for
current l7)eters)
N_ revolution
sec
(1Ql
Vb =0.232 50 +0.022
Vb =0.0684 m1s
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Vc -- 0.232 @
55 + 0.022
0, =A 1V,
0, :: 3.2 (0.0342)
0, :: 0.10944
~ :: 8 (0.0916)
~ =0.7328
OJ:: A3 V3
OJ =11.2 (0.1465)
OJ:: 1.6408
~ ::A4 V4
~ =10.4 (0.1614)
~ = 1.6786
Os:: 0.2892
Vc :: 0.1148 mls
_
@§l
Vd - 0.232 52 + 0.022
Vd :: 0.1782 mls
_
@l
Ve - 0.232 53 + 0.022
Ve :: 0.1446 mls
V,:: 0
V1::~
2
_ (0 + 0.0684)
V12
0::01+~+OJ+04+0S
V1 = 0.0342 mls
V2 ::
o=0.10944 + 0.7328 + 1.6408 + 1.6786
~
2
+ 0.2892
0:: 4.4508 m3/s
V _ (0.0684 + 0.1148)
r
2
V3 = 0.1465 mls
V-~
42
V _ (0.1782 + 0.1446)
2
4-
V4 = 0.1614 mls
~
Vs = 2
Vs = 0.0723 mls
A - 1.6 (4)
,- 2
A1 = 3.2
A - (1.6 + 2.4)(4)
2
2-
A2 =8
A _ (2.4 + 3.2) 4
32
A3 = 11.2
A _ (3.2 + 2)4
42
A4 = 10.4
As ::fHl
2
As =4
o:: 4450.8 liters/sec
2
2-
V2 :: 0.0916 mls
V _ (0.1148 + 0.1782)
@
Mean velocity:
A:: A1 +A2 + A3 +A4 +As
A =3.2 + 8 + 11.2 + 10.4 + 4
A = 36.8 sq.m.
V::.Q
A
V= 4.4508
36.8
V:: 0.1209 mls
The areas bounded by the water line Of a
reservoir is determined by using a planimeter.
The contour interval is 2m.· A, <z .20,400 sq.m.,
A 2 " 18,600 sq.m., A3 ",. 14,300 sq,m.,
A4 = 10,200 sq.m., As:: a,ODO sq.m. and
Ae '" 4,000 sq.m. Determine the following:
G) Ehd area method.
® Prismoidal formula.
@ What is the difference of capacity of the
~~Sr:f~~ using End area aM by Prlsmoldal
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Solution:
cD End area method:
2
V1 =2" (20400 + 18600) = 39000
2
V2 =2"(18600 + 14300) = 32900
V3 = ~ (14300 + 10200) = 24500
2
V4 =2"(10200 + 8000} :: 18200
2
Vs =2" (8000 + 4000)
= 12000
126600 m3
@
Prismoidal formula:
V1 = ~ [20400 +4(18600} + 143001
V1 = 72733.33
V2 = ~ [14300 + 4(10200} + 80001
V2 :: 42066.67
2
V3 =2" (8000 + 4000)
V3 :: 12000
Prismoidal Formula
•
=V1 +V2 +V3
= 72733.33 + 42066.67 + 12000
:: 126800 m3
@
Difference of capacity of the reservoir
using End area and by Prismoidal
Formula:
Diff. in volume = 126800 -126600
Diff.1n volume:: 200 m3
SECTION 2
SECTION 3
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- (315 +314) (30)
V3-
2
V3 =9435 cU.m.
V. _(A4 +As)h
2
41/
SECTION 4
Solution:
V _(A l + A?) h
12
Total volume = V, + V2 + V3 + V4
V= 2078 + 10456 +9435 +4710
V=26,679cu.m.
A,=O
A = 10 (4) + (4 + 6) (10) +(6 + 5) (12)
2
2
'
2
+ (5 +4.2) (12) 8 (4.2)
2
+ 2
@
A2 = 20 + 50 +66 + 55.2 + 16.8
A2 =208 sq.m.
- (0 + 208) (20)
V12
V, =2080 cU.m.
V
2-
2'
V4 = 4710 cU.m.
CD End area method:
2
= (314 +0}(30)
V4
Prismpidal formula:
Note:. To solve for Am: compute the
dimensions of Am using the average
values of the sections (1) and (2).
(A2 + A3 ) h
2
A =~ ~ (8+10)(10)
32+
2
+
2
FROMl&2Am
rn
+ (10 + 7) (12) (7 +4) (10)
2
+
2
+ 2
A3 = 12 +44 + 90 + 102 + 55 + 12
A3 = 315 sq.m.
Vr
-@§. + 315) (40)
2
V2 = 10,460 cu.m.
2
A = 6 (12) +(6 +10) (14)
4
+(2.5+ 2.1}(6) 4 (2.1)
2
+ 2
Am = 5+ 12.5 + 16.5 + 13.8 +4.2
Am =52sq.m.
V _(Aa +&)h
r
A =~+~(2.5+3)(6)
m 2
2
2
2
2
(10 + 8)(14) 8(10)
+
2
+ 2
A4 =36+ 112 + 126+40
A4 =314 sq.m.
L
V, = 6(A 1 + ~ + A2)
20
Vj ="6 [0 +4(52) +208]
V, = 20 (208 + 208)
6
V1 = 1386.67 cU.m.
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HYDROGRAPHIC SOVEYI.I
From (2) to (3). Use average dimensions of
sections (2) and (3).
\I
_
vr
4 (A 3 + 4 Am + A3 ) h
6
V3 = 3~ [315 + 4 (236.25) + 314]
V3 = 7870 Cu.m.
From (4) fa (5). Use average dimensions of
sections (4) and (5).
Am
A - 9(3.5) (3.5 + D(9) (7 +7.5) (11)
m2 +
'2
+
2
+
(7.5 +5.6) (12) (5.6 + 2) (9) £.@l
2
+
2
+ 2
Am = 15.75 + 47.25 + 79.75 + 78.6 + 34.2 +3
Am = 258.55 sq.m.
_;u§l (3 + 5) 7(S +4) (7) iill
2
2
+ 2
V - (A 2 + 4 Am + A3) h
r
6
Am- 2 +
V = [208 +4 (25:.55) -;. 315] (40)
Am = 9 + 28 +31.5 + 10
Am = 78.5 sq.m.
2
V2 = 10,381.33 cU.m.
\I
_
V4 -
From (3) and to (4). Use average
dimensions of sections (3) and (4).
(A 4 + 4 Am + As) h .
6
30
V4 = 6 [314 +4 (78.5) +0]
V4 =5 (314 + 314)
V4 = 3140 cu.m.
Total volume = V1 + V2 + V3 + V4
V= 1386.67 + 10381.33 + 7870 +3140
V= 22,778 cU.m.
A =4.5(10) (4.5+9)(11) (9+9)(12)
m
2 +
2
+
2
+ (9 +3.5) (11) (3.5 +2) (5) £.@l
2
+
2
+ 2
Am = 22.5 + 74.25 + 54 + 68.75 + 13.75 + 3
Am = 236.25 sq.m.
@
Difference in capacity:
Difference in capacity =26685 - 22778
Difference in capacity =3,907 cU.m.
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Pdfbooksforum.com 5·209
THREE POINT PROBlEM
Considering triangle ACB:
:; 850 Sin l?J
O CB
Sin 41'30'
A, Cand 0 are three tliangulationshoreslgnals
whose positions were determined· by the
angles W:; 150' and the sides AC ,,; flSO m.
and CO :; 760 m~ A sounding at B was taken
from a boat and the angle$ E '" 41'30' and the
angles E" 41'30' and F" 35:30' were measured
simultaneously by two sextants from the boat
to the three shore signals from the shore,
c
@
Considering triangle BC:
CB
760
Sin a - Sin 35'30'
760 Sin a
CB:; Sin 35'30'
o and@
850 Sin l?J 760' Sin a
Si041'30':; Sin 35'30'
Sin
l?J:;
1.02 Sin a
'" +a + 150' +41'30' +35'30':; 360
A
1Il+·a:;133'
B
a:; (133' -Ill)
Sin l?J:; 1.02 Sin (133' -l?J)
Sin III :; 1.02 (Sin 133' Cos III - Cos 133' Sin Ill)
Sin III :; 0.746 Cos III +0.696 Sin III
0.304 Sin III :; 0.746 Cos III
tan Ill:; 2.454
l?J:; 67'50'
Solution:
CD Distance AB:
a:; 133' - 67'50'
c
a:; 65'10'
850 _ AB
Sin 41'30' - Sin 70'40'
AB:; 1210.45 m.
A
® Distance BD:
B
C
A
BD
760
Sin 79'20' - Sin 35'30'
BD =1286.14 m.
'1' Distance CB:
CB
850
Sin 67'50' - Sin 41'30'
CB:; 1187.98 m.
B
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5-210
THREE POINT PROBLEM
A hydrographic survey was conducted to
·locate the· po.sition of soundlngs,Three
statiOl'\sXYZwere. established on the
seaShore andthesoundiogs Were observed at
pdlnt A uSitiga small bqaC The fall. data were
recorded in order to plot the position· of the
sOUndings.··
....•.
Three· shore stations A. C and Dare
triangulatron· points whose. position a.s
observed· ·from B where soundings are
obserVed and the angles were measured using
two· sextants from· the. boat ara to the Wee
shore signals, The following dala were
recorded during the sounding observation. .
·····.Djstaild~XY= 1200m..
. DistanceYZ =1809
... Angle ACO :::: 150'
m.
Angle ABC = 42'30'
Angle CSO = 35'30'
.....•...•. An Ie XYZ= 140' ....
. An9g,eXAY'=40'
••...
...
.....
Angle CAB =u7'50'
Distance AC =850 m.
Distance CD =760 m.
•.. . . AngleZAY:: 36'
Angle YXA '" 68'
Solution:
CD Angle YlA:
Solution:
CD Angle CDA:
c
z
x
A
B
e + 150' + 67'50' +42'30' + 35'30 = 360'
Angle YlA =180 - 68·36
Angle YlA = 76'
® Distance AX:
AX
1200
Sin 72"; = Sin 40'
AX = 1775.50 m,
@
Distance AY:
1200
AY
Sin 40' = Sin 68'
. AY = 1730,93 m.
e= 64'10'
® Distance AB:
ACB = 180' - 67'50' - 42'30'
ACB=69'40'
850
AB
Sin 42'30' =Sin 69'40'
AB = 1179.76 m.
@
Distance BD:
760
SD
Sin 35'30' = Sin 80'20'
BD = 1290.18 m.
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S-211
Pdfbooksforum.com
THREE POINT PROBLEM
901.76 Sin 0 :: 925.12 Sin (149'30' - 0)
901.76 Sin 0: 925.12 (Sin 149'20' Cos 0
- Sin 0 Cos 149'30')
901.76 Sin 0:: 169.53 Cos 0 + 797.11 Sin 0
104.65 Sin", :: 469.53 Cos '"
tan",:: 4.487
o :: 7T27' (angle DCB)
Three shore stations C,O and E are
triangulation observation points with CD .", 615
m. and DE ;;625 m.Angfe EOC is 125. A
hydrOgrapher point 0 wantectlo know his
at
position With. respect to· the triangulation
points, he measured angle ceo: 43' and
DBE : 42'30'.
CD Compute the angle DCB.
Compute the angle COB.
@ Compute the distance BC.
® Angle COB:
{J:: 149'30' - 77'27'
Angle COB:: 180'~· 77'2T - 43'
Angle COB:: 59'33'
@)
@
Solution:
CD Angle DCB:
D
Distance BC:
BC
615
Sin 59'33' Sin 43'
BC:: 777.38 m,
c
E
B
BD
615
Sin 0 ::Sin43'
:: 615 Sin 0
BD
Sin 43'
BD :: 901.76 Sin 0
In the accompanying figure, A, Band C
are three known control stations and P is
the position of a sounding vessel which
is to be located. If b : ;: 6;925.50 m.,
e: 6,708.40 m, angle BAG: 112'45'25", angle
alpha: 25'32'40", and angle beta: 45'35'50",
(Cllfllf\l\ Station)
B
80
625
Sin {J :: Sin 42'30'
BD:: 925.12 Sin {J
901.76 Sin 0:: 925.12 Sin {J
0+ fJ + 125' + 43' + 42'30': 360'
o + {J :: 149'30'
{J:: 149'30' - 0
ill Compute the value of angle x.
® Compute the value of angle y.
@ Compute the length of line NJ.
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5-212
THREE POINT PROBlEM
Solution:
CD Angle x:
B
AP _ 6708.40
Sin x- Sin 25'32'40"
AP = 15557.11 Sin x
AP _ 6925.50
Sin y - Sin 45'35'50"
AP = 9693.62 Sin y
9693.62 Sin y= 15557.11 Sin y
Sin y = 1.605 Sin x CD
x + (360 -112'45'25'') + Y
+ 45'35'50" + 25'32'40" = 360
y= 41'36'55" - x
@
Sin (41'36'55" - x) = 1.605 Sin x
Sin 41'36'55" Cos x - Cos 41'36'55" Sin x
= 1.605 Sin x
Sin 41'36'55" Cos x
= (1.605 + Cos 41'36'55") Sin x
Sin 41'36'55"
1.605 + Cos 41'36'55" = tan x
Solution:
CD Length of AO:
a
(0
= 360 - (245'23'22" - 93)
(0
=207'36'38"
x = 15'45'50.17"
AO
6671.50
Sin a - Sin 20'05'53"
@
Angle y:
Subt. to equation @
y = 41'36'55" - x
y = 41'36'55" - 15'45'50.17"
Y= 25'51'04.83"
AO = 19414.9 Sin (I.
AO _ 12481.70
Sin f!, - Sin 35'06'08"
AO =21705.91 Sin f!,
@
Length of line AP:
AP = 15557.11 Sin x
AP = 15557.11 Sin 15'45'50.17"
AP = 4226.47 m.
21705.91 Sin f!,
=19414.9 Sin (I.
Sin f!, = 0.894 Sin
(X
(I.
~i
+ fl) + 13 + 35'06'08" + 20'05'53" = 360
13 =97'11'21" - (1.
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S-213
Pdfbooksforum.com
THREE POINT PROBLEM
Sin (97'11'21"· ex) = 0.894 Sin ex
Sin 97' 11 '21" Cos ex • Cos 97'11'21" Sin ex
=0.894 Sin ex
Sin 97'11'21" Cos ex
= (0.894 + cos 97'11'211 Sin ex
Sin 97'11'21"
0.894 + Cos 97'11'21" = tan ex,
ex = 52'13'34.41"
AO = 19414.9 Sin 52'13'34.41"
AO = 15346,22 m.
® Angle ACO:
fJ = 97'11'21" • 52'13'34.41"
fJ =.44'57'46.59"
@
Triangulation stations C, 0 andE are
established by thePhils. Coast and GeOdetic
Survey with observation points rEJcorded as
CO =615 m. and DE =625nt Ahydrographer
at point 8 wanted to knoW his position with
respect to the triangulation stations; he then
measuted angles cao ;; 43' and DBE ';i: 42'30':
Angle EOC is 125'. Angle DCB =77'26'.
(j)
@
@
Solution:
CD Distance BC:
D
Length of line OC:
AO = 21705.91 Sin 44'57'46.59"
AO = 15338.47 m
Ave. AO
Compute the distancaBC,
Compute the distance SO:
Compute the dislanceBE.
E
c
= 15346.22 + 15338.47
2
Ave. AO = 15342.32 m.
B
8 =180' - 20'05'53"·52'13'34.41"
8 = 107'40'32.5"
BC
615
Sin 59'34 =Sin 43'
BC =777.52 m,
OB
6671.50
Sin 107'40'32.5" Sin 20'05'53"
OB =18498.33
@
" =180·35'06'08"·44'57'46.59"
,,= 99'56'5.41"
_ _O_C
12481.70
Sin 99'56'5.41" - Sin 35'06'08"
OC =21380.42 m.
Distance BD:
BD
615
Sin 77'26' =Sin 43'
BD =880.16 m.
;t Distance BE:
BE
625
Sin 65'26; =Sin 42'30'
BE =841.37 m.
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5-214
MINE SURVEYING
® Angle between strike and drift:
B
Vein . a relatively thin deposit of mineral
between definite boundaries.
c
B
Strike· the line of· intersection of the vein with
a horizontal plane.
Dip· the vertical angle between the plane of
the vein and horizontal plane measured
perpendicular to tne strike.
Outcrop . the pornon of the vein exposed at
the ground surface.
Drift . an inclined passage driven in a
. particular direction.
D
Sin fII = BC
AC
tan43'40'=. BC
CD
BC = CD cot 43'40'
2 aJ
100 = AC
AC=50 CD
.A.~ttJba$.~.$t~kePf.N, • 10'1$fW.• Mda.dlp••hf
4a~4QjlN .• A~r1ffil'lI~~Yeil1M~ilg~pf~%.
• W·•• W~~ • ~.tt@>.*Mhg.A1.w~Y~m~I.PlaQe
¢9.~PMi~Sm~dlp~)
Ij,.fr~fl~
Solution:
CD Bearing of dip:
AC
. - CD cot 43'40'
SInfllCD50
fII
@,
= 1'12'
Bearing of drift:
Bearing:: 10'15' + 1'1 'Z
Bearing = N. 11'27' W
A vein has a dip of 57' W. The bearing of a
drift is N. 37'W, having a grade of5% With the
plane oUhe vein.
..
c
D
Sin fII = BC
Dr~
Bearing =90' ·10'15'
Bearing = S. 79'45' W.
CD Compute the honzontal angle be.tween the
strike and the verticalprojectlon of lhe drift,
® Compute the bearing afthe strike.
@
Compule the bearing of the vertical plane
containing the dip.
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5-215
Pdfbooksforum.com
MINE SURVEYING
Solution:
@
Bearing ofstrike:
Bearing = 31' - 1'52'
Bearing::: N. 35'08' W
@
Bearing of the vertical plane containing the
dip:
Bearing =90' - 35'08'
Bearing::: S. 54'52' W
CD Horizontal angle between the strike and the
vertical projection of drift:
The eente~ine of amine lu nnel runs through A
and S, each 1575 m. in altitUde, with a
distance between AS equal to 1585 m. The
tunnel bears S.70' E. from A. On the other
side of the hill, 1136 m., N. 77' E. of A is a
point C at 2142 m. affitude.
.
D
<D Determine the bearing of the shortest
possible tunnel from eta AB.
® Determine the dip of the shortest possible
tunnel from C to AB.
@ Determine' the length of the shortest
.possible tunnel from eta AB.
Solution:
CD Bearing ofshortest tunnel from C to AB:
Sin", =BC
AC
tan 51' = CD
N
BC
BC= CD cot 51'
2.~CD
100 - AC
AC=20CD
Sin", =BC
AC
. . CDeot51'
SIn '" = 20 CD
,,::: 1"52'
B
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5-216
MINE SURVEYING
8 = 180' - 77' - 70'
8=33'
Bearing = S, 20' W.
Drill holes are bored through points A, Band C
until It strikes the mineral ores, Point A is 400
m. due south of Band point Cis 300 m, N.60'
® Dip ofshortest tunnel from C to AB:
E.ofB.
....
N
B
c
(j) • • Cornpute•• the.differe~ • in•• elevl3.1ion.of. the
. $lJffaceofOreatAandC;···
Solution:
567
A4:;..--------ID
CD Difference in elevation ofthe surface of ore
at Aand C.:
C
Elevation of mineral ores:
POINTS
ELEVATION OF ORES
A'
. 450 -165 = 285 m.
567
S'
C
A 4;...---I.:...-------ID
(AD)2 = (1136)2 - (567)2
DE= 984.38 Sin 33'
DE = 536.13 m.
567
tan IJ = 536.13
IJ = 46'36' (dip)
Shortest tunnel from C to AB: .
(EC)2
=(536.13)2 + (567)2
EC =780.37 m.
470 -187 =283 m.
485 -203 = 282 m.
Diff. of elevation (If the surface of are at A .
and C =285 -282
Diff. of elevation of the surface of are at A
andC=3m.
CD = 2142 -1575
CD=567m.
AD = 984.38
.
® cOIllpUtetfiflbeatinggfstfike.
@ COl1'lpufetheal1g!e ofJ@dip...
@
Bearing ofstrike:
N
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5-217
Pdfbooksforum.com
MINE SURVEYING
D
400-x_!
3
-1
3x =400 + x
2x =400
x=200
(C'D) =(200)2 + (300)2. 2(200)(300) Cos 60'
C' 0 = 264.58 m.
From point A at the opening of a tunnel. a
surface traverse is run on the side hill and an
underground traverse is run through the tunnel,
Both traverse are oriented from the same
meridian. Below are the computed latitudes'
and departures. Consider the line AF is a
straight Une on the surface and the slope AF is
uniform upward. Also assume that all points of
the under ground traverse are on the same
elevation whBe point Fis 33 meters above A
Using Sine Law:
Sin 13 _ Sin 60'
300 - 264.58
0=79'06'
Bearing ofstrike::: N, 79'06' W.
® Angle of the dip:
(j)
Find the bearing of line AF.
~ne AF.
Compute the shortest distance of a shaft
from point 4 in the tunnel of the surface
along line AF.
® Find the distance of
@
Solution:
CD Bearing of line AF:
'~
B"
10639
B" E =200 Sin 79'06'
B" E= 196.39 m
td' _ _
1_
.an Ip - 19639
dip =0'17'
E
Line AF:
Lat. =31.2 -17.4 - 3.7 + 22.8 + 12.1
Lat. =+ 45.0
Dep. =40.5 + 34.6 + 66.0 + 39.9 + 55
Dep. =+ 236.5
tan
13
=~
Lat.
236.3
= 45.0
13 =79'13'
Bearing =N. 79'13' E.
13
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S-21S
MINE SURVEYING
® Distance AF:
N
F
38m
.
_ Departure
Distance AF - Sin 79'13'
Distance AF = 240,55 meters
A line from A elevation 17.13 m. intersects the
veln at 8, elevation 54 - 85 m. The bearing
and slope distance of AS are N. 47'31' E. and
63}O m. respectively, Strike is $,71'33' E.
Dip is 50' $W;
CD Compute !he direction of the shortest level
cross cot from A to the vein.
.
..
@ Compute the length of the shortest level
cross cut from A to the vein.
..
.
@ Determine the shortest distance from A t6
·ihevein.
® Shortest distance of a shaft from point 4 in
the tunnel of the surface along line AF,
Line A - 4:
Lat. ="82.5 - 8.8 - 5.7
Lat. =-117.0
Dep. =25.0 + 50.4 +40.0 + 70.4
Dep. = 185.8
. A 4 185.8
tan beanng - = 117
tan bearing A - 4 = S 57'48' E.
.
_ Departure
Distance A - 4 - Sin 57'48'
Distance A - 4 = 219.58 m.
" = 180' - (79'13' +57'48')
" =42'59'
AO = 219.58 Cos 42'59'
AO = 160.63 meters
h
33
160.33 =240.55
h = 22.037 meters
Distance 04 = (A - 4) Sin 42'59'
Distance 04 = 219.58 Sin 42'59'
Distance 04 = 149.7 meters.
h
tan a= 0-4
(X
22.037
= 149.7
a = 8'22' 26"
0-4
d = Cos 8'22' 26"
d = 151.32 meters. (shortest distance of
the shaft)
A
Solution:
CD Direction of the shortest level cross cut
from A to the vein:
Direction Of the shortest level =47'31' - 29'04'
Direction ofthe shortest/evel =N. 18'27' E.
® Length of the shortest level cross cut from
A to the vein:
,--------
AD = ."j (63.70)2 - (37.72)2
AD = 51.33 m.
AE = 51.33 Cos 29'04'
AE=44.87 m.
EF=37.72 Cot SO'
EF= 31.65 m.
AF= 44.87 - 31.65
AF =13.22 m. (shortest level cross out
from A to the vem)
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S~219
Pdfbooksforum.com
MINE SURVEYING
@
Shortest distance from A to the vein:
AH = AF Sin 50'
AH = 13.22 Sin SO'
AH= 10,13m,
@
Length of a + 1.5% tunnel to meet the vein:
CD
350
Sin 3D' =Sin 128'
CD =222.08
CE
222.08
Sin 128' =Sin 51'08"
CE= 224.76ft.
A vein dips to the west at an angle of 52'. A
hill side assumed 10 be sloping uniformly has
an angle of depression of 22'. tram the
outcrop of the vein, the sloP'! distance along
the hillside to the top of the shaft and mouth of
the tunnel are respectively 250 ft. and 35Q fl,lf
the tunnel is driven al right angles to the strike
and the shaft is sunk vert~lly.
CD Determine the height of the shaft
® Determine the length of a + 1:5% tunnel 10
meet the vein,
.
@ Detennlne the distance from the outcrop to
the bottom of the shaft.
@
Distance from the outcrop to the bottom of
the shaft.
AE
203.03
Sin 112' =Sin 30'
AE =376.49 m.
A point B at the bottom of a winze has a
vertical angle of • 65'23' sighted with the top
telescope of a mining transit. The slope
distance to a poinl B from the inslrument at A
.is 295.87 ft The eccentricity of the telescope
is 3 inches.
zD Compute the corrected vertical angle.
® Compute the elevation of B if A is al
elevation 500 ft. and the H.l. is + 5.5 ft. and
H.PI. is· 3.3 ft.
@ Determine Ihehorizonlal distance between
AloS.
Solution:
CD Corrected vertical angle:
Solution:
-:'1)
Height of the shaft:
BE
250
Sin 30~ =Sin 38'
BE = 203.03 m.
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S-220
MINE SURVEYING
ilri~I_~~I~.INII'~'lill
;111".'ki?;$i'~'!I~~
3/12
tan", =295.87
",=0'03'
Corrected vetfical angle = 65'23' - 03'
Corrected vertical angle = 65'20'
® Elevation ofB if A is at elevation 500 Fr.
and the H.I. is + 5.5 Fr. and H.PI. is - 3.3 Fr.
h =295.87 Sin 65'20'
h=268.87
A
B
~
C
~
EI. ofB = 500+ 5.5- 268.87 +3.3
EI. ofB = 239.93 ft
® Horizontal distance between A to B:
A"r--.-_ _..,
B
c
Solution:
G)
Direction of the line of intersection of the
veins:
IJ = 63'25' + 77'10'
IJ = 110'35'
In .1 ABC:
BC = AB tan 22'
In.1 BOD:
BC = Bo tan 35'
In.1 ABO:
AB= OB Sin f1J
In.1 aBO:
B
tan 65'20' =268.87
x
x= 123.48 ft
D
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MINE SURVEYING
BD = OB Sin (fJ· Ill)
OB= ~B
.
Sm III
OBBD
- Sin (fJ - Ill)
AB _ BD
Sin", - Sin (fJ - "')
BC Cot 22' BCCot35'
Sin", = Sin (fJ - "')
Sin (fJ - "') _ Cot 35'
Sin", - Cot 22'
Sin ~IJ - "') =0.578
Sin '"
Sin 11 Cos", .; Sin '" Cos fJ - 57
Sin",
-0. 8
Sin 110'35' Cos", - Sin", Cos 110'35'
Sin",
=0.578
Cot", Sin 69'25' + Cos 69'25' =0.578
- 0.578 - 0.352
Cot III - 0.93616
Cot III = 0.241
III = 76'27'
fJ· III = 34'08'
Bearing of the line of intersection
=76'27 - 63'25'
= N, 13'02' E.
@
Solution:
Slope of the line intersection:
5
12
tan III = 153.27
III
= 0'9'21"
5
12
tan a =224.82
Be
a =0'06'22"
tana=-
08
tan a = ~~ tan 22' Sin
H=83'42'-a+1Il
III
tan a = tan 22' Sin 76'27'
a = 21'27' (slope)
H =88'42' - 0'06'22" + 0'09'21"
H = 88'44'59"
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S-222
MINE SURVEYING
45000 + 2.x2 • 1800x + 270000 =0
2.x2 • 1800x + 315000 =0
x2 - 900x + 157500 = 0
x=
x=
x=
900 ±..j81‫סס‬oo - 63‫סס‬oo
2
900 ± ..J18OOOO
2
900 ±424
2
476
x=2
x=238
1323
x=T'
x = 612 > 300 (absurd)
Usex= 238
® Bevation of discovery post: .
Solution:
CD Location of discovery post from the
location post number 2:
o (150)2 + x2 = (h cot45'~
(150)2 + x2 = h2
@
.
(150)2 + (300· x)2 =(h rot 60'~
If
(150)2 + (300· x)2 ="3
If =(150)2 +;
o&@
. If =22500 + (238~
If
h2 - x2 ="3 -(300 - x~
e
h2 = 22500 + 56700
h= 281
3h2 - 3; = h2 - 3 (90000 - 600x + x2)
31f - 3x2 = If - 270000 + 1800x - 3;
21f - 1800x + 270000 = 0
Elev. of Discovery Post =400 • 281 - 1.5
Elev. of Discovery Post =117.5 meters
O&e
h2 =(150)2 + x2
.2 [(150)2 + x2]-1800x +270000 =0
2 (22500 +x2) - 1800x + 270000 = 0
@
Distance of discovery post from cer. 1:
Distance = h cot 45'
Distance = 281 cot 45'
Distance = 281 m.
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PRACTICAl ASTRONOMY
1. Vertical circles - are great circles passing
from the zenith through the star or sun.
ods of determining time:
2. Hour circle· are great circles through the
poles.
1. Time by transit of a star across the
meridian.
2. TIme by transit of the sun.
3. Time by altitude ofthe sun.
4. Time by measured altitude of the star.
5. TIme by transit of a star across the vertical
circle through the Polaris.
6. Time by two stars at equal altitude.
Methods of determining longitude:
1. By time signals.
2. By transportation of time piece.
Methods of determining latitudes:
1.
2.
3.
4.
By altitude of the sun at noon.
Bya circumpolar star at time of transit.
By altitude of polaris at any hour angle.
By circum-meridian altitudes.
Methods of determining azimuths:
1.
2.
3.
4.
By an altitude of the sun.
By an altitude of the star.
By Polaris at greatest elongation.
By a circumpolar star at any hour angle.
3. Zenith ~ the point where the vertical'
produce upward, pierces the celestial
sphere.
4. Horizon· the great circle on the celestial
sphere cut by a plane through the earth's
center at right angles to the vertical.
5. Equator· the great circle of the celestial
sphere cut by a plane through the earth's
center perpendicular to'the axis.
6. Meridian of an'obseNer· the great circle
of the celestial sphere which passes
through the poles and the observer's
zenith.
7. Ecliptic· the great circle of the celestial
sphere which the sun appear to describe in
its annual eastward motion among 'the
stars.
8. Equinoxes· the· point of intersection of
the equator and the ecliptic.
9. Autmunal equinox· the point where the
sun crosses the equator in September.
10. Declination· the angular distance from the
equator measured on an hour circle through
the point.
11. Polar distance - is the complement of the
declination.
12. Altitude· the angUlar distance below or
above the horizon measured on a vertical
circle through the point.
13. Zenith distance - complement of the
altitude.
14. Hour angle - the arc of the equator
measured from the meridian westward to
the hour circle through the point.
15. Nadir· the point where the plumb line of
the transit when prolonged downward.
16. Celestial sphere· is an imaginary sphere
whose center is the center of the earth and
whose radius is infinite.
5-224
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PRACTICAL ASTROIOMY
(90- H) =a
C=S
sin p = sin Z sin (90 - La)
= sin Z cos L
Z· = p. secL
z
s
Parallels
Note:
At westem elongation, subtract Zfrom 180'
to obtain the azimuth of the stilr, and at eastem
elongation. add Z to 180' to obtain the true
azimuth of the star.
oflkclination
Determination of Azimuth by Polaris
at Greatest Elongation:
Derivation of the Formula for Determining
the Azimuth of Polaris at Elongation:
From the PZS Triangle, using the Napiers rule:
sin b = cos (co - b) cos (co - c)
sin z = sin Bsin c
. Z sin n
Sin =_:.L..
.
cos L
sin Z = sin p sec L
Let b = P
B=Z
(90 - L).= c
A=P
Set the transit at one end of the line to be
observed and level it carefully. Find the star
and sight the vertical hair on it. As the star
moves almost vertically (upward for eastern
elongation and downward for western
elongation) it requires slow motion of the
tangent screw to keep the vertical hair on the
star. Follow it until it seems to move'
vertically, which should be about the time
given the table of Ephemires. Lower the
telescope and set a mark in line with the. cross
hair on the ground. Reverse the telescope and
sight the star again and then set another point
along the first side. The point halfway
between these two should be the point in the
vertical plane of the star at elongation. With
the values of declination and latitude given,
silve for the value of Z, using the relation that
Z = P sec L. When the observation is at
westem elongation, just subtract the value of Z
from' to obtain the true azimuth of the star, and
if it is observed to be on the eastern
elongation, just add the value of Z to • thus
obtaining the true azimuth of the star that
particular time of observation.
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PilACTICAl ASTRONOMY
Determination of Azimuth by Polaris at
either Upper of Lower Culmination
The direction of the meridian may be
determined by observing with a transit at the
ins~ant when Polaris and some other stars are
in the same vertical plane and then waiting for
a certain lime until Polaris will be on the
meridian. At this instant Polaris is sighted and
its direction is then marked on the ground by
means of a stake. The observation to
determine when the two stars are in the same
vertical plane is done by the approximate
method by first pointing the vertical hair on
Polaris and then lower the telescope by
pointing the star to be observed. At upper
culmination the Ursa Minor is exactly below
the Polaris and at lower culmination, the
Cassiopeia is also located directly below the
Polaris. This would be repeated until the
Polaris and the star other than Polaris, are
located on the same vertical hair. The
telescope now is pointing the true meridian,
and this is marked on the ground.
Azimuth of AS = 180 + Z - H
POLARIS AT WESTERN ELONGATION
POLARIS AT EASTERN ELONGATION
Z"=P·secL
H = measured horizontal angle between star
and object.
Azimuth of AS = 180 - Z + H
Azimuth of AB = 180 + Z + H
Azimuth of AB = 180 - Z- H
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5-226
PIICI.CllISTRONOMY
POLARIS AT UPPER CULMINATION
N
Azimuth AS = 180 - H
Azimuth AS
=180 + H
1) Star between Zenith and Pole
N
NL.-_..L........;:
~
Azimuth AS = 180 - H
POLARIS AT LOWER CULMINATION
L=D-Z
L = latitude
D = declination
Z=90-H
H =vertical angle
2) Star between the south and equator
z
N
N'-----~-"""-_...... S
A
Azimuth AS = 180 + H
L=90-(H +D)
L=Z-D
Z=90-H
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PRACTICAl ASTRONOMY
(TANGENCY METHOD)
3) Polaris at upper culmination
z
SET I
Position of
Telescope
Position cf
Sun's Image
-P
D
NL-_~::.L.:::iIC:
.....J5
d-
D
L=H-P
P =90 - 0 (polar distance)
L=D-Z
Z=90-H
~
R
-b
R
4) Polaris at lower culmination
z
SET II
1---1--1.4-:*
---15
Position of
Telescope
Position of
Sun's Image
R
-P
d-
R
~
D
L=H+P
P=90-D
Determination of AzilTluth by
Solar Observation
I
Set up the transit at one end of the line
whose azimuth is to be determined. With the
telescope in the normal position, orient the
telescope due south, Sight the other end of the
fine and record the magnetic azimuth of such
line, Then rotate the instrument and point
approximateiy to the position of the sun.
Taking precautions that observing the sun
directly through the telescopic eyepiece may
result injury to the eye. Good observations
can be made by bringing the sun's image to a
focus on a white card held several inches in
the rear of the telescope. Sight the sun in the
following order and recording each observation
the values of vertical angles, horizontal angle
and time. Using the tangent method, the cross
wires shall be made tangent to the left and
lower right as shown in the following sets of
observations,
-b
D
CENTER METHOD
SET I
Position of
Telescope
Position of
Sun's Image
-+
D
R
SET II
Position of
Telescope
R
0
+
Position of
Sun's Image
+
-+
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S-228
PRAcnClllSTRONOMY
Sight the other end of the line again and
check whether the reading is still the same as
that of the previous one. It must give the same
reading otherwise the instrument is disturbed.
Take note that the interval of time between any
two consecutive sighting shall not exceed 2
minutes, if it does, discard that observation.
The date of observation must also be recorded
since values of the North Polar Distance from
the table is obtained from this date.
Solar Azimuth Formula:
cot ~ =" sec S Sec (S • P) Sin (S • H) Sin (S - L)
P+H+L
S=-2P=corrected north Polar distance
H=corrected altitude of sun
L = latitude of place of observation
CONVERGENCY OF MERIDIAN
Convergency of Meridian = is the meeting of
two tangents at each point of different
longitude as they recede to the north pole.
Angular convergence in seconds = diff.in
long in sec. x Sine middle latitude.
A = azimuth of sun if observed in the afternoon.
360 - A =azimuth of sun if observed in the
moming.
S=P+~+L
P =corrected north polar distance
Correction applied = Diff. in hours from 8:00
A.M. or 2:00 P.M. multiplied by the
variation per hour, which is to be added
when observed between June 21 to Dec.
21, and to be subtracted when observed
between Dec. 21 to June 21. .
H = corrected altitude of sun, corrected for
parallax and refraction which is always
subtracted from the observed altitude.
L =Latitude of place of observation.
Determination of Time from
Solar Observation
Tan .1 = -V Sec (S - P) Sin (8 - H)
2
Cot~
o = angle of convergency
f1 = latitude angle of AB
a =difference in longitUde between A and B
OB = R = radius of earth approximately
20,890,000 ft.
AB=O'Ba
AB
a=O'B
2
Tan ~ =
&
SSec (S - P) Sin (S - H) Csc (S - L)
12 - t = local apparent time (when observed in
.
the morning)
t = local apparent time (when ob~erved in the
afternoon)
Angle BOD = Angle BCO' = {J
BO'
Sin J3 = BC
BO'
BC= Sin B
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PRACTICAl ASTRONOMY
With negligible error
AB=BCe
AB
0=BC
ButAB = BO' a
BO'
BC= Sin B
AB
0=BC
0=
BO' a Sin B
BO'
e=aSinB
BO'=RCosB
AB
a=BO'
Let AB =d (distance between A and B)
d
a=RCosll
d Sin II
e=Rcosll
dtanll
.
o =-R- (radIans)
o = 32.38 d tan B (seconds)
o =32.38 tan B(convergency correction)
For d = 1 kilometers
If d = kilometers,
R = in feet
o =3;~~:0~~ II
1~0
(3600)
o =32.38 tan f!,
To obtain azimuth based on base meridian,
subtract the covergency correction if the line is
on the east of base meridian.
To obtain azimuth based'on the base meridian,
add the convergency correction if the line is on
the west of base meridian.
Parallax Correction:
It is assumed that the celestial sphere is of
infinite radius and that vertical angle measured
from a station on the earth's surface is the sam
eas that if it would be measured from the
center of the earth. But for stellar or solar
observations these angles are not equal.
There is an error in this observed vertical angle
due to the fact. that it is· observed on the
surface and not on the cetner of the earth. This
error is called parallax.
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S·230
PRACTICAlISTRONOMY
Combined Correction due to
Parallax and Refraction
._-------
CELESTIAL HORIZON
--
h = corrected altnude
h1 = observed altitude
hp = parallax correction
h = h1 + hp (parallax correction is added)
H =corrected altitude
HI =observed altitude
hr =refraction correction
hp = parallax correction
hrp =combined parallax and refraction
correction.
H=Hl- hr+h p
H=H 1 - (hr - hp)
Refraction Correction:
When a ray of light emanating from a
celestial body passes through the atmosphere
of the earth, the ray is bent downward as
shown on the figure. Hence the sun or star
appears to be higher above the observer's
horizon than they actually are. The angle of
deviation of the ray from its direction on
entering ihe earth's atmosphere to its direction
at the surface of the earth is called the
refraction of the ray.
h = corrected altitude
hI = observed altitude
hr = refraction correction
h = h1 - hr (refraction correction is
subtracted to the observed altitude)
CELESTIAL HORIZON
---------
Combined correction due to parallax and
refraction is always subtracted to the observed
altitude.
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PRACTICAl ASTRONOMY
ColTeCfed H= 35'16'15"
1-10"
H= 35'15'05"
P = 101'02'08"
L = 12'50'27"
28= 1S07-40
S= 77-33-50
P= 107-92-08
S-P= 29-28-18
S= 77-33-50
H= 35-15-Q5
S-H= 42-18-45
S= 77-33-50
L= 12-50-27
8 - L = 6443-23
Cot~A =...j Sec SSec (S- F') Sin (S -H) Sin(S - L)
o
o
R
R
Time
8:32:00
8:32:30
8:33:05
Cot l A
2
="
Sec 71'33'50" Sec 29'28'18" Sin 42'18'45" Sin 64'43'23'
'12 A =29-01-45
~
A =58-03-30
Mean = 8:32:48
Azimuth ofsun = 360 - A
Azimuth ofsun = 301"56'30"
Oiff. in time =8:32:48
- a;Q.Q;QO
0-32:48
Oiff. in time = 0.5467 hrs.
Correction for NPO
= diff. in time x variation per hour
Correction for NPO =0.5467 (- 43.9) =-24"
Corrected NPD = 101'02'32"
24"
P = 107'02'08"
@
Azimuth ofsun:
o
o
Horizontal
Angle
158'22'
159'04'
159'10'
R
R
~
Mean = 158'48'15"
Vertical
Angle.
35'21'
34'55'
35'36'
~
Mean = 35'16'15"
Azimuth T1 - i, c: 301'56'30"
- 158'48'15,'
Azimuth T1 - T2 = 143'08'15"
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5-232
PRACTICAl ASTRONOMY
D
Altitude (H)
36-09
D
~2S
R
36-19
R
~
Mean = 35'54'00"
Corrected H = 35'54'00"
-
1'09"
Corrected H= 35'5Z51"
:::::>:":::::::;:::::::;:;::;::;:::::;:;:;::
l,r!~W~I~~~~~P1:':
t~tt~:
•
:·.t....••...
:.·
•..•••....••.•:•..••....•
·.:
••....•
.
}r~~~;t~
<;U<>~~~
~~T." .>/11·'·:::,:,:::.:•.!.• .'•. ..•..•.•
!?j~~1cJ1;1~;~llli •· • •••
·Q\Q..
W_':~f",~·············:·········
.
..
•.••• •.•
Diff. in time =3:45:45 - 2:00:00
Diff. in time = 1:45:45 = 1,7625 hrs.
::::::;:;:: :::;:::':::::::::;
>{:~~r::~:};::::;:::::::::::-::""
.. .. ",' ••.••. '.' ....••..:.:.:<;:;:
Correction for North Polar Distance
= Diff. in hrs x Variation per hour
Con-ection for NPD 1.7625 (- 38,57") 1'OS"
:1.· ....•....••.
.•. •:...•..•.•...•
:•.•...,:'i.••..,•.•:.••.••..••.•.••••
=
·:'.·i••
li.,illiti
r.
Corrected NPD:
P = 70'36'24"
rOO"
p", 70'35'16"
Ij.;r•.!..•:! f. .·.fu.~tl.·:.:.;{., .111
..:r.o:.-, _...
..
~:,~ ,v.~:'f
.
P = 70'35'16"
H= 35'5Z51"
L= 14'53'25'
2S = 121'21'32"
S = 60'40'46"
P = 70'35'16"
S- P = 9'54'30"
S = 60'40'46"
:::-:::-:..:<:</:::.:-:-..-.-..
Solution:
CD Azimuth of sun:
. Horizontal Angle
D
D
R
R
H= 35'52'51"
S-H= 24'4755"
S = 60'40'46"
L = 14'53'23"
S-L= 45'1721"
102-59
103-39
103-57
~
Mean = 103'2S'30"
Time
D
D
R
R
3:45:00
3:45:30
3:46:00
=
Cot~A =~ Sec SSec (S- Pi Sin (S- H) Sin (S- L)
COl! A =-V Sec 60'40'46' Sec 9'54'30' Sin 24'47'55' Sin 45'17'21')
2
~
12 A= 51'42'39"
Mean = 3:45:45
A= 103'25'18"
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PRACTICAl ASTRONOMY
. Solution:
CiJ Azimuth ofsun:
MNN
N
H= 51'04'00"
L = 10'22'00"
P = 103'23'23"
2S = 164'49'23"
S = 82'24'42"
S· p= 20'58'41"
S· H= 31'20'42"
S· L = 72'02'42"
Note: Azimuth of sun = A if observation is
in the aftemoon
Azimuth of sun = 360 • Aif
observation is in the morning
e = 103'28'30" ·103'25'18"
e = 03'12"
Col~= --.J Sec 8 Sec (8 - p) Sin (S· L) Sin (8 - H)
Azimuth ofT, • T2 = 178'36'·03'12"
Azimuth of T, • T2 = 178'32'48"
@
Col ~
Magnetic declination:
Magnetic declination = 03'12" ~
=~ Sec 82'24'42' Sec 20'58'41' Sin 72'02'42' Sin 31'20'42'
A
Col 2" =2,00
~= 26'31'37"
A =53'03'13"
Azimuth ofsun = 53'03'13"
T@•• J(rJI()Wiri9d~htwet~r~i;Qrded16r.an·
p@@@Qn()flti~.stl6.
•· • ·> • .U.•,. ' < ' .
..: . ;:::".
® True azimuth of AB:
N
CiJCornpule the tnteazirrillih bfStihi' >
®. Corriput~the lrueazimuth tifiine Nt
@
Is the walchloo
how much?
.
.... .
slow or toofasfandby
'., .....
•.
B
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5-234
PRACTICAl ASTRONOMY
True azimuth of AB = 360' - 89'16'47"
True azimuth of AB = 270'43'13"
L = 41'20'
H= 46'48'
p= 82'02'
25 = 170'10'
S= 85'OS'
S-p= 3'03'
S-H= 37'OS
S-L= 43'45'
® Time of observation:
tan~= ~Cos SSec(S-p) Sin (S- H) Csc(S-L)
lan i =" Cos 82'24'42" Sec 20'58'41" Sin 72'02'42" Sin 31'20'42"
2
!.2 =15'32'27"
t =31'04'55"
f = i'04 m2O' (time of observation)
2:04:20 - 2:04:12 =8 sec,
cot~ = ~ Sec SSec {S - P] Sin (S· L) Sin (S - H)
Col ~ =" Sec 85'05' Sec 3'03' Sin 43'45' Sin 37"05'
A
Cot 2" =2.2072
Watch is too slow by 8 seconds,
~= 24'22'23"
2
A = 48'44'46"
•.
··~·~~I ;e~~Na~~~4,\~es6h~~~~~.·
.~s~Q!~~® . ~.·~~·~~~~·ltll.·~rr¢AtEl~.·No/tlJ
P.91at~l$tao¢EllS$2~Q2r~
Azimuth = 48'44'46" if obseNed in P,M.
® Azimuth if obseNed in the morning:
..
N
• .~• • • ¢6ii1Pllieitte.a#imulh~sHn.lfObsel¥~dill.
~_:lIt.'~I.~:
··®····~~~~~i~~I~~Grv~Wm.lfff.#~ •
Solution:
CD Azimuth ofsun if obseNed in the aftemoon:
N
Azimuth = 360 - A
Azimuth = 360·48'44'46"
Azimuth =311'15'14" ifobseNed in A.M.
® Time of obseNation:
tan ~ = ~ Cos 8 Sec (5· p) Sin (5 - Ii) ~sc (8 - L)
tan
f= " Cos 85'05' Sec 3'03' Sin 37'05' Csc 43'45'
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PRACTICAl ASTRONOMY
t
Diff. in time = 3:45:45 - 2:00:00
Diff. in time =1. 7625 hrs.
tan "2 =0.27357
L
2- 15'18'
Correction for NPD = variation per hour
x Difference in hours
Correction =• 38.57 (1.7625)
Correction =• 67,98"
Correction =·1' 08"
= 70'36'24"
CorrectedNPD
• 00'01' 08"
=
70'35'16"
Corrected NPD
t =30'36'
t= 2h02m24s
Time = 12:00:00
.~
9:57:36 A.M.
@
Declination at the instant of observation:
MN TN
~
\
\
T-2
\
\
\
\
I
True azimuth "
\
" T-]
\
\
\
\
\
I
0'03'12" W
True azimuth ofT-l toT-2
=178'36'00"-0'03' 12"
=178'32'48"
Solution:
CD Corrected North polar Distance:
MN TN
~
Note: .Angle Ais on the west if
observed on the' afternoon
e= 103'28' 30" - 103'25' 18"
e = 00'03'12" W. (declination)
\
\
\
\
I
\
\
A= 103 '25' 18""
\
\
\ T-]
® True azimuth:
True azimuth of T-1 to T-2
= 178'36' DO" - 0'03' 12"
= 178'32' 48"
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PRACTICAl ASTRONOMY
Azimuth ofmark:
@
TN
e = 313'48' 45"· 300'13'19"
e
Solution:
CD Corrected altitude:
= 13'35' 26"
Azimuth ofmark = 360' ·13'35' 26"
Azimuth ofmark = 346'24' 34"
TN
Recorded altitude H = 36'49' 45"
Parallax &refraction
01' 07"
Corrected H
@
36'48' 38"
Azimuth of sun:
Azimuth ofsun = 360'·59'46' 41"
Azimuth ofsun = 300'13'19"
;
(j)1f the parallaX and refr.actlon
.·is thecortecledvalu6 om
@.
@
isO'58", what.
.
Compute the true beanng ofthesu!l.
Compute the azimuth ofBLlM#1 .to
·BLlM#2.
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PRACTICAl, ASTRONOMY
Solution:
CD Corrected vertical angle H:
Note: H = 90' - zenith angle
Hl =90' - 48'33' 48" = 41'26' 12"
H2=OO'·48'49'50" = 41'10'10"
Ha =90'·48'09' 40" = 41'50'20"
H4 = 90' - 48'34' 42" = 41'25' 18"
165'52' 00"
AverageH=
165'52' 00"
4
Average H=41'28' 00"
Corrected H=41'28' 00"·0'58'
Corrected H= 41'27' 02"
Horizontal angle: 359{)2-OO
356-19-47
356-19-44
359{)2·25
1433'43' 56"
Ave. Horizontal angle =
1433'43' 56"
4
.
Ave. Horizontal angle =358'25' 59"
8 = 104'23' 24" ·102'49' 33"
8=1'33'51"
Azimuth ofBLLM #1 to BLLM # 2
=255'36' 26" + 1'33' 51"
= 257'10' 17"
® True bearing ofsun:
Azimuth of sun =360' ·104'23' 34"
Azimuth of sun =255'36' 26"
True bearing = N. 75'36' 26" E,
® Azimuth of BLLM #1 to BLLM #2:
CD Compute the azimuth of sun.
.'. . .
®compute lh@1fUeaZimulhof BlLM 1'40,110
. 8LLMNo.2,.. .'.
.
.
@ '. CompUte the probable error of azimuth. .
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S-238
PRACTICAl ASTRONOMY
Solution:
Cot ~ =...J sec S Sec (S·P) Sin (S·H) Sin (S-L)
2
N
Cot ~ = ...J 0.228140761
~2 =64'28'8.24"
A=128'56'16.4"
AZimuth ofsun =360' - 128'56'16.4"
Azimuth ofsun = 231'03'43.6"
Azimuth of BLLM1 to BLLM2
= 231'03'43.6" - 101'49'29"
=129'14'14.6"
Set 1
Time
Hor. Cirde Reading
N
Vertical Angle
8:14:36 101 - 50 - 25
47 - 06 - 12
8: 15:09 101 - 48 - 33(.tA>.ISO') 47 - 01 - 11 (sub.360·)
8:14:52.5 101-49-29
47-03-41.5
corrected H =47'- 03'· 41.5'·59.39"
H =47'02'42'11"
Diff. in time = 8:14:52.5
8:00:00
14:52.5 =0.2479 hrs.
Correction for NPD = 0.2479 (09i
Correction for NPO =2.23"
p= 51'13'46" -2.23"
p= 51'13'43.77"
H=47'02'42.11"
L = 17'16'4.80"
28 = 115'32'30.6"
S = 57'46'15.34"
p = 51'13'43.77"
s· P= 6'32'31.57"
S= 57'46'15.34"
H= 47'02'42.11"
S - H = 10'43'33.23"
S = 57'46'15.34"
L = 17'16'4.80"
S - L = 40'30'10.54"
Sel2
lime
8:15:12
8:15:51
8:15:31.5
Hor. Circle
Vertical
Reading
Angle
101- 47 - 33
101· 45 - 57
101- 46·45
43'58'01"
46'57'17"
45'27'39"
Oiff. in hrs. =8:15:31.5 -8:00:00
Diff. in hrs. =0.25875
Corr. =0.25875 (09'')
Corr. =2.33"
P=51'13'4l)"
2.33"
51'13'43.67"
Correded H =
45'27'39"
59.39"
H=45'26'39.61"
P= 51'13'43.67"
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PRACTICAl ASTRONOMY
L % 17'16'4.80"
28 = 113'56'28.10'
8 = 56'58'14.09"
P = 51'13'43.67"
8 - P = 5'44'30.33"
S = 56'58'14.09"
H =45'26'39.61"
8 - H= 11'31'34.4"
8 = 56'58'14"
L = 17'16'4.80"
8 - L = 39'42'9.2"
cot~ = ~ Sec SSec (8- P) Sin (S· H) Sin (S- L)
Cot ~ = {0.235359069
~ =64'07'12.74"
F@n!~~gellf?n~~~Qfasl)l~t9t>s~tyatitm
~.~i@wi@. ·TwZth~~QM~;.IMfPlk1WtOg~$;W.
·areoll$61Ved.·.·.·····.···.>····
• • • • • • • • ................;<>
• • • • • • • • • • • • . •. .. . . . . . .
$t~.i~p~d.: ¥~l
Blt~1~~i[4Ii~.m":)
I.~i.lil
c~~~~~~t~~~~'o;~,~~~p~~\
N
A =128'14'25.4"
Azimuth ofsun = 360' - 128'14'25.4"
Azimuth ofsun = 231'45'34.6"
CD True azimuth of sun:
231'-03'· 43.6" + 231'-45'·34.6"
=
2
True azimuth of sun = 231'24' 39"
@
Azimuth ofBLLM #1 to BLLM #2:
True azimuth of BLLM No. 1 to BLLM No.2
= 231'45'34.6".· 101'46'45"
= 129'58'49.6"
True azimuth ofBLLM NO.1 to BLLM NO.2
129'58'49.6" + 129'14'14.6"
=
2
= 129'36' 32.1"
@
Probable error ofazimuth:
Probable error ofazimuth
= 0.33725 x difference in azimuth
Difference in azimuth
= 129'58'49.6"· 129'14' 14.6"
Difference in azimuth =44' 35" =2675"
Probable error of azimuth = 0.33725(2675)
Probable error ofazimuth = 902"
Probable error of azimuth = 15' 02"
CD Compute lhe¢orrected alt~udiHor set t
true aZimufhofT1to Tz. .
® Compute the probable errOr.
@ • Compute the
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5-240
PRACTICAl ASTRONOMY
Solution:
<D Corrected altitude for set I.
Note: Vertical angle =90' -zenith angle
Position of
Telescope
o
o
R
R
Position of
Telescope
Time
8:32:07
8:32:31
8:33:09
8:33:36
8:32:50.7
-Zenith
Angle
o
<$-33-48
R
R
<$-09-17
<$-34-43
o
48-49-59
Horizontal
Angle
359-02-00
358-17-47
358-19-44
359-02-25
5=P+H+L
2
P= 69'39'10.17"
H= 41'27'7.25"
L =14'33'40.73"
2S =125'39'58.1"
5 =62'49'59"
5 - P= 6'49'11.17"
5 - H= 21'22' 51.75"
358-4().{J()
S- L =48'16'18.27"
Vertical
Angle
41-26-12
41-10-01
41-50-43
41-25-17
41-2Pr0325
Cot~= " Sec S Sec (S - P)
Sin (5 - H) Sin (S- L)
A
Cot =0.77469
2
~=52'11'6.91"
A=104'28'13.8"
Corrected altitude for set I =41'28'03.25" - 0'00'56"
Corrected altitude H = 41'27'07.25"
Azimuth ofsun =360' -104'28'13.8"
Azimuth ofsun =255'31' 46.2"
® True azimuth ofTt to h
True azimuth of T1 to T2 = 255'31' 46.2"
+
1'20' 00"
True azimuth of T1 to T2 =256'51' 46.2"
Position of
Telescope
R
R
0
0
Oiff. in time =8:32:50.7 - 8:00:00
Oiff. in time =32:50.7
Oiff. in time =0.5474166 hrs.
Correc1ion for NPO = Variation per hour
x Diff. in hrs.
Correction for NPD =29.64 (0.5474166)
Correction for NPD =16.23"
Uncorrected NPO = 69'-39'-26.40"
Correction
=
16.23"
Corrected NPO
= 69'-39'-10.17"
Position of
Telescope
R
R
0
0
SET II
Time
8:33:36
8:34:14
8:34:44
8:35:08
8:34:25.2
Zenith
Angle
47-5Pr58
48-25-57
47-46-17
48-12-50
Horizontal
Angle
359-02-12
358-20-09
358-20-54
359-03-57
358-41-00
Vertical
Angle
42-01-02
41-34-03
42-13-43
4147-10
41-53-59.5
Correc1ed H=41'53'59.5"
- 0'00' 55"
Correc1ed H =41'53'04.5"
Oiff. in time =8:34:25.5 - 8:00:00
Oiff. in time =34'25.5" = 0.57375 hrs.
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PRACTICAl ASTRONOMY
Correction for NPD =29.64" (0.57375)
Correction for NPD = 17"
Corrected NPD (P) = 69'39'·26.40"
17"
P =69'39'9.40"
H = 41'53'9.40"
L = 14'33'40.73"
2S = 126'05'54.63"
S = 63'02'57.32"
S- P = 6'36'12.08"
S- H = 21'09'52.82"
S· L = 48'29'16.59"
cot~=.ysecs
Sec(S-p) Sin (S-H) Sin (S-L)
cot~ =0.774925257
~= 52'13' 37.17"
A= 104'27' 14.3"
Azimuth ofsun = 360' -104'27'14.3"
Azimuth ofsun = 255'32' 45.7"
Azimuth ofT1 to Tz = 255'32'45.7"
+ 1'19'00"
Azimuth of T1 to Tz = 256'51' 45.7"
Average azimuth
.
256'51'46.2" + 256"51'45.7"
=
2
Average azimuth = 256'51'46"
Probable error.
Ditt. in azimuth =256'51' 46.2" - 256'51' 45.7"
Oiff. in azimuth = 0.5"
Probable error = 0.33725 x Diff. in azimuth
Probable error = 0.33725 (0.5")
Probable error = 0,169"
@
!jlt~lllf
Solution:
CD Azimuth ofsun:
Diff. in hours = 1:45.20
Oiff. in hours = 1.756 hrs.
Cotrection forNPO =- 38.24(1.756)
Correction for NPD =- 67.15" =- 01'7.15"
P = 103'24' 30.24"
- 01' 23.09"
P = 103'23' 23.09"
H = 51'05' 00"
38.92"
H= 51'04' 21.08"
L = 10'23' 5.29"
p= 103'23'23.09"
2S = 164'50' 49.4"
S = 82'25' 24.73"
P = 103'23' 23.09"
S- P = 20'57' 58.36"
S = 82'25' 24.73"
H = 51'04' 21.08"
S- H = 31'21' 3.65"
S = 82'25' 24.73"
L = 10'23' 5.29"
S· L= 72'02' 19.44"
5-242
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PRACTICAl ASTRONOMY
cot~=.y Sec SSec (S-P) Sin (S-L) Sin (S-H)
A
Cot 2" = 2.0
~=26.51'
A = 53'01'
@
Value oft:
tan ~:::.y Cos S Sec (S-P) Sin (S-ff) esc (S·L)
~= 15.53'
t = 31'03' 36"
® Local mean time:
t = 31'03' 36"
t = 2h 04m 14.4s
Local mean time::: Zh 04m 14.45 - (- 14'15.8")
Local mean time'" 2h 18 n, 30.2s
Local mean time = 2:18:30,2
@)
Standard time at 120th meridian:
Oiff. In longitude::: 124'58' 42"-120'
Oiff. in longitude = 4'58' 42"
WC()I11P(lt~lh~$fq~pt~?N()tthP()lar
. >Qisfan~ .. >....................................................................
.~>>>'h!itwillb~the)azirJ9lh.(>fth~$!Jl).
·@.wb~twm.~.·tne.(gimutnpffhEl.mal'k.
.
Solution:
CD Corrected North Polar Distance:
Horizontal
Time
Vertical
Angle
Angle
358'40' 54.7" Ave: 8:32:58.75 48'47' 08" (Ave)
Oiff. in time = 8:32:58.75·8:00
Oiff. in time::: 00 • 32:58.75
Oiff. in time = 0.5496 hrs.
Corrected for NPO = 0.5496(26.64}
Corrected for NPO =- 14.64"
Corrected NPO::: 68'22' 42.4" ,·14.64"
P = 68'22' 27.76"
® Azimuth of sun:
I it d 4.978"
OI·ff.·
, In ong u e::: -15"
Oiff. in longitude ::: Oh 19m 54,8s
Standard time:: 2:18:30.2
• 19:54,8
Standard time::: 1:58:35,4
H:::
48'47' 08"
p::: 68'22' 27.76"
L::: 14'20' 13.6"
28::: 13129' 49.3"
8::: 65'44' 5465"
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PRACTICAl ASTRONOMY
p = 68' 22' 27.76"
Sop = 2'3T 33.11"
Solution:
CD True bearing ofsun from the North:
S = 65'44' 54.65"
H:;:
S-H:;:
S:;:
L=
S-L:;:
+
48'4708"
16'5746.65"
65'44' 54.65"
51'24'41.05"
51'24' 41.05"
Cos ~ = ...j Sec SSec (SOP) Sin (S-H) Sin (S-L) ,
Z=1OI'2T50"
~:;:53.30'
A = 106'35' 34.1"
Azimuth of Sun :;: 360' - 106'35' 34:1"
Azimuth of Sun =253'24' 25.9"
@
Azimuth of Marie
358'40' 54.7"
- 253'24' 25.9"
ex. = 105'16' 28.8"
SinD
Cosl= Cos LCosh ·tanLtanh
B = 106'35' 34.1" -105'16' 28.8"
B:;: 1'19' 5.2"
Note:
l =true bearing ofsun from the norlh
NW if observed in the afternoon
NE if observed in the morning
0= 20'52' 44"
L=42'29' 30"
h = 43'16' 48"
SinD
Cosl= Cos LCosh -tanLtanh
B = 1'19' 5.2"
Azimuth of Mark =253'24' 25.9"
+ 1'19' 5.2"
Azimuth of Mark = 254'43' 31,2"
Anobsetvati\lli W8sniadetodeterminethe
Sin 20'52' 44"
Cos l = Cos 42'29' 30" Cos 43'16' SO"
Cosl=-0.19875
Z = 101'27' 50"
aiimuth btthellne .Ai3. by observing the
allilode ot sun in theaftemoon. The follOwing
da~were~erve(t.<
.' ... ' ,.'
latitUde of place I)f QbservatlQl'l == 42'29' 30' N.
Longitude of place of observation'
''
• ' ' " .,.,,' , , ' ','.'
'M~n Horizontal Angte from statian.B to the
sun =6S'54' 30" (clOCkwise) "
•
Mean altitude of Sun (corrected) = 43',16' 46"
@
True azimuth of the sun:
True azimuth of the sun
= 180'00' 00" -101'27' 50"
True azimuth of the sun = 78'32'10"
CD Compute the true bearing of sun from the
North.
@ Compulethettue azimuth ofsun,
@ Compute the true aZimuth of AB.
@
True azimuth ofAB:
True azimuth of AB =78'32' 10' - 68'54' 30"
True azimuth of AB = 9'37' 40"
::: 124'20' 30" E,
Declination of Sun'" 20'52' 44"
5-244
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PRACTICAl ImONOMY
Il1llilillli]
Solution:
<D Bearing ofstar from the north:
i:I.112
• • i·.~bs~~ijl)(\""hrnR91~ri~a.ttalns.iISUpper
..H··.¢WrnltiatlQnWllhMaijitlldeQt~3'3T
.•1ndex
~f:l'~pllQtli$~~g.~a@ffitioI11s1'01!'.
Solution:
<D Azimuth of Polaris measured from the north.
H = 10'21' 30" + 10'22' ZO" = 10'21' .55"
SinD
CosZ=CosLCosH -tanLtanH
_
Sin (-10'57' 18")
Cos Z- Cos 39'14'12" Cos 28'36'48"
-tan 39'14'12" tan 28'36'48"
Z= 136'28'04" East ofNorth
® Azimuth of star:
Azimuth of star = 136'28' 04" + 180
Azimuth of star = 316'28' 04"
® Azimuth of AB:
Azimuth of/iB = 316'28' 04 - 85'20' 04"
Azimuth of AB = 231"08' 00"
2
Z"=P'secL
P=1'15'40"
P" =3600 + 15(60) + 40
P" =4540
Z"=P'secL
Z' = 4540 sec 14'45'
Z=4695"
Z= 1"18' 15"
® True azimuth of BLLM NO.1 to BLLM No.2.
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PRACTICAl ASTRONOMY
True azimuth of BLLM No, 1to BLLM No, 2
=180+l+H
Solution:
CD Bearing of Polaris measured from norlh:
True azimuth = 180' + 1"18' 15" + 15'21' 55"
N
True azimuth = 191'40' 10"
•
P~~iS
..::,::
B
® Latitude of ObseNation:
.I "
Z I"
I
A
lan Z =
Sin t
Cos Llan D - Sin LCos t
Ian Z -
Z =3'20'25"
Bearing ofPolaris = N. 3'20' 25" E.
=43'3TOO"
Altitude
Index correction
Refraction correction
= + 00' 30"
= - 01' 01"
CorrectedH
=43'36'29"
L=H-P
P= 1'15'40"
L=43'36' 29" - 1'15' 40"
L = 42'20' 49"
Sin 45'30'
Cos 42'20' tan 86'40' • Sin 42'20' Cos 45'30'
@
True azimuth of Polaris:
True azimuth ofPolaris = 183'20' 25"
® True azimuth ofAB:
True azimuth of AB = 183'20' 25" - 62'40'.00"
True azimuth ofAB = 120'40' 25"
TheCl~~~&edll1~ridi~o~ltitl#1i • •pf•. ~• • M~tol1.·
,b;pril.•10,•• 1990.V/~s • ~~'g~\ • $tar.pe.lifi~g.S6yth, •
Refraction(;()rrecti911•• iS1!1i?:•• ··The.·deeul1at~"
oflt1~st<lrffllli~tf~~9IW~s>~$~2!f?1~,<
Polaris is obsEl!V'ed· at a..certa1n hOur angle·
equal to 45'$Q~ at a certail1place .whose
latitude is 42"20' and a declination of 86"40',· A
horizontal angle was meaS\.lred from the line
AS clockwise towards pQlarls (EastQf North)
and was recorded to be 62'40'. .
. .
CD ;:~::he bealing of Polaris measured
@ Compule!he lrue azimuth of Polaris.
;;v C(}mpule the true aZimuth of AB,
<1?QEltefTljjoeloeco~~CtedailiIW~k
@•• q~tElrll'lihe· • tflEl.1BtjtOt!e()f•• lheplll~.·.of
obsElliJafloo.•••••. <.• • • • •.• .• .• • • • • • • «.. .. .
® QelermineJhehoUrl:109IElofthestar,
Solution:
CD Corrected altitude:
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5-246
PRACTICAl ASTRONOMY
Observed altitude
Refraction CotTection
Altitude H
= 39'24' 00"
_. 1'11"
= 39'22'49"
® Latitude of the place of observation.
H
= 39'22' 49"
D
= 8'29' 21"
90· L
= 47'S110"
H + D =47'S2' 10"
90· L =47'S2'10"
L = 90 . 47'S2' 10"
L =42'09' 50"
® Hour angle of the star.
Solution:
<fJ CotTected altitude ofPolaris.
Obs. altitude =43'28' 30"
Index error =• 01'00"
Refraction Corr. =• 01'00"
Cotrected altitude:
h =43'28' 30"·01'·01'
h = 43'26' 30"
® Hour angle in hours, minutes and seconds.
Hourangte:
t = 51'20' 46.S"
B
t
= S1.3462S'
15'
t = 3h25' 23,1"
® Latitude of place ofobservation.
Sin b
Sin a
Sin B = Sin A
Sin 98'29' 21"
Sin SO'37' 11 "
Sin 101'27' 50" =
Sin A
A =49'59' 21"
Hour angle t =49.989'
,_ 49.989'
,- 1S'
t = 3h 19"' 585
p=90'·D.
P=90' ·89'04' 55.6"
p= 55'04.6"
p=5S.08'
L = h - P Cos t + ~ Sin l'
p2 Sin2 t tan h
L = latitude of place of observation
h = corrected altitude of polaris
t = hour angle in degrees, minutes and
seconds
p = polar distance in minutes
Anob$~~~i:I ~fUlUd~>(l{R6Iaris alan Mut
angle Of 51'20' 4Ek5" was reoordedto be
43'2&' 30".lndexertQrl$ + 01'00". Declination
of Polaris at this instant is+ 89'04' 55,&'.
RefraCtioncorraction is 01'00";
<fJ Determinelhe corrected altitude of Polaris, •
® CompUte the hour angle in nours, minutes
and seconds.
® Determlrte the latitude of place of
observation.
L = h • p Cos t + ~ Sin l'
p2 Sin2 t tan h
L ::: 43'26' 30" • 55.08 Cos 51'20' 46.5"
+ ~ Sin l' (55.08)2 Sin 2 51'20' 46.5" tan 43'26'30"
L::: 43'26' 30' • 34.40' +0.25'
L::: 43'26' 30"·34' 39"
L::: 42'51' 51" N
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PRACTICAl ASTRONOMY
Greenwich apparent time
= Greenwich civil time + equation of time
4h 52m DOS = G.C.T. 5m 30S
G,G,T. =4h 46m 30s
+
Tfle~~$Ur~d111tlbJ(1eqflh~s4nslJp~~J@~
Wtl~~(m •. ttl~·.m@~I~n.at·.~·.ppiJ'\t.IQ . I~tl9!lU(j~
7:3.~WWa~4TIQ·QQ'':()I'JMW¢lJ11);2005.·
i_i,llili!1
•·· • • $tI6·:l•• $~Il'li.qil!ll1'lel!lr::;fI2;OO~·.·.··················
.
0JDetenllillel.)IDe••••latill.m~·<.6f.place
of
ob~eNali()h<·i<
Oe~~nnll)~ • •~e . .• IO(;~d • • sidereaJtime••. if{he
@ .•
· · · · · • • hour.~rlgl.~0f • lhe••.slJf:l•. duJil1g.·9bserValiori
.• wa$ ••••• 32·1~··.·4?'; • t@jme.appare!Jtrlght
.... · •. as~nsiorJ.()f.th~$yn.I$.5~.4W3t)S .••.•.• <i•• .·•· •
@ •..••
p~teli1lfne.!hE!9rren\'lICh$taHdai~tifne
.·tOOeql\tllionoftirneis5tn30(>
Greenwich civil lime = Greenwich standard time
Greenwich standard time = 4h 4~ 305
• if
Inptepa.t$~MfBt~110~~~~tionol1tli~$!ar
.~~iij~.ffi~~~\t~~b~~&r~'j;e~ili~ffi~~6·
amtll~e9ftbe$tatjs1.T$~,13'.Tlle~~lcUla,ted
n~f(l1c;ij()gwrrectipnf()rffl~laltlt~(jElii~O~~O:J;</
.$. What.allilugeWill.bemeaSllredwl1ell'lli~
~rgcirrt~ifthe$e5jlbol~te~v~~e~
.. .
.@)..
Solution:
tj)
Lalitude of place of obseNation:
L=90-H+D
ObseNed H
= 47'10' 00"
Index Correction
=
02' 00"
Refraction and Parallax = - 0'18"
Suns semi-diameler
= + 12'06"
Corrected H
= 47' 19' 48"
o = +3'35'02"'
L=90-H+D
L=90'-47'19'48" + 3'35'02"
L = 46'15'14"
Vlfit1il~;~.~.
•
@ •••. 9(Jfl1Pu~ th~llpllr.,¥,gl~l.of.
. ·lilSlaffl4fo~seJVatkm, .
(3)
Greenwich slandard lime:
.
.
73'00' 00"
Greenwich apparent lIme = 15'
Greenwich apparenllime = 4h 52m OOs
.• the.staratthe
Solution:
CD Allitude ofstar:
Calculated allilude = 17' 36.8'
Refraction Corr.
- 03'
= 1T33,8'
Corrected H
@
Bearing ofstar:
SinD
Cos Z =Cos L CosH - tan L (an H
Sin 12'25'
Cos Z = Cos 42'21' Cos 17'33.8'
- tan42'21' tan 17'33.8'
Z = 89'02' 44"
® Sidereal lime:
Sidereal lime = hour angle +righl ascension
Hour angle = - 32' 15' 45"
Hour angle =- 2h Ogm 03 s
Sidereal lime =-2h D9m D3s +5h 47m30 s
Sidereal time =3h 38m 275
lft~e·.~e~matlor.·oflt1~ • staratthetfiOrnenl
Bearing ofslar = N. 89'02' 44" E.
(3)
Hour angle ofIhe star:
Sin H
Cos t =Cos D Cos L . tan 0 tan L
. Sin 17'33.8'
Cos t =Cos 12'25' Cos 42'21'
• tan 12'25' tan 42' 21'
I =77'26' 37"
. ....
5-248
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PRACTICAl ASTRONOMY
Solution:
Solution:
CD Azimuth of Polaris:
t Z
Sin t
an - Cos Uan 0 • Sin L Cos t
t=2 h 42 m
t=4O' 30'
Sin 40'30'
Ian Z =Cos 48'16' tan 89'01' 56'. Sin 48'16' Cos 40' 30'
Z= 0'57'29"
@
. @
Sidereal time:
Sidereal time = Right ascension + hour angle
Sidereal time = 1h 47 m 10.35 + 2h 42 m
Sidereal time =4h 29"' 10,:JS
Greenwich apparent time:
·&.-.
......
s
41'3S'3O" (average value)
01'00" (refraction correction)
H = 41'34'30" correction altitude
P=90-D
P == 90' - 62'14'29'
P== 27'4S'31"
L==H·P
L = 41'34'30' - 27'45'31"
L = 13'48'59" (latitude of BLLM No, 1)
. tongit.J
DI·ff.,In
uue = 120'30'
--:;sDiff. in longitude = ah 02m
Greenwich apparent time
= local apparent time + diff. in longitude
Greenwich apparent time
= 4h SOm 205 + ah 02' 00"
Greenwich apparent time = 11' 52'" 2(jS
Compute th~ latitude of the place of
obSetvatiOnP·1 from the following date.
Altitude of Polaris on Nov. 3,1967 was
observed to be 15'50'08" at Upper Culmination:
Index error of instrument::: 0
Correction for parallax" 01'00"
Polar distance of Polaris" O'OS'18"
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PRACTICAl ASTRONOMY
Solution:
.~·¢iV.~hlJlij~~~~M.~9lffiQWtb~@'11l~9t
.fr§!~~9h~**ffl¢.m®.()!'#l¢ly~l.~!t¥Wfll®
·~~·$()ijmmll$~Mittl.anq.f®MJM@gW
L..._...L...L-..z._....,.._...Js
.~!I~~~!JI~'I;~to~Ot~~.~~~~lgi~~~~·
pfm~riCIl~ij;p~~~~~W~ • -•15·®·1~m~b~t!~.
me.I~~M:ltlijflll~wirlt1
L =latitude ofplace of obseNafion
L=H-P
H = 15'50'08" + parallax correction
H = 15'50' 08" + 01'
H= 15'51'18"
L = 15'51'08" - 05'18"
L = 15'45'50"
Solution:
z
Sun
NL.-_ _~~.....1._ _....J
it,,.,·I!."
lI!irlll'~lIpllii~lJi@
'~deXftr()f¥+$p"/
[)e¢lih@~~::=faa'44'3$"
.....
Solution:
N ........-....!-,I...,...;X-
H =50'20'00"
D= ·15'30'15"
L =90'· (H+ D)
Corrected H= 50'20'00" - C,
C, = correction for refraction
Corrected H =50'20'00" - 00'00'46.5"
CorrectedH =50'19'13.5"
L =90' • (50'19'13.5" + 15'30'15'')
L = 24'10'31.5" Latitude of the place
..JS
lfj$nece$$ary•• t~{jet¢rmine.the • latituq~(11.a·
H=43'3T
Index Error =- 30"
tr~\lers~ • ~\.!l"'J~Y •. ?IJ(j • f()F1hi§•.• pyrpo~~.!llil.~I$r
Corrected H = 43'37'30"
D = 88'43'35"
P= 90- D = 1'15'25"
L= H-P
H=43'3T30"
p= 1'15'25"
L = 42'22'05" (latitude of place of obseNation)
~·.cetlabt(jcjI~F • 1J1ElJoIlQVlirlg.datawere.l;j~n;··
9rsaM~J()t~~.~asribservedon.the.rneddlary.~n
H =<41'36'(direct)
H =41'3r{revetsed) . .
Correction fotrefraclkm ;: 1'36"
Declimition of star;: 62'14'29"
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5-250
PRACTICAl ASTRONOMY
Solution:
N'L-_---l_.E..
...JS
11.""111.
l8i~1'~~~t.~i.i!
Solution:
Mean value of H =41'36'30"
CorroctedH =41'36'30" -1'36"
Corroded H =41'34'54"
Z=90-H
Z= 90 - 41'34'54"
Z =48'25'06"
L=O-Z
L =62'14'29" - 48'25'06"
L = 13'49'23" latitude of the place
ObseNed altitude = 39'24'00"
Refraction Gorr. =1'11"
H =39'24'00" - 0'01'11"
H =39'22'49"
0= 8;29'21"
H + 0 =47'52'10"
90- (H+ 0) = L
Lat = 42'07'50" N
~n.Ob#~r@tkmWa$.• lriadepo•• th~.$UI1 • at•• nd()n
'.a'}9{h~I~(',()rd~Qaltitudeisg4'~5'1a~.th~~lln
~~IMS9Wh()tth~eqllatQr·8~t~rJllinethe
l~tiNd~rRf,.ltre • p'lace•• ()t.ob~ef'll~tiqn,.p~rall~x·
,Solution:
• aDd·reJrapti(ln~92'!.s~mHiiarl1~ter.#.tl9'18"i
.declioaijoo·m$un.i$~23·02'30".
Solution:
NL..--...l-.L..-::E..
H = 15'50'08" + Parallax correction
H= 15'51'08"
P= 05'18"
L=H-P
L = 15'51'08" - 05'18"
L = 15'45'50"
.....JS
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5-251
PRACTICAl ASTRONOMY
Corrected altitude = H
Obs. altitude
Parallax and refraction
Semi-diameter
H
D
= 24'21'00"
= ·2'00"
=~
= 24'35'18"
= 23'02'30"
=47'3748"
=90- (H+ D)
= 90 - 41'3748"
H+D
L
L
L = 42'22'12"
.~'~;I.llil~;~'~~~;0~g~V:6
:ir.filB
Solution:
&ifI81111.'11
rM~9@nls.1·o1".lfm~.pql~di$«l~QHM
'--_~_~
""'S
.z..
....JS
m~rll;tlh~jQ~tCl/*of.gM¢fVM9rW*9'55'2Q·.
det¢tmlnelh¢.latilt.ideofplace.Qt.Q~$eWfl<!n.
·~
.z._"';"_.....Js
ObseNed altitude =43'3700"
Index Correction = + 30"
43'37'30"
Refraction Corr.
=. 1'01 "
H =43'36'29"
L =H- P
L = 43'36'29" - 0'55'20"
L = 42'41'09"
~
·
Solution:
CD Latitude of the obseNer.:
Altitude = 90 - zenith distance
Altitude = 90' • 76'03'37"
Altitude = 13'56'23"
Corrected H = 13'56'23"
3'40.4"
H= 13'52'42.6"
L=H+P
L = 13'52'42.6" + 0'56'05.3"
L = 14'48'47.9"
® Latitude if it was obseNed on
Upper Culmination.
L=H-P
L= 13'52'42.6"
- 0'56'05.3"
L = 12'56'37.3"
5-252
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SIMPLE CURVES
2. Inscribed angles having the same or equal
intercepted arcs are equal.
.•@
B
RAILROAD AND HIGHWAY CURVES
In highway or railroad construction, the
curves most generally used presently' are
circular curves although parabolic and other
curves are sometimes used. These types of
curves are classified as Simple, Compound,
Reversed or Spiral curves.
e
D
C
LADB=LACB
3. An angle formed by a tangent and a chord
is measured by one half its intercepted arc.
A. Simple Curve:
A simple curve is a circular are, extending
from one tangent to the next. The point where
the curve leaves the first tangent is called the
"point of curvature" (P.C.) and the point where
the curve joins the second tangent is called
the "point of tangency" (P.T.). The P.C. and
P.T. are often called the tangent points. If the
tangent be produced, they will meet in a point
of intersection called the "vertex". The
distance from the vertex to the P.C. or P.T. is
called the "tangent distance". The distance
from the vertex to the curve is called the
"external distance" (measured towards the
center of curvature). While the line joining the
middle of the curve and the middle of the chord
line joining the P.C. and P.T. is called the
"middle ordinate".
l'
LBAC=- LADC
2
4. Tangents from an extemal poiht a circle
are equal.
Geometry of the Circular Curves:
In the study of curves, the following geometric
principles should be emphasized:
1. An inscribed angle is measured by one
half its intercepted arc.
AB=BC
5. Angles whose sides are perpendicular
each to each are either equal or
supplementary.
B~AC
LACB=~ ~AOB
D
LABC=LFED
F
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S-253
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SIMPLE CURVES
Sharpness of the curve is expressed
in any of the three ways:
1. Degree of Curve: (Arc Basis)
Degree of curve is the angle at the
center subtended by an arc of 20 m. is the
Metric system or 100 ft. in the English
system. This is the method generally
used in Highway practice.
2. Degree of Curve: .(Chord Basis)
Degree of curve is the angle
subtended by a chord of 20 meters in
Metric System or 100 ft. in English
System.
a. Metrjc System:
a. Metric System:
b. English System:
By ratio and proportion:
20 2nR
0=
360
D= 360(20)
2nR
- 1145.916
D-
R
b. English System:
00
100
D=
2nR
360
D = 360(100)
2nR
D _ 1145.916(5)
-
R
(5 times the metric system)
D - 5729.58
-
R
. D 50
SIn-=-
2 R
50
R=-
. D
Srn2
3. Radius = Length of radius is stated
Elements of a simple curve:
P.C. = point of curvature
P.T. = point oftangency
P.I. = point of intersection
R = radius of the curve
D = degree of the curve
T = tangent distance
I = -angle of intersection
E = external distance
M = middle ordinate
Lc = length of curve
C = long chord
C1 and C2 = sl,lb-chord
. d1 and d2 = sub-angle
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5-254
SIMPLE CURVES
4. Length of Chord:
C
. I 2
Sln2'="R
C=2R S.1n2'I
I
5. Length of Curve:
Lc
o
R
p.e
o
o
1. Tangent distance:
I
T
tan2'=R"
I
T=Rtan2'
h_ 20
I -D
20 I ~ t ..\
Lc=O Imenc/
h_ 1OO
I - D
Lc = 1~ I (English)
2. External distance:
I
R
Cos 2'= OV
I
OV =R Sec 2'
E=OV-R
I
E=RSec2'-R
E=R(sec~-1)
3. Middle Ordinate:
I AO
Cos 2=S
I
AO = R Cos 2'
M=R-AO
6. Sub-arc: (Arc basis)
9v
fJ._Q
d1 -D
1
d1 =CCD (i'degrees/.\
~2 =M
2C
(50) (minutes)
~ _C1 D(60) (metric system)
2 - 2(20)
~=1.5C1D
I
M=.R-RCos2'
~_
M = R(1 - Cos ~)
~ =0.3 C, D (English system)
C1 0(60)
2 - 2(100)
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SIMPlE CURVES
7. Sub-chords: (Chord b, ;is)
R
R
,,
,,
,,
I
Sjn~=~·
2 2R
I
Sin Q.=~
2 2R
:R
I
C
2R=. 0
SIn2
. d
C,
'"
Sin~
C
d C1 Sin~
.
Sin ~ =- - - (Metric l
2
20
I'
I
d C Sin~
T
=100 (English)
1
Sin
20Sin~
C1 =- - - (Metric)
. 0
S102
100 Sin~
C1 = - - . 0
S10:2
,
" '\.
.
1I
II
/
I
, "';-11
18",: /
X"',r/
ex
....., »'t$
111 ) • -:.,":~.1,~
Sin~=--
2
,"
"', ,I
.A
'<:."I
Solution:
ill Distance from mid point of CUNe to P.I.:
R- 1145.916
-
6
R= 190.99
E=R(Sec~
-1)
E = 190.99 (Sec 18' -1)
E= 9.83 m.
® Distance from mid point of CUNe to the mid
point oflong chord:
M=R (1 -Cos ~)
M= 190.99 (1- Cos 18')
M=9.35m.
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8-256
SIMPlE CURVES
@
Stationing ofB:
S=R8
S = 190.99 (16) 1t
® Stationing of D:
S=R8
S =286.48 (36) 1t
180
180
S= 5133m.
S= 180 m.
Sta. of B= (10 + 020) + (53.33)
Sta. ofB = 10 + 073.33
Sta. of D= (20 + 130.46) + 180
Sfa. ofD = 20 + 310.46
@
Distance DE:
Cos 36' =286.48
OE
OE=354.11 m.
DE =354.11 - 286.48
DE=67.63m.
·.rdTJtJI¥~.I~II-I~ii~fti~nt~:··
•
·.(1).••••. C9IllP~f~ • • t~~E!xlE?rn~I.·9fsl<trce Qf.the
9J1'Y~.>
•
•~• • C;oITlputetflttmj~¢I¢~dihaWfJf.thec;(lo/~ .•·••.
(!)Cqrllpu!E!t~~~tatl(l6jn9PfJlqintAonme
• • • ~~,.~~~~,~~i~~~~&®.~f··ot6' • frQrn
\
w.
\
j""'"<Y-_, "-_
R
%n
'-'''~- \
A _.
R
PoC
Solution:
CD External distance:
2}i1:~o
20+130.46
.
Solution:
CD Long chord:
R= 1145.916
4
R=286.48 m.
!:=
RSin 25'
2
L = 2 (286.48) Sin 25'
L = 242.14m.
\ .\
,
"'-_ \ \
"'--
12'~
\
R"<:i .. 30"
, ffl
-~
I
R
/
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SIMPLE CURVES
E=R(sec~-1)
@
Area oftitlet ofa curve:
E =200 (sec 30' -1)
E=30.94m,
@
Middle ordinate:
M=R(1-COS~)
M=200 (1- cos 30')
M= 26.79m.
@
/
,/
},
Stationing ofpoint A:
'\.26.56
/
5313'/
S=Re
S =200 (12}(n)
180
S=41.89
Sta. A= (1 +200.00) +41.89
Sta. A= 1 + 241,89
A
= TR (2) _nR2 to
2
360'
R= 1145.916
3
R=381.972
T = ~ (381.972)
l~r~i11;1:~~~~~'~imP.I~@~~
I.I.I!~
Solution:
CD Angle ofintersection:
1
T= Rtan '2
!R=Rtan 1
.
2
1 1
2
tan 2=2
~=26.56'
1=53,13'
@
Length ofcurve:
h_ 2O
I-D
- 20 (53.13)
Lc3
Lc = 354,20 m,
T= 190.986
A= 190.986 (381.972)(2)
2
n(381.972f (53.13')
360'
A = 5304,04 sq.m,
/
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5-258
SIMPlE CURIES
c
Solution:
<D Length of curve from P. C:
. ,\
''\,
\'
'\ \1'I
R'
'\,
\,
/
,I'
'\~t/
o
/
a=90·24'4O'
a = 65'20'
'~'
8 =110'50' - 65'20'
o
8=45'30'
R= 1145.916
D
= 1145.916
R
5'
229.18 = OCCos 24'40'
OC =252.20 m.
229.18 252.20
Sin 45'30' = Sin ~
R=229.18m.
fa 8 =105.27
0=128'17
l/l = 180' ·45'30' -128'17
l/l =6'13'
ro
229.18
Sin 6'13' = Sin 45'30'
n 229.18
8=24'40'
LC1
LCl
LCl
=Re
- 229.18 (24'40')1t
CD=34.80m,
180
-
@
= 98.68 m.
Stationing of D:
® Distance CD:
\, 'I
'\,
~'\
\'
\\
'{P'~~,
/
/
/.
'\ '\
'\ / '
'\V'
o
Angle x = 24'40' + 6'13'
Angle x = 30'53'
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SIMPLE CURVES
Rxn
L~ = 180
_229.18(30'531n
L02 180
L~ = 148.24 m.
Station ofD =(2 + 040) + (148.24)
Station of D = (2 + 188.24)
@
Length ofcurve from PC. to A:
S=R28
S - 560.13 (27'391n
18V
S= 270.31 m.
® Length oflong chord:
.Thij~mm@lfJij~f9~~WI~~tro.mpm~(1
.9Q~Mfflp!~~OON~19qi:)~<t®!M9~~M~n~
. a4..n1..·Jfthe;d;@l(L~·frO#lilheiP·C;itijQOllffie
•.
. .
.~~~H~.ZOOfu{>···
.1• ·E~lr".~~.I~~·lrt!I~·.~~I~I··
®•• lqh~~~9J~9fli'lt~~Mli9fm~@W4~
. ·.•.• • • ·.·§4·f@miwt~m~'laogthml§ng~@ff9@.
eP.ri>p:tt)·· .
. .. . .
,1
,
\
I
,I
'
/
I
'
'
/
! /
\,
R\
\
\34" /
\V
Solution:
CD Radius of CUNe:
L
S·In 34' = 2 (560.13)
L=928.74 m.
,
I,
!R
I,
\
! /
\201 /
\V
64
lan8=260
8 = 13.83'
28 = 2T3g
R·64
R
Cos2T39'=R= 560.13m.
•
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8-260
SIMPlE CURVES
Solution:
<D Length oflong chord:
c
/
/U2
'
·11!i~"~I.~II,t~~;~41o'~.·
\,
\R
--- '-- \ ,
--" \
?~L__1
R=286.48
r.JI
Solution:
<D Central angle:
® Distance AB:
c
,
\
""', \R
A
''"'- ""', \
'"'- ""', \
~5"~~~r~
R=286.48
/0+/40.26
2~~8
tan 25' =
AB = 133.59 m.
@
Stationing of x:
S=Re
S= 286.48 (32) 1t
180
S= 160m.
Sta. ofx =(10 + 140.26) + (160)
Sta. ofx= 10+300.26
Lc 20
,=[j
'"'"
- 1145.916
DR
D= 1145.916
286.48
D=4'
240 20
-/ ="4
1=48'
® Distance from mid point of CUNe to mid
point oflong chord:
M= R( 1 • cos
f)
M = 286.48 (1 - cos 24')
M=24.76m.
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8-261
SIMPlE CURVES
@
Area bounded by the tangents and outside
the central cUrve:
T=Rtan24'
T= 286.48 tan 24'
T= 127.55
A
1R (2) 'It R2 I
",.... - 2 - 360
A_ _ 127.55 (286.48)(2) 'It (286.48)2(48)
",...a2
360
Alea = 2162.8
® External distance:
E=R(sec~-1)
E= 336.49 (sec 25' -1)
E=34.79m.
@
Length of long chord:
~=RSin25'
L =2(336.49) Sin 25'
L = 284.41 In.
I_i.
1.'.11
1:lllilllBlijl
111111
Solution:
<D Degree ofcurve:
Solution:
<D Radius of CUNe:
,
/
"
/~
,Ii'\.\
''\.,
'
'
25' \ 25' /
7\7
'\.
A
I,
'~
Sin 50' = 12~20
T= 156.91 m.
T=Rtan 25'
156.91 =Rtan25'
R=336.49m.
D= 1145.916
336.49
D= 3'24'
2.79
S·In e = 21.03
e=T3T
a= 90' -12'-8
a=70'23'
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5-262
SIMPLE CURVES
OB=R+E
OB = R+
@
R(sec~.1)
OB=R+R(SeC~.1)
Area bounded by the CUNe and the tangent
lines:
A
=. RT(2L 1t ~ (24')
2
360'
T= Rtan 12'
T= 286.36 tan 12'
T=60.87
OB= 1.0223 R
In /1 OPB
A = 286.36 (60.87) _1t (2~:~2 (24)
A = 256.26m2
o
OB
R
Sin IJ =Sin a.
1.0223R
R
5iii'"'8 =Sin 70'23'
LJ = 105'39'
0=180-a.-1J
0= 3'58'
. 21.03 _
R
11lSillitillll;
•
.!lPOl'dill~t*19tg91OQN~IJ~.2<l1l)Q·pV'1W~polllt
·~ri~I~~~~.h~ ~9()r~i#t~~ .• §f.~Q~.~~ • ~•
q)FjfJpth~di$tal1¥ofline~P')
~• • ®IY~J°rt'md~r~~Rfs.b!lple@f'I~tb~t
. . . ·•• ·,#icillliefangellt.lq.tfflj.thi'ee.lines.i\B,DI!Z
AM~P·>
@"p~lltlJisatM<itj(jn1t9S~.87tfetermilll'l
.··th¢sWIQl1ingofPT·< '
.
Sin 3'58' - Sin70'23'
R= 286.36m..
Solution:
CD Distance of line BD:
® Length of chord:
A
A
Sin 4'01' = 2 (2~.36)
x = 40.12 m.
c
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SIMPlE CURVES
I}1'IIII'JlI:I!!ill!'I~
LINE
DE
LAT
13.45
DEP
+84.27
.
84.27
tan beanng = 13.45
Bearing ofDE = S 80'56' E
DE
84.27
·t
DIS
ance
=Sin 80'56'
Distance DE =85.34 m.
1= 180' - (85'30' +68'301
1=26'
B=180'·26'
B=154'
a =80'56'·68'30'
a= 12'26'
e= 180' ·154' ·12'26'
e=13'34'
~~lr:~~~I~I~~~~~~"1
~~P'1~ld~~fa~~~~~~~~¢~~~_1
•.,: .
:~:··:::P~~~.~~~::~~im~~~~~;i;:::::::;:::;:;:::;::.· ~::-:::.":,.:::: i: :~:i:i;i;i~[:[ r~i~(;
~jllllili~li:
. . .•
.•....,.
.~ Qet¢®IM.tffl.~Og!l{qt~llti,te.
Solution:
CD Radius:
A
@
Degree ofcurve:
T1 + T2 =DE
Rtan 6'47' +Rtan 6'13' =85.34
R=374.50
D= 1145.916
374.50
D=3'04'
'\
/
\
'v'
/
T= (4 + 360.2) • (4 +288.4)
T= 71.8m.
T=
Rtan~
26
71.8 =Rtan"2
R";311m.
@
External distance:
E=R(seC~.1)
E=311
(sec~6.1)
E= 8.18m.
D.
Sta. at PT = sta. at point D. T1 + Lc
Sta. at PT =(1 + 052.87)·44.55 +169.93
Sta. at PT= (1 + 178.25)
C
~\
Stationing ofPT:
T1 = 374.5 tan 6'47'
T1 =44.55m.
Lc= 20 I
_ 20 (26)
Lc- 3.06
Lc = 169.93-
A
'\
BD
DE
Sin a =Sin B
BD
85.34
Sin 12'26' = Sin 154'
BD=41.91m.
@
{iii
@
Middle ordinate:
M=R(1.COS~)
M=311
(1.COS~)
M=7.97m.
/~
/
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5-,264
SIMPlE CURVES
@
Chord distance:
C=2Rsin l
- 1145.916
R- D
26
C=2(311) Sin 2"
- 1145.916
R6
2
C= 139.92m.
@
Length ofCUNe:
Lc =RI
1t
180
1t
Lc =(311) 26 180
Lc = 141.13 m.
R.=190.99
E= R(sec 112· 1)
E=190.99 (sec 30' -1)
E=29,55m.
® Distance of mid point of CUNe to mid point
oflong chord:
M= R(1- Cos 1!2)
M=190.99 (1 • Cos 30')
M=25.59m,
® Stationing ofB:
.li~'I~~I!~~~~~':~r:~i
@\!'1®~@6~af~t:qli~6AAf@ffl*~ij@#
m~ll6~m~~!M~~~lp!@~~gl~9f~)
Willilli~tM9¢l\tfbffi@IHfjElgp;<
Solution:
CD Distance from mid point of CUNe to P.I.
S=Re
S =190.99 (16) 1t
180
$=53.33
Sta. ofB =10 + (020) + (53.33)
Sta. of B = 10 + 073.33
t~~ • tans~~ltrr(ll~ep.g .• h~$.a(jlte¢ll«@ll~
n()rth•• and.the.ta~gerlt • tht()U91'1•• th~ • ~rnt'l~s~.
b~rlng.()f.N,50· .•~, . . • lt.nll,$a.ra,dl~~pf2QOffi; •
lJ~in$ • ·.·,;jrc • • b<!llis.·•••..·.$tliltiQilitJ9•• • pl•• P.q,•• i§.
12.+060,
.
.
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SIMPlE CURVES
S=R8
- 200 (28) 1t
S - 180
S=97.74m.
-
Sta. ofB =(12 +060) +(97.74)
Sta. ofB= 12+ 151,74
Solution:
(j)
Tangent distance:
Solution:
(j)
Middle ordinate:
T= Rtan 25'
T=200 tan 25'
T=93.26m.
® Long chord:
'Sin 25'=1:...
2R
L =2(200) Sin 25'
L = 169,05m.
@
p.e
Stationing of B:
Lc
2)
T=o
210 20
- I =4"
1=42'
M= R(1- Cos 112)
. R= 11~.916 =286.48
M= 286.48 (1- Cos 21')
M= 19.03m.
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5-266
SIMPLE CURVES
® Extemal distance:
Solution:
E= R(Sec 1t2-1)
E = 286.48 (Sec 21' - 1)
E=20.38m,
<D Radius of the CUNe:
Area of fillet ofcurve:
@
p.e
T= Rlan 21'
T = 286.48 tan 21'
T= 109.97 m.
_ 7R(2)
A-
1t
Ff2 e
2 • 360
'\ _ (109.97)(286.48)(2) 1t (286.48)2 (42)
2
•
360
2
A = 1423.69 m say 1424 m2
r-
Sin 6' = 10
R
R=95.67m.
® Angle ofintersection:
\fi • ~~f!M®~·~ij!lI~~ • •()f•• lWl)i.@~t:mij~l~t~
::.~.R!t@~M.llie·¢~r¥.m~~~tireg,"@jlh~·
:.JjJglfut•.~~~i~Smm~gh.th~.P·C· • ~@··~."S'~~~·
12~1$;.rl!$pe¢1I.y~!~; • • • • • The • • Chord••·• ijllllll*~
·Pe~~nRatKI.$is?O"t • {St~nd~rd·iij.mettk;
$y~tf'!rn}.Wt:tlI¢.tMI609.cl1ordi$ •. WQm.ml:1!l!I'$·
··lQOO'························
.
··i•• ·.~~~I'e~ieQ~j~.tiori.Of.·tb~··
$.lmp~9WV~,
.
~CQrnPllte1het8tlgentdiStaIlce..
..
/'
"
".
R'
/
"
, 1/2 112 '
/ /R
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SIMPlE CURVES
Sin l= 70
® Degree of CUNe:
T= Rtan 20'
124.46 = Rtan 20'
2 R
S' 1 50
In
2= 95.67
R=341.95
D= 1145.916
341.95
~ =31.5'
D=3.35'
1=63'
@
Tangent distance:
T= Rtan l
2 .
T= 95.67 tan 31.5'
T= 58.63 nI.
@
Stationing of8:
S=RS
S = 341.95 (16) 1t
180
S=95.49
Sta. of 8 =(10 + 060) + (95.49)
Sta. of8= 10+ 155.49
Ift.ili
IIIIII
1:11;'111 1:11.""i~
Solution:
CD Tangent dsistance:
Solution:
CD Radius of CUNe:
1\
10+
.
\\'\
R\\
"
'\
R-60"
\\
".
80 = TSin 40'
T= 124.46
,I
~ ~31~
28.0~
I
8-268
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SIMPLE CURVES
60
tane=240
e =14.04'
Solution:
<D Deflection angle at the P.C.:
2e =28.08'
R- 60 =R Cos 28.08'
0.11nR=60
R=509,70m.
® Tangent distance:
T=Rtan31'
T= 509.70 tan 31'
T=306.26m.
@
Stationing ofpainf x:
S=Re
S =509.70 (28.08) 1t
180
S=249.80
Cos 2e =219.18
229.18
2e= 16.988'
e= 8.49' (deflection angle)
Sta. of x =(10 + 080) + (249.80)
Sta. ofx= 10+ 329,8
@
1111
Stationing at B:
S= R(2e)
S= 229.18 (16.988)' 1t
180
S=67.95m.
Sta. of B=(10 + 120.60) +67.95
Sta. ofB= 10+ 188.55
@
Chord distance from P. C. to B:
Sin 8.49' = ~~
AB= 67.73 m.
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..MPlE CDIEI
II~r.~kl~If.~I~I'~~I;it~.
1:1.'• •
Solution:
CD Angle ofintersection:
,
/
••
Ill• •JI.
Solution:
CD Radius of CUNe:
I
".,
jR
R"-
I'
'-<-i
''J
A
T=Rtan.!..
2
T=2Ttan .!..
2
I
tan 2=0.5
~=26.56'
T1 + T2 = 300
Rtan 12'37.5' +R tan 30'54' =300
R= 364.75
1=53.13'
1=53'08'
® Length of CUNe:
114
R = 1;916 = 286.48 m.
bc._ 20
® Stationing of p.e. = (10 + 585) - T1
T1 = 364.75 tan 12'37.5'
T1 = 81.70 m.
I - D
L = 20(53.13)
c
4
Lc = 265.65 m.
@
Area enclosed by the CUNe:
A
Sta. of p.e. =(10 + 585) •(81.70)
Sta. of p.e. = (10 + 503.3)
=53.13'1t (286.48)2
360'
A = 380.54 rrf
@
Length of CUNe:
S=R8
S = 364.75 (8T03J1t
180
S= 554. 17m.
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S-270
SIMPLE CURVES
Sin 10' =.1Q..
2 RlO
R10 =114.74 m.
I.'.
1.
11• • .
.tqt1~A~··)<·······
O'C= Re- RlO
O·C=191.07-114.74
O·C=76.33
Solution:
CD Central angle of 10' center curve:
OC=Re- Ra
OC =191.07 -143.37
OC=47.70
Using Sine Law:
47.70
76.33
. !.m. =Sin 136'
SIn 2
ful=25'44'
2
110 = 51'28'
Central angle of 6' end curves:
@
16 +~+ 136' =180'
16 + 25'44' + 136' =180'
16 = 18'16'
@
OA=Ra
SinQ=~
2 2Ra
. 8' 10
S,n-=-
2
Ra
Ra= 143.47 m.
'. 6' 10
SIn---
2 - Re
R6 = 191.07 m.
stationing of P. T.:
- 20/1
LCl- 0
1
_20(18'16')
Lcl6'
LCl =60.89 m.
_ 20/2
Lc2- ~
- 20 (51'28')
LOL - 10'
LOL =102.93 m.
P. T. = (10 + 185.42) + 60.89 + 102.93 + 60.89
P. T. = (10 + 410,13)
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SIMPLE eURVES
tan 55'04' = 70
x
x=48.89m.
I T
tan-=2 R
T=163.80 tan 27'32'
T=85.39 m.
7+812
~IIIIIIIIIIIIII~IIIII
Sta. ofpoint ofdeviation (P. C.)
=(7 + 812) • (85.39 +48.89)
=7+ 677,72
I
I
:7Om
mounth of tunn~1
I
I
I
® Stationing ofmouth of tunnel:
b.c._ 20
I - D
_55'04' (20)
Il'lt.
4-
Sta. ofmouth oftunnel = (7 + 677.72) + (157.33)
Sta. ofmouth oftunnel = 7 + 835,05
Solution:
CD Stationing of the point of deviation:
7'
4= 157.33m.
@
Direction ofrailway in the tunnel:
90' • 55'04' =34'56'
Direction is S. 34'56' E.
j
I
,
rt----...-.---,/.-------------....~. - .r.
al
,"
~11I2,'
r;,''1/
,
U
~
~
/
,/;=163.80
~,,'
I
~y
Sin Q= 10
2 R
Sin 3.5'
,
/
=~
R=163.80m.
93.8
Cas I = 163.80
1= 55'04'
r~ilway in
rhe runnel
•
(j)yompu~ • • th~ • •f#ntral.lln~l~of lt(~ • • #~
... (;O~e.···
®.. 9()~let!)El$diusPfIt¥fl~WWW~·.
@·VVhat1s!heStallonio96HMheWf;'P,
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SIMPLE CURVES
Solution:
@
CD New cental angle ofnew cUlVe:
2W'
Sta.ofnewP.C.:
T1 "y=b
b= 431.n -166.70
b=265.07m.
x= T2-b
x =360 - 265.07
x= 94.93 m.
$fa ofnewP.C. =(10+ n1.20)-94.93
Sta. ofnew P. C. =10 + 626.27
Old central angle = 346 - 220
Old central angle = 126'
New central angle:
1= 180' - 54' -22'
1= 104'
@
Radius ofnew curve:
I
T1 =Rltan~
T1 =220 tan 63'
T1 =431.n
Using Sine Law:
431.77 _-lL
Sin 104' - Sin 54'
T2 = 360 m.
-l.-_ 431.n
Sin 22' - Sin 104'
y:r 166.70 m.
I
T2 =R2tan~
360 = R2 tan 52'
R2·= 281.26 m.
111r"
Solution:
CD Stationing of the P. C:
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SIMPlE enlEs
@
Radius ofthe approach curve:
- 1145.916
R1- 0
R - 1145.916
18
R1 =143.24 m.
R3 =R,·3
R3 =143.24· 3
R3 =140.24 m.
R2 = R3 +9.51 X
R2 =140.24 + 9.51 (12P
R2 =259.12 m.
@
Degree of curve:
0- 1145.91"
-
R
0= 1145.916
0=8'
259.12
15 20
0=4'25'
-;-=0
,,_15(8)
- 20
IIl=S'
tanS' =L
h1
t@mjb~figijr~~;)<··························
h1 =xCots'
h1 =9.51 x
SinS' =~
h2 =xCscS'
h2 =9.57 x
Cr._ ••
20m.
B'
_.l~~_ ~
I
I
-:30
: R
I
.
:
® R2 = R3 + 9.51 x
I
~'
CD R2=R3+9.57x-0.75
R3 + 9.51 x+ R3 + 9.57 x·0.75
0.06 x= 0.75
x= 12.5m.
Sta. ofnew P.C. =5 + 100
12.5
NewP.C. =5+087,5
. /A
/
I
CD R2 =R3 + ~ ·0.75
R2=R3+h1
••__
,0
,"
.
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5-274
SIMPLE CURVES
Solution:
@
CD Radius of the simple curve:
G
ldg.
f
160m J ----"'.A
JL--- ,
,:-,'30 m
~
20m
B
-.l
~
3P,I
,,'
\
Ama of ACBD:
= 20 (160) 161.25 (135.45) Sin 61'08'
A
2 +
2
1t (135.45}2(63'541
•
360'
A = 1431.70 rTf
,
D
,
: R\
,,'R
I
I
\'
,,'
:
0
,,
I
20
tan 0 = 160
0=1'08'
IJ = 90' - TOO'
IJ =82'52'
a =180' - 82'5Z· 30'
a =61'08'
AB = ~ (20)2 +(160)2
AB =161.25 m.
Using Cosine Law:
(30+ F{J2 =(161.25)2+ (RP
• 2R (161.25) Cos 67'08'
900 +60R + R2 =26001.56 + R2 -125.32 R
185.32R == 25101.56
R= 135,45m.
® Central angle of curve:
;Dli!!:~i
Solution:
CD Min. distance:
h12 =(90)2- (5W
h1 = 74.83m.
=(110}2 - (30}2
h2 = 105.83 m.
hi
B
o
161.25 135.45
Sin I =Sin 67'08'
1=63'54'
h3 == ~-h1
h3 = 105.83· 74.83
h3 =31 m.
• Min. distance between piers = h
h = h3 + 2.5 + 2.5
h = 31 + 5
h == 36 m. (clear distance between piers)
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SIMPLE CURVES
@
Solution:
Anglee:
74.83
Cos 0=00-
<D Smallest radius ofrail track curve:
0=33'45'
B
_.!:!i:!:..!!J.
25V
CosfJ-90+ 20
n _ 74.83 + 31
eos/,)110
115
fJ =15'50'
e=33'45 -15'50'
A
0
117.70
e = 17'55'
® Area ofthe road between Aand B:
AV= " (115~ + (25)2
AV= 117.70 m.
115
tan f1l =-.
25
f1l = 77'44'
a. = 77'44' - 55'
- 1t (110)2 (33'45' 1t (90)2 (33'452
A360'
360'
A = 1178,10 m2
a. = 22'44'
AC = 117.70 sin 22'44'
AC=45.48m.
VC = 117.70 cos 22'44'
VC = 108.56 m.
OVcos35' =R
OV= 1.22R
DC= OV- VC
OC = 1.22 R- 108.56
OA= R-11
(R-11)2 =(1.22R·108.56f + (45.48f
R=431 m.
p.T.
@
Area of the building:
.tan35' =~
T=431 tan 35'
T= 301.79 m.
AD+ ED+ VB.= T
AD + 105 + 25 =301.79
AD= 171.79m.
Area ofbldg. = 50(171.79)
Area ofbldg. = 8589.5 m2
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8-276
SIMPlE CURVES
@
Stationing of P. T.:
Lc=RI
L =431(70) 1t
c
180
L=526.57m.
PC =(10 + 242.60) + (526.57)
PC = 10+ 769.17
(R- SO)2 =(h-4O)2 + (R-1oo~
R2 ·1ooR + 2500 =~. 80h + 1600
+ R2 . 200R + 10000
100R = ~. 80h + 9100
tan 25' =~
h=0.466R
100R = (0.466R)2. 80(0.466R) + 9100
100R = 0.217R2 - 37.28R +9100
R2 -632.63R +41935.48 =0
R- 632.63 ±482.16
-
2
R=557.56m.
@
Tangent distance:
h=0.466R
h =0.466 (557.56)
h= 259.82 m.
T= 259.82 m.
@
Stationing of C:
" ,'-,\ '\ I
'
,
'-,~
1,• •j~lii~~!·
Solution:
CD Radius ofthe CUNe:
; + (219.82) = (507.56)2
x=457.49
+ (259.82)2 = (557.86)2
y=493.66
C e = 493.66
os
557.56
e=27'42'
IJ =SO' - 27'42'
IJ =22'18'
R81t
Lc=WO
I =557.56 (22'18')1t
'-c
180'
4=217 m.
f
Sta. ofC =(10 + 240.26) + 217
Sta. of C = 10 + 457.26
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'SIMPIE CURVES
a =90'·24'40'
a=65'20'
8 =110'50'·65'20'
8=45'30'
,
229.18
CoS2440'=OC
OC =252.19 m.
.111
Using Sine Law:
229,18
,ial:JiIIIl'
a=6'12'
aJ
229.18
Sin 6'12':: Sin 45'30'
CD =34,70m,
Distance CD:
@
, "
\
,\ I
\ I'
\
R\
c
III
0=128'18'
a= 180' ·128'18' ·45'30'
Solution:
(j)
259.19
Sin 45'30' = Sin
I
9
\J.--a /
\~-r:
\\1/
\~'
o
/
,/
,
P.T.
Stationing ofpoint D:
R(a + a)1t
i.e = 180
229.18 (24'40' + 6'12j1t
180'
229.18 (30'52) 1t
i.e =
180
I.e = 123.46 m.
i.e =
Sta. Qfpoint D = (2 + 040) + (123.46) .
Sta. ofpoint D = 2+ 163,46
® Deflection ofpoint D:
o
R_ 114.916
-
D
- 1145.916
R-
5
R=229.18 m.
105.27
tan e =229.18
e = 24'40'
d =.1 (30'S2')
2
d= 15'26'
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5-278
SIMPlE CURVES
L~
=(1 + 260) • (1 + 180)
L~=80m.
20 _bfl
D - 12
- (80X3)
I2- 20
12 = 12'
1=11 + 12
1=9' + 12'
1=21"
@
Deflection angle:
Deflection angle = ~
Deflection angle = 10'30' R
Solution:
CD Angle ofintersection of the simple CUNe:
@
Tangent distance of the simple CUNe:
I
T= R tan "2
- 1145.916
RD
Rr= 11~.916
,,
,,,
,
,,,
,,
,,
R= 381.97 m.
I
T= Rtan-
:
\
'\ ll!
/
I, /
\1'7'
2
T= 381.97 tan 10'30'
T= 70.8 m.
~I
Deflection angle at 1 + 180 to locate P. T.
_!.l !2
-2 + 2
LC1 = (1 + 180) - (1 + 120)
LC1 =60m.
A straight railroad IP intersects the curve
hjghw~routeAB. Distance oil thero).lte are
measured along the arc. USing the data in the
figure.
.
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SIMPlE CURVES
o
Sin 70' - 587.96
",'- ..........
"
.........R=600
- PI
PI = 625.69 m.
""""
o
9:Jb,/~1
"J,
I
I
,,'110'/,"
,/600
,
,,
I
: • g~~~~;.f~~~l~ •••~~,• • • • • • • ••• • • • • •. . . . .
o
~·O(lIl)P\l~!M.$tatfQoiog.9t@IDt)(>
Solution:
<D Distance 01:
tan 52' =~
T= 600 tan 52'
T= 767.96 m.
AP= 767.96 ·180
AP = 587.96 m.
, AI
tan 20 = 587.96
AI =214 m.
01= 600·214
01= 386m.
p
587.96
tan g = 600
g =44'25'
Cos 44'25' - 600
-OP
op = 840.06 m.
840.06
600
Sin a = Sin 25'35'
® Distance PX:
o
.....q ,b"t ..........
-J,' II
...
"
I
::
I
I
I
I
I
I
........R=600
.....
a= 142'48'
,
o
"
"
,
I
I
I
I
I
X
p
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5-280
SllPlE CURVES
PX
600
Sin 11'37:: Sin 25'35"
PX:: 279,79 m.
@
Stationing ofpoint x:
44'25'. 0
'4
Solution:
'~l1'37'
<D Radius:
:: 6=32'48'
N
,,",,,
Sta.
a::44'25 ·11'37
a::32'48'
Ran
AX= 180
AX:: 600 (32'48') n
180
AX = 343.48 m.
Stationing ofpoint x = (50 + 000) + (343.48)
Stationing ofpoint x = 50+ 343.48
1= 257'25'·220'45
I:: 36'40'
E= 10.20
ov =R+E
B
C
p.T.
..
.,,,
..
,.
,,
,,
,,
220'45'
\R
,.
,
JI
.
18'20/.',,\
----------------.--..
---°0
.
R
. Cos18'2O'::.B..
OV
R= (R + E) Cos 18'20'
R= RCos 18'20' + 10.20 Cos 18'20'
- 10.20 (0.949)
R0.051
R= 190.76m,
® Stationing at PT:
1
T=Rtan2'
T = 190.76 tan 18'20'
T:: 63.21 m.
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S-281
SIMPLE CURVES
D= 11~916
D= 1145.916
190.76
D=6'
_20 I
I
'"I: -
D
I
20(36.67)
'"1:=
8
4= 122.22m.
Sta. ofP. T. = (10 + 220.47) + 122.22
Sta. ofP. T. = 10+ 342.69
o
@
Stationing of PC = (10 + 283.68) - 63.21
Stationing of PC = 10 + 220.47
W=35.2O
V'P. T. =63.21 • 35.20
V'P. T. = 28.01
28.01
tan a = 190.76
a=8'2Z
ll.
Cos 8'22 = 190.76
OV
OV = 194.24 m.
190.76 194.24
Sin 53'39' = Sin B
B= 124'54'
/II =180' ·124'54' - 53'39'
/11=1'27
~PiwiM#J~·#b~ij9~P(w.M\hPt·:lh~
V'
V'
.,
.
:&
r
'ii
-',
G)
~~f:
. :~:::~~:H~)]~:\?}r<}:::::<:::;:::::::::::::::-::::·::::.:"
Solution:
Change of length of radius:
9'4~
........ ./,
~
,
......0
o
Lc, = 32.60 m.
~·.~~.~.~· ~.1.~!
•.:•. •. . •. •. •. .• . •. .• [.•·. .• .• .: .r.:
. . •.;.::.. •. .·:..·•. •. .• . •. •. . . •. .• .*.....
•..1• .•.·•·:._•.•.• •. •.
I!T.
\~
Arc P. T. to K:
Lc, =R/II
- 190.76 (9.817) 1t
Lc, 180
Stationing of point K:
Sta. Ii K = (10 + 342.69) - 32.60
Sta. of K =10 + 310,09
13"06'
old
p.T.
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S-282
SIMPlE CURVES
- 1145.916
R10
- 1145.916
R1- 4
R1 = 286.48 m.
L =201
.
c
0
- 20 (26'12')
0 - 100
0=5.24'
- 1145.916
R2 - 5.24'
R2 = 218.69 m.
AB=R1 -R2
AB =286.48 -218.69
AB = 67.79 (change oflength ofradius)
@
·1.··.·III.I!·I.i!~'.· • • • ·
Solution:
CD New angle ofintersection:
1= 180' - 70' - 82'30'
1= 27'30'
New 12, =21'30' - TOO
New 12 = 20'30'
@
New radius ofCUNe:
Distance and direction where P. T. must
moved:
Sin 13'06' = AC
AB
AC = 67.79 Sin 13'06'
AC= 15.36 m.
AD = 2 (15.36)
AD = 30.72
Therefore P. T. must be moved at a
distance of 30.72 m. at an angle of 13'06'
from the second tangent.
@
Distance between two parallel tangents:
OE= AD Sin 13'06'
OE= 30.72 Sin 13'06'
DE=6.96m.
R - 1145.916
14.5
R1 = 254.65 m.
T1 = R1 tan b.
2
.·8~&~~~N~~!~.II~ilik~.~tI~ • I~e~~i~
•~.~a~~t~~T~~dm.~1• ~~.~6.g;~~~G~
• I~··
••
llY
·pr()p()~~d • t()@:mr~M~.~h~ p~otr~I • • a"gl~ • •
Cfj~rgjQglh~9lr~ct~l1l)f.fh~.f()!\Vard.tan~am
l>Yan~ngkJCJf;rj)n$~~~iiW~Yth/lLth~
·P8Siti()n•• qft~em •• ()UhafW'#~td.·taI19~"t.al1d
the••{lirectio(l•• ()t.th~.·~~¢k: • t~(lg~nl$Mll·.t¢mam
unth!ltl9e<l·
T1 = 254.65 tan 13'45'
T1 = 62.31 m.
Considering triangle AVa:
AB
62.31
Sin 152'30' = Sin 20'30'
AS = 82.16
T2 = 82.16
82.16
R2 = tan 10'15'
R2 = 454.35 (new radius)
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SIMPlE CURVES
@
Stationing of new P. C:
VB
62.31
Sin T = Sin 20'30'
VB =21.68
h=Tl- V8
h =62.31 - 21.68
h=40.63
T2 - h = 82.16 -40.63
T2 -h=41.53
Considering triangle ACE:
42,50
EC
Sin 80' =Sin 70'
EC=40.55m.
42,50
AC
Sin 80' =Sin 30'
AC= 21.58 m,
® Angle of arc EB:
Considering triangle DEC:
tan I' =40.55
Sta. of new P.C. =(10 + 345.43) - (41.53)
Sta. ofnewP.C.:: 10+303,90
120
" =18'40'
ex =90' -18'40'
ex= 71'21'
B= 180' . 80' - 71'20'
B= 180' -151'20'
fj =28'40'
Considering OEC:
120
Case=OC
120
OC=CoS18'4O'
DC= 126.67
Considering triangle OBC:
120
_ 126.67
Sin 28'40' - Sin e
e=149'34'
Solution:
Angle X= 180' -149'34' - 28'40'
AngleX= 180' -178'14'
Angle X =1'46'
CD Distance AC:
Arc EB= 18'40' + 1'46'
Arc EB = 20'26'
@
!:f
,,
l'
-
,
32+542 " "
\,
......
9
\x
;';>00<
180
EB=42,79m.
/R:;:120
"<:7 ",=1'46'
O=IS'4()'-':>'
Stationing of B:
EB_ 120 (20'26)1t
Stationing of B =(3 + 025,42) + (42.79)
Stationing of B = 3 + 068.21
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S-284
SIMPLE CURVES
® Radius ofnew CUNe:
tan 12' .=I2.
R
i"~.
1~1I:11111I~1:
Solution:
<D New tangent distance:
1=284' - 260
1= 24'
- 1145.916
R1 5'
R1 :: 229.18 m.
tan
12'=~
T1 = 229.18 tan 12'
T1 =48.71 m.
tan 24' =~
h = 11.23 m.
T2 =48.71-11.23
T2 = 37.48m.
2
R
37.48
2 =tan 12'
R2 = 176,33 m,
@
Stationing ofnew P.C.:
5
W=Sin24'
W=12.29m.
x= T2" 12.29
x=37.48-12.29
x = 25.19 m.
Stationing ofnew P.C. =(8 + 095.21),25.19
Stationing ofnew P;C. = 8 + 070.02
".II~j
Solution:
<D Bearing AC:
E
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SIMPlE CURVES
Departure of/ine AC = 990 -420 =570
Latitude of line AC = 1570·280 =1290
.
_Departure
tan beanng AC - Latitude
@
Angle:
Angle CBD:: 40'58'
1
o = 2of angle
cao
·
570
tan beanng =1290
Bearing (AC) =N. 23"50' W.
1
0=2(40'581
0=20"29'
® Bearing DB:
Departure of line BC:: 1100 Sin 5'
Departure ofline BC =95.9 m.
Latitude ofline BC = 1100 Cos 5'
Latitude ofline BC = 1095.8 m.
Departure ofB = 420 + 95.9
Departure ofB =515.9 E. (Dep.)
Latitude ofB = 1570 - 1095.8
Latitude ofB=474.2 N. (Lat.)
Departure ofline AB = 990·515.9 =474.10
Latitude ofline AB =474.2·280 =194.20
. lAB" 474.10
tafi beanng
I' ~:: 194.20
bearing (AS) = N. 67'44' W.
·t
lAB)
474.10
DIS
ance I' =Sin 67'44'
AB =512.30 m.
~f:iMtt\ilidl"~fff1l!c~~.)
®FlodtMi~tS!M8frp1l1P,C.Il) ~P.Lii
~ • • .• $t?~QI'lJPt~;~;
Fjnq .tn•. _~llOi1jM.of.1hefn •• lf.g
. 1,. is at
.
• . •.
Considering triangle ABO:
Solution:
G)
N
Radius of the cUlve:
D
~M;;·
512.30 _ 1100
Sin a - Sin 70'44'
. _ 512.30 Sin 76'44'
SIna1100
a =26'58'
Bearing ofDB =26'58' + 9'
Bearing ofDB = S. 35'58' W.
,,·f'
,
,
··,
··
··,
·,.
",'
,,/'
>.. . . .
I' , ...
"
\ 13',' .
~17'
,,-
T1 + T2 =86.42 m.
T1 :: R tan 6'30'
T2 :: Rtan 8'30'
..."
•• --
;
......
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S·286
SIMPlE CURVES
R (tan 6'30' + tan 8'301 =86:42
R= 86.42
0.263
R=328.59
Solution:
G)
New external distance:
Distance from P.C. to the P.I.:
T1 =328.59 tan 6'30'
T1 =37.44m.
@
p.c~
Considering triangle CVD:
86.42
CV
Sin 150' = Sin 17'
CV=50.53 m.
AV= T1 + CV
AV= 37.44 + 50.53
AV= 87.97 m.
1'10'
\.Q)('1-
R-
I
"
y---- i
,,-
I"
\1'1JL!
~i,
I
/
j'
90
R= tan 14'15'
R= 354.38 m.
D
Cos 14'15' =..B.OV
354.38
OV= Cos 14'15'
OV=365.63
E1 = OV-R
E1 = 365.63 • 354.38
E1 =11.25 m.
Lc = 20 (30)
3.49
Lc =171.92 m.
New external distance:
E2 =11.25 +6
E2 = 17.25
Stationing ofP. T. =(10 + 264.27) + (171.92)
Stationing of P. T. = 10 + 436.19 '
@
New bearing:
OV'=R+ E2
. OV' =354.38 + 17.25
OV' =371.63 m.
imli~~t,mr'i~~!i!.~I~;'··
.••
®CPrnPut~lryE!oeWb~arifl~ol~E!PQM
t<l.n~~~tlin~Witht~,TIrstt<l"9~nJlin~
.~m~il'll~g • ir.• mE! • ~fl1~ • dlt~ti()milll)m~r
t~at.ttm.~egr~E!cofCUlVe~oe$fl9t.M~rlge.
CotnputElth!?l1talkmingofth¢n!ilcWRr· .
. '/"
f R
Cos-=2 OV
f 354.38
Cos 2" =371.63
\newP.T.
,,',
T1 = (10 +362.40) - (10 + 272.40)
T1 =90 m.
T1 = R tan 14'15'
L =20 I
c 0
@••••••
'
',t,'/'
""
~........
D _ 1145.916
328.59
D=3.49'
®•• • CpfupuW.m~new~~l11a'.dis~~ge
I
::
\14'lr...:
Stationing of the P.I.:
Stationing P.C. = (10 + 352.24)· (87.97)
Stationing P.C. = 10 + 264.27
- 1145.916
@
:,:
\
"
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SIMPlE CURVES
~= 17'32'
2
, =35'04'
Sin 4'30' = 2~1
,
R1 =321.31 m.
Old bearing of 2nd tangent line is = S. ,79' E.
New bearing of 2nd tangent fine is = S. 72'26' E.
T1 =R 1 tan '2
T, =321.31 Ian 15'
T1 =86.09m.
'=30·9 =21'
Stationing of the new P. T.:
L =20 I
@
c
0
New angle ofintersection =, =21'
0= 1145.916
354.38
0=3.23'
I = 20 (35'041
Lc
3.23'
Lc = 217.15 m.
New P. T. = (10 +272.40) + 217.15
New P. T. = 10 + 489.55
® Slationing of new P. T.:
Sla. of p.e. = (10 + 314.62)·86.09
Sla. ofP.C. = 10 + 228.53
0= 1145.916
R
0= 1145.916
321.31
0=3.57'
L = 20 I
c
·.d~~%~~~~~t~~~.~~6r~~,co~:p~~~~~,~~~
•
r~~iB~hJrt~~rh~i~t2~t~[61~t~b~I~~~
qUfvebayios . .•~•.. • Qir~qtj9.il.· . 9.fN.. ·.Wt1§t.. . W,
·fu~f:~.~1~]u&~i~J~ifu~~~~hg~t • ~~Q!Q9 •
@
@
3.57
Lc =117.65
Sla. ofnew P. T. == (10 + 228.53) + 117.65
Sla. ofnew P. T. = 10 + 346.18
@
···Coml'~te.th~MW •.~ngl~o.flht~~@ti6ri.(lf
. (g)•••
~8~~j~~.$faIIO~'~9.0f.6~w·~.f, • • • ••••• •••• •
T2 =59.55
Sla. of new vertex == (10 + 228.53) + 59.55
Sla. ofnew vertex = 10 + 288.08
Solution:
CD New angle intersection ofthe tangents:
o
R\
i.: ~
\\T!i21'
,:4·~.i;1{
,
,, ,,
\JJ/
Slationing ofnew vertex:
f
T2 = Rlan '2
21'
T2 =321.31 tan '2
·G9mp~t¢t~¢~Elliq~jIl9.9f·l'1~w.~rt~;,><.·•.
,,
,,,
,,
,,
0
Lc = 20 (21)
The bearings 6ftwo tai'lgents areS. 2Q'3ftE,
and S:35·OO~W. Jfthe degree ofcuryel$W
for a chOrd anD meters and
the vertex i$10 +205.50.
lhestatioriingat.
CD Find fhesflilioning at the p.r· . i
Find the lengthdf the lastsubchord;El/1d
ils corresponding sUb-angle. .
.. . ....
@ . Find the deflection dislance to the
station. .
... .
@
fourth
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S-288
SIMPlE CURVES
Solution:
@
\
---...-
Defleefion distance to the fourth station:
\~
........-}\
o
/!.-/
•
1= 20'50' + 35'00'
1= 55'30'
L=27'45'
2
0=10'
C= 10 melers
(j)
Stationing at the P. T.
R=~
2Sin~
5
R= Sin 5' =57.37 m.
T= Rtan 27'45'
T= 57.37 tan 27'45'
T=30.18m.
Station ofP.C. = (10 + 205.50) - (30.18)
P.C. = 10 + 175.32
L = 1Qill.
o
L =10(;;.5)
=55.50 m.
Station ofP. T. = (10 + 175.32) + (55.50)
. Station ofP. T. = 10+ 230.32
@
Length of the last subchord, to the fourth
station:
~ = (10 + 230.82)· (10 + 230.00)
c2= 0.82m.
!!2_Q
C
_E2..Q
dr C
c2 -
d - 0.82(10)
r
10
d2 =0.82'
'd2= 0'49'
. 0 x
SIn-=2 10
x=
10Sin~
But:
Sin~=~
10 (10)
x= 2R
100
x= 2(57.37)
x=0.87m.
Defleefion distance =2x
Defleefion distance = 2(0.87)
Defleefion distance =1.74 m.
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8-289
Pdfbooksforum.com
SIMPLE CURVES
Solution:
@
Azimuth ofnew forward tangent'
=203'45 + 18'26'
= 222'11'
@
T
.
Stationing ofnew P. T.
New I = 18'26'
0- 1145.916
-
\
+025.32 \
\
" ')3'O~'
\ \.
R'
\
I
'-C
o
3.12'
4 = 118.16
CD Change in direction of the second tangent:
1= 229'57 - 203'45'
1=26'12'
T=85.39
T= Rtan 13'06'
85.39 = R tan 13'06'
R=366.94
OV=R+E1
R:E
1
R = R Cos 13'06' + E1 Cos 13'06'
366.94 = 396.94 Cos 13'06'
+ E1 Cos 13'06'
E1 =9.80 m.
NewE= 9.80- 5
New E =4.80 (new external distance)
E=R(seC~-1)
4.80 = 3.66.94
0
=20 (18'26')
'-C -
\
~~
Cos 13'06' =
R
0= 1145.916
366.94
0=3.12'
I _ 20 I
,
Sta. ofnew P. T. =(11 +025.32) + 118.16
Sta. ofnew P. T. = 11 + 143.48
i1;;IIII:1
.
Solution:
(sec ¥-1)
CD Radius of curve:
11 = 9'13~
11= 18'26' (new angle ofintersection)
V 1=33'52'
203'45'
p.T.
'.,.....
"'"
\\
\
t
IR •I
!!
I
/
'" 7"40'(10" 19'15'';0''
R'
\,
~
\ i
.
\,\!!
The change in direction of the second
tangent is 26'12' - 18'26' = 7'46'
• ·• .·.• .·
~ 8~i:~I~~._.~:~~~~~.~~r~i.· ......
@ PE!te@i~~Ih~~@ii?llirlgt:lfA:r!'·· . .
/
",,\i//
'1f/
,/
./
,,/"R
./
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5-290
SIMPLE CURVES
tan 7'40' 30" =AC
R
AC = R tan 7'40' 30"
tan 9'15'30" -R
- BC
BC = Rtan 9'15' 30"
AC t BC =103.20
Rtan 7'40'30" t Rtan 9'15'30" =103.20
R(0.13476 t 0.16301) = 103.20
R=346.58m.
® Stationing of Point C:
I
T= Rtan"2
T= 346.58 tan 16'56'
T= 105.52 m.
AC =346lan 7'40' 3D"
AC=46.70m.
IIIT."il
.1().+2$2.34.•• AliMMf<i6~Me¢t$tb~fplV@m
'''lrt~
••1.~
(DFirid1h~radjijs6fthe6t.iNe;
• ••.•• • •<».·······"'······.
.@ Flni:llhedISl~~CE!Mit>·
.@ IM~th~.s~IIMiri99f:K·.<···················
Solution:
Stationing of P.C.:
P.C. =(10 t 158.93)· AC
P.e. =(10 t 158.93) - 46.70
P.C.= 10 t 112.23
0= 1145.916 =1145.916
R
346.58
0=3.31'
L =!@l
cj
0
=(15'211 (20)
3.31'
LC1 =92.75 m.
I
'-c1
Stationing of C:
C=(10 + 112.23) + 92.75
C = 10 + 204.98
@
Stationing of P. T.:
L =l@!l ={33'5?'} (20)
o
3.32
L =205.07
P.T. =(10 + 112.23) + 204.63
P. T. = 10+ 316.86
CD Radius of curve:
Cos 19'15' =!i.
OV
OV=R+E
OV=R+18
Cos 19'15' =R:18
R= 303.94
® Distance MK:
T= Rlan 19'15'
T= 303.94 Ian 19'15'
T= 106.14 m.
Mto P. T. = 106.14 -12.32
M to P. T. =93.82 m.
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SIMPLE CURVES
Solution:
o
G)
Distance the P. T. is moved:
93.82
tan a =303.94
a =17'09'
Cos 17'09' =303.94
OM
OM=318.08
Using Sine Law:
318.08 _.-lQ.3.94
Sin IJ - Sin 61'39'
IJ = 112'56'
El = 180' -112'56' - 61'39'
El =5'25'
MK
303.94
Sin 5'25' = Sin 61'39'
MK=32,60m,
@
Stationing of K:
Sla. ofP.e. = (10 + 252.32) - (106.14)
Sla. of p.e. = 10 + 146.18
R01t
Lc =18O
L =303.94 (15'56') 1t
c
180
Lc =84.52
Sla. ofK= (10 + 146.18) + 84.52
Sla. of K = 10 + 230,70
_1145.916
R1D
R - 1145.916
1-:
4
R1 =286.48 m.
LC2 = 100 m.
Lc
2
=2012
Do;.
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8-292
SIMPLE CURVES
n __ 20(26.2')
""2-
100
~=5.24'
Construct AD parallel to 00'
R - 1145.916
2- 5.24
R2 =218.69 m.
AD=R1·R2
AD =286.48·218.69
AD =67.79
Sin 13'06' = AS
2AD
AB = 2(67.79) Sin 13'06'
AB = 30.73 m. distance the P. T. is moved
at an angle of 13'06' from the 2nd tangent
('l)
Compound Curve consists of two or
more consecutive simple curves having
different radius, but whose centers lie on the
same side of the curve, likewise any two
consecutive curves must have a common
tangent at their meeting point. When two such
curves lie upon opposite sides of the common
tangent, the two curves then turns a reversed
curve. In a compound curve, the point of the
common tangent where the two curves join is
called the point of compound curvature (P.C.C.)
Elements of a compound curve:
Distance between the two parallel
tangents:
BC = AB Sin 13'06'
BC = 30.73 Sin 13'06'
BC= 6.965m.
® Stationing of the new point of tangency.
T1 = R1 tan 13'06'
T1 = 286.48 tan 13'06'
T1 =66.67 m.
Stationing of P.C.
P.C. = (1 + 027.32)·66.67
PC. = 0 + 960.65
LC2 = 100
Stationing ofNew P. T.
New P. T. = (0 +960.65)+ 100
P. T. = 1 + 060.65.1
R1 = radius of the curve AE
R2 :: radius of the curve EF
T, = tangent distance of the curve AE
T2 = tangent distance of the curve EF
so :: T1 + T2 :: common tangent
11 :: central angle of curve AE
/2 :: central angle of curve EF
I :: angle of intersection of tangents A'C and
eF.
t
T1 ::R, tani
T2 = R2 tan 12
2
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COMPOUND CURVES
01 =4'
Sin Q1= 10
2 R1
10
·
2'
SIn =-
R1
R1 =286.56
T1 = R1 tan !.1
2
·!••·.I.~~.il!I~I.!~~~i~ltB· • • • •
T1 = 286.56 tan 10'20'
T1= 52.25 m.
P. C. =(43 + 010.46) • 52.25
P.C. = 42 + 958,21
@
Solution:
Stationing of the P.C,C.
T1+ T2=76.42
T2 = 76.42 • 52.25
T2 = 24.17
T2 =R2 tan !2.
2
24.17 = R2 tan 7'10'
R2 =192.233 m.
Sin f?2 = 10
2 R2
<\.
. f?2_-!L
\.
Sin 2 - 192.23
~'>\.\
/;:),...·il:/····r.
Q=2'59'
2
~=5'58'
\. II /0'
V
o
L
C1
=~
0
1
= 20'40' (20)
'-<:1
4
L = 20.667(20)
c1
4
Le1 = 103.34
I
CD Stationing of the P. C.
'/1 =268'30' - 247'50'
/1 =20'40'
. /2 =282'50' - 268'30'
/2= 14'20'
1
P.C.C. = (42 + 958.21) + 103.34
P.C.C. = 103.34 .
~
\\\.-10-\-10,/
'\\
I
R\
\
/'
/R
\\DI2'\DlZ/
\.-\/'
\'-tJ/
o
@
Stationing of the P. T.
42 = 'illQl
O
2
= 14'20 (20)
LC2
5'58'
I
= 14.33(20)
'-<:2
5.966
4 =48.10
2
P. T. = (43 + 06.55) + 48.10
p. T. = 43 + 109,65
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S-294
COMPOUND CURVES
·.I~~I~lQ_m~Qelii~9jrgiW~r~~
•tlllll~irl~I~~i!l~m~i.·
•
!-p.nOOf@ffiQi~mlh~fiflltplll'l;'~i{
1~1'~I~I~~~~~~~fC;~;ijn~fh~ .
<~!IMliiji:ifIMI"i···
Solution:
S·In !1_~
2 -2R
1
R - 167.74
1-2Sin 6'
R1 = 802.36 m.
@
c
.
~
166'30'
A~
300
B
I
Radius of the first curve:
11 = 12'
12 = 15'
Considering triangle AEC:
300
_~
Sin 166'30' - Sin 6'
300 Sin 6'
BC= Sin 166'30'
BC = 134.33 m.
300
_~
Sin 166'30' - Sin 7'30'
300 Sin 7'30'
AC = Sin 166'30'
AC = 167.74 m.
Radius ofthe 2nd curve:
BC
R2 =2 Sin 7'30'
134.33
R2 = 2 Sin 7'30'
R2 = 514.55 m.
® Stationing of P. T.
L = R1 /1. 1t
180
L = 802.36(12') 1t
c1
180'
4 = 168.05 m.
. c,
1
L = R2 /2 1t
c2
180
L = 514.55(15') 1t
C2
180
LC2 = 134.71 m.
Sta. ofP.T. =(10+204.30)+ 168.05+ 134.71
Sta. ofP.T. = 10+ 507.06
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S-295
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COMPOUND CURVES
4 =2011
1
D1
11--~
20
11 =90'
~=600
D.1 =1'40'
4 =2012
2 D.1
12 = 600(~667)
12 =50'
R - 1145.916
1D
1
- 1145.916
R1- 6
Solution:
R1 = 190.99 m.
S·In 45'-..£L
- 2R
1
c1 = 2 R1 Sin 45'
c1 =2(190.99) Sin 45'
c1 =270.10 m.
R - 1145.916
2- D2,
R - 1145.916
2-
1.66'
R2 = 687.55 m.
Sin2S'
=fR;
C2 =2 Rc Sin 25'
C2 = 2(687.55) Sin 25'
C2 =581.14
P.C.C.
p.e.
P.1:
CD Length of the long chord connecting the
P.C. and P. T.
Lc • = 300
L2: (270.1W + (581.14f
·2(270.10)(581.14) Cos 110'
L= 719.76m.
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5-296
COMPOUND CURIES
® Angle that the long chord makes with the
first tangent:
(D • ltl~des1teijtoStlb~a~~te
• thecompgynd
··.···cu~ w~~~ sirnple·.C\.IfV~ • • that.sh(lll·~rld
wlml~e • • same.p,T'j•• • determlne•• the••. tolal
•
.
1erlgllt.mc~!YEl.°ftheslrJlplecurve,
• • • • >·•• •.•.
® • 1tl~ •. de~lt~ • • t<)•• SUb$titute•• thegoIl1PotlQd
{;(Jrve•• *ilh•.• ~• • si"1pleCllrvE:!•• tllat.shall.tle
.·¥!.~nt!f1!hetw<l~ng~mtljn~saswell* .
the.Pornmo~ • • tangent•• AD.• • /.~h~II$ • • t~~
TliqiIJS()fth~sil'llpleplJ~,>
@ Whatl$thestatiOfiin90ffh~rleWe,p·······
Solution:
CD Total length ofCUNe of the simple CUNe:
D1=3'30'
11 =16'20'
R _1145.916
1-
D1
_1145.916
R1- 3'30'
R1 =327.40
~=4'OO'
12 =13'30'
R - 1145.916
581.14 _ 719.76
Sin e - Sin 110:
r
Rr
e =49'21'
Dt.
- 1145.916
4
R2 =286.48
Sin 8 Sin 110'
270.10 = 719.76
8=20'39'
The angle of the long chord makes with the
first tangent line is
45' +49'21' = 94'21'
@
Angle that the long chord makes with the
2nd tangent line is 25' +.20'39' = 45'39'
o
Given tbeJollnwing compound curve With the
vertexV,iriaceessible.
Angles VAD and VDA
are equal 10 1~'20; and 13'30' respectively.
Stationing of Ais1 -+ 125.92. Degree of cUlVe
are 3'3Q' for the first curVe and 4' 00' for the
second curve,
o
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COMPOUND CURVES
T1 = R1 tan !1
2
® Radius of the simple curve:
v
T1 = 327.40 tan 8'10'
T,=46.98
old r.C'
T2 = R2 tan h.
2
T2 = 286.48 tan 6'45'
T2 =33.91
AD= T1 + T2
AD =46.98 +33.91
AD = 80.89
VA
80.89
Sin 13'30' = Sin 150'10'
VA = 37.96
VD _ 80.89
Sin 16'20' - Sin 150'10'
VD =45.73
T= VD+ T2
T=45.73 +33.91
T= 79.64
/
T=Rtan"2
T
R= tan 14'55'
79.64
R = tan 14'55'
R=298.96
D = 1145.916
298.96
D=3'5O'
h = T- VA
h = 79.64 • 37.96
h =41.68
Length ofcurve:
L = / (20)
o
_ 29'50' (20)
L - 3'50'
L=29.833(20)
3.833
L = 155,66
T1 + T2 =80.89
T1 = Rtan8'10'
T2 =Rtan 6'45'
R(tan 8'10' +tan 6'45') =80.90
R= 308.89
® Stationing of the new P.C.
T, = R tan 8'10'
T, = 308.89 tan 8'10'
T, =44.33
Stationing ofnew P.C.
P.C. =(1 +125.98) - (44.33)
P.C. =1 + 081.65
Problem
A compound cUNe connects three tangents
haVing an azimuths of 254', 270' and 280'
respectively. The length of the chord is 320 m.
long measured from the P.C. to the P.T. ofthe
curve and is parallel to the common tangent
having an aZImuth of 270', If the stationing of
the PT. is 6 + 520.
CD Deteimine the total length afthe curve.
® Determine the stationing of the p.e.e.
@
Determine the stationing of the p.e.
Solution:
CD Tota/length of the curve:
320
x
Sin 164' = Sin 5'
320 Sin 5'
x = Sin 164'
x= 101.18 m.
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S-298
COMPOUND CURVES
~_.1­
4 =20/,
Sin 164' - Sin 8'
320 Sin 8'
y= Sin 164'
y= 161.57m.
D1
1
20(16)
~1
3.15'
4 = 101.59 m.
I
::
1
80.785
-S·In 5' =
R2
R _80.785
2- Sin 5'
R2 =926.90 m.
lJ.z =1145.916
R2
lJ.z = 1145.916
926.90
lJ.z = 1.24'
I _
~2-
~
8
320
p.T.
Total length ofcuNe = 161.29 + 101.59
Total length of CUNe =262,88 m.
'• .AIl1ll'i"'>'" P.C.C
@
P'C
lJ.z
=20(10)
~2
1.24
4 2 = 161.29 m.
I
5'
p.e
20 12
@
Stationing of the P. e.e.
p.e.e. = (6 + 520)· 161.29
p.e.e. = 6+358.71
Stationing of the P. e.
p.e. =(6 + 358.71) -101.59
p.e. = 6 + 257.12
I
I,
R,I
,
's.'':
kl?- /'"
I
Sin 8' = SO.59
R1
R - 50.59
1- Sin8'
R1 = 363.50 m.
D - 1145.916
1R
1
D - 1145.916
1- 363.50
D1 =3.15'
's'
.~
./",'
'
•
~~~~;~~~U~~ ~:b1~'j~ • ~Tgh~~y • t;y.
cgnneq1lng•• four .liiPg~~ts~th • ~• • COlTlPOum:t··
ClllY~·~(lIl~stillg·.9f.thfElEl~irnpl~·clJl'Vl!s
..•.• J~$
__llliilt.ifl
tir~ll:ll.1rye.,,~~qi§l~nCIJEl9::;~O~lTl·~flij
CO#2Ql)m.·········
G)¢QIl'tptJl~tIjraqiusl#the~td~lJrve...·
.• •
@Cwpute.thl!tadiu$.(jf.lhe.sel:Ondcurve.••••••
@lfgqisilI12t152.BO,What .isthl'l
$ta!k@O!l(jfth~P.r,
.
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COMPOUND CURVES
Solution:
4
=R2 /2 1t
180
L =217.81 (55'54') 1t
""""2
180'
4 = 212.50 m.
cD Radius of third cum:
2
264'30'
2
4
=R3 !a1t
180
L = 115.21 (72'34')1t
""'3
180'
4 = 145.92 m.
3
3
Sta. ofP. T. = (12 + 152.60) + 355.91
+ 212.50 + 145.92
Sta. ofP. T; = 12 + 654.43
11 =264'30' - 220'15'
/1 =44'15'
12 =320'24' - 264'30'
/2 = 55'54'
13 =360' • 320'24' + 32'58'
13 = 72'34'
T1 +T2 =303
R1 tan 22'7.5' + R2 tan 27'57' =303
0.407 R1 + 0.530 R2=303
T2 + T3 = 200
R2 tan 27'57 + R3 tan 36'17 = 200
0.530 R2 =0.734 R3 =300
R1 =4R3
0.407 (4 R3) + 0.530 R2 =303
1.628 R3 + 0.530 R2 =303
0.734 R3 + 0.S30Rz =200
0.894 R3 = 103
R3 = 115.21
@
Radius of2nd CUNe;
R1 =4 (115.21)
R1 = 460.84 m.
@.
Vl/hal.Shotild"birthe • raii\us•• {)fthflpthet
siIl1PI~.ClIN~fu~t~l@l.~.16eA.T.?" • "·." · " "· "
® C0Il1Putelhe~f~l16nlrt!lofmeP'9'C.
@) "·Whatis.thlll~ngllJ.pttn~"tMg~lltftomthe
P.I.J/)@!p:r.pfthEiCOmPPUl'lclCUrve?
Solution:
cD Radius of second CUNe:
0.407 R1 + 0.530 R2 =303
0.407(460.84) + 0.530 R2 =303
R2 = 217.81 m.
® Stationing of the P. T.
_ R1 1,1t
LC1 - 180
I
=460.84 (44'15') 1t
~1
180'
LC1 =355.91 m.
"
..II .""/ / ./ .
.T.
9~/R2=136.94
Rl=286.54"
i/,'.
'"~~\8'O'
"
,:
~
"
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S-300
COMPOUND CURVES
v
D1 =4'
. D. 10
sln::...L=-
2
R1
Sin 2' = ~~
R1 = 286.54 m.
tan 8'30' = 2~~54
T1 =42.82 m.
AV= 80·42.82
AV=37.18 m.
AB
37.18
Sin 143'32' =Sin 19'28'
AB=66.31 m.
T1 + T2 = 66.31
T2 =66.31 ·42.82
T2 =23.49 m.
$POi¥lP~t~ttWl$lijtj(jijj6go/P;g'B>/
·.~• • ll~~~~:~~f~~th&ir~~,8~·····<
Solution:
<D Stationing of p.e.e.:
tan 9'44' =J2
R2
23.49
R2 = tan 9'44'
R2 = 136.94 m.
@
e.
Stationing of P. e.:
S=R18
=286.54 (17')(n)
S
.
180'
S= 85.02 m.
Stationing of p.e.e. =(10 + 163) + 85.02
Stationing of p.e.e. = 10 + 248.02
@
Distance from P.I. to P. T. of compound
curve:
VB
37.18
Sin 17' - Sin 19'28'
VB =32.62 m.
Distance from P.I. to P. T. = VB + T2
Distance = 32.62 + 23.49
Distance = 56.11 m.
1= 282'50' • 247'50'
1= 35'
11 =268'30' - 247'50'
. 11 =20'40'
12 =180' - 145' - 20'40'
12 = 14'20'
Sin' Q1 = 10
2 R1
Sin 2' = 10
R1
R1 =286.56 m.
T1 =R1 tan 10'20'
T1 =286.56 tan 10'20'
T1 =52.25m.
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COMPOUND CURVES
S - Rj / 1 1t
1- 180
S _286.56(20'40') 1t
Solution:
<D Radius offirst CUNe:
,180
S, = 103.36 m.
(6+421) ....
Sla. ofp.e. = (10 + 010.46) - 52.25
Sta. of P. e. = 9 + 958.21
Stationing of p.e.e. = (9 +958.21) +103.36
Stationing ofP.e.e. = 10+ 061.9
\
.
;;;,\
180
S - 192.22(14'20') 1t
r
180
S2 =48.09 m.
P.T.
I
.
® Radius ofsecond CUlVe:
T, + T2 = 76.42
T2 = 76.42 - 52.25
T2 =24.17m.
T2 = R2 tan 7'10'
24.17 = R2 tan 7'10'
R2 = 192.22 m,
® Stationing of P. T.:
S - R2 12 1t
i
:
"
:
\ II: /
/.
/
'Rl
. //
\
\
1//
\\ \ t¥
\\!.
.~
v
A
B
2-
StationingofP.T. = (10 +061.57) +48.09
Stationing ofP. T. = 10 + 109.66
AB=(6 +721)-(6+421)
AB=300m.
T1 + T2 = 300
In any triangle the angle bisector divides
the opposite sides into segments whose
ratio is equal to that of the other sides.
I.L _I2.
In a compound curve, the line connecting the
P.I. at point Yahd the P,C,C, is an ap91e
bisector.. AVis Z70rnetets andBV =Mm~
The statibnlrig otA 1S6 + 421 and that otBis
6 t 721. Point Als along the tangenl passing
thru the P.C. while pointS is along the langElnt
passing IhTU theP.T. The P,C;C,dsalonglln&
AR
.
.•.......
......•
(1)
@
@
Compute the radius of the first curVe
pas;lingthru IheP.C.· . }
Compute the radiUs of the second curve
passing lhru the P.T.
..
. ...
Determine tf'e length of the long chord from
P.C. to P.T.
270 -90
T, = 3T2
T1 + T2 = 300
3T2 + T2 =300
T2 = 75 m.
T, = 225
Using Cosine Law:
(90)2 =(270)2 +(300)2 - 2(270)(300) Cos A
A=17'09'
Using Sine Law:
270
90
Sin B - Sin 17'09'
B= 62'11'
/1 = 17'09'
12 = 62'11'
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S-302
COMPOUND CURVES
T1 =R1 tan !.t
2
17'09'
225 = R1 tan -2R1 = 1492,15 m.
® Radius ofsecond curve:
T2 =R2 tan b.
2
75 =R2 tan 62;11'
R2 = 124,37 m.
® Length oflong chorp from P. C, to P. T.:
Solution:
M
Sin 8'34.5' = 2
(14~h.15)
C1 = 444.97 m.
Sin 31"5.5' = 2
(1~.37)
C2= 128.45 m.
P.e.c.
~
P.e.
.
L
H= 180' - 31'5.5' - 8'34.5'
H= 140'20'
Using Cosine Law:
L2 = (444.97)2 + (128.45f
- 2(444.97)(128.45) Cos 140'20'
·L=550m.
P.~
G)
Length ofBD:
60
tan ex = SO
ex = 50'12'
H:: 75'20'· SO'12'
H=25'08'
60
EB = Sin SO'12'
EB= 78.10
BC =78.10 Cos 25'08'
BC= 70.70
CE= 78.10 Sin 25'08'
CE=33.17
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COMPOUND CURVES
FE =33.17 • 12.00
FE= 21.17
FE
Coso= EO'
Solution:
12+320.30
,,_21.17
Cas 0 - 48.00
0=63'50'
Angle CEB = 90' ·25'08'
Angle CEB =64'52'
13 =64'52' -63'50'
13 = 1'02'
CD Stationing ofpoint G:
30
tan8=--
CD=FO'
CD =48 Sin 63'50'
CD = 43.08'
BD=BC· CD
BD = 70.70·43.08
BD=27.62m.
58.64
8 = 27'06'
a+8=47'36'
a =47"36'· 2TrJ5
a=20'30'
@
Angle GEA:
Angle GEA = (90' • a) + 1'02'
Angle GEA = (90' • 50'12') + 1'02'
Angle GEA = 40'50'
@
Deflection angle BAG:
tan 20'30' = ~.;2
AC =64.52 tan 20'30'
AC=24.12m.
LC=24.12-8
LC = 16.12 m.
CH=30-8
CH=22m.
Deflection angle BAG = ~ (40'5O')
(LHl =(CH)2. (LCf
(LHt = (22)2 - (16.12f
LH=27.3m.
LH=AG =27.3 m.
Deflection angle BAG = 20'25'
lri{haMlJrestJQWn'AVi$*ifaiflht.road~~F
.~·.~C;\jfVf,l~~tr~W~,.···Th.~ •. ~{llllS • Qfm~.9UW~d
st[eetis301lt • • ··~ • ArcUlarcu&e·(lra.n1..radlu!l
l~t~~l~.~~Jg~~~2~ffl~i6d¥~i~~~~~
rn·Th~ • ang~iA~F.I$.equat.tCl.4t36'·'orn~
stiitioning.~t·A.i$ • • lZ.+·~~Q;3Q'.··· ·Pefle¢Uoo·
.....
ang~ofpointJ<Jl'/:)mF·i$20·27'.·
(j) • Fin~the.~tati9rUng.ofP.OlrJt.(3.
··@.)•••••• flfl~.1~e.$tlationing.mP9mt.tK i •.
@.·•• Findthe.stattcmin!JQf.pOil'lt.K,
••
Stationing at G:
G=(12 + 320.30) + 27.30
G= 12+347.60
@
Stationing of point E:
C f!, 16.12
as =22
f!, =42'53'
GE = 8(42.883) 1t
180
GE= 5.98m.
Stationing at E=(12 + 347.60) +5:98
Stationing at E = 12 + 353.58
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S-304
COMPOUND CURVES
@
Stationing ofpoint K:
Solution:
CD Radius of 2nd curve:
(12+353.58)
~n
c~s
K
iii = 62~54' + 26'37'
iii =89'31'
S=Rliln
180
- 30 (89'311 n
S180'
.
\
\
;;;"". \\
".
.~
~/
l/
S=46.87m.
Stationing of K = (12 + 353.58) + 46.87
Stationing ofK =(12 + 400.45)
AH= 125.70 Sin 44'36'
AH=88.26m.
VH =125.70 Cos 44'36'
VH=89.50m.
AI =286.50 Cos 44'36'
AI= 204 m.
IF =286.50 Sin 44'36'
IF= 201.17 m.
,/
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COMPOUND CURVES
Solution:
FJ + IF = VH + 155.60
FJ + 201.17 =89.50 + 155.60
FJ =43.93 m.
CD Central angle offirst clJIYe:
EJ=AI+AH
EJ =204 + 88.26
EJ = 292.26 m.
JG=R2- EJ
FG= R2- R1
FG =R2 - 286.50
FJ =43.93
JG = R2 • 292.26
Considering triangle FJG:
(JG)2 + (F.f)2 =(FG'1(R2 - 292.26'1- + (43.93'1- =(R2 - 286.50)2
Ri- 584.52R2 +85415.91 + 1929:84
= Ri- 573R2 + 8208225
11.52R2 =5263.5
R2 = 456,90 m,
@
Central anlt1e of 2nd CUNe:
JG = R2 - 292.26
JG = 456.90 - 292.26
JG = 164.64 m.
FJ
tan 12 =JG
43.93
tan 12 =164.64
12 = 14'56'
AC =200 Cos 50'
AC= 100
Be = 200 Sin 60'
BC= 173.20
CO=EF
EF = 100 Sin 20'
EF=34.20
BE = 100 Cos 20'
BE=93.97
GO=AC+CD·1oo
GD = 100 + 34.20 -100
GO =34.20
OF=BC-BE
OF = 173.20 • 93.97
OF=79.23m.
GO
tanS-- OF
34.20
ta n 8= 79.23
® Central angle offirstcuNe:
11 + 12 =44'36'
I, =44'36' -14'56'
11 = 29'40'
•.
·.~~~~~~.~u~~~~~ ~I~I • ~r~;}~ •
t;.
~(e·T¥e9lntYi$th~m@{~nq!~~¢~AAAf:
lh~taO~~ls(fM·):AJ191~VA!:l¥$O·~P9"O~.
8=23'21'
11 =28
11 =2 (23'211
11 =46'42'
\fflA • ~ • 4g'· • • • ()l$~M~AEl.i$~oqng~rid.tM·
raditlsoUh~$CQiTdrcU,.w;~#ioom/
.@ •••• Q~~etmine • IM.p.~6M!~6gl~
Ctll'ile.«
. .
• •()ftheJrf$1
··~•.• • O~fel'flllne.·ltie • • C$lltra!·.M91l!l(lf!tleZnd
c;qtv~,<
·~· • . • Del~lnether.;l~il.l$.()t·.t@·fIf$f¢OlVe.·.·.·····
@
Central angle of 2nd CUlve:
/1 + 12 =70'
12 =70 - 46'4Z
12 = 23'18'
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5-306
COMPOUND CURVES
@
Radius of first curve:
- 1145.916
R,- D
1
- 1145.916
R1- 4
R, =286.48
R - 1145.916
2-
5.5
R2 =208.35
Sin 46'42' = 76·~3
VB = 110 Cos 42'36'
VB=80.97m.
AB = 110 Sin 42'36'
AB= 74.46 m.
AD+AB=R2
AD + 74.46 = 208.35
AD =133.89 m.
OF = 108.87 m.
OF= R,-100
108.87;;; R1 -100
R1 = 208.87 m.
Cos 42'36' = AD
AE
AE= 133.89
Cos 42'36'
AE= 181.89 m.
EG= R,-AE
EG = 286.48 -181.89
EG = 104.59 m.
• ~Clfry~
~~~fd~~~b5~~16~J~~=~.b~t • [~Q~~~~.·
gClp~j~ting~t • ~• • 4·•• cyry~.·ilnd 5·~Q·
••
.
.<l.. .
9UW~,· • ·.lftpe··~M~~t~tm~.bElginnjrg • 6f·m~·.
··Ctihi¢.tqlh~pOl~fofirf~t$ectiAnoflhetan~~ht.
is110Jfl~
long....· . .
••...••:::.
FG= R,-R2
FG = 286.48 - 208.35
FG = 78.13
E
Solution:
CD Central angle of first curve:
G
Using Sine Law:
78.13
104.59
Sin 47'24' = Sin e
c
e = 99'48'
11 = 180' ·47'24' -99'48'
11 = 32'48'
@
Central angle of 2nd curve:
11 + 12=42'36'
12 =42'36' - 32'48'
G
12 = 9'48'
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COMPOUND CURVES
@
Di$fance of tangent from end of 5'30' CUNe
to the P.I.:
~
113'
E
A
31'
36'
B
Tji-T.=180.40
- 1145.916
R1- 3'
G
EF
78.13
. Sin 32'48' = Sin 47'24'
EF=57.50m.
- 1145.916
R2- 5'
R2 = 229.18 m.
VC+VB=DE+EF
DE= 181.89 Cos 47'24'
DE= 123.12 m.
VC + 80.97 = 123.12 + 57.50
VC=99.65m.
,.t.ifilliirl'~~
~~~;~II~~~~~;I~,i;'>
@lf~tn.P"Li~ITIClyed15tri·jJutfrQrnt~e
• • ••• •:~lt~~t61~!¢.~s~~lli1~~~o~~h~wr
Solution:
tan 15'30' =11
R1
T1 =381.97 tan 15'30'
T1 =105.93 m.
T2 =R2 tan 18'
T2 = 229.18 tan 18'
T2 = 74.47 m.
AB= T1 + T2
AB = 105.93 + 74.47
AB= 180.40
AV180.40
Sin 36' = Sin 113'
AV= 115.19 m.
Sla. ofP. C. = (26 + 050) • (115.19 + 105.93)
Sla. ofP. C. = 25 + 828,88
® Slationing of P. T:
L _W1
1- 0
CD Stationing of the P. C.:
26+ 05
P.I.V
R1 = 381.97 m.
67'
1
L1 =20 ~31)
L1 =206.67
Sla. ofP. T. =(25 + 828.88) + (206.67 + 144)
Sla. ofP. T. = (26 + 179.55)
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S-308
COMPOUND CURVES
@
Stationing ofnew P. T.:
Solution:
<D New tangent distance:
=x
· 36' 15
SIn
x=25.52 m.
T3 = 74.47 + 25.52
T3 =99.99
tan 18' =99.99
R3
R3 =307.74 m.
L - 307.74 (36)'lt
T1 =R1 tan 18'10'
T1 =100 tan 18'10'
T1 =32.81 m.
10 .
h = Sin 36'20'
h= 16.9 m.
180
L3 = 193.36 m.
3-
Sta.of new PT
= (25 +828.88) + (206.67 + 193.36)
Sta. ofnew PT = 26 + 228.91
New tangent distance = 16.9 + 32.81
New tangent distance = 49.71 m.
@
New radius ofCUNe:
T2 = T1 + h
T2 = 32.8 + 16.90
T2 =49.71 m.
T2 = R2 tan 18'10'
49.71
R2 tan 13'10'
R2 = 151.40 m.
@
Stationing of new P. T.:
Stationing of p.e. (30 + 375.20) - 32.81
Stationing of p.e. = 30 + 342.39
0= 1145.916
R
0= 1145.916
151.40
0=7.57'
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5-309
Pdfbooksforum.com
COMPOUND CURVES
L =201
C
I
'-c
D
L - 20 (36.33)
I
'-c
7.57'
Lc =95.99 m.
C-
= R1 /1 1t
180
=190.99 (42') 1t
180'
.
4= 140 m.
Stationing ofold P. T. = (0 + 168.15) + 140
Stationing of old P. T. = (0": 308,15)
Stationing ofnew P. T.:
New P. T. =(30 + 342.39) + 95.99
New P. T. = 30 + 438,38
@
Stationing offhe P.C.C.
T1 =190.99 tan 21'
T1 =73.31 m.
VB Sin 42' = 20
VB= 29.89 m.
••••
~!!~~~~~lilllllll~,t>
Solution:
<D Stationing of old P.T.
T2 = T1 + VS
T2 = 73.31 + 29.89
T2 = 103.20 m.
.
tan a. =.I2.
R1
ta
- 103.20
na.- 19O.99
a= 28'23'
Cos 28'23' --OS
B1.
190.99
OS= Cos 28'23'
OS = 217.09 m.
R - 1145.916
2-
~
R - 1145.916
24
R2 =286.48 m.
GC=R1 +20
GC =190.99 + 20
GC = 210.99 m.
O'G = R2 - GC
O'G = 286.48·210.99
O'G =75.49 m.
00'= R2 -R1
00'= 286.48 -190.99
00'= 95.49
O'G
Cos 0 = 00'
- 1145.916
R1- D
1
- 1145.916
R16
R1 =190.99 m.
75.49
Cos 0 = 95.49
0=37'46'
fJ =42'·37'46'
£l =4"14'
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8-310
COMPOUND CURVES
LCl
I
_ R1 (J 1t
180
_ 190.91 (4'14) 1t
b=350m.
. a= 550 m.
c= 762 m.
a+b+c
s=-2-
-
180'
LC1 = 14.11 m.
'-<:1 -
550 + 350 + 762
s=
2
s= 831
(s- a) =281
(s- b) =481
(s. c) =69
Stationing ofP.C.C. =(0 + 168.15) + 14.11
Stationing of P.C.C. = 0 + 182.26
@
Stationing of new P. T.
_ R2 01t
.LC2 -
180
1_ _ 286.48 (37'46') 1t
N
~
180'
Lez = 188.83 m.
Stationing of new P. T. =(0 + 182.26) + 188.83
Stationing of new P. T. = 0 + 371.09
1
Sin ~=
(b- b)(c- c)
2
be
. ~ _ (48)(69)
Sin 2 - (370) (762)
Sin~=0.1234
~=20.65·
A= 41.3'
LA =41'18'
Bearing ofCA = S 3'42' E.
@
N
Angle ACB:
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S-311
Pdfbooksforum.com
COMPOUND CURVES
_1145.916
R2 4'
R2 = 286.48 m.
- 1145.916
R1 6'
Sin §i = ~ (s ~ a) (s - c)
2
ac
. §. _
(281)(69)
(550) (762)
Sin 2 -
Sin ~ = ~ 0.0463
R1 =190.99
Sin ~= 0.215
T= 190.99 tan 12' + 286.48 tan 18'
T= 133,68 m,
~= 12.45'
L.B =24'54'
L.C = 180' - (45' + 24'54')
@
It - D1
L = 20 (24)
1
6
L.C= 110'06'
Angle ACB = 110'06'
@
Stationing of P.C. C.:
bt_ 20
L1 =80
Bearing CB:
Bearing of CB = N 73'36' W
Sta. of P.C.C. = (10 + 420) + 80
Sta. ofP.C.C. = 10+ 500
@
Stationing of P. T.:
0._ 20
12 -
Dt.
- 36' (20)
L2- 4'
L2 =180
Stationing of P. T. = (10 + 500) + 180
Stationing of P. T. = 10 + 680
Solution:
CD Length of the common tangent of the curve:
common tangf:nt
\: ./
t1
" 1",'"
..
) i 18" I
1/
if!
The length of the common lan~entof a
compound curve is equal to 6M2 m.The
common tangent makes an angle of 1Z'and
1$' respectively to the tangents of, the
compound curve. If the length of the tangent of
the first I;urve (on the side of P.C.) IS equal to
4tOZm.
CD Compute the radIUs of the second cUNe,
Compute lIle radius of lhe first curve.
@ Compute the stationing ofthe P.T. if PC is
at 20 + 042.20.
@
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S-312
COMPlII. CIIVEI
Solution:
<D Radius of the second curve:
Solution:
CD Length of the long chord:
T2 =68.62- 41.02
T2 =27.60
. T2 = R2 tan 9'
27.60 =R2 tan 9'
R2 = 174.26 m.
® Radius ofthe first curve:
T1 =41.02
T1 =R1 tan 6'
- 41.02
R1 - tan 6'
R, = 390.28m
@
,,
" '~"\ V'
, \, I
\~
"" I
470
L
Sin 9' = Sin 165'
L=m,61m.
Stations of the P. T.:
L - R, /1'1t
,- 180
L - 390.28 (12) 1t
,180'
L1 = 81.74 m.
L - R2 /2 1t
2-
® Radius of first curve:
470
S·In 6' =
2R1
R, = 2248.19 m.
180
L - 174.26 (18) 1t
2-
.or.
,,
,,
,
180
L2 =54.75m.
Sta. ofP. T. =(20 + 042.20) + (81.74) + (54.75)
Sta. of p.r. = (20+ 178.69)
@
Radius ofsecond curve:
...fL _
470
Sin 6' - Sin 9'
~ = 314.05
· 9 314.05
S
,1n=2R
2
R2 = 1003.nm.
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COMPOUND CURVES
The common tangent of a compound curve
makes an angle of 14' and 20' with the
tangent of the first curve and the second curve
respectively. The length of chord from the
p.e. to p.e.e. is 73.5 m. and that from p.e.e.
to P.T. is 51.3 m.
ill Find the length of the chord from the P.C..
to the P.T. if it is parallel to the common
tangent.
@ Find the radius of the first curve.
@ Find the radius of the second curve.
A compound curve has a common tangent
84.5 m. long which makes angles of 16' and
20' with the tangents of the first curve and the
second curves respectively. The length of the
tangent of the first curve is 38.6 m. What is
the radius of the second curve.
Solution:
Solution:
ill Length of the chord:
38.6 + T2=84.5
T2= 45.9 m.
T2 =Ttan10'
P.c.c.
~"
45.9 = R2 tan 10'
lEI)'
l~
7"
P"Cc
~~
L
R2 = 260.3 m.
L2= (73.5)2+(51.3)2- 2(73.5)(51.3) Cos 163'
L= 123.5 m.
@
Radius of the first curve.
73.5
S·In 7' =
2R 1
R1 = 301.55 m.
@
Radius of the second curve.
A compound curve has a common tangent of
84.5 m. long which makes an angle of 16' and
20' with the tangents of the first curve and the
second curve respectively. The length of the
tangent of the second curve is 42 m.
Sin 10' = 51.3
2R 2
CD What is the radius of the first curve.
R2 =147.71 m.
@
@
Find the radius of the second curve.
Find the length of curve from p.e. to P1.
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S-312-B
COMPOUND CURVES
S.olution:
.
~ "
,:
""..
"
'ilLO;,'LO"/
~'
/
/
• Rz
j-;y
y'
<D Radius of the first curve.
84.5 = T1 + 42
T1 =42.5
T1 = R1 tan 8'
42.5 = R1 tan 8'
R1 = 302.4m.
<.?)
Solution:
<D Length of the common tangent of the
compound curve. Use arc basis.
- 1145.916
R1- 6'
R1 = 190.99 .
· - 1145.916
R2 4'
Radius of the second curve.
R2 = 286.48
tan 10' = ~
1
R2
AS = R1 tan
R2 =238.19 .n.
@
<.?)
@
tangent of the compound curve. Use arc
basis.
.?) Compute the sta. of PC. if P.I. is at sta.
12 + 988.20.
@ Compute the sta. of P.T.
Sta. of P.C. if P.I. is at sta. 12 + 988.20.
_d_ = 133.68
Sin 36' Sin 120'
d= 90.73
Sta. P.C. = 12 + 988.20 - (90.73 + 60)
P.C. =12+856.87
Given a compound curve 11 = 24', 12 = 36',
0 1 =6',° 2 =4'.
(f Compute the length of the common
12
+ R2 tan '2
AS = 190.99 tan 12' + 286.48 tan 18'
AS = 40.60 + 93.08
AS = 133.68 m.
Length of curve from P.C. to P.T.
L= 302A(16)1t + 238.19(20)1t
180
. 180
L =167.59m.
'21
Computethesta.ofP.T.
P.T. =(12 + 856.87) + 190.99(24) 1t
180
+ 286.48(36) 1t
180
P.T. = 13 +116.87
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5-313
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[.-..
COMPOUND CURVES
1P1~.IQrigchO,f(iQf~¢fupQu~l:lJr¥eiS12()m,.
19M.vmlChl1@~~$ • aQ~~I~mmrltPrn.tfj~
•
.111l'glllllcf.·ltIe·filSt.~·.pa/j~iflgtll@tlgh • ttlEi.
1.:l.G.l$d?Q·•• ff()IJ1.ttJ¢tanQellt(lf.·ttIe•• ~fl~M
1:1I:'Tl,."Jii··
i:I'II"'I
l;!r;
lillilllillii
~rye~il'lgthtWl~mmeI'NJ'.lft~~.(X)mm9~
.t.jltlg!lti~i$pij@~Ii<i@llQlIgc:OOrd····
i
•.•. . . •
.~··• • ~~~#l~~iolj(lll)$~f:@l1@~·
.
i:tfId~cllrV~»
.
Solution:
<D Length ofthe common tangent:
Solution:
<D Length of chord from PC to PCC:
,
14'
p.e.
'~
.........TiO:iii..·..·....···
380 ,
I
\,
\
R'
i
/
i, '/'R
y-V
\~
~_-EL
@
Length of chord from
@
pce to PT:
Sin 17' - Sin 163'
~= 120
Diff. in radius of 1st and second curve:
71.27
S·In 7' =
2R(
R1 = 292.40 m.
Sin 10' =~
2Rz
Rz = 345.53
Oiff. in radiUS = 345.53" 292.40
Diff. in radius = 53.13 m,
14\
17;/'
~\r
'\~\I
'!>
Sta.ofP.C.C.
L - R1 / 1 7t
1 - 180
L =380(28) 7t = 185 70
1
180
.
p.e.e. =(20 + 000) + (185.70)
p.e.c. = 20+ 185.70
-.fL_~
@
I
@
,
/
'- L,,/ /220
T1 =380 (tan 14')
T1 =94.74 m.
Tz = 220 tan 19'
Tz = 75.75
T1 + Tz = 170.49 m,
\ II': /
Sin 10' - Sin 163'
C1 = 71.27m.
•
.r.
,/
\
Sta. of P. T.
L - Rz/z 7t
2 - 180
L = 220(38) 7t = 145 91
z 180
.
P. T =(20 + 185.70) +145.91
P. T. = 20+ 331,61
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S-314
COMPOUND CURVES
@
1 - 180
L - 290 (42) 1t
1180
L1 =212.58 m.
L - RZ I2 1t
Z- 180
L _ 740 (28'36')1t
z180
L2 = 369.38 m.
~cornpoundcllNep~~se$thtllllcOllln-lQIf
lan9~~tA~havip~a,en9thMi3®m.1'~~
r~diH$.of.thll.prst.~uryEl.jsequ.llt9f9qrn
Stationing of the PT:
L - Rl /1 1t
•.<ll)d·
·.a•seC6t1dcurveiS740rn.
• ceilfral.angtepf.4g...·.If··.the•
. • radll.ls.I)f•• th~ •..
Sta. of PT = (20 + 542.20) + (212.58) + (369.38)
Sta. of PT = (21 + 124.16)
Solution:
0)
Tangent of second CUNei
·fh¢P9mrl1()nt~1'l$~WA~9fafqmP9qnq.cutv.~
•
tn~~~~aral'1gly~iththetjlO$eht$OfJhe
C()lIlP()im<.iClJrvElpf~g';3Q>ari93Q'99·
·.rib1~~~~)4~· • ±~~~dgr~&aj~Q6~i(e • .~fth~fi~~··
,,
,,
'\.
l:uo/~js4'30'while.thatofthe
\
I
I
'- 'I
Rj=290 "rn'-,
'
"'~!.J: :'"
/
,\:'1-=.1 /'Rz=740m
~~I,' ,
"
,,'/
if/
T1 = R1 fan 21'
T1 = 290 tan 21'
T1 = 111.32 m.
Tz =300-111.32
Tz =188.68rn.
® Central angle of second CUNei
tan 2=Ji
2 R
t
z
2_188.68
an 2 - 740
~ =14.30'
Iz =28.60'
Iz = 28'36'
• Se()()nd.curve.• is
• sameP.;T·\VhiteWe.direClloo
~JtYit~~h]~~p~~tbu~i~~.lW~II~®Pai~6~.·
• ollhe.·tangent$
rem~jQ$tM$;1Im~,>
..'
.
(j)FlMtti~t~diGs°flt1es[mple(;i)(Ye.•·.
®flMlh~st~tioningottl'lenewp.C·· .
mfi@jhestatiQl1fngofPJ\
Solution:
0)'
Radius of simple CUNe:
- 1145.916
R1- 0
1
- 1145.916
R1 - 4.5'
R1 = 254.65 m.
- 1145.916
Rz- 5'
Rz =229.18m.
fan 15' =Ji
R
z
Tz = 229.18 tan 15'
Tz = 61.41 m.
.
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COMPOUND CURVES
T, = R, tan 12'45'
T, =57.62 m.
@
Stalioning of P. T.
L =RI1t
180
c
Lc
= 234.91 (55'30j1t
180'.
4=227.55
Sta. of P. T. = (10 + 311.05) + 227.55
Sta. of P. T. =(10 + 538.60)
~~~~~~~~,~!wt~'h~ff:n~~:~f~6~ah~··
AB = T, + T2
AB=61.41 +57.62
AB = 119.03 m.
VB'
119.03
Sin 25'30' = Sin 124'30'
VB = 62.18
New tangent of the simple CUNe:
,T= VB + T2
T= 62.18 + 61.41
T= 123.59
lliiill.till
r....ltjlit~
@FIMt@r@m$9f1~~~lfnpteCtJl'\I~>
@Fin~tfJ~$t<l~!M~99n~¢~ewP.p .•,.' '.
@fIMJM~t~tlMl@QftM.newp·r······"·
Solution:
CD Radius of simple CUNe:
T
R=-I
tan -
2
123.59
R = tan 27'45'
R= 234.91 m.
,
I
,
I
I
\'\'
'",
,
I
' \
, ... "{I \ ,\
"
Old P.C. = (10 + 362.42)· (57.62)
Old P.C. = 10 + 304.80
AV
119.03
Sin 30' = Sin 124'30'
AV= 72.22 m.
'13
01
19° I.
' l~t
\ , \ t 64 o,'/
'·~'I
'-c:"\~\ i//
...
,
"
\
"
R - 1145.916
1-
0
1
- 1145.916
R1- 5'30'
R1 :: 208.35 m. ,
,
"
"
R
,
I
',,' 13' \
II
I
':/
;"
,
/
/
I
""'...
® Stationing ofnew P.C.
Sla. ofV= (10 + 362.42) + (72.22)
Sla. afV= 10 +434.64
Sla. of new P.c. :: (10 +434.64)· (123.59)
Sla. of new P.C. = 10 + 311.05
:
\'
j
I
I
I
Rz
,', '
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S-316
COMPOUND CURVES
- 1145.916
Rz- 3'30'
Rz= 327.40 m.
® Stationing ofnew P. T.
0= 1145.916
279.61
0=4.10'
Lc = 20 I
o
Lc =20 (64)
4.10
Lc =312.20 m.
Sta. ofnew P. T. ='(12 +259.91) + 312.20
Sta. ofnew P. T. =12 + 572.11
T, =R1 lan !J.
2
T, =208.35lan 13'
T1 =48.10m.
!2.
Tz =Rzlan 2
Tz =327.40 lan 19'
Tz = 112.73 m.
AS= T, + T2
AS =48.10 + 112.73
AS= 160.83
T3 =Rlan 13'
T4 =Rlan 19'
T3 + T4 =AB
Rlan 13' +Rlan 19' =160.83
R=279.61m.
@
ili.l.:iil
Solution:
CD Radius of simple CUNe:
Stationing of new P. C.
T3 =279.61 tan 13'
T3 = 64.55
AV
160.83
Sin 38' =Sin 116'
AV=110.17m.
Sta. ofnew P.C. = (12 +434.63)· (110.17)
• (64.55)
.Sta. ofnew P.C. = 12 + 259.91
M
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COMPOUND CURVES
T1 + T2 ::; 107
T1 ::; Rtan 13'20'
T2 ::; Rlan 17'2'3
Solution:
CD Stationing of P.C.
Rlan 13'20' + Rtan 17'25' =107
R= 194.30m,
® Stationing ofnew P. C.
T1 = Rtan 13'20'
T1 = 194.30 Ian 13'20'
T1 =46.05 m.
Sta. ofnew P.C. =(1 + 97S)· 46.0S
Sta. ofnew P.C. = 1 + 928.95
@
Stationing of new P. T.
Ran
Lc = 180
L = 194.30 (61.5') n
c
180'
Lc =20a.56
o _1145.916
. Sta. ofnew P. T. = (1 +928.9S) + 208.56
Sta. ofnew P. T. = 2 + 137.51
1 - 763.94
0 1 = l.S'
n.. _ 1145.916
V"L - 208.35
~=S.5'
Acomp~@d.cUr'lel~la@ •.OlJt4aQro.Wpmth~·
.mp··.19tryet=>.9·q:M0r9~la~N~@f.193.~4.m· •
·lhen·from·loo·fl;C,C;·srio1hertUrvewaslaidotll·
4
=20/1
01
/ _ 480 (1.5)
120
1
toth~F·T·gsqw·19n9wii~~f:~djp~W
/1 =36'
tntElrsElctionof.th¢l::ln~El(lt~js1Q"'4$2.25,.·,·
L =20/2
.4Qa.&51l'1·JfmEl§tlltr9rl69Hfm~.P9irt()f
.
Q)OelerlllinethestallQnir90fl~~P.C .., . < . ,.',
.®pelerrnjnethelength9fth~I&@¢h()f~fr()m
@
Ihey.c.to.lheP.T.• • • . • • • • • • • • • • • • .• • • • • • • • •.• • • •.• • • • • • • • .• •.•.
OetermlnetheaOglel/1aLlheloogchord
makesWithlheJal'l9flnt. . .
c2
~
/ _250 (S.5)
2-
20
/2 = 68'45'
T1 = R1 Ian !l
2
T, = 763.94 tan 18'
T1 = 248.22 m.
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S-318
COMPOUND CURVES
T2 : 208.35 tan 34'22.5
T2 : 142.53
AB: T, + T2
AB: 248.22 + 142.53
AB = 390.75
VA
390.75
Sin 68'45' = Sin 75'15'
VA = 376.59
Sta. of P.c. = (10 + 432.25) -(376.59) - (248.22)
Sta. of P.C. = 9 + 807.44
@
Length oflong c~rd from P.C. to P. T.
~
~
PoCo
L
PoT.
.
Sin 18' =&
C, =2 (763.94) Sin 18'
C1 =472.14m.
Sin 34'22.5' =
~
fk.
A reversed curve is formed by two
circular simple curves having a common
tangent but lies on opposite sides. The
method of laying out a reversed curve is just
the same as the deflection angle method of
laying out simple curves. At the point where
the curve reversed in its direction is called the
Point of Reversed Curvature. After this point
has been laid out from the P.C., the instrument
is tlien transferred to this point (P.R.C.). With
the transit at P.R.C. and a reading equal to the
. total deflection angle from the P.C. to the
PRC., the P.C. is backsighted. If the line of
sight is rotated about the vertical axis until
horizontal circle reading becomes zero, this
line of sight falls on the common tangent. The
next simple curve could be laid out on the
opposite side of this tangent by deflection
angle method.
Elements of a Reversed Curve:
= 2 (200.35) Sin 34'22.5'
C2= 235.21 m.
I! = 180' ·18' - 34'22.5'
I! = 127'37.5'
L2 = C12 + ~2. 2 (C1~) Cos 127'37.5
L2= (472.14)2 + (235.21)2
- 2(472.14)(235.21) Cos 127'37.5
L= 643.30m,
@
Angle that the long chord makes with
tangent at the P. T.
643.30
235,21
Sin 127'37.5 = Sin l1J
l1J = 16'50'
a ='180' - 127'37.5' - 16'50'
a =35'32.5'
R1 and R2= radii of curvature
0 1 and O2= degree of curve
.
V, and V2= points of intersection of tangents
e = angle between converging tangents
12 -I, = e
P.C. = point of curvature
P.T. = point of tangency
P.R.C. = point of reversed curvature
Lc : LC1 + LC2 : length of reversed curve
m =offset
P =distance between parallel tangents
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5-319
REVERSED CURVES
Four types of reversed curve problems:
1. Reversed curve with equal radii and
parallel tangents.
2. Reversed curve with unequal radii and
parallel tangents.
3. Reversed curve with radii and converging
tangents.
4. Reversed curve with unequal radii 'and
converging tangents.
Sin 9'34' =.1Q.
AB
AB =60.17 m.
® Radius of reversed curve:
2T=AB
2T=60.17
T= 30.085
T= Rtan 1
2
30.085 = R tan 4.78'
R= 359.78 m.
@
Twopar#Ueltangehls1Qrfi'ap~d#f~
conneyt~d • by.~ • ~ver%~~@f\ie .••• T~~¢P9t~
length from the P.C,tothep.%~q(lills12.0 nk
(j)
c.()rJ'1P\Jte•• • the•.• • llll)gtll•• ·i:lf•• taJlgEin~.'&j~~
COrnin!)ngjr~9tior><
@
~&~eT'lne • the•• eq~~I • r0~iU$.9~.~hT.~Merld •
@
Compute•• th~ • statiori~gpfl~e • P.R·9·jfthe.
stati9rm~otbat\fleqElgj~ni9~oLthEl
t<l~gentwlthCOmn·l(ml:lif¢9ti(m.i~ .• ~t4@>··
Stationing of P.R. C.
L=Rln
c 180
L = 359.78(9'341n
c
180'
Lc =60.07 m.
Stationing PC = (3 + 420) ·30.085
Sta. PC = (3 + 389.92)
Stationing of PRC = (3 +389.92) + 60.07
Stationing of PRC = 3 + 449,99
Solution:
CD Length of tangent with common direction:
.·IH•• ?.@ilt9ad·.·layplJ(,.A9~·.¢@lerlr~}M!W(!
,
Pilr<llleLtrClcksaresonl)ect8cuv.ithatllYllr:!iEl~
I
£ur'Je.o1 unequal.radji·•• Thecentralary91~qfthe
R'
Ll-:-,/-'-/----'
Sin
i=.~
2 120
~=4.78'
/=9'34'
first•• 9\jrye • is•• 16' .and!l1e.di.st<lnp~ • • t>et\\l~eg
parlill~Wlraqk$ .iS2T6Q • m,•• W~t'9Qjh9 • • cifIHe
p.q.j~t~~420<llldthElraQjilsoftheseqlJ.rJQ
¢uWelsZ90m.
.
.... .. .. .. .. . . ..
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S-320
REVERSED CURVES
Solution:
CD Length oflong chord:
"WO.Plifl3lj~I~t36000~~have • ~irectiollS()frtl~.
east~lld~te~(lQ~W~p~rt .• ~re • connecledbya
15+420
p.e
8,,,.r4if.
rfW~~9¢~o/~ft~~n~~~~lradiusof&QO.m·
Trye~,y.titt~tiJ!V~lsoltthe upperlan~111
600rfj./})
....
Sin 8' = 27 60
L
L = 198.31
® Radius of first curve:
a= R1 - R1 Cos 16'
a = R1 (1 - Cos 16')
b = R2 - R2 Cos 16'
b =R2 (1 - Cos 16') .
Solution:
a + b =27.60
R1 (1 • Cos 16') + R2 (1 - Cos 16') = 27.60
(R 1 + R2) (1 - Cos 16') = 27.60
R1 + R2 = 712.47
F?1 =712.47 - 290
R1 = 422.47 m,
@
Stationing of PT:
L - 422.47(16) 11:
C1 180
LC1
CD Length ofintermediate tangent:
(-V R2 + x2 )2 =(400)2 + (700)2
= 117.98
{8oo)2 + x2 =(400)2 +(700)2
x= 100 m.
2x= 200m.
L~=2~~g)11:
L~
=80.98
Sta.
Sta.
Sta.
Sta.
ofPRC. = (15 +4W!0) + 117.98
of PRC. = 15 +537.98
of PT= (15 + 537.98) + 80.98
ofPT=15+ 618,96
@
Distance between the centers of the
reversed curve:
D = 2 ~ (800)2 + XiD = 2 ~ (800)2 + (100)2
D = 2(806.23)
D = 1612.45 m.
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5-321
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REVERSED CURVES
@
Stationing of P. T.
700
Cos f!, = 806.23
f!,
=29'45'
100
tan 8 = 800
e = 1'08'
/ =29'45' • T08'
/=22'3T'
Rln
Lc = 180
L - 800 (22'37') n
c180'
Lc = 315.79 m.
® Radius of curve:
a= RCos 30' - RCos50'
a = R (Cos 30' . Cos 50')
a= 0.223 R
b = R - R ccs 50'
b =R (1 • Cos SO')
b= 0.357 R
a + b = 116.50
0.223 R + 0.357 R = 116.50
R= 2OO.86m.
@
Slationing of PT:
L - 200.86 (20) n
1180
L1 = 70.11 m.
L _ 200.86 (50) n
2180
L2 = 175.28 m.
Sta. ofPT= (10 + 620) + 70.11 + 175.28
Sla. of P. T. = (10 + 020.40) + 315.79
+ 200 + 315.79
Sta. of P. T. = 10 + 851.98
Sla. ofPT= 10+ 865.39
Two tangent~cqnv~rge~t,m angle of 30',
The ditecliririofthesecohd tangent is due
east. The distance of tlie Itc. from the second
tangent is 116,5(l nt The bearing of the
common tangentis$,AO'E.
CD Compute the central angle of the first
. .curve.' ', .. c··,.·. . C'.··· .'C·..··. : ..
® If a reversed curve ista connect these two
. tangents, de~tminetheqOmmon radius of
the curve,·.· ,'. ' • c": . "
..
@ Compute lhestationingof the P.T. if P.C.
is at station 10 + 620, '... •
.
Solution:
Q)
Central angle of the first curve:
Givenbrokenlin~s,l\S • •:'•• $r.~.l'rl,,~C¥~~'~m;
andCD=91.5rn.alTange4a~~h()~9'~
rever$ecurV~i$to:cl:mnectlh~~elht~llneS
thusfo,rmll1g.thecenter.li oll i:lfa·new.rQl'!l:L'. "..
find•• the•• !en9\h.of••th~.tdM~9I)radlu~.of
thereverse.c~rve., .••••••••'•.•.••••••••.••: ,••••••.••••••::•..• ,•• •. ,.:•.•• >
® If•• the.•.P.Q••• j$~t,$,\a.10.tOQq,\"hat.:i~ • the
(i)
slatklJjirl~ofe·T',/.
@\A.'hatis.the.tolal<lreaiflcIB~~~ • i9th~ • r9ht
Of·..~~y.ln.thi~.sectionof.the.rQad •. (A.t?.D).jf
. theroadwidthis45rn.····
.'
Solution:
CD Length of the common radius of the reverse
CUNe:
/1
/1
=50' - 30'
=20'
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S-322
REVERSED CURVES
T1 =Rlan11'
T2 = Rlan 32'
h+ T2 =91.5
R(tan 11' +tan 32')= 91.5
R=111.688m.
@
A = 3~0 (86) [(119.188)2 - (104.188j2]
A = 2514.63 m2
Sta; of P. T.
P.e. to PRC. = L1
L, =111.688 (22')
R1 = 111.688 +7.5
R1 =119.188m.
R2 =111.688- 7.5
R2 = 104.188 m.
Total area = 2514.63 +35.89(15) +21.710(15)
Total area =3378,63 m2
1~0
L, =42.885 m.
- 111.688 (64')n
L2 180'
L2 = 124.757 m.
Tota/length of reverse curve:
L, + L2 = 167.642 m.
Stationing ofPRC. = (10 + 000) +42.885
Stationing of P.R.C. = 10 + 042.885
Stationing ofP. T. = (10 +042.885) + 124.757
StationingofP.T. = 10+ 167.642
@
Area included in the right of way:
~
~\R=1l1.688 m
:
I '
---
C
\ .-.....
705m
7.5m E~'s;~'~~--.::~~
,
35.89
\
/
R=1l1.688 m,'fJ"
'I
T1 = 111.688 tan 11'
T1 = 21.710 m.
T2 = 111.688 tan 32'
T2 =69.79'
AtoP.C. =57.6-21.710
A to P.C. =35.89 m.
P. T. to 0 =91.5 - 69.790
P.T. tQD=21.710m.
n
A = 360 (R12 • Rl) "
,,=22'+64'
,,== 86'
Solution:
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8-323
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REVERSED CURVES
G)
Stationing at P. T.
-~
R2 -2
Sin 30'
S·In 22 .5' -- -.fL
2R,
381.97
R2 =2 Sin 30'
R2 =381.97 m.
L - 381.97 (60) 1t
c2 180
LC2 =400 m.
C, =2(190.986) Sin 22.5'
C, = 146.174m.
Using Cosine Law:
Sta. at the P. T. = (0 + 520) + (150) + (400)
Sta. at the P. T. = 1 + 070
® The angle that the long chord makes with
the first tangent = 61'10'
(485.025)2 =(146.174)2 + Ct.2
- 2(146.174) Ct. Cos 127.5'
213882.41 = Ct.2+ 177.97 Ct.
C2 =381.97 m.
Using Sine Law:
381.97 _ 485.025
Sin e - Sin 127.5'
e =38'40'
a = 180' -127.5' - 38'40'
a= 13'50'
_ R1 11 1t
Lc, - 180
- 190.986 (45) 1t
Lc,180
LCl = 150m
@
The angle that the long chord makes with
the second tangent:: 43'50'
Poe.
................... ,
-
","
,
.. ,
,.
Typ~ ofP~~eiTlel1tt;Il~~ 311{Portand d~k~h~~
>,.. .
_ R2 12 1t
Le2 - 180
Sin 30' =~
2 R2
Number oflaties~Two (2) lanes .
'.
Width of Pavement= 3.05 m. per lane .•.••....•
Thickness of Pavement:: 20cm5. .
// .' .
Unit Cost::: P460,OO persquare meIer
.
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S-324
REVERSED CURVES
(J,)•• P~P~@.th~fi)~i9*9ftl)ellrstcl1rve.
@{•• • Q()lT1put~.thElT"qj~s<if~ .• seC()ndcuf\le.
,·@•••• <;6ITlR~yf·.·me • • • ¢P,St•• • l)f•• Jhe<•• • c()ncn~te
Area Df SectDr:
A:: hI
7t
360
.....• P~¥!:lfl)fl~I~I~)M¢9ryli!s{reYersed)fr()rn
A1 :: ;~ [(125.78)2. (119.68f]
. ·.•. .·Jh~g.q;.t9!h~p,T·,.·.p.1l$~dpn • lhe • giy~t1
. ···l1lghwaY¢~$HnQeKaM$pi:ltjficatiOil$;
A1 :: ~~~) [(125.78)2 - (119.68)2]
Solution:
(j)
A1 :: 365.86 sq.m.
Radius of first curve:
A2 =7ti5~) '[(162.05)2. (155.95)2]
A2 :: 1015.68 sq.m.
A:: 365.86 + 1015.68
A:: 1381.54 sq.m.
T-IO
Total Cost:: 460 (1381.54)
Total Cost:: P635,50B.40
1
T1 =4(122.40)
T1 =30.6 m.
T2 = 122.40 - 30.6
T2 = 91.8 m.
• rWj:)t<j~~nt~jnt€lr~eSts~t~llMg~·()f.4&'40'
30.6
·langef\tisS4$'20''fI••'r1Je.ti@l~s·ofthe.wrve.
tan 14' =~
R1 = tan 14'
R1 :: 122.73 m.
9rE!·~o.b~.9()~r~cted.~Y<3.r~Y~~E!dBuryM!The
l~ng~ntW##hl'Aifrdm@~ • PdlrllPflm¢r$etlion
2j~&.i~04~&6~~.0 • ~h~PbW~ijJ~hW • ±h~e6~B~.
thf9119tlthE!P.q.i$24qm·~~gmE!dj~t~nce
ft(im.ihe.P9iMqf.inl¢j'seGUOllpft#genl$IQth¢
P.C.Qf:thereverse9cul}'ei$®0.4~il1;
.•.•....
® Radius of second curve:
tan 30' ::I2:
R2
91.8
R2 :: tan 30'
R2 = 159m.
Solution:
® Cost of concrete pavement:
(j)
Radius Dfthe Dther branch of the curve:
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5-325
Pdfbooksforum.com
REVERSED CURVES
240
tan 8 = 360.43
R1 81t
Lc, =180
e = 33'40'
- 240 (37'34)1t
Lc, 180'
Lc, = 157.36 m.
a =46'40' - 33'40'
a= 13'
240
AE = Sin 33'40'
AE =4:32.89 m.
AD =432.89 Sin 13'
AD=97.38m.
DE=432.89 Cos 13'
DE =421.80 m.
DF= DE· 48.60
OF = 421.80 • 48.60
OF = 373.20 ; AB = 373.20
BC= Rr BF
BC = R2 • 97.38
(R, + R2~ = (AB)2 + (BC)2
(240 + R2)2 =(373.20)2 + (R2 • 97.38)2
57600 +480 R2 +
.= 139278.24 +
R2 = 135.10
@
Stationing of P. C.
@
R2 A1t
LC2=18O
A=OO' ·5'46'
A. = 84'14'
- 135.10 (84'14')n
Lc2 180'
LC2 = 198.62 m.
Sta. ofP. T. = (10 + 577.36) + 198.62
Sta. ofP.T. = 10+ 775,98
Rl
Ri -194.76 R2+9482.86
Stationing of P.R. C.
BC = R2 - 97.38
BC = 135.10 - 97.38
BC = 37.72
AC = 135.10 + 240
AC =375.10
BC
tan 0 = AB
37.72
tan 0 =373.20
. 0=5'46'
. tan 13' =
Sta. ofP.R.C. =(10 +420) + 157.36
Sta. ofP.R.C, = 10+ 577.36
:a.~o
tRill'
radiuspfll1~fitWill.ll'\tE!Js.Z8$.4~m.U...i.• •.•·• •.• • • ·
0)[)etem,ltJelher@iliSQnMtrtd·cUN~,>··><
Pell~rll1lo~tH¢smtl<ltllogM~iffi¢,>.·• ·•• <'"
@
Oeter!1ll11~lhe$t~ti~tdng~fPm»··
®.
FG = 1'1.22
Cos 13' =48.60
EG
EG =49.88m.
Solution:
CD Radius of 2nd CUNe:
,, 60·
,,
,,
,
:,'
.
,
t
AG= EA· EG
AG =432.89-49.88
AG = 383.01 m.
8 =56'20' • o· 13'
8 =56'20' - 5'46' • 13'
0=37'34'
I
"
",'
1/
'
,RF285.40
tJ~ I.'>,'
J~
I
--: Iv.:"
" ,
~/
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S-326
REVERSE" CURVES
R1 = 285.40
11 + 30 = 12
12 = 50
11 =50·30
I, = 20
AB.
100
Sin 30' =Sin 20'
AB = 146.19
T1 = R1 tan !1
2
T1 = 285.40 tan 10
T1 =50.32 m.
T2 = R2 tan 25'
T1 + T2 = 146.19
TW6jl<ltM¢ltaiIW~Y4Ppm;ij~~M¢f¢~IJ~
•
II~l~~e1Ii~~~lt~dl~~r~ t~~.11ad\~~ •
(j)
.Oeferrni@1~~.(;Elotr:~I~@@9t(ti~rij~fli~
9JfVE i . Y > <
:· ·• irJ~~iI1;i~.i~I~;~£~;I~~~6t!~'.·.
Solution:
T2 = 146.19 - 50.32
T2 =95.87 m.
R - 95.87
2 - tan 25'
R2 = 205.59 m.
® Sta. of PR C.
o - 1145.916
1 - 285.40
0 1 =4.02'
D - 1145.916
2 - 205.59
~=5.5T
CD Central angle of the reverse CUNe:
G
LC1--~
01
_ 20 (20)
4.02
=99:50
LC1LC1
o
L~=2~.~~)
L~
= 179.53
P.C. =(10 +432.24) - 50.32
P.C. = 10 + 381.92
PRe. =(10 +381.92) + 99.50
PRC. = 10+ 481.42
® Sta.ofP.T.
P. T. = (10 + 481.42) + 179.53
P. T. = 10+ 660.95
N
400
tan IX =2 (1100)
=10'18'
~---OG =;j (40W +(220W
IX
OG = 2236.07
2236.07 Cos IJ =2x + 200
x+ 200 =R
x= R- 200
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REVERSED CURVES
2236.07 Cos 8 = 2 (R- 200) + 200
2236.07 Cos 8 = 2R -400 + 200
2236.07 Cos 8 = 2R - 200
1118.03 Cos 8 = R- 100
R-1oo
Cos 8 1118.035
1100 -100
Cos 8 = 1118.035
1000
Cos 8 = 1118.035
8= 26'34'
o = 26'34' - 10'18'
0=16'16'
@
Solution:
CD Radius of first curve:
Sla. of the middle of intermediate tangent.
L=R01t
180
= 1100 (16'16') 1t
L
180'
L = 312.30 m.
"----..;-..::::::-c
AB
150
Sin 30' = Sin 20'
Sla. = (20 + 460) + 312.30 + 200
Sla. =20 + 972.30
@
Sta. of the P. T.
Sla. of P. T. = (20 + 972.30) + 200 + 312.30
sta. of P. T. = 21 + 484.60
AB - 150 Sin 30'
- Sin 20'·
AB =219.29 m.
T1 + T2 = 219.29
T1 =R1 tan !J.
2
T2 = R2 tan !1
2
·~W:ffi~S~1~s~gli~~~{~~I~pi~ • IOi6;~~iH~.
di$lanoeoflhisb'\t~rs~¢liOnJr9tnttieB.tQMh~
curvelsA50m,Tfjedefl¢C;tjona~glei:ltttie
C()rnmOnIElngenLfml11th~?a(;~J~JlMrW.is
2(FR; .al'ld•• trye~i'mqmoIJ~~p6mrripli.~~9#~t
i~.32Q' •.•••••• T~e • d~grEle • ()f.·CJJry~ • 9ft~¢sM9I)d
simple Cl,lrveis$' and the~tatlgf\i~g~nW
pQint.·. oC.intersection.·.of • • the•• fj(~t¢t@~ • ·is
4·+450,··••·
CJ) Determine the
. . ••.. <.••..••• >..
radius oltha fltstculVe.•.
® Determine the stationing of P.R,C,·· •.....
@ Determine the stationing oUhe P.T.
12 = 11 + 30'
SO' =/1 +30'
/1 =20'
R1 lan !J.
2 + R2 lan !2
2 =219.29
R,lan 10' + R2 tan 25' = 219.29
- 1145.916
Rr
~
- 1145.916
R2 6
R2 =190.99 m.
R, tan 10' + 190.99 Ian 25' ::: 219.29
0.1763 R, T 89.06 = 219.29
R, = 738.68 m.
5-328
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REVERSED CURVES
® Stationing of P.R. C.
o - 1145.916 _ 1145.916
1R1 - 738.68
0 1 = 1.55'
T, =R, tan !1
2
T, = 738.68 tan 10'
T1 = 130.25 m.
Solution:
<II Central angle of 2nd CUNe:
Sta. ofP.C. =(4 + 450) - 130.25
Sta. ofP.C. =4 + 319.75
_ (20) (20)_
1.55 - 258.06 m.
L, -
Sta. ofP.R.C. = (4 + 319.75) + 258.06
Sta. ofPRe. = 4 + 577.81
® Stationing of P. T.
L
-~
2- ~
L _20(50)
26
L2 =166.67 m.
Sta. of P. T. = (4 + 577.81) + 166.67
Sta. of P. T. = 4 + 744.48
Arl~y~fseGilr\i~$t@~n#NiHgraoi9~
AD = 520 Sin 30'
AO=260 m.
EA = 400 Cos 30'
EA = 346.41
DE =346.41 - 260
DE=86.41 =FB
Rl•• :;.40grrt,anqR2::?OOrrt·.l()f\g.I~t()GCltln~t
lI1elW()t;~g~nt~&\I'alidvs.\'IjmanglEipf
intflr~l3Clh:m . .• ()f•.•• l/:lfl·.t~llgeril~.eqll~I.· • t()~q\
A¥·#520rit
.
.
°1
~.
® Central angle of 1st curVe:
11 + 30' = 12
11 = 61'29' . 30'
/, =31'29'
® Distance VB:
F0 1
S·In 12 - 400 + 200
FO, =600 Sin 61'29'
FO, = 52721 m.
EO, = 400 Sin 30'
E0 1 =200
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S-329
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REVERSED CURVES
FE= F0 1 -E01
FE =527.21 - 200
FE =327.21 =DB
In triangle VDA:
. VB+DB
Cos 30 =~
' - VB + 327.21
Cos 30 520
VB = 123.12 m.
x+h=d
346.41 - 400 Cos 11 + 200 - 200 Cos 11 = 281.91
600 Cos I, = 264.50
11 = 63'51'
® Central angle of 2nd curve:
11 = /2 + 30'
12 =63'51' • 30'
12 =33'51'
® Stationing of P. T.
_ R1 /1 1t
LC1 -
Arever~.CulV~I$#>HIl~Mlhe.1\VcjlanQMt.
liflesAI3•.~~Cl.99 • haMingdir~pwqf(llleEalit.
·f.ltlt~iillilll
as
ofA(P.p.)lsatlO+1;20:5P,WSDI$30Prri.
long<ihd h 8 be<lriIllPfS.20\E.>· ."
~. ·~l~~
• f~:.~lf::.~~~l:~~~go'~c~;
.• ·. · "
Findthe-sfalioningpfP;T.
" ,,' "
@
180
- 200 (63'51') 1t
Lc1 180'
LC1 = 222.88 m.
_ R2 /2 1t
LC2 -
180
- 400 (33'51') 1t
L~ 180'
L~ = 236.32 m.
pR.e. =(10 + 120.50) + 222.88
PR. C. = 10 + 343.38
Solution:
CD Central angle of first curve:
P. T. =(10 + 343.38) + 236.32
P. T. = 10+ 579.70
parallel,tangents 20m" apart, are lope
conMCted bya reversed CtJrve.The,ra~iUs¢.
tile firsfcurve at the P.C. has a.ra~iusQf
800 m, and the total 'length afthe chOrdfrom
Two
Sin 70' =-.!L
300
d= 281.91 m.
h =200 - 200 Cos 11
y=400 Cos 11
x= 400 Cos 30' - Y
x=34641-400Cos/ 1
the P.C. tethePT. is 300 m. Stationing ofthe
p.e. is 10 +620.
CD Find the central angle of each cuNe, "
® Find lhe radius of the curve pas$ing thru
theP.T.
'
® What is the stationing afthe P.T.
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5-330
REVERSED CURVES
Solution:
<D, Central angle ofeach CliNe:
J"W9•• parallelt13pg~Jlts .• artlconllecll:d•• •llYA.
t¢Ver$eQ(;uMl·havingequalradij.of~611.m.{i>i
(j)lft~e • centrl~larygle • ofthecu(\lei$~'
>colTlPuletb~dislarl(:e •• betweel1fuJr!lII~[
~nQents.<>
• ~. •.•.GQ(nputetM .IIU@h.. QfchQtd.ft9m.me.ft¢.
A6tt@~.T.>
.~ • • • ltg.C.•.• i~~t • ~~a··.$.t • • 96MO'•• WQ!lt.l~~h~
> s~t~lIlng9{l'h~frt<
Sin i=1Q..
2 300
.
.
Solution:
<D Distance between parallel tangents:
i_
2- 3'49'
1= T3B'
@
Radius of the CUNe passing thru the P. T.
a =800 •800 Cos 7'38'
a =7.09 m.
b =20·7.09
b= 12.91
x= 360 Cos 8'
x= 356.50
b :: R2 • R2 Cos 7'38'
12.91 =R2 (1 • Cos 7'38')
R2 = 1456.~5 m
@
Sta. of P. T.
_ R1 /n
LCl -
180
y = 360 •356.50
y= 3.50
2y= 7.0m.
® Length ofchord from P. C. to P. T.
Sin 4' = 2 (3.50)
L _800(7'38)n
cl -
.
180'
LCl = 106.58
L - 1456.85 (7'38') n
~180'
LC2 = 194.09
Sta. of P. T. = (10 + 620) + 106.58 + 194.09
Sta.ofP.T. = fO+ 920.67
L
L':: fOO.35m.
@
Sta. of P. T.
R8n
Lc =18O
L :: 360 (8') n
c
180'
Lc:: 50.27
Sta. of P. T. = (3 t 960.40) + 50.27 + 50.27
Sta. of P. T. = 4 + 060.94
.
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REVERSED CURVES
The common tangent CO of a reversed curve
is 280.5 m. and has an azimuth of 312' 29'.
BC is a' tangent of the first curve whose
azimuth is 252' 45'. DE is a tangent of the
second curve whose azimuth is 218' 13'. The
radius of the first curve is 180 m. The P.l.1 is
at sta. 16 + 523.37. Bis at PC and E is at P.T.
a) What is the stationing of the P.C.
b) What is the stationing of the PRC.
c) Whatis the stationing of the P.T.
Solution:
b) Stationing of P.R.C.
_ 380(59' 44')
Lc1 180
Lc1 = 187.66
n
PRC = (16 +420) + (187.66)
PRC = 16 + 607,66
c) Stationing of P.T.
L._ = 164.41(94'16') n
---z
180
L~=
270.50
I
\
\
Rl=180m~.~r!
/
"
,:ID52'
'Vi
a) Stationing of P.C.
CO= 280.5 m.
280.5 = 180 tan 29 52' + R2tan 47 08'
R2 = 164.41
T1 = 180tan2952'
T1 = 103.37
T2 = 280.5 -103.37
T2= 177.13
P.C. = (16 + 523.37) ··(103.37)
P.C. = 16 + 420
P.T. = (16 + 607.66) + 270.50
P.T. = 16+ 878,16
S-330-B
RMRSED CURVES
Two parallel tangents 20 m. apart are to be
connected by a reversed curve with equal
radius at the p.e. and P.I. The total length of
chord from the P.C. to the P.T. is 150 m.
Stationing of the p.e. is 10 + 200.
(j)
@
@
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@
L= 150
2
L= 75m.
Find the radius of the reversed curve.
Find the length of cord from p.e. to PRe.
Find the stationing of the P.T.
@
Solution:
(j)
Central angle ofeach cUfVe:
Length of chord from P. C. to P.R.C.
Sta. of P. T.
Lc= RI (If)
180
L - 280.93 (15'20') If
.c-
180'
Lc=75.18m.
P. T. = (10 + 200) + (75. jd) + (75.18)
P. T. = 10+ 350,36
. I
20
S'"'2 = 150
f
=7'40'
1= 15' 20'
R- R Cos 15'20' = 10
R (1 - Cos 15'20') = 10
R= 280.93 m.
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5-331
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REVERSED CURVES
L - R2 /2 rc
~ - 180
- 314,90 (78') rc
Lc2 -
~a~v;~[~~~~~~e~!~iRre1~n~·~j
L~
tit4lii»IIB
180'
=428,69
Total length of curve:
Lc=Lc; +L~
Lc = 167 + 428.69
Lc= 595,76
$;4Q.. . E, . ~nd·a:dl.S@nl:¢ . t'lf•• ~OOm ..•.•. lftnefli':$t·
Sta. ofP. T. = (12 + 340) + 595.76
Sta. ofP.T. = 12+ 935.76
BlriYjlr'·'
li.llIlI.t1l~J
ij~~Mrelioft1Z·fMQ;
..
.......
Solution:
G)
Radius of 1st curve:
1
T1 =4"(340)
T1 =85m.
T1 = R1 tan 13'
R1 =368.18 m,
® Radius of 2nd curve:
T2 = 340·85
T2 = 255 m.
T2 = R2 tan 39'
R2 = 314,90
® Sta, of P. T.
_ R1 /1 rc
LC1 - 180
L - 368.18 (26') rc
c1 180'
LCl = 167.Q7 m.
:'.11111.
Solution:
G)
Common radius of curve:
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S-332
REVERSED CURVES
R2 + (100)2 = (R - 100)2 t (400)2
R2 +{1oo)2 = R2 - 200R t (100)2 t
Solution:
(400)2
G)
Distance between parallel tangents:
200R =400 (400)
R=800m.
o..-".-..,-~..".;;=--/-fl
,
i IR
® Central angle of the CUNe:
1 =460
200
tan (J = 1600
-----------tu
B= TOO'
(AB)2 =(200)2 + (1600)2
AB = 1612.45
p.T.
1
BD = (1612.45)
x =460 Cos 12'
BD=S06.23
700
Cos a =S06.23
Y=200 Cos 12'
x=449.95
y= 195.63
a =29'45'
a =460 - 449.95
a =10.05
e =29'45' - 1'OS'
e = 22'37'
b = 200 -195.63
b=4.37
® Sta. of P. T.
Re1t
LCl
=180
I
_ SOO
_cl LCl
Distance between the parallel tangents
= 10.05 +4.37
= 14.42m.
(22'37') 1t
1S0'
=31519
® Stationing of PR C.
Rl en
Sta. of P. T. =(10 +340.20) +315.79
+200 +315.79
Sta. of P. T. = 11 + 171.78
LCl
=180
L - 200 (12') n
C1 180'
LCl =41.89
Sta. of PRC. =(2 +360.20) +41.S9
Sla. of PRC. = 2 + 402.09
Ar~'Jers~¢lJry.ehasafadius()flheCur\le
pa~SII19.th"()l!grythe.P.C,.~ualt()200m.;lnd.·
thatqfthe•• §~P9rd.Cll[Ye • paSSirg • throU~llt~~
P.T,)~49Pm·19r9;ltthecentr~langle(jf
c9tvEli~12'
.
..
.
theperpendiclilar distance between
. the tWo parallel tangents.
..
.
® If the statloning of the p.e. is2 + 360.20.
fOO lheslalioning of the P.RC.
@ Plna the stationing of the PT.
@ ·Plnd
@
Sta. of P. T.
R2 en
LC2=18O
L - 460 (12')n
c2 180'
LC2 =96.34
Sta. of P. T =(2 + 402.09) + 96.34
Sta. of P. T. = 2 + 498.43
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REVERSED CURVES
The perpendiciJlar ~istan~El·between two
parallel tangentS of areversedcurved is 7.5 m,
and the chorndistance frotli the PoCo to the
P.T. isequalto65m./'··
.
Solution:
0)
SYMMETRICAL
PARABOLIC CURVES
In highway practice, abrupt change in the
vertical direction of moving vehicles should be
avoided. In order to provide gradual change in
its vertical direction, a parabolic vertical curve
is adopted on account of its slope which
varies at constant rate with respect to
horizontal distances.
Central angle ofthe reversed curve:
Properties of Wrtlcal Parabolic Curves:
Forward
fa!'lg~Tlf
· 8 7.5
SIn =-6.5
8 = 6.63'
28 = 13'15' (central angle)
® Radius of the curve:
Cos 13'15' = R- 3.75
R
R- 3.75 = 0.913 R
R= 140.87m.
@
Slaioning of P. T.
R(28)n
LC =-100
L
=140.87 (13'15')n
c
180'
Lc = 32.58
Sla. of P. T =(4 + 560.40) + 32.58 + 3258
Sla. of P. T. =4 + 625.56
1. The vertical offsets from the tangent to the
curve are proportional to the squares of the
distances from the point of tangency.
K_.YL
(X1)2 - (X2)2
-h.._.J:L
(X3)2 - (~(
2 The curve bisects the distance between
the vertex and the midpoint of the long
chord.
From similar triangles:
BF m
.BF=CO
-=L
L
2
2
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5-334'
PARABOliC CURVES
From the first property of the curve:
BE Q)
(~)2 =If
r=~
r=2k
BE= CDL2
4 L2
8. The maximum offset H= 1/8 the product of
the algebraic difference between the two
rates of grade and the length of curve:
1
From the figure: H=BE = 4CD.
BE=CD
.
Therefore the rate of change of slope is
constant and equal to:
4
1
H=4 CD
CD
But -= BF
2
L
CD =(91· g2)2
BE=H~)
H=.!CO
4
1
BE =-BF
2
3. If the algebraic difference in the rate of
grade of the two slopes is positive, that is
(g 1• g2), we have a·summit" curve, but if it
is negative, we have a "sag curve".
4. The length of curve of a parabolic vertical
Location of highest or
lowest POint of the Curve
curve, refers to the horizontal distance from
the P.C..to the PT.
5. The stationing of vertical parabolic curves
a) From the P.C.
is measured not along the curve but along
a horizontal line.
6. For a symmetrical parabolic curve, the
number of stations to the left must be equal
to the number of stations to the right, of the
intersection of the slopes or forward and
backward tangent.
7. The slope of the parabola varies uniformly
along the curve, as shown by
differentiating the equation of the parabolic
curve.
.
y = kx2
Q1.=2kx
dx
The second derivative is
~ = 2k
where ~
dx2 = rate of change of grade or
slope.
1-51
x
(x,-y,)
.T.
(XrY,)
p.e
U2
U2
The slope of the tangent at P.C. is 91'
From the equation of the parabola,
y = k x2
Q1. = 2 kx
dx
where Q1.= 9
dx '
x = $,
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PIUBOUC CURVES
~=2kx
dx
91 = 2 k (51)
ill 9,=2k5 1
The slope of the tangent at P. T. is g2-
~­
dx -92
At the P.e.
x=(L-5 2)
2Y._
dx- 91
2Y.=2kx
dx
ill 91 = + 2k(L - 52)
X = L - 51
~=-2kx
dx
92 =- 2 k(L - 51)
® 92 =·2 k(L • 51)
At the P.T.
x = 52
2Y._
dx - 92
*=-2kX
Divide equation ill by ®
9.1_ 2kS1
92 - • 2k (L • 51)
9.1_~
92 - • (L· 51)
- 91 L + 91 51 = 51 g2
5, (g1 • 92) = 91 L
c'
location of the highest point of the
curve from the P.C.
b) From the P. T.
Divide equation ill by ®
9.1_ 2k (L • $v)
92 - • 2k (52)
• g1 52 =92 L - 92 52
52 (92 - g1) = g2 L
location of the hi9hest or lowest point
of the curve from the P.T.
....><
-----t---U2!---
E11~[(JI~
Q) • Whllt.ls.lh~lellillh.OflheCllrv.e? • .• • • •.• • • • • • •')•• .
@.. '•• GQmputethE!elevatiofloftll~J()westpQll'lt
The slope at the P.C. is g(
~=2 kx
dx
btlhfi'cui"Ve/ '
@).··.·.¢OmPute·tl'le¢levatiollat~tatlOll.10.rQQQ.·.·
.•
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5-336
PARABOliC CURVES
Solution:
,,) Length of curve:
n = 92.:Jl1
r
_ 0.4 - (- 0.8)
n0.15
n=8
L =20 (80)
L =160m.
@
A symmetrical vertical suminitcurve has
tangents of ... 4% and - 2%, The allowable
rate of change of grade is' 0.3% per meter
station. Stationing and elevatiOn of p.r. is at
10 +020 and 142.63 m. respectively.
..
Elevation of lowest point of curve:
CD Compute the length of curve.
.
@ Compute the distance of the highest point
of curve from the P.C. . .' '. . . .
. ".
@ Compute the elevation Of the bighest point
of curve.
.
p.
Solution:
"'%
CD Length of curve:
'
C
10+000
10+020
EI.240.60",
H=8
0.3=
4 - (- 2)
n
n = 20 sta,tions
S=Jl1..L
g1- g2
S = - 0.008 (160)
. - 0.008 - 0.004
S= 106.67
L
H=S(g1- 92)
160
Rate of change = ~
Length of CUNe =20 (20)
Length of curve = 400 m.
® Sta. ofhighest point of curve:
'.
(- 0.008 - 0.004)
H=0.24
~_--lL
S
_Jl1..L
g1 - g2
_ 0.04 (400)
S1 - 0.04 • (- 0.02)
S1 = 266.67 m. from P.C.
1-
(80)2 - (53.33)2
0.24_~
® Elevation of highest point of curve:
(80)2 - (53.33)2
Y = 0.11
E1ev. A = 240.60 + 26.67(0.004)
Elev. A = 240.71 m.
Elev. of lowest point of curve =240.71 +0.11
Elev. of lowest point of curve = 240.82 m.
@
Elevation of station 10 + 000:
.J:2... _.0.24
(60)2 - (80)2
Y2 = 0.135
E1ev. 0 =240.60 + 20(0.008) + 0.135
Elev. 0 =240.895 m.
L
H=S(91· g2)
H=
a400 (0.04
H=3
+ 0.02)
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5-337
PARABOLIC CURVES
--.!i- _.--L(200)2 - (133.33)2
_ 3 (133.33f
Y-(200)2
y= 1.33
p.T.
EJev. at highest point
. = 142.63 + 133.33 (0.02)-1.33
Elev. at highest point = 143.97
L
H ="8 (92 • g1)
200
H="8 (0.02 + 0.04)
H= 1.5 m.
....h-_~
(66.67)2'" (100)2
Y1 =0.67 m.
Elev. B = 124.80 + 0;02(33.33) + 0.67
Elev. 8 = 126.14 m.
® Elevation of sta. 12 + 125.60:
-l'.L_~
(75)2 - (100)2
Y2 =0.84
Elev. 0 =124.80 + 0.04(25) + 0.84
Elev. 0= 126.64m. (eJev. at 12+ 125.60)
Solution:
CD Length of curVe:
r=92J1
n
0.6=
2- (-4)
n
n =10
@
A verticalsljrnmilpataboliC cuNehas itsPck
at station 14 + 75Dwllh elevalionof76.30rn~
L = 20 (10)
L=200m.
The gradepflhe back tangentis3A% an<l
forward tangEmfof- 4.8%. lfthelengt~ ·of
curve. is 300m. •
..
. . '.'
Elevation of lowest point of curve:
m Compute the.location of the vertical curVe
'. tumingpaintfrom the Pol.
'.
.'
..
S-~
2-
g2' gl
S - 0.02 (200)
2 - 0.02 =004
82 =66.67 m. from P. T.
Compute the elevation of the vertical curve
turning point in meters.
'.
® Compute' the stationing of the vertical
curve turning point.
.
(2)
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5-338
PARABOLIC CURVES
Solution:
CD Location of vertical CUNe turning point:
A vertical summit parabolic curve has a
vertical offset of 0.375 m. from the curve to the
grade tahgent at sta 10 f 050. The curve has Ii
slope of + 4% and- 2% grades intersedingat
IhePJ.The offset distance of the cutveatPJ
is equal to 1;5 m. If the stationing of theRC.
is at 10 + ObO.
CD .Compute the required length of curve.· ..
® Cornputelhehoiizont(\1 distanceofth¢
8-~
1 - gl . g2
vertical curve tiimingpoirit from lhepolnl of
intersecltpn ofthe grades.
. .. .
@Coinputethe elevation of the vertiCal CUrve
. Jurningpoint if the elevation of P.T is
86.4(!.iTl. .
8 - 0.034(300)
1 - 0.034 t 0.048
81
=124.39 m.
Solution:
CD Required length of CUNe:
x = 150 -124.39
x =25.61 m. from the P.I.
@
Elev. of vertical CUNe turning point:
L
H=S (g1- g2)
300
H =8"' (0.039 + 0.048)
H= 3.075
.--1- _ 3.075
.-L_....!!-
(124.39)2 - (150)2
y=2.11 m.
0.375 1.5
(50)2 =(U2)2
(50)2 - (Wi
L= 200m.
Elev. of vertical CUNe tuming point
=76.30 - 25.61 (0.034) • 2.11
Elev. of vertical CUNe turning point
=73.32 m.
® Stationing of the vertical CUNe turning
point:
(14 of- 750)· (25.61) = 14 + 724.39
@
Highest point of curve:
.
..
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PARABOLIC CURVES
g1 L
g1 - g2
S --1-
_ 0.04(200)
S1 - 0.04 +0.02
S1 = 133.33 m.
x = 133.33 - 100
x= 33.33m.
® Elev. of highest point of curve:
--..L-_~
(66.67)2 - (100)2
y=0.67
Elev. A= 86.42 +66.67(0.02) - 0.67
Elev. A = 87.08 m.
By ratio and proporlion:
L
2H (91-g0'2
T=-L-
2
H =IJl.LJhlb
8
H = [0.06 - (- 0.04)) L
8
H= 0.0125 L
Elev. of B = 100 - 0.06 (20)
Elev. of B = 98.80
y= 98.80·98.134
y=0.666
----L- _J:L.
(~_20)2 - (~y
0.666 _0.0125 L
(~'20r- (~y
0.666 _4 (0.0125) L
-
2
A symm.etril;:al parabolic summit curve
connects .tWo grades of +6% and· 4%.. It is to
pass through a point "pH on the curve atstalion
25 + 140 having an elevation of 98.134 m. If
t~e elevation of' the. grade intersection is
m. ~\'ith a stationing of 25 + 160.
too
'4 -20L+ 400 -13.32L =0
G) Compute the length of the curve.
@ Compute ihe stationing of the highest point
L2 -133.28L + 1600 = 0
L= 120m.
. ofthe cur/e.
@ Compute
the elevation of station 25 + 120
(~-20Y
L
(~-20)2=13.32L
L2
® Stationing of highest point of curve:
. on the curve,
Solution:
CD Length of curve:
_-.9.LL
S1-
gt- g2
S - 0.06 (120)
1 - 0.06 - (- 0.04)
S1 =72 m. from p.e.
Sta. of highest point =(25 + 100) + 72
Sta. of highest point =25 + 172
8-340
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PARABOUC CURVES
@
Elevation of station 25 + 120:
Solution:
CD Length of curve:
H=0.0125L
H =Jl.0125 (120)
H= 1.5
H + 5 = 152.74 -146.24
H+5 =6.5
H= 1.5 m.
L_J.i..
(20)2 - (60)2
L
_ 1.5 (20)2
H="8(~-g1)
Y- (60)2
y=0.167 m.
Bev. of A = 100·40 (0.06)
EJev. of A = 97.6 m.
Bev. ofB = 97.6 - 0.167
Elev. of B = 97.433 m.
L
1.5 ="8 (0.035 + 0.03)
= 1.5 (8)
0.08 .
L'= 150 m.
L
® Clearance at left edge:
Elev. A =146.24 + 0.03 (6)
Elev. A = 146.42
.1L_J.i..
(69)2 - (75)2
Y1=1.27
• ~I:.5~~%:)ddi(:~Suhd:~~n gJj~~p~s~··bri~~~.
\\I~9~~tlndersjde .is at elev; 152;74, and
.~me$~ii6ther'road acrossthegraljes at right
augle$;< . <•.•. '. ..' . ...•. . ..
<DWhatis the longest parabolic curVe that
. "ganbe used to connect the two grades and
i1ttM same lime provide at leas! 5 in.. of
.c1¢aran,ce under the bridge at its center
t:~1~:
underside of the bridge is level and is
12 m. wide. find the actual clearance at the·
left edge of the bridge.
@)
If the underside of the brtdge is level and is
12 m. Wide, find the actual clearance at the
right edge of the bridge.
Elev. B = 146.42 + 1.27
Elev. B = 147.69
h1 = 152.74 -147.69
h1 = 5.05m.
@
Clearance at right edge:
EJev. C= 146.24 + 0.05 (6)
Elev. C = 146.54
JL_J.i..
(69)2 - (75)2
Y2 = 1.27
Elev. 0 = 146.24 + 0.05 (6) + 1.27
Elev. 0 = 147.81
h2 = 152.74 - 147.81
~
=4.93m.
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PARABOliC CURVES
L2
ID
36 _4 (4,40 2 +400)
L-
J\~f~d~Vd~*4~f@ngattherate(}L·4%
inw@ct$.~99tl)~tgl'Me.a$~ndio~atthf1.ralE!
()f.+8%at~taUon.2·.+ . 0pg, • ElleVaJi9rt.100.ni.••.• A
~----
vert~al·~(iW~i$M • poilW~cLlh~ • !Wtlsuch.lhat
lhe.cll@.wm.clearabOulderJocatedal.statkln
1f~~Q,~IElV#I~h161,34rn. '
'.'
.G) •••
L= 116+.y(116f- 4 (1)(1600)
2
- 116 + ...J7056
L2
IQ~~~i~~ • 1D~• • n~~ssary • • length•• •Of•• lhe.
L= 116+84
® Determiriethe·slatkliloflhelocalionof.a·
l!l.li1l1~llm.1
2
L = 100m.
@
Station of the location of a sewer:
_.E.1..!:.....
Solution:
G)
L2
36L = L2 - 80L + 1600
L2_116L +1600=0
51 - g1 - f/2
5 - - 0.04 (100)
1 - _ 0.04 - 0.08
51 =33.33
Length of the CUNe:
Sta. = 2 +000 -16.67
Sta. = 1 + 983.33
@
Elev. 1 +980 = 100 + 20 (0.04)
Elev. 1 + 980 = 100.80
Y= 101.34 - 100.80
y=O.54
Elevation of station where sewer is
located:
p.
.---l:'- _--.tL
(~_ 20)2 - (~y
0.54 _ (~_20)2
H -
(W
L
H ="8 (g1 - f/2)
L
H="8 (g1 - g2)
. L
H ="8 (0.04 +0.08)
H =- 1.5 (sag cUNe)
----L-_J.&..
L
H="8(0.12)
8 (0.54) _
L (0.12) -
100
H=S(-0.04-0.08)
i~' 20l (4)
L2
(33.33)2 - (50)2
Y= 0.667
Elev. of A = 100 + 16.67 (0.04) + 0.667
Elev. of A =101.334 m.
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PARABOLIC CURVES
Sta. of P.C. = = (2 + 700) - 80
Sta. oW C. = 2 + 620
On a ~aUrQad a • 0.8% grade meters a +0,4%
grad~$tation 2 + 700 whose elevation of
300 m, The maximum allowable change in
gradaper station having a length of 20 m. is
0.15: ....
.
Sta. of lowest point =(2 + 620) + 106.667
Sta. of lowest point = 2 + 726.667
® Elevation of the invert of the culvert·
El. 300.214
@ .' Compute the length of curve.
® CQmpute the stationing where a culvert be
...•...• lo¢~ted.
. ....
® Atwhatelevation must the invert of 1he
cJtvert b~ set iflhe pipe has a diameter of
(l.~m: and lhebackfill is 0.3 m. depth.
•. Ne9lect thickness of pipe.
Solution:
INVERT
26.667
H=~(g2·g1)
H=
1~0 (0.004+0.008)
H= 0.24 m.
LcW~Sl
Point
C
£1.300 m
2+700
CD Length of curve:
one station = 20 m. long
r = rate of change per station
r=JlLIl1=O
n
.15
.
r= 0.4· (- 0.8) = 0.15
n
1.2
n = 0.15
n = 8 stations
L =8 (20)
L = 160m.
@
Stationing of the lowest point on the curve:
5-~
1-
gl - g2
S - ·0.008 (160)
1 - _ 0.008 . 0.004
51 = 106.667
-lL_--L(8W - (53.333)2
0.24 _--L(8W - (53.333)2
y= 0.107 m.
Elev.
Elev.
Elev.
Elev.
E = 300 +0.004 (26.667) + 0.107
E =300.214
of invert = 300.214 • 0.3 - 0.9
ofinvert = 299.014 m.
The grade ora symmetrical parabolic curve
from station 9+000 to the vertex V at sta.
9 + 100 minus 6% and from station 9 + 100 to
9 + 200 is minus 2%. The. elevation at the
vertex is 100.00 m. His required to.connect
these grade lines with a vertical parabolic
cur,e that shall pass 0.80 m. above the vertex.
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PARABOLIC CURVES
CIi Depth of cut at station 9 + 080:
10e,20 • • • • • ~
:?9.65
105.:m9+140
98.36
9+040
102A9
9 + 160
99.00
9+000
102.00
9+180
98.00.
9+080 ., 102,18" '9 +200 ,98.00
" I
19+10(Y 101.80
9+000
9+020
"I-l-l-+-+-+-
STATIO!o1NG
Solution:
L
H='8(grg1)
H= 1:0 (-0.02+0.06)
H=0.80
.Jl_ 0.80
(60)2 - (80~
Y3 =0.45
)02
~~ 101
::~I'"
'"
.:.~
"
" :;g
~ ~
~
Elev. of 9 + 080 = 100 + 20 (0.06) +0.45
Elev. of 9 + 080 =101.65 m.
Depth of cut =102.18 - 101.65
Depth of cut = 0.53 m.
® Depth offill at sta. 9 + 140:
CD Length of the vertical parabolic curve:
L
H='8(g1- g2)
2x
H =8 [- 0.06 - (- 0.02)]
2x
0.80 =8
(- 0.04)
0.08x =0.080 (8)
x= 80 m.
2x= 160 m,
;,f-·+-+-+-+-+-++-f
L" 160m.
STATro.V1.'V(,
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PARABOLIC CURVES
A_-lL
@
(40)2 - (80)2
A_0.80
(40)2 - (80)2
Y4 = 0.20
Elev. of st~. 9 + 140 = 100 - 40 (0,02) +0,20
Elev. of sta. 9 + 140 = 99.40 m.
Depth of fill = 99.40 - 98.63
Depth of fill = 1. 04 m.
Lowest point of curve:
S-~
2 - g2 - g1
S - 0.01 (300)
2 - 0.01 + 0.05
S2 = 50 m. from P. T.
@
Elevation of station 10 + 000:
L
H ="8 (g2 -91)
300
H =8 (0.01 + 0.05)
H=2.25
_1.-_ 2.25
A gradeQf ;5%is followed bya grade af
1', 1Q~,thegrades intersecting at station
10 +050 of eleva,lion 374.50 n'\.,. The Change of
graders testridedioO,4%in 20m.•
(100)2 - (150)2
y= 1.0
EJ. B = 374.50 +0.05(50) + 1.0
EJ. B= 378 m.
A vertical summit curve has its highest point of
the curve at a distance 48 m. from the P,T.
The back tangenthasa grade of + 6% and a
Solution:
CD Length of curve:
forward grade of· 4%. The curve passes thru
point A on the curve at statiori25 + 140. The
elevation of the grade intersection is 100 m, at
station 25 + 160,
,
p.
Solution:
CD Length of curve:
r = g2 - g1
n
0.4 = 1 + 5
n
n = 15
L =20(15)
L = 300m.
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PARABOliC CURVES
S-~
g1 - g2
Rate of change =-.-n-
-0.04L
48 =_0.04 - 0.06
L = 120m.
0,3 -
2 - 92 - g1
@
@
_U1L
n
Stationing of P. T.
Sla. of P. T. = (25 + 160) + (60)
Sta. of P. T. = 25 + 220
n = 20 stations
L =20 (20)
L = 400 m.
@
Sta. ofhighest point of curve:
g1 L
S --.1 - 91 - g2
0,04 (400)
S1 - 0.04. (- 0.02)
S1 =266.67 m. from p.e.
@
Elevation of highest point of curve:
L
H=- (91 - 92)
8
400
H=g (0.04 +0.02)
Elev. of A on the curve:
L
.
H=a(g2 - g1)
120(0.06 +0.04)
H=
8
H= 1.5
.....L_J.:§...
(40)2 - (60)2
y= 0.67
Elev. ofA = 100 - 20(0.06) - 0.67
Elev. of A =98.13 m..
H=3
_H__ ---l.(200)2 - (133.33)2
Y
Elev. of highest point
= 142.63 +133..33(0,02) -1.33
Elev. of highest point =143.97
A symmetrical vertical summit· curve. has
tangents of +4% and· 2%, The allowable rate
of change of grade is 0,3% per meter station.
Stationing and elevallpnof PT, is at10 + 020
and 142.63 m. respectively.
~i) Compute the length of curve.
.'31
.
Compute the distance of the highest point
of curve from the P.C.
.' .... . '. .
Compute the elevation of the highest point
of curve.
Solution:
CD Length of curve:
= 3(133.33)2 = 1 33
(200)2
'
A vertical symmetrical sag curve has a
descending grade of· 4.2% and an ascending
grade of +3% intersecting at. station 10 +020•
whose elevation is 100 m.The two grade
lines are connected by a 260 ni, vertical.
parabolic sag curve.
d).
At what distance from the P.Ccis the
lowest point of the curve located? .' . :
What Is the vertical offset of the para\J<llic
~~~:n:~r~hd::,oint of interj)ection of the
If a 1 m. diam. culvert is placed at the
lowest point of the curve with the top of the
culvert buried 0.60 m, below the subgrade,
what will be the elevation of the invert of
the culvert?
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S-346
PARABOliC CURVES
Solution:
CD Lowestpoint ofthe curve from P. C.
21.67
Illjll
@
Nearest distance ofcurve from
pt. ofintersection of grades
L
H=a (g1 - g2)
Solution:
CD Sfa. ofhighest point of curve:
260
H =8 (- 0.042 - 0.03)
H =- 2.34 m. (sag curve)
@
Elevation ofinvert.
El.I02.27B
O.6m
s;:...9.t!:.-.
gl-g2
S = 0.03 (120) = 78 26
0.03 + 0.016
.
Sta. of highest point = (5 + 21p) + 18.26
Sfa. ofhighest point = 5 + 234.26
invert
----L- _ 2.34
(108.33)2 - (130)2
y= 1.62 m.
Elev. ofB =100 + 21.67(0.03) + 1.62
Elev. of B = 102.27 m.
Elev. ofinvert ofculvert.
Bev. = 102.27 - 1.6
Elev. = 100.67 m.
@
Bev. of highest point of curve:
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PARABOLIC CURVES
L
H=a(gl· ~
120
H ="8 (0.03 + 0.016)
H=0.69
-1L-_ 0.69
(41.74)2 - (60)2
Y1 = 0.33 m.
Bev. of highest point = 27 -18.26 (0.016) - 0.33
Elev. of highest point = 26.378 m.
@
Depth of cover over the pipe:
(ft11
;1'lifJllli
.·.·.···.·.··Qt#l$~~~ii}}.<·· . .
...................\ • ».•. .•.
Solution:
ill Length of curve:
1'.T.
p.c.
...lL_ 0.69
(48)2 - (60)2
Y2 =0.44
Elev. of B= 27 -12(0.016) - 0.44
Elev. ofB =26.368 m.
EI. 26.368 m
Elev. of A = 223.38 . 0.03 (75)
Bev. of A = 221.13 m.
y= 221.13·220.82
y= 0.31
L
H=a(gl'~
H=~ (0.02 t 0.03)
H=0.00625L
-lL_.--L(~ )2 - (~_ 75)2
INVERT
Depth of cover:
h = 26.368 • 26
h= a.368m.
(4)O.00625L_~
L2
0.025
- (~. 75)2
0.31
-L-= (~. 75)2
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S-348
PARUDUC CURVES
(~. 75)2 = 12.4L
L2
4- 75L + 5625 = 12.4L
L2 - 349.6L + 22500 = 0
L=264.55
r=!l1:J12.
L
_ 2- (-3)
r- 264 .55
r=0.0189
r= 1.89% < 2% ok
@
Stationing ofhighest point ofcurve:
S1-_..Bib.g1 - g2
S - 0.02 (264.55)
1 - 0.02 -(- 0.03)
S1 = 105.82 from P.C.
'''WI
III'.
Solution:
CD Length ofcurve:
Sfa. ofhighest point = (4 + 135). 2~.55+ 105.82
Sfa. ofhighest point =4 + 108.545
® Elevation of highest point of curve:
p' =200 -20 (0.06)
p'= 198.80
y= 198.80 -198.13
y= 0.67
H=0.00625L
H = 0.00625 (264.55)
H= 1.65
H
_----L-
1
L
h=4(g1-~2
L
h=8(gl-~
L
h =8(0.06 + 0.04)
(132.275f - (105.82f
1.65
----L(132.275f - (105.82f
y= 1.056 m.
h=O.10L
8
h= 0.05L
Bev. of B = 223.38 • 76.455 (0.02)
Elev. of B= 222.85 m.
h=0.0125 L
Elev.· of highest point of curve at C
= 222,85 -1.056
= 221.794m.
(~- 20)2 {~)2
-y-=-h-
4
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PARABOLIC CURVES
(~_ 20)2
(~)2
0.67
=0.0125 L
Solution:
CD Stationing ofhighest point of CUNe:
(~- 20)2 = 13.4 L
Hilhatpmntofarnoc
L2
'4- 20L + 400 = 13.4L
L2_113.6L+1600=0
L=240
2
L';' 120m.
@)
Stationing and elevation of P. C.
Sta. of P.C. = (35 + 280) - 60
~"ta. of P.C. = 35 +220
Elev. of P.C. =200 - 60 (0.06)
Elev. of P.C. = 196.40
@
Stationing and elevation of P. T.
Sta. of P. T. = (35 +300) +60
Sta. ofP. T. = 35 + 360
Elev. of P. T. = 200 - 60 (0.04)
Elev. of P. T. = 197.60
5+592
5+432
5-t682
Using squared property ofparabola.
o
1.._~
x2 - (90 + x>2
1.._~
8 x2- (160.x)2
o
4.7x2+x2y=y(8100+180x+x2)
4.7x2 • 180xy- 81OOy= 0 (multiply by 3)
14.1x2 - 540xy - 24300y = 0
8 3x2 + yx2 = Y(25600 - 320x +x2)
3x2 + 320xy - 25600y =0 (multiply by 4.7)
14.1x2 + 1504xy • 120320y ,;, 0
0&8
<··.·•• elfMilj6:ri~)
()~;®ffitUt
. .·4$;Q06«>
················!Q;~Qqt)<.·
14.1x2 - 540xy - 24300y =0
14.1x2 + 1504xy-120320y =0
• 2044xy + 9P020y =0
96020
x= 2044
x=46.98m.
Sta. ofhighestpoint =(5 + 592) - (46.98)
Sta. ofhighest point = 5 + 545.02
® Elevation of highest point of CUNe:
o 4.7x2 - 180xy· 8100y =0
4.7 (46.98)2 -180 (46.98) y- 8100y= 0
y=0.63 m.
Elev. ofhighestpoint of CLiNe =48 + 0.63
Elev. ofhighest point of CUNe =48.63 m.
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S-350
PARABOliC CURVES
@
Elevation of P. T.
Solution:
CD Length of vertical parabolic CUNe:
$, = 160 - 46.98
$, = 113.02
$-~
1-
g1 - g2
11302 =0.05 (200)
.
0.05 - f12
0.05 - f12 =0.088
g2=-0.038
g2 =·3.8%
Elev. P.I. = 45 + 100 (0.05)
Elev. P.I. =50 m.
Elev. P. T. = 50· 100 (0.028)
Elev. P. T. =46.2 m.
Bey. of A = 78.10 - 5
Elev. ofA= 75.10 m.
Elev. of B = 70 + 0.08 (5)
Bev. of B = 70.40 m.
y= 73.10·70.40
y=2.7
L
H=a(g,· fh)
L
H=a(- 0'(~4. 0.08}
H=· 0.015L (sag curve)
---L.-=...!L
(~. 5)2 .(~Y
(~. 5)2 =45L
L2
"4. 5L + 25 = 45L
L2. 200L + 100 =0
[= f99.50m.·
® Stationing of catch basin:
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PARABOliC CURVES
s _.1l.tl-.
gl- g2
S - - 0.04 (199.5)
1- -0.04-0.08
Sl =66.5 m. from P.C.
1-
Solution:
CD Length of vertical parabolic curve:
EL120m
~
~
-8
S
-_.-----r---HI
IS.Sm
;c
Sta. of the point where catch basin is placed:
= (7 + 700) - 33.25
.
=7+ 666.75.
@
Elevation Of the point where catch basin is
placed:
--.i'.- __ H_
(66.5f - (99.75)2
H= 0.015 (199.5)
H=2.99
--.i'.-_~
(66.5f - (99.75)2
y= 1.33 m.
Bev. of the point where catch basin is placed:
Elev. 0 = Bev. C +Y
Elev. C=70 + 33.25 (0.04)
Elev. C=71.33 m.
Bev. of 0 = 71.33 + 1.33
Elev. of 0 = 72.66 m.
L
2H 2' (112 • gl)
T=-L2
L
H='8(g-rgl)
L
H='8 [0.03 - (- 0.04)]
H=0.00875L
S
_-921:...
gr gl
_ 0.03 (Ll
S2 - 0.03. (- 0.04)
S2 =0.429 L
2-
L_L
(~2 - (~)2
•••,
.~J~I~IC~'Jl~~~~ll~lr~il~l..
~H4%aMthattifttiep·nl$f$%H
@()¢t~rtnjl'l~Jti¢l~n~lthq((~~y~rtl¢M
;:llerliili.;
CUrve.
Y
_0.00875 L
(0.429 Lf -
(~)2
y= O.00644L
Elev. ofA = 120·5.5
Bev. ofA = 114.5 m.
Elev. of B = 105 + x (0.03)
L
X=2'- S2
x= 0.5 L· 0.429 L
x= 0.071 L
Elev. of B = 105 + 0.071 L (0.03)
Elev. of B = 105 + 0.00213 L
Elev. ofA = Elev. of B + Y
114.5 =105 + 0.00213 L + 0.00644 L
L = 1108.52 m.
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5-352
PARABOLIC CURVES
® Stationing ofpoint 'p.:
x =0.071 L
0.41
H
(54~ = (60)2
x =0.071 (1100.52)
x= 78.70m.
I:f=O.5rem.
L
2H (gl- gV"2
Sta. ofpoint .p. = (5 + 800) + 78.70
Sta: ofpoint .p. = (5 + 878. 70)
1:.=-,-'L2
@
L
Elevafjon of P. T.
H=a(g1- gV
Elev. of P. T. = 105 + 110;.52 (0.03)
120
0.506 ="8 (0.02 - ~
Elev. of P. T. = 1~1.63 m.
0.02 - 92 =0.0337
92 =-0.014
92 =-1.4%
® Stationing of highest point of curve:
,
SI=
Illtll
ii~.I.
~
91- 92
_ 0.02 (120)
SI - 0.02 - (- 0.014)
SI = 70.59 m. from p.e. ,
Sta. ofhighestpoint = (4 + 100) + 70.59
Sta. of highestpoint;: 4 + 170.59
@
Elevafjon of highest point of CUNe:
Solution:
CD Grade of forward tangent:
-,1
rr
Elev. of B =22.56 + 54 (0.02)
Elev. of B = 23.64 m.
Elev. of e;: 22.56 + 60 (0.02) -10.59 (0.014)
Bev. of =23.61 m.
e
..J:L;:--L(60)2 (49.41)2
0.506_--L(60f - (49.41)2
y= 0.343 m.
y =23.64 - 23.23
y=0.41 m.
Elev. ofD = 23.61 - 0.343
Elev. of D = 23.267 m.
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PARABOLIC CURVES
1-----------------A vertical highway curve is at times
designed to include a. particular elevation at a
certainstalion where the grades ofthe forward
and backward' tangents have already been
established. It is therefore necessary to use a
curve with unequal tangents or a compound
curve which is usually called "unsymmetrical"
or asymmetrical parabolic curve where one
parabola extends from the P.C. to a point
directly below the vertex and a second
parabola which extends from this point to the
P.I. In order to make the entire curve smooth
and continuous, the two parabolas are so
constructed so that they will have a common
tangent at the point where they joined, that is
at a point directly below the vertex.
Let us consider the figure shown below:
Solving for Ll:
2H _(g1 •gz) Lz
L1 - L1 + Lz
2HL1+2HLz::: L1 Lz(g1- gz)
L1[Lz(91 • gz) • 2H] ::: 2HLz
Applying the squared property of parabola, in
solving for the vertical offsets of the parabola.
./
~/1
i
I
!
i
~~-+----+--+---"""~p.T.
jL_J::L
(X1)Z - (L 1)Z
~_J:L
(xz)Z - (Lz}z
Location of the highest or lowest point of the
curve.
a) From the P.C. When¥<H
l1 ::: length of the parabolic curve on the left
side of the vertex.
Lz ::: length of the parabolic curve on the right
side of the vertex.
g1::: slope ofbackwarcj tangent
gz ::: slope of forward tangent
I-~I
I
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S-354
PARABOLIC CURVES
Let g3' be the slope of the common tangent
of the parabolic curve.
Likewise, the location of the lowest or
highest point of the curve could be
computed from the P.T. of the curve, this
holds true when
B
1-g~/2
IJ!!/I$!!E:..------~E
Considering the symmetrical parabola
~ is greater than H.
Considering the figure shown, let us
assume that the highest or lowest point of
the curve is found on the right side of the
parabola.
b) From the P. T. when
AVF, thelocalion of the highest point of
¥
>H
the sag is obtained from the relation.
CD 5
1
=.9i.h
gl- g3
SUbstitutintg these values and solving for
g3, we have:
Considering the right side of the parabola,
VFCD.
v
Solving for g3 in equation ®.
2H =L2 g3 - L2 g2
51 = location.of the highest or lowest point
of the curve from the P.C,
@
_2H +L2 92
g3L
2
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PARABOLIC CURVES
Solution:
CD Height offill needed to coverthe outcrop:
ilI
<frr,)U2
I
When
¥
> H. the highest or lowest point of
the curve is localed on the right side of lhe
curve.
CD When ¥>H
.
Use.
-.lL_A
_.9.t!£
2H (from the P. T.)
52 -
(40'1 - {20'f
_ 1.56 (20)2
Yl - (40'1
® When ¥<H
Use:
2H _ L2 (91 - gz)
L1 - L1+L2
H= [1 L2 Ish - fb)
2(L1 +L0
H- 40 (60) [0.05 - (0 0.08)1
2(40 +60)
H= 1.56
YI =0.39
51 = 1h11!.
2H (from the P.C.)
B EL=1O.6Im
p.e
Ijill=2.21 m
EL=108.40m
EL=llOrn
\
Outcrop
AriunsYmmefrtc~lpataqoll~.9ilr@.nalla
fOrWar9·.tang®19f·•• S;%~rida~~C~fangflot~f:
t5~ ..... Th~lenglh .• tif.(;prvl'.·qrtth~ . le~ •. $id~9f
i~,liiliilii
13111v~ijpnQtlOMQro.
'.' . ".
.....
Bev. ofB =110 + 0.05 (20) •0.39
Elev. orB =110.61 m.
Depth offill at the outcrop =110.61 -108.40
Depth offill = 2.21 m.
® Elevation of CUNe at sta. 6 + 820:
. .
0)·•• Q9@PW~··t!'Ie.hEli9htqfnt@~e¢~.t(lcqV~
!!li_!~~!~t~~!~!t!~~
..... .....
.~··oflhecU~.
• • • qPmp~f~.~e • ele't@~n.9fltlE! .• l'ii~MM.paiilt
Efev. ofC =110 + 40 (0.05) -1.56
Efev. of C = 110.44 m.
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PARABOLIC CURVES
@
Elevation of highest point of CUNe:
Solution:
CD Length ofCUNe:
b1.9J. =40 (0.05) =10 < H
2
2
.
Sl --~
2H from P.C.
x2=24.24
6O--4--'--L2-
S _ 0.05 (40)2
1- 2(1.56)
Elev. B =230 - 30 (0.07)
Bev. B =227.90 m.
Y1 =227.90 - 227.57
Y1 =0.33
Sl =25.64m.
-lL_-lL
(4Of - (25.64)2
iLA
_1.56 (25.64)2
Y2 (40)2
Y2 =0.64
x2 - (L 1f
Bev. ofE= 110 + 0.05 (25.64) - 0.64
Elev. ofE= 110.642m.
- (30)2
H= 1.32 m.
2H _ b2JfI.c.922
L1 - L1 +L2
2 (1.32) _ L2 (0.07 +0.04)
60 60+ L2
2.64 (60 + L~ = 60 (0.11) L2
158.4 + 2.64 L2 =6.6 L2
~.96 L2 = 158.4
L2 =40m.
0.33 _-lL
(3Of- (60f
H- 0.33 (60f
@
Stationing of the highest point of the CUNe:
\91 = 60 (~.07) =2.1 > H
1.1'li~~
Therefore the highest point of hte CUNe is
on the right side.
8-<=
fl2l-i
2H
~_ == 0.04 (40)2
VOl
2 (1.32)
S2 =24.24 m. from P. T.
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PARABOliC CURVES
Sia. ofP. T. =(6 + 300) + 40
Sia. of P. T. = 6 + 340
Sia. ofhighest point = (6 +340)· 24.24
Sia. ofhighest point = 6 + 315.76
@
Elevation of the highest point of the curve:
..YL_.l!..-
.
(X2}2 - (L2f
-l'L._ 1.32
(24.24)2 - (40}2
Y2 =0.48
~
Bev. of C =230 -15.76 (0.04)
Bev. of C= 229.37 m.
Elev. of highest point =229.37 - 0.48
Elev. of highest point = 228.89 m.
!II!~!
Solution:
CD Elevation of curb:
2H 1320
160 = 280
H=3.77m.
...lL_-...!:L
(60}2 - (120}2
_3.77 (60f
Y, - (120}2
Y, = 0.94
Bev. of B =30 - 60 (0.04) •0.94
Elev. of B= 26.66 m.
Bev. of curb =26.66 ·4.42
Bev. of curb =22.24m.
® Length of CUNe on the forward tangent:
Elev. of A = 30 • 0.04 (50)
EJev. of A =37.6 m.
Elev. of B =22.6385 + 4.42
Bev.ofB=27.0585
Y2 = 27.60 - 27.0585
Y2 =0.5415
--h-=.1L
(L2 - 6O}2 (tj
2H _ 0.11 L2
160 - (160 +L2)
H=~
160 + L2
0.5415 _
8.8L2
(L2 • 6O}2 - (160 + LV (l..2}2
0.5415 (160 + L2) L2 = 8.8 (L2 - 60)2
86.64L2 +0.5415Li = 8.8 (Li - 120 L2 + 3600)
. 8.2585Ll-1142.64L2 + 31680 = 0
Li -138.359L2 + 3836.048 = 0
L2 = 100m.
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S-358
PARABOliC CURVES
@
Stationing of highest point of curve based
on second condition:
AC•• 3!o/;,•• grGlde • /ll~~t~)r+§%i.~W!e/Mar.(In
UtJd~rp~S5,.···mQrdett(lmamt~ln.meml~jmUm
·91El~rllll~~'()W~d~llclEl(th~q~~g~@d.~I.~b~
$aroe.tlrtla·.intrt1dtlP~.avertipal~s«t9n • Ptlrv~·
fn.tI)eW"'®.li!l~, • jt.~ •. ryege~§aWIR •.~e • <i.•.ClJrve.
Ih1!lnil:l~.~Og.rn·pn9ne.sid~.ofthe·Verl~Xofth~
.$lt~igm.9rade.and1Qri • ro,•• dn.the•• qiher,.t~e
$1~~i()m()f.the • ~~giflning • 9f.tt@9t1rv~(2()().I'lI.
skle)·iS.10.+.000M¢·it$.ele~ali¢n·i$·228·m.
ill
Oetermll'le·.·.··lne.·•• ele'1ation·•• af•• • statloll·
@
Iflh~ • uphill.~dse.af.th~uriqer$~~Clf·the
. bridge • • i~N • • staUon • • 1O·•• t··•• f2g/~im2 • • ~t
2H_~
160 -260
H= 3.385
>.. . . •.•· /.. /
?•• • • •
elevatiClrl.229.2Q~·.m.,·Wttat·is.lne~rtiC<:lI·.
!::l.Jll. _160 (0.07)
2 -
1o•.f64Q..•... •.• • • i
2
®
clearanceUnd~th~~n~lll(l~ntilsP9rrit1 ••..•
Dl:llerrtllne • • ttt~.·.ttaliQning • • ()f.ltle.fowest
painl ofthe Curve, . . .
.
¥=5.6>H
Solution:
ill Elevation at station 10 + 040:
The highesfpoint of curve is on the right
side.
fl21l
S2= 2H
S - 0.04 (100)2
2-
2 (3.385)
S2 = 59.08 m.
Sta. ofhighest point of curve
= (12 + 200) + (40.92)
=12+ 240.92
8 2H
300 = 200
H= 2.67
--lL_~~
(40)2 - (200)2
Y1 =0.107
Etev. of sla. 10 + 040 = 228·0.03 (40) + 0107
Etev. of sla. 10 + 040 =226.907 m.
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PARABOliC CURVES
@
Clearance under the bridge:
•
:~ea[~~~~~o~~~;~~tY ~I~lal·
IQqryfJl3et.a~fdlallgEllltot:a~~M~bllCf(
..1:'2-_~
(80)2 - (100)2
..1:'2-_ 2.67
.(80)2 - (100)2
Y2 = 1.71
Elev.
Elev.
Elev.
Elev.
Elev.
Elev.
of P.I. = 228 • 200 (0.03)
ofP.l. = 222 m.
of C = 222 +0.05 (20)
of C = 223 m.
of 0 = 223 + 1.71
of 0 = 224.71 m.
Vertical clearance h = 229.206 - 224.71
Vertical clearance h =4.496 m.
@
litill
the • • !?rojEict .• ~~91~e~fij~Ci~~ijtq~ijm~.·lhe
vertiC<ll.par~l~cUrve.jnsUc.h~·weYlhatth~
1I1'~1I.18r~1
ill
o.eterrnme•• • th~ • • 19tal•• len~th • • of•• t~e • • new
patab9Iip¢9t\1~,.>.
® QetEl!1Tlin¢tM.·~t~tiQni~~.afJd;l:lt$vati®.()f
tffitne\',lp.r,<i
@ [}ElWnnineW~~I~¥ati()ll9ft~ell)~elil.B9lnt
QfthecW:ile~>
. ..
Solution:
G)
Total length of new parabolic CUNe:
Stationing of lowest point of CUNe:
h21
= 200 (0.03) = 3 m > 2 67
22
. .
The lowest point of curve is on the right
side.
_fl2ii
S2 - 2H
S _0.05(10W
2 - 2 (2.67)
S2 = 93.63 m.
Sta. of lowest point of CUNe =(10 + 3(0) - 93.63
Sla. of lowest point CUNe = 10 + 206.37
L
'£!:!L 2. (g2· g1)
L 2
H1 =
L
~
8
H1 =2 m. < 2.67 (it will hit the boulder)
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S-360
PARABOLIC CURVES
Construct an unsymmetrical parabolic CUNe
?!:!2. L2 (g, • fh)
L1
L1 + L2
~ _ L2 (0.03 + 0.05)
100 5.34
1.50 L2 = L2+ 100
L2 =200 m.
Total length of CUNe = 100 + 200
Total length of curve = 300 m.
® Stationing and elevation of the new P. T.
Sta. ofnew P. T. = 10 + 300
Elev. ofnew P. T. = 100 + 200 (0.03)
Elev. ofnew P. T. = 106 m.
AWfVIard • ~llge~f&f~~W~~d~~ignEldl~
iQIW1~qt~~k.tM9Mt9fc$rci~t~Pf~$~~
.lJIl9~fP<iS~.Clf<m~.~pij~M~·~~IQ~tS~nt8~
IIJItJ.1Utitlll'
gp°tltqlJrvelll'i~Q(lth£l$jd~9ft~~~ac~
tang~rt.Wlle • <i.• 19QI11·••~~~ • li~@tQeslij~
of.ttl~f%'f!ltdI2!nQ~nt' • The$taij@ir\!fand
el~yat!9Q9fthElgt<!d~jlll~r~~9tiqwjs
1f;:$30.2l)aod19Qm;r~$M¢~¥~ly.)rhe
@nWlm~9fme~ri~geffllls~t~tali9{l
·12+.S75·20·•• ·Th~.e~'1CltiQnoflhellllderstd!'l.()f
thepndgei$117A8.
.
..
@
Elevation of lowest point of CUNe:
,'
.
(2)
.
.
.
,
c
.
-
++---·-::;OiIJP.·T.
ll-
,-
.
~~
e.
...
.'
,
~
Solution:
CD Clearance on the left side of the bridge:
kI11
= 100 (0.05) - 25 H
2
2
-. <
The lowest point of CUNe is on the left side.
ill .
81 = 2H
S _0.05 (10W
1 - 2 (2.67)
81 =93.63m.
---L.- _-.!i...
(93.63)2 - (100)2
---L.- _ 2.67
(93.63}2 - (100)2
y=2.34
Elev. oflowest point of CUNe
:; 100 + 0.05 (6.37) + 2.34
= 102.66m.
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PARABOUC CURVES
2H _ Lz (g, - 9,)
L1 - L1 + L2
H == L1 Lz (gz - 91)
2(L, + L2)
H== 200 (100) (0.06 - 0.03)
2(200 + 100)
H==3m.
Sight distance = is the length of roadway
ahead visible to the driver. For purpose of
design and operation it is termed stopping
sight distance and passing sight distance.
-lL_-.!i(Xlr (Lzf
..li.-==_3_
(60)2 (100)2
Y1 == 1.08
12- __3_
(50)2 - (1oof
Y2 == 0.75
1.)
·.· ~@#~~~·.~~~~.P,~~i:~i· .
Stopping Sight Distance is the total
distance traveled during three time
intervals.
Elev. of A == 100 + 0.06 (40)
Elev. of A == 102.40
Elev. of B == 102.4 + 1.08
Elev. ofB == 103.48
a. The time for the driver to perceive the
hazard.
b. The time to react
c. The time to stop the vehicle after the
brakes are applied.
Clearance on the left side = 117.48 - 103.48
Clearance on the left side == 14 m.
Based on the National Safety Council,
average driver reaction time is 3/4
seconds.
® Clearance on the right side ofthe bridge:
Elev. ofC = 100 + 0.06 (50)
Elev. of C = 103
Elev. of 0 = 103 + 0.75
Elev. of 0 = 103.75
Clearance at right side = 117.48 - 103.75
Clearance at right side = 13.73 m.
® Clearance at the center.
.l:'L __3_
(55r (100)2
Y3 = 0.91·
Elev. of E = 100 + 45 (0.06) + 0.91
Elev. of E = 103.61 m.
Clearance at the center = 117.48 -103.61
Clearance at the center = 13.87 m.
pr:--
Ht, ofdf'fwr"3 qe
ill OPPOSJt e dlf~CIlOll
H,.
if objU1
h -6' \
.
'-'---'h =3.7S'
1
I: :Ton;NG
,
DISTANCE:
(brahng dlstana)
S
S=Vt+D
A car moving at a certain velocity V after
seeing an object ahead of him, will still
travel a distance Vt before he starts
applying the brakes. The braking distance
depends upon the speed and type of
pavements, in this case we have to
consider the coefficient of friction (f)
between the tires and the pavement. The
time t (sec) is called the perceptionreaction time as is approximately 3/4 of a
seconds.
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5-362
SIGRI'DISTMCE
Using wort<: - energy equation in solving for
the braking distance d.
w1J2
D=2gfW
0= 2~f if nis moving on a horizontal plane.
0= 2g
(~G)
if moving at a certain grade
Safe stopping distance: S =d + 0
'Safe Stoppmg Distance
\f.!
S =Vt +2g (f ± G)
<D Distance traversed during perception plus
brake reaction time.
S=stopping distance in meters
.t = perceplion.rea~on time in seconds
V= velocity of vehicle in meters per
second
f =coefficient of friction between tires and
pavement
g= 9.81 meterslsec2.
d =Vt
V =running speed in kph
t = reaction time
t = perception time + action time
t = 1.5 + 1.0
t =2.5 sec.
d = distance traversed during perception
time plus brake reaction time in meters
d=Vt
@
Distance required for stopping after brakes
are applied (braking distance)
Positive wort<: - Neg. work = ~ ~ (Vi· vlJ
1W
O· OF =29(0- IJ2)
DfN=W\f.!
2g
N'=W
2.)
ell$jll_~
Passing Sight Distance is a shortest
distance sufficient for a vehide to tum out
of a traffic lane, pass another vehicle, and
then tum back to the same lane safely and
comfortably without interfering with the
overtaken vehicle or an on incoming
'vehicle traveling at the design speed
should it come into view after the passing
maneuver is started.
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SI8IT IISTaCE
Stopping Sight Distance
B. When S>L
t=---
~
o
CD Stopping Sight Distance:
(Summit Parabolic Curve)
h1 = 3.75 ft. (height of drivers eye
above the pavement)
h1 =1.14 m.
h2 =6 inches (height of object
above pavement)
A. When S<L
s
:
s
----1
~
when h1 = 1.14 m.
h2 =0.15m.
L =in meters
S =in meterS
when h1 = 3.75 ft.
h2 = 6 inches
A=gl-92
L = in feet
S = in feet
1------L-----<01
SIGHT DISTANCES
o
Vertical Curves
<D When
when h1 = 1.14 m.
h2 = 0.15 m.
L = in meters
S = in meters
when h1 = 3.75 ft.
h2 =6 inches
A=gl- g2
L = in feet
S = in feet
Sight distance is less than the length of
curve.
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5-364
SIGIT IISTIICE
L
2H "2(g,-gV
b. = L
2
~ = -V 2OOAh, L
S=d , +d2
L
4H ="2 (9dlv
S = ,,1'""2-00-:-,L-+
L
H~8(g,-gV
S=
Express the slope In pefCflflt not decimals
·.l.·•..
@
Using the squared property of the parabola:
.hL_J:!..
(~)2
~_J:!..
d;.2 - (~)2
·d2_h,L2
1 - 4H
d 2 _ h, L2 (800)
1 4AL
2_200h1 L
d1 A
,r-
1
L
11.'~.!
.• !i.i.:.'.·.~.•i!.:r•~.'•~.;.,.,.i.·I.f;
Ex:
g, = +2%
g2::: - 3%
A=g,- g2
A=2-(-3)
A=5
d _'" 200
- h,-L-
,- V
L..J2il;"+
S2 = 1~ L('';2h, + &)2
H=AL
800
2
1
2
s=~1~L(m,+~)
A
100 =g1"!n
LA
H=8100
d, -
.y O~
-V 200Ah L
-V O~ .{2ii;
A
~_J:!..
d;.2 - (~)2
d 2= h2 L2
r 4H
d1-= h2 L2
AL
4 800
d~2 _ 200hzL
r- A
:.i•.
:-";':'::-::>:/:<::::: •
When
Sight disla,nce is greater than the length of
curve.
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SlIIT DISTMeE
SIGHT DIST ANCES
5--,..---1
~---L·-------t
a.
b.
:iil!li!~~II!_!I:i
when h1 = 12 =h
S·3616
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Using the previous relation of parabolic
CUNes.
a.
Where
S = length of passing sight distance
L = Length of curve
h1= height of drivers eye
h2 = height of object
C = vertical clearance from the lowest
point of underpass to the curve.
H=C-~
2
Y=
t(~)
(g2" gl)
L
y ="8(g2" g1)
c.
By ratio and proportion:
S
_2__
S
H+y-S.
"2 (g2 - g1)
S
Passing Sight Distance for
Vertical Sag Curve at Underpass
CD When passing sight distance is greater
than the length of curve.
2S
2H +~(91- g0 = S (g2" g1)
L
S(g2 - g1) =4H +.2(g2· g1)
L
.2 (g2 - g1) =S(g2 - g1) -4H
AASHO Specs.
when C = 14 ft.
h1 =6ft.
h2 = 1.5 ft.
A= g2" gl
82
L=2S_
A
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SIGar DISTANCE
Passing Sight Distance
(Summit Parabolic Curve)
® When the passing sight distance is less .
than the length of curve.
A. when
(j)
Using the squared property ofparabola.
when
h,=1.14m.
hz =1.14 m.
L =in meters
S= in meters
when
h1 = 3.75 ft.
hz= 3.75 ft.
A= g1" g2
L = in feet
L
8(g2 -g1) _J=L.
(~)2 - (~)2
@
4L (92' 91) _4H
8 L2
- S2
S=in feet
B. when
l-t:=-.= --.--8'---.=---I:j
:K:".'.'.'.":"'.'.'~
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SlIIT DllTlleE
B. when
-when
h, = 1.14 m.
~=1.14m.
L =in meters
S =in meters
when
hI =3.75 ft.
h2 =3.75ft.
A=g1"g2
L =in feet
S =in feet
'.:::.;.:.:.;. :.:.:::.
+
:::m:i~i~t:
~I-~-p'-,C- -lI!I~:~~?-P.-'T.-'
. .-. .
-h-l,'
L = in feet
S = in feet
Sight Distance for $19 parabolic curve:
A. when
L = In meters
S=in meters
C. Max. velocity of car moving in a vertical
sag curve.
:lill&I:~1
L =lnfeet
S =infeet
A=g2" g1
!:!!!~~!!~!j!!!i
:;:::;::::::::~:~:~:~:~:~:~:~::::::::;:::::;::::::.:::.;.
v = velocity In mph
A= 92" g1
L =length of curve In feet
L =In meterS
V=ln kph
L='in meters
S =in meters
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SIIIT IISTIICE
SIGHT OISTANCE
(Another Formula but
some results)
@
'Mlen
VlllIl
<D When
A
h~
Ll/~~
JI
Considering triangles ABC and CDE:
By ratio and proportion:
H
h
[=sc
5 = sight distance
L = length of curve
h = height of drivers eye above the
pavement
By ratio andproporlion:
L
2H 2(g1- Q2}
h=
L
H=a(g1-gV
U~ng the squared property ofparabola:
h
(~)2 =(~)2
L .
8(g1-92) h
-L2-=52
1
L
H=a(91-9V
L
a(gl-9V
h
-L-=28-L
4
L
2
4
4 2"-4
4
Lh =a(91-gV 52
4
~ __h_
8
- 25-L
28 (91- Q2)- L(91- 9v =Bh
L(91-9v =28(g1- 9V- Bh
11~11;:I~:!II~l.!~;~II!
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5-370
SIGHT DISTANCE
AssumeS<L
L =_ _..:.A=S_2_ _
1oo(fu, + ~)2
A=g1-~
A=5-(-3.4)
A=84
L=
(8.4) (83.32)2
100 (..J 2 (1.37) +..J 2 (0.10))2
L = 131.92 m. > 83.32 ok as assumed.
® Elevation ofhighest point of curve:
Solution:
G)
Stopping sight distance:
,-_AL-
S
g1-~
S - 0.05 (131.92)
S = Vt + 29
(~ G)
V= 60000
3600
V= 16.67 m1s
_
(~)
(16.67)2.
S -16.67'4 + 2 (9.81) (0.15 + 0.05)
S= 83.32m.
@
Length of curve:
0.05 + 0.034
S1 =78.52 m.
1-
L
H=8(g1-~
H_ 131.92 (0.05 + 0.034)
8
H= 1.39
-.l.- _-.!lL
(53.4)2 - (65.96)2
y= 0.91
Elev. of highest point
=42.30 -12.56 (0.034)·0.91
Elev. of highest point = 40.963 m.
~----L----..j·
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SIGHT DISTANCE
.!!:I~~J~.~mt.~~~~~~~#~k
If;".)_•
.·~. ·•. COmpul.Ei the.lel'l~ •. l)tthe.$fgljt(1i$t<ilO~~ .•·.. ·.
Solution:
11~111:1'.J~I~m~gll~
~:~I(111
CD Slope of forward tangent:
Solution:
Alfl
CD Upward tangent grade:
L =395
Alfl
L=395
130 =A (100)2
395
A =5.135
126 _A(100)2
- 395
A=4.98
A = gl + g2
5.135 =2.5 + g2
g2 = 2.635%
lID Distance of lowest point of CtlNe from P.
S=-.9.1..L
gl" 92
S - "0.025 (130)
- "0.025· 0.02635
S= 63.29m.
® Length of sight distance:
Ass: S<L
AS2
L=122+3.5S
5.135 S2
130 = 122 +3.5S
5.13582 = 15860 +4555
82 .88.618" 3088.61 =0
8 = 115.38 m. ok
A=g2+ 2
4.98 =92 + 2
fI2 = 2.98%
c:
@
Sight distance:
A82
L = 122 +3.5 8
4.9882
126=122+3.58
15372 +441 S =4.98 S2
S2. 88.55 S" 3086.75 = 0
S= 115.32m.
® Distance of/owest point from P.C.
_..9l.L
8 1 - g1' fI2
8 - ·0.02 (126)
1 - • 0.02. 0.0298
8, = 50.60m. from P.C.
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5-372
SIGHT DISTANCE
Solution:
G)
AVl
AV~liicatMtyehasad¥~Mlldfrlggf<l{j~()f
L =395
c>1.~/o$~artlng.frOllllheF>.g:~ndana~~~lgg
.••g~d$ • ~f • ±~,~%.passln~lhl'\.ltbe~.T. .• ]M••
·cllJ\'~·h$.<ll~rg~t·liislfJ:Oqel)f.tBQtn,····
Length of curve:
L = 5 {1(0)2
395
L = 126.6 m.
.
<D•• • Comp~t~.m~ler@h#tMyertff:alfurv~, • •••• • ·
@C°triPJ.lle·them~.yelQ~tYofm~aar1h~r
• .•. Q?Uld•• PWJ~··tfjfUthecul'Ve .• .• ·•.• • • .•.• • • ·•• •.• • • • • •.• • • • • . •.•
@·.q()mpute•• th~dlstarce • Oflp~ • I()west•• p()idt.
Pfttie¢(ll'V~@m IheR,C,
.
(2)
Sight distance:
AS2
L= 122 +3.5 S
Solution:
G)
5 S2
Length of curve:
126.6 = 122 +3.5 (S)
S<l.
15445.2 + 443.1 S = 5 S2
S2 -88.62 S- 3089.04 = 0
AS2
L = 122 +3.58
A = gr g1
S= 115.39
A=3.5-(-1:2)
A=5
_ (5) (180)2
L -122 +3.5 (180)
L= 215.43 m.
(2)
@
Minimum visibility:
115.39
' VlSI
"b'l'ty
Mm.
~/, = -2Min, visibility = 57.70 m.
Max. velocity:
AVl
L =395
5Vl
215.43 = 395
V = 130.46 kph
@
Distance of lowest point of curve from P. C.
S=~­
g,-g2
S =- 0.012 (2J5.43)
- 0,012 - 0.038
Avertical surnmtt curve has a back langentof
+2% and a forward. tangent of· 3%
interseCting at station 10 + 220.60 m. and
elevatiOn of 200 m, The design speed of the
CUMlis 80 kph. AssumJngcoefficrent of
fiictiQOis 0.30 and a perceptkm reaction lime
of2Bsec.·
S = 51.70 m.
. . ..
<D Compute the safe stopping sight
@
Compute the length ofauNe.
@
Compute Ihe elevation of highest point of
. . curve,
The deSign
$P~d of
a vertical sag curve Is..
equal to 100kph, The tangent grades of the
curve are, 2% and +3% respectively,
Compute the length of curve in meIers,
~ Com~utelhe sight distance in meters.
@ C;ompuie lhe length of minimum visibifity·
inrn~rs ..
(j)
Solution:
CD Safe stopping sight distance:
V= 80 kph
V= 80000
3600
V= 22.22 mls
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SIGHT DISTANCE
V2
s= Vt+ 29 (f+ G)
_
.
(2222f
S - 22.22 (2.5) + 2 (9.81) (0.30 + 0.02)
S= 134.19m.
® Length ofcurve:
For safe stopping sight distance:
h1 = 1.14m.
h2 =0.15m.
AssumeS < L
AS2
L = --====,-=:::--
100(W1 +~)2
A = g1 - f12
A = 2- (- 3)
A=5
L=
5(134.19}2
~ndl~~%'WHmml~lm~:tl~eio~l.
ifJP~~!9Pp;ng~lgflt~~I~nc~·
'.' '• js.eq#~I
" • ·.tq..
1~$.ml'M<
~• • [1~r@gt~.~~¢.·n@t~~r~~s~@d¢9~~.·t;.
·····~v(Jig . ~II~fQllff .• W~.RElr~pt[onre.~Ctfcm
~J'\1~Rfth~dtiv~f!$Z.P$~9;M(l;lfte
poElfliCj@I~Mctionise9ualfQo.29'<:
®·•• GPfllP9!~!.Mt~tllQt¢tlNll.·.· .• ·.• ·•·.•. .• .•. • • '
ot(:#~.><
100 (',./2 (1.4) + -V 2 (0. 15})2
L = 212.64m.
S < L ok
@
Elevation of highest point:
163.78:: V (2) +2 (9.81) (0.25'" 0.03)
0.182 V2 + 2 V-163.78:: 0
V2 :: 10.989 V· 899.89 =0
V:: 25 m/s
V:: 25(3600)
1000
V= 90kph
S _.1h11- g,- g2
S - 0.02 (212.64)
1 - 0.02 + 0.03
S1 = 85.06m.
® Length of curve:
Assume S< L
AS2
L::--~-~-
100(m1+~)2
L
H ='8 (g1 - f12)
H - 212.64 (0.02 + 0.03)
8
H= 1.33 m.
J
__ ~
(85.06)2 - (106.33}2
y= 0.85 m.
Elev. of highest point of curve:
Elev. B = 200 - 0.02 (21.26) - 0.85
Elev. B:: 198.72 m.
i
@•• Pl)mp~t$lti~~~rJl~9Btlh~h~1l~(PQi~1.
h1 :: 1.14 m.
h2 :: 0.15 m.
A::3-(-2)
A:: 5
L::
\~
5 (163.78,"
.
100 (~+ -V 2 (015))2
L:: 316.76m.
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S-374
.SIGHT DISTANCE
® Stationing ofhighest point ofCUNe:
® Stationing ofhighest point of CUNe:
1.68
22.75
SI=~
gl - g2
S-~
S - 0.03 (316.76)
0.03 +0.02
SI = 190.06
1-
Sta. ofhighest point of CUNe =(20 +040) - 31.68
Sta. ofhighest point of CUNe =20 + 008.32
, - g, - fJ2
S - 0.025 (182)
1 - 0.025 +0.D15
SI =113.75 m.
Sta. ofhighest point =(12 + 460.12) +22.75
Sta. ofhighest point = 12 + 482.87
® Elevation ofhighest point of CUNe:
lIil."ili
11-=.1.
Solution:
L
.
H=g(g,-gi>
H =182 (0.025 +0.015)
8
H=0.91 m.
-L_ O.91
(6825r - (91)2
y= 0.51 m.
Elev. A =150·22.75 (0.015) - 0.51
Elev. A= 149.15 m.
CD Passing sight distance:
hI = 1.14 m.
h2 =1.14 m.
AssumeS> L
L=2S-
200({hl+~2
A
A = gl - fJ2
A =2.5-(-1.5}
A=4
182 =2S-
~levatl()tlof24()m,above.sealal/el,
200(fu+~2
2S= 182 +228
S = 205m. > 182 ok
·A·l/~rti~I • $a.g.para®li¢.Cl1rv~.@¥al~n91h • of
1·Mm.Witnt~ng¢n19tade~pfSJ.50/0aJld
f~·~%irlll3f$epti~!1.at • $talioll.·1?±Mo,gg•• all~
4
·q)•• •• C@lplJt¢.ttle.l@gth.btth~.$i9ht.{llStaMe.
r
~• • qofljPlJte•• th~ • • rn~*im(Jm.~p~trth~t.a
·•• · •·•·•·•• WOl:lI(!•• ttav~ltl:i.av()jd.coIU$ibn.· ·
.ca
·r$l•• ••Oflherorve.
¢Pmpute.the.stillionjnS·.oflh~ • lowest.p()int
...
....
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SleHT DISTANCE
Solution:
CD Length ofsight distance:
AssumeS> L
L =2S. (122 + 3.5S)
A
A=g2-g1
A=2.5 -(-1.5)
A=4
141 =2S- (122 +3.5S)
.
4
564 = 8S - 122 ·3.5S
4.58=686
S= 152.44 m. > 141 ok
@
Max. speed:
AI/l
L=395
41/l
141= 395
V= 118 kph
@
g) • •.• P(lmPU!lttl'i~~glfl.9f!li~p~IW~>
®11:iil.'~$$I~
@··.p()ttlpul~ • t~~e~~Pl!~¢l!Qntirli~9f
· •·•· · • ·.~~~t~i • ~• ~~Ii~jgl~~~. · · •· l· · ·i~1
··· .
····~ehtth~d.3~.·····»·····
Solution:
CD Length of sag curve:
Assume: L > S
AS2
L= 122 +3.5S
A =2.98 - (- 2)
A =4.98
_ 4.98 (115.32f
L - 122 + 3.5 (115.32)
L = 126m. > 115.32 ok
@
L=AI/l
Stationing oflowest point of curve:
p.
Max. speed of the car:
395
126 =4.98 I/l
395
V=100kph
@
Perception reaction time:
S=Vt+
I/l
2gf
V= 100000
3600
V= 27.78 mls
_
(27.78)2
115.32 - 27.78 t + 2 (9.81) (0.38)
t= 0.42 sec.
S1--~
g1 - g2
S - - 0.015 (141)
1 - _0.015 - 0.025
S1 =52.875 m. from P.C.
Stationing of lowest point of curve
=(12 + 640.22) -17.625
= 12 + 622.595
AParap6I1C$a9Ctlry~ • ti~s • ~.sijltdi$tlllJ¢~(lf
11S·.rn;•• >Th~··tllllg~llt~@cl~(jflh~29iy~~r~
·2%alld*3%.
'.' .
(i) CQmplitelheIM9thl)1th¢Cu~i< . .
.@).. 9Qmp~~t~~m~x·speedthata~r·t;(l91d
• l'llClVeajoMthiS:@/"i~.@Jlr<lv~llt$l<idl:liDQ; •
Th~$19htdlslance.ofa~~parab°llC¢1l~·I$
155,~2 .• m~··.I()rg • • Witb.·QWcl~.t<!rgel'll~·.gf.<~%
and +2;9a%respectiveJy.··
.
@
·····~I::tirhjl
• 1f~lfus,~r&.t~ei~~~~~·
ltielowElSlpOltifMtheCOtve>'" . '. ". ..
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S-376
SIGHT DISTANCE
Solution:
CD Length ofcurve:
AS2
L= 122 +3.5 5
A=g2-g1
A =3-(- 2)
A=5
:.. 5 (115)2
L -122 +3.5 (115)
L = 126.07 m > 5 ok
@
HEAD LAMP
SIGHT DISTANCE
Sight Distance Related to Height of the Beam
of a Vehicles Headlamp
Case 1:
Max. speed:
L=AVL
395
5Vl
126.07 ~ 395
V= 99.80 kph .
@
Elev. oflowest point of curve:
L =Length of curve
S = head lamp sight distance
h = hI. of headlamps above road surface
fil = angle.the beam tilts upward above the
longitudinal axis of the car.
r =~ rate of change of grade
Change of grade =g2 - g1
s-AL
1- g1 g2
5 _·0.02 (126.0n
1- _0.02- 0.03
51 =5Q.43 m.
L
H=a(ffr g1)
H = 1~.07 (0.03 + 0.02)
H =0.79
-..L-_JUL
(50.43)2 - (63.05)2
y=0.51 m.
Bev. oflowest point= 120 - 0.02 (50.43) +0.51
Elev. of/owest point = 119.50 m.
To get the offset y, the change of grade
from A to B is multiplied by the average
horizontal distance ~.
L
y= (92 - g1)2"
.If we multiply by
f
-~
2L
y-
1
y=2"rL2
y=h+Sfil
~
2 =h+Sfil
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HEAD lAMP SIGHT DISTANCE
Case2:
~en
Solution:
CD Head lamp sight distance:
V2
Safe stopping distance = Vt + 2g f
V= 120000
3600
V=33.33 m/s
3) . (33.33)2
S=33.33 (4 +2(9.81)(0,15)
S=402.47m.
r=~
1
y=2"r52
® Length of CUNe:
1
2(g2" gl) 52
y=
L
y=5,,+h
(92 " 91) 52 =5 " + h
2L
L
S" + h = (gr g1) 2"
L = 2 (S" + h)
Y2- gl
Tn
180
,,=
,,= 0.01745
pe$lgti¥~IRcll¥:;1~O~plt
. • •. ..........>..
•8te~~~~.%I1~~i~w;6~d• ~aW~~fit
~Mlir¢$j$Q.t$/
.
mG~fup~~tMh~dl~PP$tQht~l~ta~.'
• ~.
If.jh~hElC!dl,,,mP!i·.~r~.M~~ID,··llbclV~.JM
';iila.,II.
@
lfthe•• g;q;i~<~f.~ati(l" • • tO•• f··.540)l.n~·
elflValion • 1QPlll~;(;()mp(lI~·th~~lElvll~()l'jlif
ttte·lowe$tp<:iiNofthe¢UJYe•••·· · ·
.
L =2 «402.47)(0.01745) +0.75}
0.02 +0.03
L = 310.92 m. < 402.47 m. ok
® Bevation oflowest point ofCUNe:
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S-378
HW lAMP SIGHT DISTANCE
s-~
1-
Solution:
g1 g2
S - - 0.03 (310.92)
1- -0.03-0.02
Sl = 186.55
CD Max. speed:
A=92-g1
A =3.2 • (4.4)
A=7.6
AV2
L
H="8 (92 - g1)
L=395
H = 31~.92 (0.02 + 0.03)
153 = 7.6 V2
395
V= 89.17 kph
H= 1.94
--.L-_~
(124.37f - (155.46)2
@
y= 1.24 m.
Max. head lamp sight distance:
Elev. oflowest point of CUNe:
Elev. ofP.l. = 100·0.03 (155.46)
Elev. ofP./. = 95.43 m.
Elev. ofA =95.34 +0.02 (31.09)
Elev. of A =95.96 m.
Bev. of B = 95.96 + 1.24
Elev. of B = 97.2 m.
/,,- - ".\ L
Th~l.andTfah$pO~Udr\ • .()ffl~~.{kr9}·.requjr~s
ltlat.• •~~rli • ·.m~st . l}Vlit~h··.q~· • ~~ij,~.··~~M • Hgh~s
i~j~~l~V~~~SP~i~~~~.~W~\uf~gih1J~
Ug~PI\'lMyflMAAs'¢QMilr9~g~My¢m~
S"+h-~
2
-~
., - 180'
•
., =0.01745
S(0.01745) + 0.90 =(0.022 + 0 028) (153)
2
~~gffM~~~~m~~~~~~ri~0~.~t~.~~~B:··
S= 167.62
t1lr()~gry~~CaI~W~¢¢Qo/l:l1531lf'
Icmg•
•. h~'JlM9tffl1~l~ng~~t$ • •.9t.• +A~i%
+3;z%r$P~¢Uvl:llii»···
. • ~~d·
·¢Wm#hl<®~~~~g@~(ll1afle4Ver~~
. .·.JhI~ • • ¢Q@~@PijW~rit\$llgj!l9Wltl$·
·•• ~effl~jent.()flrl(:tlprb~tVI~M*es • • ~nd
p~\i~~tj~lt1§7>
~§QmPtJt~JheI'M~<h~MI~mpsI9ht
@
Perception time:
V= 89170
3600.
V= 24.77 m1s
S=
Vt+~
29 f
•.. . . . d(s~~~19~m~¢pIli$i()rJw~~th~C4rv$·
(24.77)2
167.62 =24.77 t + 2 (9.81)(0.15)
. . . . . ·•.• (lI>jl@~he<!l(!.l'f·ijrrW~Mth~.lj.me· • ~k~rJ
JrQmlh~!li$m~M~W9Pj~Ctl$¥j$i~IM(jIh~
. effecti\'ely.
dnv~tQt~lo$ti=mltMWlllK~$l:im;applled
.
.
t= 1.65 sec.
@•• • If!ldrl~r~ppr~¢ljjl1g.~(s·q~rfe • $El~san
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HEAD lAMP SIGHT DISTANCE
S(lJ+h=~
2
_ 0.85 (tt)
AP#r9Q•• truckappt6aq~~~ • Cl~9MfflM~s
p(/rve1l(a.$pe~()f10Qkph
.••• Th$JeMthPffh~.
(lJ
cUlV~I$18Qm.Jpn!:FWiW9t~~~!~~M~~~m
240.19 (0.0148) + h =(0.02 +0 03) (180)
2
·.30J0.and•• ±2~·.r~$p/¥:~¥ely, • • • l'h~ • lrit¢r$$~~Io~
Offh~.graqe • •tao9~nt~ • j~.m.10.t.43(JWlllt~r
elevafion • (jf.24g·60.·%••·•• • T~edrl¥~rhliS.l.o
sWiICh.90 • me•• ~eam • I~t1fs • al.m9ht•• tlm~tra\!~r
\'IiththElb~~m • lig~t.lll~~jng.~R<:lPQle • 9ftil\.9f
180'
= 0.0148
(lJ -
h= 0.495m.
® Max. design speed:
AV2
O;a5·•• ab('l'e••tnerlongJtlJ~inal.·a~is.(lf.th~ • • car.·.
Tt1ednvWspElrpeptiqnre~9ti()nJjmwls
0]8 sec.
.....
....
L=395
A=2-(-3)
A=5
CD Assuming a coeff. of friction otO.18,
compute the length of Ihe head lamp sight
distance...
.'
..•. ..•..
.
® How high was Ihe head lamp above the
pavement at this instant?
.. .
@ What is the max. design speed thatacar
coukf maneuver on this ClINe?
180 = 395
V= 119.2kph
5V2
(3~~ry·.m~f()II9Wlhg.d#t~for • aheM'amR~IQhr.
di~tli~;i
Solution:
CD Head lamp sight distance:
s= Vt+£
2g f
.
Oe$iSnv~l~titY¥1j9.lql~............)iii • • • .• • • ).. . . '.'
q9Elfficierltgffrigio~~~np~V~l'1let:ltClmi .
tir~$~9A4i
pef<:ep~()l'Itll'lJ~l$q.8P$~l'Id~.i)
V= 100000
3600
V= 27.78 mJs
~r~I~9ttiltl:lfth~b~mll~ht::=1'abClveth~>
_
(27.78)2
S- 27.78 (0.78) + 2 (9.81) (0.18)
$laUonlflggfJM);::=10+~2Qlitl'!I~.g{)QI'1'l.·· ".
S= 240.19m.
@
Gra(lei)ftl~Cktan~rI#·2,~~·
Gra9ElpfJ9w~r9l<!~g~nlFt?,W~ • • ·•• .• .· • • • • • ·•.•.• •.• .• i
Height ofhead lamp:
lon9lt\ldll1ala~i$Qtthecar.)·
Lel'lm~ ofcu·~•. ~:28P·IJl<.·· . ·.·.•. • • • · • ·.••····.i·•••·••• <•••..·• ·».··•.··•·•.••
$. qo01J:llltettle~®9J~tte$ighJgl,~!M~:
.•. . . . ·.·
® ComputetM•••nefgnlof••
head • lamp.
tn;e •
@
al>oVEltb~I'O<Jd§Urfll~··i
qomPlltethe.el~aticJnof.tI1$I()W~t.pOint
ofthecutve,'
.... .
Solution:
CD Head lamp sight distnace:
V2
S= Vt+ 29 f
V=110oo0
3600
V= 30.56
_
(30.56)2
S - 30.56 (0.80) + 2 (9.81) (0.14)
S= 364.45m.
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5-380
HEAD lAMP SIGHT DISTANCE
® Height ofhead lamp:
.t@•• g~~ig#.We@/)1.*v!¥ti~I • MG.·P~f~/:)(jli~.
•
·b~~:.~~1~~.~~~~~~~\I~I~~I~f ~~~.
1;11&,1,"11
·•• • •
·~l~~~~:~~ilii~.~t~:t.·.~~j~~~w~;~~~··
S9+h-(g2-g.,)L
-
2
• •·>Q,{qfll;<ipqV$tfflro@M~·>.· . . .•. . . . / . . • .
l' (n)
9= 180'.
9 = 0.017
·~ • ••• pP$PQt¢1lla··M91~.tfW·t~~ipMml19b:tUltsc
~"pV~I~$lc@~ill.lclIM@1'l~()fll1~r!m
Solution:
h + 364.45 (0.017) = (0.021 + 0 029) (280)
h= O.80m.
@
2
CD Length of CUNe:
AV2
L =395
A=Y2,-g1
A= 2.8 - (- 2.2)
A=5
Elevation oflowest point of CUNe:
P.c.-'~-----,f-f-
L=~
__-l
395
L= 126.58m.
® Head lamp sight distance:
S=vt+
S --.9.1..f::....
1- g1 -Y2,
_ - 0.029 (280)
51 - • 0.029.0.021
S1 =162.40 m.
L
H=a(Y2,- g1)
280
H =8 (0.021 + 0.029)
V2
2gf
V= 100000
3600
V=27.78 m/s
. _
(27.78)2
S - 27.78 (0.75) + 2 (9.81) (0.16)
S= 266.67m.
@
Angle the head lamp tilts:
H= 1..75 m.
---..r...--.. _ 1.75
(117.6)2 - (14Of
y=1.23
Elev. oflowest point of CUNe
= 200 • 0.029 (140) + 0.021 (22.40) + 1.23
= 197.64m.
(g,-g,)U2
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HEAD lAMP'SIGHT DISTANCE
I"--"·\L
Ij2
8= vt+ 29 f
S0+h-~
-
2
266.670 + 0.70 =(0.028 + O.O~) (126.58)
V= 120000
3600
V= 33.33 mls
_
(~)
(33.33}2
8 - 33.33 4 + 2 (9.81)(O.15)
5 = 4()2.47 m.
0=0.00924
" =0.00924 (180)
n
0=0.53'
@
Length of curve:
L= 2 (h + 50)
gz- g,
_r.fttl
Tf!6$llg@rtioaf¢llrve .. hllsab~cktM~$ot
graqij•• 6tS~% • ~nd • ~f°tWa@l~rnJMt.gra(jll.of·
0-
+2$•.!s·to•• f)e.··designed·.on.the·tlasi$ . th8lthe
.~~n~·11_~J~~ • ~4&~1~et~J • ~I.~~V~~~
•
sp~4ifi$@ij~.
.
$tdppll:\g • • m$~all()e \Vllh .•·•• th~ • ·•• '(jllgWli)g
0= 0.017 rad
L=2 {0.75 + (402.47) (0.017)J
(0.02) - (- 0.03)
L=310.98 m.
.:,':}-:>:>:>::<:}::::::« •...,-'. -•• -::-:::.---:::," • -,'. :::<:<:-:::<:-:::<:'::':,::::>...--::
Veloclty;"120 kph
.' •... ..•.••..' . ...•
:-em::::±~~
. < < < .' ..•.• .... . ...•••'.
The:hea(jlamps are O.7Sm. above the road
surfacealld theirbeamstiltsi.ipward ·at an
angltlof one degree abQve tlle Ion!;liludinc:if
axisofthEH::aL
". . ' ....>
> ' . •.
ai1dUre$isO.1~,
@.·.ROlllPIJ~trenead.lal11PS~hlcJjsl~®e
@
Max. velocity:
AVl
L=395
A=2-(-3)
A=5
_ (5) Ij2
310.98 - 395
V= 156.74kph
.••••·• ·•.•
®¢O@Jmt~the~ogth.mln~$ag¢(lt·
~)99mptttt!.thel1'lex; • •Vel~ly.ofttiEl.tiSrtl'lat·
COl1/dpa$sthruthesagcurve.inkph.
180
The length of the sag CllNe having gra4Elsof
and +3;5% is e(JualtQ 310 m... , ....
, 2.5%
Solution:
CD Head lamp sight distance:
0.25..' .. ' .
' .....
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S-382
HEAD lAMP SIGHT DISTANCE
Solution:
G)
Head lamp sight distance:
.;r8••1t",
~~@Ms>················································
L =2 (h + SfIJ)
fh.. gl
1.2n
fIJ = 180
fIJ
=0.021 rad
Solution:
. G)
L=A\f2
310 =2[0.80 + S (0.021)]
(0.035)· (. 0.025)
0.80 + 0.021 S =9.3
S=404.76m.
@
395
A =1.8· (- 3.2)
A=5
L = (5) (156f
395
L= 284.81 m.
Max. speed ofcar:
L=AV2
395
A=3.5 • (. 2.5)
@
6V2
V= 142.86 kph
Pemepfion reaction time:
V2
2gf
V= 142.86 (1000)
3600
V= 39.68 mls
S=VT+
V2
2gf
(39.68f
404.76 =39.68 t + 2(9.81X025)
t'= 2.11 sec.
V2
2gf
V= 150 (1000)
3600
V=41.67m1s
_
.
(41.67)2
S - 41.67 (2.1) + 2(9.81) (0.40)
S= 308.76
310= 395
S=VT+
Min. safe stopping distance:
S=Vt+
A=6
@
Length ofsag curve:
@
Height ofhead lamp:
.
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IUD lAMP SIGHT DISTANCE
® Height of head lamp above the pavement:
L=2(h+ $0)
L>S
gz- g,
_S2 (!l2 - g,)
_ 1.1 (1t)
1Il- 180
III = 0.0192 rad
284 81 =2 [h + 308.76 (0.0192)]
.
0.018 - (- 0.032)
h + 5.928 = 7.120
L- 2 (Sill +h)
200 = (164.69)2 (0.02 + 0.03)
.
2[(164.69) (~':b1t + h]
164.29:.9) 1t + h = 3.39
1
h= 1.19m
h=O.80m.
@
Max. speed that a car could safely travel:
A1f2L=395
A=grg,
A=2-(·3)
A=5
2OO=51f2395
V=125.70kph
ilitC.!
~lilllmij;···
. . .• . . ):.• • .>
li.,i'iI~(lIll.
Solution:
CD Head lamp sight distance:
1f2S = lit + 2g (f + G)
V= 80000
3600
s-
4
Solution:
CD Length of curve ofa sight distance of 130 m.
h, = 1.5 m.
~= 100mm
~=0.10m.
V= 22.22 mfs
_22.22 (3) +
ii.'"liiiBIB
(22.22f
2 (9.81)(0.20.0.03)
S= 164.69m.
S=Sight Distance
1----L=Lmght of Curv,-
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S-384
HEAD lAMP SIGHT DISTANCE
Assume sight distance is lesser than the
length of CUNe:
A=2.8-(-1.6)
A=4.4
AS2
---=-=-----=--100(m1+~)2
L=
L=
4.4 (13Of
100 ({3 + {Qi)2
L = 156.574m. > 130 m. (ok)
® Stationing of highest point of CUNe:
S _.E.1..!:1-
S
Bevation ofhighest point of CUNe:
Elev. B = Elev. A- Y
Bev. A = 100.94 +0.016 (56.934)
Bev. A = 101.85 m.
Elev. B = 101.85·0.45
Elev. B = 101.40 m.
g1
-
g2
~ 0.028 (156.574)
1 -0.028
+ 0.016
S1 = 99.64 m.
·A.·yem~J<1I • c91'V~(;(lOflEl¢ts.~'f~%Qr<!~~ • • @.~.
.4%•• graqe.aS.$h~n- • • Ihesta«(ltiQtth~.pp!nt.
of.W~i~aLinterseq~9"'0fthl:l.ta~~iltslll • at stll'
72tooqanf:lfry~l3.Il!l¥l.Itip~()M~epoil'ltof
InterseGij(lflJ~.1OQ.m· • • 9~tertnl~tM·I~I1~thof.
cUrV~f9rtisightdis~~ce.ofjSory ..Jhe®lght
qfotijeptl.lt1d.·ob$ervE!rbelng•• 1.Sm;~b9vet~e
P~Y~flJE!IlLjfpr • the.ri~hlsKl~·pfm~sul11l'nnof
lI'le~~o/elsanQbj~ln~wn~~l'IeiQN(lf
'..' '
. .
Q:~QI'll"
Stationing of the highest point of CUNe:
= (10 +040) + 99.64
=10+ 139.64
@
Elevation of highest point of CUNe:
i
!~~~,
!
Lf2=;;;;;-=t~;8.287
1-----L=156.574--~1
L
H=e;(g1- fl2)
Solution:
CD Stationing of the highest point of the cUNe:
H= 156.574
8
i
H=O.86m.
---l-_~
(59.934f - (78.287)2
. y= 0.45 m.
3,=93.75
I
Hig~.rt Point
~~.....%
Elev. of P. T. = 100 + 78.287 (0.028)
- 78.287 (0.016)
Elev. of P. T. = 100.94 m.
1-----~L=16875'------I
I
I'
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HfAD lAMP SIGHT DISTANCE
Sta. ofhighest point = (71 + 915.625) + 93.75
Sta. ofhighestpoint = 72 + 009.375
375
,-+--+-75
L
H=a(g1- fh)
H =~ (0.05 + 0.04)
H=0.0112SL
EL
~;;J.-~.375
f----L=168.75-----I
® Length ofnon-passing sight distance:
Jl..._ 0.6
(75f - (x)2
x =47.43
Using the squared properly ofparabola:
H
h
Non-passing sight distance = 75 + 47.43
Non-passing sight distance = 122.43 m.
(~)2 :: (~)2
@
L
a(g1-92)
(~Y
Lh -
h
= (~)2
S2 (gl - {h}
8
_82 (91 - 921
L-
8h
L = (150)2 (0.05 + 0.04)
8 (1.5)
L = 168.75 > 8 = 150 ok.
Sta. ofP.C. :: 72 + 000 _16~.75
Sta. of P.C. = 71 + 915.625
S
_51l.L
1- g, - g2
s - 0.05 (168.75)
1-
0:05 + 0.04
S1 = 93.75
Stationing ofnon-passing sight distance:
(72 + 009.375) + 47.43 = 72 + 056.805
(72 + 009.375) - 75 = 71 + 934.375
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5-386
HEAD lAMP SIGHT DISTANCE
_--.91.L
Solution:
51 - (gl - 92l
S - 0.05 (168.75)
1 - (0.05 + 0.04)
S1 =93.75m.
168.75
Sta. ofP.C. = 5 + 00 ·-2Sta. ofP.C. =4 + 915.625
CD Length ofCUNe:
AssumeS< L
H
h
(~)2 =(~)2
L
a(g1 - 92~ __ h_
(by - (~)2
2
L=
2.
se (g1 - fJ2)
8h
A =91-92
A =5- (- 4)
A=9
L=
9 (15W
100 ('12 (1.5) +..j 2 (1.5))2
L =168.75 > 150 (ok)
L = (150)2 (0.05 + 0.04)
8 (1.5)
L = 168.75> 150 (ok)
@
Location ofhighest point of curve:
Sta. ofhighest point of CUNe
=(4 + 915.625) + 93.75
= 5 + 009.375
® E/ev. ofhighest point ofcurve:
L
H=a(g1-92l
H = 16~.75 (0.05 + 0.04)
H=1.90m.
-.!L_1..
(~)2 - x2
~_.-L
(84.375)2 - (75]2
y= 1.50 m.
E/ev. ofP. T. =20 - 0.04 (84.375)
E/ev. of P. T. = 16.625 m.
E/ev. ofB = 16.625 + (0.04)(75)
. E/ev. ofB =19.625 m.
E/ev. of highest point of CUNe =19.625 - 1.5
Bev. ofhighest point of CUNe = 18.125
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SIGHT DISTANCE
SIGHT DISTANCES
@
VVhen
(Horizontal Curves)
CD When
r
I
..
R..........
,
\
,
,,
,,
oM,'
/R
I
,,"
\0.\ "
,,'R
--_ '''I,' ,-'
,
-,'
,
'~i'''
S = sight distance
l = length of curve
(AC'f =M2 +(AD'f
(AD'f =R2 - (R - M'f
(AD'f =R2 - (R2 - 2RM + M2)
(AD'f = 2 RM - M2
L+2d=8
Sol
d=-
2
(AC'1 =(AD)2 + M2
(AD)2 =(AO'f - (R - M]2
(AO)2 =(ftE)2 + R2
(AD)2 =(ftE'f +R2 - (R -M]2
lADy =d2+R2_ R2 +2RM-M2
AC =~ S (approximately)
(AD)2 =d2+2RM - M2
(~)2 =M2 +2RM _M2
(AC)2 =(AD)2 +M2
(AC)2 =d2+ 2 RM • M2 + M2
82
4
-=2RM
(AC)2 =d2+ 2RM
8. L)2
(AC)2= (2
M = clear distance from center of roadway
to the obstruction
S = sight distance along the center of
roadway
R = radius of center - line curve
L = length of curve
D = degree of curve
R-M
Cos Iil =T
R-M=RCoslil
M=R-RCoS0
M=R(1-COS0)
+2RM
S
2
letAC =S2 =(S-Ll\2RM
4
4
S2 =82. 2SL +L2 +8 RM
8RM=2SL-L2
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S-388
SIGHT DISTANCE
Solution:
Solution:
,,
,, ,
I
,'R
"R........ ,I
/ ,\at " -",' R
...
,
..
" '!' ,\
....., I,'
M=~
8R
82
M= L (28- L)
8R
R=8M
R= L (28- L)
(190f
R= 8(9)
R =450 m. (min. radius ofhorizontal curve)
8M
R_ 550 [2 (600) - 550]
8 (40)
R = 1117.19 m.
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S-38~
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REVERSED VERTICAL 'IUBOlIC CURVE
gl =grade of approaching tangent
g3 =grade of receding tangent
g2 =grade of common tangent
L1= length of first curve
L2 = length of second curve
L = total length of curve
L=L1 + L2
H1=difference in elevation between Aand B
H2 =difference in elevation between Band C
r1=rate of change of grade of first curve
r2=rate of change of grade of second curve
r, =.92.:.91
L
1
f2
=.9J..:...92.
L
2
L=L1 + L2
L =92.:.9i + .9J..:...92.
f,
f2
L1 =.92.:.91
f,
L2 =.9J..:...92.
r2
Elev. B = Elev. A +
[91f- (-E2f)]
r)
Elev. B - Elev. A = (91 +
L1
- (91 + 92) L1
H12
Elev. C= Elev. B +
[93f- (-92f)]
i2) L2
Elev. C• Elev. B = (93 +
Hr
- (93 + 92) L2
2
H = H1 +H 2
L, =S2.:..9.1
r1
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S-390
REVERSED VERDCll POIBOliC CURVE
~
~= 2(0.5)
rliWl
rlll'III"~
16 - n~2
~=1
H1+~:;=18
~~(-to) + 1 - 18
2gi =·18 +20
Y2 2 =1
Y2 = 1% (grade ofcommon tangenQ
® Length of first and second CUNeo assume
one station is 20 m.long:
L1 =!l2..:.ll1
(1
1-2
L1 :;: -0.5
L1 = 2 stations
L1 =2 (20)
L1 = 40 m. (length of first cUNe)
Solution:
<D Grade of the common tangent:
L _fJLJJ2
2-
B
I------L.-
r2
4-1
L2 = 0.50
L2 = 6 stations
L2 = 6 (20)
L2 = 120 m. (length of 2nd cUNe)
_
Total length = 40 + 120
Totallength= 160 m.
@
Elev. ofB:
Total length of the CUNe = 40 + 120
Total length ofthe CUNe = 160 m.
rJ:L!Jl
H1 = 2(1
-~
H1 - 2(-0.5)
1-4
H1 =---:T
H, =3m.
Elev.ofB=100+3
Elev. of B = 103 m.
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SPIWCURVE
I
Elements ofa spiral curve:
1.
2.
3.
4.
5.
6.
S.C. = spiral to curve
C.S. = curve to spiral
S.T. = spiral to tangent
Ts = tangent distance
Tc
I
7.
8.
T.S.
9. Rc
10. Dc
11. L.T.
12. S.T.
13. Es
14. L.C.
15. Xc
16. Yc
17. X
18. Y
19. Sc
20. i
21. Lc
22. L
= tangent distance for the curve
= angle of intersection of spiral
easement curve
= angle of intersection of simple
curve
= tangent to spiral
= radius of simple curve
= degree of simple curve
= long tangent
= short tangent
= extemal distance of the spiral
curve
= long chord of spiral transition
= offset from tangent at S.C.
= distance along the tangent from
the T.S. to S.C.
= offset from tangent at any point
on the spiral
= distance along tangent at any
point on the spiral
= spiral angle at S.C.
= deflection angle at any point on
the spiral, it is proportional to
the square of its distance.
= length of spiral
= length of spiral from T.S. to any
point along the spiral
dL = Roo
dL
ds='R
Q::!1:
L Lc
But 0 is inversely proportional to R:
. --6.
L
~y
,,
,,
,,
,,
'\
,/
,,
,,
\A/
, ,
V
D =1145.9.16
R
K
K
RL =RcLc
R-&.h
- L
ds = dL L
Rc Lc
ds = L dL
Rc Lc
=!S
R
D
y
R
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5-392
SPiRAl CURVE
At S.C.: L = Lc
L3
Xc =6RcLc
L3
L_~
XC -
..!1.
6Rc
(spiral angle at any point on the spiral)
For metric system:
20
20= RcDc
9-
20
Rc=Oc
(spiral angle at S.C.)
dx = dL sin s
sin s = s for small angles
T.S.
'~dt
I
S
sidt
dL:
dL
T.s.1
h
c=-
2S
:Y
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5-393
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SPiRAl CURVE
h =dx
S=dL
_(dxf
c - 2 dL
dy =dL - c
(dx)2
dy =dL - 2 dL
dx = dL sin s
dx = s dL
s2 dL2
dy =dL - 2 dL
s2 dL
dy=dL·-2
AB=RcSc
AB-&.h
-2Rc
AB=b:
2
AB = b (approximately)
By ratio and proportion:
b:
AG 4
b:=Rc
2
AG=~
L5
dy =dL - 8f\? L 2
c
From the figure shown:
1
Ts =b+(Rc+p)tan:2
1
1 (Rc+p)tan:2
sin:2=
OB
1
(Rc+p)tan:2
OV=
1
sin-
2
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S-394
SPiRAl CURIE
- 3
8.
Ye
(distance along
= l.c -~
40Rc
tangent at S. C. from T.S.)
9.
Ts
= ~+ (Rc +~) tan ~
10.
Es
=
11.
Ie
=
12.
P
=
13.
e
=
(tangent distance for spiral)
&
1
.
(Rc + 4) sec '2- Rc (extemal
distance)
I - 2 se (angle of intersection of
simple cUNe)
X L2
~--4 24Rc
0.0079
R K2 (super-eIe~at'Ion)
where K= kph
14.
e
=
0.00; K2 (considering 75% of
Kto counteract the super-elevation)
15.
Le
16.
i
ie
17.
0
Dc
SUMMARY OF FORMULAS
FOR SPIRAL CURVE
1.
_2.
L2
180
.
2Rc l.c x --;- (spiral angle at
any point on the spiral)
S
=
se
= D40Le (spiral angle at S.C.)
2ft
se
=
4.
Xc
= ~ (offset distance from
X
=
tangent at S.C.)
L3
Xc l.c3
6.
= ~ (deflection angle at any point
7.
on the spiral)
LS
= L- 40 Rc2 L 2 (distance along
e
tangent at any point in the spiral)
I
Y
spiral)
L2
=. 2L (deflection angles val}' as
e
the squares of the length from
the T.S.)
.
= LLe (degree af CUNe vanes
directly with the length from the
T.S.)
are basis, metric system
L
180
.
x --;- (spiral angle at S.C.)
3.
5.
= 0.0~6 K3 (desirable length of
A spiml80 m,lorig conriec(sa tangent with a
6>30' circular cuMtlHhe stationing of the T,S.
is 10 + 000, and lhegauge of beltact Qn the
curve is 1.5rn.
G)
@
--
Determine -the elevation -of the outer rail at
the trti&pOiot;if Ih~ velocilY¢ the fastest
train to pass OVer the ctltvels60 kph.
~~~:~rilhe spiral angle at the first
® Determine !he deflection angle at the end
pOint
_.
@ Determine the offutMrom the tangent at the
second qualterpoinl
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SPIRAl CURVE
Solution:
@
CD Elevation of the outer rail:
Spiral angle at the first quarterpoint:
L=20m.
L2 180
s=-. 2Rc LeTt
_ (20)2 (180)
s - 2(176.3}(80) Tt
5=0.81' ,
s= 0'49'
® Deflection angle at the end point:
At the end point Lc =80 m.
_ Lc 180
sC-2RTt
c
_ 80 (180)
Sc - 2 (176.3) Tt
Sc = 13'
i=~
3
. 13'
1=3
i = 4,33' deflection angle at the end point.
tan"
=-W-
@
Offset from the tangent at the second
quarter point:
V"
tan" =-
fT
V= 1000 K
3600
V=0.278K
g = 9.8 m1sec2
e'
tan" =1" = e
e =(0.278 1\)2
9.8r
0.0079 K2
e=
r
R= 1145.916
D
R= 1145.916
6.5
_ 0.0079 (60j2 (1.5)
e176.30
e =0.241 (outer railj
0.241
e=-2
e = 0.1205 (at midpoint)
L = 40 m. at the second quarter point
_iL
XC -
6Rc
L3
X=Xc L 3
c
_
(80}2
Xc - 6 (176.30)
Xc =6.05
X- 6.05(40f
- (80}3
X= 0.756m,
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5-396
SPIWCURVE
• t!l~·\MgM~·9f·#.·.~W~f#A~··Mi • ~ij]lAA~.m·.
·~ll'I'~gil.ltltl.~[1
• II~rJI~.~I.~·~~ll~.~ • ~~·.
W.P~W:fulh~ttij@g~()@itl1Pl~#jjlY~;·.··
•.• • •·• • •
IllIi• •fli
Solution:
<D Degree ofsimple CUNe:
U
0.004~
.
see=-R0.004~
e=-R-
l0)2
0.10 = o. OO4
R= 196m.
D= 1145.946
R
D= 1145.916
196
D = 5.85" (degree ofcUNe)
® Length ofspiral:
L = 0.036 1(3
C
R
. 3
L =0.036 (70)
C
196
Lc =63 m. say 60 m. (use multiple of 10 m.)
® SUper-elevation of the first 10 m. from S. C.
on the spiral:
1
el = 6(0.10)
el =0.017 (at 10m. from T.S. on the spiral)
es =5 (0.017)
es = 0.085 (at 10 m. from S.C. on the spiral)
e =0.085 (9)
e =0.765 m. (super-elevation at 10 m.
from S. C. on the spiral)
.·~ • • • P~~~triM~ • •lli~ • • ()ff$~lft()m.t?@~@~\t@.
•~.·• ~~IJ~i~I;lpil;~.lh~ •t~~g¢hl.·
Solution:
<D Radius ofthe new circular CUNe:
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SPIRAl CURVE
130.57 =
~ +[Rc +~~~] tan 25'
90.57 = tan 25' [Rc + ~67]
194.22 =R,,2 +;:6.67
Rc2 + 266.67 -194.22 Rc = 0
Solving for Rc = 192.84 m.
® Distance along the tangent at the mid-point
ofthe spiral:
whenL=40m.
L5
y=L- 40R 2 42
c
_
. (40)5
Y- 40 - 40 (192.84)2 (80)2
y=40-0.01
y=39.99m.
® Distance that the curve will nearer the
vertex:
Old external distance:
(for old simple curve)
cos 25' =280
OV
OV= 308.95 m.
E= 308.95 - 280
E=28.95m.
For the new curve:
(External curve)
@
a~rnU@lmZr~8M¢·tP+r$~Qve.IWWlja~.~.
i{fll-tl:l.
Es -(
4 sec !2 - Rc
- Rc + &)
(j) ·P~(~lMI@41$~~(lij·~tHMMw
&_J:L..
!'ill!lJli~lIt'
4 - 24 Rc
@
."I1'I~.·.tW91~~@~!$.#.·~$@pl~··.¢~rmM~~
Central angle of the circular curve:
Ie = I- 2 Sc
La 180
Sc = 2 R 1t
c
_ 80 (180)
Sc - 2 (192.84) 1t
·..·.· . .·¢tlct~~tM)Tl~W~fi'®)!IJ~¥.~*< •. •·••.•• • •
Solution:
<D Distance the new curve must be moved
from the vertex.
Deflection angle at the end point of the
splraf:
1= §.c
3
. 11.80
/=-31= 3.96'
.@
Offset from tangent at the end point of the
spiral:
L2
Xc=~
_ (80)2
Xc - 6 (192.84)
Xc = 5.53rn.
I P
2 DE
P
DE=-I
Cos-=Cos-
2
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5-398
SPiRAl CURVE
Valueofy:
x=b+y
super-elevation at quarterpoints.
e4 = 0.149 (10)
e4 = 1.49
1
el =4(1,49)
tan!=l
e1 =0.3725
h
2.5
=Cas 50'
h=3.89m.
@
2 p
Y=2.5 tan 50'
~
y= 2.98m.
@
1
='2(1.49)
~
= 0.745
3
e:J =4(1.49)
Distance from T. S. to P. C.
x = 30 +2.98 m.
x=32.98m.
e:J = 1.118
Deflection angle at the end point:
@
T~ • ta!ig~rl&·.MViri9.~~imutn~.of.~4Q.· • ~nQ
.~?~.·?r~R~Q®t~~.pt.~~··.~Q.·m;$irliil.~~
.•
··i'i~~[l~~l~~fu f:~~;.·.~~~~~~~~l~~i~
.
.
1111I.III1iilli
j
........• P()fllf($i~n ••/·./·•• • • • • • • • • ••• • •·•• ·••••• • • • • • • • • • • .•.• • •.•. • •.
·.~·P~f#®ih~~¢~t~l't@gi$@rJ~;
.
Solution:
CD Super-elevation at quarter points:
S=...h- 180
2 Rc 7t
80
180
S = 2 (190.99) 7t
S= 12'
, S
.
i= 4'
R = 114~.916 = 190,99 m.
@
0.0079 J(2
External distance:
- Rc + &)
Es -(
4 sec 1
2· Rc
R
_ 0.0079 (60)2
e- 190.99
~ = 0.149 mlm
3
. 12
1='3
R= 1145.916
D
e=
1=-
width of roadway
1= 282·240
1=42'
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S-399
SPIWCURVE
CD Length oflong and short tangent:
L2
Xc=~
_ (BOf
Xc - 6(190.99)
Xc =5.58
_ J:L
Yc-4- 40Rc?
Es = (190.99 + 51 ) sec 21'-190.99
(BOt'
Yc =80 - 40 (190.99f
Yc = 79.65 m.
Es = 15.0Bm.
XC -
8
_lL
6Rc
_ (80f
Xc - 6 (190.99)
Xc =5.58 m.
X
tanS=h
5.58
h=tan 12'
h=26.25m.
Long tangent (L T) = Yc - h
LT= 79.65 - 26.25
LT=53.4m.
Short tangent (ST):
SinS-&'
- ST
5.58
ST= Sin 12'
ST= 26.B4 m.
® External distance:
Es = [Rc+~] Sec~-Rc
Es =( 190.99 +5;a) Sec 21' - 190.99
Es = 15.OBm.
@
Length ofthrow:
p_&
-4
- 1145.916
Rc- '0
_1145.916
Rc 6
Rc = 190.99 m.
S=..h- 180
2Rc 1t
S- 80 (180)
- 2 (190.99) 1t
S=12'
_5.58
P- 4
p= 1.395m.
® Maximum velocity:
I
_
'-C-
0.036 'Ifl
Rc
BO = 0.036 ve
190.99
V= 75.15kph
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S-400
SPiRAl CURVE
~
~ f4f
/ ~f] l
J
,f
f
;~li~~~I;~i~~tl~lri~~;~~'I~~
·li\~I!l
•
•
II,rll&lil.'fi-~~
~
!m=.!I~
~~.,., ~c·a., ~ ·m.e,·~.p~r.i:.•. ~·.~ 2•·. , ~ '.·~a ~e' ~,.~@idriB~tft~
,.'~ ~~fi:;~~1
o•.•. :.•. ·.• .• . •.••,•.,..•.•a.•.•.•.,.·•·. .• .•l.'•.,•.•,• .:.• U._.O.•·.•."• .•,•e. •. .m• .• .·•..·.,•·. t.• .• .o..• .• •. •.r.••a.•. •a'. . .•.•.•.•.•.m•.e.•,ll,•. .•. •. i.•.•. . •. •.
'.·• . • •.• .•. •.• • .,.••,.•.•. .••.•. •. .•
i ,tt.,
•.•.
."t..LlI
•
@s~rti¢nt¢i,l~.<.'.'• · ·,·., · · · ,· · · · .
Solution: .
(j)
Solution:
(j)
Centrifugal acceleration:
80
C=75+ V
80
C=75+SO
C = 0.484 mlsec2
® Radius ofcurvature:
L - 0.0215
cCR
va
120 =0.0215 (SOf
Velocity of car:
80
C=75+ V
80
0.50= 75 + V
V=85kph
® Spiral angle at the S. C.
Rc=t14~916
Rc - 114;916
Rc = 229.18 m.
O.484R
R=269.86m.
S=~
Lateral friction on the easement curve:
4 - 0.0215 V3
2Rcn
c
@
\Il
R= 127 (f+ e)
.
_
(90)2
269.86 -127 (f+ 0.07)
f=0.166
-
CRe
_ 0.0215 (85)3
4 - 0.50 (229.18)
4 = 115.23 m,
'
~,
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S-401
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SPiRAl CURVE
s = Lc 180
@
2Rc 1t
S - 115.23 (180)
c - 2 (229.18) 1t
Sc= 14.40'
c
0.02151)1
CR
I _ 0.0215 (100)3
"'C - 0.457 (360)
4= 130.68m
I
_
"'C-
Length ofsharf tangent:
@
Length of transition curve to limit
centrtfugal acceleration:
@
Length ofshorf tangent oftransition curve:
;;:r=0;'"
."
·.Aiiii"'-~~""s'C.
s.c.
L2
_.:::.c.....
Xc-6Rc
X = (115.23f
c 6(229,18)
Xc =9.66
S..
T.S.
= 9.66
S·In 144'
.
S. T.
S. T. = 38.84 m
Sc-- h..ill!Ql
2R 1t
c
s - 130.68 (180)
c - 2 (360)1t
Sc= 10.4'
7..•··. ,..•·~:.·,•. '.•:.•[m~.lrlllll.'
p
•..A.••.nOQ3
••.1.6
.•0
•.•.•:. .•
.~
".
.•
."' ".' '."".. '",",' :":>::::::::::::>::::::::::;::;::::::::::;;<;:;::;:':::":>":'
.
11111
Solution:
CD Centrifugal acceleration:
80
C= 75 + V
80
C= 75+ 100
C = 0.457 nv'sec3
iT.
Sin 10.4' =
7.91
S.T. =Sin 10.4'
S. T. =43.82 m.
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8-402
SPIRAl CURVE
Solution:
G)
Solution:
Offset distance on the first quarter point:
CD Length ofspiral curve:
L3
x:::-6Rc Lc
1
L :::4"(80)
xc -k
6Rc
L:::20m.
Lc :::80m.
p:::2tifc
p:::&
4
L2
_
(20)3 ,
x- 6 (280) (80)
x::: 0.06m.
@
Rc
5'
Rc ::: 229.18 m.
Le2
1.02::: 24 (229.18)
Length ofthrow:
L2
xc:::tR;
Lc ::: 74.90m.
-~
@
6(280)
xc::: 3.81 m.
Xc -
Length of throw:::
CR
0.0215 Vl
74.90::: 0.50 (229.18)
Length ofthrow ::: 4
Length ofthrow::: 0.95
Velocity of car so as not to exceed the
min. centrifugal acceleration:
L ::: 0.0215 Vl
c
4'x
3.81
@
::: 1145.916
V::: 73.63 kph
@
Length of long tangent:
Max. velocity:
::: 0.036 J(3
Lc
R
c
80 ::: 0.036 J(3
280
K::: 85.37 kph .
s:::~
'tlle~h~I.l'#rve-otlln~sementcu~ispn.a
l~tI~I:.£oi.~~~:.~*~~$·,~··I~~
··~""PQffiPUte-.ttie • reqijlte~lfm9thOftlffispiral
•••· • • • ~·· • • • ?
• • • ·.H • •·••••••••••.••••••••.••••.•••••••••••••••••••••••••••••••.•
~• • P~~~@i~lh~~¢!o/()f.tb~carpa$$in9
Jh~11:!i~9\ltV~$QJh8tjtwiO.nQt~xce~d
11.".
c 2Rc 1t
S ::: 74.90 (180)
c 2 (229.18)(n)
Sc= 9.36'
Xc :::4 (1.02)
Xc =4.08m.
tan 9.36' =~
4.08
tan 9. 36 ' =/1
h::: 24.75
Long tangent::: 73.60 - 24.75
Long tangent::: 48.85 m.
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5-403
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SPIRAl CURVE
_JL
11'_111.1
the••lell9lhof$pif1llcllrVe·f$
.
.
Ihepa$$~llger$·· • •
aOmdCiog.
(I) • • CPmpme.·ijj~ • Y~@)ityQt.th~.apprQl@lng
~i~l\Pl1;<
®··.·.sdttipute.ltlareq~lfe~~iu$.Qtlb~centr~J·
CQN~()flM • ~Setnerltc~f\I~ • t()•• timil·the
XC - 6R
e
_ (80f
Xc - 6(266.61)
Xc =4m.
Ian 8.6' =~
h=26.45
LT= 79.30·26.45
LT= 52.85m.
cenlriftlgala~letation. ... . ...........•....•.•
the.len9th•• of.tM.I()n~ • t;tn~~nt«
tb~.spirlIl •.8Jrve•• if.lIJe•• ~istail~ • !llong • th~
@ •.••• C?ffiPllte••
tangentfrOm'T',S, IQS,CLis 79.301Tl;1(M'IQ,.·
Solution:
CD Veloctiy of approaching car:
80
C=75 + V
l"aditl$6f2()Qrildilttll~~¢l.lM!l'<
92·•• • (:Qll'lPut~umlfll'l9Ih*flhr!lW#I16~m$····
80
0.5161 = 75 + V
V= 80kph
® Radius of central curve:
L =0.0215
~~:,:~I~~'~$~~~1~~··.~'~·i,~~·0·Wiltl~
va
CRe
80 =0.215 (80)3
0.5161 Re
Re = 266.61 m.
e
@ ColTIPlI~·tb~.f~~gtl1.()f~El.logg~h!l$lt()1
~~~~~~.~~~I.~t~~~61;,~
@. ~~~~t~'~~~· YaIQ~ • • Bf•• t~~ • • •~tH(J9~1
.. . .
•
~fll.~citli~me1l!l'Sl~.
Solution:
CD Length ofthrow:
S=~
c 2R 1t
e
@LengtIJ oflong tangent:
1146'-~
. - 2 (200) 1t
Lc =80 m.
L2
_..::e-
Xc-6R
e
_K
XC - 6 (200)
s = Le 180
2 Re 1t
_ . 80 (180)
Sc - 2 (266.61) 1t
Sc =8.60'
c
Xc = 5.33
p=~
4
_5.33
P- 4
p= 1.33
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5·404
SPIRAl CURVE
@
Length oflong tangent:
"'h~~sigfl • S~~ •. Of<a • t,ar.~$~jngthru~~
~~f!tIl.~rye.i$$(!tlallQ~Ol<pll· • • Tll~.flldtU$
~~_--~s.c.
•
~11:_r~·(Jftl1e$pU!II;!JfV~~ ~~~
Q)i • ¢wPll(e.~ev~ue·pf~.l'lltemC8ll~i
.
·@ •••••
tan 11.46' =~
h=
5.33
tan 11.46'
~I~~n.~;;.I~.i
~ ~.~,~=$1of..'0~j
Solution:
G)
Min. value ofcentrifugal acceleration:
h =26.29 m.
80
C=75+ V
@
80
Long tangent = 27.20 -26.29
C=75+80
Long tangent =52.91 m.
C= 0.516 m's3
Centrifugal acceleration:
L =0.036
c
Rc
ve
80 =0.036
ve
200
@
Length ofspiral curve:
3
I _ 0.0215 V
CR
'-C-
l.c =0.0215 (80)3
0.516 (260)
Lc =82,05m.
1/ = 76.31 kph
L = 0.0215
c
CR
ve
80 = 0.0215 (76.31)3
C(200)
C =0.597 m'sec3
@
Llmgth ofthrow:
p=&
, 4
x,c -_-=6 Rc
L2
X = (82,05)2
c 6 (260)
Xc =4.32
p= 4.32
4
p= 1.08 m.
'.
• puNe.
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COMPOUND CURVES
@
The tangents of a spiral curve forms an angle
of intersection of 25' at station 2 + 058. Design
speed is 80 kmlhr. For a radius of central
curve of 300 m. and a length of spiral of
52.10 m..
Stationing of the start of central curve.
Sta. @ S.C. =Sta. @ r.s. + Le
Qta. @ S.C. = (1 + 965.36) + 52.10
Sta. @ S.C. = 2 + 017.46
® Length of central curve.
S
CD Find the stationing at the point where the
spiral starts.
@ Fin.d the stationing of the start of central
curve.
® Find the length of central curve.
e
=..s-.
x 180
2R
e
TT
S = 52.10 x 180
e
2(300)
TT
Se =4.975'
le=/-2Se
Solution:
CD Stationing at the point where the spiral
starts.
Ie = 25 - 2(4.975)
Ie = 15.05'
2+058
1=25'
Length of central curve:
S=R I ~
e e 180
S =300(15.05') 1;0
S= 78.8m.
2
Le (R + Lc
T =-+
- - )' tan-I
s
2
e
24 Re
2
T = 52.10 + [300 + (52.10)2] ta 25
'2
24(300)
n 2
T, =92.64 m.
Stationing @ T.S. = (2 + 058) - 92.64
Stationing @ r.s. = 1 + 965.36
A simple curve having a degree of curve equal
to 4'30' has central angle of 50'50'. It is
required to replace the simple curve to another
circular curve by connecting atransitioncul'\Ie
(spiral) at each ends by maintaining the radius
of the old curve and the center of the new
central curve is moved away by 5 m. from the
intersection point.
CD Determine the central angle of the new
circular curve.
@ Compute the tangent distance Ts of the
spiral curve.
® What is the maximum velocity that a car
could pass thnu the curve without
skidding?
S-404-B
COMPOUND CURVES
Solution:
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@
CD Central angle of the new circular curve.
Tangent distance Ts of the spiral curve.
~c
=P =4.52
T = Lc. +(R + ~) tan
'2
'4
~
2
T, = 166.2 + (254.65 + 4.52) tan 25'25'
2
T.
@
=206.26 (tangent distance)
Maximum velocity:
L = 0.036V
,
3
R,
3
166.2 = 0.036 V
254.65
V = 105.54 kph
R= Rc
R= 1145.916
4.5"
R=254.65
0' A =5 Cos 25'25'
0' A=4.52m.
p=~
24R c
2
Lc = P (24) Rc
Lc2 =4.52(24)(254.65)
Lc = 166.2 m. (length of spiral)
s
c
= Lc 180'
A simple curve having a radius of 600 m. has
an angle of intersection of its tangents equal to
40'30'. This curve is to be replaced by one of
smaller radius so as to admit a 100 m. spiral
at each end. The deviation of the new curve
from the old curve at their midpoint is 0.50 m.
towards the intersection of the tangents.
(j)
@
@
Determine the radius of the Central curve.
Determine its central angle.
If the stationing of the intersection of the
tangents is 10 + 820.94, determine the
stationing of the T.S. of the spiral curve.
(j)
Radius of the central curve.
2 Rc n
S ~166.2 (180')
2 (254.65) n
Sc :: 18'42'
Solution:
c
Ie::; 1- 2 Sc
Ie = 50'50' - 2(18'42:)
Ie'" 13'<'6' (central angle of the
new circular curve)
Cos 20'15 = 600
OA
OA = 639.53 m.
BC =0.50 m.
AC = 639.53 - 600
AC = 39.53
AS = 39.53 - 0.50
AB = 39.03m.
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COMPOUND CURVES
® Central angle.
Tan 20'15' = ...!.600
T =
,
S-2 +(R c + Xc)
Tan 20'15'
4
~=P
4
P=~
24 Rc
p=
(100)2
24 (598.73)
P =0.70
p=~
24R c
T, = 1~0 +(598.73 +0.7) tan 20'15'
p= (10W
24R c
T, =271.14 m.
p= 416.67
Rc
S = Lc 180'
c
2R c n
a'A =Rc +39.03 .
S = 100(180)
c
2(298.73) n
Considering Triangle ADO'
Sc =4.78'
Cos 20'15' =
Rc +P
Rc +39.03
le=I-2Se
R + ~16.67
Ie = 40'30' - 2(4]8')
c
Cos 20'15' =
Rc
Ie = 30'56' 24"
Rc +39.03
(central angle of new curve)
Cos 20'15' = R/ +416.67
Rc (R c +39.03)
0.94 R/ +36.62 Rc =R/ +416.67
0.06 Rc2 - 610.33 Rc +6944.50 =0
@
Stationing of the T.S. of the spiral curve.
Sta. ofT.S. = (10 + 820.94) - (271.14)
R = 610.33± 587.13
c
2
Rc = 598.73 m. (radius of the central curve)
Sta. ofT.S. = 10 + 549.80
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S-404-D
COMPOUND CURVES
19000 - R = (8W
•
c
24 R
c
A simple curve having a degree of curve equal
to 6' is connected by two tangents having an
azimuth of 240' and 280' respectively. It is
required to replace this curve by introducing a
transition curve 80 m. long at each end of a
new central curve which is to be shifted at its
midpoint away from the intersection of the
tangents.
CD Determine the radius of the new central
curve if the center of the old curve is
retained.
@ Determine the distance which the new
curve is shifted away from the intersection
of the tangents.
@ Compute the length of throw.
4583.76 Rc - 24 R/ =6400
R/ -190.99 Rc + 266.67 = 0
R = 190.99 ± 188.18
2
c
Rc = 189.59m.
@
Distance which the new curve is shifted
. away from the intersection of the tangents.
h = R1- Rc
h = 190.00 -189.59
Solution:
h = 1.40 m. (amount the new curve is
CD Radius of central curve:
shifted away from the intersection
of the tangents)
@
\ i '
\
\
,
R~"
J
I
Rc\
\ Ie
\
/
/
\~'/
", 0 20 ,,
.\*~'
o
p=~
24 Rc
R1 • Rc =p
R = 1145.916
1
D
R = 1145.916
1
6
R1 = 190.99 m.
,
'Rc'
,
length of throw:
p= (ly
24 Rc
'';'1
p=
{80f
24 (189.59)
P = 1.41 m.
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COMPOUND CURVES
® Spiral angle.
A simple curve having a radius of 200 m. has
a central angle of 50'30'. It is required to be
replaced by another curve by connecting spiral
(transition curve) at its ends by maintaining the
radius of the old curve and its center but the
tangents are moved outwards to. allow
transition. Part of the original curve is
retained. The new intersection of the tangents
is moved outward by 2 meters from its original
position along the line connecting the
interseCtion of tangents and the center of the
curve.
CD Determine the length of the transition
curve (spiral) at each end of the central
curve.
® Compute the spiral angle.
@ Compute the central angle of the central
curve from the S.C. to C.S.
CD Length of the transition curve:
,
'
,,,' ,\Rt::L
'.f
'
"
'
Rc'"
I".
,
,
\
R',
ang~nI,
&'
\
, I
,,~/
"
,'11
'
P= 2 Sin 64'45'
P= 1.81 m.
P = (L c )2
24R c
L~=
P (24) Re
.
L~ = 1.81 (24)(200) .
Le= 92.95 m. (length of spiral)
S = 92.95 (180')
-c
2 (200)n
Sc =13'19'
@
Central angle:
le=I-2Se
Ie = 50'30' - 2(13'19')
Ie = 23'52' (central angle of the
new cUNe)
From the given compound curve, it is required
to replace it with a transition (spiral) curve
100 m. long starting at A and ends up at B.
The degree of curve of the first curve is 4'
while that of the second curve is 10'. Central
angles are 6' and 15' respectively for first and
second curve.
Solution:
"
' "
\0ld To
S = Lc 180'
c
2R c n
Qld'fa1l;8i!nl
S-404-F
COMPOUND CURVES
CD Determine the radius of central curve.
Determine the length of short tangent CB.
® Determine the length of long tangent AC.
@
Solution:
CD Radius of central curve.
R = 1145.916
1
4
R1 =286.48 m.
R. = 1145.916
2
10
R2 = 114.59 m.
0 1 O2 = R1 - R2 - P
0 1 O2 = 286.48 - 114.59 - P
L2
0 1 O = 171.89 __c_
2
24 Rc
R1 =Xc + R2 Cos 21' + 0 1 O2 Cos 6'
L2
R1 = _c_ + R2 Cos 21'
6 Rc
(
L2
+ 171- _c_) Cos 6'
24R c
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@
Length of short tangent CB.
X =~
c 6 Rc
_ (10W
Xc - 6(146.46)
Xc =11.38m.
L3
Y=L __
c_
c c 40Rc2
- 0
(1oW
Yc -10 - 40 (146.46)2
Yc =98.93 m.
S = ~ 180
c 2 Rc 11
S = 100
180
c 2(146.46) 11
Sc = 19.6'
SinS = ~
c CB
CB=~
Sin 19.6'
CB= 33,92m,
® Length of long tangent AC.
286.48 = (100)2 + 114.59 Cos 21'
.
6Rc
+ (171.89 - (100)2 ) Cos 6'
24 Rc
8.55 = (100)2 _ (10W COS 6'
24 Rc
6 Rc
8.55 = (100)2 (1 _ Cos 6')
6R c
4
Rc = 146.46 m.
X
Tan Sc =----'. CD
CD =
11.38
Tan 19.6'
CD=31.96m.
AC=Yc-CD
AC = 98.83 - 31.96
AC = 66.87 m.
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A ::~
m
Derive the prismoidal correction formula for a
triangular end areas using the prismoidal
formula.
2
v.::h[~"~
c 3
2
2 + ~]
2
_ bl +b2
bm- 2
hm:: ~
2
v=h[~"~~
~]
c 3
2
2
2+ 2
!ii]
V =h[~.~"~ ~"~
c32
4
4"44+2
Solution:
V :: hIb lh1 " b l h2 " bill + b2h2]
c 3 L4
V~ :: 1~ [b1 (~1 " h2)" ~ (hI" hv]
L
Vc :: 12 (b 1 " bV (hI" h2)
(prismoidal correction to be subtracted
algebraically from the volume. by end
area method.)
Let us consider the triangular prismoidal shown
below:
Derive the Prismoidal Formula for determining
volumes of regular solid.
Solution:
//~:/
/f==A
L
Vc ::"3(A I -2Am +A2)
A-~
1- 2
A-~
r 2
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5·406
UITHWORIS
V --~A
A 2- .1!1.&
3
VOLUME OF EARTHWORK
fu_~
A2 -hi
{A;=b1 ~=!Y2
..fA; hz fA; h,
~
A2 = h,2
(1)
End area
_ (A, +.4.2) L
V2
(2)
Prismoidal Formula
V=~.hlAl
3h, 2
3
, _ 11,3 A, - h,3A,
V3 h,2
V= A12 (hz3 • h,3)
3 h,
V= A, 2(hz - h,) (hz2 + hzh, + h,2)
3 h,
,A,h 2
h 2\
V=3h,z,(h2 +hzh, + n
'V - At! h22 A, h h2 .~
. - 3 h,2 + 3 h, + 3
(~)
L
VP =6(AI +4A",+A2)
y;;;
Am =area
V =&!!
+ fu.!!
+ &J).
3 A,
3 A,
3
V= ~ (Az + ~ A, A2 + A, ) (frustum ofa
pyramid)
(3)
of mid-section
Volume with Prismoidal Correction:
(Applicable only to three level section)
~
~=_2_
h,2
A,
4Am =~(hl +2h2h,
+h,~
A - A A, h,2 2 A, h2
4 "m-'+
h, 2+'h ,
w:;
4 A", =A, + A + 2 A, _r-
V=VE-Vcp
VE = volume by end area
Vcp = prismoidal correction
L
Vcp r1 2 (e, . Cz) (0, • 00
2
'V A,
4 A", =A, + A2+ 2 ~ A2 A,
2 .y A, A2 = 4 Am - A, - A2
- r.-:'1/
fu &.
A, A2 = 2 A",. 2 - 2
h
fu
~
V= 3" (A, + 2 Am - 2 • 2 + A2)
V= ~ [lA, +4A", .~. A2 + 2A2
l
V = ~ (A, +4 A", +A2) Prismoidal Formula
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•••
I!I,.,·"
••
•
~ ·Et&nI.~~~t.~~~9fi1 SI~e •
® Distance of left slope stake from center of
the road:
2· hi
_..!fL
1.5 hL + 3.5 -100
2·hL =0.15hL +0.35
1.15hL = 1.65
hL = 1.43
Distance of left slope stake = 1.5 hl + 3.5
Distance ofleft slope stake = 1.5 (1.43) + 3.5
Distance ofleft slope stake = 5.65 m.
@
'i!l!t• ••
Solution:
G)
Diff. in elevation· of right and left slope
stake:
Bev. ofleft slope stake = 152 • 1.43
Bev. of/eft slope stake = 150.57 m.
Bev. ofright slope stake = 152 • 2.76
Bev. ofright slope stake = 149.24 m.
Diff. in elev. = 150.57 -149.24
Diff. in elev. = 1.33 m
Distance of right slope stake from center of
the mad:
1.5hr
- ·]SI,;i·······!
JlJ'I,
...,j..
2·hL:
r·····_···
i hr.,
· -
I"
----~;-.·~~·;:~~r
h,·2 _..!fL_
3.5 + 1.5 h, -100 - 0.10
hr' 2 =0.35 +0.15 h,
0.85 h, = 2.35
h,= 2.76
Distance of right slope stake = 3.5 + 1.5 h,
Distance ofright slope stake = 3.5 + 1.5(2.76)
Distance of right slope stake = 7.64 m.
compute the side s1ope()f both sect!o(ls•.
® Compute the value ofx at statkln 10 +200
it it has a cross sectiona' area of 14.64 m2.
@
Compute the volume between stations
10 + 100 and 10 + 200 usfng end area
method with prismoidal correction.
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S-408
URTIWHD
Prismoidal correction:
L
Solution:
<D Width ofbase:
Vp = 12 ('1 - CV (~.~)
100
Vp =12 [(10.95 - 12.9)(1.5 - 1.2))
.
LL _
I
I
Vp =- 4.875 m3
2.~ :
V. =VE-V:
V: =1390.125 - (- 4.875)
1=2.35 6 I sn.
Vcp= 139Sm3
.45--1---4.
B
2+
2.35 =6.45
B
2+ 1.05 =4.5
1.35= 1.95
5:; 1.5
2B +1.0(1.5) =4.5
B.=6m.
® Value of x:
t-----Dl"12.99---~~
,,
I
:2.6
________J
f-----fr-----+----(il~
@~m!tlllwlQth()fthebase.>
.
p
@
• . 2•.•.•.•.•.• • • •. •.•.•.hasailareaOf
c.••. om
.• . •. . •. •. . . •. t•.•1t. • .•.•.e
. • .• •.•. '•. b.•6.•.•• . •. "•.• 16.82m2.
•. a
. . . .• I.OO
.•. . . •.•.•.•.•. O
.•.•. •.f•.•.•. c• • • •tIt
. •.•. •.•.•.•.••a..•• .•.'•.•. .• .•.S
.•..•..I..•. a
..•.•••. J.•f.• •....it.•
.•.••.•..•.i.OO
..••.•..•.•.••..••..• ..S
·~· • ·.·CQf:llf.Wle•• t~~.·.VOlul'l'le • bf.llW~~n.Aand • .B
With Piismoidal.CotrectiOit
t.
Solution:
CD Width ofbase:
@
Volume between sia. 10 + 100 and
10+ 200:
A - ~ 1.5(6.45) 4.5(1.5) ~
1- 2 +
2 + 2 + 2
Al =13.1625 m2
A2 = 14.64 m2
~ _(At+A,) L
E2
~ _ (13.1625 + 14.64)(100)
E2
VE =1390.125 m3
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<D ~~.~~(M at thecentef ()f
B
6.3=2.2S+2"
~ =t~ ill !he .glJt s10pe
B
7.2=2.8S+2"
@
0.9 =0.6 S
S= 1.5
C>etermille .1he.·.·~lti~
betw~en $ta;
.1 +1001tld1 .. 20Q by apPlYing ptismoidal
c:ol'ftlC1jQft.
.
.
..
B
6.3 = 2.2(1.5) +2"
SoIufIon:
CD He9It of cut at tM center of sta. 1 + 1()():
B=6.m.
@
Value of cut at station B:
2.2(3} 6.3x ?11& ~ = 1682
2+2+2+2
.
6.75x=9.32
x= 1.38 m.
@
Volume using Prismoidal correction:
- ~ 6.6(2) 4.8(2) ~
A,- 2 + 2 + 2 + 2
Stalion 1 + 100
A, = 16.80 m2
v. -~~
\I
_
VE-
A = 100sq.m.
2
E-
(16.80 + 16.82X20)
2
5h
2+
VE = 336.20
L
Vp = 12 (e, - C0 (0, - ~)
~
2
+
5(~)_
h(5+h)
2 + 2 - 100
5h +5h + 2ff. + 5h + Iil + 2.5h =200
3h2 + 17.51'1·200 =0
...-_---
_ -17.5- ~(17.5f.4(3)(-200)
h2(3)
20
Vp = 12 (2· 1.38X11.40 - 13.5)
II =- 2.17
VE- II
WOO .
h- -17.56
h = -17.5 + 52.2
8
if=
V= 336.20- (2.17)
V = 338.37 CU.m.
h= 5.7Im.
@
At. $taljpn.·.1.+•• 1QO•• maP9rtjon.of~~~tl~ij'.
slrelch•• ha~ • <ire,g•• ·Qf.1(}O$q··.m~~~~.m • ~~·
while.that~f.S~tion • 1"'••~.~~.'8~$i$~$A?
melers•• j'l(;Ut••·•.. AI·.~U1lioi.llt~09;~··~·
SUrface•• lo!tleIeft•• ()fllle~ll"~I~'~t·i'll1 ••.
an•.
Height of the right slope stake at sta.
t + 200:
Slalion 1 + 200
~1'I~'II.tllII
sta.k~ • j~·.3 • till1e!!•• tlil;Ih~tf¥l!lttl<:it·.(jf··.lflj.·.l!i!ft
slope.!lfake; • • ~.~l)tf!tllUljt$@li19.t .... ~'
Thewidthofll1ero~yj$Wm.Wilha~
slopeof~;l.
. .. .
.
A =240sq.m.
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5-410
EARTHWORIS
5x 2.89 (5 + 2x) 2.89 (5 + 6x) ~
2+
2
+
2
+ 2 =240
5x + 14.45 + 5.78x + 14.45 + 17.34x+ 15x =480
43.12x = 451.1
x= 10.44 m.
3x= 31.32m.
@
Volume between sta. 1 + 100 and 1+ 200:
CD Compute the area of station 1+040..
. ® Find the $re8 of sfallon 1 + 100, .
'.
@Oelermine lhediff. in volume of cutand·fift
using end area melhod~
. .
Solution:
CD Area of station 1+ 040:
62.64
Volume by end area:
1/ _ (A, + A2 ) L
2
VE-
1/
_
(100 + 240) (1oo)
2
VE-
1
1
1
+'2(1.22)4.63 +'2(4) 0.42
Afill = 11.47
100
Cp =12 (5.78 - 2.89}(27.34 - 93.52)
100
Cp =12 (2.89) (- 66.18)
Cp =,- 15.94 cu.m.
Corrected volume:
V= VE - Vp
V= 17,000· (-1594)
V= 18,594 cU.m.
1
A~= '2(4)1.84 +'2 (1.22) 6.76
VE = 17,000 cu.m.
L
Cp = 12 (C l . G.1) (D, - D0
@
m2
Area of station 1+ 100:
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URr....IS
1
Solution:
1
Acut ="2 (4.5) 0.98 +"2 (3.05) 5.48
CD Value of x:.
1
1
+ "2 (3.05 5 + 2" (4.5) 0.5
~3.7
Acut = 19.31 m2
r-=-!~~-_~I ".,.. .2......5""""'~;;.,;.;;;;~'-rI
I
@
Diff. in volume of cut and fill using end
area method:
0.8:
I
I
x=3+1.8
x=4.B
® Area offill:
Area offill = 2.5 (D.8)
2
Area of fill = 1.0 m2
1.22 _ 1.22 + 3.05
x 60
x=17.14 .
6O-x=42.86
D+ 19.31
Vcut =
2
(42.86)
@
Area of cut:
Area of cut = 3 (1.8)
2
Area of cut = 2.7 m2
. Vcut =413.18m3
VRlI =
11.47+0
2
(17.14)
VRlI = 98.3 m3
The following is a set of nOtes. of an
earthworks of a roadcQr'istrtlctitinwhich' is'
underfal(en by the BureauofPubilcWOl'I<$. '., .
Diff. = 315.51 m3
:-' "
_,.,:- 'c-.'.: . . :-.:
-,-.,-.-:,-.'.-"'-'-"_:'-':-:':'.-:'.:-<:'::-::::::<:>':'::::-::::::>~>::':::::
StaliooCross$4!CllOn . ....
Jt02O •.•..
Given the folloWing section of an earthworks
for a proposed road conslruct!t>n ana hiUy
portion of the route. The width of the fOadbase
for cutis 6 m. for alklwance of d~lnage canals
and 5 m. fbffin. Sideulopes rot cut is 1:1 and
for fill is 1.5:1.
.
3.7
-0.8
0 .
CD Compute the value of x.
® Compute the· atealn fill.
@ Compute the area in cut.
x
+ 1.8
.".
~~••.. ~••..,. +~:~~~>~' .•.
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5-412
EARTHWOIIIS
Solution:
Assume a level section with an average value
of cut and fill for each stretch.
CD Area section f + 020:
CD Determine the volume of cut.
® Determine the volume of fill.
@ If the shrinkage factor is 1.2. determine the
volume borrow or waste.
,,
3.0:
:
t--- ;;i---..+--4.S---I-----4.5-1-;--;.l
- 4(4.5) (4 + 2)(4.5)
A1- 2 +
2
+
Solution:
Average depth of cut:
(2 + 1.5){4.5) ~
2
+ 2
A1 = 31.50 m2
® Area of section 1 + 040:
- 7800
C - 850
C=9.18m.
_~
(5 + 4)(4.5)
A2- 2 +
2
(4 + 2)(4.5)
+
2
2(4.5)
+
2
A2 = 45.75m2
@
VOlUme between stations:
(A 1 + A2) L
V=
2
Average depth of fill:
8500
f= 1200
f= 7.08
1-----37.54.------1
V- (31.50 + 45.75)(20)
-
V=
2
772.5 m3
Side slope =1.5 : 1 Cut
In determining the position of the balance line
in the profile diagram, a horizontal grade line
is drawn such that the length of the cut is· 850
m. and that of fill is 1200 m. The profile area
between the ground line and the grade line in
the cut is 7800 sj:l.m. while that of fill is 8500
sq.m. If the road bed is 10 m. wide for cut and'
8 meters wide for fill and if the side slope for
cut is 1.5 : 1 while that for fill is 2 : 1.
_ (10 + 37.54) (9.18)
A2
A = 218.21 sq.m.
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EARTHWORKS
CD Volume of cut:
Vc :; 218.21 (850)
Vc = 185,500 cU.m.
Solution:
(1)
Slope = ~.~ = 0.016
Side slope = 2 : 1 Fill
_ (8 + 36.32) (7.08)
A-.
2
A = 156.89sq.m.
(2)
(2)
Distance in which the fill is extended:
0.048x = 1.2 + 0.016(50 - x)
0.064x'= 2
x = 31.25
@
Stationing of the point where the fill is
extended:
Sta. = (7 + 110) + (31.25)
Sta. = 71 + 141.25
Volume of fill:
V, = 156.89 (1200)
V, = 188,000 cU.m.
@
Volume of borrow:
Vol. of borrow == 188000 (1.2) - 185500
Vol. of borrow =40100 cU.m.
428-A CE Board May 200
,h>i/'C;':;>:$"''''''
"kM.',"o/.. }·'·h';'>N'
",<'
,,' ",."•., .. "
'«l;<
The center height of the road at sta. 7 + 110 is
2 m. fill while at sta. 7 + 160 it is 12 m. cut.
From sta. 7 + 110 to the other station the
ground makes a uniform slope of 4.8%.
(I)
@
@
Slope of the new road:
Compute the slope of the new road.
Find the distance in meters from station
7 + 110 in which the fill is extended.
Compute the stationing of the point where
the fill is extended.
At station 95 + 220, the center height of the
road is 4.5 m. cut, while at station 95 + 300, it
is 2.6 m. fill. The ground betweens!ation
95 + 220 to the other station has a uniform
slope of - 6%.
CD What is the grade of the road?
How far in meters, from station 95 + 300
toward station 95 + 220 will the filling
extend?
@ At what station will the filling extend.
(2)
S-412-B
EARTHWORKS
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Solution:
CD Grade of road
Slope of road = 2.3
80
Slope of road =0.02875 say 0.029
(2)
From station 0 + 040, with center height of
1.40 m. fill, the ground line makes a uniform
slope of 5% to station 0 + 100, whose center
height is 2.80 m. cut. Assume both sections to
be fevel sections with side slopes of 2 : 1 for
fill and 1.5 : 1 for cut.
<D Find the grade of the finished road
surface.
(2) Find the area at each station.
® By end area method, find the amount of
cut and fill.
@ Between these two stations, is it borrow or
waste?
Roadway for fill is 9.00 m. and for cut it is
10.00 m.
Distance from 95 + 300 where filling will
extend:
Solution:
CD Slope of roadway:
0.029(80 - x) + 0.30 =0.06x
0.089x = 0.029(80) + 0.30
x = 29.44 m.
® Station where filling extend:
(95 + 300) - (29.44) = 95 + 270.56
- 1.20
Slope of roadway =. 60 . .
Slope of roadway =. 2% (downward)
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EARTHWORKS
@
Area at each station:
1.40-0.02 x = 0.05 x
0.07 x= 1.40
x= 20
6O·x=40
L
Vol. of fill =2(A 1 + A2)
Station 0 + 040
20
Vol. offiH ="2 (16.52 + 0)
1--------18.40-------1
Vol. of fill = 165.20 cU.m.
L
Vol. of cut =2(A 1 + A2)
~--+--lOl--t--4
Station 0 +100
40
Vol. of cut = "2 (39.76 + 0)
_ (14.60 + 9) (1.40)
A-
2
Vol. of cut = 795.20 cU.m.
A= 16.52 sq.m. (fill)
_ (10 + 18.40) (2.8)
A2
A = 39.76 sq.m. (cut)
@
Volumes of cut and fill:
@
Since the volume of cut is excessive than
the volume of fil" it is then necessary to
throw the excess volume of cut as waste
by an amount equal to
795.20-165.20 = 63.00 cU.m.
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S-414
EARTHWORKS
@
Area ofsection 0 + 040;
r----·-1.5 x
The following data are the cross section notes
at station 0 + 020 and 0 + 040. The natural
ground slope is almost even.
0
I
I
I
I
Xi
I
Base width
Cut = 9m.
FiII=8m.
Side slope
Cut=1:1
I
Fill =1.5 : 1
I
I
!
__x_
8 - 1.5x + 8
8x = 3x + 16
5x= 16
x= 3.2
A - 3.2 (8)
- 2
~
?
Station 0 + 020
+3.0
+1.5
4.5
0
0
0
?
?
Station 0 + 040
- 2.0
-1.0
4
0
0
0
?
CD Compute the area of section 0 + 020.
Compute the area of section 0 + 040.
@ Compute the volume of borrow or waste
from station 0 + 020 and 0 + 040
assuming shrinkage factor of 1.20.
A = 12.8 m2 (fill)
@ Volume of borrow or waste:
@
Solution:
CD Area of section 0 + 020:
1.5
1.0
x·= 20·x
x = 30 ·1.5x
2.5x = 30
x= 12
20 - x= 8
Vol. of cut =
(A 1 + A2l L
2
I f
_ (20.25 + 0) (12)
va. a cut 2'
II
1.5
h
4.5 =h+ 9
4.5h = 1.5h + 13.5
h=4.5
A - 4.5 (9)
-
Vol. ofcut= 121.5 m3
(A 1 + A2l L (1.20)
Vol. of fill =.
2
Vol. of fill = (12.8 + 0!(8)(1.20)
Vol. of fill =61 .44 m3
2
A = 20.25 m2 (cut)
Vol. of waste
Vol. of waste
=121.5 - 61.44
=60.06 m3
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EARTHWORKS
In a 20 meter road stretch, the following crOSS7
section of the existing ground and
corresponding subgrade cross-section notes
were taken.
1
A1 = '2 [6(- 5.5) + 7(1) + 18(2)+ 12(1) + 4(0)
Existing Ground Cross Sections
Sections
o
Left
Center
~2
-1
5
o
o
1
4
-3 -2
10 +300 13.5 10
1
7
1
o
1
5
10 + 280 16.5
9
+ 0(- 1) + (-S){- 2) + (-16.5){-5.5)
Right
2
+ (- 7)(- 5) + (- 6)(- 5)]
1
12 "18
- [- 5(7} + 18(- 5.5) + 1(12) + 2(4) + 1(0)
+ 0(- 5) + (- 1)(- 9) +- (- 2){-16.5) +0(- 7)
0 - 1.5
917
+ (- 6)(- 5.5) + (- 5){6)]
Subgrade Cross Sections
Sections
Left
o -5.5
Center
1
A1 ='2 [(247.75) - (- 69)]
Right
A1 =158.375 sq.m.
-5
-5
- 5 - 5.5 1
- 3 - 7.5 - 7
10 + 300 13.5 -7- 6
-7
-7 - 7.5 - 1.5
10 +280 16.5 -7-
6
o 6-7-"18
(-7,-7.."i)
Solution:
A
1
Area of station 10 + 280:
=.1.
Area at station 10 + 300:
o 6-7-"17
CD Compute the cross sectional area at
station 10 +280.
@ Compute the cross sectional area at
station 10 + 300.
@ Compute the volume between the two
stations.
(1)
@
[~ x2 ~~ Xs x6 X7 ~ Xg x10:U~ ]
2 Y1 Y2 Y3 Y4 Ys Y6 Y7 Ya yg Y10 Y11 Y1
A = 1. [2.-_7_. 18 QiQ:1.:1- 16 .5 ...:L.:!2.1 2 - 5 - 5.5 1 2 1 0 - 1 - 9 0 - 55 • 5 - 5
]
(1.-1.5)
I.
5-416
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fIIT. . .
1
A2 =:2 [6(. 7.5) +7(-1.5) +(17(3) + 9(1) + 5(1)
+(- 7)(- 2) +(. 10X· 3) +(-13.5* 7.5)
+(- 7)(- 7) +(- 6X-7)]
- [7(- 7) + (17)(- 7.5) +9(·1.5) +0(5) + 1(- 7)
+ 1(-10) +(- 2X-13.5) +(- 3)(- 7)
+(- 6)(- 7.5) + (- 7)(6)1
Q)Coinp!JltJJI)$~~9fc~t~1isla~(Jff
'19f040. / , > •.•..••.•••••••.••.•••..•..••••..••••..•••.••••••.•.••.•......•.
®.. . 4~mPllte.~tl • ~llffl'l~ • 9f•• ~t.~t • $tatiO~
10tQ60'
<
@•• ·.ewnPUt~.f~~.%llJffiff.Qf.!)C)1'I'9wor.~ti~~
frl)lll·••. $:latjon • • 111•• f.~P.to.10 • •.• f(laO·.
.glfI$kt~_li!I:l~~gorQf~$~4. .'
1
A2 =:2 [(194.75)- (-156)}
A2 = 175.375 sq.m.
@
Solution:.
ill .Area of cut at station 10+ 040:
A1 =1 [!t!2.!.l~~!6.!.L!t]
2 Y1 Y2 Y3 Y4 Ys Ys Y7 Y1
Volume between two stations:
L
V=2(A1 +A;il
fQ - 7.5
- 5 - 3 3 5 8.3 0]
2A1 = LOo -2.5-2 -2 -2.5 0.820
20
V= 2"(158.375+ 175.375)
V = 3,337.50 CU.tn.
(10,1)
2 A1 =[0(0) + (. 7.5X- 2.5) +(- 5X- 2) +(- 3X- 2)
+3(- 2.5) +5(0,82) +8.3(0)]
- [0(- 7.5) + 0(- 5) + (- 2.5X- 3) +(- 2)(3)
+(- 2)(5) + (- 2.5X8.3) +(0.82XO)]
2 A1 = [(31.35) - (- 29.25)]
2A 1 =60.60
A1 = 30.30 sq.m.(cut)
@
Area(){cut· at sfa, to+{)6();
A2 =1 [!t~!.l&~~!t]
2 Y1 Y2 Y3 Y4 Ys Ys Y1
0 3 -3 - 8.1 - 5 - 2 0]
2A2 = [OO.50.5-0.8:-2~O
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2 ~ = (0(0.5) +3(0.5) +(- 3X- 0.80) +(- 8.1)(- 2)
+(- 5X- 1.5) +(- 2XO»)
- (0(3) +0.5(- 3} +(0.5)(- 8.1) +(- O.BO)(- 5}
+(- 2X- 2) + (-1.5}(0»)
Volume of cut from station 10 + 060
10+080
L
V2 ='2(A,+A2l
to
" _ 20 (1.70 +0)
V22
2A2 =27.60 - 2.45
A2 = 12.575 sq.m. (fill)
Area olcut:
A =1 [~~.&~~]
3 2 111 Y2 Y3 Y4 Yl
L3 56.45 3]
2A3 = lQ501.41 0.5
2 A3 =[3(0} +5(1.4) +6.4(1) +5(0.5»)
- [0.5(5) +0(6.4) + 1.4(5} +1(3))
2 A3 = 15.9 -12.5
2 A3 = 3.4
A3 = 1.70sq.m. (cut)
@
Volume ofborrow or waste:
Considering station 1 + 080
1 rO 5 8.5 3 - 3 - 9 - 6 - 2 Ql
A4 ='2LO 0.5 0.151.51.5 o::-t5::-t5Q.J
(-3.15)
(0.1.5)
V2 = 17 cu.m. (cut)
Volume of fiff from station 10 + 040
10+ 060
L
V, =:2 (A, +A2l
v, -20 (0 +212.575)
1/ _
Vl = 125.75 cu.m. (fiU)
Volume of fiff from station 10 +
10+ 080
L
V2 =:2 (A l + A2l
060 to
. . V = 20 (12.57i +25.40)
2
V2 =379.75 cu.m. (fiU)
(3.15)
Total volume ofcut = 320 + 17
Tota/volumeof cut = 337 CU.m.
2 ~= [(0(5)+5(0.1Sf+ 8.5(1.5) + 3(1.5)
+(0)(- 3) +(- 9X-1.5) +(- 6)(-1.5) + (OX- 2)1
- [0(5) +0.5(8.5) +3(0.15) +(1.5}{- 3) + 1.5(- 9)
+0(. 6) +(- 2X-1.5) + (0)(-1.5)1
2 A4 =40.5 + 10.3
A4 =25.40 sq.m. (fill)
Volume of cut from station 10 + 040, to
10 + 060
L
V, ='2 (A 1 +A2)
_20 (30.30 + 1.70)
V1. 2
V1 =320 cU.m. (cut)
Total volume offill = 125.75 + 379.75
Total volume of fiN ", 505;50 Gllom.
Volume of fill reqUired from sta. 10 + 040 to
10+ 080
Vol. offill =505.50 (1.25)
Vol. offill = 631.875 m3
Therefore there is a need of borrow since
vol. of fill is greater than that of the volume
of cuI.
Va. ofborrow = 631.875 -337
Vol. ofborrow = 294.875 fn3
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S-418
URTHWORIS
A = (10 + ~.412) (11.603)
A= 385.29
.[h$•• cenl~r#ne • • Qf•• a.·.·prl:lP05ed • toad·•• crOSs
~~c,Uon9fOss~a,stn~llvalleYbelweE!rt
s~l~on • • 10•• t.p??.(ele'9li~tI • 12~'()O • m·)·.~M·
$tatjonJQ",qaO(elev~llOn1Z2.5Qm.l.The
® Vol. offill from (10 + 022) fo (10 + 037)
+
V= (A1 +0) (15)
2
V= (385.29 + OJ (15)
2
• Sfationing.at•• th~·txlttom(it • tte.• valley.• iS•• 10 •
·.037•• •.•.• •.• . (el~v.. l11.2rn.),.The.grade.line .of.fhe·
·prqpOsedroagPa#e$.lI'ie·.grOUlld·POinls 3t.lhe
edges()f~viJl~Y{$ta·Wt022)an(t \(10
t.~}l!nl.tme~llqn.~ttlny.ofthese.slation$···
ate~hree}evel$et66n~.Wldthof toad base;;:
.1Qm·.·lNjtryside~qpeof.?;1~ ••.• Assume•• that.the
.~!l'~.qM~yaij.eY·.~'9pe.·qlreetlytq.the • lowest
poiIltftofutfleedgM'
..
Q).• Find the .cross si!ctiollal
.••••..• statiOO·1Q+63f
..
V= 2890m3
® Vol. offill from (10 + 037) fo (10 + 060)
V= (385.29 +0)(23)
2
area of fiU at
V= 4431 m3
@}Compute thevolumtt of fiU from station
•
{11)H)22}~(10+ 037)·
. ...
.
® Compijte tIleVQtume of flU from station {10
+03mQ{1ll+000l. . ..
.
ELJ1I.2
The location' survey Qf thllproposed road
passes lhrough srough terrain; and crosses a
small valley between two points along the
. center line of the proposed road: One of the
points is at station 40 +536.00 al1d at
elevation (150.42 m.), the other point is at
station 40 + 584.00anl:L ·at . elevation
(149.82 m,). The iowestpoinl~t the bottom of
fbe valley is 23 m. from the highest point arrd
has an elevaUonof its bottom equal to
140:64m. The road ~sesthr'OOglt these
three points. All sections an this proposed
roadway are three level sections having a
width of roadway equal to 12 m. with side
slape of 1.5 : 1. Assume shtlnkage factor to be
1.30.
Solution:
CD Area of fill af 10 + 037
L_O.5
15 - 38
y=0.197
f+ y= 123 -111.2
f = 123 - 111.2 - 0.197
'f= 11.603
.
CD Compute the cross sectional area
station 40 + 559.
@
at
Compute the volume of fill needed starting
from the highest point of road to the lowest
point of the valley.
.
Compute the volume of fill needed from
station 40 + 559 to 40 + 584.
.
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Solution:
CD Cross sectional area at 40 + 559:
EL 140.64.
Q)..
~%rlte • •the•• •VOlum1•• • US~g.· • ~~~fI1~1
® C0Jl'lP'*~.th~Yol~l:lllSinsF;tI<J~l'l$.·~h
Prisnl6ld~llffilt~lftln'i«i
.®.. . CqmP~ • ~~e.x~l1llle • usill9l;tld~W~.~h·.
L_O.6
23 - 48
y=O.29
f = 150.42 • 0.29·140.64
f=9.49 m.
Area =(12 +40.47) (9.49)
2
······ClJrv~hlrf!.r.ol'ffl«io~iflhtlrOcidlsl:irt~~'
ClJr":eWhiCh•
• turo$wthe.right~ittl.·.·lfI~
giv¢h¢t~$S$e¢ti()illh
..... .
Solution:
CD Volume using Prismoidal Formula:
Area =248.97 m2
® Volume offillfrom 40 + 536 to 40 + 559:
V= (0 +248.97) (23) (1.30)
2
@
STA5. +000
V= 3722.10 m3
@
Volume of fill from (40 + 559 to 40 + 584):
V=(0 +248.97) (25) (1.30)
2
V= 4045.76 m3
® STAS +020
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8-420
Use average values of dimensions of AI
and A2 Am (mid-section)
As, =A, - [5 (1:.5) + ~l (2)
!Q@ §(ill ~ §@
A, = 2 + 2 + 2 + 2
As, = 150:25 • 85.5
As, =64.75 m2
A, = 150.25 m2
6(14) 7(41) 7(17.25) ~
~= 2 + 2 + 2 + 2
e'=3 0 ,
A2 =259.375 m2
_ 12(6) ~ 6(15.375) 3.75(6)
Am - 2 + 2 ~
2 + 2
1
44.5
e'=3
e, = 14.83 (positive the excess area is
away from the center ofcurve)
Am =201.375 m2
L
Vol. =6(A, + 4Am + A2)
Vel - ~o [150.25 + 4(201.375) + 259,375]
Vol. = 4050,42cu.m.
@
Volume by End area with PrismoidaJ
Correction:
Va. = VE- Vp
v; - (A, + A,) L
E2
_ (150.25 + 259.375) (20)
VE. 2
VE =4096.25 m3
~=~.
58.25
3
= 19.42 (positive)
92=e2
L
Vp =12 (C, - ~ (0, • D-J
Vol. = VE + Vc
L
Vc = 2R(As, e1 +AS2e2)
20
Vp = 12!(5 • 7) (44.5 - 58.25)]
- 1145.916
R0
Vp =45.83 m3
R
- - 6- -
Va. = VE - Vp
Vol. =4096.25 • 45.83
Vol. = 4050.42 m3
@
_
[7 (17.25)· 6 (4.5)] (2)
A"2- Ar
2 + 2
As2 = 259.375 - 147.75
AS2 =111.625 m2
Volume with curvature correction:
.. '1140.916'
R= 190.99 m.
Vc 2 (1~~.99) [64.75(14.83) + 111.625(19.42)]
Vc =163.78 m3
II
_
(A 1 + A2) L
2
(150.25 =259.375) (20)
VE2
VEII
_
VE =4096.25 m3
Vol.=VE+Vc
Vol. =4096.25 + 163.78
Vol. = 4260.03 m3
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h,-hl
_12hL + 12 +2h, -12
9·h/
_1
2hL + 12 + 18 - 6
54-6hL =2hL +30
BhL =24
hL =3m.
.
,.
...,
:
..
..
...
.,
..
.
~
• '7
--,
?
<D•.• •ComPuw•• the • volullle • ti~1\¥een • thelWp
·".,.,$taticln~.·lJSlogPnsOl?k1alfort1'IlI~, ....•,.,. • ,•.,•..,.•.•.• ',
®····q(linPute..• • tht~ • ·•• Pri§ml)@ll • • • (:()rt~CtjP"
"•.• • belWee(llh~lWo. slati()OsJh.CU,m.• \ . •.",•••.••.,•.••.•••••••
®.·•.• CQmputelhe.culVaturecorrectjC)".~Il.·
Al =(3+9)(36)_~_~
2
2
2
Al =126 m2
STA5+ 140
the tvl0~ta~0l"is..W.ther(Ja~j~.0l"i.a • 5de9((!e
~!Ve.Which.tums ,tQlhe.right.Qf.
s®t1OMi6cu,m.
.
. • ~·.¢m~ .
Solution:
CD Volume using Prismoidal formula:
STA 5+ 040
~_1-
12 + 2h, -12
6h,. 36 = 12 + 2h,
4h,=48
h,= 12
hr - hi'
h -4
2
12 +2h, 12
6h,- 24 =12 + 2h,
4h,= 36
h,= 9 m.
~=-
_1-
2hL + 12 + 2h, -12
12·h/
_1
2hL + 12 + 24 - 6
72- 6hL =2hL + 36
8hL =36
hL =4.5
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S-422
1
e, =3" (O,)
e, =3"1 (36)
e, =• 12 (neg. towards the center of cUNe)
A =(4.5 + 12}(45}.~.~
2
222
A2 =206.75 m2
A~ =206.75 • [6 (~.5) +7 (~5)] (2)
Am (mid· section)
A = (3.75 + 10.5)(40.5}. 7.5(3.75}. 21(10.5}
m
2
2
2
Am = 164.25 m2•
L
Vol. =6(A, + 4Am +A2)
100
Va. ::"6 [126 + 4(164.25) + 206.75]
. Va. = 16495.83 m3
<il Prismoidal correction:
L
Vp = 12 (C, . Cz) (0, • ~
100
Vp =.12[~5·~(~:45))
Vp = 150m3
@
CUNature correction:
As, = 126. [~+ 5 (~2)] (2)
As, =48m2
AS2 = 74.75
1
m2
~=3"D:!
1
6:! =3" (45)
~ =-15 (neg. toward$ the center of ClJNe)
- 1145.916
R0
R= 11~916=229m.
L
Vc =2R (As,
e, + As2 ~)
100
Vc =2 (229) [(48)(- 12} + 74.75 (- 15))
Vc =• 370.58 m3
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5-423
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w•• comPllt~ • • .t6&VQ1t!~ .• ~• • mt.lW(f
$tatl<lI1~l.lSillggnsll'\Oic!~lfOfJTlll~;>«
••••
®•• ComJll1t~tb~ • • vqtplJle • PeM'eenffiK.W&
. .··$tatiOf!:;ifJsil1lJ.~lldareaWilhPfj$ITlQi<l~
CQIT~..<
tije/VoflJrrffl • •bAAveen • ttJ¢ • ~·
stalions.if··tbe··roag.js.oll.a•• cUfV~~~ich-
@ ••••• QomPt!te • •
. tllf~ •.·.to •.• lh~ • I~ft· •. Wllh•• ·tb~ •.• giVI:\tl • (1{~$$·
.$~$itithli$·liI·fEldittsQf.200rrk< ..
Solution:
G)
VoIiJme by Prismoidal Formula:
Am section
I·
Note: Use average dimensions of
sla. 1 + 020 and 1 + 040
,
I
2.3:
I
,:1.0
:
f;;;-614.~B_n_l--B_n-4.5I--~l
_2.15(3) 1.35(6.225) ~. 3(1.8L
Am- 2 +
2
+ 2 + 2
Am = 13.974 m2
L
Vol. =6(A 1 + 4Am + A2)
,,
I
2.3:,
Vol. = ~ [(13.1625) +4(13.974) + 14.64)
,,
[;~;~-31~..l..-
Vol. =279",3
STATION 1+4J20
@
STA 1 +020
B
Volume by end area with Prismoidal
correction:
2" + 2.3S = 6.45
B
2"
B
+ S=4.5
1.3S= 1.95
S= 1.5
.
2"+ 1.5 =4.5
,,
,
,
:2.6
,,,
B=6m.
I
- 2.3(3) 1.5(6.45) 1.5(4.5) M!l
A1- 2 +
2 + 2 + 2
_ _ _--+-3-.+.-----;.~
A1 = 13.1625 m2
STA1+040
-",,~C.Y(tT/
=-~'}
--+-__
3---61;
,.
,
,:2.6
,
,
:
;F3
--
\I
_
VE11_
VE-
(A 1 + A2) L
2
(13.1625 + 14.64) (20)
2
VE = 278.025 m3
8-424
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L
Vp :: 12 (C, - ~ (0, - ~
20
.
Vp :: 12 ((1.5 -1.2) (10.95 -12.90)1
1
&2 :: 3' (12.90)
&2 :: +4.3 (positive away from
center ofcurve)
L
Vc :: 2R (As, e, + AS2 ~)
Vp :: - 0.975 m3
@
Vcp :: VE-lip
Vcp :: 278.025 - (- 0.975)
Vc :: 2(200) [3.4125(- 3.65) + 1.44(4.3)]
Vcp :: 279m3
Vc :: - 0.313 m3
Volume by end area with curvature
correction:
V:: VE+ Vc
V:: 278.025 + (- 0.313)
20
V:: 277.712 m3
~_:"1i:iltl
'The earthWQrtsdata of a propciSedhlpClYis
----D,slO.95l---
===~~c,.c·
As, :: 13.1625 _[!3)J1) + 1.5 ~4.5)] (2)
... " t+498,0:3artd 2... 94S,@··
.•. • .
Stationing of limitsm free haul ••.. . ..• .•..
As, :: 3.4125
1
e, :: '3 (10.95)
Free=h~:I~~~:~~~~3~ 12
a, :: - 3.65 (neg. towards the renter of cUNej
.
AsSume fhegmundsurface to be unifoonly
.
.
. sloping. .
.
STAnON.
-:-------,
G).];:~ . Q'l¢rbau4'lol1!!D~'.'.'
® COl1lJlut~~Mvolu~Qfw~st~.
@
COInpuletttevolume of-borrow.
G)
Overhaul volume:
Solution:
As.1 :: 14.64 - [(3J1"J+ 1.l}69(2)
"As.1 :: 1.44
.
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h
50
26.88 =300
h=4.48
a
SO
241.97 =300
a =40.33
(j)
SoIuIioIf:
Urrit of economical /J8UI:
Overhaul voIll718 - (4.48 + 40.33) (215.09)
-
2
LEft ::
Overhaul volume =4819.10 m3
@
+
LEH::.lI'L
@
V = 2620.92 m3
Volume of borrow:
C
70
208.03 = 300
C=48.54
\ I I Alb
en
l.EH =42.0 @J} + 50
21
Volume of waste:
V:: (40.33 + SO) (58.03)
2
@
C,C FHD'
Free haufvolum.:
h 41.13
x=~1.97
h= O.17x
47.85 _-L
208.03 - SO - l
y= 0.23 (SO- x)
hX_~
_ (48.54 + 70) (91.97)
vo. UI onow-.
2-
2
2
0.17~ :: 0.23 'SO - xi
2
2
O.86x=50-x
Vol. of borrow = 5451.06 m3
x=26.88
5O-x=23.12
hx
FrH haul volume ='2
The giv~n data off a proposed M<lt'l~ - Cavite
Coastal road is tabulated below. Thetrae tia~
distance is 50 m. andth~ cost of borrOW 1$
h = 0.17 (26.88)
h=4.57
y=O.23 (23.12)
y= 5.32
P420 per cu.ffl.'\oostof~is-P.350per.
CU.m. and the cost of haul Is P21 perineter
'i
Fret haul val. =4.57 6 .88}
station. The ground sutface is assume to be'
uniformly sloping.
Free haul 't'd z 11.42 cu.m.
@
Overhaul voIum8:
10+ 160
10 +401.97
10+610
CD Compute the limit of economical haul. '.
ct Compute the free haUl volume.
@
Compute the overhaul volume.
- (41.13 +•. 57) (215.09)
V,2
V1 =4915""
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5-426
EARTHWORKS
The!following data are results of the earthwork
computations of areas, free haul distance and
limits of economical haul by analytical
solution instead of graphical solution (mass
di39ratns). iM cross sectional area at station
1 +460 is40sq.m. in fill and at stanon 2 + 060
the cross sectional area is 60 sq,m. in cat.
The- balancing point is at slation 1 +- 760 where
area Is equal ro zero, Assume the· ground
surface to be sloping upward unifomlly from'
station 1 + 460 to f + '760 and then with
sflghtlysteeper slope to 2 + .060. Assume fr~
.hau/distanCe ;: 50 m. and limitofecbfiomical
MLiI;:;4$Om. '.
....
,
Stafloning of lhelimits offree haul distance .
;:; (1 +732.47) and (1+ 782.47)
.
StatiOning the ijmits of economiCal haul
, =(1 + 512.26) and (1 + 962.26)
(DOIHe@in~tbeoverhaut
Overhaul volume
=(4.494 o.452 ) (179.79)
+t
=4040cu.m.
@
Volume of waste:
V J40.45~ + 60) (97.74)
V=4909
® Volume of bofTOW:
V = (40 + ;3.032) (52.26)
V = 1908 cU.m.
vOlume.
.~.· • !¥tel1tlinethev0k.lmeafwaste,
@ [)e(eI'fuine1lJevoliJineofborfuw~
Solution:
CD Overhaul volume:
Here under' shows a table of quantities' of
earthworks ofa proposed Highway to conned
Sago City and Danao City, The length of the
free haul distance is specified 10 be 50 m. long
and the limit of economical haul is 462,76
long. Assume the ground surface to be slOping
uniforrtlly.
10 + 020
x 40
27.53 =300
110 +"115.65
x=3.67
10+ 297,92
10 + 320
10 + 347.92
10 + 578.41
i47=~
y=4.494
a
40
220.21 =300
a =29.36
b
60
179.79 =300
6'=35.958
10 + 620
Cut
80.00
54.57
5.90
o
Fill
(D Compute the overhaul volume.
@
.
lnitialooirll
Umltof
economical haul
L1mitoffteehalll
Bafanclnoooint
. 4.60 Limit offree haul
43.15
Umitof
economical haul
End ooinl
50.00
Compute the volume ofbofTow.
® Compute the volume of waste.
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S-427
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Solution:
Solution:
CD Ovemaul volume:
CD Umit of economical haul:
LEH = fb..f + FHD
Cb
LEH =
6~;~) + 50
LEH=450m.
® Stationing of limits of freehaul distance:
Overhaul volume =
(54.57 +5.00) (182.27)
2
Ovemaul volume = 5510.93 m3
tID Volume of bOfrow:
~~me
@
""-" (SO +43.15) (41.59)
01 ~/UW=
2
Volume ofborrow = 1937.05 m3
!!..28.6
x- 200
h=OJ43x
Volume of waste:
....L_~
"
I
vo,ume 0
f~""
W"",8 =
(80 +54.57) (95.65)
2
Volume of waste = 6435.61 m3
SO-x-250
y= 0.08 (SO· x)
Vol. of excavation = Vol. of embankment
hX_~
22
0.143 x (x) _ 0.08 (SO - x) (SO - x)
2
The profile of the ground surface along which
the center line of the rQadWay issJoping
unifOrmly al acertaiJfgrade. At sta.5 '" 400 ih~
cross sectional area is 20.89 I'nL in fill and the
finished roadway slopes upward producing a
cross sectional area of 28.6 m2 In cut at station
5 + 850 The stationing of the balancing point
isS +650.
Free haul distance =SOm.
Cost of haul. =PO.20p¢fmeter$tatioo
Cost of borrow =P4 perro.m.
CD Computethelimltof economicalhauL
Compute the stationing of the limits of
freehautdistance.
@ Compute the freehaul volume.
@
-
2
0.143'; =0.08 (50 - x)2
0.378 x = 0.283 (SO - x)
50-x=133697x
x= 21.40 m.
Limits offreehaul distance
= (5 +650) +21.40
=5 +671.40
=(5+671.40)-50
=5 +621.40
limIts of freehaul distance
= 5 + 671.40 and 5 + 621.40
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S-428
EARTHWORIS
@
Freehaul volume:
Freehaul vol.
hx
="2
Solution:
CD Overhaul volume:
h = 0.143 (21.4)
h=3.06
c:
h I I _3.06 (21.4)
2
rree au vo. -
Freehaul vol. =32.74 m3
"".' rha I vol
(3.845 + 40.18)(180)
U
ume=2
vv8
Overhaul volume = 3962.25
® Volume ofwa$fe = (40.18 ~60X98)
Volume of waste = 4908.82 cu.m.
@
Volume of borrow =
(33+40)52
2
Volume of borrow = 1898 cu.m.
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5-429
Pdfbooksforum.com
EARlHWDRIS
CD Volume of waste:
Vol. of waste:: 350 - 200
Vol. of waste:: 150 m'J
@Overl7aul volume:
Overl7aul volume :: 910 • 350
Overl7aul volume :: 560 m3
Volume ofborrow:
Volume ofborrow = 350 + 520
Volume of borrow = 870 m3
@
Solution:
STATION
10+000
10+040
10 +080
10 +120
10 +160
10 +200
10 +240
10 +280
10 +320
10 +360
10 +400
10 +440
10 +480
10 +520
10 +560
10+600
VOLUME
+200
+100
+150
+140
+110
+190
t50
·40
·120
-90
-80
. ·200
·220
-110
·320
-280
Mass Ordinates
-+-200
+300
+450
+590
+700
+890
+940
+900
+780
-tti9O
+£10
+410
+190
+80
-240
-520
(D•• • ·~~ • lheQ~lJlv~rnelrieu'lll·> • • •.·.•·
~ • • • (»~t~.tbe~~gth(itovt~\lll(lt@te~if
••
·~ ~~:l-r~.~=~(~~~G~~~·············
Solution:
CD Overl7aul volume:
·130
Overhaul volume =600 - 200
Overhaul volume:: 400 m3
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5-430
EARTHWORIS
@
Length of ovemaul:
@
LEH= Cb C + FHD
Ch
Mass ordinate of inital point of limit of
economical haul:
450 = 500 (20) + 50
Cost of haul:
Ch = P25 per cU.m. I meter station
171100 = 25 (201.40) Vol. of overhaul
Ch
20
Total cost of haul = P105750.OQ
105750 - 25(L) (400)
Vol. of overtlaul =680 m3
L = 211.50 m.
Mass ordinate of inital point of limit of
economical h8uJ
•
-
20
® Total cost of borrow:
Vol. of borrow = 200 + 130
Vol. of borrow; 330 m3
=800-68G
= 120m3
Cost of borrow = 330 (500)
Cost of borrow = P165,OOO
+800
Th~ • cO$t.of.l)OrroW.pet.Cl.nn·••I$.·A5po.atld.lbe
cq~t.of.bt:lul.~r.m~tel"~tiPl1i$.P25· • 9q$t·.Q(
~~",\)~li® • • i$ • i!PPt9Xki)~teJy • P6~.pElr • CU.m·
Th~.·fO*! • bl:lyl•• djstanl:El.ls50••Itl·•• IoIl~ • ~.IKi •
leJlgltlm.()~mallli$.~q®I!P.?Qt4Q·tl1< • • lfthe
.mass • ~rgliJ~t~.Of • lh~ini~~.poJtlt •
.~~l1l •. djsmn:C~ •.• j$ .• t80Q • • 'tld•• • ttl~>roa$$
·oroinates•• l'fthe • ~urnrnit.ma,sS • diagram.fJ'Ql1l
m3••.
me
oftM.ftee
lQt{)(}QJo10+6()Ollre·60I'I'I~alld.
-60
,
o.Vtrhl1M1
Was"vQlw,",
140m3
respecllvely.
$ • • • CQfuputelhe.length.ofeeonomicalhaut
®•• • COItlPot~ • th~lTlass • • otditl~ • of•• lhe.jrlitial
·PQint.. Qf.lhe•• limit.of.ecQtlOItl!cal•. t18ul.lf.the·
• • • total¢()~of.haulih9.jS.P17H~.
® Computethefotalcostofwaste.
Solution:
CD Limit of economical haul:
LEH= CbC+ FHD
Ch
LEH = 5O~~20) + 50
LEH=450m.
® Cost of waste:
Cost = 650 (120 + 60)
Cost = P11T,()(JO
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EARTHWORKS
@
Length of overhaul:
192000 = 120(400) x
,
20
The following are the data on a simple summit
mass diagram.
STA
0+000
0+ 500
MASS ORDINATE (m3)
-80
-130
Initial point of limit.of freehaul distance = +600
Inmallimit of economic haul = +200
Freehaul distance = 60 m.
limit of economical distance =400 m.
Cost of haul = P120 per cU.m per meter station.
(j)
@
@
@
Determine the volume of waste in m3•
Determine the volume of borrow in m3.
Determine the overhaul volume in cU.m.
Determine the length of overhaul if the
total cost of hauling is P192,000.
Solution:
(j)
x=80m.
Volume of waste:
+600 :f---++-P:dU.~!lIbO.,.
Using the following notes on cuts and fills and
a shrinkage factor of 1.25.
(j)
@
@
Find the. mass ordinate at station 20 + 040.
Find the mass ordinate at station 20 + 120.
Find the mass ordinate at station 20 + 180.
STATIONS
20 + 000
20 + 020
20 + 040
20+ 060
20 + 080
20 + 100
20 + 120
20 + 140
20 + 160
20 + 180
VOLUMES
CUHm3)
FILLim 3)
60
70
30
110
50
50
40
60
20
30
solution:
CD Mass ordinate at station 20 + 040:
+2oo~--------.f--~=-",=--">'"
0+000-:-{---------\-..:..'0+5oo
VOLUMES
STATIONS
CUT
-80
CORRECTED FILL
(m'l
-130
Volume of waste = 200 + 80
Volume of waste = 280 m3
@
Volume of borrow:
Volume of borrow = 200 + 130
Volume of borrow = 330 m3
@
Overhaul volume:
Overhaul volume = 600 - 200
Overhaul volume =400 m3
20 + 000
20 + 020
20 + 040
20 + 060
20 + 080
20 + 100
20 + 120
20 + 140
20 + 160
20 + 180
,
110
50
50
20
30
MASS
ORDINATES
m'l
1.25 60
1.25 70
1.25 30
=- 75
=- 87.5
=- 37.5
+ 110
+ 80
+ 50
1.2540 =-50
1.25(60\ =- 75
-+ 20
-+ 30
20 + 000
20 + 020
20 + 040
20 + 060
20 + 080
20 + 100
20 + 120
20 + 140
20 + 160
20 + 180
Mass ordinate at station 20 + 040 =- 200
I
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S-430-B
EARTHWORKS
@
Mass ordinate at station 20 + 120 = • 10
@
Mass ordinate at station 20 + 180: =·35
@
2+040
The grading works of a proposed National
Road shows the following data of an
earthworks:
Borrow
_h1_=~
208,03 300
h1 =47,85
Free haul distance = 50 m,
Cost of borrow = P5 per cu,m,
Cost of haul = P0.25 per meter station
Stationing of one limit of Free Haul
= 2 + 763,12
Stationing of one limit of Economical Haul
= 2 + 948,03
~=~
23,12 300
h2 = 5,32
~=~
241.97 300
h3 =41,13
~=~
Assume the ground surface has auniform
slope from cut to fill.
STATION
2 + 440
CUT 1m2)
51 m2
2 + 740
0
3 + 040
Overhaul volume:
26,88 300
h4 =4,57
AREA
FILL 1m2)
Overhaul volume = (h 3 + h4 ) (215,09)
2
Overhaul volume = (41,13 + 4,57) (215 ,09)
2
Overhaul volume = 4915 ~3
Check:
(h + h )
Vol. = _2_ _
1 (184,91)
Balancing
Point
69 m2
CD Compute the length of economical haul.
Compute the overhaul volume,
@ Compute the volume of borrow,
® Compute the volume of waste,
@
2
Vol. = (5.32 + 47.85) (184,91)
2
Vol. = 4915m 3
Solution:
CD Limit'of economical haul:
LEH = Cb 20 + FHD
Ch
LEH = 5(20) + 50
0.25
LEH = 450m.
@
Volume of borrow:
Vol. of borrow = (47,85 +69) (91 ,97)
2
Vol. of borrow = 5373.35 m3
® Volume of waste:
Vol. of waste = (51 + 41.13) (58,03)
2
Vol. of waste = 2673.15 m3
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TRANSPORTATION ENGINEERING
TRAFFIC ENGINEERING
3.
Highway Safety and
Accident Analysis
1.
4.
A (100;000;000) .
R~ ADT x N x 365 x L
It; = sum of all time observations
n = no. of vehicles
d = length of a segment of the road
}Js = sp'ace mean speed
R = the accident rate for 100 million
vehicle miles
A = the number of accidents during period
of analysis
ADT = average daily traffic
N = time period in years
L = length of segment in miles
5.
2.
d = length of a segment of the road
ti = time of observation
n =no. of vehicles
. '" A (1,000,000)
RADTxNx 365
6.
R =t'Je accident rate for one million
entering vehicles
ADT = the average daily traffic enteririg the
intersection from all legs
N =time period in years
q= KIJ$
q =rate of flow in vehicles/hour
K =density in vehicIes/hourlmile
IJs = space mean speed
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5-432
YRI.SPURllnH _ _• •
7.
10.
N(l••. ~fV#l:iC~$""knt
=.VokC!ftratliCm~U(
·.·~V~.~ot
.... 1UJtJtr
Average density = no. rX vthiCJts per kml
Spacing of vehicles'"
1~
Jit =time mean speed
it
.ve. ens y
Wi = sum of a" spot speeds (kph)
Note: 1km= 1000m.
n =no. of vehicles
8.
11.
S = ave. center to center spacing. of CIJfS
in meters
V = ave. speed of cars inmeters
t =reaction time in seconds
L = Length of one car in meters
C - 1000 (V)
-
S
L 1- =sum of the reciprocal of spot speeds
fJ1
C = Capacity of a single lane in
n =no. of vehicles
vehicles/hour
fJs = space mean speed
9.
12.
Ht =time headway in sec.
C " capacity in sec.
C __10OOV
- S
. V" average velocity in kptr
S =spacing between cars
SO' Vt+ L
t =reaction time in sec.
L =length of one car in meters
K =density of traffic in vehicles/km
q = flow of traffic in vehicles/hr
fJs =space mean speed in kph
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5-433
TRANSPORTATION ENGINEERING
13.
[)at~()natt!lfficaccidentreg6r#dona~~r
irytersectlQnlarthep~$t5y~ars~~~~n
a%liclelllr~t~()fM69p£(rn!mRnentfmn~
vehlclrS(J@MV)·lftneaveragj·ditlty.traffjc
entefing.theinters~!¢n·.IS •. ~Q4;.Md •. ~~e .tplflll
nUlTlbE!r.()fac~idents9urifl9tl'le$YEl(Jrp~riQd.
Solution:
K = density of traffic in vehicles/km
R= A(1000000)
R = sum of vehicle lengths
1 length of roadway section
.
14.
ADT (N)(365)
4160 = A (1000000)
504(5)(365)
A= 3826 (number ofaccidents)
Data ona jraffi£accld$Olreeordegona.¢e(tain
IntElrSe911onf()rJhepast4Ye~r,sN~'~(Jn
~ccjdentrateofP~200p~rflljll!on.enterln9
vehicles•• <ARMYJ. • • • • If•• th.e.·.tolal•. numberj)f·
aC9ident~.ls8Qf, • find • the(JyerCl~e • <:I~ny .fI(lf!iS
entering • •lhe • • jnlersecUon • durlng··the·4 .• y~ar
as2 = variance about the space mean speed
III = tim€: meas speed
Ils = space mean speed
15.
P.H.F. = peak hour factof
peliOd.
..
.
Solution: .
R =A (1,000,000)
ADT(365) N
9200 = 802(1 ,000,(00)
AOT (365) (4)
ADT= 59.71
The•• accjdent.·rate • ata • shaq>. highway••cUTVe
was•• 124D per.·millionpassing·vehicles.·.· .l~the
liist•• ~.years, • there•. had•• been.. 2607.accidents.
What was the averagedailytrafflc?
P.HF
..
= sum of flow rate in one hour
60
max. peak flow rate x 5
Solution:
_A(1000000)
R - ADT (365) N
1240 = 2607 (10000001
ADT(365)
ADT= 1152
5-434
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TRANSPORTATION ENGINEERING
Find lheaceldfinl rate at a road intersection per
mlllkm entering vehiCles If the average datly
lrafficis 348 and 1742 accidents had occurr~d
j r. tt,· last 4 years.
.Q/ution:
A (1000000)
R::: ADT (365) N
_l1i2 (1OQooOO)
R - 348 (365)(4)
R= 3429
Solution:
..
fatal +:.!..!inc:.I·U:.LIY--,-__
Seventy ratio - fatal + injury + property damage
21 + 293
Severity ratio =21 + 293 + 961
Severity ratio = 0.246
Solution:
.
"
fatal + 0.i!:!!y-__ ~
Seventy raTio =fatal + injury +-property damage
24 + 324
Severity ratio;:: 9T7 + 23+324
Severity ratio =0.275
Solution:
..
Seventy ratio
fatal + injl!!YoL
_
fatal + injury + property damage
13+x+293
0.24863 =13+7+ 293 .;. 961
31501 + 0.24863x =306 +x
0.75137x =901
. x;:: 12
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5-435
TRAISPORTATIOI EIGIIEERIIG
~~ta • ()~·~ffip • I'~~S!!lglhr~Minte~~iol1·
indl!1ate$ • • th~! • ·Y~iCle$.·m9M~ • ata•• $pa!1(t
m~r~I¥lM()f4()fllPh~iefhede!1~lyis22
y~hlcles.p~h9Y~.®rmlle.qptopqtelheffltepf
ftQWlh'll:lbi¢I&$PE!tb®t:.>
"..
.. ,
'..' .
Solution:
q=KJ.ls
q= 22(40)
q = 880 vehicles per hour (rate off/ow)
Solution:
Severity ratio =.
. Injury + Fatal
Injury + Fatal + Prop. damage
_ 318+14+(x+y)
0.26 - 318 + (14 + x +y) + 1006
_ 332 + (x+ y)
0.26 - 1338 + (x + y)
347.88 + 0.26 (x + y) = 332 + (x + y)
0.74 (x + y) = 347.88 - 332
x + ;r = 21.46 say 21
HoW·m~hY.~~j¢l$$P~$lhru • ll•• Cerl~IlPointl~
Compute the tate of flow in vehicles per hour if
the space mean speed is 30 mph and the
dens~y
is 14 vehicles per km.
. .
Solution:
K =14 vehicles per km
J.ls =30 mph
_ 30 (5280)
J.ls - 3.281 (1000)
J.ls = 48.28 kph
q=KJ.ls
q = 14 (48.28)
q = 675.92vehic/eslhour
a..• hi~nW~Y.·.~V~ry • • p~yr • • • jf•• • th~ • ·.#W~itY • • • is
~.~~~w!~,.~rQ.~p~~ 'JT1~~Il."~flllf#J • • iS
.•
Solution:
q=K~s
_
q- ~s
:>
50000 (3.28)
5280
~=31.06mph
q=K~s
q = 48 (31.06)
q= 1490
The rate of flow at a point in lhe highway is
1200 vehicles per hour. Find the space mean
speed if the density is 25 vehicles per mj(e.
Solution:
q = K fJs
1200 =25 fJs
fJs = 48 mph
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5-436
TRANSPORTAnOI ENGINEERING
t.~,.
Solution:
tMav~9~ • $PE!~~~ • efv~hiote$ • • i~• a•• sl!l9!e
.~jghW~y.i$§Q.m;¢l!~f~rt9PEmler· • • J'tl~··YQl~m~
{)flt!1ffj~ • ~ • QO()•• v~h[c/e:sp~r.h()ur. . •. Delel1Tl~e
1b¢W~~g~ • s~q.Qt~e • cars.• u~ll~l~i$lare
illkPO·/
... .....
Solution:
800
No. of vehicles per hour = 40
No.
of vehicles
per hour = 20 (density)
. 0 f ve h'Ie/es = 20
1000
SpaCing
Spacing of vehicles = 50 m.
Density = 100(J
50
Density = 20
600
Speed of car = 20
Speed of car = 30 kph
Delerrnine.theltPproPlial~.$pa¢log.9t\lehlcl~s
9~llt~r.toc~nterin • ~·.cerlaill.l<ln~.if.the:ilvefflge
speed·Bflhe•• C?rs • USmS.that>p?rtICUlarlaneis
Solution:
1000
10kpr~ndtO~'loIUrneoftraffic)~890
vehiclesPerh(lUt
Solut/cm:
80 = Density
.
1000
Denslty=80
Density = 12.5
No. of vehicles perkm = 8~g
No. of vehicles per km =20 (average density)
.
1000
Vol. offraffic = 12.5 (50)
Spacing of vehicles = 20
Vol. of traffic = 625
Spacing of vehicles = 50 m. center to center
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TRANSPORTATION ENGINEERING
® Capacity of single lane in vehicles/hr:
Compute
average
the
speed inkphthat a
passenger car should travel hfacertaiil
,treeWq¥ if the spaclngofl~e cars moVing in
the same lane is 40 m, center 10 center.
Volume of traffic at this instant is ,2000
vehicles per hour.
60000
C= 15.87
C = 3781 vehicleslhr
® Average density in vehicles/km:
Vel = Vehicles/hr = km
. Vehicles/km fr
3781
Vel. = Density
Solution:
Density = 63 vehicles/hr.•
S . f h' I
1000
pacmg 0 ve Ie es = No. of vehicleslkm
40 =
1000
No. of vehicles/km
.
1000
No. of vehicles per km = 40
No. of vehicles per km = 25
~ I
.ty vehicles/hr
e OCI = vehicles/km
Velocity = km/hr
Velocity =2000
25
Velocity = 80 kph
. ··In•• an•• qbservaijon.p(}st•• sho~s • thaf$V~h1Cles
P~s •. through.the•• postal.,nlitryal$Of8••
9•• • sec, • • •10•..•.sec.•••·•. 11 • seP • • ~od • •·13 • • • sec.
r~~p~jv~ly·ThespeedS(lf ltle ...eblcle~wre
80kph,•• 1~kph.·.70' kpb,.~(}·kp~ • anli50.kph
resp'ectiveJy.
. ."
sec,
@i()ornpute.thetirneW='lfj.sp~d,.............i.
@.• • cgrnputEl•• I~~ • $®q~ • • m~n • • $pe~if.tM.
• 91~ta~~~.1~lbY.!Wl·.¥eri¢lflsjsf5Pm·
@ AtthedenSjt¥of~affi9.1s • 2(lNl~niel~§~f
lffili•• 9PrnPlll~ • • lh~ • f~tfl9f.1I9W.9ftfaffjgm·
...
... .
vehipl~hdur:
··th~ • $p~a()fa.car.movingon • asiogl~Jao~.ts
60 kph/lfthelengthoflhecari$4.2in.an~
thfl·v!lltlepfthe~tinleis(H.~,
(i)•• • ¢9t'i1@!~ • the•• <iVerage.center•• t()centerOf
1,.1r411~1'!11I1~
~·····~I~ep~e~tage . de~itY • 9f•• ~~tfW • iD
Solution:
CD Average center to center spacing of cars:
S=Vt+L
60000
s:: 3600 (O.?) + 4.2
S= 15.87m,
Solution:
CD TIme mean speed:
. 80+76+70+60+50
f.i.t=
5
f.i.t = 67.2 kph
® Space mean speed:
nd
fJ.s=
_
rt
5 (250)
fJ.s - 8 +9 + 10 + 11 + 13
fJ.s = 24.51 mls
_ 24.51 (3600)
p·s 1000
~l" ~
88.24 kph
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5-438
TRANSPORTATION ENGINEERING
@
Rateofflow:
q=Kl-Ls
q = 20 (88.24)
q = 1765 vehicleslhour
·TW/)•• $efs.onltu~ents • ar¢COI~¢I§gtrafllctt1la
~tlW(jS~A~og~.·A;lOd • 13l?t.• ~ • N9~~~Y2OQm·
at>art·•• • ()t)se~tiQn~tp.··~~PW!l~lltSvehiCl~
P~~$ • ·II'I<lt.$~9~9r • m
• •I~t~l\'al~Rf~o18·.~~q,
9p9.5e¢.•• 1{1,23~c, .• 11.~~sElc,~n(ll~;64S@;
r~~~ctively .•• lf.II}~·.$~9.of.~.··¥eftfcl~$,~e
8(},72i64'~;:Ind48kPl'lresj:lec;tiv~ly;
..'
GP••·•• cmnMte;thedeoslly.hf.trafflclnvehicles
p~r~R1·<
~)., compu~tfiliurnemearspaedillkph. '.•
@ •C911lP\ile!tlesPacemeailspeedirlkpt"
Solution:
CD Space mean speed:
1:S
I-Ls=nT
Solution:
88+86+83+82
I-Ls =
4(3)
CD Density of traffic:
5
K=-(1OO)
200
K = 25 vehicleslkm.
@
Time mean speed:
80+72+64+56+48
I-Lt=
5
I-Ls = 28.25 m/s
I-Ls = 101.7 kph
@
q =4 (3600)
3.
q =4800 vehicles/hr.
I-Lt= 64 kph
@
Space mean speed:
nd
I-Ls = It
5 (200)
I-Ls =8.18 +9.09 + 10.23 + 11.68 + 13.64
fls
I-Ls
= 18.93 m/s
=18.93 (3600)
1000
f1.s = 68.16 kph
Flow of traffic:
@
Density of traffic:
q=,usK
4800 = 101.7 K
K = 47 vehicles/km
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TRANSPORTATION ENGINEERING
Solution:
"L,d/t n=5
J1t=-n- d=1 km
1 1 1 1
1
96"+"72 +90+102+108
..
5
J1t=
J1t = 0.0108 km/sec.
.
.
J1t =0.0108(3600)
J1t = 39.23 kph
Solution:
nd
Space mean speed J1s = lJ
n = 4 vehicles
. d::one.km
_
4(1)
J1s - 1.6 +1.2 + 1.5 + 1.7
J1s = 0.667 km/min.
J1s = 0.667(60)
J1s=40kph
Solution:
5
J1s = 1
1
1.
1
1
-+-+-+--+34.20 42.40 46.30 41.10 43.40
fls
= 41.05 kph
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S-440
TRANSPORTATION ENGINEERING
Solution:
LS
From tbefoUowingdata of a freeway
surveillance; th~i'l~ are 5ve!'iicles counted for
length of200m. and the fdllowingdiStance "$"
are thedis\ance that each vehicle have
.lravel~dWMnobser\ied
on the two
phcitographslaken 2seconds apart. Compute
theftow Qftraffic if the density of flow Is 25
\iehicles perkro~ Express in kmlhr. '.
a
J1s=(jf
25.6 + 19.8 + 24.2 + 23.6
J1s =
4(3)
J1s = 7.77 mls
_ 7.77(3600)
J1s - 1000
J1s = 27.96 kph
. .' ....•. Vehlc;le . ' .• Oistance"S" (ro.)
< •• ··1
'·24.4 m.·· .
• ·•· •·. ···.·."l>~1~: '
'.' 5·' ' 2 2 . 9 ro.
Solution:
LS
space mean speed J1s,= fIT
24.4 + 18.8 + 24.7 + 26.9 + 22.9
J1s =
5(2)
J1s = 11.77 m/sec.
_ 11.77(3600)
J1s- 1000
J1s = 42.37 kph
Flow of traffic:
q = J1s k
k = density of flow in vehicleslkm
q=42.37{25)
q = 1059 vehicles/hour
Solution:
CD Flow of traffic in vehicles per hour:
q= 5(3600)
4
q=4500
@
Compute the space mean speed from the
followin9 dafa of an observation of four
vehicles with' the corresponding distance that
each have traveled when abservEld on two
photographs taken 3seconds apart.
Vehicle
Distance (meter)
1
25.6 m.
2
3
19.8 m.
24.2 m.
23.6 m.
4
Space mean speed:
LS
f.ls=NT
96 + 98 + 97 + 95 + 94
5(4)
f.ls =
f.ls = 24 m/s
f.ls
=86.4 kph
® Density of traffic in vehicles per km.
q= f.ls K
4500 = 86.4 K
K = 52 vehicleslkm
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TRANSPORTATION ENGINEERING
ii,l;lllili
ill c~ut@b~~~~®~@vi;@~~$/Ilr.•.• • •.
~i• •f.'fllll~
Solution:
G)
Trafflc intensity:
A,
p=f.l
A, =arrival rate
Solution:
G)
A, := 200 vehideslhr.
Density in vehicleslkm.
4
K= 0.200
0.833 =200
f.l
K = 20 vehicleslkm.
f.l =240 vehicleslhr (service rate)
@
Space mean speed:
IS
® Average waiting time:
f.ls=N7
3600
f.l=-t-
_ (24.4) + ~3.6) + (25.2) + (24)
f.ls 4(2)
240 =3600
t
t= 15 sec.
f.ls = 12.15 m/s
_ 12.15 (3600)
f.ls 1000
f.ls = 43.74 kph
@
Flow of traffic:
q=f.ls K
q = 43.74 (20)
q= 875
@
Total delay time:
1
t=f.l-A,
1
t = 240 - 200
t = 0.025 hrs.
t= 1;5 min.
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.5-442
TRANSPORTATION ENGIIIEERING
Vehi¢les~rtivei~fa~topSj~Il • • iltan•• av.erag~
rate¢t.3(J()p~thQur. • • Av~ras~.w~tirlg.~eat
the • $ijlp'•• ~'90i$ • • • tp~q~d~Per • ~HWl~.
~$~~mitl9~()lh.~r1~~ • • ~~~ • • <!~tWll.lr~~~t~.
El:~p(menlicl'lydislfjtj~ted.
' ""
,
illiallIfr~;
Solution:
CD Traffic intensity:
3600
,u=w
Ast~p$lgn~itll~~11edafl~$COltl~tQf~~~A'
·~Il¢ • • Q~ig;J~'J\Y~Jlll~W~l'treYElhi9fEl:$9~~n~
"*t1/l<llJr.~IT@~!~l'lm'ef~e.fat~m~OOw~r
hotlt.Jfth~t$ffiC@ell$lWlSQ:8Q.
'(j)
@
ir
.·'Col'rlPlJt!lfQrtb~.~~rag~.Waltillgti@')
•
$~~ds~Uh¢~tqp~j9tt'>.)
.Whatis •~ • aVElrag~(IE!laytlll1eperv~h19~
.'.', ~1.,mi~·in$tli\ntlry,$e@~(\$, .• ,."."'." • "'•• • • • '••.• • • >•• '.',.,.,••
~. C()mPut('Jmetclt.lld¢laytl/l'lEljfI'SElcQtif:l.~;·'"
Solution:
CD Average waiting time:
'A
,u
p=~
,u = 360 vehicleslhr. (service rate)
p=~
0.80 = 300
~
.u
.u= 375
'A = 300 (arrival rate)
'·t· t'
3600
,
Average
wal mg Ime = 375
p= 300
360
Average waiting time =9.6 sec.
P = 0.833 (traffic intensity)
® Average delay time per vehicle:
'A
m=--,u (.u - 'A)
300
m = 360 (360 - 300)
® Average delay time:
'A
(0=--
,u(,u-'A)
300
(0 = 375 (375 - 300)
m = 0.014 hrs.
(0
= 0,0107 hrs.
m = 0.83 min.
(0
= 38.4 sec.
® Total delay time:
® Total delay time:
t=_1-
t=_1_
,u-'A
,u-'A
1
t= 360- 300
t = 0.017 hrs.
t= 1.02min.
'"
1
t= 375 _300
t= 0,0133 hrs,
t=48sec.
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TRINSPORTlnON ENGINEERING
@
Max. queue (no. of vehicles)
Max. queue = 1580 vehicles
@
Max. queue length:
No. of vehicles perJane = 15:0
No. of vehicles per lane =395
Total length of queue = 395(5)
Total length ofqueue = 1975 m.
Time
Cummulative
Demand
10:00
11:00
12:00
1:00
2:00
3:00
4:00
Time
10:00
11:00
12:00
tOO
2:00
3:00
4:00
4000
7500
10000
12000
14000
16000
18000
Cummulative
Capacity
2960
5920
.8880
11840
14800
17760
20720
Demand-Capacity
4000·2960
= 1040
7500 • 5920
= 1580
10000 - 8880 = 1120
12000 - 11840 = 160
14000· 14800 =. 800
16000 -17760 =-1760
18000·20720 =·2720
Max. queue occurs at 11:00 A.M
Solution:
CD .Volume of traffic across the bridge:
V=2900-10C
C= 50+ 0.5 V
V= 2900 - 10 (50 + 0.5 \I)
V=29oo·500-5V
6 V=2400
V= 400 vehicles/hour
® Volume of traffic across the bridge if 25
centavos is added to the toll:
C=50+0.5 V
C=50 +0.5 (400)
C= 250 cents
S-444
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TUNSPORTAnON ENGINEERING
NewC=250+25
New C=275 cents
h1 =1800 (0.75)
h1 =1350
~ =3600 (0.75)
~ =2700
V=29OO-10C
V= 2900 -10 (275)
V= 150 vehicles/hour
@
Vol. of traffic if C = 50 + 0.20 V:
V=29OO-10C
V= 2900 ·10 (50 + 0.20 V)
V=2900-500-2 V
3 V=2400
V = 800 vehicles/hour
Length ofqueue =2700 -1350
Length of queue = 1350 cars
@
Time to dissipate the back up:
1350 + 6000 t =2700 + 3600 t
2400 t =1350
t =0.5625 hrs.
t = 33.75 min.
Total time to dissipate
=45 + 33.75
= 78.75 minutes after the accident
@
Solution:
CD Length of queue:
Average delay per vehicle:
Total delay = area of shaded section
t·
1350 (0.75) 1350 (0.5625)
,oa eay lme =
2
+
2
Ttl d I
Total delay time =885.94 vehicle-hour
'cI 885.94
Average deayperve
I
hI e= 4725
Average delay per vehicle =0.1875 hrs.
Average delay per vehicle = 11.25 minutes
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TRANSPORTADON ENGINEERING
@
It.tl'uck~Sl$$r12:90rioon~SOk#10rl
InE!nQnbbOlJod.l'Qiliiltllbli.prq\lifl~.ofC@u,
P9roPIe@Y.b~f:kil)g.tI1<lffJi9~way .• • ForttlflatelY··
Time queue will dissipate is 1:24.9 PM
·ttm•• l@l~~nt • Slt~J$ • N~tpeY~M.anQ~rpa$$,
.• ~e~l1.~r.off·raltlP~r~~OPli~ramp. • •••1"his
we~nsthlltIJ1Pst'{eniclesYlills~~e
Time the queue dissipate:
3600 x + 650 = 1550 (x + 1)
2050 x = 1500 - 650
x =0.4146 hrs.
x= 24.9 min.
@
Average delay per vehicle:
.~~~t~~6d~~t~I~~ .•Ib"at&~I·
··ttm·~ttiurofthrougn~bl¢~$ • Si~.Il~ff~r
oflJslngthllse.~rrlPst9~o.~rQlJnt1.tll~ifJcJde6t
sit#·Afferlh~tl1.lck~./l'Ii~h~p,f/1~ramp
• ~pa£itieS.~r~ • g9y~m~Y~~<lp~jgn.~t.lh.~
enq~ft\'lEl •. Pff·rampllfl~lh~lltio~ty • given.·.tlJ
qr9$~.lraffjc,wttlqlt.did.l)ot·.h~Ve.a • ~op.·$lfln;
J"h~rnmpsselVi~@t~.f9I;.dEllCltffitl~tfaffj(;~
~pml(lmateM~50vph.Ate~¢UY11:QQeM
thEl • hi~~W'aY'/tClSrepP#nedt()thr9ugh • trarl¢
w~llcapacitYof36()O@I'l.The flOW rate
ltlls6rrte·onhedaYl$.1~$QvPh.
.
at
Longest vehicle delay = area of shaded section
Area =900 (1) +.900 (0.249)
2
qUe!le(/i$sipatE!1/
@lG@Pl1tetheave~~¢laY~rVehICl¢/ ..
Solution:
2
Area =562.05 veh-hr.
Longest vehicle delay =562.05 veh·hrs
cv)What~s.tl1~~fl$E!$t8~W~~Qeue? .
~N~ppro~jmilJeIY~batlim~#()Elstlle
562.05
·
Average delay per vehIcle = 1935.95
Average delay per vehicle =0.29 hrs.
Average delay per vehicle = 17.42 min/vehicle
CD Longest vehicle queue:
TtieinterSeotioIlQfSD$AandOrtig:a~ A'fflll~e
m~ynot
hiveqtialifJed as a MzardOUs
(jiiverS· perceive it .as
il1~r~c"on.~tmany
unsafE!•. At~llrii~fOb.$ervers s~e~t4(}~(lutsat
the
iJ1terseCflbrt
lnfolTl1atlort
. ..arJdcollected
.... . . !he. following
,
h1 =650 (1)
h1 =650
11;>= 1550(1)
h2 = 1550
Longest vehicle queue =1550·650
Longest vehicle queue = 900 vehicles
G)
@
@
@
~·~taIConflict$,with54 being
of rear.end
. . ..
.
COnflict type. . ..
Average hcitmyapproach vdume
=1205 vehicles.
Total ti.rrte to cOllision (TIC) ~verity
=190 lor the 94 conflicts.· •.
Tatal risk of colUsion(ROC} $llverily .
201 for the 94 conflicts.
.=
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S-446
TRANSPORTlnON ENGINEERING
Solution:
CD Average hourly conflict per thousand
entering vehicles (AHC):
AHC _
Total no. of conflict
Number of OOseNation hours
94
AHC= 40
AHC = 2.35
Solution:
CD No. of crashes prevented 10 yrs. from now:
AHC per thousand entering vehicles = 2.3~J~OO)
AHC per thousand entering vehicles = 1.95
® Total conflict severity (TCS):
TCS = TTC + ROC
TCS =190 + 201
TCS=391
N =(EC) (CRF) Forecast ADT
baseADT
EC =expected no. of crashes over a
specified time
CRF =crash reduction factor
ADT = average daily traffic
N= (11) (0.26) (~)
N = 3.5 per year
® Overall average conflict severity (OACS):
res
OACS = Total conflict
® Total number of crashes prevented on the
3rd year:
OACS- 391
N= (ECi (CRF) (1 + r)0
OACS=4.16
N=(11) (0.26) (1.02}3
N= 3.035
-94
® Fatal benefit after 1year:
Crashes prevented = (Ee) (eR?) (1 + r)0
Crash prevented after one year
The.~~llaPltys~gil'E!e~$taff.~elieY~s.lh~t
in~t<I!lmgstMfl}911S<itapr~ViQ~l\I¥
u~antWll~.iqt~r~ctiop·· Will.fadlJc~pwScn~.
(~C9klent$L~¥?§~r~~nt.lOt~~bll~yei:lr
(lastwar).·~re¥fflre.t1·.right<ingle.oOllisiQl'lllt
the•• lhWseqti®.Wliose appraacbyql~me.W~
·3273vel'liCle~@rClaY, • •.·.Ten•• year$ftQIl1.{l()W,
theapp,.6aChaveragepaUytrafflcJADT)is
forecasted to be 4000 vehicles per day.
= 11 (0.26}(1.02}1
= 2.91 crashes
Fatal benefit =2.917 (0.005) (2854500)
Fatal benefit = P41,633.88
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TRANSPORTIDON ENGINEERING
® Response time if he was driving at speed
limit:
Distance the car traveled from the first
wamingsign
=300-90
=210m.
A.•• ~~l3g~~m~tM~ • • t!llm~~ijlt~M~lirIg~t
~.~lwfl1r~;£~~~~II~li~
·.~~~~.jlo~~~~~~~~~~~~~J..·
.300<TiLbefooHhebaftlcildealldlheseoond
. ~'9n~~1~9m.~~f9re • th~.~rtlAA~;<TI1~·
·.il~@m~~f@n~b~l@".PElgi~99m{~~lI1e
.~mM@,.t@~t~I~9.tol~~~~~n~M~,·
··~I~~W~~tLm~~.~~~~'~~~~~.·til~~
Q.~>.<
Speed limit = ~6:
•
.
.·m•• • ·~{M·@Iqpty9t!ItlAA9kW§l$.49.mh,.Wfl~t
·• • · · • ·•• w~t@mjjj~IVlll(l¢ily9¢fO~.~.·~Qm?9f
(j)•• • ~~~I~~ld • ~ve·b~n • •mij••t~5~n~e • tiM~ •
. . .•• ··.#i:lI1"ltl'1~mstW~mlrJ!l~!96Jotti~ir\~I!l@W
.•••·.• ·~r~.jl~~~~' •·.~• ~j.00~ •~rt~0~ • jl·~0.·
Speed limit =24.44 m/s
210
Response time = 24.44
Response time = 8.59 seconds
@
Response time to th~second sign:
speed = 25.56 m/s
90
f= 25.56
f= 3.52 sec.
··@j.·•• ·Pm~®j~tner#PoO$E!«IlWWtli#·~MI1d·
Wijilling~@i.<
. ... .. ..... . ..
Solution:
CD Initial velocity:
VI
Arn~rj$Wmoij~l®g~JO<;alrQa~lrl~~kp~(
V2
~I---90m---.--j
. ""'1"'' ' '
Skid marks
started here
vi = V,2 - 29 (Ji) S
4()()()()
V2 = 3600
V2 = 11.11 m/s
(11.11)2 = V,2. 2(9.81)(O.30){90)
V, = 25.56 m/s
V - 25.56 (3600)
11000
V, = 92kph
Illtlr.
.~~• m9vln.9at~~ptlW~~r.themanfif$!.~~~*
~~$~~~1~~Alffl~~~}W~$?49m.frOmJh~
~A~~ri~i~~~~>~~~ar~~C~I~Otirne~f
• • • • • • O.6$~ndfkh(lloYf.fr?'Tl.lb~.i::~$jnQ.'oVjU
l'1~tffl'M1e~h~~eQl~M<la9cel~rat~?>
.~ • • • lf•• tl)e.~t~ari • acceler~te.al.therateof
.•••.•·•·• ~,phV$~.pU~~Il$ . ~.0)~x:.~PE!~d.of140.~pb;
. ···hpWf<ii$tWIlIJI®gplog'l;'heniUeaCh~
th~q()~SIQSf<>
• .• •. . .
~.·.·HOWm\.lCh.timedjd·hebE!attbt1!·.trall1?
5-448
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1IAISPORTInOI EIGIIEERII.
Solution:
CD Distance from crossing when he begins to
accelerate:
Velocity of car = 8~:
Velocity ofcar = 24.44 mls
D= 240 • 24.44 (0.6)
D=225.34m.
@
Velocity of car when it reaches the
crossing:
V,2 + 2aS
- 140000
V,- 3600
vi·
V, =38.89 mts
vi =(38.89f + 2 (8.5X225.34)
V2 =73.10mls
V2 = 263.2 kph
® Time he beat the train:
V2= V, +at
73.10 = 38.89 + 8.5 t
t= 4.02 sec.
Velocily of train = ~:
Velocity of train = 22.22 nVs
t= 150
<,1).C¢mPAtedh~t~l@mbef;;ofvehiC1e$.;in
=.11t• •~I~(~
Solution:
CD Total vehicles in queue:
. 22.22
Total vehicles in queue = 36 + 32 + 34 + 34
t = 6.75 sec.
Total vehicles in queue = 136
Time he beat the train =6.75 • 4.02
Time he beat the train = 2.73 sec.
@
Total delay:
Total delay = 136 (15)
Total delay = 2040"
Total delay = 34 min.
;lllli~!
ass~mpliprj$mllt~phyehicl~CPUll~~ditl
lhiswaYYlillY{alt~~~rl)~l)d~.Tli~J(*~1
appro"cov01UmeotV~bi~le~d4rjn~the
1;0 min. period was4t·;;·,·.,.,. ';" .,.;.,;;
® Average stopped delay:
2040
Average stopped delay = 41
Average stopped delay = 49.8 seclvehicle
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8-449
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TRANSPORTAnON ENGINEERING
Vehicles entering the queue after 30 min.
=900·450
= 450 vehicles
AIl.incid~t.reSpClns~.pr-oQ-ram.AAlled.e~9Fis
·'$tatiA~ed.aIQn9 • the.·frfleWaY.duri~g • pea~.traftic
~Qurs, •.• • naffic • i~.l@ni~(lredJJy.®t~(l% • • ~q
fh$•• IP9ati(ln~t·(lf • ~~ • • jtlqkte~t .•.tba.t•• Cil~$$$
@
Reduction in delay for cars on the freeway:
II
ttmt•.
• llJ~• 1~p.1Q • lltl~.IlP.·ls.Kn(l~~ • wilhI6•• ~.fe~
·.mil1li~es, • • .• •,.raffie•• b~~s • up.qyic~I~ • \l4t•• the
!
600011
=I125 I
·ERve.·.vetdde~as.abl~.tq • r~~Clt·the·s~m~ • qt
tt1ElIl)cid~nLisl~mll~'l!lM)n15~~lfj#()n~1
• mTriUtes,·•• isable•• Ul•• pl.lStl~jS~b1(ld • c~t(l.·~~El
sb()uldef•• inCfe~singttW.~OWfr()Ill • 1f3()()yptlJl'.
3600Vp11. ••• Bec:aIJ~.lhi$nEl~sEl~k:er<l~js
·the•• sa·~·.as • ttl~.d~ermmi~tiCattlVaI • rate·Jh~
sh°ckwav€.. ~ill • n(lt.lTI()ye·.tarlblJlr•• bac~ . ••••ThEl
w~.and.amblJlanc.e@ntearl(j~atlhjn~
·fUlly•. jn.30·miQutes,•• • s()•• m<lt•• ~~e.·.ff~~way.·.ls
~m.!tl
@
upforftlllflow,
....
13600(0.5"1)
>Ih.-----ll--t
1350
900
Al""''''-__~+!--
h,=lf37 )
--o±-,--.~8 5 L-t
05
i--I----1l.25--0.4375---·
----{).6875----
.
Determine thena.of vehicles enleringlhe
qUeLle after 30 min,
=2475
6000 t+ 1350 = 3600 (0.5 + t)
2400t=450
t = 0.1875 hrs.
t = 11.25 min.
h = 1800 (0.50 + 0.1875)
h= 1237.5
..
What is the reduction in delay for cars on
the freeway If it has a max. flOw .of
· 6000VPh.
..•
•.....
@Al. P15O(Jper vehlcleMur,what is the
•. value of the time savings? ..... .
@
Reduction of delay with ERUP
= area of shaded section
Solution:
CD No. of vehicles entering the queue after
30 min.
Reduction of delay
_ 2475 (0.6875) 450 (0.25)
2
2
(450 + 1350)(0.25) {1350 + 2475)(0.18751
Vehicles entering
the queue after 30 min.
2
.
2
Reduction of delay = 183.74 vehicle-hours
@
h1 = 1800 (0.5)
h1 =900
h:J = 1800 (0.25)
h3 =450
h-;, =3600 (0.25)
h2 = 900
Value of time savings:
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5-450
TRANSPORTInON ENGINEERING
6000 t + 1350:: 3600 (t + 0.75)
@
2400 t:: 1350
t = 0.5625 hIS.
t :: 33.75 min.
I
-
7ioc!
t I deiay-
Distance the driver can travel in 4 sees.
1350 (0. 7~ 1350 (0.5625)
2
+
2
Total delay:: 886
Savings:: (886 -183.94)(1500)
Savings:: P1,053,090
Clearing the intersection requires going a
distance of D + L + Wbefore the light turns
red.
AgriverlravelingiOhis4.811(SWafaspeed
Iimlt.ot48.kphwa$att~$tEl(!for.wnl'\lng • a.·red
·ligtll.ilfan.i{lte@$-tlim~ti~.18m,.wiQe,
• • TM
dr~rcl~trml~.lfjra~~lo/,.()n.·W.f!grcWrtd~ • 1t:l~~
thEl•• t/'(lfflc .Sigrj~$w~rer()t.s~lpropetly .••• Th~
yetlawlighlwas<mforlhestatidard • •·4
sllC(:H'!ds,··•• 'fhe.SQV<tijvElrs..•reactlgrl. ·time. i$
aSsumed 10 be 1.5 sell. Comfortable
decele1'a6()nis~t~tat~6t3J11r$?
.
g)••••• GolTlPuteth~.·ml~,~I$tMql'ln¥di'!dW$tqp
as soon •asn~ WestheyeHoW traffic
signaL>
®c;ompute th~~I$t~th¢~tlv~r~nMY$t
1
xy = V1 t1 + V1 (4 - t1) +2a (4 - t1f
xy ::
1
13.33 (1.5) + 13.33 (4 • 1.5) + 2(0)(4 - 1.5)2
xy = 53.32
® Length of di/lema zone:
Xc
··.·i.i<>
. •••••• jn••
lI'le••4.sePOrtdstMt•. they!lI~~lig~tVol~s
m.
--
-_ .. - .
. Qnat acon$laorSJll:!l!dof4&kph.. . .
@
l"lctwlongiS!vediletrllua zone .•. this
il1lerseplionapP@lcfi? . .
(D
Min. distance:
V2 =V1 -at
If the driver was any closer to the
intersection than the distance 49.61 rn. just
calculated, he must be able to drive
W + L = 18 + 4.8 ::: 22.8 m. farther than
49.61 to clear the intersection.
Solution:
48000
V1 :: 3600
V1 = 13.33 rnls
0= 13.33 - 3 t2
t2 = 4.44 sec. time to stop
Xc = 49.61 + 22.8
Xc =72.41 rn.
Xy < Xc =53.32 <. 72.41 a dillema zone exists.
1
0:: V1 t1 + V1 t2 - 2a t2
D = 13.33 (1,5) + 1333 (4.44) - ~ (3)(4.44)2
D= 49.61 m.
Length of dillema zone =72.41 - 53.32
Length of dillema zone = 19.09 m.
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TRANSPORTAnON ENGINEERING
• clear
.I
th·
time ,0
e mtersect'Ion =72.41
13.33
time to clear the intersection =5.43 sec.
A man isdrivlng at Ii speed of 48kph and is
approac;hlng .Em intersecuoo v.tIietl.1$ .1$ Ill,
wide, The length of hisqlr is 4JJ 11'\, .Jheyellow fight . was . on for. the standard
4 seconds. ··lflhe drivers reaction tfme· is
1.5 sec, and he deCelerates al arate of 3m/&'
as soon as he sees the yellow light s~nal
wason.. .
CD Compute the min. slopping distance. ...
Deten-Nne the length of time. that ftlered
light Wi!lS o~ .tor the VehiCletG, etear the
jntersec1IOi1~ •.... . ..
. . ...•....•..
® ltall r~clearance Interval iS2 sec; long, .
. determine .lhe spee\l at which a vehicle
can clear the intersection.·
.
. .
@
Solution:
CD Min. stopping distance:
1
D:: V, t, + V2 tr "2 a ti
Ttme the red light was on =5.43 - 4
Ttme the red light was on :: 1.43 sees.
@
Speed at which a vehicle can clear the
intersection:
V2 =V,- at2
0= V, - (3) t2
_Yi
t2 - 3
Max. stopping distance = V, t, + V, t2 • ~ a ti
1 (3) (~)
Xs = Vd1.5) + ~
3 -2
9
Xs -- 1.5 V, +~.!~
3 -2 3
Xs --1.5 V,
+~
6
V2 :: V,· a t2
V - 48000
, - 3600
Distance to clear the intersection =Xs + 18 + 4.8
Distance to clear the intersection = Xs + 22.8
V, :: 13.33m/s
0:: 13.33 - 3 (t2)
t2 :: 4.44 sec.
D:: 13.33 (1.5) + 13.33 (4.44)
Xc =1.5 V, +~
6 + 22.8
-i
(3)(4.44j2 .
when tred =2 sees. (all red clearance intervaQ
Xc =(ty + tred) V,
Xc=(4 +2} VI
D=49.61 m.
@
Length of time the red light was orr for the
vehicle to clear the intersection:
t :: time to clear the intersection
- time of yellow light
Distance to clear the intersection
=49.61 + 18 +4.8
=72.41 m.
.
V2
6 VI =1.5 V, +
+ 22.8
T
. V2
t : 4.5 V, - 22.8
V12 - 27 VI + 136.8 =0
V, = 26.98 mls
V - 26.98 (3600)
,1000
VI :: 97.14 kph
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S-452
TRANSPORTATION ENGINEERING
QonlP~tet~p~.khl!urfact()r(PHF}ifttle
~()~rtYVf,lI~roe()ftrafllCi$1aOOvehiel~p~r
h6utand•.tpe.highest5min.'1OllJlilll.is25ft
rhe peakhouffa(){Qr.(PfiF} =O.OO.an(Llhe
V<llume•• dflraffiel$•• Z400·.vebicteSlhr.•·.·.V\lha!1$
fhehigliest() mill. VOluftlEl offtaffic?
Solution:
Solution:
For one hour = 60 min.
~
= 12 (there are 12-5 min in one hour)
1800
PHF = 250(12)
0.60 =.1.400
x(6~)
x=400
PHF= 0.60
the.njgh~I.1(lmlrl~t~ • ¥()'Wn~9ttrafficj$ • §OO
vehj~les .•••••lf.tflEl••peCil<hQlJrtaptt.)rJ~.q.60,'Wh<l1
i$llJeVQIl.JmeiriYl#1i(;lesl!lpur? .
Solution:
060=~
.
500 (~~)
{j)DelermjneJhepeaK.h(jQtv~urne·• • • ·
~• • • qeterrnlflE!lhepeak·hourfsct()1"••••.••.•.•••••• <>.>
®.·..OIfueappmaCh.
Det¢l'lllloefbede$~n .• hou~y\l~lume(DHV)
. .
Vol. = 1800
Solution:
CD Peak hour volume:
Vol. := 375 t 380 t 412 t 390
Vol. = 1557
@
Peak hour factor:
Vol. !iurin~~~a~k:.:.:ho~u:...r
PHF =
What is the peak hour fa·ctor (PHF} If the
volume of the traffic ls1500 vehicles per hour
and the highest 5 minute volume is 210.
PHF= 1557_
60 (412\
15 - I
Solution:
PHF= 0.945
PHF=_1fJJO
210
(W)
PHF= 0.595
_
~~ x (Vol. during peak 15 min. within peak hour)
@
Design hourtyvolume:
DHV = Peak-hour Vo/.
Peak-hour factor
DHV= 1557
0.945
DHV= 1648
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5-453
TRANSPORTAnON ENGINEERING
Solution:
V2
The•• peak•• hoUf.fr#°r.f()r·!llraflic.~urlng • rtlJ3h
.hour.l$eqU~ltQ • • 9·6Q•• Ylilh•• ahtg~~t • • 5•• rnln.
voN!lffi.·.()f•• 259.·.vehicl~ .•••••••• 11J~ • . spa(.;e•• mean
$Pf$:I.ofthe.traffl~·i!l90.kPh.
S=2gf
V= 70000
3600
V= 19.44
(19.44f
48 =2 (9.81)(
f=0.40
Solution:
CD Flow of traffic:
PHF = Flow of traffic
traffic (~~)
0.60 = Flow Of:;;ffic
250 (15)
q = 1800 vehicles/hour
Find the length of the skid mark if the average
skid resistance is 0:15 and the velocity oHM
car When the brakes were applied was 40 kph.
Solution:
V= 40000
3600
V= 11.11 mls .
V2
S=2g f
® Density of traffic in vehicles per km.
?=I-lsK
1800=90K
K = 20 vehicles/km.
® Spacing of vehicles:
.
1000
Spacmg=:20
Spacing = 50 m. center to center
_ . (11.11)2
S - 2 (9.81) (0.15)
S =42m.
Applylnlffull brakes at a speed of 60 kph, the
car traveled 40 m. until it stopped. Determine
1heaverage skid resistance.
Solution:
112
S=-
A car traveling 70 kph requires 48· in. to stop
after the brakes have been applied. What
average coefficient of friction was developed
between the tires Rnd the pavement?
2g f
60000
V= 3600
V = 16.67 mls
40 =i16.67)~
2(9.81)f
f= 0.35
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5-454
TRANSPORTATION ENGINEERIIG
Solution:
CD Speed of the car in kph:
Fultbtak~sW~~pPljeaWttenthe
V2
cats speed
S=2g(f+G)
wa$••~• l(p~.lfth~·aVElrage·$~ld reslslallcei$
V2
P.Z4,.flri~l$lellQtbOfl$slddll1afk.
. 48 = 2 (9.81) (0.35 +0.02)
V= 18.67 mls
V= 18.67 (3600)
1000
V= 67.21 kph
Solution:
V2
s=2g f
v- 60000
- 3600
V= 16.67 mls
@
V2
S= 29 (f- G)
._ (16.67f
S - 2 (9.81}(024)
S=59m.
Ttlebraktl$;;rgsuddeIllY~Jlplied to
Speed if the road grade is 2% downhill:
V2
48 = 2(9.81) (0.35 - 0.02)
V= 17.63 mls
V= 17.63 (3600)
1000
V= 63.46 kph
stopaf;af
Is.mnnl~at48.lq1t) .• • • Q~~ttlllll~ • the .bra~lrl~
dj:;t~m;~ • ifJ:tt~.·cp~ff/fjf.fticUol1.·belwee11 • the
lires~lhe~sul'f<lceisO.~$.
..
.
Solution:
V2
S= 29 f
48000
V= 3600
V= 13.33
_ (13.33f
S- 2 (9.81) (0.35)
S= 25.9m.
A bus is running at a speed of 50 mph downhiH
on a grade of • 2%. The cooff. of friction
between the road surface and the tire is 0.34.
After suddenly appiyingfullbrakes, how far
will the bus travel until ~ stops?
Solution:
CD V= 50 mph
V= 50 (5280)
3.28 (3600)
V= 22.36 mls
V2
S = 29 (f- G)
--~~
A car runs 4S m. from Ihe the brakes were
suddenly applied until it stDpped. The road
S- 2 (9.81) (0.34 - 0.02)
S= 79.7m.
grade is 2% up hill and the cooff. of friction
between the fires and the road surface Is 0.35,
CD What was Ihe speed of the car in kph, just
before the application of the brakes.
® Compute the speed if the road grade is 2%
downhHl.
@
When it is moving uphill:
@~
S == 2(9.81) (Q34 +0.02)
S= 70.8m.
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TRANSPORTATION ENGINEERING
a••••••
tM'I~;
Solution:
····················/i·••••••••••••••••••• {} •••••
;> . •.• • •
Solution:
V= 40(5280)
3.28 (3600)
V= 17.89 m1s
V= 80000
3600
V= 22.22 m/s
If?
S=2g(f-G)
If?
s = 2g (f- G)
_
(17.89f
70 - 2(9.81) (f- 0.05)
f - 0.05 = 0.23
f= 0.28
_ _Q?22)2
. S - 2 (9.81) (=0.::1..
3 .-0-.04-)
S= 96.8m.
A~rist(aVelingat40ITlphonl:lnuphnlg~de
.1j•• ~W~; • ·.lt·.ttle•• I:)l:~~·.i'lr~sQ~(loolY~pl~d:'.W
• Willtmv~.$~.m.16~n~9P;~~~ffl~~;C#·
6ic®~.~~lWelltttffil.·~r~aM~toad~~,*f
Solution:
_ 40 (5280)
V- 3.28 (3600)
V= 17.89 m/s
.
ilt!lfli
Solution:
V= Va tat
V=Va +
fo
I
ctdt
100000
Va = 3600 =27.78 mls
If?
s = 2g (f+ G)
when V =0 (stops)
_--.J1I~
56 - 2(9.81) (f + 0.05)
0= 27.78 + (-T
f= 0.2if
c(2)
27.78=?-'-~
t = 4.71 sec..
S-456
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TRANSPORTAnON EIGINEERIIG
'1'hEl~rof.a~ttraveti?g70kpllrEl(@res.4a··
.met~fOM9P~ft~rthe.f)rak~$ • h~vebetm
·~t!Ife~;.>MJhalaN~~e.we~· • m.~llwa~
dev~ll)rmQ • • • b~tw~en ··tM.·.·.• ~Ir:El$ • • •~n~1he
pave/1Wlt···· .
Solution:
Solution:
i;fia~k9;jjti_
v- ~(J()oo
- 3600
V= 11.11 mls
V2
S =29 f
_ill.:.1.!l
8.5 - 2 (9.81) f
f = 0.74 (average skid resistance)
Eft = 0.74(100)
0.85
Eft=: 87.09%
V2
s= 2gf
V= 70000
3600
V= 19.44 mls
_(19.44)2
48 - 2(9.81) f
f= 0.40
Flng••thElt(j~ldis@1ceihata~rtrav~ledftl:lrn
Whel1aJ;;lrlsliav~iQg~t~Pwh,~~~rtHe
bra~~s • ~re§Udq~nIX .• ~pli~d,· • t~~~r~I'sml
tr~V~I.3o • m·.·•• ~f()(e • • It • stpPS·•• )•• IfVl1a;t.i$m~
coeffiCJerrtl)ffriction·t>etweenthll·ijre~aoo.tne·
. .. . ... ... ... ....
.theyUnm·lhedrtv~.~~.th~'.ha~~(d~b~?,*·
·Wl!Is.tt.lY~Hog .~t.7Slit>h.P~t'(;~ijo/l·~m~.i$
~ • ~~ .•a~~th~.~"e(E19~$Kid.resi$ltlrl~isR~,
·ASSllll1~·lIJ~lm~CiJrha$.i:ln.·effl¢iency.()f80%.
toild$9rf~?
Solution:
Solution:
V2
s=·~
29 f
V= 60000
f = 0.60 (80)
f= 0.48
V= 75000
3600
V = 20.83 mls
3600
V=: 16.68 m/s
_.(16.68j2
30'·2 (9.81) f
f=: 0.47
V2
S=Vfof-29 f
'2)
(20.83)2
S =: 20.83 ( of- 2(9.81)(0.48)
=: 87.70m.
.s
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TRINSPORTlnON ENGINEERIIIG
,a,tI\JCK)Vllslra~li@dOWrthinat5()f(ptI·rn~
.b@k~~ • • a(e•• $1l~9~lyapPIIl!d • • an4 • • the.trU~~
stopp®••11l.a.d@llrpe·.Qf32.rik • • lf.· tne .cQeff.• Qf.
frlctlon.belW~~.ll'\etlr~$alld.lhe.road·s~rfage
jsQAtWilatis!hEIgrad~9f1!lef()ad? .
AVellicll!isryKlVingata(s~ofll(}Mtt"kln~.
~ryl~¢nlffl$~~bllv!ngM~()f~f4, • lfth•
c~ffi~Il~.Off1i~tion.j~O.~O,~omP~tfl·.·th~
~raKlng(U$ltin~,<
.
..
Solution:
Solution:
\.R
5=29(f+G)
50000
V= 3600
V= 13.89 mls
(13.89)2
32 - 2 (9.81) (0.4 + G)
G=-O.09
V2
5= 2g(f+G)
80000
V= 3600
. _
=22.22 m/s
(22.22)Z
5 - 2(9.81)(0.3t<l.04)
5= 74.01 m.
Ci)tnpute•• W~tpt~ldi~t~M~lhat • :a•• ear•• ~~
traveledfrOll1tnetimellledriversees.a.hazalrl
Wh~ • he.i~ • • tfav~liog· • at•• ~ • • ~P~d.9t7§.·kPh,
f>~rcepfiCln • ijme··js.2.$~ndSj.ant1.m~.~~e~g~.
··Skld.• reSi$tancei$~ualt()O~@ ..••• As~lIme • the
c;arhasilneffic:1~Qf~Q%,
Solution:
f = 0.60(0.80)
f=0.48
75000
V= 3600 = 20.83 m/s
V2
5= Vt+2gf
?
(20.83)2
S = 20.83(...) + 2(9.81)(0.48)
5= 87.73m.
.
.A•.cat.@veUrygat.80•.l<ptl.re~~res • 4a.l'Il.to••S~bP.
a~~.the.~~es·hW~·.~rt···lWpfleq ••• P¢ler{l)ln~
·tM.sl8Pe.m.@~ro~d.surt~if • t~~.~'{~e.
eoefllclentQffrlCUpnfs{);50:
Solution:
V2
5 =2g(f+G)
80000
V= 3600
V= 22.22 m/s
_
(22.22)2
48 - 2(9.81)(0.50 + G)
0.50 + G = 0.524
G = 0.024
G = 2.4% (upward)
. ....
. ..
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S-458
TRANSPORTlnON ENGINEERING
A•vefliclernQving.~t • 60@IF~loflg.~n.i~cl@!
$lJrfaca.·~ • $tQPpedby•• applyjngbr:al<~$~d
ln~t!~~kill9"djstanl:elV$s30m.lfthe
.M¢fid~nt·.ot • ff,iCtion.hF(},$O'9ompute •. th~
.S1()p~pfthElinCl~dsudace.· .
. .
A vehicle moving a 4{). tcph was stopped lly
and the length of the sk~d
mark was 12.2 m.. If the average skId
resistance of the !eveLpavement is known to
be 0.70, determinetne brake effiCietlCyof the
test vehiCle. ..
.
.
applying brakes
Solution:
v2
Solution:
S=2g(f+G}
v- 6000Q
- 3600
V= 16.67 mls
_ (16.67)2
30 - 2(9.81)(0.5 + G)
G=- 0.0279
G=·2.79%
".----s:---~
_40(1000) _
V_. 3600 - 1.11 m/sec.
V2
S = 2g(f+ G)
_ (11.11)2
12.2 - 2(9.81)(f + 0)
f=0.516
1
Brake efficiency = x 100
At/lKlk.W<ls·.trav~llng • doWnhi!lllt.59kpl1.Tbe
.bfake~ .• <lfe·•• $lld<je(ily•.• ~pplle~~(iglh~.tfyq<
slopped.•• irl.lIQlsfanceof.3g.Jn,lfthecQElff_.of
frietion•• ~lYIeen.lhe.lire$.and.lh~rQadWllfa~
.
0.516
Brake efficiency = 0.70 x 100
Brake efficiency = 73.7%
is.0.40,what.islhe·grade·oftheroad?··••.· .
Solution:
V= 50000
3600
V= 13.89 m/s
V2
s= 29 (f+ G)
_
(13.89)2
32 - 2 (9.81) (0.40 + G)
0.40 + G = 0.317
G =- 0.0927 say· 9.27%
A.lnltkdfiVtlrapproachedahal*d.ata~~d
of58mpn.••••• 'What•• wa$ • lhe(}~tanceltav~l~g.
dUrlns~rcepti(lOreactiontll'llejfm~f'I§V
(percepli()n,identiflcation. emotioll<lnd
VQlition}tlmeJS2.6 SeC.
Solution:
D=Vt
V= 58(5280)
3.28 (3600)
V= 25.93111/5
0= 25.93 (2.6)
.0= 67.43 m.
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S-459
Pdfbooksforum.com
TRANSPORTATION ENGINEERING
D=VI
A.~.~riV(ir.~ppr~~fJI3.~.l'li3z(lrd.~tll~pElEl(!·.()f·
aQ.kfJh' • ~t • islh~Qjsta,*efnl~re<l • @ilil'lg·
. th¢•• p~rt¢Pti(m..fi!l~9W~Ii.lilM • • jf.••• ~b¢.· • el~V •
(f¥!r¢apti/)n, • • ld~UflcMi()tl; "•.• ~tI1@()6.<arld·
.
61 = 1112.8)
V =24.79 m/sec.
V= 21.79 (3.281)(3600)
5280
V=48.7 say 49 mph
MmIM)~r@i$2.4$ec
Solution:
d= VI
d =ilO(1000) (24)
3600
.
d = 53.33 meters
;I.
.
i
. . .
Solution:
AbU$drlvera~proa¢tle$aha~@:l~ta.ll~dl)f
.&O•.·mph•• ~o(f • • travefll•• a~i$tlJ.tlM • ()f.~.~$Ill.
•duting•• • a•• giv~n • • p$r¥ptiqJHetlsti()n>.~ill'le.
Cl)rnPule•• • tb~ • • • dtjV~r$.· • (I'I§V} • • pet~P:ti~ll.
identlficatioo;emoUorriffldyolltionlll'll&'
..
Solution:
0= VI
_ 80(5280)
53.33 - 3600(3.28) I
1= 1.49 sec. (PIEV)
V= 52(5280}
3.28 (3600)
V= 23.25 m/s
s= VI
S =23.25 (2.6)
S= 6O,45m.
Atar~rl¥~r~llw~g~t'I.SpeM
• 9f.65•• mp~
apptq~Cheda·.h"~rQl)l1dtr~~I~d • • 7~~~m,
.!t~I.~mf~~ii~~J~I~~@
Acar driver approaChed ahazard and traveled.
a d~tance of 61 m,durlog thepercep!ion·
reaction time 012.8 s. What was theCal"S
speed of approach iI'l mph? .
Solution:
V= 65 (5280)
3.28 (3600)
V= 29.07 m/s
Solution:
-......
v
S= VI
1= 72.2
29.07
1= 2.5 sec.
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5-460
TUNSPORTAnON ENGINEERING
_ 45 (5280)
v1 - 328 (3600)
}\.carYlcl$·tr~li~·~~Il~~f§pmp~,
• • Tbe
~l'ly~t • ·~~ • ·i;\••·m~.··"'9~~ .• ~9.·% • • ~~ati • ®~.
st~pPe~ • •·~nlf,i~ • ~~ • • ¢@~I~ • • lh~ • • ¢<!( • 10.
~~I&(~~ • Wl~ .• ~t . . 19mt~(:~~ . . • fi"ti.. ttl~
.~i~Jml~.·~:ti.'~~.I;·
Solution:
V1 = 20.12 m/s
vi = V12. 2a S2
0= (20.12)2·2 (7) S2
S2 =28.92
S1 =100·28.92 ·10
S1 =61.08 m.
S1 = V1f
61.08 = 20.12 f
t=3.04sec.
_ 50 (5280)
V1 - 3.28 (3600)
V1 = 22.36 m1s
S1 = V1f
S1 =22.36 (2)
S1 =44.72
vi = V12. 2aS2
0= (22.36)2·2 (10) ~
S2=25m.
x =80·44.72·25
x= 10.28
Solution:
.
While•• qO\liM~IAS • mph•••l@•• dtlV~rMt~.l
r()CIcl.l:llclqK.10P.rJl·.~~Y· • •~.!ijlplil¥lll'@.~@~~
ai1~.tb~.Gat~~l¢r~t~.Ui1jfo/"mty.t.7.tllt$~
.• • lt
1@.~r.$topp~d.1q.m·.ffgrJjtti~ • I'l:l~~ • • I;lI()ck.
What.W~s
dnvel'1'• tM•• tM@I@QM~MQ®.t@~(lf.thl! .
So/utioh:
_ 50(5280)
V1 - 3.28 (3600)
V1 =22.36 m/s
S1 = V1 f
S1 = 22.36 (3)
Sl =67.08m.
82= 110-67.08
S2=42.92 m.
vi = V12. 2a S2
.
o=(22.36)2 - 2a (42.92)
f1 =5.8 nYs 2
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TRANSPORTADON ENGINEERING
Adr1\@"l'lOliced}lroad•• bll)Cj(.WhiletdlVe!lOO.at.·
4O.(llptl,•• H7~PPl~tl)e~r~k~$call$irg~~r.
tp·~cc~l~ate~f\i1Brmlwat.6mrs~; • • • Jtie·.qJr.
~tQPfJ@.·.1Z • ill· • ft(lJ'll.lhEl.rpag·.P1Pc~ .• • • H9w.f~~.
wa$tneb'ockwllenth~driVer $a~IJt~l'$11
A$$~1M·1I~r¢~pllCm~n:tEl()f3sec.<
. .
Solution:
V j =40 mph
---- ---V,
_ 50(5280)
V1 - 3600(3.28)
V1 = 22.36 m/s
81 = V1 t
S1 = 22.36(3)
S1 = 67.08 m.
= V1 2 • 2aS2
(0)2 = (22.36)2 -2(6) ~
82=41.66 m.
vl
Distance from wall upon perception:
S = 67.08 +41.66 + 12
8= 120.74m
Stop
a=-6m/s 2
~{:?:;j1i'l>
_ 40 (5280) _
V, - 3.28 (3600) - 17.89 m/s
51&111\1'1£":
• •4Blllit:
8, = V, t
8, = 17.89 (3)
8 , = 53.67m.
vl = V,2·2a82
0:= (17.89)2·2 (6) 82
82= 26.67 m.
Solution:
Total 8 =8 , + 82 + 12
Total 8 = 53.67 + 26.67 + 12
Total S = 92.4 m.
A.•• driMr.tRW@@•• ~I~p • mph.~~~~ • ~ • W~I@{.~ •.
~tl~h~l~tanC~~heaq·rtl~9dyerapPlie~ltle
•.
bra~~~ • • • tmrnMj~{~ly • . (p~meptl9tlWn~ • • ·f~
3•. ~e999d$) •. <1m:l.Mgjl'l$.sl~Qgth~v~N(;I(!.~t •
8, = V, f
8, = V, (2.2)
vi =V,2 - 2a 82
··I).rnI$~.(d~~r~tin9)· • • • fflpedjSlaM~.m(jm·
o= V,2 - 2 (6.4) 82
was.lhegtr.ll'qmt~~w~II~PQI'l·per~Uon7
82 -12.8
8, +82 +5=112
.the•• slgppiOgpoipf·tomelllall.fs•• t2rtt;.M\y:far·
Solution:
-~
2.2 V, + ~12.8 +5-112
1/I
---break is
applied here
Car s lOp here
28.16 V, + V,2 + 64 = 1433.6
V,2 + 28.16 V, -1369.6 = 0
V, =25.52 mls
V - 25.52 (3.28)(3600)
1. 5280
V, = 57.06 mph
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5·462
T0II5'ORTITION ENGINEERING
o=(aQ2 ±2aS
·a:~I~~r~~ig.II~\S:~~
Solution:
w
v=o
•.
a2 + t2 = 2aS
2S
a=f2
F=ma
fW=~a
9
f=~
9
- 2S
f
-t2g
f2(7)
- (1.4Y (9.81)
f= 0.728
± at
_40 (1000)
0- 3600 • a (1.8)
V2 = V1
a = 6.17 mfsec2
F=ma
W
9
fW=-a
f=5!.
9
f= 6.17
9.81
f= 0.63
02••·..·p,•• m~icfe • rnoV~g·aI6(1l<Phwass~oppl':d
·•• • • •bY.applYiOO • l)fakes•• and.the.lengtt\Qflhe
•. • • §~%W~·22·Zm .• . lf.thf3·d~t~~~ftpm.tr~
• • • • • •PointWl1ere.itstoPs•• to•• the\lehltle.poslt~n
. .··wh~n •.• • ttle•• • ·~liyer: • • jl1iliaUy•• JeJ)ctMv"'Js
~,2.nl;flfld.lheperceplJontime.
@AY911icla••l1l?vihg··at••ah•• il'lmal.·.13~edof§O
·~.~.~~~.~t9f1P.~ • • 'oVittlIQ.3sec,•• {tll'~klng
tOll~) .• • ~117r • Jhe • ~PPHcaWm • • Of·.brake~.
c:qIl'lPl.lte•.•·•th~ • • • aver(lge.~9~fficief1t • • • of
fti¢tlqm~l.()r·skld.resi$tan(:e.··.·
~H~~\S~W~~~~~p~glry1,4$~~.~Yf\lIIY
JaWmmg··•• tM.~n!l~¥ • ·•~fl~th~ • • skjd • • JTlark
·riI~~w~]m··.·.~ffirm!r~m~.~v~~e~l<id
resl$tanceonth~l~el~ve~1lt $i,J~; •...••.•.••...
Solution:
@•• CoJ1lPll~ • thelo.iliLclistance.thalaCarhad
. ·•• • t@y~led • jrwn.·.the.tlmetl1e~rIYer~es.~
. . • •~f~rp~r.~~ .• is.fravelin9~t.aspee9of
~v~t~~fJ • ski(tJElsf~t~l'lCe<isiP.60.·· Brake
. ..•. %l•• ~ph, • • • P£!rce~ton • • tTrlie.·l~@5$t!c.llrd
. eftiCiefll::yis85%.
Solution.:
CD Perception time:
~ph
V2 =V1 - at
0= Vl • at
V1 = at
V22 = V1 2 ± 2aS
• · •••••••.•••••.••..•.•••••..••.•.
60000
3600 (t) + 22.2 = 34.2
t = 0.72 sec.
.
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TRANSPORTIDON ENGIIIEERING
@
Skid resistance:
Solution:
V= 50(5280)
3600 (3.28)
V= 22.36 mlsec.
8=80-10.28 =69.72m.
S1 = 22.36t
V22 = V1 2 - 2a S2
0= (22.36}2 - 2(10) S2
S2=25 m.
V2 =V, ±at
81 =69.72 - 25 = 44.72 m.
60000
0= 3600 • a(3}
S1 = Vt
44.72 =22.36 t
t = 2 sec. (PIEV) time
a =5.56 mlsec2
F=ma
W
/-IN=--a
g
W
p,W=-a
g
a 5.56
/-l=-=g 9.81
#=0.57
@
.J\•• dri\leitl'a~~l1g • at5Qriipfj • ~.~Om,fI"dfJl.~.
W'~lI~~~t1,ffthedrW~r*p!I~~(tl~.bra~~
Total distance the car had traveled:
~ph=16.67 m/s .
liJli~ij[ii
Solution:
V2
S=Vt+29 f
f= 0.60 (0.85)
f= 0.51
_
(16.6?i
S -16.67 (2.5) + 2 (9.81}(0.51)
8= 69.45m.
1----69.72---1--1
10.28
1-----80-----'
_ 5O(5280L
V1 - 3600(3.28}
V1 = 22.36 mls
A,.QnVo/•• tl'ave1ih9<ttsn.rnpn.is6Qfu,•• trorn.a
W~Uah~$(i·JfmedrIM~LliPplje$tb~braR~s
lTtnedlatelYal1d..~g!&.S~"9.tlJe • veh~e<lt
··1Q•• il'Il~c2.(d~k¥atl9Q),flot1Jlj$.{~I~ • • tjme·
W~Efn'lhe .diSlance.• i)f·lhe·•• carfrcimthe .waU
When if stopis10.2$tll. .....
.
81 = V1 t
S1 =22.36(2}
S1 =44.72 m.
S2 =69.72-44.72
82 = 25 m.
V2 2 = V1 2 - 2a 82
o=(22.36}2 - 2 a(25)
a = 10 m1sec2 (deceleration)
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5-464
TRANSPORTIDON ENGINEERING
Solution:
W·
A••~ff'itlfav~liljg@t~() • mllh~~~~~t~i#·
(WA@~nij( • •~ • • ~rn.lq·.~$~M¢l1l • lilfi~M· • • • Tij~
.qriy~t.~®~ppl,@ • • tb~ • W#~~ifflltl~l1j~~fY
(BleV}lllM~Z~oij~®4~I~s~~
·~~I.rl~~'1~11'1~g
• ~~I·.
~itllM~.!#llM~~·1Q~~.m«tIoV.ifltrw~m~
li~fmm(@~~ijJXij~Clri'?··/
.
Solution:
_ 30(5280)
V1 - 3600 (3.28)
V1 = 13.41 mls
V2 = V1- a t
0= 13.41· a(2)
a =6.71 m/s'2.
S1 = V1 t
V - 50(5280)
1 - 3600(3.28)
V1 = 22.36 m/s
S1 =22.36(2)
S1 =44.72m.
V22= V1 2 - 2a S2
0= (22.36)2 - 2(10)S2
S2=25m.
Distance of car from boulder upon perception
S = 44.72 +25 +10.28
S=80m.
F=ma
IJN=ma
W
IJW=-a
9
a
9
IJ=-
6.71
IJ = 9.81
IJ = 0.68
A.Vehi¢lernovingat.~O.mph.Vla$stoppe9.after
theappHca(ionwlhebrakes,lllhesldd
resisl<ince.is•• O;e6,.l::QrT1pl:llEl·ltil!lstopping·~l'fle
(bfaklrigtlrne),
..
Solution:
W
v=o
A.yehicle.rr@vingat~n.inlli~l.sPee~ • Pf.~ • l'llpb .
Wasstop~ • Wilhlf1g•• s~¢.• (bfaklng.llr!l~) • ::lfter
theaPPHc<ltiol"lofl>t~k~$.qol'tlP~teltlll
aVflragKOoeffl¢ifln!()fJriCtlooal or skid
resi$tan~;· ....
N=W
_ 30(5280)
V1 - 3600(3.28)
V = 13.41 mls
1
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5-465
Pdfbooksforum.com
TRANSPORTADON ENGINEERING
JiN=ma
W
JiW=-a
g
a
0.68 = 9.81
a =6.67 mls
2
V =V • a1
2 1
0= 13.41·6.671
1= 2.01 seconds (breaking time)
Solution:
~~1
• ~:•.••~~~d~be~~~~~~i.l~~'6t.
re$i$t~llc!!lfth~.initi~I.$pe@.~~·.3$mpl:k.····. .
_ 38 (5280)
Solution:
_ 35 (5280)
V1 - 3.28 (3600)
V1 = 16.99
V- 3.28 (3600)
V= 15.65 m/s
w
N=W
W
F=-a
g
F=p,N
W
g
-a=p,W
N=W
F=p,N
F=ma
W
p,N=-a
g
W
p,W=-a
g
8.
p,=-
9
V2 = V1 +at
0= 15.65 ± a (2.5)
a =6.26 mls2
6.26
P, =9.81
p,= 0.638
a
g
p,=-
0.70 =~
g
a = 6.867 mls2
V2 =V1 +al
0= 16.99 ± 6.867 t
1= 2.47 sec.
S-466
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TRANSPORTATION ENGINEERING
Solution:
v
v
-flo-
-flo-
=Vt
.. Lag Distance.
~::j
~~aking DiS~
S= VtI3.6+ Y j 2_ Y ;()~3,6)2(/-+O)
6‫סס‬oo
V1 ;:: 3600 ;:: 16.67 m/s
V12 • vi
'_
Solution:
S - V1 t + 29 (f· G)
_
(16.67)2 -.
S -16.67 (0.75) t 2 (9.81) (0.15 ().02)
S;:: 121.41 m.
v _§0000
, - 3600
V, = 16.67 m1s
V - 40000
r 3600
V2 = 11.11 mls
..
(V1 2 • Vi)
Brakmg dIstance - 2 (g) (f ± G)
C()mPtlf~lheint¢@edialeSlg~
'1jI$Ia@710l'a
lreE!y.'~y·~ilh • a• ®~~%$~E!d.0.a9~p~fftH~.
perRepfI9~ • ti~ • l~ • a$Stltl'@1•• ~<>.~Z#l$~~g~·
\¥ith?ski(j•
• r~j~I®CEl • ptMQ.As~ometlmk!io·.
E!ffl¢IE!11QYI9bE!@%. .. . ~ ~
Solution:
v
v=o .
................
Y
-flo.-.- .
· d' t . - Jl~&?f.:n!:1.1L
Bra kmg
IS ance - 2 (9.81)(0.15 t 0.05)
Braking distance = 39.33 m.
·d=Vt
·············r·=-·.. · · · . ·. ~
lAg Distance
£-..eaking Distance
S=VtI3.6+Vj 2_V,j12 -(3.6)2(/-+0)
80000
V;:: 3600 ;:: 22.22 mls
s;:: Vt t
V2
2(9) f
f=0.70 (0.6) (60%brakeefficincy)
f= 0.42
t
(22.22)2
S - 22.22 (2.5) 2 (9.81) (0.42)
S:: 115.48 m. (stopping sightt.:stance)
Intermediate sight distance
:: twice the stopping sight (,.Stance
= 2 (115.48)
:.: 230.96m.
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5-467
TRANSPORTAnON ENGINEERING
The.driver.ofa. vehicl~traveling~t~Q.kph • up.~.
~ram~ • • requjre$ • • ~ • rn.•••• le$s • • to•• stQP • ~fter • • h~
9G2 + 50.34G -2.25 =0
G2 +5.59G - 0.25 = 0
G=0.044
G=4.4%
.app'~:;.lhe-.br'al<esthan·tntl'~riv¢rtr<!';'illing",t
0=
Iflhe·•• coeffltlen16ffilCtlpll'.betW~o··the.·tires'
and • •p<lv¢rrient·•• is • O.pO, • whal.is • lh~perqnt
.gt~eand
lheg$de: • 'M1at•.•is•.• th~
. • pr;)kil'i9•• dj$laHce,~9Wn
.
o
·ltle • $f!me•. lnitJal·.$~e¢,.down.lhl:!.$~.grad~·
Solution:
25.17 +9
0.5 + 0.44
= 55.27m.
TWp.Mrs.~r~aPProachingfl~.@herfrofl)Wlf
ClPPO~il~WrElctlOn~~t • $Beedpf.1g0.~pttard
~q~phte~lffi9iiyely .• • !\ssuming~reacti@lim~
a•
oq.Q.$~on~ • l'IM••
a•• £OeffICIElnt••offrletlon•• of
9:QQ.Wilh.a•• bmke•• effiCi~¢y.pf5Q% .••• 'C9rnpyte
!h~ • rninirTlulTI·:;i9I'lt(:lim~nq~ • t~Yifed • I()••. ,aYoid
a Mad oOcoftislonQflhe.two•• cars,
Solution:
----
V=120 kph
8‫סס‬oo
V= 3600 = 22.22 m/s
----
V=90 kph
.....An,>,
v=1;~~0 =33.33
90000
V=--=25
3600
D = 2 (9.81) (f+ G)
_
(22.22)2
D- 9- 2 (9.81) (0.5 + G)
D-9= 25.17
0.5+ G
(22.22f
_
D- 2 (9.81) (0.5 - G)
D= 25.17
0.5- G
D-9= 25.17
0.5 + G
D= 25·!L+ 9
0.5 + G
25.17 = 25.17 +9
0.5- G 0.5 + G
25.17 (0.5 + G) = 25.17 (0.5 - G) + 9 (0.25 • G2)
12.585 + 25.17G =12.585 - 25.17G + 2.25 - 9G2
For the 120 kph car:
f =0.50 (0.6)
f= 0.30
Ij2
S1 = V t + 29 (f + G)
_
(33.33)2
S, - (33.33) (2) + 2(9.81) (O.3)
S1 = 255.44 m.
For the 90 kph car:
_.
-l?~
S2 - (25) (2) + 2 (9.81) (0.3)
82 = 156.18 m.
Sight distance to avoid head on collision
= 156.18 + 255.44
= 411.62 m.
'j,
5-468
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TRANSPORTATION ENGINEERING
·A . • •eajero....lnterpoomr.. . skidd,d>inlp.·.lin
intersection of M8gsaysayAv~nueandQt.iirina
Avenoeandslrucklibystanderandeontinued
ul'\tll.lthit•• cme•• ofthe•• colVllln.sllpport.·pf•• the
$~y, • • a~$edqn.th~dam~g~ • tgt/)~fra~tof'
.mElcar; • • me.poliser~Plll'l.e~tilil~ledt!¥ltlhe.car
wa$d()ln~.8 • • kPI"laltM•• lJ1am~nt.()flmpaCt'on
IhElg)Il!mn·.• ~4PP()rt" • • • Th~l~mglb • Oflhe••·skid
ma~$VJasreeor~dt9.M.40.1ll .•••.r~~.raa(j·has
ildpWnhiU•. 9@dll,oftS%, .• • Atest@r•• ~kid(jed
·.j~Tll .• oh.thesame • ~ctipn.·Af • ltte···roal:twren
Compute the passing slght distance·for·the
following data: '..'
.. '
.' .
Speed of the pa$sing car :::~. kph
Speed Of the overtaken vehicle; 88 kph
n~ ofiili.~al maneuver =4.3 sec, .
Average acceleration ~ 2.37 kphlsec,
TIme passing vehicle occupies the left lane
= 10.4 sec;
Oi5la~ between
!he passing Vehicle at the
.end oHIs maneuyerand the opposing" ....
thr•• b@~~·isappli~.,ff:Pll'l.a.spe® •. qi~Okphto
vehiCle:;; 76 m.
the•• h~I~, •.• Deterrnine,1hEl.prob~bJe.s~d(lfthe
Car InvQlved inJhe a¢eidentwnen lhe brakes
...
Solution:
'tVel"¢aPPljoolhkPh. ....
0ppOSlfli I'(n:c"!{ tlPpeat:> \<then
pa~lfIg whrd t JPproocM.1 A
F1RST PRASE
Solution:
m =96 • 88 =8 kph
V= 40000
3600
V= 11.11 m/s
s =2 (g)(f- G)
_
(11.11)2 _
. 14 - 2 (9.81)(f - 0.05)
f-O,05 =0,449
f = 0.50 (coeff. of friction)
8000
V2 = 3600
V2 =2.22 m/s
._~-=J(l
S- 2 9 (f- G)
_
V12 - (2.22)2
40 - 2 (9,81) (0,50 - 0.05)
V12. 64 = 4576,95
V, =68.12·kph
d1
=.!.L (V_m +_at1)
3.6
d1 = 4.3 [96 _8 +
3,6
d1 = 111.20 m.
2
?]1H:~]
2
Vt2
d2=3:6
d =96(10.4) = 277 33
2
3.6
. m,
d3 = 76 m.
2
d4 =3(d2)
2
d4 = '3 (277.33)
d4 = 1.84.89 m,
Total passing sight distance
=d1 + d2 + d3 '" d4
= 111.20 + 277.33 + 76 +184,89
= 649.42 m.
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TRANSPORTATION ENGINEERING
eo!upule•• the.minirnum.pa$Sing131gm.<list<i"qI!.
forthefollowlllQd13ta;
. . .. ....
. ..
.
,.,.,'
.. ,.,'.',','.' .. , ... .... ... -, ...
_
'
,
... _.,-.- ... -, ..
_._ .. '.'.','
....
,
.. -... '.'.-
.. ... '.'.,.,.
,'
~....
Speed of the passing car :: 90
•.
Speed of the overtaken vehicle:= 80kph
.
Time of initial maheuver =4sec; •. . . . . ••. i.·.··.· .
..
~';~=:=::telt~
Avehicl~.tray~la • di$tal)ce'9f40·n1·•• llefQr~
·~~:~~f~fld~~~t~~r~f::~ • oi$l~l~~~ •
A~er • C<lUl$lom • • lfb~WyehlCle$skldthrQQ9h
14•• Il'I..•• bElfor~$wRPi~9 .• c;ofllPute • t~~#*i~t.·
Sp~oflh~ll\oVi"Q.¥el'li(JJ~·.A§§~lffl@C!I~11
lipeffiejellldfQ,62.}
. . . ..
... . . .
Solution:
Distance belweenthe passingvehlde altful·.•.
end of its maneuiler and IheQpPOSing ••••.. •.
vehicle:: SO m.
..
Solution:
Minim_ pauing si,hl di.lt6llcr-------l
0l'posi.., I'thirl"
np~
ptusi", whid." <>pp
• ,.,111m
hn A
Vj
Vr-O
.~ ....:.I:~
After collision:
(80m)
Min. passing sight distance =~ d2 + d3 + d4
but d4 =td2
i
(Wa + Wb) 2 2
-zg(V4 - V3 )
- (Wa + Wb) f S2 =
o-vi
- f8 2=-_
2g
2
V3 = 2g fS 2
Min. passing sight distance = d2 + d3
vi = 1(9.81)(0.62)(14)
d3 = distance between passing vehicle at
the end ofits maneuver and opposing
vehicle.
d2= Vt
r/... = 90(1000)(9)
V3 = 13.05 m/s
Ul
3600
d2 =225m.
Min. passing distance = ~ (225) + 80
Min. passing distance =380 m.
Momentum before impact = momentum after
impact
WaV2 (Wa + Wb)
-- =
V3
9
9
Wb= 0.75 Wa
WaV2 (Wa +0.75Wa)
-g- -·v3
9
V2=1. 75V3
V2 = 1.75 (13.05)
V2 =22.84 m/s
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5-470
TRANSPORTATION ENGINEERING .
Before coflision
·Wa fS1 =
~; (Vl· V1 2)
vi-
V1 2
·0.62 (40)~(22.84)2. V1 2
2(9.81)
(22.84)2. V1 2= - 486.576
V1 =31.75 mls
V _ 31.75 (3600)
11000
V1 = 114.31 kph
·0.62 (40) =
Momentum before impact = momentum after
impact
Wa V2 (Wa + Wb)
--=
V3
9
9
4oo0V2 = (4000 + 2000) V3
9
9
V - 10.85(6000)
2 - 4000
V2 = 16.275 mls
Before collision:
- Wa f51 =
• f51 =
Acargo truck having a welghtof 4000 lb. skids
lhrougha distance of 461)t .before colliding
with a parked Toyota land cruiser having a
weight of 2000 lb. Atter comsJon bOth vehicles
skid through a distance equal to 10m. before
stopping. If Ihe coefficient of Uiction between
tires and pavement is a.a,compute the inilisJ
speed of the cargo truck.
.
Wa (Vi -V1 2)
vi-2g
2g
V1 2
(16.275)2. V1 2
2(9.81)
2
(16.275)2. V1 =- 541.512
VI = 28.40 m/s
V - 28.40(3600)
11000
V1 = 102.23 kph
- 0.60(46) =
Solution:
A >cargo truck of weight 6000 lb. hits a
Mercedes Benzhavingii weight of 1600 lb.
and both the vehicles skip.togetherfhrough a
dlshmce of 5 m, before &'lmln{f to stop,
Compute the initial speed of the cargo tlUck if it
does not apply brakes before collision. Ass.
coeff. oftnction ;;:0;60.
After Collision:
(Wa + Wb)
(Wa + Wb) 2
2
· - - - f82 =
(V4 . V3 )
9
2g
O· V3 2
. f82 = - - 2g
V3 2 = 2g f 82
V32 = 2(9.81)(0.6)(10)
V3 = 1085 m.s
Solution:
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TRANSPORTATION ENGINEERING
After coalition:
After collision:
Momentum before impact =momentum after
impact
W1 + W2 (W1 + W2)
. -g-- =
'g
V3
6000V2 = (6000 + 16oo)V3
g
g
6000V2 = 7600V3
V - 7600 (7.6?}
2-
6000
V2= 9.72 mts
V - 9.72 (3600)
. 2- 1000
V2 = 34.99 kph
·,tvian•• h.aVjn9.~.Weight·Qt8000Ib.htts~ • pa~ed·
TOYQt!l~r • • af•• wei9ht•• • 2QO() • • lb.•.•• !lrld • both
• Y~iCl¢~Skidt()~ltier • tlir9lig~.ti.(jJs~r~of
&••·rrJ••• ·~efQre8'!l)lng • • to~toP ••••.CompllI7•• m~
V~IQCity9f.lrllP~qt.wm~ • • v!ln • !lppl~$.br!lk~s
and • skidsthro~h.a.~istar!~of • 4• ro·.·befora
cl:l1Jj$i()Il'.·.•.• ·N>$l.llIl~>9~(;ifll1t·.llf.ffi¢tion.js
TVf~¢clrsAandEloftiq¥aIWl'~hl,al'e
apprO~shi~g!lmjrll$M~qtiOr()fJwo
perpeMicularroads, • A.frQmlhe~st • ~nd •.
e
fr0I1'l•• th~ • • solJlll~.g • • c~lid~.wltheClch • gth~r ..
rh~@ti!lI • ~Kid • dj$tarces.of~and.~l>efore
collisl0r! am3PIll.and•• 2qpl..re5~c!tvely.
A~theCOIliS1qn • • ~.$kl<j.~lstah~ofAarlde
afe•.14•.• m.•. ~od.~itm .• respe¢ti\lalY.ASkids • a
directjpnof.~.M)· • Vj.'i\lhil~.~skids~direc:tjpn
of•• ~ .•• 3q'.• ~ .••• If.t~ra,V~t~9~.~k!9~~$tflnge.pf
fheptlYernentisfolJnd•• tobep,60·c{llnput~.t~e.·
sp@dpf A1~kphb~fQrfJh~appli$S~Qfak~. .
Solution:
Q.5Q.
L
Solution:
. . V~I
.
.. _,__ ~_
. . ··'0 m~ r 2{m
.:
Velocity of Impact
VAN
VJ ....-
VBf~--+-.l
~~tl
.
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5-472
TRANSPORTATION ENGINEERING
L
.At collision:
(Alo~fvl.West-East direction only)
momentum before impact = momentum after
impact
WA =WB
VAz =- 8.25 + 10.295
VAz = 2.045 m/s
. WA(VA/- VA1Z)
- WA (0.60)(30) =
29
- 60.30 (30) =
(2.045)2 - VA12
2(9.81)
4.182· VA12 =- 353.16
VA12 = 357.342
VA1 = 18.90 m/s
- 18.90 (3600) = 68 50 k h
VA11000
.
P
After collision:
Work done in skidding = change in kinetic energy
Positive work -negative work
= change in kinetic energy
o. WA (~ (14) =
W (V 2 V 2)
A
~g -
Aa
Compute lheminifuum reqUire<jslgbtdlstanpe
tg •.~yoid,.a • .• c()Ui~ipn •• for·.t~Yo~,#iJY •.• traffic·.Wlm
$ln~le • lane.'Nilh.'a•• car approaqhifl9~~fYi.m~
opposite djr¢clions.if.bolh.CCjfs.i:J!'El.roolJm9,<lla
speed.Of8tfkPlf••·.T6tal.perceprt6n·and·.t~~et\()1l
HmeJs2, Ssec',60efficiOOI(lftrtCli0l1 is 0040
ilndbrake efficienqy is 50%."
Solution:
0- VAl
- 0.60(14) = 2(9.81)
VA3 = 12.84 mls
ForB:
0- We (0.60)(36) =
V 2
·0.60 (36) =.~
VB3 = 20.59 mls
Wa(Va/ Va})
2g
V=80 kph
V=80 kph
.~f~
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TRANSPORTAnON ENGINEERING
80000
V= 3600 = 22.22 m/s
f= 0.4 (0.5)
••,1.~
f=0.20
Cf)mp~t~merE1911it~1~n9thoflh~saf~
V2
.s = Vt + 2 9 (f + G)
.
(22.22?
s =22.22 (2.5) + 2 (9.81) (0.4 + OXO.S)
.re~cijon• lirnEt9fff1edtiv.er'fS?·6~ep,~l\clJ~e
C(j~fflpltmt9fm~Wm~~t\¥~~tlJtre$~~d
S = 55.56 + 125.85
S= 181.41 m.
p~v~rnentl$li.40,AS~umethe$k)peQf
foal:t'NClytobebQtiz(lg@,\.
Sight distance to ?void, r;01l1~>n
= 2 (181.41)
= 362.82m.
.'
.'.
Solution:
,"
.C(jmpO.~• • llle•• ··reqUjr~d • • $~fe· • $t6ppiN~.· • $1~ht
dl~~nqElf()r··~·.fMl • 'f/~Ylt<tffi9jr • ~• ·$ifl.g~~fl.e·
T"'oLane
to .WaldB9111$iQry.wrtbllcar~ppr93et)ing~~m
thEl.Qppo~itedjreclio~jfb()tbc~~sClr~mQyln~
at•• ~• SPl¥ld • •ot80.l(pb, • • • • Tot~lper~~pli(jm·
re<Wjon.t!nie.bft~e·iiriyer • I$.g.sf¥J· • . COeffi9i~rt
ofJr:lcti<m·.Ilel\v~n • tl1e.tires·and·the.pavet'Ylenf
.
i$O.l)o.$I()p¢()ft'()~dWClyi$+2%.
...
V= 60000
3600
V= 16.67m/s
Solution:
S = Vt + 29 (f + G)
Single Lane
......
V=80 kph
V=80kph
-...
_
(16.67i
S -16.67 (2.6) + 2 (9,81) (0.4 + 0)
S = 78.73 m (stopping distance)
80000
V=.3600 =22.22 m/s
V2
S = V t + 29 (f + G)
_
(22.22)2
S - (22.22) (2) + 2 (9.81) (0,5 + 0.02)
S = 92.85 rn.
-sate stopping sight distance == 2 (92,85)
Safe stooping sight distance = 185.70 m.
Vehicles often travel city streets adjacent to
parking lanes at56 kph or faster.. At his speed
and selting detectIon through responseinitiation lime for an alert driver at 2 sec. and
f =0.50, how far musUhedriver be ~way from
a sUddenly opened car {joot to avoid $trlking
it?
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TRANSPORTADON ENGINEERING
Solution:
I----s;--~
56000
V= 3600
V= 15.56 m/s
(15.56f
8 = 15.56 (2) + 2 (9.81) (0.50 + 0)
8= 55.78m.
()r¢Whj¢fejs•• fljuowirJ!rari()th'*~fl~ • tWo-lane
80000
v1 = 3600
v1 = 22.22 mls
V12 • vi
8 = V1 t + 2 (g) (f + G)
80
8=-(6)
16
8=30m.
(22.22f- vi
30 = 22.22 (0.5) + 2 (9.81) (0.65 + 0)
(22.22)2 . vi = 240.90
V2= 15.9 mls
v:2 = 1000
15.9 (3600)
v2 = 57.2 kph
C\)mptJt~th¢headingllght$i9htdl~taoce.fora·
tWp•• hi9hway • ~t.rtlght~q9qfdir~ • I().·.th~· • s<lfe
fr~e't\'ay • with • a.desjgn • • SI¥lElcl.of•• • 75 • • kph.
A$SUl11E!•• tirne()f~tcaPtj® • t().be3.seG.artd
.f0l'~ch.1e.I<pH.()f.sP¢E!l:i·lf·b9thY~~!PIe~.lire
effi:clency. . .
M¥in9tu~ • ~1IhUm~.ot(m~.~tletl9m~Pll(;ing
·lra~l~g.at8(J.kp~ •.<l~l:i.WE!.I~~.%1t.crashEl~.W
th<lt~pet'ld·intQ.ltte • rllar.Qf.'ID.9nij@1~dPllr:l<ed
!ruck,.at.What•• $peE!tiWjQ·.~ • follp~l~gY~hl~e
blt•• th.e;•• ~recki!g~?.A$SIlJrl~· • ac.c~r • •'ength•• I~
skid.r~~istClnce
• l()be • pJ50..•••.• Use •. 80% ·.br<jke
.
Solution:
~f'rHa~~~og~~~lsO;q • ~.• ~Il~.~.¢effiC~W
Solution:
v= 75000
3600
V= 20.83 mls
\fl
s= Vt+2gf
(20.83)2
S= 20.83 (3) + 2 (9.81) (0.48)
S= 108.59m.
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TRANSPORTATION ENGIIIUIING
Solution:
~icarhavtngaigrt1ssiW~ight6f~f<NiS
lffllVill~~ta cerl~jTl~~~Qn$p~~~IO'}9a.
OitmWay¢lIrve,•• • •. Ne~~MN~.·.fli¢t~~··b~tlen
trnl••tlre$.apd•• the.paVm'~gf.find*M.tl)r~.{tlat
wUI•• ~ • t().pull.lt1e~W~~Y:ff9m·~~nter9f
the.·qu~.iffh~.cul'Ve • ~~.laIIirll~tfal:t()f'·.of
0.00.·
.
F=~V2
g r
50000
V = 3600
.
=13.89 mls
- 1200 (1389)2
F70
F=3307N
Solution:
_ Centrifugal force.
Impact factor - weight of vehicle
F
0_30 = 50
F= 15kN
il• • •
Solution:
A.tarls•• runlllhg • at ·m.ax•••~IJ6W~Pte • $pe~d
alOl'lQ•• ~ • hjgh\\!a.y.Cl1rve.~t~.~njmPClctf<lclpr(}f
p.gQ.••••• ,f.the·.Wejgl'll • ofW~.:eat~N.tij$~~diS
4{l)kN,•• • \Vhat • i~.th~· • serlll'if1.lSlJl.twCEl.···()n•• ;t
N.e9I$ctfrlcI1Qrybetw~E!Tlffi~III'(j~.al1#]M
p;:lYemtmto
. _. .
..
Solution:
.
<
Centrifugal ,orce =
wV2
gf
V2
wVl
g r
F=-60000
V=3600 = 16,67 mls
Impact factor = gr
Centrifugal force = 40 (0.2)
Centrifugal force = 8 leN
A Qlr weighing 1200 kg runs at 50 kph around
an unbanked curve with a radius of 70 m.
What force of fnction on the tires IS required to
keep the car on the circular track?
F = 1000(9.81)(16.67)2
9.81 (100)
F= 2778
.Because some railroad track follows a rllfer
bed, tightcu~sare' required. The tightest
curve has aradius of 21 ~.4 t m. The center to
cenler distance between raUs is called the
effective gage and is equal to 1.5 m.
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S-476
TRANSPORTATION ENGINEERING
®
·r()•• Ylhalhelgbt.tll~flheQU1$fd~iail • b~·
raisel:!•• if.<In.~q'lJilillriI.Jlnspe~()f.64kph.is
@
tObetillJlffllllned?ii
Note:
Length of spiral in feel is usually 62 limes
'e' in inches.
®..••• What•. is•• ~h~ • COmf(lrtabltl•• speed<llong • the
nvetbe(j?
@ •••
Length ofspiral:
e = 125 mm = 5 inches
·WMt•• i$.m~.1®9t/) • pf.tb~ • ~a~$illOn.·$pjl'al
with•• ·.·the.·•.• 125 .•·•• mm•.• • elevation jf••·.the·
·¢<lUllibrl@lsp~i$.USedf
Ls =62 (5)
Ls = 310 ft.
Ls = 94.51 m.
Solution:
CD Height of outside rail:
tMde$ign•• $pee~ • fQra.HqnZ®lal.¢l)rve • of.a
niUroadjslimitedJo6Qkph, '.. . '.'
(j) .' c:ompulelhertmOireg~upe(eleValiof11obe
pmvlde~at .• h9ri4or~~I • CWV~ • qf·a.hm•• mad
naVill9atMhJsPfC9lYeqf40Prrt .....
® 'vVhal.is.the.miJvradjusofflOnzonl~lc~rve
In.ahUly.Joad·.lhal.isJW1Uirediflhe.lalerat
fril)t~nfaclor.is.tak13l'lfs.O.1$and.asuper
elevationofO.06arn;<
® What•• js•• the•• all()wablerat~of • changeof
centrifu9<i1•• <icceleraliqn.f()r•• a•• horizont~1
lane=wa
gW
curve&fahlllf(lad.
Ian e =~
y2
Solution:
9
CD Required super elevation for a hill road:
a=R
V2
v2
lane=-
e= 225 R
gR
_ (60f
e- 225 (200)
. e
Sin e = 1.5 =Ian e for small angles
~-~
e= 0.08
.
1.5 - gR
64000
V = 3600 =17.78 mls
@
V2
1.5 - 9.81 (213.41)
e = 0.226 m. = 226 m
Comfortable speed:
when e = 75 mm (3 inches)
e
y2
1.5 - gR
0.075 _
y2
1.5 - 9.81 (213.41)
V =10.23 mls = 36.8 kph
Min. radius of horizontal curve in hill road:
R= (e + ~ (127)
R= 0.008 (65)2
0.068 + 0.15
R= 155m.
~_-i!.?~
@
...'
@
Allowable rate of change of centrifugal
acceleration of horizontal curve in hill road:
80
c= V+ 75
.
80
C=65+75
C = 0.571 mlsec3
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TRANSPORTATION ENGINEERING
A•• carweig~ing.?O • ~N • lTl9ves.a.cElMain • $Peeg
arOU.nd.aClrcufSr•• 9tlrve•• CallSi~9a.centlif\lgal·
f()r(;El.'Nhl(;~ • tElnds.toP1.lIl.ttJe·~ra'Nayfr9IlltM
c®t~·.()f • tne. .CQfile••. at•• ~ .• l1lagl'lltud~.f!Qlll;lllQ
15k!''t
.
@} .90(tlpUtem~.@Ui ...a~nl¢enjrlfyg~l.ral@(
.~ • •·.WnatisttJe.rnax.••sp¢M•• that.tnis•• C<lrO(llJld
tIi~nelJvef.ol1.the.Cl1t\(e
• $Othat•• ltWill1'l6t
9verttlrr1.ffthe.cllN~hasaradj@0f?8()rr1,
·~ • • • What\VoUI~ • • be.lh~de!$~p$tlper~~\lati()ll.
~ertliet~tq·prevElI'I!$lkllog9r(:l',lettUO'\irg.it.
IWfriction fa¢toris 0.12. .. .
. .
A..lVio.l!lne • hjgtW8Y.fl~vio~.avAd!h.Qf3.€lIJI..
()I'IAACh.lalle•• Witha~$l$l't.sp~d(lf • 10Pkph
h~sa40(}i11·r<W~$.~Qri~()ntalcg~ye
C011®¢tiIl9.~~n~llts.\VIm.t@:lrillg$9fN·.Z5'l;~
al1dS>78'E.
...
p~termln~I~~$O~rel~~ti(mw-telftJe
jict~~f<!BI()ri§9.t~,\·
~q()rnp~temeJ~ij9tb°t$pjt~tffthe
.@
•·• • • .• •·.qlffeWllce•• ill.ijr;~¢~~~Jh~.~~~jO~
... ··aI'\Q$:lg~gflhe~~~W~yi$UIl'lit~~tq
112Pg·•• •••• • • • • • Ui•• • ,.....i•• • • • • • • U•• • •·, • • • U<·•• • • · · · · .....
•.·.
@bQmpllte.i:fu'.!lpiralili1g1~pflt1e$;C
Solution:
Solution:
CD Centrifugal ratio:
CD Super elevation:
Centrifugal ratio = ~
V2
R= 127 (e+0
Centrifugal ratio = ~~
_
(100)2
400 - 127 (e + 0.12)
Centrifugal ratio = 0.25
e= 0.08
@
Max. speed:
Centrifugal ratio =
~
® Length of spiral:
De __1_
V2
L -200
0.25 =9.81 (280)
3.6 (0.08) _ J...:
L
- 200
V:= 26.20 01/$
V =26.20 (3600)
1000
V= 94.34 kph
L= 57.6m.
@
@
Super elevation:
R=~­
127(e+0
_
(94.34)2
280 -127 (e + 0.12)
e =0.13 m/m
..
Spiral angle:
S
c
=iL 180
2 Rc
1t
S =~180
c
2 (400)
Sc = 4.13'
1t
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S-478
TRAISPORTATIOI EIGIIEERIIO
An easement CUlVe has afengthofspiral equal
10 60 m.having a central curve of a radius of
400 m. The design velocity ofthe·car allOwed
·to passthru lhisporllonls100kph.
<D. Compul&therateof increese ofcenlripelal
. acceleration.
.
..
..
®!fthe ftiCtion factor is 8ClUal to 0.14,
. . cOmpute tlie $lipereievathn rate in mlm .
~ =~~~j~ofonel~ofrolidwaY
•. ...
. if the
A car having a weight of 40 kN is moving at a
certail:! speed arotllld a given curve.
Neglecting friction between the tire and
pavement arid aSStlming a centrifugal ratio of
0.30.
(j)
l<N . . . .
.•.
® 11 ~e degree of cUrve is 4'. compute the
max. speed hI kplrlhat a car could move
dlff~renoein9ral1ebetweenlhe
. centerline andth& ~e of the roadway is
11220.
.....
.
Compute the force that wUt lend \() pull the
.tar.~ from the center.of the curve in
aroum the clirve.
.
. . .•.
® Compute the value of the super elevation
to .be actually provided for this speed if the
.skid reSistance Is 0.12.
.
.
Solution:
Solution:
CD Centripetal acceleration:
L - 0.0215 V.J
-
CD Force that will tend to pull the car away
from center of cUNe:
RC
F
Centrifugal ratio =W
60 =0.0215 (100)3
400C
F
0.30= 40
C = 0.896 mlsec2
@
Super elevation rate:
F = 12 kf'/
@
V2
Centrifugal ratio = 9 r
.V2
R=127(e+~
1145.916
r=--4
r=286.48
_ .(100)2
400 - 127 (e + 0.14)
\fi
e = 0.06 mlm
0.30 =9.81 (286.48)
V= 29.04 mls
V= 104.53kph
® Width of one lane:
De __1_
L - 220
Max. speed:
@
Super elevation:
V2
0(0.06) __
1
60 - 220
R= 127 (e+fi
0= 4.55m.
286.48 - 127 (e + 0.12)
e = 0.18
_ (104.53f
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5-479
TRANSPORTATION ENGINEERING
I'I_~'_1.1E"
··pelWeel\••. tn~~r:1erU.® .• al\lith~rqM~e; .
Ra~jllsofcenltal$lrVels420nt . ..
. ... ..
\1!qOl1lPu~.~heW1dthrequ'~.fclreaWIa~e.<.·
• ·•
:iI8'RllI,.lr,
Solution:
CD W(dthfor each lane:
De __1_
Lc - 250
D (0.06) __1...
60 - 250
D=4m.
___Ii
;01.1
Solution:
CD Super elevation:
Vi
R=127(e+~
_
® Velocity of car:
Vi
(80}2
•
350 -127 (e + 0.12)
e= 0.024nYm
® Impact factor:
V2
Impact factor =9 R
420 =127 (0.06 +0.15)
V= 105.84 kph
@
V= 80000
3600
V= 22.22 mls
Centrifugal ratio:
Centrifugal ratio =
~
_(22.22f
Impact factor - 2 (350)
V= 105.84 (1000)
3600
V= 29.4 mls
... Irat·/0 --~
Cent•lliuga
.- 9.81 (420)
Centrifugal ratio = 0.21
Impact factor = 0.705
@
Angle of embankment:
Vi
tane =-
gr
tan e = 0.705
e =35.20'
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S·480
TRANSPORTlnOIL ENGINEERING.
'JlII'.-
'1!llllllli
C~Il1Pute • lffll.·.impact.faclOrf9'•• ah9ri~ontal
~tl~ •. raqius.t)t'fflQ•• tll·)t1h~(J~sjgl).$~~~~.
12Q1<Ph.
'.
.
.'.
Solution:
Centrifugal ratio or impact factor
_ Centrifugal force
- Weight of vehicle
wv2
t
Impact factor =
Solution:
CD Radius of CUNe:
V2
Impact factor = fT
V2
R= 127(e+~
_
(90f
R -127 (0.08 + 0.10)
R=354.33m.
.V = 120(1000)
3600
V= 33.33m1s
(33.33f
Impact factor = 9.81 (400)
Impact factor = 0.283
® Degree of CUNe:
D = 1145.916
354.33
D=3.23'
@
Impact factor:
Ij2
Impact factor = 9 R
V= 90000
3600
V= 25 mls
.
Solution:
(25f
v2
Impact factor = 9.81 (354.33)
Impact factor =gr
Impact factor = 0.18
V2
0.15 = 9.81 (500)
V= 27.12 m1s
- 27.12(3600)
V1000
V = 97.63 kph .
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S-481
Pdfbooksforum.com
TRANSPORTATION ENGINEERING
574. CE Board Nov. 2005
peterniirl~• lb~lllax .••• spe~iin1q)hthata • #ar.
Solution:
CD Force to pull the car away:
. F
Centrifugal ratio =W
.
CQuldrilnar9l!Ma5q~g~~cuo/~)fth~
F
0,30 =40
.1f~ ~~~~:~~.~~~Ad~h~4~~~I~.I~j •
® Max. speed that a car could move around
the curve:
Solution:
·
F= 12kN
1f1 .
Centrifugal ratio =
Impact factor =gr
1145.916
r= D'
1145.916
r=-D1145.916
r=-4-
1145.916
r=-5-
r= 286.48 m.
V2
0.60 = 9.81(286.48)
V=29.04 mls
V= 104.53 kph
r= 229.18
1f1
0.14 = 9.81 (229.18)
V= 17.74 mls
@
v= 17.74(3600)
1000
V= 63.9kph
.1\. carh1!l\lfi'lg~weighlof4(tkN
"
Super elevation:
V2
R= 127(e + ~
_ (104.53)2
286.48 -127(e + 0.12)
e+ 0.12 =0.30
e=O.18m'm
istT!ovirt!il~tt!:
9Wal.n<speed '. around· a given' .¢~tVe;
Neglecting friction between the lirti an4 ·lh~.
pavement and assuming centrifugal .ratio of·
0.40
~
..
CD·' Computi:llhe force that wiU tend topul/the
car <may from the center of lhe curVe. .
® If the degree of curve is 4'; compute the
· max. speed inkphthat a car could move
around the ci.irve.
... .'. .
@ Compute the value of the super elevation
·to be actually provided for this speed if the
·skid reslslance is equal to 0.12.' .
..'i~I~~lt~I~~~
Solution:
V2
R=127(e+~
V= 50 mph
V- 50(5280}
- 3.28 (1000)
V= 80.49 kph
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5-482
TRANSPORTAnON. ENGINEERING
---.!§Q~
V2
R- 127 (0.10 + 0.16)
R=196.19m.
R=127(e+~
D= 1145.916
229.18 - 127 (e + 0.12)
e + 0.12 = 0.22
_
R
D= 1145.916
(80)2
e= 0.10m'm
196.16
D=5.84'
l\~jmpleandihorlZ9htalCUrveroadh~$a
d~.9fcuf\l~.9fa.2' .•••• ()~t~nllillet~e.9¢$i9n
$p~d • o/l.IhI$C~flJell1.·.tnph • if ·e•• ·.'=;•• O.O&ahd.
fWO:1~/·
'Solution:
!N.curvefflad•• •74.m·.·.W•• tM!lls•• ·.~~ • • a••s@l?r
eIElyalic)ll•• of·O.12.•• aild~.·t1eslgfl·.spe~ • of.~Q
kptLC•• OeterminetM¢®f1l¢ieril • offncIiAIl
betWeenthetiresatlq.the~ent....
.
Solution:
V2
R= 127(e+~
_
(80)2
R= 1145.916
3.2
74 -127 (0.12 + ~
R=358.1 m.
V2
0.12+f=0.68
f= 0.56
R=127(e+~
V2
.358.1 = 127 (0.08 + 0.12)
V=95.37 kph
V= 95.37 (1000) (3.28)
5280
A4~ • hP@()rJ!~lcplY~rdadj$gesijrJ@r.~ • s~d
V=59.2 mph
of.59.fllpl1'.·.S~pElr.eleY<ltionis.Q.~
the(;(;)E@s.iei1t2f!~t~U®t[O!L~
Solution:
rhede9~.·of.curve9fa$irnPle··CUl'Vej$
• 5'.
COO1Pl.lfe·the.desired.super~levation·tequired.jf
lhedeslgnspeedofacarpassing lhrothe
curve. is 80 l<phanl:ijhe skid resislanceis
equaltoO.12.
Solution:
R
= 1145.916
D'
R =1145.91 6.
5
R=229.18m.
_ 50 (5280)
V- 3.28 (1000)
V= 80.48 kph
R= 1145.916
4
R= 286.48
V2
R=127(e+ry
(.130.48)"
28648 =-"'--'-.
127 (0.06 -;- f)
{=0.118
.••• [ffllerrnine
.___ ..._._
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~ 483
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TRANSPORTAnON ENGINEERING
PEltel1lliM•• • lrell@9'lh•• l)fth~spifal.9~rve
#~$lgll~l1tilr.~·m~·caf.$,~~.tlt.90kph • lf.fh~
g~9ri$()f • lhEl • qllltr<:lICl.lrye • i~.&\<q~e·.~f'(;
~a$i$.
.
.
·filld.tlj~.I~@th9t • ~ • • trM~~l96P41'f~.a~.m~
~~~$ • Af•• ~· • • ¢~l'ClI • ¢(r&El•• n~0t\9 • •~• • r~l:fi@o~
1~~ • l'Ikhi~lald.$ilib • tffill¢elr~ • tr:aYE!@g•• l ij•• il.·
• $p@(j9tlQ.kpti~».Mt,$kjd.6t'oVertl.lffi • • • •· ·
Solution:
L =0.036 j(-l
c
Rc
L =0.036 (70}3
c
195
Lc = 63.3m.
Solution:
R= 1145.916
5
R=229.18m.
L = 0.036 j(-l
c
Rc
L =0.036(9W
c
229.18
Lc = 114.51 m.
At~atmax, • • $~e~jllkph • ¢OlJld••aGarpa$s
lhfOl.l~h.·.a • • ~pir~t • • l!l~;StJm~nt • • (;lm!'e.>~jthollt
()YerfiJ£lllnQ.ifWew~m$9fWe.cen~al.etlrve.ls
1~Om·ijn~the~llP~r¢levadqnJ~OJZPef
fllElter()f~,?
.
.
..
Solution:
Solution:
0.079 i<!
R= 1145.916
5
R=229.18m.
e=-R-
0.12 = 0.~7~ J<2
K=52.3 kph
L =0.036 j(-l
c
Rc
=0.036(9W
229.18
Lc = 114.51 m.
I
'-C
A spiral easement ClUve has a length of 1(«) rn.
with a central curve haVing a radius O! 3eO 1")
Determine the offset distance fram tt>~ far ~',!!t
[0 the 2n0 quarter poin~ of the spiral.
S-484
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TONSPORTATION ENGINEERING
Solution:
A$Plr~I.1ool"rl· • longc6nrttlcl$~.*ngl!ntVlitha
-i.·~~~.·8r(:UlarC\jwe,·· •. AMm~~pil'<ill.ary9Ie
~tlhEi$;C>
...
Solution:
L3
x=-6Rc Lc
_ (50)3
R= 1145.916
x- 6(300)(100)
x=0.69m.
R=286.48
,
S
e
.Vj.II1~1I
4
=lL 180
2Re
1t
100 (180)
Se = 2 (286.48) -;-
Se= 10'
Solution:
L = 0.036 J<3
R
120 =0.036 J<3
260
K= 95.3kph
• flrtQttledegf~9t<:Ufffl9ia~ottal51mple
·AlIl'V~·ffhasa • ~~ir91.ClJrvNlf • 1QPlTtl~ng.{)~
1Wq • . ~ld~sqnWhiC1)a<:attr~elil'lQatr?kph
WlIllllitskiQ.l.Jsearcbasls. ...
Solution:
~tls • ·tftEl•• 9ffsetdlstijnll~fr901fH~.~~.ngef}ttA
.m~~nd·9uafterp0ir1t.l)f~ • $Pi~l99rv~lh~a
has•
.~.lerigth.·of8Qll1ililld.ll.·c~1ralraqius
• of
300m'
.
Solution:
L3
6 ReLe
x=--
-~L
x - 6 (300)(80)
x= 0.444 m.
x= 44.4 em.
L= 0.036 J(.3
R
100 = 0.036 (75)3
R
R= 151.875
0::;: 1145.91~
151.875
0=7.55'
.
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8-485
Pdfbooksforum.com
TUNSPOITIDOII EIIGIIIEERIIII
Solution:
CD Radius of curve:
R- 1145.916
1i1'lllililB;i
~ ~~~~.~:~ ~,tfi.",~.~,"i, ~.l
,•.• .•,.• '•.•,.• . ,.•.• ,•. ;• .•. •.•. • .• ....•.,:•... •. ".•,.,• .,•••.,,••..• • ."• .,•. .•,•,. ..-•.•.
. . . . :\AIl.goo..
.
-
@
Length of throw:
Xe=lL
6Re
_ (80f
Xc - 6 (190.99)
::::::.:;;:}:::<{««</:::>::~:<:::;::::::::. :.;::.:::<:
Xe=5.58
Solution:
-~
P -4
CD Offset distance at S. C.
I
Es = (Rc +P) sec:2- Rc
p_ 5.88
- 4
13.20 =(230 + P) Sec 18' - 230
p= 1.30
p= 1.395m.
@
Xc =4 (1.3)
Xc =5.2
@
6
R= 190.99m.
External distance:
I
Es = (Rc + P) Sec - Rc
2
42'
Es = (190.99 + 1.395) Sec 2-190.99
Length of spiral curve:
-.!L
Xc-6Rc
Es,= 15,08 m.
_-!:L
5.2- 6(230)
Lc=84.7m.
® Design speed:
L =0.036KJ
• •[~'lRi
Rc
=0.036 K3
84.7
20
K= 81.5kph
A$Pllt.~pjfi!I~~~Il'iEl~t¢ij~.~~ • ~.$i • ~r@
fl)ril$¢E!ntra1cWVf:l"
...
...
Il.[liiili
Solution:
CD Length of throw:
@f•• • lle~~$~e.fi!dm~af~.~n~al<lUrye'·.'
~
••
• • •
I~ I~ • •m0~1~ • 6t.~~r9~.; ~ ~I~I
• • •••••• •
.·@••••• ff.t®¢elllrl,lIMg~ • ()ft@~~tt~I@WI$·
• .•·•• • • • 4Z·; • l:OtllPllfeltte.~~I~~~[$t~M~¥·the·
,. ~ntral~l'¢ofll~pir~l~liS~~lltC9lV~{<
I
Es = (Rc + P) Sec:2 - Rc
.
42'
15.98;:: (190 + P) Sec 2-190
190 + P =192.30
P=2.30m.
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·S-486
TRUSPORTlnOI EISIIEElIlI
@
Length ofspiral:
L2
X --=:c.....
Length ci throw:
@
c-6Rc
&=p
4
><;; =4 (2.3)
><;;=9.2
L2
9.2=6(190)
.
4
p= 1.16
@
4= 102.41 m.
@
p=&
4
p= 4.65
External distance:
I
Es = (Re +- P] Sec 2. Rc
Es = (229.18 +- 1.16) Sec 20' - 229.18
Es = 15.94m.
Max. speed:
- 0.036KJ
Lc- R
c
102.41 = o.~~KJ
K= 81.46kph
11J!)'.~gtl)«#splr.~twaqm· ·W®·.~~(IlU$M
.th~.fffilll\l,y~Il~I1Q·2()O·m
.•·..··>·.·•.•.· ·.·• • • ·• .•.•. . . . .
@)()ompQl.¢.th~wiri,lI~IfJ:at.~.·.~d~~~
~.¢.)<
~• • ··p¢ppW~JMoffset~~tant~Ait.lhe<erid .
~($,¢,)
@C()mptM!hetengthQftb~.<
.
Solution:
CD Spiral angle at S. C.
S=~
c
2nR
S - 80 (180)
c - 2n(200)
S(;'" 11;46' .
Solution:
@
CD Offset distance at S.C.
- 1145.916
Rc 5'
Rc = 229.18 m.
L2
Xc
=6t
_ (80f
Xc - b (229.18)
Xc = 4.65
Offset distance at end point (5. C.)
_4
_K
Xc 6Rc
XC - 6(200)
Xc =5.33
@
Length of throw:
p=&
4
p= 5.33
4
p= 1.33
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S·487
Pdfbooksforum.com
TRANSPORTADON ENGINEERING
·i~'!IIL~i~~~~·~=I~i·
ill:!lil
Solution:
Solution:
ill Length ofspiral:
ill Length of spiral:
L =0.0215
RC
c
P=~
Vl
4
Xc =4 (1.333)
L = 0.0215 (100)3
Xc =5.332
360 (0.6)
c
Q
Lc = 99.54m.
XC =6Rc.
L2
® Offset distance at S. C.
5.332=~
Xc=~
X - (OO.54?
c- 6 (360)
4=80m.
® Max. velocity:
Xc =4.59
4= 0.036 Vl
Rc
@
80 =0.036 Vl
200
Length of throw:
1
P=-X
4 c
V= 76.31 kph
p= 4.59
4
P=1.15
@
Spiral angle at S. C.
S=~
c
2rr: R
_ 80 (180)
Sc - 2 rr: (200)
Sc = 11.46'
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S·488
TRUSPORTADOI EIGINURIIG
It••111
1;11-;.
Solution:
Solution:
<D Length ofspiral:
I
_
t-e-
0.0215 y6
RC
L = 0.0215 (100)3
c.
340 (0.79)
4=80.04m.
@
Length ofthrow:
L2
-~
Xc-6Rc
(80.04)2
Xc = 6(340)
Xc =3.14
400
r= 3.281
p=l4 xc
R= 121.95m.
V=48kph
p=3.14
P=0.785m.
V= 48000
3600
V= 13.33 m/s
External distance:
tan '" =
grW
4
WV2
@
I
Es =( Rc + P) Sec 2" . Rc
Es = (340 + 0.785) Sec 20' • 340
Es =22.66m.
V2
tan '" =f1
_ ( 13.33)2
tan",·· 9.81 ( 121.95)
"'=8.45"
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5-489
Pdfbooksforum.com
TRANSPORTIDON ENGINEERING
~.~.~~v~;~l9l~~&~~
·rla:~~.~I=t.~.o.t~=l~ho~.·
Solution:
••Ir.
~f#jijffiOWitl@ijts.l@iM9;>• <····· .}.. .
Solution:
~2
tan a -grW
v2
tan a=-
fT
30(5280)
-328(3600)
V= 13.71 m/s
(13.71)2
tan a =9.81 (122)
e =8.92'
v-
·1.~.~I$.ti.$~~~I~··
~~: • llll.·~.~~~~~r~I~~ll(!Bti • 9f••
Solution:
\fl
tan a =gr
_ 40(5280)
V- 328 (3600)
V= 17.89 m/s
tan T =i1L89}2
9.81 r
r= 265.6m.
wV2
fan (9+B) =-.gr W
V2
tan (a +6)=gr
tan 6 = 0.30
6 = 16.7'
V2
tan (15 + 16.7) = 9.81 (120)
V= 26.96 mls
V = 26.96 (3600)
1000
V= 97kph
<,•• •.• •}.
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5-490
TRDSPORTIDOI ElIIIEElIII
__tti!
Solution:
Solution:
Vl
tane =gr
•
lir• •
lfJ.
tan 6 =9.81 (150)
V= 12.44m1s
- 12.44 (3600)
V1000
V=44.n kph
- 44.n (1000)(3.28)
V. 5280
V= 27.81 mph
603. CE Board Nov. 2005
lfJ.
tane=gr
Sin e = tan e for srnaH angles
V- 45(5280)
- 3.28 (3600)
Solution:
lfJ.
0.10 _(20.12)2
1.5 - 9.81 r
tane =gr
•
lfJ.
tan 6 = 9.81 (150)
V= 12.44m/s
V_ 12.44 (3600)
-
1000
V=44.nkph
- 44.77 (1000)(3.28)
V-
V=20.12m/s
• 5280
V= 27.81 mph
r = 619. 10 m.
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S-491
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TRUSPORTlnOIl EIIGIIIEERIIUI
Solution:
For small angle tan e = Sin e
Solution:
1.5./'1
~O.15
tan e = I,Pgr
0.15
I,P15 = (9.81) (420)
V=20.30 m1s
V =20.30 (3.28) (3600)
5280
V=45.39mph
For small angles Sin e = tan e
I • •til
Solution:
I,Ptane=gr
_ 50 (5280)
V- 328 (3600)
V= 22.36 m1s
_ (22.36)2
tan e - 9.81 (287.5)
8 = 10.05'
0.15
S·In e =
1.5
tan e=0.15
1.5
WV2
tane=grw
V2
tane=fT
0.15
v2
15 =9.81(420)
V=20.30mls
V =20.30(3600)
1000
V=73.70kph
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5-492
TRAISPORTlno. EIGIIEERII•
• ,al
:eleValll)j'\ is 10'.<
...
.
.
Solution:
10"
WCos8
WCos 10"
LMA=O
WSin 10' (0.68) + WCos 10' (0.75)
+ W Vl- Sin 10' (075) = wVl- Cos 10' (068)
gr
.
gr
.
0.118.+ 0.739 + 0.0000885 Vl-= 0.000455 VlV=48.35 mls
V=48.35 (3600)
1000
V= 174kph
LMA=O
WCos e (0.70) +
wVl-.
.
gr
Sin 8 (0.70)
+ WSin 8 (0.6) = ~; (0.60) Cos 8
- 120 (10001'
V- 3600
V=33.33 mls
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S-493
TRINSPORTlnON ENGINEERING
AY¢hi91Wirtlp~s • • to•• overlUfll9n.··a.sup¢t
el~a~~hl~~WIV7~1.~ • ~Il~.91~9mp~·
1]ll'l·~~jQ!Jt.ot.ils • ~mr . of.gf~vi~.~ .• O,6S·W··
artdit$tre8dls.t46j1t.. ·• The$uperelevliuOl'l~
10·,.C(lmp~le.~.l<adillSof.lhe.GUr'IEl.assl/ll'lio/J.
n()s~lcl!lil'lg9Pqur~.
. .
Solution:
10'
WV2 /gr
-t~~
r-~rG~~
0.65'
"'Sin 10'
~ 60 (52.80) _
V- 3.28(3600) - 26.83 mls
LMA=O
WSin 10' (0.65) + WCos 10' (0.73)
+ wV2 Sin 10' (0.73) =wV2 Cos 10' (0.65)
9r
9r
0.113 +0.719 +(26.83j2 :.~~ ~O' (0.73)
_ (26.83)2 Cos 10' (0.65)
9.81 r
~7.67 = 0.832
r
r=45.28
'E«~'l.
[)is@:JCEl•• bl:l~~ • • ftont•• • • ~I~ • • i§•• • 1;2r',.ff
1(~~wJili:~r4_r~~~i~t
Solution:
S-494
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TII.SPORTID. ElCIIEnll.
DM=O
F, Cos 0 (0.8) = WSin III (0.8) + WCoS 0
(0.6) + F, Sin 0 (0.6)
F, Cos 9.31' (0.8) = 15000 Sin 9.31' (0.8)
+ 1500 Cos 9.31' (0.6) + F, Sin 9.31' (0.6)
0.79F, = 1941.31 + 8881.45 + O.097F,
0.693 F, = 10822.76
F, =15617
F1 = 25
v2
lane =gr
,
v2
tan 7 =9.81 (265)
V= 17.87 mls
V= 17.87(3.28)(3600)
5280
V= 39,96 mph
F _ WV,2
,- g1500 '1. 2
15617.26 - 9.81 (1do)
V, ::: 35.01 mlsec.
V. - 35.02 (3600)
11000
V, = 126 kph
.:ffi• •h'm• •~·.~··
·l)$tWeen.!M··tirasand.ttle.md·I$(J.6Q·.·wnatlS
111~~Xifflllmll~!lt~tlc#lr~roiilid·
... . .
IMCtl~Wim<MiOOrldmg1>··
Solution:
A~jghwaY~.tVti.·~~t<@WI$M~~m,~6~~~·.
ar~~~()f~~~a~l)l'lmetrl~o~~tQt7n • rlnd•
.~s~~tv~~.~,tt~~;ll~I~~m~tr~.
pre$$tire.b~lVfflenthatir~.~Jlll>lh$tQadWay;··.·.·.·.
Solution:
V= 65(1000)
3600
V=81. 06 m/s.
Vl
tane=-
gr
_ (18.06)2
tan e- 9.81(100)
e=18.4'
tan
0 =
0'
= 31'
0.60
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5-495
TRANSPORTlnOIi EIiGINEERIIiG
V2
tan (0 + 8) =fT
V2
tan 49 . 4'=fT
V2 = 9. 81 (100) tan 49. 4'
V= 33. 83 mls
Power=PV
4000=300 V
V= 13.33 mls
V= 13.33 (3600)
1000
V=48kph
V= 33.83 (3600)
1000
V= 121. 79kph.
Solution:
w
Solution:
W=1200kN
P = WSin 8 + 1doo (3500)
P = jOOO (0.03) + 0.004 (1‫סס‬OO)
P=340kN
Power=PV
3500 =340 V
P=F+ WSin8
P=0.005 (12000) + 12000 (0.02)
p=
~M
k'N
V= 10.29mts
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5-496
TRANSPORTATION ENGINEERING
SIGHT DISTANCE
Metric System:
-Metric System:
s>[
----~.s----_
where:
L :: length of curve in meters
S :: sight distance in meters
A:: g1 - g2
·«······vtA
>f~95
L :: length of curve in meters
S :: length of sight distance in meters
V :: velocity of car that could pass thru
the curve in kph.
L
V :: velocity of car that could pass thru
the curve in kph.
English System
English.System
··>S.~·L··>
-----s:---,--_
where:
L :: length of curve in feet
S :: sight distance in feet
A:: g1 - g2
«v2A
L=~···~··
····...,46,50
V :: velocity of car that could pass thru
the curve in mph.
L :: length qf curve in feet
S :: length of sight distance infeel
V:: velocity of car thaI could pass thru
the curve in mph.
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5-497
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TRANSPORTlnOI ENGINEERING
•
I
Metric or English System
A vertical sag curve has tan£j$nt .grapes of
-1.5% and + 3.5%. Compute the following to
have a minimum vl$ibilityofB9m....
(j) Length of sight distance In meters.
Length of curve in meters.
® Max. velocity of a car that could pass tnru
1M vertical sag curve inkph.·
@
Solution:
where:
L = length of crest of vertical summit in
(m) orft
S =sight distance (m) or ft.
h1 = height of eye of average driver above
roadway (m) or ft.
h2 = height of object sighted above
.roadway (m) or ft.
A = algebraic difference in grades in
percent.
CD Sight distance:
S = 2 (89)
S= 178m.
® Length of curve:
Assume: S< L
AS2
L= 122 + 3.5S
A =3.5 + 1.5
A =5
_ 5 (178)2
L - 122 + 3.5 (178)
L = 212.64 m. ok as assumed
S>t>
@
Max. velocity of car:
AV2
L =395
_V2 (5)
212.64 - 395
V= 129.6 kph
Metric or English System
AS2
L = 1400
S>L
A descending curve has a downward grade of
- 1.4% and an upward grade of +3.6%. The
length of CUM is 220 m, long.·
.
(j)
~~:~~~~~~~~~~s?f minimum viSibility
@
Compute Ihe max. design speed of the car
passing thru the sag curve in kph.
What is the stationing of the lowesl polnl
of Ihe curve if the p.e. is al station
12 + 12O.60?
L = 2 S. 1400
A
where:
L =length of curve in feet
S = stopping sight distance in feet
A= g1 - g2
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5-498
TRUS'ORTlnOI EIOIIEERIIO
Solution:
G)
Minimum visibility of curve:
Assume: S<L
Solution:
AS2
L = 122 +3.5S
A= 3.6· (- 1.4)
A=5
V= 5O(5280)
318{36(0)
V=22.36m1s
5S2
220 = 122 + 3.SS
26840 + 770 S= 5 S2
52-154S-5368=0
S - 154 ± 212.57
-
Vl
2
S= 183.29m.
Min. visibility = !!!~29
.
Min. visibility = 91.64 m.
® Max. design speed:
L=AVl
395
51fl
220=395
V= 131.8kph
® Stationing oflowest point ofcurve:
5 = V, + 29 (f+ G)
_
{22.36f
S - 22.36 (2.5) + 2 (9.81)(0.3 +0.02)
S= 135.53 m.
S := 444.54 ft.
AS2
L =1400
A=gl-~
A=2- (- 2)
A=4
L- 4 (444.54)2
- 1400
L=564.62ft.
L = 172.14m.
S=..911.g1'~
S = ·0.014 (2201
-0.014 - 0.036
S=61.60m.
Stationing = (12 + 120.60) + (61.60)
Stationing = 12 + 182.20
5Q • • rnph•• 1S • tbe.~e$19~ • • $pe~~ •. ()nA··.Wlrtl~at
SUWgiit.ClJrve(;()nl1~ctifigaf2~.flr~d~mth.~.·
••• 2%.grade•• on.a.hIShWaY,•• • UmllQ.~~,t~pijQn-.
~GUol1.tjme.()t2,5.~~.<l~a·COElff·Bffrit;liQr
of•• P·3Ul•• • deterJ1)im;•• • the • mln.• • f~9th • ot.ltte
IlEirfiCalcurve.
' . '. ...
~1.1{i1
:.,.__1:11
®P9rnp\lte.lliete!lgtIJoflh~.~tl;ll~;· • • • •·
® • • • qol'tlPijtl'i•• tl'Ift•• ~~~t.piffe~r~·.·.f~m • ttm·
• cuNelo·
.' beginnlng··.0f1he.
MnzohtallYtromK<
. a•. Pdll)t··@.·.m,·
. .
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5-499
TIIISNITln.1 ERIIIEElIII
Solution:
@
L= 395
S2
R=
2({h, +~)2
A=2.2 +2.8
A=5
3558 = .
(13Of
2 ({1.5+~)2
({1.5 +{h2)2 = 2.375
m+~=1.54
. ~=0.316
@
Length ofcurve:
. A V2
(j) Height of object abovrt the road surface:
L= 5 (99.79f
395
L = 126.05 m.
@
Length ofsight distance:
AS2
h2 =,O.10m.
L= 122 +3.5S
Length of wrrrnif CUIVe:
'L _Ry(91" g;)
5 S2
126.05 = 122 + 3.5S
100
L ~ 3558 (2.6 + 1.8)
100
L = 15S.56m.
15378.10 + 441.175 S = 5 S2
~ • 88.235 S· 3075.62 =0
S= 114.98m.
® Height difference from the P.C. at a
horizontal distance of30 m. from P.C.
.x2
y=g, x+ 2R
Y = 0.026 (30) + ~
2(3558)
y=O.854m.
A.Vfjltitalsum!llitcurve••hlls.fangent.gradel;pf·
+5M>~llq.· • 3\IW~ .••Tb~horl~ont~l~j~t~~~9m .
~~rtr~~t~e1fi~~~···tb~ • . 'tl'rt~X.~f .•.m~ .•
(i)(¢ornputefhelElnglhof.lliEl.sull1rpife:t.n'Ve.• i••
~Co/Wll~ili~rad~~qf~$Qm$t~o/~ .••• > .
··@-.··•• GOfl;Ipllt~.1l'lEltange~t.()f1tle.~ulTllTltt~r'J~·······
~1.~a~~~··1~.I·
Solution:
CD Length of summit CUNe:
_A!::..-
··(j)<gc@@l~tIj·.r!llIK·~J~l!It~¢ar¢tIDl4·
S, - g,. fJ2
~·.· • .C<>mp@~ • thl! • • IMlg~tt • ()f.t~.·$~·~rtlca,·
113.64 = 0.05 + 0.038
·•• • • • • • Patl$.!'i• ~.sagttiwi .• • • .• • • • • • ·>•• • • • •.• • • • /•• • • • .•.• •.
~• • ~u1e.th~Ien9~·lfuij$~.~.~~fan~.·.·· • ·
Solution:
CD Speed of car:
R=
V2
6.5
.
V2
1532=6.5
V= 99.79kph
O.OSL
L=200m.
® Radius of summit CUNe:
L = Ry (fb • 92)
100
200 = Ry (5 + 3.8)
. 100
Rv = 2272.73 m.
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5-500
TRANSPORTIDON ENGINEERING
@
Tangentlength:
T =& f91.:9z}
2 100
T = 2272.73 (5 + 3.8)
200
T= 100m.
:J~1t~i~III~:I~~~~~~I_.
i':• • •1
Solution:
~~~!~'~~fJl.l;r~1
lilll1ll_
Solution:
CD Design speed of verlical sag curve:
If2
Rmin. =6.5
1500=..t
6.5
V=98.74kph
® Length of vertical sag curve:
CD Length ofcurve:
AS2
L = 122 +3.5S
A=grg1
AIf2
L= 395
A=!h-g1
A =32 - (-1.8) =5
A =3- (- 2)
A=5
L=5 (;~4f =123.42 m.
5 (115)2
L -122 +3.5 (115)
L= 12M7m.
® Design speed:
2
AIf2
123.42 =122 +3.5S
15057.24 + 431.97 S =5 SZ
S' -86.394 S· 3011.448 = 0
S= 113.04m.
L= 395
51f2
126.07 =395
V= 99.80kph
@
Length of sight distance of vertical sag
curve:
AS2
L=
122
+3.5S
.................
58 ..
@
Min. radius ofsag vertical curve:
Ij2
,
Rmin.:: 6.5
_ (99.80)2
6.5
Rmin. = 1532.31 m.
Rmin. -
p6It1pute·•• thernax.•·.velE)Cityt'·W•• a~t.C()Ul~
·p~sthWCJs~~P9~tl()fi(;ClJlYem~r9asigbt
qlstanceof$QQfkWfjell!tleltiPge/lt~tifd¢~~
.1.5%<\nd+2,5%.l!:xpress in mph.
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S~501
Pdfbooksforum.com
TUNSPORTlnOIi ENGINEERING
Solution:
5------1
I
.
·¢PmPti.~~·J'lj~*,.W!~Mm~·~rMltwW~
·.I;.~ r~I~III~1(~il~16~'.' •.
~1$tM~Pf.1$2;~m;>:·············
AssumeS> L
Solution:
L = 2S _(400 + 3.5,s1
A
A= g2- gl
A =2.5 - (-1.5) =4
L'; 2(500) J400 + 3.5(500)}
;"---,5---1
4
L=462.5 ft.
AS2
L=400 + 3.SS
S>L ok
vQA
.
L =46.5 (Relation of L, Vand A)
· . 5 = vQ (4)
462 . 46.5
V= 73.33 mph
A=3.5 - (-1.5)
A=5
S = 182.93 (3.28)
S = 600ft.
_ 5(600)2
L - 400 + 3.5(600)
L = 720 ft.
.
vQA
L= 46.50
AilR
.
r
...
Solution:
720 = vQ(5)
46.50
V= 81.83 mph
¢(jlllPul~treC~pa¢lW6,fa#lrigl#I~Mjn
y~hjcl~$p~rMQ(ifJM$~ElgpttM¢ar
AssumeS< L
A=f12-g1
A = 3- (- 2)
A=5
AS2
L=122+3.5S
_. 5 (178)2
L -122 + 3.5 (178)
L = 212.64m.
m~Vil'9l1l.the.W,\gl~.I@e.~ • $olq:ib>.~~S~Clt
c:arJM.8rn.WithllI'll1l¢tioo til1le O.&$@./· .
Solution:
Spacing of cars = V f + L
50000
S = 3600 (0.8) +4.8
S= 15.91 m.
.
'f f'
I Iane= 50000
Capaclyo
smge
15.91
Capacity of single Jane =3142 vehicles/hour
5-502
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TUIISPORTADOI EIGINEERING
It.I.111.'d1~1
Solution:
Assume: S<L
Solution:
S>L
AS2
L= 3000
A=3- (- 3)
A=6
S =160 (3.28)
S= 524.8 ft.
- 6 (524.8)2
L - 3000
L =175 (3.28)
L =574 ft.
A=2-(-2)
A=4
3000
L=2S-A
3000
574=2S-
4
L=550.8 ft.
S = 662ft.
L= 167.94m.
S=201.83m.
tfnd•• ~h~.M#iMSi9ht9i$~~(,'tJ.9fa.·ggq.·m·
f)~lerflli~ • ft)~sig~t4i$~Ilt.e • ()f~>vertigal
.IQI'Ig·.,!~~lqll,t • • ~~m@t~H~ • ~yI119 • • ~'19$1lt
g@dAAQf±2%an<lY~%?
•.. . . /
.
.. solution:
A = 2- (- 3)
A=5
S<L
L = 220 (3.28)
L=721.6
AS2
L= 3000
L =5 (721.6}2
3000
L =867.84 ft.
L = 264.59m.
P~t~l)QljQ$a9~f\l~~~~m;190g/and
¢(,",lEil:;ijt@.~·."~%gr<l®Wilbi.l~o/.~r<ld~
Solution:
. Assume: S < L
AS2
L = 122 +3.5S
A=3-(-2)
A=5
5S2
225 = 122 + 3.5S
27450 + 787.5 S =5 S2
S2 -157.5 S- 5400 =0
S= 186.9m.
.
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5·503
Pdfbooksforum.com
TRANSPORTAnON ENGINEERING
1.1• •
Solution:
\!LA
L= 395
_\!L (5)
214.59- 395
V= 130.20 kph
130.20 (1QOOK3.28)
V=
5280
V= BO.9mph
Assume: S> L
_ S (122 + 3.5S)
L-2 -
A
A=2.3-(-1.7)
A=4
L =2 (150) _[122 + ~5 (150H
L = 138.25 m. < S= 150
\!LA
L = 395
_ V2 (4)
138.25- 395
V= 116.84 kph
V- 116.84 (1000)(328)
5280
V= 72.6 mph
ilfr.ItJl1l11i
Solution:
Assume: S< L
AS2
L = 122 + 3.5S
A = 3 - (- 2)
A=5
._~~­
L - 122 + 3.5 (179.4)
L = 214.59m.
•••i.;
Solution:
Assume: S< L
AS2
L = 1400
A =2- (- 2) =4 .
S =130 (3.28)
S =426.4 ft.
_ (426.4)2 (4)
L - 1400
L =519.48 ft. =158.4 m.
IlIl~••'1It~.
Solution:
S>L
L =28-
1~0
A =g1- fh
A = 2.2 - (-1.8) =4
8 = 100 (3.28)
S =328ft.
1400
L =2(328) --4
L = 306 ft. = 93.3 m.
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5-504
TRANSPORTAno. ENGINEERING
@
·IM.@r~·r~.~Mij.·~··(t,t~l.~~
.
.~$i@r~~, • • • 9*•• ~9~~~ic1~~ • ffl~~ • ·~~··
Perception time:
V= 89170
3600
V= 24.77 mJs
V2
r~~~[;~lil~.II!iiryj~l=
8=\11+29 f
_
(24.77)2
167.62 - 24.77 t + 2 (9.81)(O.1~)
t= 1.65 sec.
1f$~~W:~~¥~~~~fflIl~9fli'tefAr~~~m~ .
•19llg@m~I~~·mW~~~~t~#.~j~9f
•~.6.1rll;t~r.!~••~a!.l.
~'&~~~~~II~j;r~~Wb.·~~~}'·i~
I~• • • •
11111_••,
?·?#~@)\%~#?(j!c1.P9l~i®~!M~@iMNI#
eIEiWiliooof240;£O.m, The·riverfias·.t6.sWilch
~y~~tli;~·1~·<
~P9mPM~·1M~~{~~Mllijtip~19~t
1~).·.Jtl:l~y~t~p.P@~(#1i~g~P!IIY~~~~)m
~ji:!.iililil.
Solution:
<D Max. speed:
A =g2-g1
A =3.2 - (0 4.4)
·;.~%;!ijllt~$fflAA~~V~M'1'tleill@j~iclll.
Qt.~~~ • g@1#J~il!~~~~19t~~~~.~r.
·~~~!I~~~~~~W~~i~~4~~'~,~.
• •·.1jle
~®vel~.I®gjttldin~I.~$(lfllffi~··
4ijY~MI'¢¢p~®@i¢@~~ijtl:!J~Q;7~~·······
(j)~~l'llIn9 • •·.~ • • @eff?QfftlcUo6 • ()f • 04~,
<..•.
i$fflMt¢#'@~r@tIMlfjij~dJamp$i9~1
<li~tall¢~.<
~• • _~~®~.;~I~~~I~· • I~ • • jbQve•• ~e
•.~• • Whali$~~~~X . . ~e§iQA.~e(I.(flllt • ~·eaf
A=7.6
·····.······.··~I(fm ..6El~verPhtbl$.~~?·.>···
Solution:.
AVl
r:p_Ii~ad/~fT1e ~i~~t distance:
40=395
153= 7.6 V2
395
V= 89.17 kph
@
Max. head lamp sight distance:
80 + h = (gr fl1) L
2
_1M
0-
180
0=0.01745
V2
8 (0.01745) + 0.90 = (0.022 + O°28) (153}
8= 167.62m.
2
S=Vt+29 f
V= 100000
3600
V= 27.78 mJs
.
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TRANSPORTlnOI EIGIIEEBIIG
_
(27.78f
S - 27.78 (0.78) + 2(9.81XO.18)
PAVEMENTS
s= 240.19m.
@
Height ofhead lamp:
Sill +h =(fh -2fJ1? L
_0.85 (n)
180
III =0.0148
240.19 (0.0148) + h =(0.02 + °203X180)
III -
o Without dowels or tie bars:
h=O.945m.
@
The critical section Is at the edge of a
contraction joint. it will crack
approximately 45' with the edges.
Max. design speed:
L=A\P
395
A = 2- (- 3)
A=5
M=~
f
5\P
_6M
- brJ2
180= 395
M=~
v= 119.2kph
b=2x
d=t
1....
_6~
f -2xt2
_1
t
>.. :-.-:-:.>:-:-:~ ...
(thickness ofpavement at
edge and at center)
f = allowable tensile stress of concrete in
W= wheel load in lb. or kg.
•.gq•• @~I~~~n~9·1!:tm~J~'fllI$1~!··
~rvf;i!;:>->:
=~
•
With dowels or tie ban:
Purpose of dowel is to transmit the
stresses due to the load from the adjacent
pavement.
At the edge of pavement:
92. - g1
n=-r
_0.6 - (-1.2)
n- 0.18
n = 10 stations
Length of curve = 10(20)
Length of ClINe = 200 m.
~
M=-
2
6(~X
f--- 2x t1 2
t1
_f3W
='4 2i
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TUlSPDITln.1 EIG.IEEI.I.
At the center ofthe pavement:
~
M=T
6(~)
f= 2x t22
t2 =
Cracks
. Cracks
~ (thickness at thecente~
By ratio and proportion:
A1 A2
-'=(t+rl
W
rr? T
-;2= (t+r)2
1W
(t+~=:nT
t+
r=o.564-vf
t =0.564
t1 =
-vf.
r
0.5641f- r
P
T= K10 91O S
P = wheel load
S = sUbgrade pressure
K = constant value from table
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TIlANSPORTATION ENGINEEBIIIG
t _ expansion pressure
- average pavement density
t = thickness of pavement
contact area oftire
~D
Subgrade
_ (EB)1/3
SF. - E
p
EB = modulus of elasticity ofsubgrade
Ep = modulus of elasticity of pavement
w
SF. = stiffness factor
A
d=-._..:..:...._D-E- (D·A) .
F
d = bulk sp.gr. of core
A= weight ofdry specimen in air
[) = weight of specimen plus paraffin
_. ~[1.75
t- " W
GBR·
1-]1/2
rm
t =thicknessofpavemenUn em.
W =wheel load in kg
CBR = California Bearing Ratio
p = tire pressure in kglcm 2
coating in air
E = weight of specimen pIus paraffin
coating in water
F = bulk specific gravity of paraffin .
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5·508
TRANSPORTAnON ENGINEERING
G=-WL
Pc Pf
-+Gc Gf
G = absolute sp.gr. of composite
aggregates
IG·r/\
V=~x100
G
v= percentage of voids
G = theoritical or absolute sp.gr.
d =bulk sp.gr.
Pc =percentages of course material by wt.
Gc = sp.gr. of course material
Pf = percentages of fine materials by
weight
Gf = sp.gr. of fine material
G-d
n=-
G
n =porosity
G= absolute sp.gr.
Gsb = bulk sp.gr. of total aggregate
P1 = percentage of total weight of
coarse aggregate
. Pz = percentage of total weight of
fine aggregate
d1 =bulk sp.gr. of coarse aggregate
d2 = bulk sp.gr. affine aggregate
d = bulk sp.gr.
d=~
wa - Ww
d =bulk sp.gr.
wa =weight of specimen in air
Ww = weight of specimen in water
Gse =effective sp.gr. of aggregate
Pmm = total loose mixture
Gmm = max. sp.gr. of paving mixture
Pb = asphaff (percentage by total weight)
Gb = sp.gr. of asphalt
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TRANSPORTATION ENGINEERING
Gse - GSb)
Pba= 100 ( G G
Gb
sb se
Pbe = asphalt absorption
Gse = effective sp.gr. of aggregate
Gsb = bulk sp.gr. of aggregate
Gb = sp.gr. of asphalt
A flexible pavement carries a static wheel
load of 53.5kN. The Circular contact area of
thelirt'l Is 85806 mmhmd the trartsmiltedfoad
is dIstributed acro$sa wide area of the
subgrade at an angle of 45'. The sUbgrade
bearing valuei$ 0.14 MPa, while that of the
base Is 0,41 MPa. [)esign the thickness of
pavement and that of the base.
'. .
Solution:
Pbe
PbaPs
=Pb • 100"
Pbe = effective asphalt content
Pb =% weight of fine aggregates
Ps = sum of % weight of fine and coarse
aggregates
Pba =asphalt,absorption
VA = air voids
Gmm = max. sp.gr. ofpaving mixture
G-mIJ= bUlksp.gr. ofcoJIlfl~te!imix._ ...
VMA = 100 _ Gmb Ps
Gsb
VMA = percentage of voids in mineral
aggregates
Gmb = bulk sp.gr. of compacted mix
Gsb = bulk -'p.gr. ofaggregates
Ps =sum of %weight of fine and
course aggregates
Flexible Pavement:
A -W
1 - f1
A =nr2
A A1
-;2 = (t + ,2)
n,2
W1f1
7= (t+ 1)2
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5-510
mlSPIRTlng EIIGIIIEERIIG
(t+r)2:0,564~
t =0,564
~ 5;,~~O
At the edge: (with dowels)
r
-165
W
2
6M
f -b~
M=-x
t= 184 mm
6(~X
A=nr2
f=-2 x /1 2
85806 =nr2
r= 165 mm
t1=0.564~r
. t1 =0.564
-V 535M . 165
~
t1 = 39,mm (thickness of pavement)
t2 = t - t1
t2 = 184 - 39
t2 '= 145 mm (thickness of base)
t1
=-{iff
/r'i
t ='" 3-(53-50-0-)
.1
2(1.38)
t1 = 241 mm
At the center:
M=~x
4
6M
f= bd2
f- 6 (wf4)x
- 2xt22
A.liQid.PilVelTleNiSto~!!p~m¢<jtt'laWMttl
toadpt5a.5kN,~~$lgll~JhickM~$PftM
paveTe"t•• • • • tM••.•a.llp',l!al:ll~.· • tel'ls"e$tre$s•• Af
concrt!te • i§••. 1,~8Mea. $l.tfficient•• doWels·.are
used. acrossJbe joJnlf,
t2=~
,...---
_'" /3(53500)
t2 - 'i 4(1.38)
t = 171 mm (at the center)
Solution:
bElt~rTTline.thethicl<rles~6f.~figl~p~'1etnMtQf
Ih~Pf9pQSed Na9taMl'l•• mad·lgC#rl'Y •.~.·.nmx.'
Wheelloa(f.of60RN'N@Ie¢l•• effe¢t9fa~s.
fC'=20MP? .. AI19VJabletensile~lrmof
c<lrlcretepavementisO,06.tcL ..... . .
Solution:
t=~
r---
t.:
3(60000)
0.60(20)
t=387.3mm
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TRANSPORTATION ENGINEERING
A.53..5.1<NwtiElEll~Mfl8!l • ~·rnaJ(.tire.presstir:e
ofQ.a?MF~·.Th~.pre$$Urejst9.M • ~niformlY:
djstrm~ted(ly~rthEl."re<lo/tirecontact.prlh~
r9a~~y. • • A§surnj~gij~u~9rade • p~s~ureiS
nQt•• • to·•• ex:ceM•• •O·H•• ···MP~, • • d~t~fm~r1fJ • • • th~
req1.lire9·•• • lhi9kl1~ss •.• Qf•• flexible • • • pa.vernlOnt
struCtu%a$P%lio~t9m~pliryClpleofm~.sooe
pressuredi$ttibt.itiOri.<·
. .
A•• flexlble•• "avemen~ • havlflga•• thlcknessof46··
mr:n·•• carr~s·a.sWtr¢ • Wl1eel • IQad • of.. . ~· • • •
llIe
cir:()l.dar•• cont<ict•• <Jtll:Clf.tuw~h~s • ~r1e9Ui'ffllellt·
radiu$•• of.150.rn"l; • ~·.thEl·.l~·.·'~'.·~.~ssurnM·
lo~etrans@tt~da(;rQ$$aWi~eare?9f
sUbgr:ade.afanMgle.pf45·,gompute.the"'II11~··
of.the • Vlheel•• IQa<l~.W".ij.lhe.·be@ng • stre$$•• qf
thebasElisOA~MP~/
..,..
Solution:
Solution:
I
_
t = 0.564
-vf
t - 0.564
r
A=~
p
r
_TW
'V f - r
~ ·150
46 = 0.564
W= 50723 N
W= 50.72kN
A=53500
0.62
A = 86290 mm 2
1t,2 = 86 290
r = 165.73
t = 0.564
~5~.~0 - 165.73
t =182.92 mm (thickness ofpavemenQ
Determine the pavement thickness in em using
an expansion pressure of 0.15 kg/cm2 and a
pavement density of 0,0025 kgfcm3. Use the
expansion pressure method..
Solution:
t = expansion pressure.
average pavement densIty
t= 015_
0.0025
t= 60 em.
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TRANSPOIITlnON ENGINEElliNG
~te1llline.theth@fue$.o1lti~
• di@renl1WeS
~t~VemeFltl.l$irigtl1etol~Md~t~W·
.
••
®. RI9ht•• ·.p~v~tnenf WI~h • • aW@~I • t~W
caPa¢jly(.if•• 54-•• "~ • lf.ttl~ • ~lIfif:/~~I~ • ~Mll~
-. ~ steelll>tenmon..•
·stte$S.ofc9~Cf~te.is.M$fAPa'fileglE!¢t.me
A~Q
fl~~ible.·.p;3~Ill~~~ • l,Ifith•• a.W~~~I~Il~.·9f
Unit weight of cOncrete = 2400kglcu.m ••..•.•.......••
onJhe~lls#Qft6eW1yem~Mil!qM[W
Coefficient offriclio~ between pavement . . •• .'
and$ubg~e F1.5·
. .
effectm~o~~.>i
.@
A q:lmehl coFlCrelepavemenlhas athickness
af1Scm.alldhastwolariesof 7 meler$wilh~ ..
fOIlgitudirialpitll.Oesign the spaCing ofth~ tie
baran!l th$~ofbar.
. .. ..•. . • . . .•••• ...•.
$4.kNw1l'l1•• l:ln .•~lO-Wal:ll~~al'il'lgpl'f!~$l.1te·
·····9.15•• MP.~ • • Lt~S.·tN~.prm:ipl~(W~r.~
qIS~i~ut~fl~n • • YIt1~¢t¥ • ~~ •. a~$om.~~tc.
M•• • lt~l1~rnitt~d • • ffl:ros~ • • ·~ • ·Wide•• .9rf!~()f
~vtlgra~at~n • ~I~t>f4$"~ry~tll~t • the·
eqUivale!ltfad~~~.tfJ~.epnta~t • ~.on~~
tlre~I~~llalto1R$ll1Itl'/
® PilY$rnenl•• §u~jec{~~.I(.i •.• lln·.~p~m~Bn·
·pres$l.1re•• pfO·50•• ~g12f1l2Wilh~nay~e·
d~ns.lty
O.O$ • • • 1(9Jcm~.·
pavrllirnt•• • •
..pf•• •
.
Expr~sslnmm.
.
.AUowablebondslress in concrele:i24kgfri~
us~ 16rrimtfiam. sleelbars.
.
...
Solution:
..
Solution:
CD Rigid pavement:
t=~
---
t ='" /3 (54000)
'I
1.6
t= 318mm
® Fle>..fb,'e pavement:
t =0.564
t =0.564
-{f
-v
1m
r
54000
0.15 -165
l-3.5
t= 173.4mm
® Expansion pressure method:
t = expansion pressure
Ave. density of pavement
t = 0.50
0.05
t=10cm.
t= 100mm.
nDL x Bond Stress
~.
...-...-...-
...----------L
p
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TRANSPORTATIOII ENGINEERING
Consider one meter length ofslab.
W= 0.18 (3.5)(1X2400)
W=1512kg=N
F=IJN
F = 1.5(1512)
F= 2268 kg
dLy
Solution:
•J.'.'. , ". , :."•. :. .
B /
-'_':-."'-';".'~.~ :.:~.-"
.. .
:
.
rf~·::···::·
.
••.•
w
As fs = F
As (1600) = 2268
As = 1.42 sq.cmlmeter
AS=~(1.6f
As = 1.256 sq.cm.
.
1256
Spacmg = 1.42
Consider only half of the section
(Using Principles of Mechanics)
Spacing = 0.88 mUse 80 em on centers
W = (~)
Note: The length ofbarmust be at least
twice the computed value.
W=720L kg.
N= 729L kg.
Length ofbars:
As fs = (n DL) (Bondstress)
1.42 (1600) =n (1.6X24)L
L = 18.83 say 19 mm '
Use L = 2(19)
L=3Bcm
F=fJN
F= 1.5(720L)
F= 1080L kg.
i'1••'W
21ookg/cu,m.AlIoW~~leJeri~ll~stf~6f
concrete•• isO.8.kgfcm2and.l&ato/ste~liS800
k9lCmZ·•• • unitwelght··9f:.*~llsT50q.kgtcu,m
st~~lbars.~a¥jng • a•• diam~terof.1.~.<:m ••••••T()tal
reintwcemeQtis4kg,tm~andi~equally
dl5IribUledill.b()JhQi~clion5·.·.Forpl~i119!m1~nt
cOncrete(wUb@tdoWel~)' ..
.
b;==3m
b~3QOCl1t .
~ (3)(2400)
T = 300(20)(0.8)
T=4800 kg.
T=F
4800= 1080L
L=4.44m.
. L = 2fx 104 = 2(0.8)(10)4
uD
1.5(2400)
L =4.44m.
A concrete paVement 8m wklei:1nd1l:l9mrn
thick is to be provided wilhaceliter
longitudinal joint usIng 12 tnm o bats; Th~
unit weight of concrete is2,400kg/m 3 .
Coefficient of friction of
t~e
stab ali the
subgrade is 2.0. Assuming Em "llowable
working stress in tension for steel bar$at
138 MPa, determIne the spacing of the
longitudinal bars in mm.
.
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TRANSPORTAnON ENGINEERING
Solution:
Solution:
~.'
., . . ;"--:.', -?f;t>'
',.;..'.-H'
.,.-··t.'· ~'1·~'
". :,. ij.
'.:
"s.'
4,
:,-
..
~~~~.
.
,":.
,
-'
:~,: ~'.~'
.
"~:
.',,'
.'
,
'
w
.'
/-":"""-T
.
w
T
/ ...~-T
0.1_,.=<-'--~"......
W::: 0.60 (0.15)(4.5)(2400) 9.81
W =0,15 (4) (8) (2400) (9.81) =14126.4 S
T= fs As
T= 138~ (12)2
T=15600N
N= W= 14126.4 S
F=IJN
F = 2 (14126.4) S::: 28 252.8 S
T=F
15600 =28252.8 S
S= 0.552 m
S=552mm
A 12-mm0Jlart~. ~~:@.CJ~J~e IQn9lliJ91~!!!
bars' of a concrete pavement. It isspac~at
600 mm'oocenters. The Width of roadway is
meters and the coefficient of friction of the slab
an thesubgracie ts 2.0. Thickness of slab Is
150 mm. ,If the allowable band stresSlS
0,83 Mrs, determine the length of to
longitudinal bars.
Solution:
S=600
:[ifH1H$
8=600
W=9535.32
W::: N::: 9 535.32
F= IJ N
F =2 (9 535.32)
T:::F
T= 19070.64
T:::ndx U
19070.64 = n (12) x (.83)
x=609.5 mm
2x = 1 219 mm (length ofbars)
The width ofexpanslon joint gap is 24mmin
a Cement coricrete pavement If the lay1tt9
tbmperature ls12'C and the maximum slab
temperature, is 50'C, calcUlate the ~paCing
betWeen the expatwionjoltit$; , ASsume
!iOefflCienfoHnei'ifiafeXl58nmOnof cont~1tt
he9.5 x 10. 6 per C'. The.expansion joint gap
should be twice the allowable expansi{ln in
concrete.
Solution:
.,
t 24
ExpanslOn In concre e =2'
Expansion in concrete =12 mm
Expansion in concrete =0,012
tl =0.012 m.
tl =KL (Tr T1)
0.012:= 9.5 X 1(J6 (50 -12) L
L =33.24 m.
(Spacing between expansion joints)
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TRANSPORTATION ENGINEERING
@f(OJTrlh~• result<1fWtPr()gtorSOTp;lction·
·.·t'*tl3~~L fhilfllW1PI~flog ptg(~~iJ'l~
ppei'alionsll.irtdiQateS.l1'IarlhematerlafS
cOmPtlt~tl1emodulus.ofsu~grade. reactlo~if.a
forCEi9f.50Ql:lI~ ••.•is••. appJli¥'4.0der • ~ • Clrcylar
pIClt~h~~i1l9aradll.lsofRm·pr()(iuAAsa
@rnpact~dpn th~'PadWl3Y \'Iilth~v~~
.···VQi9rmtoof().5Z,•• I~eW)di$!Ultled§~mpl~ .•
PtthaO'l<ltMalt~k~Jrorn.th~.IlQtrow~f
h<lsa~idJati()(t72·W~at$tirlnkag~
ta¢tO(ShOl.lld·.beU~e(l(n(;()mpUtln9.~0(I'()V.i
d(:l~~(lfQ;t~in9hllflijffl"fh(:lPlate<>
·<Jfldembaflkment·.quantiti~.
Solution:
Solution;
F=SA
5000 = S (n)(9)2
S = 19.65 psi
CD MOdulUS ofsubgrade reaction:
F=SA
5000 =S(n) (9)2
S = 19.65 psi
Modulus subgrade reaction - 0 shre~~
e eClOn
Modulus subgrade reaction = 19.65
0.12
Modulus of subgrade reaction = 163.75 psi
.
Stress
Modulus ofsubgrade reaction = 0 efl ect·JOn
.
19.65
Modulus of subgrade reaction = 0.12
Modulus ofsubgrade reaction = 163.75 psi
® CBR of soil sample:
CD U\?On completion of grading I:Iperationsa
stibfJrarle waslest~dforbearjhg capacity
.'. by loading •onl~rg~b~a.rin~rptates .
p
Stress =;;\
58
. •. . • ~~fu~~t~~~~gl~~.~~s::fi~e~~p~
circular plata having a radius of 9 Inches.
produces a defleetiOnof 0.12 inChulider
lhepl~te.
'"
Stress=--
~(5~
4
Stress = 2.95 kg/cm 2
..
® The sofl sample wa$(lbtiHn~d·fromthe
project. site after' completioh ofgradlllg
operatIons and . the CBR lest was
c?nducled at field density. The'sampte
~th the same surcharged imposed upon it
IS. then sUbjected to apenettatlont~tbya
pIston plunger 5 em. in diameter at a
certain speed. The CBR value of a'
standard crushed rock for 2;5rnrn
penetration. Is 78.68 kglcm2. Compute the
. CBR of the soil sample when subjected to
aload of 58 kg it produces apenetratioh 01
2.5mm.
.
CBR = 2.95 (100)
78.68
CBR=3.75%
@
Shrinkage Factor:
=(el-~)100
SF
..
(1 + e1)
SF
. .
= (0.72 - 0.52) (100)
(1 + 0.72)
SF
=11.63%
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5-516
TRANSPORTATION ENGINEERING
The soil· sample was obtained from the project
site and the CBR test was conducted at field
density. The sample with the same sUbgrade
imposed upon it is then subjected to a
penetration test by a plslon plunger 5 em.
dlam. moving at a certain speed. The CaR
value bf a$tand~rd crushed rock for 2.5 mm
penetration is· 70.45 ~g/cm2~ Compute the
CBR of the soil sample when subjected to a
toad of 55.33 kg it produces a pen~tration of
2,5mm.
.
.
COl1lpu~the • • CBR.ofa • • soil•• safllple, • ff. • • the
$ample • j~ ••• $U~lected • • tQ•• .~ • I();!\d • • Q~ ••• a • pls!qn
plung~rgirt@~lljryd;arn.andprod~~~ • ~
penetratiOl1.e>fO,1g·.illch.•.•TtieCBR.v8Iue.ofa
standardptJ,lshtl9rocKJof<lO·WWc~
pel1etratlonW1Q¢O.p$ial1~ • • forfM·.A·2I)..lnc,h
1ratlQO • 1$.1500•• pSi.•••.•ThelOad•• appllEldl$
PeM
1600 lb.
. ..
.
Solution:
p
Stress=-
A
1600
Solution:
p
Stress =-
Stress=--
Stress=--
Stress =50.9%
~(2)2
A
55.33
~(5)2
Stress =2.82 kg/cm2
2.82
CBR= 70.45 x
CBR=4%
100
CBR= 5~~0 x 100
CBR=50.9%
l'h~.C~R.valu~()t~.~ta°gar<f9N$h~r{}Ckfot
Thestandaro CBR value of a standard crushed
rock for 5mm penetration IS 105.68kg1cm2. '
.Compute the CBR of the soil. sample whose
r~IHHs follows:
,., ..'.
Load applied::: 752 kg
Penetration =5mm .
Dia.of piston plunger =4 em.
Solution:
p
Stress =:4
.
75.2
Stress=--
~(4)2
4
Stress:: 5.98 kg/cm 2
598
00
CBR = 105.68 x 1
CBR= 5.62%
it020jnChP€lllelrationis.m1u~l.lo • 150Ppsi.••• A
soil•• sampl~·.W<l$~~ed.bY.llpply.ing.a • ~a<fof.~
pi$tQn.plung~·.\YtJish.~j'()dlJ(;El$<:I •. P;Ell'letr~tionpf
{),20.inct:.ifth~di.lIli~ter.()f.th~p'stonplunger
is.equal•• to • 1•. 98.lOchl:!s••••·lllhe•• cQfllPut~d .CBR
ofthe.·~ •.$.~~.js41~,-j;9ffiPllt~tl:lel()M
appJied.to.the·plul'\ger.
Solution:
CBR = ~~~s x 100
7 - Stress (100)
4 1500
Stress = 705 psi.
p
Stress =;4
p
Stress=---
~(1.98)2
p= 2171 lb.
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TRANSPORTATION ENGINEERING
..
The.COIJlPuted•• B~R..of~~Msarn.~.~.whICh
lNas.te$t~ • ~Y.·~BPMI'lQ9J~9f.7S,3 .• ~~ • QI1•• a
piSf()l'lpluDg!lr,V·mi(;hp~~!lfrat~5§rllfltiS'
equalt9a·92%.• • • rheqaR\I~uepfa.staO~~rd
..
rffi
_,~
.
..
CrU$heq•• rpC~Jot.~.·Swm·Penetrati(m.IS1Q5.&&
··kg/cI'rl2.·.·•• ••CPI'rlPt;lt~ .• th!l••.di<1.rJ'l.()f•
. • th~.plston
~
.
.'
plun~r.
Solution:
CBR = Stress
105.68 x
Yo
:~i~$'
100
362 - Stress (100)
. - 105.68
Stress =3.83 kg/cm2
® Cpmput~@:la~$plut~ • s~tiMg~vityof
." ··lhebnurryl'lous·.mIXfi1fe ,•• • ·» • • • • •·.• \•• • • <•• • • • •.• • • •
® COn1pUI~.lhe • blllk.~P(lc~c • 9ravnY • pf•• th~
J.X)1'l1@CledspeC1 1m , >
@ ··Compple•• th~.P9ro~flY • 1:)flhe·.~mp'We~ • •
specimen.
.
..
p
A
m
Stress=-
3.83 = 75.3
'!!:.cf2
4
Solution:
D=5cm.
CD Absolute sp.gr. of the bituminous mixture:
G
100
Es..ElEa
+ +
Gs Gf Ga
100
ti¢.ITIBute•• tMll)odqI4S!ir·.~Wti¢ily.t)f.lhe
• $Ub'
grad~ifth~1'Jl99\:1lw;otellJ§ticityofJhe
·pav~Iii~~rf$·1~OMR~·WitliK$fitf#e[~]aCior()r
0.50("
.
...
® Bulk specific gravity:
d=~
Wa-Ww
1140
d= 1140 - 645
Solution:
E 1/3
Stiffness factor = (r)
E
0.5 =(120)
d= 2.303
p
1/3
_b.
0.125 -120
Es =15 MPa
~.
@
Porosity:
Porosity =(G - ~ 100
P
·t - (2.368 - 2.303)(100)
oroSI Y-
Porosity =2.74%
2.368
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5-518
TUNSPORTATION OGINEERING
A plant mix is to be made usfnQlhe foil.
percentages by weight Of !he lQtal mile •.,
••
··~ ~;~~~lc:etlti'e
••
$P'jf·•• • 0f•• •
th~'
(i)••·•••lf.thflteSI•.mePilT1~riw..~h$· •.t1~O • gr,ln.<lir
.·•• • • am:t.·was•• fouM•• • t().wel~h.·F3~ • gt•. Wb~m
".. ~usp~mjEld • in • Water;··(:oWPul~ • tf)El.pUI\(
· ·.·• • ·~P.W,·ilf~he.C9rnPl~~s~~rt •.•.• • .• . •. .•·.•.
@•• ·.Qpmpl.lle•• the•• potpsl!yof.thEl•• FompattEld
specimen.···
. .
sand(sp.gf. tt 2.em .' ';9%'·· '.. .
Filler (sp.gr. =2.70)
14 %
AsPhalt Cement (sp,gr; ., 1;01)
.7 %
WI. of the wmpacted spedmen in air .
'. '=1265.5 gr '.' . .... ..
. .. .
Wt. of the compacted specimen when . . .
suspended inWater =720 gL .
(j) Determine the absolute sp.gr; of the total
. compacted specimen. '.'
.
. .
® Delermine the bulk $p.gr. of Ihe compacted
specimEln.
....
.•... "•.'
@ Determine the PQfOsily Of the compacted
specimen,'
Solution:
Solution:
CD Absolute sp.gr.
CD Absolute sp.gr.
G=
100
~Et.E.a
+ +
Gs G, Ga
100
G = 79
14
7
2.68 + 2.70 +1.01
G = 2.404
G=
100
~frEa.
+ +
Gs G, Ga
100
G= 78
14
8
-+-+-.
2.60 2.70 1.02
G = 2.324
® Bulk specific -gravity: .
® Bulk specific gravity:
d=~
d=~
Wa• Ww
1265.5
1265.5 - 720
d=2.32
Wa.Ww
1130
d= 1130 _635
d = 2.283
@
d=
Porosity:
_(G-d)(100)
PorosityG
. _ (2.324 - 2.283) (100)
Porosity 2.324
,
'
Porosity =1.76%
@
Porosity:
Porosity - (G - d) 100
-
G
.
(2.404 - 2.32) (100)
POroSIty =
2.404
Porosity =3,49%
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TRANSPORTATION ENGINEERING
Solution:
A@or~or • fOI1lP~8ed"'~sph~lrC()flcn~l~
pav~m~nl.~ te$tedfm.$p~lfl~gfa¥ilY·· • • TM
•
follwting)lil'ligtJISwerep!:>t<lirllQ, ..
..
..
Weight.·Of.dry.SP~i@~.in.pir.~.2QQ7.5• 9~nlS.· .
··W~igtJt·()f.~~ffiElll.plll~paraffin.CCl~tihg.ifl!1ir
¥20~·5g~rll$\
.• WEligHI()tspecjillel1·plusparjffircl.l<;l~il1g.1n
Wilter'!'113&'()~~rtl$\
• •.•.• • • • • .
Btllks~t;lfjcgra\lity.ofthe.parafjil1·=.{J.903···.··
Solution:
d=
W=VxO
W= 11500 (0.075)(2.47)(9.81)
W = 20898.98 kN
(requiredwt. of surfacing)
20898.98
A
(O-A)
O-E- F
No. of batches required = -go-No. of batches required = 232.21 batches
d = bulk specific gravity of the core
A = weight of dry specimen in air
0= weight of specimen plus paraffin coating in
air
E = weight of specimen plus paraffin coating in
air
F = bulk specific gravity of paraffin
d
G= 6
8
41
45
-+-+-+1.02 2.75 2.66 2.77
G= 2.47
2007.5
(2036.5 -1135) _(20360~~~07.5)
tj = 2.309
d = 2.31 (bulk specific gravity of core)
t@·.QtYffiMspf.a • SfJmple••(lf•• ~~gregiltesi$
.·~.Wlq9· . . -rn~l1las!Jjn~~!1~rcIl~d9ry.g>n~lti9D
i$~9009'ThElYOluw~ptag~r~gate$
~~q9dillg • ·tr~ \'pllJrlle•• pt@sprbed • • \',I!1ter•• i~
730Ctilh . • .
. '.' .
.'.
(i) • • • C9nJPYl~th~~pp!1rE!tlt.$Repiftc.graVlty • qf
JtJ~saMPI~"lg9reg<:!te~.>
@.•. .• (j<;lIR~l!1t~ll1tlR~pent~~abSorplion.
@¢~lqlll~tetr~~lliK$p,gr.¢th~$ample
~gr@ale$.
.
Solution:
• G)
Apparent sp.gr.:
W= VxO
1980 = 730 (1) Gs
Gs = 2.71
® Percentage absorption:
.
(2000 - 1980)
%absorptIOn =
1980
x 100
%absorption =0.01 =1%
o
@
Bulk specific gravity:
VVI. of water = 2000 - 1980
VVI. of water = 20 9
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S-520
TUNSPORTADON ENGINEERING
Vol. ofabsorption water = ~ ;: 20 cm3
=
=
Bulk vol. 20 + 730 750 cm3
Fr9lTlt~i9ill~~®ta'~I'\(MfliW~'~!:ltElftlf~
W = Vol. x D x GB
1980 = 750 (1) GB
mIX®Sl@fQr~$hij»¢Q6~··········
.' .•. >.••..••.•. .•.•.••.<..
Ma1Maf$<~~lk /§P~9p~l!it
1~~;rtl1........ ·~·~· . ·······.······1!03.············ ··5.3···········
GB= 2.64
fjl'le~ggt~~ .•••.•••·~,~9?······.············ · · · · · · . ·.·4M<
gQ~~S~~gq1~.A7A)
Max.~p@Ijlbg~~ll¥~~.Ji~gmi~\ir:~
• • i/
~lilM:::M3~/<i.<
~WK$~&:i¢g(~WYW@!OOMtl'#@il@i'¢·.·
• •. .•.•.
~~JR2~<442>
PPillPl1t~th~~ff(lc~y~$p~l?if'9<9fi:lVltyi:lf
. ..
agg~t~>····
Solution:
G;:Pavn-Pb
E.mm.Eb.
Gmm -Gb
#6Qgij,lfu$«
'.' .
Solution:
G=
100
Ed.+&+&
Gd Gs Ga
100
G= 17
73 10
2.80 + 2.60 + 1.02
G=2.28
A
d= B- C
100
d=114;-60
d = 2.04 (bulk sp.gr.)
Pmm = total loose mixture
Gmm ;: max. sp.gr. of paving mixture
Gb = sp.gr. of asphalt
Pb = asphalt (percentage by total wt. of
mixture)
G = 100 - 5.3
se
100 5.3
2.535 -1.03
Gse = 2.761
A tabulation shown are . materials and lIs
properties whiCh .are usectll\· a COMPacted
paving mixture:"
'.
.
. MaterialS
. $1):9f. 'BuIK
Coarse aggregate ....
sp.gr.
. 2.69
weigh!
462
46.0
V= (Q,:,g1 x 100
Fineaggtegate .
Asphalteemenl
V= (2.28 - 2.04) 100
Compute' the asphalt absorption of the
aggregate expressed as percentage by weigh!
of aggregate. Max. sp.gr. of paving mixture
G
2.28
V = 10.53% (percentage of voids in the
laboratory molded specimen)
Gmm =2.54.
2.7';.
•. % of
. 1.2
7.8.
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TRANSPORTlnlN ENGINEEBIIIG
Solution:
Pbs = 100 (G se - Gsb)Gb
,
Gsb Gsa
Gse = effective sp.gr. ofaggregate
Gsb = bulk sp.gr. ofaggregate
Gb =sp.gr. of asphalt
Pbs = absottJed asphalt
G - Pmm-Pb
.se- Pmm Pb
Gmn - Gb
100·7.8
Gsa = 100 7.8
2.54 -1.02
Gse = 2.906
G- P1 + P2
sb- P1 P2
-+G1 G2
G - 46.2 +46.0
sb- 46.2 46
-+2.69 2.72
Gsb = 2.705
P = 100 (2.906 - 2.705) (102)
bs
2.705 (2.906) .
Pba = 2.61
47.3 +47.4
, Gsb = 47.3 47.4
2.689 + 2.716
Gsb = 2.702 (bulk sp.gr. ofaggregate)
- Pmm-Pb
Gsa - P
rrm Pb
Gmm - Gb
100- 5.3
Gsa = 100 5.3
2.535 -1.03
Gse = 2.761 (effective sp.gr. ofaggregate)
Asphalt absorption of the aggregate:
Pba = 100 (Gsa - Gsb) Gb
Gsb Gsa
P = 100 (2.761 - 2.702) (1 03)
ba
2.761 (2.702) .
Pbs =0.81% by wt. of aggregate
Effective asphalt content:
PbsPs
Pbe=Pb-16O
Ps =47.3 +47.4
Ps =94.1
=53- 0.81(94.7)
Pbe·
100
Pbe =4.53 (effective asphalt content)
rilll~~tl~~jt~~Bfllnlt~I~~~~e~
· ·•. • • Ma~~al~ • • • • Perc~n~~].~.fotal • • ~~~.• •
.• • • • • • • • • • • • • • <•• • • • • • • • • • • • • '• • MI*.tW.W#19N...
•·•·•·. . ataVity
f¥lpn~Wt:elllehtl.01'03()
Mll)ernlflll~r.
fi/lElilgStegale>
... 7:<)' . •'. . '. <MOO
g.§~{pcilksP·gr.)
·.. ··:3().o
.Coal'$~~gg~gate . ' . ·~2~e11{bl.llk$p.gr.)
Solution:
G - P1 + P2
sb- P1 P2
-+G1 G2
Max.•• speCific.gravity•• qf.lhe••p~Vjng • nlixture
mfTI"'2.478
>
BUlk~pecifi().grayity.qflhe.sqrnp?cle~paving
• •.• • • m~ure.satr\ple·.qIllP.5 • 2.$a4<.>•• ·
COmputethepercent<lg~.ofvoids.inlhe.
.. comp<!ctedmineralaggregales.
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5-522
TRANSPORTIDON ENGINEERIN.
Solution:
Percentage of voids filled with asphalt:
100 (VMA· Va)
VFA=
VMA
P1 +P2+ P3
Gsb = P1 P2 P3
VFA = 100 (14.44·3.6)
14.44
VFA =74.58
-+-+-
g1
g2
g3
56+30+7
VMA = % of voids in the mineral aggregates
G bP
VMA = 100---lD........-§.
Gsb
Ps =30+56
Ps =86
VMA = 100 _ 2.384 (86)
2.668
VMA = 23.154% of voids in the
mineral aggregates
-rti~pmpornotl$bYv.'~lg~}~hCl$P~¢lfl¢
9fl:tyiti~~·.·Bt.~ch.()fj~e • l;(m~tillJt!I'lI$9f:~.
Pattll:ulafsltijel$Phaltpavlo~~~m~~
MltiWs:>
":':-:--::'::-:-:"::,:::.:
AsPha!t(lemern
limestOne dust
Sand
",-<,:,"".,",::.>:'.,-,»"::,,,:::.'.:
':.>:-.. , ..:.::::-:....,':,..
A•.•9'lilldticalisPecimel'l • ofm~ • • n@M'~.~a$
qOITlPll~ • th~ • p~r~ni~ • •()f•• vqiclS••. fiUElcl•• W~h
asph~lt~JhemaXrmum$p'9totP~Vl?9
ll1il(twe(;mtJl5Z.§~§,6~!Ksp,gr,ef
·.~°tnpaCIMmIX • §fflb5?·<442;<•• ·Percenl~g~ .
V>'e~ht.<lf • #~p~~ltceJ'llent • i~5.SWhjl~thi1tpf.
fio~ • aggrl:!g~l~s.!lh~~()~I'$~ • ·a~gregaleS~f~
47.3•• ~Od.47.4f~~J)i1i~i~M • ··Th~b1JIK()f$P,gt.·
Qfaggreg~le~G$b7 • f·7Q3 • andthe.• effeptiv~
specincgraVltyofaggregateGse '" 2.761. ..
Solution:
Airvoids(VN = (Gm~::mb) 100
V = (2.535 - 2.442) 100
A
2:535
VA = 3.67 (air voids)
Percentage of voids in the mineral aggregate:
VMA =100- GmbPs
Gsb
VMA = 100 _ 2.442 (47.3 +47.4)
2.703
VMA = 14.44
·m~~.in • • lhe•• labpralory.and • W~lmtedmalt'
~iJ(UrrVr.ltetWithJh~f?llO't.'j'1gres\!lts,)
W'eigbtpMry·specilTlElni9.~r::.1t1·95.9rM1s .• •
W~i9nt(lf$alU1'ated;$l.1rlaoo;.({ry$Pecloumkl .• · .·•
~rR11Z·09g@Rls.<.
VVeight9fsaNratedsp~oim~in\'later.~.61.29.
C9mp\JtemebulkspeGlfiC9r",vilYqflh~
(:(lmpa~eospeCimell.·
'. . . ."
Solution:
A
Bulksp.gc.= B~C
A = weight of dry specimen in air
A = 111.95 grams
B = weight ofsaturated, surface-<iry
specimen in air
B = 112.09 grams
C = weight ofsaturated specimen in water
C = 61.20 grams
A
d=B_C
111.95
d= 112.09-61.20
d = 2.20 (bulk sp.gr. of
compacted specimen)
.
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TRANSPORTAnOIL ENGINEERIIiG
Solution:
<D Length of heel:
Tb#.tfo/ma.ssQf~~ttml~ot~ggr.~g*~~.
a =length of heel
Frog n. heel spread
c9ffll~@l.j$1~'~9t~~.n\ElV91t@~~n~~·
aOOt~gaf$s ... exqludlrt9thffYtllum~(lf~$Qrtled
9 =length ofheel
336.11
length of heel = 3025 mm
19Q$gr-!'lll'l$;~m~~~j$~~~Il@tted4ry
.".at~ri$440.6qn3,P~lcw~t$th$btJtt(
~P~~9@yity·ofthe$!lrnP/El<Qf~g~flll'.·.·····
® Total length of tumout:
=3025 + 1820
=4845mm
Solution:
wr. of water = 1226.8 - 1206
wr. of water = 20.8 kg
Vol. of water
Vol. ofabsorbed water = Density of water
@
Angle subtended by heel spread:
1
8
Vol. ofabsorbed water = 2~.8
Frog no. = 2Cot 2
Vol. of absorbed water = 20.8 cm 3
Bulk volume = 20.8 +440.6
Bulk volume = 461.40 cm 3
9=2 Cot 2
1
8
e
18 =Cot~
Bulk specific gravity: GB
8
MD
GB=-VBW
Va = total volume of aggregates
including vol. of absorbed water
2=3.18
.
e = 6'22'
1206
GB = 461.40(1)
GB= 2.614
A•• turnouf··haslllength••·.of•• ·heel.·equ~l.to
SOfS 'mtnandah~( spre~dof336.t1 .•·.• lf the
lenwh.oflheto~j519~5ml1'l,.<:Ohlptile.thefrog
numberotthetumoul.
I~JijiP@tha$a.frognti/tltl(!r~fWWlth{a.l~m
Wl'iElElI~pr~adequaUo~36,11·mm;·
Solution:
• .• • •·•••.••.•.••.•.••.. . •••••.. •
~CQfhPUtethelen9thofh~ .•.•••
F no = Heel length
rog . Heel spread
>H•• • Wtll~t$~etotall~rlQt~Aftt\~tumout
Frog no. =336.11
.·~ • • . lftbl>length ·of~he.tw • ~.~ql@JtQ182l)mlll,.
~}¢i>n'lpt.rle~heangle.$ubtl,mQed.by.the·h~1
.
:~d.(·······
3025
Frog no.
=9
.
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5-524
MISCEllANEOUS
.F~t<lt(;~srieS~ • tl)9$eth",t•• te$ult•. jll~tle"st
()@•• ®attl,.Whileetcl$l1~$tfult.~s.ult •. inil)j4rie$
~lltrlQ~<ltl$are.cl~~ifielfa5p~r50JI~Ul'ljUty
bl1t¢r~~t¥$ttlatr~$~tt • • h$ttl1~!f.d~lt:il'lor.
·k@rleSbml'tVi)lVedq~~@lifpro,*nrare.
~~~fied~~proPfflY~rnage· • • ThjslJle~()(.1gf
~Umll1Mfjl'lgcra~he~ • • i$ • • Co:mmo~W • I1$ed•• tq
m<ilkflj.complilrjsl)n$~J • diffe~rtl()c"U9nsby·
as~ignlJl~ • <lweight•• $()ale.lP~ch.·craslr.Pllsed
l'n~ • fj9ur~ • • ~b()r-'~ • • V@tcl~.jray~l@l • • • ~1
A~n$tant • • $pe~d$ • ·• QQ•• • t\¥Q••·•• I~@nlgh'f~Y
tl~tWeeIlStlcUOl'l$l<aiJd)'witBtheir.p~~ol1
~M$fl~~~.()~i~~~rW1.lJ1~tl3m?f.ti~ . ~y
ph9tpgrap~Y+lm9~t'V~r.l~t~· at•• ffllli t• x
ol;tsery~s.4.va~iflespas~i~glhl"Oughp~rt • )(
gyriMjJ•• P~.~ •. ()t"~~ng~,·· .•. ~ .veloWies
(lf~vehililes~sMea~reoare·4&,4~,40"aod
30mPhJespect~ly.<
WI.jt~~v~ritY; . • .A.typil:<ll.. WEll9hillgsgale.h"¥~
beefiusedwhichis·asfoHaws,
.
:::::;:"-.::>,,:":::::-::::-;-:
PetSonalinjury=a.. ... . .
Property damage only ~·1 ...
. ...
.
30 mph
X
.
year at apartfcular slte, compute its severity
Solution:
Severity number =12 (1) +3 (3) + 1 (5)
Severity number = 26
The density of traffic in acerta:in obsElwation
point Oil a highway was recorded to be 30
.vehlcles per km. IfthEl space mean speed·of
Ihe vehicle is 50 kph, haw many vehk:les will
be passing every 30 seconds.
Solution:
K=!L
fJ.s
30 =!L
50
q = 1500 vehicles/hr.
.
1500 (30)
No. of vehIcles per sec = 3600
No. of vehicles per sec = 12.5 vehicles
40 mph
45 mph
45 mph
• • • . •
IY
___.Vir.ctWn ofjlow
lfon~ fataLcrash,Spersonal trijuriesand 5
property damage crashes occurted durifjga
nUmbeI"•..
.
15' 20' 30' 20' 35' 20' 85' 20' 45'
..
Fat~ity =12
..••.... .
l..----.----
~
300'
® CQmputethe~ty¢t~fflc.
@iqoMW\Eltl1efiffi6.M~~spee~·
@Computethespa~mean~peed·
Solution:
<D Density of traffic:
4
K=300
,5280
K = 70.4 vehicles per mile
@
TIme mean speed:
30+40+45+45
fJ.t=
4
fJ.t= 40 mph
@
Space mean speed: .
nd
fJ.s= 2)
300
300
300
300
Ll " 30 (5280) + 40(5280) + 45 (5280) + 45 (5280)
3600
3600
3600
3600
Lt = 6.82 + 5.11 +4.55 +4.55
l:t = 21.03 sec.
_ 4(300L
.
fJ.s - 21.03 - 57.06 fps.
_ 57.06 (3600)
5280
fJ.s -
fJ.s = 38.9 mph
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MISCELlANEOUS
OfwoSetsfJfSlll~~~areCOllectiggifaffl¢daW
• ~; • $~~~&~;~ • ~Bt~a$h~i'jh~~!'.
ililfi.__l
:1~1~7i~~i.·
1~1~11Bj:l;f
111I1[tlll~~I:I~: 1~_illilli11
So/won:
CD Density of traffic:
5
K= 600
5280
K = 44 vehicles per mile
··~·~r~~()t~~~fWW~I~13IIYtlll~
So/won:
CD 24 hr. volume of traffic for Tuesday:
24 hr Vol.
_ 400i29}+S35122.0S}+650118.80)+710117.10l'lil50{18.S2l
5
.
= 11,959
® Time mean speed:
I-lt=
50 +45 +40 +35 +30
5
1-4t=40mph
® Seven-day Volume oftraffic:
7day Vol. of traffic;" 11959 (7.727)
7day Vol. of traffic = 92407.19
@
@
Space mean speed:
nd
f.4s='iJ
5 (600)
1-4s 8.18 +9.09 +10.23 + 11.68 + 13.64
f.4s = 56.8 fps.
56.8 (3600)
f.4s - 5280
1-4s =38.7 mph
Average annual daily traffic:
MOT= (MEF) (AD?)
MOT=(1.394} 92~7.19
MOT= 18402
.the tadiusofthe summit curve having a ~e
stoppil1g distance of 130 m~ is e~ualto
35Sa m. The tangent grades of the 5ummit
curve is +2.6% and" l.$%.lflhehelghf of the
observer above the road surface is eqUal 10
IU5. . .
A•traffjc•• engloeer••utgerltly••needs.·to••del.errnine·
tt\e·.AADI.()rl•• ~ • rUtal.·primary.rg®•• that.ha$.th~
y()lurn~.·.<:llstribution • chl!lr<tpt~rt5tic5 • • shoWl'l•• ln
lhe·Qi"en.tll,ble.•/She.cofl~ted.fhe.data.$hown
belowona.Tuesctayduririg.·themonlh•• ofMay.
(j)
@
@
..
Compute the height of the object above the
road surface that the observer could see at
the other side of Ihe curve.
Compute the length of the summit curve.
Compute the height difference from the
beginning of the curve to a poinf 30 m.
horizontally from il.
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5-5]6
MISCELIINEOUS
Solution:
® Length of curve:
CD Height of object above the road surface:
S2
R =---=:.-.-...,..-
L=A V2
395
A =2.2 + 2.8 =5
L= 5 (99.79f
395
L = 126.05m.
2 (...fh1 +~}2
3558=
(130f
2(fu+~)2
(fu + -Yh;)2 =2.375
fu+~=1.54
~=0.316
~
® Length ofsight distance:
AS2
L= 122 + 3.5S
5 S2
126.05 =122 + 3.5S
15378.10 + 441.175 S =5 S2
SZ· 88.235 S - 3075.62 =0
S= 114.9Bm.
=0.10m.
® Length 9fsummit curve:
L=/\,(91-&7)
100
L=3558 (2.6 + 1.8)
, 100
L= 156.55m.
® Height difference from the p.e. at a
horizontal distance of 30 m. from P.
x2
e..
y=g1 x+ 2R
Y-- 0.026 (30) + H..
2(3558)
.~• ¥~~I.$IM@I@~#~$~9~otQrt@M<)f
+~~anli;~.a% .••.• ~. . hllnz9fl1Sl~js~n~frorrl
~.C.t~r11C~@~~bE!.vartex.(}fJhe
;.·!··f• ••IIII~ili~~.·
y=0.654m.
Solution:
CD Length ofsummit curve:
s-~
•
1- g1' Y2
.~~~1.2r~~~J~;~t~~~~:~; ~.
0.05L
113.64 =0.05 + 0.038
L = 200m.
(j) • • • q®iM~e.ib~ • @I~ •. ~p~® • ~hllr~ • •9at(X)UI~.
® Radius of summit curve:
L=Ry (gl'!b)
/pa~~thNlhEtS®.c.1JlVe,.<r
• leOfllll•• ?t:.th~ • • sa~• •yertlcal.
qy'!'E!,,« ......<
.~..9(jrnept~lhT
@Col1'lputethelengltlQfth$sighldislSf\G8.•.•.•.
Solution:
CD Speed of car:
R= VZ
6.5
1532 = V2
6.5
II" 0070 knh
~oo
ZOO = Rv (5 + 3.8)
100
Rv = 2272.73 m.
® Tangent length:
T=&~
2
100
T =2272.73 (5 + 3.8)
200
T= 100 m.
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5-527
MISCEllANEOUS
Solution:
CD Design speed of vertical sag curve:
V2
tnesiqbt•• di~~ncll.<if • ~ • •liaijvett1caI9~1$
Rmin. = 6.5
~uattcl11qm·)Htll:ir~M~~tsrad~~()Hh$
. '.
',' .' . .
1500=~
cu~<te~2%<Jn(\+3%.··
6.5
V=98.74kph
<$c6mMtetootength6ft~~l:OtIJ~.<>
®••pom~te • ·tne•• d~i9n • SP~~6f61e¥ert~I •
~~99I.1rv~«>
~ • • compt!t~th~.mlninlum.radius.oflh~sa9'
® Length of vertical sa9 curve:
AV2
L= 395
Yfll'tiqaLcllryEk, .• "..,<,<>i""
A=92-9j
A = 3.2 - (- 1.8)
A=5
L= 5 (98.74f
395
L= 123.42 m.
Solution:
CD Length of CUNe:
A S2
L = 122 + 3.5S
A=grg1
A=3· (. 2) =5
_ 5(115)2
L -122 + 3.5 (115)
L = 126.07 m.
® Design speed:
® Length of sight distance of vertical sag
CUNe:
AS2
L = 122 + 3.5S
. 5 S2
L=A ~
395
123.42 = 122 + 3.5S·
15057.24 + 431.97 S = 5 S2
SZ . 86.394 S· 3011.448 =0
S= 113.04m.
5V2
126.07 = 395
V= 99.80kph
® Min. radius of sag vertical CUNe:
V2
Rmin. = 6.5
R . = (99.80}2
min.
6.5
Rmin. = 1532.31 m.
A sampl~of.cdlnpMfedasphaHk;C9ncrele
wa~cutfrQmthe roadway for laboratory
ancilysjs.ThesamPle Was found· to weigh.
.1020gralTlsin air. Since the sample ¥iasqiJile
A yertteat~ag OUfV& ha¥~ la1l9etl~gr<J~~9f
• 1~8% and -+3.2%. ffthe· Pl'Iiriifl1iim riidius01'
the sag CUNei$ 1500 m. Jong
. ..,. ..
Q)
Compute the design speed of the vertical
sagCUN~.
. ....
. .... ". '.
® Computelhe length of the vertical sag
, curve,
. . . . '..
.
@ Compute lhe length of sight dfstance of the
vertiCal sag curve
.
POl'Ou$,itWas completely CDafed With paraffin .
havifjg ClsP.9L ofM5..• The coaled sample
~ighed 103M grams in air. .The weight of
the coate(! sartlple iowater was 575.3 grams,
(D .. Determine the
@
volume of paraffin,
~:;,~~~~~~Up~; :c%ht .of pavement
® If this asphatllc concrete costs P1QOOO per
fon in place, determine the cosl per stalion
for sUrfacing 8 m. wide and 200 mm. thick
with side slope of 1:1.
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S-S.28
IISCEllI.EIUS
\
)
Solution:
ill Volume ofparaffin:
~ ofparaffin =1035.3 - 10.20 ,
. ~. ofparaffin = 15.3 gr.
15.3
@
Unit weight of pavement sample in grams
per cu.em.
Vol. ofasphaffic sample =
1035.3 - 575.3 -18
(1)
Vol. of asphaltic sample =442 cu.em.
.tJnit weirtht = 1020
442
.~
Unit weight = 2.31 grlce.
@
.le1~;;-eltlr~~~~~:~~~:I~y
. . . . . .•.
. .•
wE@flt()f1¢~Imtx.
Volume of paraffin =0.85 (1)
Volume of paraffin = 18 au.em.
•
<:i:i1;BJI~
•••~li.~e._~& •1.1.~.~;jtt.~w
~iii••:II1i
® YYf1~tpercen1agemlh:1QtalV~ti~dQthll
Cost per station:
.. .... . ..... . .
<#Jgr~gat$9fl¢1.lPy?
Solution:
ill Total volume of mixture plus air:
Vol.=~
Gsyw
----8.4-4----
(
Volume =
2
W = Vol. x Density
Unit wt. = 2.31 gr/cc
.
2.31 (9.81X10W
1000
Unit wt. =22.66 kN/m 3
Weight = 32.8 (22.66)
Weight = 743.248 kN
1 ton = 1000 kg
1 ton =9810 N
1 ton:: 9.81 kN
IAI •
ht _ 743.248
..elg -
Vol. = (1) (1)
Vol. = 518 CU. em.
(8 + 8.4XQ.2O)(20)
Volume = 32.8 cu.m.
Unit wt. ::
1190·672
9.81
Weight =75.76 tons'
@
Totalpercent of air voids:
0.45 (1190)
Vol. of coarse aggregate = 0.60 (1)
Vol. of coarse aggregate = 205.96 eU.em.
_ 0.45 (1190)
Vol. affine aggregate.- 2.70 (1)
Vol. of fine aggregate:: 198.33 cu.em.
.
Vol. ofmmeralfiller=
2.76(1)
Vol. ofmineral filler =21.56 eU.em.
0.05 (1190)
Vol. ofasphalt c:ment =
(1) (1)
Vol. ofasphaff cement =59.50 eU.em.
V:: 485.35 eU.em.
Vol. ofair = 518 •485.35
Vol. of air = 32.65 eU.em.
.
Total cost:: 10000 (75.76)
Total cost =P757,643.22
0.05 (1190)
.
Percent ofair voids =
32.65 (100)
518
Percent of air voids = 6.3%
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MISCELlANEOUS
@
Percentage of total volume occupied by
aggregates:
Vol. ofaggregates =518 -32.65 - 59.50
Vol. ofaggregates:: 425.85 cU.cm.
._1
425.85 (100)
Percentage VUlume =
518
Percentage volume = 82.2%
@
Volume ofasphalt:
~. of asphalt =151 -143.81
~. ofasphalt = 7.191b
\I
f
7.19
v'oI. a asphalt = (1)(62.4)
@
Percent air voids:
Vol. ofsolids =0.835 + 0.115
Vol. of solids =0.95 ft3
Vol. ofair =1- 0.95
Vol. of air =0.05 tt3
Vol. ofasphalt =0.115 ft3
I
(100)
P,ercent air· VOl'ds =0.05 1
Percent air voids =5%
® Percent voids filled with asphalt:
. (1 - 0.835) 100
Percent aggregate voids =
1
Percent aggregate voids =16.5%
Percent voids filled with asphalt
=0.115 (100) = 69 ....L
0.165
.t 7.
@
~.¢I~~i~~I~w~1.~>;
. • · . •·
1_.•I··.• =·I.II'~~·
·.i·.•.•••
Density to which the mix be compacted to
provide a mix with a 75% voids filled with
asphalt:
0.115
.
Volume af voids = 0.75
Volume ofvoids =0.153 tt3
Total volume ofmix =0.153 + 0.835
Total volume ofmix =0.988 tt3
151
Densl.ty =0.988
Density = 153 pcf.
S:;luff(m~
C1J
VUIU/llC.ii aggregate:
Absolute sp.gr. ofaggregates
100
=-6-1-2-9"':"::':::-'-7--3-2.8-0 + 2.7 + -2.7-1 + -2.7-3
.
151
Absolute sp.gr. ofaggregates =2.76
Wt. ofaggregate per cu.ff. =1.05
Wt. of aggregate per cu.ft. =143.81 Ib
.
143.81
Vol. ofaggregate =2.76 (62.4)
Vol. ofaggregate = 0.835 ((J
'
,1tw4iillifB
••
'
~;~
~:Xt~:r~f~;~~·80~Pb
3{NlIO'ltiefOf Iane$"'2>
4, .W!J~el~llfltlltk,.~ • ~,1rn' • • • • • • ,.,.,.,'
5;~l!s()f¢l:lNe;12501ll;>
'.' .'
'"
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5-530
MISCELlANEOUS
~••••••t.kll;.~~.··.I~ • ·~I~~~I:
0••llt~~[I; • ·~.J~~~~~i_,
··~.· • • ~~m~Jm¥E!$~·~~ffl9·9~#\~
Solution:
CD Width of mechanical widening required:
nL2
Wm=iR
Solution:
CD Volume oftraffic across the bridge:
V= 2900-10 C
C= 50+0.5 V
V= 2900 -10 (50 + 0.5 V)
V=2ooo-500-5V
6 V=24oo
V = 400 vehicleslhr.
@
n = no. of lanes
L = length of wheel base in meters
R = radius of curve
_2 (6.1)2
Wm - 2(250)
Wm =0.149m.
@
V
p
V=29OO·10C
V = 2900· 10 (275)
V = 150 vehicleslhr.
@
9,5-{R
80
W = 9,5~
Wp =0.533m.
p
@
NewC=250+ 25
New C= 275 cents
Width of the psychological widening
required:
W,=--
Volume across the bridge:
C= 50 + 0.5 (400)
C=50 +200
c= 250 cents
Volume of traffic:
V=2900-10C
V =2000 -10 (50 + 0.20 V)
V=2900-500-2 V
3 V=24oo
V =800 vehicleslhr.
Pavement width required:
Pavement width = 7 +0.149 +0.533
Pavement width = 7.682 m.
#.;tj)ol•• ~rlpOO • .~~S • 1OW9•• v~h@~~@!Y· • • •·1l\~· .
9~mmt.t()QjsP3.00ptlrvehi{:le;St@iestl<!v~
.11._
~~!ji.~~l~~r~~~r~~~'n~~~r~~~~
Q)•• • Oetenl'line.lh~\I()lu@(lft@ljc:l~~o~~.ttl~
.bodge•....•....•... ..··i
i'<i
@•• 'ta .t()II•• ()f2S•• l::tlnt~·isLWdil •• what·is··tI'lEl
volumEl~Gro$$thebndgeU"/
~• • A.••190l • bQlh .i~ • tobeadded:,·l~~$ • t~(l~{;il'lg
the•. traveltlrtlEl • to • cro$$th¢lm~S~' • Tl'le.
new.c:ost·•• func:tiotl.·.ls••. C;::.SO•• +·.O.2Q.··Y.
Oeterminethe•• volurneoflra1ficth,it.WO~lfl
i"ri"\~~
fha.hrinnp..
~~()vmt~atf~n~MhfOCm~$~lIit911~t~Q
pentsjth~t[<mJcvolllJ1iEl~iIl~crea~~6yj(J()o
v~tii(;~~Pf!rd<lY. .lUs$lsir~t9Inq~~~jh~
t<)nJoap9InLrWheteieveO\,le~M!lP~
.. . .. ...
..
l1lij)(lIl)iZ#~.Y
charge tomaXlroize
revenue; .... ...• . . .. ..... •..•.....•.•..
\D Determine the toll
@ •petermli'l~the
traffic volume per day after
thEI. toll inerea$6, .
.
.
total re~nUe increase with·
tool. ..
.
@ . O~terminethe
thEi new
Solution:
CD Total charge to max, revenue:
x = the increase in toll in centavos per
vehicle
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MISCElliNEOUS
300 + x = Rew toll per vehicle
1000 x
,
10000 - "5iJ' = volume of vehicles per day
R = (300 + x) (1 ‫סס‬oo- 20 x)
fiR
dx
.
.
=(300 +xl (- 2?) +(10000 - 20 xl (1) =0
20 (300 + x) = 1CXX1O - 20 x
6Poo + 20 x = 10000 - 20x
40x=4000
x = 100 centavos •
Total charge =300 + 100
Total charge =400 centavos
Total charge = P4.00
@
Solution:
CD Crash rate per million entering vehicles:
_A (1‫סס‬OO00)
RMEV - AD7{365)
RMEV = 23 (1000000)
6500 (365)
RMEV = 9.69 crashes/mil/ion entering
vehicles
® Rate of total crashes per 100 million
vehicle miles:
_ A (100000000)
RMV¥r- ADT(L) (365)
'_ 40 (100000000) _
RMVMr- 5000 (365) (17.5) - 125.24 crashes
Rate of fatal crashes per 100 million
vehicles - miles:
RMVM = A (100000000) x%
F
ADT(365) L
_ 60 (100000000) (0.05) _
.
RMVMF 6000 (365)20
- 6.85 crashes
@
Traffic volvme after toll increase:
V= 1(}{)()O - 20 x
V= 1‫סס‬oo- 20'(100)
V = 8000 vehicles
@ . Total revenu~
increase with newtoo;
R = 8000 (4) -10000 (3)
R = 32000 - 30000
R=P2000
AnlJrbaJl••~rtedalstreEl(segrJlenfO.30km!o@
~1i~1.IIBtI
•
ll1i\l~n.Vetlipl~~))flr.l@fqf.~ • • ~• ¥~~~ ))EItip(I•• P~
Vo'hk:h•• 120f~l/plve(j • c1fl~t9 • ~rp • i~jlJo/·.~~(jg~§ •
¢~Q$ed·.pmp¢rtY.·d~i1i~~e • pnly·•. • ·rn.ld¢(\tifYing·.
h~iardP:uslocaliql'l~,8ql'lsicl~{!f)aL$>sil'l~lfl
d£!allr()r.lIlNW.~@$h·j~·~\1iVa~~tt9·.~
• prmmW
'
.'. . .'
d®lageCta~Ms;
&>•• • IUsobselYe~lhat4p~r~Ms~ttedon
.•..• • • ~ • • 1],?mIJ~~I¥#1~Il •. ~f:.a.h!9hW~Ylrjmle •.
.·.• .y~~r .• • •. Th~ .• aver~ll.~~¥Jr~ffl~.(APn.·Qn·
'.' ·the.sec!i&iwas·5000vehicles) De.lerinine
. .•• t~~ • • rat~·.qttQt~I~~~t#l$ • eer1pq.Il1111lo0
. yehjples!'I'11~,(BMVMtJ//" .. .
@ • 1l1ereilr~
• ~ • crashe$.·9ccurrlng.·i"a • 2q
rnilesf.lclFqrt • of~ • • ~j9'HW~y.in .•. 9~~ • • Yf.llir.
The.aVf.lrage•• daiIY•.• traffl9.·.on•• m~ $eCtlon
was·6000vehj(:les;•• • ()etElrrninem~.rate • of
• • fatal.crash~s • •
mllll()r•• Ve~lcle • "
milEls, • • if•.• S%·.·.9f .·ffi e. • ·¢rashf.lll·•. ir\VQlyed
fatalitles,(RM\lMF).·
.
•
per·1qq••
@ ·CalciJjaliitheiral¥icba$~lnYehiclespE!f
ktrl(ll3@<
@CIl.I~I~I~• 19E!.~ • ·ye4r~Ver~g£! • • cr~$~ • • r~t~
(Ayg}«
.\ID..
One~~hO(jOf.sUfT\In~rizingcrashesi$tg
.·•• • • U$~Cori1pll.rl~on$ • atdi~erent • lb~ti8n~pY.
··<lssignin9 • ~ • ·weight•• SCillfl·.t9Mch•• H~§P··
. baseanit$severity.Atypi~IWeighing
scale•• hav~.been • used .which•• isilS•• follp)Vs;
falality • ::•• 12,personat•• i~jufy • =3,.property .
4amageonlY=h .lfoneJatalcrash'i$
personal injuries and 5 property damage
crashes occurreoduring a year ala
particul~r site,/compute. ltsseverity
number,/"
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5-532
MISCEllANEOUS
Solution:
CD Traffic base (TB):
@
Space taken up by the vehicle in feet of
each lane - mile:
.
Space taken up = 112.35 (15)
Space taken up = 1685.25 ft.
@
Average spacing between vehicles:
TB = L (MOD (N) (365)
100000000
TB =0.30 (15400) (3) (~
100000000
TB = 0.051 vehicles/km.
S - 5280 • 1685.25
-
3 year average crash rate (A VR):
AVR = 3 (death + injuries) + property damage only
AVR = 3(120) + 255
AVR=615
@
@
Severity number:
SN=1 (12)+3(3)+5(1)
SN=26
112.35
S=32ft.
The fenglh ofan average vehicle passing lhru
an eXPress way Is 14 ft. If the average
spaCing between~hlcles tS16 ft.
.
<D CQmptM the jam density In vehtcles per
mile per lane, ... .
® If them flow speed Is 104.85 mph,
Thedenslly of traffic In an expressway was
recorded to be 16>05 vehicles per mile perfatle
and the free flow speed of 70 mph. . If the.
average speedofa drtverls 60 mph. . . ..• >..
CD Cqrnpute. the Jam density In vehiCles per
mUe p~rlaOE!; ..•.
... .•• . •.•.
® ·If·lhe average.• space·between the front
bumper ofafbllowingcar an rear bUmpefof
.the leading car must account for the
number of vehicles in a lane (112~35 per
mile) andlenglhof each vehicle 1515 ft,
compute the space taken up by the
vehicles In feet of each tane·mile. .. ..
®. Compute the average spacing between the
. vehicles.
.
@
computelhe speeda1 max. flow.
Compute· the maximum flow In vehicles
per hour,
Solution:
CD Jam Density:
5280
0=14+16
Dj = 176
@
Speed at max. flow: .
-§
Smax- 2
S
max
= 104.85
2
Smax = 52.43 mph
Solution:
CD Jam density:
D= SrD
} Sf'S
D= 70JJ§~
j
70·60
q =112.35 vehicles per mile
@
Max. flow:
SrDj
q= 4
q= 104.~ (176)
q= 4613 vph
, .
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MISCEllANEOUS
Solution:
CD Time headway between vehicles 1 and 2:
Vehicle Crossed Line
No.
1
2
3
4
5
6
7
8
·.1'h~it~l)ul~te~ • •~~~.areth~f$$\Jlls.(lfaGQ
·~Rri~:t~i~~~~.~~~ir~#:lll
·:111111
Vehicle No.
1
2
3
4
5
6
7
8
Crossed Line at (seconds)
6.52
11.26
14.59
19.33
28.30
39.93
43.76
58.16
(sec)
6.52
11.26
14.59
19.33
28.30
39.93
43.76
58.16
11.26 -6.52 =4.74
14.59 - 11.26 =3.33
19.33 -14.59= 4.74
28.30 -19.33 =8.79
39.33 -28.30 = 11.63
43.76 - 39.93 =3.83
58.16 -43.76 =llJQ
51.64
Time headway between vehicle 1 and 2
= 11.26 -6.52
=47.74 sec.
tlte",
Mv~~nt• ~$ • ltle•• getftlltlqll.ofth~~\I~n1 • belr~.·
@S~tY~(t • • PJt¢rtlallm~~Mit~n59PQ~.·~.~~r.·
Time headway
(sec)
® Average headway for 8 vehicles:
51.64
Average headway =-7Average headway = 7.38 sec.
@
Traffic flow rate in vehicles/hr.
1
Flow rate q=average headway
_ 1 (3600)
q- 7.38
q =488 vehicles per hour
liil.l!
4~ry.•<'Dh~«Jl'iElt9.J::har~w.p'~11~!i • inJ~paO)$ .
.49m,n~1~$?JI1~~lffe~.wm~~tdl~ta~~I$.
·r~lisible.; • ··~~tlm~ma!I~.t~~~$;3Q • mlplJt~$W
.giJY~ • !Il!:l.2()l'lJ'I~.f@'rly(j~r~9me.tPttle.~im9rt
~!Wy.OU.P'lUst.~rrt\le~t •
the•• ~itPQrt.90 .m1ryUl$$
~wlylo(;h~qkjn.WI1enYPl.llantlm$iW
W~~¢i.sS9ijttt:lk$$Zqmj119t~$t~9~tYPl.lf
W9gClge•• ~nd • • arM~jti()ral~{) • JJ1inu~~ • lo.th!:l·
dg.vnlovmhptel®.miles.frorn.the airpOlt..• •,.he
airtare.for.the.r911••stop.ftlghl",·.iS. $500•• Wh~$
!l~if11e·.wtIer~as.tlle.lickE!tJor.flight • Bw1tll·.the
er
trsnsfer atJapani$$3aO.Assumean'oth
costs arelhesal1'le at $60andtha! you valUe
your tlmeat$50 per hour.
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••SCllIAIEOUS
li• •JlII
Solution:
CD Max. length of the queue:
Solution:
<D Which flight is cheaper.
forA:
t = 3 + 30 min. + 90 min. + 20 min. + 40 min.
t=6 hIS.
Cost of flight A =$500 + $60 + 6($50)
Cost offfight A = $860
ForB:
t = 4 + 30 min. + 90 min. + 20 min. + 40 min.
t = 7 hIS.
Cost offlight B=$360 +$60 +7 ($50)
Cost offlight B=$770
Fight B is cheaper
@
How much cheaper.
Flight Bis cheaper by 860 • 770 =
m
Arrival CUNe:
After x = 15 min, Y2 = 15 (45.25)
Y2 = 679 vehicles
® VaT when both flights would be equally
expensive:
500 + 60 + 6 (V01) = 360 + 60 + 7 (V01)
VOT= 560 - 420
VaT =$140
Departure Curve:
.
, After x = 15 min, y, = 15 (22.33)
y, =335 vehicles
Max. queue length =679 - 335
Max. queue length =344 vehicles
@
irr8ail~
Y~htpl~B~rllbUr(f2,~~Y~ltIil'il ••• Jl'i~··.AAW
rlilt~<)~th~~ee\\l~yat#'l~.Ull1epfdl:lY.is.2715
V~hl¢!~$P~FPq~f(4?@v ..~Jrpln)<Jl'le
Illocl$geisrernovEl(laft~r1Srnil1t1tes, . .
(j)
@
.
VVh~tjs!hernax,f~ngthof~q@Il~?
vy.hat\'iaf•• thelq~~~titltne.. ·.~ny • •·single
V~hlCjeV&sm·th¢.qge,*? • •·
® AtWhat~lTledldthetl~ElIJe claar?
Max. time in queue:
when x =15 min, y =15 (22.33)
Y= 335 vehicles
'
lime the 335th vehicle enter the queue t,
335 74 .
. =45.25 = . min.
Max. time in queue =15 - 7.4 min.
Max. time in queue =,l.6 min.
® lime the queue cleared:
x = time the queue was cleared.
Equation for arrival CUNS Y3 =45,25 x
Equation for departure CUNe = Y4 + 335
where Y4 =60 (x -15)
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MISCEllANEOUS
Y3 = Y4 + 335
45.25 x =60 (x-15) + 335
14.75 x- 900 + 335 =0
x = 38.3 minutes
Solution:
CD Max. length ofqueue:
,
I
, I
I
,
Time the queue was cleared is 8:51.3 AM
I
I
I
,
I
I
,
,
,
,
,
,
,,
,
,
I
,
I
I
I
!...-Normalp_
,
r
=65 \'cJilmin
I
,
I
:'
max. qucMe:
len,," :
I
I
I
L- =60uWmiJt
Normtfl flpw
,,
I
,,
I
,
,
I
I
I
,
,
,
I
I
I
I
,
I
,,
,,,
,,
,,,
,,
,,,
,
I
,
,
,
I
Y,
I
meLt. ,,"GU! t
Im,m :
,
I
I
I
x
Y1 = 24 (20)
Y1 = 480 vehicles
Y2 =46 (20)
Y2 = 920 vehicles
x
Minutes after 8: 15 ll1I1
Blockage
Max. quel1e =920 - 480
Max. queue =440 vehicles
@
~_III
trnfflca~rp~llt . •.• l'h~I(WJlc;v'~$¢l~®l • bYmft
MMDAaf!erZ(lll1inuteS./
....
. ..
• ($•• • PeterO)i~ • th~ • • tM><.• •lellgrh•• 9ftheqtieu~
b~forethe~OCk!l~~¥lasrernOV~d?
. ...•.••..•
·~toCkag~wascleared·
. ..
.
•@·What.liri1~W<ls.thl:lqueue·clElllre(j.
..
• ®•·',Valt~dfpr>lheJp~gque~~tl~foreJhe
8el~rmiQ~ • • t~~ • • ·titlle.itha,t•• th~ • • V~hICI~s·
TIme the vehicles waited for the queue:
440
t1 =46'
t1 =9.56 min.
TIme the vehicles waited for the queue
=20-9.56
= 10.44m.
® TIme the queue was cleared:
Y3 =46x
Y4 = 65 (x - 20)
Y3 = Y4 +440
46 x =65 (x - 20) +440
19x.-1300 +440=0
x = 45.26 min.
TIme the traffic was cleared = 6:45.26 AM
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S·S36
MISCEllANEOUS
·1:.~.II~1-~I.~·rl-I~11
1
1>1••llllli'lli
Y3 =45.25 x
Y4 =60 (x· '15)
Y3 =335 +Y4
45.25 x =335 + 60 (x -15)
x= 38.3 min.
Y4 =60 (38.3 -15)
Y4 =1398
Ila_ii
Solution:
<D Total delay to traffic because of lane
blockage:
,
":
,
,
,
,
,
,
,
"
,
I
I
I
I
I
I
I
I
,
I
,
I
,
I
!
yi=1398
:'
:
..-NanM~f!/tw
,
,
The area of the shaded section represents
-the total delay on the freeway because of
the traffic accident.
:
:
,
.1,,=1733
=60w'f'"Ur
I
",;~:e:
I
I
I
I
I
I
,~,=
't
10 1
20 2S
30 3S '0
A_
35
_
l'1I",a-
1733 (38.3) 335 (15)
2
• 2
_335 (23.3). 23.3 ~ 398)
Atea = 6582.25 veh-min.
x
Minul.. after 7 :00 am
Y1 =45.25 (15)
Y1 = 678.25 vehicles
Y2 =22.33 (15)
Y2 = 335 vehicles
Ate~ - 6582.25
- 60
Area = 109.70 veh-hrs.
Total delay = 109.70 veh·hlS.
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MISCEllANEOUS
® No. of vehicles ahead ofit in the queue:
,
I
,
,
, I
I
,,
,
I
I
60 (x-15) +335=543
x = 18.47 min.
time the 7:12 AM arrival will leave the queue.
Waiting time for this vehicle is 18.47 - 12
= 6.47 min.
/ JI
,
,
,
,
,
,
I
I
,
I
I
I
,
,
I
I
I
,
r,
__
L-N._l~w
,
,
,
,
,
,
,
,
,
I
INU. 'f'IelCI
1LoP
,
I
I
'
I
I
From 7:00 tWo to 7:12 tWo
t= 12 min.
h1 =22.33 (12)
h1 = 268 vehicles
~ =45.25 (12)
h2 = 543 vehicles
Length ofqueue at 7:12 AM = 543 • 268
Length ofqueue at 7:12 AM =275 vehicles
No. of vehicles ahead of it in the queue
= 275vehicJes
® Time the driver have to wait in the queue:
Solution:
<D Velocity of car when it reaches the
intersection:
2
1 -2aS
vi= V
80000
V1 = 3600
V1.= 22.22 m1s
=(22.22)2 -2 (4.6) (40)
V2 = 11.21 m1s
~ _ 11.21 (3600)
21000
vi
V2 =40.4kph
® Location of man when the light tums red:
V2 = V1 • a t
V2 =22.22 - (4.6) (4 -1)
V2 = 8.42 m1s (his velocity when the
light tums red)
Distance he will have traveled:
Vi=V12-2aS
(8.42)2 =(22.22)2 - 2 (4.6) S
S=45.96 m.
He is 45.96 m. after entering the intersecting
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MISCElIUEI.
@
Time the light have been red when he
clears the intersection:
t=
o
V
® Monthly expansion factor- (MEF) for
August:
28450
MEF = 4550 (12)
MEF=0.521
0= (15 - 5.96) +4.2
0= 13.24m.
- 13.24
t - 8.42
@
Average Annual daily traffic for month of
May:
MDT = MEF x ADT
28450
t= 1.54 sec.
MEF = 12 (1700)
MEF= 1.394
MDT = 1.394 (1700)
MDT= 2369.80
f'()r•• 1Q.YAAr~;fb~.Plif~~!l~;M*~emj~
.ba$e.adO"m,.a~.ps( • ~~rt\Ceb~ • ~.S~an: .•
. .1\1>~Omettte~~5roul~ • caifi$$~n*~ri!S~4f·
1Q()(lpa~~l1ge~p~r9aY·.~~~rn~lJn~m~
r~~gelaslicityof .0.33' .
..
.
(j)·•• ·HoWll1aflyrider$Wlll.~ • • dfiV~P~~ydu~
• • • •·•• • t6k1Ct~ • 1rlf(lre$? • • • • • • • •·••·•• / • • • • i•• • • ?>•• •.• • •.•. • .
{!) .YlJhlitwoutd•• l>fl•• £Iiff~rer)c~ • •~• • its.rElv~ne~
due\()ttleinCr~il'lf.1lr!'l?>
@
Solution:
CD Daily expansion factor (OEF)
Wednesday:
75122
OEF= 11413
OEF= 6.582
for
AS!lOl'1'ling • • h~lf.pf.!h~ • Ras~lfflgtlf~,on . • ~
ty~ical .• dayroq~d9~~gthB • A.fW~I1~ • •
eM
.• • peaKperiodsY.'h~.th~P!*J~~n9~~~a~
.head.. w<lys.were3Qmln.~ • VlHlll.ypuldb~
the.ridef'$bip • Jf•.• m~· hM~VI~y$ • \'I'~re.
••
•
jn{\reased·•• to6Q•• • O'lin·•• ··.~tw'eer ··bu5~S?
Assume • • a • peak • M~dW~y·.el~sUClty·.·9f
-0.37.and an off.-peaKheacIWayelasticity
of-O.46.
.
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MISCELlANEOUS
Solution:
CD No. or riders driven away due to increase in
fares:
Solution:
CD No. ofpassengers on the average:
138700
Q,-Qo) fu
Shrinkage ratio elasticity = ( P, _Po ·Qo
_(Q, - 1000) (SO)
·0.33 - (75 - SO) (1000)
Q,·1000 =-165
N= 7(3729)
N = 5.3 passengers
@
138700
x= 7(30)
x = 660.5 BTU/passenger-mile
Q, =835
No. ofriders driven away = 1000 - 835
No. of riders driven away = 165 passengers
@
Energy intensity:
® Energy intensity:
138700
x = 28.7 (7)
Difference in revenues daily:
50 (1000):: SOOOO
75 (835) =62625
. x = 690.4 BTU/passenger-mile
Diff. = 62625 • SOOOO
Diff. = P12, 625
® No. of ridership if headways were
increased to 60 min. between buses:
..
(Q, - Qo) So
Peak headway elastICIty = (S, • So) 00
_ (0, • 5(0) (30)
- 0.37 - (60.30) (500)
=315
1
..
(0, _5(0) (30)
Off peak headway elastICIty = (60 _30) (500)
Offpeak headway elasticity =- 0.46
o
0, =270
Total no. of ridership = 315 + 270
Total no. ofridership = 585
.~~sle6Q~~a~~:~~~~7IW~~~~··
..
. . . .. ...
.. .
.
Pi'l(g;illl~O{
i~llll(i~fAI~j
~~r~·carrie9.Wtlllt~lll~th~en~r~¥
jl)tell$jtyo.1lt\etl1J~~~1>
®•• ·W~#t*~.theerletgyirter~jti~~9fW~i~~
····WitijTn~~rs.·inplt!~in~ • ihe•• grtv~c~·m~ftJ~l.
.~C:(lmmlYofftle • blJsise<lUalt8~~;7lr1nes
··patgllll(jll,
11I~~I!i~r~lm~~~~I.
Solution:
CD Peak hour volume:
Peak hour volume =465 + 480 + 510 + 490
Peak hour volume = 1945 .
@
Peak hour factor:
1945
Peak hour factor =00---
15(510)
Peak hour factor = 0.953
® Design hourly volume:
1945
Design hourly volume = 0.953
Design hourly volume =2041
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MISCEllANEOUS
@
Spacing between construction joints:
~~:~~I;~~~~~~~~~~~~~
~eJ~ .• • • "f~El$~m~@.'fflm • tNt~afl1e~~~A~
\mJ)t¥$~4uM~Jthlt~~l}~~I~~~4W~
P~Mlt~~9Vt~$t • bya.~j$t9~PI9~~$9m·
4i~~$rfl1PViIl9a:~rttlin~~eq, • • T~~~~
·\,aIU~Qf:~ • ~tttndatdcnJSb~df~tQrZ.~ •
pen~®ti®is7Q.45.l@cmz .• • 1'1"u=•. sWpI8;Wa~
mm
AA~~tqaloa~Qf83~9· anqUPl'O!:it.ICesa
..
poo~filti9nQf2,5ttlrfJ;
• ~• • PrililPot~[th~ • pB~vl¥l~eefthei$()it~n1Ple.
~Q$@t~J$$9i{f()r~stlbgraq~qfa
. . .•. .•·.·.~~¥~J1'\~~t;~e~®I!l~~mlC\(~~~l#.tp~
......>p~'l~~ntvpe,~~'IhI~~lo/lcl~t~O(}()~9
· .· ••.•.•.• • ~~~ • ~.t~pres~\!reot.~··~~Jsil11P9sed
•.•.•••. ()nJb~sV~ra~e.· • • Us~ • tt§;.• CQrp$•• of
·•••••.• .!;figine~sfQl1l'lul<t • ·.•· .••.•.•. . . .•.••••..•••..•.••..<•.<••
•~• • lfum•• mWemehti$~d~ •. UP•• pl(;9P9"ete,
. . . •.•. #~termme • • • ···the•• • .• sp~llj~W • • • • p~WJeen
...·.•Q9r@r99ljqryjoin~~ral'l~ItI·Wl~I1Qta
·•.• .• ~~tlrffis~~'w:~ • t~~~~h§~Wa~0~6j
· •· • ·• ~~~:t.~M~d:=.·~~~rd1d;6.~i~¥$
0.2758
-'+::---......;.~=--
W= ~ (0.2758) (4) (2400)
W= 1323.84 L
N= 1323.84L
F=p.N
F= 1.5 (1323.84 L)
F= 1985.76 L
T=F
T= Ac ft
27.58 (400) (0.8) = T
T=8825.6 kg
8825.6 = 1985.76 L
L=4.44m.
.< . . p~ytlm~nt •.• ii$ .•. . O,~5> • ·•. NI~~~I~.·t~n.#il~
Solution:
CD CBR of soil sample:
p
Stress =A"
<P .A rigJd l?w~ment'With an alkiwa~leteO$j\e
.. ..str~~$.ofQol')creteequal to·fc'Witha
. s~lfledcompressive strength of concrete
83
Stress=--
~(5)2
.• .0(28:$ MPa.•.·Neglecleffect Of dowels,
.. USErOlders Theory;' . •... •.. ••.. . ... ..
Stress = 4.23 kg/cm 2
(g) •.. AfleXlple' pavement w"h
CBR = 4.23 (100\
70.45
O~sjgl'!ttte thickness of Ii pavemeriHo carry a
wtieetJoad of 30 kN based 6rittieftillOWing
conditions. and type of pavement.··
....
I
CBR= 6%
® Thickness of pavement (US Corps of
Engineers):
t={W [1.75._1]1/2
CBR Pn
-1-]
t = -Y 3000 [1.75 _
6 8n
t= 27.58 em.
1/2
an· allOwable
. . subgrade pressure of 0.14 MPa<and the
..·max, time pre$SUre equlllto· {t8~ MPa.
This pressure is assumed to be uniformly
. distributed over the area· Of tire. contact on
the roadway according to the principle of
.Cdne pressure distribution. ...
.
@ A pavemenl with a maximum caR value
of 6% for the subgrade SdnSupportin9 this
load, The tirepresstire is equ.al to
4kglcm 2. Use U.S. Corps o Engineers
Formula.
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MISCEllANEOUS
Solution:
Solution:
CD Rigid pavement: .
t_A
CD Time after gapers block starts until the
queue will be cleared:
f3W
-\{T
I
I
,----
I
I 3 (30000)
- 'V 0.06 (28.5)
I
t -'"
I
I
I
I
I
t =229.42 mm.
\1
I
I
I
I
I
I
I
® Thickness of flexible pavement:
: 100.
r
:
t=0.564~
I
I
,
I
W=AP
30000 =1t r2 (0.82)
[" 107.91 mm.
t=O.564
I
I
I
I
I
I
r
~3~-107.91
I
: 1350
I
r
t=153.17mm
x
45
® Thickness of pavement based on
Corps of Engineers:
t = {W
54+x)
lOOyplfl
[lZ§..
_1]
CBR P1t
Us
30(45) = 1350 cars
60 (45) = 2700 ears
100 x+ 1350 =60 (45 + x)
40 x= 1350
x = 33.75 min.
1/2
W= 30000
9.81
W=3058kg
Time after gapers block starts until queue
will be cleared =45 + 33.75 = 78.75 min.
t = ...j 3058 [1 ;5 _41~ 112
@
t= 25.46 em.
t= 254.6mm.
Total delay of traffic because of gaper's
block:
Total delay of traffic = area,ofshaded section
.
,
I
,,
,,,
,
,,,
I
I
·~• ~~rr$ • ~'9C~ • QQCI1r$.When.trafflciri•• QO~ .
~itectj~nsl()'NstQ .•l99K•• ~tal'! • im;id~~t.·9~.~h~
9Ppo~1t~sjd~ • of.·the.lPad~aYs • rnedfan.•• • • J'~e
·9~P~~i~ppKr~u~~.r~W~y.~~qtW • ft9m
,
HiO(33.75)
:=3375~
IOO~",
:
2700 ----
,,,
1()O.YElfji¢lIlS.. ~ •. l1lill,·.tq.·(3j)·.v~hiqlEl~~~mir·
J()r~9Il1jrute~.lfJhe~rrlyatrat~staY$~t
!5QvehIQlElllpermin.
.
I
I
I
,,
.. ...
~• • •~~~~~.~~~~i~pat$ • brock•• $mrt$.Wjll
~VVh~ti$lheI9talV~hjpledeJaytolraffic
•.. ~i:lQ§Ei'.Of • lhe•• ~~rs.bIOCk • in • vehjcles.
m i n f / i ) ) •.• ..i<
®.··.Whatis.theaverage.delay.per.Vf:lhicle?
4 25
130
45
33.75
78.75
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IISCElll.EOUS
A_
(2700 • 1350) (45) (2700 - 1350) (33.75)
2
+
2
Area = 53156.25 veh·min.
h, =30 (40)
h1 =1200 cars
Average delay per vehie/e:
53156.25
' perveh'e/
Average de,ay
J e =~
Max. length of queue = 2400 - 1200
Max. length of queue = 1200 cars
_
a-
f11 ..
@
Average delay per vehicle
= 11.25 minlvehic/e
(
~=60(4O)
~ =2400 cars
® Average delay per vehicle:
Total vehicle delay =area of shaded section
100 x+ 1200 =60 (x +40)
4Ox= 1200
x=30min.
Total vehicle delay
_ (2400· 1200) (40) (2400 -1200) (3O)
·l'~~ • • ~U#~rNOO~~Y, • ffl/g~~I!16 • • t«.hay~ • • ~.,
-
._i.~11B
CQIl1PlJt~.,• t~'~.,>Il1~Xlmgm'.', • I~ng!t).',.il>f>$~.
~jl.illli
2
Average delay per vehicle = 10 min.
@
Longest time any vehicle spent in the
queue:
,,
,,
,,,
@ .'Wfl<tt~th~Ig~~$ttl~,<tnyve~lcl¢~p~W,
,
,
',' ',' ",,"' ",' "
IIlifu3q@4~?i
,,,
,,,
,
,,,
Solution:
CD Max. length of queue:
,
,,,
,,r
100 vpm :
,o,
,
,
,,,
,
,,
o
,,
o
o
o
,o
TIoo
:1
:
100 'P"!
,,,
,,
.,1
j
'
,
~l= 200
~:
x
,
IS4+X)
=4200
,
,,,
,
,,
40m
2
+
Total vehicle delay:: 42000 veh-min.
.
42000
Average delay per vehicle = 4200
Tfl~ • ~.sr~&~tIl~'M~Q¥PfI1' • • ·,W~1ro~fflC
thenrasll~~etAQn1ln;
,,',
"
(j)
,
o
1-----40>------
60 t= 1200
t= 20 min.
Longest time any vehicle spent in the
queue
=40-20
= 20m/n.
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MISCELlANEOUS
Aclosed traverse has the following data.
LINES
A-B
B-C
C-A
BEARING
N.30'30'E
S.20'30'E
---
DISTANCE
46.50 m.
36.50m.
---
® Find the distance of line C-A.
® Find the bearing of line C-A.
@ Find the area of the closed traverse.
Solution:
Given the data of a closed traverse:
LINES LAT.
AB
446.56
BC
Y
CD - 58.328
- 2.090
DA
DEP.
30.731
75.451
---
2A
1372.324
x 2158.023
148.621 -8668.766
DMD
30.731
- 42.43
---
® Find the value of x..
® Find the value of y.
@ Find the value of z.
® Distance CA:
Une~
AB
BC
CA
Bearing Distance Lat
Dep
N30'30' E 46.50 +40.07 +23.60
S20'30' E 36.50 - 34:19 +12.78
--- - 5.88 - 36.38
-
--
Distance AC = ~(5.88f +(36.38)2
AC= 36,85m.
® Bearing CA:
tan bearing = Dep
La!
.
36.38
tan beanng = _.5.88
Bearing =S. 80'49' W.
BearinoCA:
Lines Lat
Dep DMD Double Area
AB +40.07 +23.60 +23.60 23.60{40.07)
=+945.65
BG - 34.19 +12.78 +59.98 59,98{-34.19)
= 2050.72
CA - 5.88 - 36.38 +36.38 36.38{-5.88)
= 213.91
2A = +945.65 - 2050.72 - 213.91
A= - 659.49
Solution:
® Value of x:
LINES
AS
BC
DEP
30.731
75.451
DMD
30.731
x
30.731 + 30.731 +75.451 = x
x= 138.913
® Valueofy:
@
2A = DMD x LAT.
2158.023 = 136.913 (y)
y= 15.762
~
Area of transverse ABC = 659.49 sq.m.
.Check:
44.656 + 15.762 =- 58.328 - 2.090
60.418 =- 60.418 ok
z
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S-544
MISCElUNEOUS
@
Value ofz:
LINES
AB
BC
CD
DA
LAT.
446.56
15.762
- 58.328
- 2.090
0
DEP.
30.731
75.451
-'63.743
- 42.439
0
DMD
2A
30.731 1372.324
136.91~ 2158.023
148.621 -8668.766
42.439
z
Using Cosine law:
(DE)2 = (43)2+(27.77)2 - 2(43)(27.77) Cos4S'
DE=30.52m,
® Bearing ofline BC:
(BC)2 = (95)2 + (88)2 - 2(95)(88) Cos 45'
BC= 70.33m.
Using Sine law:
95
70.33
--=-Sin (} Sin 45'
e=72'46'
a, = 90' - 72' 46'
a,= 17'14'
Bearing of BC = S. 17'14' E.
2A=DMDxlAT
z =42.439 (- 2.090)
z= -88.698
@
" -A lot is bounded by 3 straight sides A, B, C.
AB is N. 45' E. 95 m.long and AC is due East,
88 m. long, From point 0, 43 m. from A on
side AB, a dividing line runs to E which is on
side CA. The area ADE is to be 1n of the
total area of the lot.
CD Determine the distance DE.
® Determine the bearing of side BC.
@ Determine the distance AE.
The deflection angles of two intermediate
points A and B of a simple curve are 3'15' and
8'15' respectively. If the chord distance
between A and B is 40 m.
@
Find the radius of the curve.
Find the length of curve from P.C. (0 A.
Find the length of chord fr()m P.C. to B.
CD
Radius of curve:
CD
®
Solution:
Solution:
CD Distance DE:
(AE)( 43) Sin 45'
2
!'IE =27.77 m.
Distance AE:
AE=27,77m,
=..!.
7
(95)(88) Sin 45'
2
· 5' 20
SIn
=1f
R= 229.47 m.
MISCEllaNEOUS
®
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Length ofcurve from P.e. to A:
An unsymmetrical parabolic curve has a
forward tangent of - 8% and a back tangent of
+ 5%. The length of the curve on the left side
of the curve is 40 m. long while that on the
right side is 60 m. long. P.C. is at station 6 +
780 and has an elevation of 110m.
<D Compute the elevation of the current at
station 6 + 800.
® Compute the stationing of the highest
@
S=R0
S = 229.47 (6'30') 1t
180
S = 229.47 (6.5) 1t
180'
S= 26.03m.
@
Length of chord from PC to B:
point on the curve.
Compute the elevation of the highest point
on the curve.
Solution:
<D Elevation at sta. 6 t ,800
poel ~~JJp.r.
6+~
EI. 110m
L,=60
~ = Lz (9, -92)
L,
L, + L2
2 H = 60 (0.05 +0.08)
40
40+60
H= 1.56
y _ 1.56
(2W - (4W
y= 0.39
C
Sin 8'15' = 2 R
C = 2(229.47) Sin 8'15'
C= 65.86m.
Elevation of the curve at sta. 6 + 800
Elev. A = 110 + 0.05(20) - 0.39
Elev. A = 110.61 m.
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MISCELlANEOUS
@
Stationing of highest point of the curve:
Check the location of the highest point of
the curve.
,
When L, 9 < H it is on the 40 m. side
2
5 == g, L/
2H
1
,
When L 9, > H it is on the 60 m. side
CD Compute the length of curve if the rate of
change of grade is restricted to 0.20% in
20m.
@ Compute the elevation of the P.T.
@ Compute the slationing of the P.T.
2
5 == g2 L/
2H
2
A parabolic sag curve has a grade of the back
tangent of - 2% and forward tangent of + 3%.
The stationing of the P.C. is all0 + 300 with
an elevation of 100 m..
L, g, = 40(0.05 = 10 < H = 156
2
. 2
.
.
Solution:
Use:
. L2
5, =.fU.-L
C
2H from P..
= 0.05 (40)2
5
,
5,
2 (1.56)
=25.64 m.
Stationing of highest point of curve:
Sta. (6 + 780) + (25.64)
Sta. = 6 + 805.64
@
CD Length of curve if the rate of change of
grade is 0.20% in 20 m.
g1- g2
r==-n-
Elevation ofhighest point of curve:
0.2 = 1- 2; 3
I
n = 25 stations
Length of curve =25(20)
Length of curve = 500 m
p.r.
Y2
_ 1.56
(25.64)2 - (4W
Y2 = 0.64
Elevation B = 110 + 0.05(25.64) - 0.64
Elevation B = 110.642 m.
@
Elevation of the P.T.
Elev. P.T. = 100 - 0.02(250) +0.03(250)
Elev. P.T. = 102.S0m.
@
Sationing of the P.T.
Sta. of P.T. = Sta. of P.C. + 500
Sta. of P.T. = (10 + 300) + 500
Sta. of P.T. = 10 + 800
MISCELlANEOUS
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At station 95 + 220, the center height of the
road is 4.5 m. cut, while at station 95 + 300, it
is 2.6 m. fill. The ground between station
95 + 220 to the other station has a uniform
slope of - 6%.
.
At sta. A 12 + 200
Section: Level and trapezoidal
Base width: 12 m.
Side slope: 1.5:1
Depth of cut: 2.25 m.
What is the grade of theroad?
<ID How far in meters, from station 95 + 300
toward station 95 + 220 will the filling
extend?
@ At what station will the filling extend.
At sta. B 12 + 220
Section: Level and trapezoidal
Base width: 12 m.
Side slope: 2:1
Depth of cut::: 1.8 m.
G)
Determine the volume of cut in cU.m. from·
Sta Ato Sta B.
<ID If the volume of cut from A to B is 650
cU.m., find the depth at section A?
@ If the volume of cut from A to B is 650
cU.m., what is the top width of the section,
A.
G)
Solution:
G)
Grade of road
Solution:
G)
Volume of cut in cU.m. from Sta A to Sta
B.
Slope of road::: 2.3
80
Siope of road::: 0.02875 say 0.029
<ID Distance from 95 + 300 where filling will
extend:
"~::}~~~
~
~-12----l
- (18.75 + 12) (2.25)
A12
A1 ::: 34.59 m2
80
0.029(80 - x) + 0.30::: 0.06x
0.089x::: 0.029(80) + 0.30
x::: 29.44 m.
~-12
@
Station where filling extend:
(95 + 300) - (29.44)::: 95 + 270.56
- .u..9.2 + 12) (1.~
A22
A2 = 28.08
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5-548
MISCEllANEOUS
~
Depth at section A.
Given a side slope of 2:1, a road width of
10 m. and a cross-sectional area of 31.7
sq.m., find the value of x in the follOWing
cross-section notes.
9.8
0 7.4
+ 2.4
+ 1.2
x
Solution:
_ (3d+ 12+ 12) d
A12
A1 = 1.5d2 + 12d
v=
(A1 + 28.08) (20)
2
A1 = 36.92
36.92 =1.5d2 + 12d
d= 2.373
(A1 + A2) L
2
31.7 = 9.8(x) + 7.4(x) + 5(2.4) + 5(1.2)
222
2
8.6x = 22.7
x=2.64m.
v=
- (34.59 + 28.08) (20)
V2
V= 626.7m 3
Determine the side slope from the following
cross section notes of an earthwork.
7.7
0 8.0
+ 1.8 1.2 +2
Solution:
@
Top width of the section A.
I
,
jJ:2.0
I
:
w
181
. I
1----7.7
x
1.8S-t-2" = 7.7
x
28 + 2 = 8
W = 1.5d (2) + 12
W= 8(2.373) + 12
W= 19.12 m.
I
I
G____
_
'
1-185-1 XI2±X/2--1--2S==J
0.6S = 0.3
3
8=2
8 = 1.5
Side slope = 1.5:1
.
8.0
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MISCELlANEOUS
Determine the side slope from the following
cross-section notes of an earthwork.
8.6
0
9.0
+ 1.8 +1.2'
rn
Solution:
Solution:
<D Area of cross section:
hL -2;5 _.12hL+ 3.5 - 100
100hL- 250 = 16hL+ 28
84hL = 278
hL= 3.31 m.
2.5 - hr
8
3.5 + 2hr = 100
250 -100hr = 28 + 16hr
116hr = 222
hr = 1.91 m.
Side slope is 2:1
A =~(3.5)(3.3f) +l(2.5)(2 x 3.31 + 3.5)
1
1
+ 2' (2.5)(3.5 + 2 x 1.91) +'2 (3.5)(1.91)
The cross section notes of the ground surface
ata given station of a road survey shows that
the ground is sloping at an 8% grade upward
to the right. The difference in elevation
between the ground surface and· the finished
subgrade at the center line of the proposed
road is 2.5 m. Width of subgrade is 7 m. with
sideslope of 2:1.
A= 30.935
A= 31 m2
® Distance of the left slope stake from the
center of the road.
xL = 2hL .;. 3.5
XL = 2(3.31) + 3.5
. xL = 10.12 m.
@
<D Determine the area of the cross section.
® Compute the distance of the left slope
stake·from the center of the road.
@ Compute the distance of the right slope
stake from the center of the road.
Distance of the right slope stake from the
center of the road.
XR=
3.5 t 2h r
xR = 3.5 + 2(1.91)
xR:=' 7.32 m.
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S-550
MISCELlANEOUS
Given the following data of the cross section of
an earthworks.
Sta. 2 + 100
2.75
1.5
0.5
9.5
0
5.0
sta. 2 + 120
1.:Q.
2.25
9.0
0
0.8
5.6
Width of base is 8 m.
<D Compute the area of station 2 +100.
o Compute the area of station 2 + 120.
@ Compute the volume between stations
using average area method.
The longitudinal ground profile diagram and
the grade line shows that the length of the cut
is 950 m. and that of the fill is 1320 m. The
road bed is 10 m. wide for cut and 8 m. wide
for fill. The side slope is 1:1 for cut and 2:1 for
fill. The profile areas between the ground line
and the grade line are 8100 sq.m. for cut and
9240 sq.m. for fill.
<D Find the volume of cut.
o
@
Find the volume of fill.
If the shrinkage factor is 1.30, find the
volume of borrow or waste,
Solution:
<D Volume of cut:
Solution:
<D Area of station 2 +100:
I
I
f
2.75:
I
I
,
I
1:-;;--
.
I
Average depth of cut:
4
9.50
C = 1800 = 8.53 m.
5.0
_ 4(2.75) 1.5(9.5)' 1.5(5) 4(0,5)
A1- 2 + 2
+ 2 + 2
Average depth of fill:
A1 = 17.375 m1
o
950
f= 9240 = 7 m.
1320
Area of station 2 + 120:
f'
f
1---~-13.20----l
I
2,25:I
:
I
:0.8
r~;---
I
9
n;~--:i
5.6
- 4(2.25) 1m lli:§l 4(0.80)
A2 2 + 2, + 2 '" 2
A2 = 13.4 m2
@
Volume between stations:
(A1 + A2)
V=
2
L
V= (17.375 + 13.4) (20) = 307.75 cU.m.
2
Acut = (27.06 + 10) (8.53)
2
Area cut = 158.06 m2
Volume of cut = 158.06(950)
Volume of cut = 150,157.86
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MISCElUNEOUS
® Volume of fill:
Area of fill =
Area fill = (36 + 8)(7)
2
Area fill = 154 m2
Volume offill = 154 (1320)
Volume of borrow:
Vol. of fill required = 203,280(1.3)
Vol. of fill required = 264,264 m3
Vol. of borrow = 264,264 -150,157.86
Vol. of borrow = 114,106.14 m3
(6.6)
® Volume of cut:
•
1-----13.2:0------,---1
The longitudinal ground profile diagram and
the grade line shows a length of cut and 880
m. and a length of fill of 1400 m. The profile
areas are 8432 sq.m. for cut and 9240 sq.m.
for fill. The widths of the road bed are 10 m.
for cuI and 8 m. for fill. The side slopes are
1:1 for cut and 2:1 for fill.
.
CD Find the volume of fill.
® Find the volume of cut.
@ Find the volume of waste or borrow if the
shrinkage factor is 1.25.
2
Area offill::: 139.92 m2
Vol. of fill = 139.92(1400)
Vol. of fill = 195,888 mJ
Volume of fill :;203,280
@
(34.40 + 8)
(29.28 + 10)
(9.64)
2
Area of cut = 189.33 m2
Vol. of cut = 189.33(880)
Vol. of cut = 166,610 m3
Area of cut =
@
Volume of barrow:
Vol. of borrow = 195,888(1.25) -166,610
Vol. of borrow = 18,250 m3
Solution:
CD Volume of fill:
8432
Depth of cut = = 9.64
880
9240
Depth offill = = 6.6
1400
The areas in cut of two irregular sections 40
m. apart are 32 sq.m. and 68 sq.m.,
respectively. The base width is 10 m. and the
side slope is 1:1. Find the corrected volume of
cut in cU.m. using the prismoidal correction
formula.
.
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S-552
MISCEUJlEODS
Solution:
l-------,D 1=lS.1Q------f
f----10+2CI - - - - - i
Two irregular sections 50 m. apart have areas
in cut of 32 sQ.m. and 68 sQ.m. respectively.
Side slope is 1:1and base width = 8 m. Using
the Prismoidal correction formula, find the
corrected volume of cut in cU.m.
Solution:
A = (10 + 2C, + 10) C
2
'
32=(10+C,)C,
C/ + 1OG, -32=0
\------,D1=19.2lS--------i
\------eI =2.SS----1
A1=
(8 + 2C1 + 8) C1
2
. (16 + 2C 1) C1
32=
2
2C12 + 16C2 - 64 = 0
A = (10 + 2C2 + 10) C
2
2
68=(10+C2 )C2
C/ + 10C2 ·68 =0
Cz =4.64
L
Vc = 12 (D j ·D2 )(C, -Cz)
40
Vc = 12 (15.10 -19.28)(2.55·4.64)
Vc =29.12m3
Ve, =Ve • Vc
Ve , =2000-29.12
Ve ,
=1970.88 m3
C1 2+8C1 - 32 =0
- -8 ± 13.86
C1-
2
C1 =2.93
01 =8 + 2(2.93)
01 = 13.86
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MISCELlANEOUS
O2 = 2{5.17) +8
Solution:
O2 = 18.33
A
V= '4 [rh1 + 2rh2 + 3r h3 +4 h4]
r
rh 1 =24.8 +23.3 + 16.2 + 1.5
L
Vc = 12 (C1- C2) (0 1- 02)
rh2 =22.3 + 20.2 + 18.3 + 15.5 + 12.8
Vc = 12 {2.93 - 5.17) (13.86 -18.33)
+ 13.5+ 19.2+21.5
rh2 = 143.3
Vc =41.72
Ih3 =0
50
Ccorrected vol. =
I~=
(Ai + A2) L
2
. Vc
19.3 +21.9 + 16.3 + 17.9
rh4 = 75.4
Vcor = (32+ ~8)(50) - 41.72
V= 26~20) [75.8 + 2(143.3) + 3(0) 4(75.4)]
Vcor = 2458.28.
V= 66400m3
A square piece of land 60 m. x 60 m. is to be
leveled down to 5 m. above elevation zero. To
determine the volume of earth to be removed
by the Borrow Pit Method the land is divided
into 9 SQuares whose comers are arranged as
follows with the corresponding elevations, in
meters, above elevation zero.
Find the
volume of cut in cu .m. by unit area basis.
A= 29.8
E = 26.5
1=24.2
M= 21.2
A (24.8)
20
B= 27.3
F =24.3
J=21.3
N = 18.5
C = 25.2
G =26.9
K=22.9
0 = 117.8
0 =28.3
H =23.3
L=20.5
P = 16.5
C(20.2)
20
D (23.3)
B (22.3)
20
A square piece of land, 90 m x 90 m, is to be
leveled down to 5 mabove elevation zero. To
determine the volume of earth to be removed
by the Borrow Pit Method, the land is divide
into 9 SQuares whose comers are arrange as
follows with the corresponding elevations, in
m., above elevation zero. Find the volume of
cut, in cU.m., by unit area basis.
A= 29.8
E =26.5
1= 24.2
M= 21.2
A (24.8)
20
8 = 27.3
F = 24.3
J = 21.3
N= 18.5
B (22.3)
20
C =25.2
G =26.9
K= 22.9
0=17.8
C (20.2)
20
20
E
(21.5)
F(19.3)
G(21.9)
1
1(16:3)
K07.9)
L (15.5)
E
F(19.3)
G(21.9)
P (11.5)
H(18.3)
20
1
(19.2)
J(16.3)
K(l7.9)
20
M(16.2) N (13.5) 0 (12.8)
D (23.3)
20
(21.5)
20
~
(19.2)
H(18.3)
0=28.3
H= 23.3
L= 20.5
P= 16.5
L (15.5)
20
M(16.2)
N(13.5) 0(12.8)
P(11.5)'
S-554.
MISCELlANEOUS
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® Distance from 5 +420 where filling extend:
Solution:
v = ~ [rh1 +2l:h2 + 3rh3+4r h4]
rh 1=24.8 +23.3 + 16.2 + 1.5
rh 2=22.3 +20.2 + 18.3 + 15.5 + 12.8
+ 13.5 + 19.2 +21.5
rh2= 143.3
h3=0
rh4 = 19.3 +21.9 + 16.3'" 17.9
h4=75.4
r
r
v= 3~30)
[75.8 +2(143.3) + 0 + 4(75.4)]
V = 149,400 m3
Center height of the road at STA 5 + 320 is
4.25 cut. At STA 5 + 420 it is 1.80 m. fill. The
ground slopes uniformly at • 5% from STA
5+ 320.
Q)
@
@
(100 - x)(0.0105) + 0.75 = 0.05x
0.0605x = 0.75 + 100(0.0105)
x= 29.75m.
(from 5 + 420 where filling will extend)
Find the grade of the finished road.
How far in meters from STA 5 + 420
toward STA 5+ 320 will the filling extend?
How far, in meters from STA 5 + 320
toward STA 5 + 420 will the excavation
extend?
@
From STA 10 + 060, with center height of
1.4 mfill, the ground makes a uniform slope of
~% to
STA 10 + 120 whose center height is
2.8 m. cut.
Q)
@
Solution:
Q)
Distance from 5 +320 where excavation
will extend
100· x:: 70.25 m. from 5 + 320
where excavation will extend
Grade of the finished road.
@
Find the grade of the finished. road.
If the roadway", fill is 9 m. 'Wide and the
side slope is 2:1, find the cross-sectional
area of fill at STA 10 + 060 assuming that
it is a level section.
If the roadway for cut is 10 m. and the side
slope is 1.5:1, find the cross-sectional area
for cut at STA 10 +120 assuming that it is
a level section.
5%
.
d
1.05
Slope of finished roa way = 100
Slope of finished roadway = 0.0105
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MISCEllANEOUS
Solution:
<D Grade of finished roadway:
s= ~ (100)
MASS DIAGRAM
60
S= ·2%
Using the following data of a mass diagram,
@
Cross sectional area of fill at sta. 10 +060
Area =
(14.6 + 9) (1.4)
2
Area =16.52 m
<D Find the volume of waste.
@ Find overhaul volume.
@ Find volume of borrow.
Area at sta. 10 + 120
Solution:
2
@
Length of economical haul =450 m.
Free haul distance = 50 m.
Mass ordinate at initial point of mass diagram
(sta ) =• 100 m3
Mass ordinate where length of economical
haul intersects the mass diagram = 60 m3
Mass ordinate where the free haul intersects
the mass diagram = 800 m3
Mass ordinate at the final station
(0+600) =• 200 m3
t..----18.4'O--------I
WaSle
vol. of waste
=100+60
=160m'
Area = (18.4 + 10) (2.8)
2
Area = 39.76 m2
600
-200
<D Volume of waste.
volume of waste = 100+ 60
Volume of waste = 160 m3
S-556
MISCELlANEOUS
® Overhaul volume.
V= 800 -60'
V= 740m3
@
Volume of borrow.
V=200+60
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CD Determine the volume of borrow in m3.
Determine the volume of waste in m3.
Determine the volume of overhaul if the
length of the free haul plotted horizontally
intersects the mass diagram at a point
where the mass ordinate is 300 CU.m.
length of free haul distance is 50.
(i)
.@
Waste
V=260m3
The length of economical haul intersecting
horizontally the summit mass diagram is
450 m. The free haul distance is 50 m. The
vertical distance between the free haul
distance and the limit of economical haul is
3 em. The scale of the diagram is 1 em. = 100
m3. Determine the volume of overhaul in m3.
Solution:
Solution:
CD Volume of borrow in m3.
Volume of borrow = 120 + 180
Volume of borrow = 300 m3
Vol. of overhaul = 3(100)
Vol. of overhaul = 300 m3
The mass diagram for an earthwork starts
from sta (8 + 000) where the mass ordinate is
+ 60 cU.m. and ends at sta. (8 + 320) where
the mass ordinate IS· 180 cu.m. The length of
economical
haul, plotted
horizontally,
intersects the summit mass diagram at a point
where the mass ordinate is + 120 cU.m. The
length of economical haul is 200 m.
(i)
Volume of waste.
V= 120-60
V=60m3
@
Volume of over haul:
V= 300-120
V= 180m3
The mass diagram for an earthwork starts
from STA (12 + 010) where the mass ordinate
is
+ 60 cU.m. and ends at STA (12 + 340)
where the mass ordinate is - 210 cU.m. The
length of economical haul, plotted horizontally,
intersects the summit mass diagram at a point
where the mass ordinate is + 140 cU.m.
MISCElWEOUS
G)
@
@
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8-557
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What is the volume of the borrow, in
cu.m.?
What is the volume of waste.
If the summit has a mass ordinate of
+240, what is the volume of over haul.
Solution:
Wast~
Solution:
"'
....
+115-1--~:"""-_-
~
+50
o 0+000
Volume of waste = 115 - 50
Volume of waste = 65 m3
0
G)
Vol. of borrow:
Vol. of borrow = 140 + 210
Vol. of borrow = 350 m3
@
Volume of waste:
Vol. of waste = 140 - 60
Vol. of waste = 80 m3
@
Vol. of overhaul:
Overhaul volume = 240-140
Overhaul volume = 100 m3
Using the following data on a single summit
mass diagram:
MASS ORDINATE (m3)
STA
0+000
+45
0+500
-175
Initial point of limit of economic haul = + 142
Length of economic haul = 300 m.
G)
@
The following are the data on a single-summit
mass diagram.
STA
Mass Ordinate (m 3)
0+000
+ 50
0+400
- 240
Length of economical haul = 270 m.
Initial point of limit of economical haul is + 115
What is tli:) volume of waste in cU.m.?
Find the volume of waste .
Find the volume of borrow.
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MISCEllANEOUS
Solution:
,
+700:
+2
soo
ll+800
-so
ll+OOO
-l1S
Solution:
CD Overhaul volume:
Overhaul volume = 700 - 230
Overhaul volume =.470 m3
@
Volume of waste:
Volume of waste:: 230 + 80
Volume of waste:: 310 m3
@
Volume of borrow:
Volume of borrow:: 230 + 130
Volume of borrow:: 360 m3
Volume of waste = 142 • 45
Volume of waste:: 97 cU.mo
Volume of borrow = 175 + 142
Volume of borrow = 317 cU.m.
A single-summit mass diagram has the
following data:
STA
MASS ORDINATE (m3)
~ 80
0+ 000
0+800
-130
Initial pointof freehaul distance :: + 700
Initial point of economic haul:: + 230
Freehaul distance:: 60 mo
Limit economical haul:: 400
CD Determine the overhaul volume, in cU.mo
Determine the volume of waste.
Determine the volume of borrow
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@
0
-130