8-292
SIMPLE CURVES
n __ 20(26.2')
""2-
100
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
S·293
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
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
S-295
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.
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
5-297
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.
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,
.
S-299
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'
"
,:
~
"
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.
5-301
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'
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
5-303
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
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.
,/
8·305
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'
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$ectiAnoflhetan~~ht.
is110Jfl~
long....· . .
••...••:::.
FG= R,-R2
FG = 286.48 - 208.35
FG = 78.13
E
Solution:
CD Central angle of first curve:
G
Using Sine Law:
78.13
104.59
Sin 47'24' = Sin e
c
e = 99'48'
11 = 180' ·47'24' -99'48'
11 = 32'48'
@
Central angle of 2nd curve:
11 + 12=42'36'
12 =42'36' - 32'48'
G
12 = 9'48'
5-307
COMPOUND CURVES
@
Di$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)
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'
5-309
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'
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:
S-311
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.
@
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.
S-312-A
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.
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
5-313
[.-..
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)$~f:@l1@~·
.
i:tfId~cllrV~»
.
Solution:
<D Length ofthe common tangent:
Solution:
<D Length of chord from PC to PCC:
,
14'
p.e.
'~
.........TiO:iii..·..·....···
380 ,
I
\,
\
R'
i
/
i, '/'R
y-V
\~
~_-EL
@
Length of chord from
@
pce to PT:
Sin 17' - Sin 163'
~= 120
Diff. in radius of 1st and second curve:
71.27
S·In 7' =
2R(
R1 = 292.40 m.
Sin 10' =~
2Rz
Rz = 345.53
Oiff. in radiUS = 345.53" 292.40
Diff. in radius = 53.13 m,
14\
17;/'
~\r
'\~\I
'!>
Sta.ofP.C.C.
L - R1 / 1 7t
1 - 180
L =380(28) 7t = 185 70
1
180
.
p.e.e. =(20 + 000) + (185.70)
p.e.c. = 20+ 185.70
-.fL_~
@
I
@
,
/
'- L,,/ /220
T1 =380 (tan 14')
T1 =94.74 m.
Tz = 220 tan 19'
Tz = 75.75
T1 + Tz = 170.49 m,
\ II': /
Sin 10' - Sin 163'
C1 = 71.27m.
•
.r.
,/
\
Sta. of P. T.
L - Rz/z 7t
2 - 180
L = 220(38) 7t = 145 91
z 180
.
P. T =(20 + 185.70) +145.91
P. T. = 20+ 331,61
S-314
COMPOUND CURVES
@
1 - 180
L - 290 (42) 1t
1180
L1 =212.58 m.
L - RZ I2 1t
Z- 180
L _ 740 (28'36')1t
z180
L2 = 369.38 m.
~cornpoundcllNep~~se$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.
.
S·315
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
,', '
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
5-317
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.
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
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.
.
.... .. .. .. .. . . ..
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.
5-321
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'
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:
8-323
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
.
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$®0.4~il1;
.•.•....
® Radius of second curve:
tan 30' ::I2:
R2
91.8
R2 :: tan 30'
R2 = 159m.
Solution:
® Cost of concrete pavement:
(j)
Radius Dfthe Dther branch of the curve:
5-325
REVERSED CURVES
240
tan 8 = 360.43
R1 81t
Lc, =180
e = 33'40'
- 240 (37'34)1t
Lc, 180'
Lc, = 157.36 m.
a =46'40' - 33'40'
a= 13'
240
AE = Sin 33'40'
AE =4:32.89 m.
AD =432.89 Sin 13'
AD=97.38m.
DE=432.89 Cos 13'
DE =421.80 m.
DF= DE· 48.60
OF = 421.80 • 48.60
OF = 373.20 ; AB = 373.20
BC= Rr BF
BC = R2 • 97.38
(R, + R2~ = (AB)2 + (BC)2
(240 + R2)2 =(373.20)2 + (R2 • 97.38)2
57600 +480 R2 +
.= 139278.24 +
R2 = 135.10
@
Stationing of P. C.
@
R2 A1t
LC2=18O
A=OO' ·5'46'
A. = 84'14'
- 135.10 (84'14')n
Lc2 180'
LC2 = 198.62 m.
Sta. ofP. T. = (10 + 577.36) + 198.62
Sta. ofP.T. = 10+ 775,98
Rl
Ri -194.76 R2+9482.86
Stationing of P.R. C.
BC = R2 - 97.38
BC = 135.10 - 97.38
BC = 37.72
AC = 135.10 + 240
AC =375.10
BC
tan 0 = AB
37.72
tan 0 =373.20
. 0=5'46'
. tan 13' =
Sta. ofP.R.C. =(10 +420) + 157.36
Sta. ofP.R.C, = 10+ 577.36
:a.~o
tRill'
radiuspfll1~fitWill.ll'\tE!Js.Z8$.4~m.U...i.• •.•·• •.• • • ·
0)[)etem,ltJelher@iliSQnMtrtd·cUN~,>··><
Pell~rll1lo~tH¢smtl<ltllogM~iffi¢,>.·• ·•• <'"
@
Oeter!1ll11~lhe$t~ti~tdng~fPm»··
®.
FG = 1'1.22
Cos 13' =48.60
EG
EG =49.88m.
Solution:
CD Radius of 2nd CUNe:
,, 60·
,,
,,
,
:,'
.
,
t
AG= EA· EG
AG =432.89-49.88
AG = 383.01 m.
8 =56'20' • o· 13'
8 =56'20' - 5'46' • 13'
0=37'34'
I
"
",'
1/
'
,RF285.40
tJ~ I.'>,'
J~
I
--: Iv.:"
" ,
~/
S-326
REVERSE" CURVES
R1 = 285.40
11 + 30 = 12
12 = 50
11 =50·30
I, = 20
AB.
100
Sin 30' =Sin 20'
AB = 146.19
T1 = R1 tan !1
2
T1 = 285.40 tan 10
T1 =50.32 m.
T2 = R2 tan 25'
T1 + T2 = 146.19
TW6jl<ltM¢ltaiIW~Y4Ppm;ij~~M¢f¢~IJ~
•
II~l~~e1Ii~~~lt~dl~~r~ t~~.11ad\~~ •
(j)
.Oeferrni@1~~.(;Elotr:~I~@@9t(ti~rij~fli~
9JfVE i . Y > <
:· ·• irJ~~iI1;i~.i~I~;~£~;I~~~6t!~'.·.
Solution:
T2 = 146.19 - 50.32
T2 =95.87 m.
R - 95.87
2 - tan 25'
R2 = 205.59 m.
® Sta. of PR C.
o - 1145.916
1 - 285.40
0 1 =4.02'
D - 1145.916
2 - 205.59
~=5.5T
CD Central angle of the reverse CUNe:
G
LC1--~
01
_ 20 (20)
4.02
=99:50
LC1LC1
o
L~=2~.~~)
L~
= 179.53
P.C. =(10 +432.24) - 50.32
P.C. = 10 + 381.92
PRe. =(10 +381.92) + 99.50
PRC. = 10+ 481.42
® Sta.ofP.T.
P. T. = (10 + 481.42) + 179.53
P. T. = 10+ 660.95
N
400
tan IX =2 (1100)
=10'18'
~---OG =;j (40W +(220W
IX
OG = 2236.07
2236.07 Cos IJ =2x + 200
x+ 200 =R
x= R- 200
5-327
REVERSED CURVES
2236.07 Cos 8 = 2 (R- 200) + 200
2236.07 Cos 8 = 2R -400 + 200
2236.07 Cos 8 = 2R - 200
1118.03 Cos 8 = R- 100
R-1oo
Cos 8 1118.035
1100 -100
Cos 8 = 1118.035
1000
Cos 8 = 1118.035
8= 26'34'
o = 26'34' - 10'18'
0=16'16'
@
Solution:
CD Radius of first curve:
Sla. of the middle of intermediate tangent.
L=R01t
180
= 1100 (16'16') 1t
L
180'
L = 312.30 m.
"----..;-..::::::-c
AB
150
Sin 30' = Sin 20'
Sla. = (20 + 460) + 312.30 + 200
Sla. =20 + 972.30
@
Sta. of the P. T.
Sla. of P. T. = (20 + 972.30) + 200 + 312.30
sta. of P. T. = 21 + 484.60
AB - 150 Sin 30'
- Sin 20'·
AB =219.29 m.
T1 + T2 = 219.29
T1 =R1 tan !J.
2
T2 = R2 tan !1
2
·~W:ffi~S~1~s~gli~~~{~~I~pi~ • IOi6;~~iH~.
di$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$' and the~tatlgf\i~g~nW
pQint.·. oC.intersection.·.of • • the•• fj(~t¢t@~ • ·is
4·+450,··••·
CJ) Determine the
. . ••.. <.••..••• >..
radius oltha fltstculVe.•.
® Determine the stationing of P.R,C,·· •.....
@ Determine the stationing oUhe P.T.
12 = 11 + 30'
SO' =/1 +30'
/1 =20'
R1 lan !J.
2 + R2 lan !2
2 =219.29
R,lan 10' + R2 tan 25' = 219.29
- 1145.916
Rr
~
- 1145.916
R2 6
R2 =190.99 m.
R, tan 10' + 190.99 Ian 25' ::: 219.29
0.1763 R, T 89.06 = 219.29
R, = 738.68 m.
5-328
REVERSED CURVES
® Stationing of P.R. C.
o - 1145.916 _ 1145.916
1R1 - 738.68
0 1 = 1.55'
T, =R, tan !1
2
T, = 738.68 tan 10'
T1 = 130.25 m.
Solution:
<II Central angle of 2nd CUNe:
Sta. ofP.C. =(4 + 450) - 130.25
Sta. ofP.C. =4 + 319.75
_ (20) (20)_
1.55 - 258.06 m.
L, -
Sta. ofP.R.C. = (4 + 319.75) + 258.06
Sta. ofPRe. = 4 + 577.81
® Stationing of P. T.
L
-~
2- ~
L _20(50)
26
L2 =166.67 m.
Sta. of P. T. = (4 + 577.81) + 166.67
Sta. of P. T. = 4 + 744.48
Arl~y~fseGilr\i~$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
S-329
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.
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
.
S-330-A
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)
@
@
@
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.
5-331
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:
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
5-333
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