478
CIVIL ENGINEERING
Surveying
O b je c tiv e Q u e stio n s
^
1.
M atch L is t-I with L is t-II and select the correct answer using the codes given below the lists
L ist-I
A. V ertical cliff
B . Steep slope
C. Hill
1.
2.
3.
4.
D. Overhanging cliff
L ist-II
Contour lines of different elevations unite to form one line
Contour lines of different elevations cross one another
Contour lines are closely spaced
Closed contour lines with higher values inside
Codes:
2.
A
B
C
D
(a)
4
3
2
(b)
1
3
1 .
4
(c)
1
2
4
3
(d)
4
2
1
3
2
A s s e rtio n (A): A series of closed contours indicate either a valley or a hill without any outlet,
when their elevations, respectively, increase or decrease towards the centre.
R e a so n (R ): Contour lines of different elevations can unite to form one line only at a vertical
cliff.
Of these statem ents:
(a) both A and R are true and R is the correct explanation of A
(b) both A and R are true but R is not a correct explanation of A
(c) A is true but R is false
(d) A is false but R is true
3.
4.
W hat is the angle of intersection of a contour and a ridge line?
(a) 30°
(b) 0°
(c)
(d) 90°
180°
Consider the following statem ents about the characteristics of contours:
1. Closed contour lines with higher values inside show a lake.
2. Contour is an imaginary line joining points of equal elevations.
3. Closely spaced contours indicate steep slope.
4. Contour lines can cross each other in case an overhanging cliff
Which of these statem ents are correct?
(a)
2, 3 and 4
(b) 1 and 2 only
(c)
1 and 4
(d) 1, 2 and
I.E.S M A S T E R
m$Mu(e fa EnQ^eer*
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.pin
ENGINEERING
Contc u rin g
rvp ica l L a n d F e a t u r e s a n d
475
their Cont
our Forms Slopes;
' * *t«
A slop
or steei
Avery steep slope is termed as
AY'f h scrap is known as era
(a ) A g e n t l e s lo p e
scarp
(b) A steep slope
High-lying F o r m s
g
(c) A hill
These are characterised bv elevatprl
j
o
'V elevated grounds, for example hill, hillock and plate,:au.
Hills are elevated ground usuallv witv, Q
in shape and increasing the contour values''inwards* '' ^ C°Unt°UrS of hllls are bit circular
Low-lying F o r m s
The most common among the low-lying forms are ravines, valleys, etc.
Ravine
Ravine is a through like depression of the earth's surface, elongated in one direction with the
bottom inclined towards one side.
A ravine can be imagined as a depression washed out in the ground by flowing water.
Valley
A valley is a broad ravine with a gental sloping bottom.
The countours of a valley are in a shape of V.
If the ground is low as compared to the surrounding land and the sides slope generally, it is
i
called as a d ep ressio n .
The contours are quite few and far apait.
,
■'
If the valley floor is very narrow and has steep sides on a level terrain, it is called as gorge
and in mountains as canyons.
Due to the steepness of sides, the contours aie cio
riV IL EN G IN EER IN G
List-I
List-II
A. Traversing
1. Rays are drawn to locate the station on which the table is set-up
2. At least two rays are drawn from two different stations to the
details to be located
3. Rays are drawn in the direction of details through the station
point on which the table is set-up
4. Rays are drawn on the map by setting up the table over each of
the stations towards the subsequent station
B. Resection
C. Intersection
D. Radiation
Codes:
A
B
(a) 4
3
(b) 2
(c) 4
(d) 2
469
Plane Table
1
1
3
C
2
D
1
4
2
4
3
3
1
The process of determining the location of the station (on the map) occupied by the plane table
is called as
8.
(a) intersection
(b)
three-point problem
(c) traversing
(d)
resection
W hich one of the following surveys employs alidade?
(a) Contour survey
(b) Archeological survey
(c) Plane table survey
(d) Reconnaissance survey
Pick out the correct statem ent .
(a) U-frame is used for orienting the plane table
(b) The drawing sheet used in plotting a plane table survey needs no special care in fixing on the
board.
(c) An alidade is also called a sight rule.
(d) A simple alidade and Indian pattern clinometer serve the same purpose.
. If the plotted position of an instrum ent station is not known, the most accurate orientation
of the plane table can be achieved by
(a) a trough compass
(b) backsighting
(c) observations of two well-defined points
(d) observations of three well-defined points
The most rapid method of orientatation by the three-point method of plane tabling is the
12.
(a) tracing paper method
(b)
graphical method
(c) trial-and-error method
(d)
both (a) and (b)
It is necessary to go to one of the plotted stations in the method of resection
(a) by trough compass
(b)
by a back ray
(c) by both (a) and (b)
(d)
by three points
I.E .S M A S TER
1
i □___ L
insstme for Engineers
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C IVIL ENGINEERING
Surveying
466
iii. H the plane table is not properly oriented at each station,
iv . If> the plane table is not properly clamped after orientation
sights and the error will occur.
v.
the plot obtained will be inaccurate,
the plane table will rotate between
If the object is not sighted accurately and bisected propeily, the e n o i
vi. I f the alidade is not properly pivoted on the point, the lay s diaw n
T his error is cumulative if the survey is extended from a sm all p a it to a
g
p
vii. The error will occur if the tripod is not planted firmly into the giound.
3. E r r o r d u e to I n a c c u r a t e C e n tr in g
If the plane table is not exactly centred over the station, an error will occur. The magnitude
of the error can be determined as below.
Shows the case when the plane table is not centred, and the plotted position p of the station
P is not exactly over the station P. Let A and B sighted after pivoting the lalidade on p. The
plotted angle between A and B is ApB, but the actual angle in the field is A PB. The angulai
error y caused by inaccurate centring is equal to the difference between the angles APB and
ApB.
Angular error in centering
= Z APB - Z ApB = a + (3
Y
Let Pc & Pd are perpendicular to the line Ap & Bp.
sin a -
Pc
^p-
a = sin
Pc
AP
Pc
AP
Pd
sin P =
BP
a —
P
sin
'P d '
, BP,
Pd
P
BP
A ngular error in centring, y = a + P
............. .
InsMute for Engineers
i f s /g a t e / p s u s .
o a r c ll. INGVV D e l h i • 1 1 U U U >
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CIVIL E N G IN E E R IN G
Measurement of Area & Volume
Example 4,
445
Determ ine
th e area of the traverse ABCDEF of the below figure by the departure
latitude
method.
ar>d the total
100m
REFERENCE
STATION
—
•-+.-----------------100m
I---- *Xtr>
■200 m--------- -J*—i00m— *|
T otal latitu d es of the Doint^ ft e n n
»
.
below
points B. C. D. E and F with respect to the refer
Point B
L ', = 0 - 100 = -1 0 0 m
Point C,
L\,= -1 0 0 + 0 = -1 0 0 m
P oint D.
LV, = -1 0 0 + 100 = 0
Point E.
L ’ , = 0 + 100 = 100 m
P oint F.
L ’5 = + 100 + 0 = 100m
ence point A are a
The algebraic sum of the departures of the two lines meeting at the stations B C D E and F are
as below
’ ’
u r are
Point B.
100 + 200 = 300 m
Point C.
200 + 100 = 300 m
Point D.
100 -
100 = 0.0
Point E, - 1 0 0 - 200 = -3 0 0
Point F, - 200 - 100 = - 300
2A = - 100 x 300 - 100 x 300 + 0 + 100 x (-300) + 100 x (-300)
or
2A = - 120000 m2
A = 60000 m2 (Negative sign is neglected)
la titu d e s and d epartu res of the lines of a closed traverse are given below. Calculate the
The
area o f th e tra v e rse
CIVIL ENGINEERING
M ining Dial
•
The mining dial is a simple form of theodolite with a built-in compass. It is mai
mine surveying.
y
• * It consists of a large magnetic compass fitted with either a telescope 01 plain s’g
vertical arc.
B ru n to n ’s C om pass
*
Brunton’s compass or Brunton’s pocket transit is an instrument with is a combin
compass and a clinometer.
P an tagrap h
A pantagraph is not a surveying instrument. This is an instrument used in the office to
enlarge or to reduce a plan already drawn.
Eidograph
An eidograph is an improved version of the ordinary pantagraph. It is used for the same
purpose as a pantagraph.
Sou n d in g S e x ta n t
A sounding sextant is similar to a nautical sextant. However, the index glass of a sounding
sextant is generally much larger as compared to that of a nautical sextant. The sounding
sextant is generally used in hydrographic surveying. As the small boats used in hydrographic
surveying may be rocking when the object is sighted, the sounding sextant is more useful than
a nautical sextant.
E
I E.S M A S TER
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Office: F-126. Katwaria Snrni New Delhi • 110 010
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CIVIL ENGINEERING
Minor Instrum ent
481
There are two types of sextan ts :
(1) Nautical sextant, and
(2) Box sextant.
N au tical s e x t a n t
It has two glasses called Index glass and Horizon glass. If the angle of inclination between
Index glass and Horizon glass is angle 0 then angle betw een two sides objects in the ground
will be 20.
Used o f N autical S e x ta n t
1. The nautical sextan t m easures angle in the plane of the two objects and the telescope. It is
unlike a theodolite (or a compass) which m easures the angle in a hoiizontal plane. Therefore,
the nautical sextant is a more versatile instrum ent.
2.
An angle can be measured while the observer is on a ship 01 a boat.
3.
It is specially useful for navigation and astronom ical puiposes.
4.
The angle measured between the two objects at d ifferent elevations can be reduced to the
horizontal angle, if required.
Box Sextant
The box sextant is a small, pocket instrument used for measuring angles. It consists of a box about 75 mm.
in diameter and 40 mm deep.
Uses o f a B ox S ex tan t
1.
The box sextan t is generally used for m easuring angles in chain surveying, if required.
2.
The box sextan t can be used like an optical square for settin g the perpendicular in chain
surveying if the vernier is set at 90°.
3.
It can be used for locating inaccessible points in chain surveying by intersection.
4.
It can be used for locating details by rad iation method in trav erse surveying.
5.
It can also ’be used for checking approxim ately the angles m easured by m eans of other
instrum ents.
6.
The box sextan t is extrem ely useful in recon n aissan ce.
Advantages of a Box Sextant
1.
It is a very compact and handy in stru m en t.
2.
It can be used in any position.
3.
It can be used even in a moving boat or a ship.
4.
It is a pocket instrum ent, and can be easily ca rried from one place to the other.
5.
It is quite inexpensive.
Site Square
The site square is an in stru m en t used to s e t out two lin es a t rig h t angles to each othei.
l.E.S M A STER
InsStute for Engineers
lES/GATE/PSUs
O ffice: F* 126. Katwaria Sarai. New Delhi - 1 10 01G
Website: www.iesmaster.org. E-mail: ies_master? vahno n> in
Phone: 0111101310(5. 783881010(3. 971 185:1908
Minor Instrument
Hand Level
•
A hand level is a small levelling instrument which is held in hand while levelling.
•
It is used for approximate determination of elevations in reconnaissance, preliminary surveying,
for locating contours on the ground, and for taking short cross-sections in profile levelling.
Abney Level
•
An Abney level is an improved version of the hand level. It can be used as a hand le\el for
levelling, and as a clinometer for measuring slopes. It is a quite light and compact instrument.
•
The vertical scale is extended type
Indian P attern Clinom eter
•
The Indian Pattern Clinometer, also called the tangent clinometer, is a simple instrument used
for determining the difference of elevations of the two points by measuring the inclination of
the line of sight.
•
It is specially useful for plane tabling.
Ceylon Ghat Tracer
•
The C'eylon ghat tracer is a simple instrument used for measuring the slopes.
•
It is specially useful for setting out a grade contour on a given gradient in the preliminary
survey of a road in a hilly area.
S e x ta n t
A sextant is an instrument used for measurement of the horizontal and vertical angles.
The distinguishing feature of a sextant is the arrangement of
. .
” , ..
i
. •. . .
. rr
. i. ,
,
s utm oi two mirrors which enables the
observer to sight two different objects simultaneously.
e n g in e e r in g
Contouring
479
AJtonetry may be depicted moat accurately by
»
haChU,'e3
W
relief .hading
*
Unt,ng
»
\latch L ist-I with L is t-I i and selppt tk,
contour line.
j ^j g
a
n I ec a ns we r using the codes given below the lists:
Equally spaced contour lines
L is t-II
B- Contours a re alw ay s p erp en d icu lar to
C. C ontours in crease in elev atio n from
inside to outside
«
8l° Pe
* aaAA]
‘‘
e
4. A depression
5. Uniform slope
Codes:
B
2
3
3
1
A
(a)
(b)
(c)
(d)
7.
Contour inte rval on a map s
(a)
(b)
(c)
(d)
8.
1
1
5
5
C
4
4
2
4
vertical distance of contour lines above the datum plane
vertical distance between two successie contour lines
slope distance between two successive contour lines
horizontal distance between two successive contour lines
Which of the following ch a ra cteristic featu res may be used while plotting a contour plan?
1. Two contour lin es having the sam e elevation cannot unite and continue as one line.
2. Contour lines close togeth er indicate a gentle slope.
3. Contour lines cross a valley line a t right angles.
Select the co rrect answ er using the codes given below.
(a) 1, 2 and 3
(b)
1 and 2
(c) 2 and 3
(d)
1 and 3
A s s e r tio n (A ): E xcep t in the case of an over-hanging cliff, two contour lines cannot merge
9.
or in tersect at a point on the map.
R e a so n (R ): In tersectio n of two contour lines means one point on the surface of the earth will
have two d iffere n t elev atio n s. T h is is not possible.
6-
(d)
u
7.
I.E.S M A STER
institute for Engineers
lES/GATE/PSUs
(b)
8.
(d)
9.
(d)
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m
ENGINEERING
a
high land, h av in g fla t narrow top with
7----------------
on the o th er side is called as e s c r a p m e n t
?
^
° n ° " e side and gentle (dip) s,ope
The contours w ill be clo ser tow ard*
1
---------------------------------- ----------------------------- s^cepjudeanci far apart towards the gentle side.
(j) E s c r a p m e n t
Cliff
Cliff are the steep ro ck faces along the sea coast and may be vertical where the contour lines
concide w ith e a ch o th e r, an o v e r h a n g in g c l i f f where the contour lines intersect each other.
476
CIVIL ENGINEERING
ii i
-f '
m
m
(d) /I depression.
(e) A gorge or canyon
Valley Line and Ridge Line
•
•
The slopes of ravine intersect along a line referred as the axis of the ravine, the line of
d isch arg e, or a valley line in case of a valley.
Counter part of a ravine is a ridge—a convex form of terrain gradually declining in one
direction.
•
Two ravines are generally separated by a more or less pronounced ridge.
•
The line along which the slopes intersect is referred as the axis o/ ridge, the watershed or
w atershed line. The watershed line is generally wavy.
Ravine represented by contours
Scraps
indicated
b/ teeth
Metres
1050
1000
950
900
850
800
If
llttvtnt*
i
4 l - s h t i p a i W illey
Saddle
•
•
•
The lowest points on the watershed are known as passes.
Pass is narrow low land passing through high mountains on either sides.
Sometimes this narrow low land is cut back by the streams. This steep-sided depression is
called as a COL. When this depression is broad and low, it is known as sad d le.
•
»
A part of the land in form of tongue, which cuts out from a hilly area is called as spur.
The contours are similar to that of a valley, with a difference that here the counter values
decrease towards the vee.
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insatuU fa Er>j«iocr5
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474
Surveying
C IV IL EN G IN E E R IN G
Characteristics of Contour Lines
1-
All th e points of a contour line have the sam e elevation. I he elev atio n s o t e contours are
shown eith er by in sertin g the figure in a break in the resp ective contoui 01 pi intec c ose to
the contour. W hen no value is represented, it ind icates a flat te n ain. A zeio m etei contoui
cave p en etratin g a hillside.
3.
A contour line is a closed curve. They may close eith er on the map or ou tside the map. it
depends on the topography.
4.
Equally spaced contour represent a uniform slope and contours th at are w ell a p a rt representsa gentle slope.
5.
A set of close contours with higher figures inside and lower figures outside in d icate a hillock,
w hereas in case of depressions & lakes, etc., the lower figures are inside and th e higher figures
are outside.
6.
A w a t e r s h e d or r id g e lin e (line joining the highest points of a series of h ills) and the th a lw eg
or v a lle y lin e (line joing the lowest points of a valley) cross the con tou rs at rig h t angles.
7.
Irreg u la r contours represent an uneven ground surface.
8.
The d irection of the steepest slope is along the sh o rtest d istan ce betw een th e contours.
At a point the direction of the steep est slope on a contour is. th e re fo re , at rig h angles to the
ENGIN E E R IN G
Contouring
■JTO r a lly th' contours m e not risible on the
473
grounds excepts in the case of shoreline*.
• The virtu al distant< between consecutive contours Je term< I i contour in terval
•
It
desirable to hu\e a constant contour interval throughout the map
• In special cases, a variable contour interval may also he provided.
• A variable contour interval i> as far as pc»’ *ible avoided since it gives a false impression of the
relative steepness of the ground in different parts of the map.
• Generally contour intervals are taken 1 to 15 m
• The smaller the contour interv ii the
m o re
preci*el> the terrain relief is predicted on the plan
• The contour interval depends upon the following factors.
(1) Scale of the map
lull Nature of the Country
■p
<n>
Purpose of the map
(tv)
Time
(v) Funds
L Scale o f th e M ap: If scale is sm all, the coutour interval is kept large so that there is no
overcrowding of the contours. On the other hand, if the scale of the map is large* the contour
:
interval can be kept sm all.
1 Purpose o f M ap: The contour interval selected should be small so that the map serves the
intended purpose, but at the same time it should not be too small otherwise the cost of the work
would be too much The contour interval should be kept small when the plan is required for
the detailed design
*
Nature o f G ro u n d : For a flat ground, the contour interval is small, but for a steep slope the
contour interv al is large. If the ground is broken, the contour interval is kept large so th at the
contours do not come too close to each other.
*
Time: Contour interval is kept large when time is less
4 Funds: Contour interval la kept lartie when fund- un lew
(M ir# f 134 k#nn in« $ mm i V # N b
Oil <101 ISM
STUI
I10 0 U
C h a p te r
C ontouring
In tro d u ctio n
The relative position of points in it plane are represented by a map The value of the n a p t*
even more if the relief (variation in the elevation of earth * surface) i$ also included along with
th eir relative position*.
T h ere are two m ethods by which the conformation of the ground may he presented on a map
( l ) By delineating the surface slope* b\ shading intended to given an impression of relative relief
The relative elevation* of the point* are not indicated in thi* case.
ib) By plotting the contour lines (imaginary line parsing through points <d equal elevations) on
map* These line- are arranged such that the form of fh« earth s surface can be portrayed with
greater accuracy and thoroughness and can In* readily he interpreted
C ontour* are used by engineer* in a many w<i>*
Use o f C ontours
ia> Proper and precise location of engineering work* -uch an road* can al* etc
ib> In location of w ater supply water distribution and to solve the problems of steam pollution
tel In planning and designing of dam * reservoirs aqueduct* transm ission lines etc
idi In selection of Mte« for new industrial plants
te) Determining the intervisibihty of station*
(0
J
Determining the profile of the country along any direction.
ig) To estimate the quantity of cutting filling, and the capacity of reserv o ir*
D e fin itio n o f C ontour
1T
” ? * d' f,ntd " * "
,ln'
point* of
.I m o w t *
A contour l.nc m ., . 1 * be defined . . the m ter«c.u.n of . U „ , *urftK.. , „ h
>f ^
earth.
Contour line* on a plan illustrate the topography of the ground
When the contours are drawn underwater, they are term ed
or b a ih ym e tric t u n e*
'
- a
M
contours.
Reason <R): The problem bee,™
re sp e c tiv e of (he orientation "f’th mi,et‘™ in a te as the ,b
. , fu
e „
'
Plane' table
Consider
the following
following
state
' ab' e'
ree
Consider
the
statemen,
8
r
1.
Two-point problem is solved by
_.
- d l uttersect at one point
8 t0 plaM ‘ able survey:
2, Three-point problem is solved hv n
me‘hod'
3. In two point problem, auxiliary
y station is required
Which of these statem ents are correct?
(a) 1 and 2
(c) 2 and 3
&
.
^
(d)
Which one of the following instrume
•
1 and 3
1, 2 and 3
of horizontal a »d vertical distances d i r e c t “ Pla" e table surveyiag for the measurements
(a) Plain alidade
(b) Telescopic alidade
(c) Tacheometer
(d) Clinometer
11.
(c)
12.
(b)
13.
(b)
14.
(b)
15.
(a)
16.
(a)
17.
(d)
18.
(a)
19.
(a)
20.
(c)
21.
(c)
I.E.S M A S T E R
institute for Engineers
Office- F-126. Katwaria Sarai, New Delhi - 110 016
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ub 'n l n i l - 11013106. 7838813406. 9711S53g08-----------------
C IV IL ENGINEERING
470
f t
The m ajor source of error is sm all-scale mapping by plane table is due
(a)
inaccurate centring
(b) long sight
(c)
shrinkage of drawing sheet
(d) few observations
Which one of the following errors is more severe in plane-table surveying.
(a) Defective sighting
(b) Defective orientation
(c) Movement of board between sights
(d) Non-horizontality of board when points sighted are at large differences of theii elevation
15. A s s e r tio n (A): The solution of a three-point problem in plane-table surveying is aided by
Lehm ann’s rules.
R e a s o n (R ): The application of Lehm ann’s rules reduces the trian g le of error and is a
controlled trial-and-error technique.
16.
Consider the following statem ents:
A s s e r tio n (A): In plane table photogrammetry. the areas to be mapped are taken from either
end of a base line.
R e a s o n (R ): The position of the detail point with reference to the base line is obtained by
intersection of rays drawn to it from each end of the base.
Of these statements
(a) both A and R are true and R is the correct explanation of A
0)) both A • and R are true but R is not a • correct explanation of A
•
(c) A is true but R is false
(d) A is false but R is true
17.
In a plane table survey, the plane table station position was to be fixed w ith respect to three
reference points. It was found that one of the reference points was not visible due to some
obstruction. It was, therefore decided to make use of the other two points only. Which one of
the following statem ents is true regarding the determ ination of station position?
(a) The work can be done faster
(b) Two settings of the plane table will be needed but the work will be accurate
(c) Only one setting of the table is needed, however the work will be less accurate
(d) The work will be less accurate and time consuming
The fix of a plane table from three known points is good if
(a) the middle station is the nearest
(b) the middle station is farther than the other two stations
(c) either of the extreme stations is the nearest
(d) the middle station is close to the great circle
19. A s s e r tio n (A): If the plane table station P lies on the great
stations A. B and C, then the position of point y
problem.
•
u i
can be determ ined b ^ h e t l X - p o :
J
j
mw
•i ■'»1
468
Surveying
C IVIL EN G IN EER IN G
O b je c tiv e Q u e s tio n s
It is required to produce a sm all-scale map of an area in m agnetic zone by directly plotting
and checking the work in the field inself. Which one of the following surveys will ie most
appropriate for purpose?
(a)
Chain
(b) Theodolite
(c)
Plane Table
(d) Compass
For locating an inaccessible point with the help of only a Plane table, one should use
(a)
traversing
(b) resection
(c)
radiation
(d) intersection
The method of plane tabling commonly used for establishing the in stru m en t station is the
method of
ytT'
(a)
radiation
(b) intersection
(c)
resection
(d) traversing
M atch L is t - I (Statem ent) with L is t-I I (Situation) and select the correct answ er using the
codes given below the lists:
List-II
L ist-I
A. Accurate centering in plane table surveying i. Inaccessible points
is necessary for
2. Open country with good intervisibility
B . E xact orientation is more important, than
accurate centering for
3. Large scale maps
C. The intersection method of plane table
surveying is particularly employed for
4. Small scale maps
D. Plane table survey is useful for
5. Hilly regions
Codes:
A
(a)
(b)
(c)
(d)
jp f
X
3
4
5
3
B
4
3
4
1
C
1
2
3
4
D
2
5
1
2
The method of orienting a plane table with two inaccessible points is known as
(a) intersection
(b)
resection
(c) back sighting
(d)
two-point problem
M atch L 18* ' 1 (Methods.) Wlth L i s t -H (Procedures) and select the co rrect answ er using the
codes given below the lists:
E.S M A S T E R
Institute for Engineers
lES/GATE/PSUs
CIVIL ENGINEERING
-
Plane Table
Pc
sin
\
AP
+ sin -i
46
Pd
BP
The above figu re show s the c o rre c t position of points A & B as a' & b', therefore the error
potting is a'a & b ’b.
We know th a t
,
a'a = pa sin a = pa x a = pa x
Pc
AP
Pd
bb' - pb x sin P = pb x p = pb x
If R .F (r) =
BP
1
n x 100
[i.e. scale is 1 cm = n m eter]
pa = AP x r
pb= B P x x’
L e t Pc = Pd = e m tr.
Pc
aa.i _
= pa
= AP x r x
= er
AP
AP
Pd
e
bb' = pb x — = E p x r x — = er
It is observed th a t d isp la cem en ts of the points from th e ir co rrect positions is re.
If the lim it of p recisio n is 0 .2 5 , for in sig n ifica n t effect in plotting,
i.e.
re < 0 .0 0 0 2 5 •
e<
For r =
0.0 0 0 2 5
r
1
100
0.0 0 0 2 5
Sim ilarly for =
1/100
1
(i.e.
1 cm = lm )
= 0 .0 2 5 m = 25 mm.
1000
e = 2 5 0 mm
H ence it is observed th a t ce n te rin g erro r is quite im p ortan t when the scale is large.
Example 1.
In settin g up th e p lan e tab le a t a sta tio n P, the corresponding plotted point p was not
accu rately cen tred over P. I f th e d isp lacem en t of P w as 20 cm in a direction at rig h t angles
to the ray, d eterm in e th e d isp lacem en t of th e point from its true position on the plan if
(a) r =
5000
(b)
* - 500
(c)
50
S o l. We know th a t,
aa' = r x e.
In th is case, e = 20 cm = 2 0 0 mm
1
(a) aa' = 200 x ------- = 0.04 mm
5000
■E.S M A S T ER
Institute for Engineers
lES/GATE/PSUs
1
(b) a a ' = 2 0 0 x ------ = 0.4 mm
500
(c)
aa ' = 200 x — = 4 .0 mm
50
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Remove the tracing paper and move it
through the plotted points a. b and c
Fix the point p on the plan with Q
table.
'
Plane Table
465
1 6 Plan Untl1 the tllree rays simultaneously pass
,,
n°pdle point to obtain the position of p on the plane
Pivot the alidade on p, with erltm
bisected. The plane table is now oriented.^’ ^
^
^
Pla" e
Until the 8tati° n A
Strength o f F ix in T hree-P oint Problem
The following points should be considered while selecting a plane table station.
1. \Vhen the plane table station P is inside the great circle, the station should be selected near
t etc n lmc o
f gii at ti langle. When p is at the orthocentre of the great circle ABC, the
strength ot tix is a maximum.
2. Avoid the plane table station near the circumference of the great circle, as its position is
indeterminate.
3. When the plane table station P is outside the great circle, select the station near the middle
known station.
4. If one angle is small or when the station P is in line with the two known station, the larger
the angle to the third known station, the better is the fix. But the angle should not be larger
than 90°, and the two known stations on the line should not be near to each other.
5. If the point P is within the great triangle, and the middle station is much nearer to p then
the other two stations, the fix is good.
Errors in Plane tab le Surveying
1. In stru m en tal E rro r
i.
The error will occur if the top surface of the plane table is not perfectly plane.
ii. If the fiducial edge of the alidade is not straight, the lines drawn would not be straight and
in error will occur.
hi. If the fittings of the table and the tripod are loose, the plane table will not remain stable
iv. If the magnetic compass is sluggish, there will be an error in orientation done with the
magnetic compass.
v. If the sight vanes are not perpendicular to the base of the alidade, there would be an error in
sighting.
vi. If the level tube is defective, the plane table will not be horizontal when the bubble is central.
The plot obtained will be inaccurate.
vii. If the drawing paper is not of good quality, it will be affected by the weather conditions. It may
expand or contract. The plot obtained will not be conect.
2. E rro r o f M anipulation and Sighting
i.
If the plane table is not properly levelled and made horizontal, the r i g h t . v a n e s b e inclined
to the vertical. There would be an error and the pomts located will not
conect.
ii. If the plane table is not accurately centred, the error will occur. The error is more important
in large scale plotting
E.S M A S T ER
insMiie for Engineers
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Office: F-12G. Katwaria Snrai. New Delhi • 110016
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464
3.
If the point P lies on the great circle passing through the « t a t i ^ e ^ a y s w ^ U n t e r s e T t
of the point p cannot he determined by three-point problem,
as
at one point, irrespective of the the orientation of the plane table,
oi exa
I
If it is suspected that the point is on the great circle, change the orientation
a second trial. If three rays again meet at the same point, the point ie- on
And if it is confirmed that the station P is on the great circle, replace oik of the
A, B and C with another point D such that the point p does not fall on t u new
4.
If the point P is outside the great circle passing through A, B and C, the point d o
the intersection of the two rays drawn to the nearest points and the point P
side of the ray to the most distant point. For example p,
*
d by
me
5.
If the point P lies outside the great triangle but inside the great circle, the ray to the middle
point lies between p and the point e obtained from the intersection of the i ays to t k ot ei two
(extreme) points. For example p,.
6.
If the point P lies on or near one side (say, AC) of the great triangle, the point p will lit between
the two parallel rays drawn to points A and C, and it will be on the same side of each of the
rays. For example pv
7.
If the point P lies on or near the prolongation of one line (say, AC) the point p lies outside the
parallel rays and on the same side of all the rays. For example p(>.
8.
If A, B and 0 are on one .straight line, the great triangle gets converted into a straight line
and the great circle will have abc as its arc, and an infinite radius. A ray drawn to the middle
point is between the point p and the point f obtained from the intersection of the rays to other
(extreme) point.
M ech an ical M ethod for T h ree-P oin t Problem
The three-point problem can be solved by a mechanical method
using a tracing paper.
1.
Set up the plane table at station P w'hose location is required
and orient it approximately
using a compass or by eyeappi
judgement
2. Stretch a sheet of tracing paper over the plan and fasten it on
the plan.
3.
4.
\
(a) Tracing Paper
a\
Select any convenient point p, on the tracing paper to represent
the station p.
Pivot the alidade on p, and sight the station A and draw a
ray. Similarly sight the stations B and C and draw the rays.
IP S
mas
re?
Fnsit Je Id Eng'neere
IFS/GATE/PSUs
b
/c
P
7
Plan
(b)
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B a s ic r u le s
R ule 1. The point p is on the same side of all the three resection lines a'a? b'b and c'c.
When the surveyor faces such a problem the corresponding to stations A, B and C, the point
P *ies e*ther t0 t,le r*Sht of all resection lines or to the left of all resection lines. In above, the
point p lies to the left of all the lines.
Rule 2. I he distance of p from a resection line is proportional to the length of that line. Hence.
pi
pa
p2
pb
p3
pc
Now keep the alidade along p'a and rotate the plane table to sight the station A. Clamp the
plane table. This orientation of the plane table is more accurate than the first trial.
8. Now pivot the alidade at b, sight B and draw a ray. In a same way alidade at c. sight C. and
draw a ray.
9.
Determine the intersections of the three rays drawn. These rays will form a smaller triangle
of error than the previous one.
10. Now select a point p'\ satisfying the two rules, and repeat above steps.
The above procedure is repeated till the triangle of error reduces to a point p.
In this way, the location of station P is fixed on the plane table.
Su pplem entary L eh m ann’s Rules
1. If the plane table station P is outside the
great triangle ABC the triangle of error will
also be outside the great triangle abc formed
by joining the plotted points a, b and c.
As per basic rule 1, the point p lies on the
same side of all rays, therefore, the point p
shall be outside the triangle of error. For
example p,.
2.
If the plane table station p is inside the great
triangle ABC, the triangle of error will also
be inside the great triangle abc. As per basic
rule 1, the point p shall be inside the triangle
of error. For example p2.
I.E.S M A STER
insiilute for Engmeerc
lES/GATE/PSUs
Office: F-126. Katwaria Sarai. New Delhi • 110 016
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Phone: 011-41013406. 7838813106. 0711853908
462
CIVIL ENGINEERING
Surveying
N o te :
F o r b etter resu lts th e p oin t P s h o u ld be ch o sen q u ite f a r o f th e p o in t C.
Resection After Orientation by Three Points^
The three-point problem consists in locating the position of a plane table station on the drawing
sheet by means of observation of three well-defined points, whose positions are already been
plotted on plan.
Let A, B, and C, be the three well defined points and let their plotted positions be a. b and c.
It is required to fix ground station P on plan as p.
Three point problem can be solved by several methods.
1.
Graphical method
2.
Trial and Error method
3.
Mechanical method
4.
Analytical method
5.
Geometrical - construction method.
J v G ra p h ic a l M ethod
•
There are several graphical methods to solve the three-point problem.
V
But the most simple and the best one is the Bessel's solu tion by the inscribed quadrilateral.
•
This is based on the geometric principle that in any inscribed quadrilateral, the angle made
between one of the sides and one of the diagonals is equal to the angle made between the
opposite side and the other diagonal.
T ria l-a n d -E rr o r M ethod o r L e h m a n n ’s M eth od
•
T h is method is very comrp^nly nspd in fipJH measurements
as it gives very accurate^r&snits^
•
The position of a plane table is estimated by judgement. Let
it be p\
•
The alidade is kept against p a and the table is oriented.
•
Now, pivot the alidade on point b and sight B. Draw a back
ray. Pivot the alidade against point e and sight C. Draw a
back ray. If the orientation of plane table is correct, the
three rays will intersect at one point p, otherwise a triangle
or error is formed.
•
This triangle is reduced to a point by trial and error.
L eh m an n ’s Rule
There are two basic rules and few supplemantary rules. Although the basic rules are sufficient
for determining the position of point P but the supplementary rules also facilitate in the quick
solution.
f.E.S M A S T E R
Institute for Engreers
lES/GATE/PSUs
Office: F-126. Katwaria Sarai. New Delhi - 110 016
Website: www.iesmaster.org. E-mail: ies_master$yahoo.co.Ja
Phone: O il- 1101310G 7838813106.971 1853908
^
riV IL ENGINEERING
,
Plane Table
in this case the two point problem is used to nl
whose positions are alread y plotted on p |an
P°
461
‘
3 Statlon C by sighting: to station s A & B
procedure
1.
Choose a su itable statio n D near C
two obtuse.
,
6 angles CAD and CBD are neither two acute nor
Set up the plane table at station D C W n t
t, ui
or altern ativ ely , by eye-ju dgm ent Su rb t h a t l'* ° approximately using a m agnetic compass
Clamp the p lane table
Ch ^
lmG ° b 15 made a PProximately parallel to AB.
2.
Pivot the alid ade on o and eight station A. Draw a ray through a
“
n
^
6 “ d S- h ‘ B
»• intersecting the ray
I t tio n D it'ap p Vreo x ^ a t aePPr05dmate P0Siti° '’ ° f “ “ «I0 ’“ d P° ' " ‘
*>—
orientation a .
Now tra n sfe r th e point d , to th e ground using a plum bing fork.
3' n r ' M hV l i ade pi™ te d fa t *>• “ Sht the station C. Draw a ray d , c , to represent the distance
DC . M aik the position of c } by approx, estim ation.
4.
Now, sh ilt the p lane tab le to station C and cen tre it such th at the point c, is above C.
Orient the plane table by b ack sig h tin g on D. Thus the orientation at C is the same as it was
at D.
5.
Now, pi\ot the alidade at point a and sight station A. Draw a ray ac., to intersect the ray
at point c.,.
Hence, c^ rep resen ts the approxim ate position of the station C. because the orientation is still
approximate.
6.
Pivot the alidade a t point c.2 and sight station B. Draw a ray c2fe, through c.,. In general, the
ray c 2b 2 will not pass th rou gh the correct position b. because the orientation is approxim ate.
This point fa, gives th e approxim ate position of station B with respect to the orientation made
at D.
As the length a b is the true representation of distance AB, the error in the orientation is equal
to the angle fa, ab betw een the lines a b and crfa,.
To elim in ate th is erro r in the o rien tation , place the alidade along ab,. Fix a ranging rod at a
point P (a random point) a t some distance from the plane table and in line with ab,.
8.
Place the alidade along line ab and turn
the plane u n til the ran g in g rod at P is
bisected. Clamp the plane table in this
direction.
Now the o rien tatio n of the plane table is
correct and the line ab is exactly parallel
to AB.
9.
Now to find the tru e position point c of
the station C, pivot the alidade on a and
sight A. Draw a ray cci through a.
REQUIRED
TURN
Sim ilarly pivot the alidade on fa and sight
B . Dr a w a r a y c b t h r o u g h fa. T h e
in tersection of the rays c a and cb gives
the tru e position of c.
.E.S M A S T E R
Institute for Engineers
lES/GATE/PSUs
O ffice: F-12G. Katwaria Sarai. New Delhi •.110 016
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CIVIL ENGINEERING
Surveying
Kin 's lire drawn from these stations to the station to he plotted.
I hi intersection of the rays from these two stations gives the position of th. station to l„
plotted on the drawing sheet.
Sometimes, this method is also called as g ra p h ica l triangululion.
•
1’his method is most commonly used for plotting details. It is preferred when the distance
between the stations is too large, the stations are inaccessible, or the ground is undulating
l‘ °r example is of broken boundaries, which can be very conveniently plotted by this method
4 . R esection
•
This method of orientation is employed when the plane table occupies a position not yet plotted
on the drawing sheet.
•
Resection can be defined as the process of locating the instrument station occupied by the plane
table by drawing rays from the stations whose positions are already plotted on the drawing
sheet.
•
1 he point representing the resection of two rays will he the station to he located provided the
orientation at the station to he plotted t> correct, which is seldom achieved.
•
1his problem can he solved by any of the methods such as resection after orientation by back
ray. by two points, or by three points
•
This method is employed when surveyor feels that some important details can be plotted easily
by choosing any station other than the trinngulntinn stations.
The position of such a station is fixed on the drawing sheet by resection.
R esection After Orientation by Back Ray
This method is very useful when one of the plotted stations is accessible from the station to
be plotted, problems.
Procedure
1.
2
3
I
5
6
7.
Let a & b be the plotted positions of the two ground
stations A & B and station C is to be plotted.
Set up the table at station A with point a being over
station A.
By keeping the alidade over 06, bisect the station B.
Now as the table is oriented, bisect the station C and
draw a ray ac.
Shift the table at station C and place the alidade along
ca and rotate the table till it is oriented and clamp it.
With the alidade pivoted on b. sight the station B and
draw a back ray be'.
The point of intersection of ac and be gives the required position of station C on the plane table
R esectio n After Orientation by Two Points
•
E
The two-point problem consists of locating the position of a plane table station on the drawing
sheet by observation of two well defined points, whose positions have already been plotted on
plan.
I E S MASTER
ifVSATE PSU»
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j w mi iioimK >
■a.lero.fcu*'
it'iuk ■
riV IL ENGINEERING
Plane Table
S„, the table on station B, place the alidade along the plotted line „ and rotate the till until
the line o atg i k, bisects the station A. Clamp the board along this line of sight. The line
ba truly represents the line BA on the ground.
3. R esectio n
The method of resection will be discussed later in this chapter in details.
Methods o f Plane Table Surveyin au
The methods of surveying with a plane table are
^
Radiation,
• Traversing.
.# Intersection and
Resection.
/
1 . Radiation
In this method the instrument is setup at a station
and rays are drawn to various stations which are
to be plotted.
The distances are cut to a suitable scale after
actual measurements.
This method is suitable only when the area to be
surveyed is small and all the required stations to
be plotted are clearly visible and accessible from
the instrument station.
The scope of the method is increased when the
distances are measured with the help of a
tacheometer.
2 : T raversin g
• This method is similar to compass or theodolite traversing.
• The table is set at each of the stations in succession.
• A foresight is taken to the next station and required the distance is cut according to a suitably
chosen scale.
• This method is most suited when a narrow strip of terrain is to be surveyed, e.g. survey of
roads, railways, etc.
• This method can be used for traversing both the open as well as close traverses.
3. In te rs e c tio n .
•
In this method two stations are selected such that
all the other stations to be plotted are visible from
these.
• A line joining these two stations is called base
line. The length of this base line is measured
very accurately.
SB
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xc
/
*b
Base line
a
b
8
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-- _________ _
3 . O rientation
It is the operation of keeping the plane table parallel to the position it occupi
station.
In such a case all the lines plotted will be parallel to the corresponding lines
g
If the position of board is different at successive stations, the relative positions
p
details will not remain same as the relative positions of the details on t e groun
As a result, the plotted work of the previous stations cannot be connected to t
he
successive stations.
During orientation the table is rotated and the plotted position of the insti ument stat’o
also
disturbed and shifts relative to the ground stations.
Therefore operations of orientation and centring are therefore interrelated. Oiientation of plane
table can be close using a trough compass, back sighting or by resection
M ethods o f Plane Table Orientatio
1. By Trough Compass
A trough compass is placed on the top right side corner of the plane table in such a way that
the magnetic needle points exactly towards the N —S direction.
•
•
•
Draw a line along the edge of the compass.
Shift and set up the plane table on the next station. Place the trough compass along the N
S line.
Rotate the table till the magnetic needle coincides with N—S line previously drawn.
N ota:
This method o f orientation arid cannot be employed at stations where local attraction is
suspected.
2 . By B ack Sighting
This is the most accurate method of orientation.
A
Traversing with a plane table
In this method plane table is set on a new station and the alidade is placed against the line
joining the new station with the preceding station and the table is rotated until the line of sight
bisects the previous station.
To achieve this, let the plane table is shifted from station A
to B and let the line ab has been
plotted with the plane table at A.
SH
Insllute for Engineers
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Plane Table
The portion of the plane
7 r [he 7
7
......
oc' u' " ed »>■ • *.
--------------------- - x M i e i i i .
* » , M ,h
.......S* * « . _________ (C
Vae of a plumbing fork is justified only if the plotting i* done at large
i.
* *cal«? an<* 1*1C rayH ^**mK
Whereas small-scale mapping which
is usually done with u plane table, the use of plumging fork is a sheer
7 pawing Sheet
The drawing aheet used should be of the beat quality to withstand
rubbing and scrubbing.
Plumb Doto
empormry A djugtnien
»• * " ,; Z e ° n t t h ‘
’Himbtngfo>*
B ° “ rd * hOU,d b'
•» t t . . V ertical A > l. of the
This can be achieved by placing a spirit level over the plane table
2.
The F u d icial E d g e o f th e Alidade should be a S traig h t L ine
This can be checked hv drawing a line along the ruling edge, reverse the ..lida.le and place it
against the end* of (he line.
Again draw a line which ahould coincide with the previous line
If the two line* do not coincide, the edge are corrected
3. The Two V an es shoul d he Pe r pe n di c u l a r to the Base of Alidade
Set (ho alidade on the corner edge of a huilding or on a suspended plumb bob
Set the alidade vane* along any of the above two case.
The plumb line and vane should coincide. If they do not coincide adjust the hinge* till tin* vane
coincides with the plumb line.
E etting up th e P lane TablP
k
Following operation* art* included in netting up of the Plane Table
C entering flt in the operation of bringing the plotted station point exactly over th«* ground station. A
plumbing fork in used for checking the centering
.fixuct centring i* i mportant for large-scale mapping
-----------------
--------------------------------------------------...tr in v of about 30 cm is permissible
or small-scale mapping, an error m e
>evelling *
.
.he operation of brtng.np the plane tabie in a honron.al pian.
**vel the board with the help of a spiritjeve
~ ~-v
Office* F-126 K jtv an i .<bnii NV« Itellu 110 01*
off Ka«iI i*» mmln+f9Kao«i »
Fhno» oil noi no* iK im m 07i mtm
W+*d*
I E.S MASTER
NWiih
mrswi
toi E
f<v*«
IE!
iTfcfSUi
CIVIL ENGINEERING
Tripod
An open frame type light tripod is usually used in the simplest form of plane tables, levelling
o the board is achieved with the tripod legs and checking the horizontality of the board with
the help of two spirit levels fixed at right angles to each other in a block of wood.
Alidade
An alidade is a straight-edge ruler having some
sighting device, it is used for sighting the objects
and drawing the lines.
Plain Alidade it is a straight-edge ruler about 450
mm long, made of a metal or wood. One of the
edges is bevelled and graduated. The alidade is
provided with a sight vane at each end. The sight
vanes have hinges at the lower end so that they
can be folded down on the ruler when not in use.
narrow slit and is used as an eye vane. The object
wire at its centre. The two sight vanes is open and
The two sight vanes provide a definite line a sight
The bevelled edge of the ruller is also known as the fiducial edge. The line of sight of the
alidade is in the same plane as that of the fiducial edge or in a plane parallel to it.
A fata:
N o n a days Telescopic A lid a d e are also much in use, in place of p la n e A lid a d e . When the
po in ts too high or low are to be sighted , the accuracy an d the range are considerably
in creased by p ro vid in g a telescopic alidade.
4 . T ro u g h C o m p ass
G enerally it is 15 cm long and is provided to plot the
m agn etic m eridian (N - S direction) to- fa cilitate
orientation of the plane table in the magnetic meridian
~
5
0 -j
5° '
/
_______________ ________________ ^
5
0
5'
T r o u g h com/><tss
At the extrem ities of the trough compass, there are
graduated scales with zero at the centre and m arking upto 5° on eith er side of the zero line.
Longer sides of the trough compass are parallel and plane such that they can be used as a ruler
for drawing the line or for placing the compass such that it coincides with a line already drawn
on the drawing sheet.
^ 5 . L e v e l T u b e /S p ir it L evel
•
The essential condition in plane table surveying is that the board should be level, This level
tube is eith er tubular or of the circular type.
•
It is placed on the board in two positions mutually at right angles and the bubble is centred
in each position to make the board horizontal.
^ 6'. P lu m b in g F o rk
A plum bing form is a U-shaped piece
table over the station.
One end of the frame is pointed and is kept over the drawing sheet touching the plotted position
of the instru m ent station while the other end of the frame carries a plumb bob
""""
l.fc.S M A S T ER
'
O ffice: 1-126, Katwaria Sarai, New Delhi - 110 016
irrsotute fa Engineers
edsite: www.iesmaster.org, E-mail: ies master@vahoo.co.in
[ES/GATE/PSUs----------- ---------------------------------------------------------------------------Phone:011-11013406, 7838813400 071 iHsagna '
r iyiL ENGINEERING
Plane Table
455
D isadvantages
X It is not suitable for, work in i
,•
' ’
,
f
c
climate and in a densely wooded country
2. The absence of measurements (field
3
a untiyto some different scale.
° ( S Causes ^convenience, if the survey is to be replotted
3
k
Plane table is heavy and akwa*A *■
• ,
t ■
c d t0 car.r y and the accessories are likely to be lost
It does not glve very accurate results.
A ccessor ie s used in Piane Table Surveying
®
s
<4)
T r° ush c° mpass <5)
s p iriti- '
Clinometer
U-frame
Trough
compass
Alidade
Drawing
sheet
Plane
table
Tripod
Spirit level
\ \
\>\\\
V
P lain table with accessories
J.. Plane T able B o ard
The drawing board is carefully made of well-seasoned wood so that counteract the effect of
warping and damages due to weathering.
The upper surface is kept smooth.
• The table at the centre of the underside, is attached to the tripod with the help of a screw and
wing nut. By means of the wing nut, the table can be clamped in any position.
Plane tables are available in the following different sizes.
Designation
Size (mm x mm)
B0
1500 x 1000
B.
1000 x 700
B,
700 x 500
B„
500 x 350
I.E.S M A S T E R
Institute ftx Engmeets
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Office: F-126. Katwaria Sarai. New Delhi - 110016
Website: www.iesniaster.org E-mail: ies_mastei$yahoo.co in
Phone: 011-41013406. 7838813406. 9711853908
Plane Table
Introductioi
The plane table is an instrument used for surveying by a graphiealjnethod in
.
r
work aqd plotting are done mmulLanequsly.
The main feature of plane tabling is that the topographic featuies to be mapped
view. Hence no chance of missing of any important detail.
It is suitable for small and medium scale-mapping ( 1: 10,000 to Jj2^5Q ^0qp)^ h ere the gi eat
accuracy is not required. It is also used for plotting the topogiap ica map.
Before commencing a plane table survey, the instrument stations aie fixed to co\u the entire
area.
These stations may be fixed by surveying a trigonometrical framework, establishing a network
of control points on a pattern to suit the scale at which plane tabling is carried out.
The elevations of these points are determined with the help of Levelling.
A surveyor starts filling in details from any of these control points, one by one. and traverses
all the control points.
The finished maps so produced are known as top og rap h ic m aps.
This graphical method of producing topographic maps is known as c a r to g r a p h ic surveying.
It should be noted that all the measurements made are plotted directly on the drawing sheet
instead of recording in the field book.
The principle used in plane table surveying is that an unknown point of interest can be
established by measuring its directions from known points.
A d van tages
simultaneously. Therefore there i , no risk of omittmg
The error and mistakes in plotting can be checked by drawing the check lines
C3r Irregular objects can also be plotted accurately as the lay of land is in view
It is most rapid and useful for filling in details.
No great skill is required.
6.
It is less costly in comparision to theodolite survey.
7.
It is very advantageous in areas, where compass survey is not mKoki
magnetic fields.
* ‘
ehaMe e*
area ^ t e d *
c,wn ENGINEERING
*53
^
The method oi com puting areas by subdividing a plot into triangles is suitable for
ja) work ot small nature
(b) work of big nature
(C) road work
&
(d)
The method su itable for computing th * .
from a straig h t line is
a ,e a when the boundary hnes departs considerably
(a) Mid-ordinate rule
(c) Trapezoidal rule
9.
canal work
(b)
Average ordinate ride
(d)
Simpsons rule
The area o f an irre g u la r nlnM„a
of a
gUre can be accurately obtained with the help
(a) Pentagraph
(b)
paraUax bar
(C) Plamme‘er
(d)
any of the above
10. If n is the number of sides and L the length n f
r
,
,
gtn or the sides of a regular polygon, its area is
nr2 .a.. 180c
1ftn°
__
n
n. n o
2 ISO0
(a) —L" cotJ —
0}) —L sec ----n
4
n
n . •>
■>180°
n-I.T2,tan 2-----180(c) —L" cosec"-----(d)
4
n
4
n
11. Simpson s rule for calculating area is applicable only when the ordinates are
(a) odd
(b)
even
(c) either (a) or (b)
(d)
hone
12. Volume of earthw ork can be calculated by
(a) Mid-ordinates
(b)
Average ordinates
(c) Prismoidal rule
(d)
Hund’s rule
^ O b je c t iv e Q u estio n s
A n sw e rs
1.
(b)
2.
(c)
3.
(c)
4.
(c)
5.
(a)
6.
(b)
7.
(b)
8.
(c)
9.
(c)
10.
(a)
11.
(a)
12.
(c)
I.E.S M A S T E R
CIVIL ENGINEERING
452
O bjectiv e Q u estion s
1.
Which one of the following methods estimates best the area of an irregular and cur
(a) Trapezoidal method
(b)
Simpson’s method
(c) Average ordinate method
(d)
Mid-ordinate method
What is the volume of a 6 m deep tank having rectangular shaped top 6 m
4 m x 2 m (computed through the use of prismoidal formula).
2.
(a) 96 nr*
(b) 94 m *
(c)
(d) 90 m*
92 m*
4ma
ndarv.
bottom
Which one of the following methods of computing area assumes that the short lengths of the
boundary between the ordinates are parabolic arcs?
(a) Average ordinate rule
(b) Middle ordinate rule
3.
(c) Simpson’s rule
(d) Trapezoidal
rule
In the given formula formats, L is the length of the base line split into n equal segments
each of length *d\ (),, 0 .......... . On. , are the ordinates at the sequential ends of the segments
and M,. M0......... Mu are the mid-ordinates of successive segments. Which of the following
pairs of rules and the formulae for computation of the area standing on the base line are
correctly matched?
4.
1 . Mid-ordinate Rule
A=
O, + Og +... + On
xL
n
2.
Average ordinate R ule............... A - —[M, + M2 f... + Mn|
n
3.
Trapezoidal R u le................A - d
4.
Simpson’s Rule ................ A = |[(O, + On+1) + 4 (0 , + O ,... + On) + 2 (0 :i + 0 5 +... + On_,)]
( 0 i.._Qa±l' + ()., + O.j +... + On
2
Select the correct answer using the codes given below:
(a)
1 and 2
(b) 1 and 3
(c)
3 and 4
(d) 2 and 4
If the cross-section areas of an embankment at 30 m intervals are 20, 40, 60, 50 and 30 m2
resp ectively, then the volume of the embankment on the basis of prismoidal rule, is
5.
(a)
5300 m3
(b) 8300 m3
(c)
9300 m3
(d) 9400 m3
Excavation is to be made for a reservoir measuring 20 m long, 12 m wide at the bottom and
2 m deep. The side slopes aie to be 1. 1 and the top to be flush with the ground which is level
in the vicinity. As per piismoidal formula, the volume of excavation will be
(b) 618.66 m3
(a) 610.33 m3
6.
(c) 625.00 m:!
T T m astef
Insltute for Engineers
IE S /G A TE /P S U s
I
S
(d)
633.66 m3
Office: F-126, Katwaria Sarai, New Delhi - 1 10 016
Websjte: www.ieamaster.org. E-mail: ies_m aster^yahoo.co.m
- hone: 011-41013t0(>, 7838813-tOfi. 9711853908
ENGINEERING
Measurement of Area & Volume
= 9° + (4 5 6 .6 7 -3 2 8 .3 3 )
x(361 -3 2 8 .3 3 )
= 90 + 1 273
= 91.273 m
Rule calculate the volume of a 5 m deep pit whose top and bottom dimens
irrespectively 10 m x 20 m and 20 m x 40 m.
P r is m o id a l
.
in prismoidal Rule ordinates required should be even but here ordinates pio l
Sol. bm
t^e average of the sides and then calculate the area.
even so
A = 10 x 20 = 200 m2
A, = 20 x 40 = 800 m2
y__ ~[A t + 4 Am + A2]
3
£ ^ [2 0 0 + 4 x 4 5 0 + 800]
v _ £ ^ x 2800
v
3
V = 2 3 3 3 .3 3 nv
%
|
450
CIVIL ENGINEERING
Surveying
Example 11.
In a proposed reservoir, the areas containing within the
contours are:
Using the method of Prismoidal rule and End area formula,
calculate
(i) capacity of the reservoir when it is full at 100 m level:
(ii) elevation of the water level when it is 60% full.
Ignore the volume below RL 65 m.
S o l.
V= 5
32 + 2 + 26 + 24+18 + 15 + 13 + 7
Contour (in m) Area (in ha)
32
100
26
95
24
90
18
85
15
80
^
13
75
7
70
2
65
2
V= 600ha - m
6
V= 3
+ 4(7 + 15 + 24) + 2(13 + 18) +( —
]x 5
V= 456.67 + 145 = 601.67 ha - m
Capacity of Reservoir when it is 60% full
V=
601.67
100
x 60 = 361.002 ha - m
Capacity of reservoir upto R.L of 95 m
= 3 ( ^ ~ ^ J + 4 *7 + 15 + 2', , + 2*13 + 18)
= | (l4+ 184 + 62) = 456.67
•J
Capacity of Reservoir upto R.L. of 90 m
I.E.S M A S TE R
insttue la Ergtncera
IFS/GATE/PSUs
Office: P-126. Katwaria Sarai, New Delhi •1 10 016
Website: www.iesnm8 ier.org. E-mail ies_ma8terrtynhooc
Phone: 011-11013100. 7838813106. 9711853908
ENGINEERING----------- M easurem ent of Area & volume
449
ohoVfi the e xisting ground levels on a 15 m square grid for a plot of land w hich is to
figuV° ^.aled to a uniform formation level of 100.00. Calculate the volume of earth require
H
8
i a
l
Sol.
Number of times a particular corner height is used in various squares is marked in the circles.
I h , = 2.40 + 1.20 + 1.80 + 1.40 = 6.80
Eh., = 2.10 + 1.70 + 2.30 + 1.30 + 1.90 + 1.60 = 10.90
Sh;i = 0
Eh, = 1.50 + 2.70 = 4.20
V = — (Eh, +2Eh. + 3Eh, + 4Eh,)
4
(15x15) (6.80 + 2x10.90 + 0 + 1 x 1.20)
CIVIL ENGINEERING
Surveying
448
E xam p le 8.
Figure shows the three cross-sections of an embankment at an intreval of 30.0 m. Calculate
the volume between the end sections by (a) Trapezoidal rule, and (b) Prismoidal formula.
fHlnv-H
SSCT10N-1
SEC TION-2
Sol.
m
—-
A
A, = 2.0 (11 + 2 x 2) = 30.0 m2
A2 = 3.5 (11 + 2 x 3.5) = 63.0 m2
A3 = 5.0 (11 + 2 x 5.0) = 105.0 m2
Trapezoidal rule V = D
V = 30
'A , + A
—+ A„ + A3 +...
\
(30.0 + 105.0)
+ 63.0 = 3915m:i
Prismoidal formula
D
V = —[(A, + An) + 4(A3 + A., +...) + 2(A3 + A5 + ...)]
o
= — [(30.0 + 105.0) + 4 X 63.0] = 3870m3
3
E.S M ASTER
Institute for Engtieens
lES/GATE/PSUs
Office: F-126, Katwaria Sarai. New Delhi - 110 016
Website: www.iesniaster.org, E-mail: ies.master^yahoo co in
J-honc 01M1Q1340C. 7838813406. 9711853908
___.
c iv il e n g in e e r in g
Measurement of Area & Volume
447
CIVIL ENGINEERING
446
S o l. Calculated using any one of the methods
Let the independent coordinates of A be (0, 0). The coordinates of other poii
as below.
Line
Latitude
c t - n f in n T n r lA n p n d e n t C / o o rd ifirtv y -
Departure
y
X
_________ AB
A(l)
+154.8
-157.2
B (2)
_________ CD
|
-109.0
-228.7
( 1)
(2 )
x
0
-157.2
y
*0
+154.8 "
'
.i r i O
+ 1 5 4 .0
. a r* o
+20/.
3
r\
C(3)
+53.3
D(4)
1 +228.7
+109.0
A(5)
0.0
n. un
u
-98.3
+175.4
DA
-157.2
52.5
210.5
_________ BC
r\
o .u
0.0
+ 207.3'
We know that 13.1G a. 2A = I P - IQ
2A = [0 x 154.8 - 157.2 x 207.3 + 53.3 x 109.0 + 228.7 x 0.0]
- [0 x -157.2 + 154.8 x 53.3 + 207.3 x 228.7 + 109.0 x 0.0]
2A = - (32587.65 - 5809.7 + 8250.S4 + 47409.51]
= 41219.105 m-
E x a m p le 6 .
ad eembankment
m o a r iK m e iiL
a m ».ide at the formation level. The centre line of the embankment
A road
is 11
mw
m above the ground surface. If the ground slope is 1 in 22 at angles to the centre line,
is 3 m
the side slopes are 2 : 1 , calculate the side widths and the area of cross-section by various
and
formulae.
S o l.
I-
I~ '
B
n
n
m
w-
GS
In above figure,
- 5'5m’ m
\
w«
= 22, n = 2 and h = 3m.
m
Office: F-126. Katwana Sarai. New Delhi * 110 016
I.E.S
MASTER
InsaluteforEnQficcrs
lES/CATE/PSUs
Website: www.iesmaster org E-mail: ies_master@yahoo.co.»n
Phone: Oil- 11013406. 7838813406. 9711853908
444
Surveying
CIVIL ENGINEERING
Example 2 .
The following perpendicular offsets were taken from a chain line to an irregu lar boundary
Chainage
0
8
20
35
60m
47
Offsets
14.5
24.5
30.8 ----------27.4 ».......
28.4
--------------------... *—18.4m
—
— i
determine the area between the chain line, the boundary and the end offsets.
Sol.
8m, d2 - 12 m, d3 = 15m, d, = 1 2 m and d5 = 13m.
Gi\en, dj
Ai'ea = |(14.5 + 24.5) + ^ ( 2 4 .5 + 30.8) + ^ ( 3 0 .8 + 27.4)
2
2
^ ( 2 7 .4 + 28.4) + — (28.4 + 18.4)
2
2
= 156 + 331.8 + 436.5 + 334.S + 304.2 = 1563.3 m2
Example 3.
Determ ine the area of the traverse shown in figure below using the meridian distance method.
] 00 m
45" * 1
Reference'
100m
Parallel
100m
Sol.
100m
M eridian distances of the various lines are determined as below,
Line AB,
m ] = 50 m
Line BC,
m., = 50 + 50 + 50 = 150. m
Line CD,
m3 = 150 + 50 - 50 = 150 m
Line DA,
m 4 = 150 - 50 - 50 = 50 m
Line
AB
BC
1
CD
DA
M.D.
(m)
Latitude
<L)
P ro d u ct = m x L
50
-10 0
-5 ,0 0 0
150
+100
150
+100
50
-10 0
*
+15,000
+15.000
-5 .0 0 0
“ = + 20.000 m3
Area =
E.S MASTER
Insfilufe for Engineers
•lES/GATE/PSUs
20,000 m2
Office: F -126 . Katw aria Sarai New Delhi - 1 1 0 0 1 G
Website: www.iesm asier.org. E-m ail ies_mastef^yahooco.*n
Phune: 0 1 1 - 1 1 0 1 3 1 0 0 . 783881310G 97116 539 0 8
444
Surveying
CIVIL E N G IN E E R S
Exam ple 2.
he following
perpendicular offsets were taken from a chain line to an irregular bound
i—-------------ary.
Sol.
Civen, d, - 8m, d, - 12m, d:1 = 15m. d, = 12m and d5 = 13m.
Area = §(14.5 + 24.5) + ^ (24.5 + 30.8) + ^ (3 0 .8 + 27.4)
2
2
12
IS
+ ^ (27.4 + 28.4) + ^-(28.4 + 18.4)
2
2
156 + 331.8 + 436.5 + 334.8 + 304.2 = 1563.3 m2
Exam ple 3.
CIVIL ENGINEERING
443
Measurement of Area & Volume
Volume from Spot
• T his method of estimating the earthwork quantities is also known as the unit a rea method.
•
In this method the area is divided into regular figures such as squares, rectangles or triangles
and the levels of corners the figures are measured before and after the construction.
•
If the depth of the excavation are op 6 ,f c, and d v respectively, and A is the area of the figure
obcciy then the volume is given by
y = °\
+ c/,
4
A
Exam ple 1.
The following perpendicular offsets were taken at 5 m intervals from a traverse line to an
irrregular boundary line 2.10; 3.15, 4.50; 3.60; 4.58; 7.85; 6.45; 4.65; 3.14 m.
Compute the area enclosed between the traverse line and the irregular boundary from the first
to the last offset. Use
(a) Average ordinate rule; (b) Trapezoidal rule
(c) Simpson’s rule
Sol. (a)
Given
A
(O, + 0 2 + —+ On.i)
n +1
L = 40m, n = 8.
a - — 12 10 + 3 15 + 4.50 + 3.60 + 4.58 + 7.85 + 6.45 + 4.65 + 3.14]
A- y
= 177.87 m2
(b)
A = d Qi + Qnq. + O + o 3+.... + o n
2
= 5
2 ,1 0
+ 3.14 + 3 15 + 4 50 + 3 eo + 4.58 + 7.85 + 6.45 + 4.65
2
= 187.0 nv
Sim pson s r
as the number of offsets is odd, the Simpson’s rule can be applied,
(c) There are 9 onsets.
_ d [(2 l0 + 3.14) + 4(3.15 + 3.60 + 7.85 + 4.65) + 2(4.50 + 4.58 + 6.45)]
.
5rg 2 4 + 7 7 . 0 0 + 31-06] = 188.83 m2.
A" 3
____________
Office: F-126. Katwaria
Website.
.*
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D e lh i- H 0 « E
jeg mafiter@yahooco.in
S u rve yin g
442
CIVIL ENGINEERING
Area in excavation
wh.
A= 2
w
A = 72 (n - s)
+nh
A = k2____ L
2 (n -s)
Centre line in embankment
w, = b - w
f b + h \( ns
hl = U r h Jl n - s ,
b
dl =
ns ( A _ h
n -si ,2n
2
Area in embankment.
-nh
. _ w.hi _
A
> "
2
2 (n -
v
s)
End A rea Form ula
This is also called as trapezoidal form ula.
V = A l A + A, + A, +... + A„_,
where A,. A,, A.,......A„ are the end consecutive areas and L is the distance between them
P r i s m o i d a l Form u la
This is also called as Simpson's rale for volumes.
V = —[(A, + A„) + 4( A, + A, +... A,,.,) + 2(A2 + A5 +... + A„_2)]
3
Note :
__•______________________________________________
It is necesary to have an od d number o f cross sections to use th is form u la.
In case o f even num ber oj cross section s, the end strip is treated sep a ra tely a n d the volum
o f the rem aining strips is detei m ined by the p rism oid al form u la a n d the volum e o f the la
strip is calcu lated eith er by the end area or P rism oidal rule.
In case i f only P rism oidal rule is to be a p p lied , the a rea h a lf ivay betw een the sections
in terpolated by averagin g the dim ensions o f the en d section s a n d not by av erag in g the e
areas.
_______________________
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Phone Oil - 1 1 01 3uw 7838 Hl;U0 ti U?11853908
CIVIL ENGINEERING
Measurement of Area & Volume
s h ,= (h - h,) n - ^
n
hl - s + n n| h - A
2n
n
hi - s + n h 2n
u __
From equation (1 )
d=
d = ^ + -£ il ' h - i )
2 n +s
2n )
n ,
b
h+—
h, =
n- s
2n
• b
sn
b '
h+—
d,i = ~ +
'l n - s
2n )
Substituting the values of d,, h, and h., in equation ( 1 )
b)
A=
3.
2,
+ n2(bh + sh‘ )
( i r - s 2)
T h ree-level S ection
b V ns
d = l h + £ Kn + s )
f n,s
d.= (h+ 5 jvni -
s
Area ABODE = Area ACF + Area FCD + Area FDE + Area FEB
= - —h. +hd + hd. + —h„
2 L2 1
1 2 A = £ (h I +hs) + | (d + dI)
44
440
Surveying
CIVIL ENGINEERING
P la n im eter a r e u sed to fin d out a r e a o f irreg u la r s h a p es.
It is a mechanical integrator used for the measurement of areas of figures, plotted to a scale.
M e a su rem e n t o f V olu m es
•
The com putation of volumes of various quantities from the measurements done in the field is
required in p la n n in g and design of many engineering works.
Volume of E arthw ork
Selection of suitable alignment of road, canal and sewers. In estimation
of pavement materials.
Volume o f Regervoior — Design and planning of reservoirs.
•
'Fhe direct computation of the volume from the measurement of length, w id th and depth is not
fesible for large engineering works.
•
In such a case computation of the volume o f earthw ork is generally done after com puting the
areas o f various cross sections.
•
Spot levels o f the ground are taken to estimate the volume of the earthw ork.
•
The estim ation of the volume of w ater in a reservoir, the contour map is generally used.
M e a su rem e n t o f A rea o f C ro ss-sectio n
For estim ation of the volume of earthwork, cross-sections are taken at rig h t angles to a fixed
line (generally, the centre line) which runs lon g itu d in a lly through the earthw ork. Spacing of
the cross-sections w ill depend upon accuracy required and the character o f the ground.
Follow ing type of cross-section generally occur in field.
1. Level section
2. Two-level section
3. Sidehiil two-level section
4. Three level section
1. L ev el S e c tio n
A = (b + sh) h
2 . T w o -le v e l S e c t i o n
Area ABED C = Area ACF + Area CDF + Area DEF + Area FEB
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Measurement of Area & Volume
439
4 . Sim pson’s One Third Rule
•
If the irregular boundaries are curved Simnenn'o „,i„ ■
calculate the area of the grven tract. '
'
.
According to this rule the short length of boundaries between the two, adjacent ordinates is a
parabolic arch
Area of trapazoid ABDC =
r
.
,
., ,
OVer the tra|,az0,dal ,rule “
* 2d
2
2
*
Area of segment CIDHC = - x Area of Parallelogram CDP"E
3
2
2
= -x(ABxHI) = "
(X -
Oj + o,
~2~
x 2d
= y [ > 0 , - ( 0 , + 0 3)]
Area o f first two divisions
A = ^ ^ x 2d + y ( 20._,-(0 1 + 0 3)]
A = ^ [3 0 ,+ 3 0 ,+ 4 0 ,-2 0 ,-2 0 ,]
A = | [0 ,+ 4 0 ,+ 0 3]
Similarly, area of next two divisions
A - ^ (03 + 4 0 ,+ 0 5)
Area of last two divisions
d
A = |(0 „-2 + 4 0 n-.+°n)
Adding up
i-eaA
J + 40s +20i +404+... + 20n_3+40n.1+ 0n]
Total area
^ =
~ -[(0
3
A
_ -[(O, + On) + 4(0, +0, +... + 0,, ,) + 2(Q, +0. +... +0,,-u)]
•
lm;<
=innlicable only when there are odd number of offsets (even no. of segments).
Simpson •s ruie
u
en number of offsets then also Simpson’s rule is applicable but in such a case, the
’
to t^ a re T of tnsct is divided into two parts.
0)
Area up to the secon
m
Simpson
s 1 ul
determined from the trapazoidal rule.
Area of the last segment is aei
(2)
A iea 01
a
lout elements i.e. it has odd number of offsets. This area is calculated from
m 0f both the areas.
Total area is obtained as a sum
AIota :
a r e a h as to be cau clu ated
Two
If in th e q u estio n it is
om
ffsets
e l a re evenaree emn n u m ber,
in
o/offsets become odd in
^
-------------~
^
r
t
r
r
n
^
r
T
T
^
offsets
an
d
h
en
ce---a d ja cen t offsets
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A = R-
ttot sin a
360“ ~2~
Area of a tract with approximate methods
1. M id-ordinate Rule
•
If offsets /i,, hv
h n are measured at the mid-point of each division.
Area = average ordinate x length of base
_ fl\+ ,h + —+ h„ ^
n
/ii + /i2 + ... + h
j
n
= (/?! + h 2 + ... + h n) d
where n = number of divisions.
2 . Average Ordinate Rule
•
II offsets Op 0.r ..., On are measured at the end of each division and are spaced apart at equal
distances d.
Area = average ordinate x length of base
_ °1 + 0 2 + -- +Q,. n .. L
n +\
_
+ —+ On ncj ndl.0
n +l
77+1
3 . Trapezoidal Rule
.
•
The trapazoidal rule is more accurate than the mid-ordinate rule and average ordinate rule.
The end ordinate are considered even though they are equal to 0
O, + O., .
Area of first trapezoid = — - —~d
L*
O.} + o , ,
Area of second trapezoid = —^ —- a
0 ;,_i + o n ,
=
2
”
Area of last trapezoid
By adding up
o, + o9 o 2 +o 3 ,
A= d
2
2
A = d
, o„_, + 0„
2
/
+o , + o j + ... + o „ .1
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Measurement o f Area & Volume
437
(3) T r ia n g le
A = ba«« * half of the perpend,cular height.
Huight
------------------------------------------------ ----(4) P a r a l l e l o g r a m
Bam
A = base x perpendicular height.
•----Height
/
i f l l
(5) Trapezoid
A - h a lf o f the sum o f p a ra lle l sides x p e rpe n d icu la r height.
A
(6) T r a p e z i u m
(7) R e g u l a r polygon
A = le n g th o f p e rim e te r *
n
A
h a lf o f the perpendicular distance from the centre o f sides.
n
lique tr i a n g l e
;re a. 6 and c are th e atdea. and a = ^
. _J55l«o-ix2R“nt xRc“ I
one side.
13
Measurement of Area & Volume
Introduction
One of the objective of many of the survey is to obtain quantities such as ai eas and volumes.
M easurement of Area
The method selected for computation of area depends upon the shape of the tract and the
accuracy desired. Foremost reasons for making land surveys is for the determination of aiea.
Prevalent methods of measurement of areas are:
•
•
1. By Field Measurements
2. By Plan Measurements
If the plan is enclosed by straight lines, it can be divided into geometrical figures e.g.:Triangle. Rectangle. Square etc. The area of these figures can be determined by using appropriate
form ulas.
But. if the boundaries are irregular, then approximate methods are being used.
N o t* :
P lan im eter is used to determ ine area o f irregular shape.
Com putation o f Area of Geometrical Figures
(1) R ecta n g le
T
width
i
H--------- length
A ^ length x width
H
(2) S q u a re
1
length
|
«— length H
A = length x length
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Field Astronomy
13. Which one of the following
(a) Astronomical survey
x » •i
*
(c) Aerial survey
435
18 ' equired ,n observations of stars?
(b) Cadastral survey
(d) Photogrammetric survey
14. If the equation of time is —13'28
tu
u n
hour Greenwich Mean Time ~
inen the Greenwich Apparent Time corresponding to zero
(a) 13’ 28.5"
00 3 day 1S
,
(b) 46’ 31 5"
(c) 23h 46’ 31.5" of same dav
(d) 23h 46’ 31.5" of previous day
15. Consider the following statements:
A ssertion (A): In a spheri 1
circumference of the great circle113ng e’ l *le 8um
sides is always less than the
Reason (R): The sum of the tv
angles, but less than six right angles^ ^ ^ 3 Spherical trian^le is greater than two right
Of these statements
(a) both A and R are true and R is the
correct explanation of A
(b) both A and R are true but R is not
a correct explanation of A
(c) A is true but R is false
(d)
A is false but R is true
16. At a given place of observation, the declination of a circumpolar star is
(a) greater than the latitude
(b) equal to the latitude
(c) less than the co-latitude
(d) greater than the co-latitude
17. The process of determining the location of the station (on the map) occupied by the plane table
is called as
(a) intersection
(b)
three-point problem
(c)
(d)
resection
traversing
18. Which one of the following is the angular distance between the observer s meridian and the
vertical circle passing through a star measured along the celestial hoiizon?
(a)
Right ascension
(b)
Azimuth
(c)
Declination
(d)
Hour angle
Answers
ctiv e Questions h
I.
6.
(b)
3.
(d)
4.
(b)
5.
(b)
(c)
2.
(d)
8.
(c)
9.
(a)
10.
(b)
7.
(b)
(a)
14.
(d)
15.
(b)
(b)
II.
(b)
12.
(a)
13.
16.
(d)
17.
(d)
18.
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1
5.
e n g in e e r ing
How many sidereal days are there in a solar year/
(a) 365.2840
(c) 360.2500
6.
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Surveying
434
'
(b)
366.2422
(d) 365.0000
Consider the following statements:
A sidereal year can be defined as the time interval
1.
between two successive transits of the sun through the meridan ol any of the fixed stars
2.
between two successive vernal equinoxes
3.
between two successive passages of the sun through perigee
Which of these statements is/are correct?
7.
8.
9.
10.
(a) 3 only
(b) 1 and 2
(c) 2 and 3
(d) 1 only
The declination of a celestial body is the arc of the declination circle intercepted between that
body and the
(a) prime vertical through that body (b)
azimuth of the body
(c) equinoxes of the Earth
equator of the Earth
(d)
The difference between the apparent solar time and mean solar time is know n as
(a)
real time
(b) average time
(c)
equation of time
(d) sidereal time
The standard time m eridian in India is 82°30c E. I f the standard tim e a t any in sta n t is 20
hours 10 m inutes, the local mean time for the place at a longitude o f 20°E would he
(a)
4 h PM
(b) 4 hlO m PM
(c)
1 h 20 m PM
(d) 0 h 20 m PM
W hich one of the follow ing methods would give accurate results in d e te rm in in g the direction
o f the observer's meridian?
(a) Observation of circumpolar stars on the same vertical
(b) Observation of circumpolar stars at culmination
(c) Extra-m eridian observation of a circumpolar star
(d) Observation of the Sun at equal altitudes
11 .
G iven th a t 6 denotes declination, 0 the latitude o f the place o f observation and a the a ltitu de
o f a s ta r at the prim e ve rtica l, then
(a) sinO = sino cosa
(b)
sin0 = sin5 coseca
(c) cosa = cosS siii0
(d)
sin8 = sina cosO
12. The declination of a star is 2 1°15'N at a latitude of 43°30'N Th»
i
upper
culmination is
* * ZCmth d,8tance a * the
S
3
(a) 22° 15'
(b)
21°15’
(c) 6I°45'
(d)
43°30'
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O b je c tiv e Q u e stio n s
W hich one o f th e fo llo w in g represents a circu m p o la r star?
(a) Upper culm ination above horizon, lower culm ination below horizon
(b) Both upper and low e r culm inations below horizon
(c) Both upper and low er culm inations above horizon
(d) A ltitu d e a t upper culm ination is m inim um
M a tch L is t - I (T e rm ) w ith L i s t - I I (D e fin itio n ) and select the co rre ct answ er u sin g the code
given below th e lists:
L is t-I
A. Apparent solar day
B. Sidereal day
C. Tropical year
D. Sidereal year
L ist-I 1
1.
The tim e in te rv a l between two successive upper transits o f the firs t point of A ries over the
same m eridian
2.
The interval between t wo successive lower transits of the centre of the Sun across the same meridian
3.
Tim e in te rv a l between two successive passages of the Sun over the m eridian o f any one of the
fixed stars
4. Tim e in te rv a l between two successive vernal equinoxes
Codes:
3.
A
B
C
D
(a)
1
2
3
4
(b)
2
1
4
3
(c)
2
1
3
4
(d)
1
2
1
3
Flam steed gave num bers to stars observed by him in each co n stellation according to th e ir
(a) B rilliance
(b) Altitudes
(c) Co-declinations
4.
(d)
Right ascensions
W h ic h o f the fo llo w in g coordinate systems is the most convenient r i o f th e s ta r on ce le stia l sphere?
' ° 9Pe cify the position
(a) L a titu d e and longitude
(b) A ltitu d e and azim uth
(c) D eclination and rig h t ascension
(d) D eclination and hour angle
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CIVIL
e n g in e e r ing
D e te rm in a tio n o f LMT F ro m L ST
1.
Compute the L M T of the tra n s it from the GMT
2.
C onvert the given LST to the mean tim e inte rval.
3.
Now L M T = L M T o f tra n sit *■ mean tim e in te rv a l
E x a m p le 3 1 .
D eterm ine the L M T at an in s ta n t a t a place in longitude 7 5 'E corresponding to the LST of
lOh 25m 15s. The GM T o f tra n s it o f V on the same day is bh lK in zu..
Sol.
75
Longitude 75° = — h r = 5f»r Om Os
15
L M T o f tra n s it o f V = GM T tra n s it of V + 9.8296s x lo n g itu d e h o u r
= 6h 18m 20s + 9.8296 x 5s
= 6h 18m 20s + 49.15 s
= 6h 19m 9.15 s
LST = sidereal tim e in te rv a l = lOh 25m 15s
= 10.120833 hr
Retardation = 9.S296 x 10.120833 s
= 102.13 s
Mean tim e in te rv a l = lOh 25m 15s - 102.-13 s
= lOh 23m 32.57s
LM T = L M T o f tra n s it + mean hours
= 6h 18ra 20s + lOh 23m 32.57 s
= 16h 11 m 52.57 s
C o m p u ta tio n o f GM T o f T ra n s it o f V e rn a l E q u in o x F r o m G S T a t GMM
The given Greenwich sidereal tim e (GST) at G M M (i.e. Oh G M T) shows the n u m be r of sidereal
hours w hich have elapsed a fte r the tra n s it o f V. The next tr a n s it o f Y
N ext tra n s it o f V w ill take place 24 sidereal hours a fte r the p re v io u s tr a n s it Hence the
sidereal tim e in te rv a l a fte r G M M o f the next tra n s it can be found out. The corresponding mean
tim e in te rv a l is found by applying the retardation.
The required G M T is equal to the computed mean tim e in te rv a l as G M M is zero.