"'

290

Disadvantages
not recorded, it is a great inconvenienc.:e if
SURVEYING

@]
(I) Since ootes of measurements are
duced to some different scale.
the map is required to be repro
(2) The plane tabling is not in!en
ded for very accurate work .
. (3) It is essentially a ttopical
instrumen!.
n and in wet climate.
Calculation of Area
It is most inconvenien! in rainy seaso
(4)
nt to transport,
(5) Due to heavyness, iE is inconvenie
es, there is every likelihood of these being
(6) Since there are. so many accessori
lost. 12.1. GENERAL the tract
PROBLEMS surveying is to determine the area of
One of the primary objects of land land in plane surve ying
of earthwork. The area of
vantages of plane table surveying over other
methods. surveyed and to determine the quantities urem ents of area
I. (a). Discuss lhe advantages and disad ontal plane. The units of meas
ng a point by plane_ table survey. means the area as projected on a horiz units are sq. metr es or
Explain with sketcbfs. the follow ing methods of locati while in metric units, the
(b) of tbe foUowing methods : in English units are sq. ft or acres,
Also discuss the relative merits and appli cation the relation between the two systems.
(ii) Inter section hectares. The following table gives
(l) Radiatioo. VALI!NTS
(A.M.I.E.) RE MEASURE WITH METRIC EQUI
(iii) Resection.
.. ·; TABLE 12.1. BRITISH VNITS OF SQUA
us iccessories of a' pJane table. e yards Square Jeer Square links Merrie j
2. Describe briefly the use of vario Sq. mih AtrtS Square clwim Sq. polts OT· Squar Equiva/e,us
us methods of orienting lhe plane table.
1

Pen:hes
3. Diseuss with sl:etcbes, the vario 258.99 ha
I

be carried out at a scale of I : 5000

. Show that at 4'
640 6,400 102.400 3.097.000 - -
4. (a) A plane rable survey is to y station is DOt neces sary. What error I
43,560 100,000 0.40467 ha I
table over dle surve 4.840
this scale, accurate centring of the plane point is 45 em out of the vertical lhrougb the station? I 10 160
if lhe
would be caused in position on a map
2
16 484 4,356 10.000 404.67 m
g paper method. I
(b) Define three-point problem and
show how it may be solved by tracin 625 2
I 30.25 272.25 25.29 m
hes, Lehmann's Rules.
5. Describe, with lhe help of sketc 20.7 1
1 9 0.836 m
is it solved ?
6. What is two-point problem ? How 2
gle of 2.3
solved by m Bessel's method (i1) Trian 1 929 m
7. Wbat is three-point problem ? How is it 2
error melhod. I 404.67 cm
eliminated ?
errors in plane tabling ? How are they
8. What are the different sources of
ting plane table by backsighting. is the Acre.
9. (a) Describe lhe inethod of orien Note. The sWldard of square measure
rable surveying.
'intersection' methods as applied to plane SQUARE MEASURE W1T1I BRITISH -
EQUIVALI!NTS
(b) Distinguish between 'resection' and iency? TABLE 12.1 (a) METRIC VNITS OF -
chain surveying in point of accur.icy and exped
(c) How does plane table survey compare with Sq. tentinutres British
AT< (a) Sq. marts
(A.M .J.E.) SqUllrt kilometre H<ctam Equiva/e~ts
(/r4) (m'J (<m'l
vantages of plane !able surveying with those
of chain (tm'J
10. (a) Compare tha advantages aod disad 100 10.000 1.000.000 - 0.3861 sq. mile
surveying. I I
trial method giving 100 10,000 - 2.4710 acres
(b) State three-point problem in
plane tabling aod describe its solution by (A.M. I.E.).
I
the position of lhe point SOUght. 1,000.000 1076.4 sq. ft.
the rules which you will follow in esrimaling I 100 I
1 10.000 10.764 sq. ft. I

I 0.155 sq. ft. J

ure is the Are.

Note. The standard of square meas (291)
r' 293
292 SURVEYJNG CALCULATION OF AREA

12.2. GENERAL METHODS OF DETERMINING AREAS (z) Mid-ordinate rule ; (ii) Average ordinate rule
(iiz) Trapezoidal rule ; (iv) Simpson's one-third rule.
The following are the general methods of calculating areas:
(1) MiD-ORDINATE RULE (Fig 12.1)
1. By computaJions based directly on field measurements ~
The method is used with the
These include :
assumption that the boundaries be-
(a) By dividing the area into a number of triangles
tween the extrentities of the or!iinates
(b) By offsets to base line
o,
~
(or offsets) are straigltt lines. The o, ,o,
(c) By latitudes and departures :
base line is divided into a number
lo,
(i) By double meridian distance (D.M.D. method)
(il) By double parallel distance (D.P.D. method)
of divisions and the ordinates . are i
measured at the ntid-points ~f each ~
1 '\t '
·.· _ ~ •
I·'
d->1
r 4 n
8
(d) By co-ordinates.
division, as illustrated in Fig: 12.2. ------~ L=nd
2. By computaJion based.· ~n ''fneosurements scakd from a map.
' . The area is calculated by the AG. 12.2
3. By meclwnical method : Usually by means of a planimeter.
formula
12.3. AREAS COMPUTED BY SUB-DIVlSlON lNTO TRIANGLES Area = /!. = Average ordinate x Length of base
In this method, the area is divided into a D J• ~ + Oz + O, + .... +On) d = d EO
= 0,+0,+0,+ ...... +On "-"'-\0, ... (12.3)
number of triangles, and the area of each triangle n .
is calculated. The total area of the tract will then where o, 0, ... = the ordinates at the ntid-points of each division
be equal to the sum of areas of individual triangles. EO = sum of the mid-ordinates ; n = number of divisions
Fig. 12.1 shows an area divided into several triangles. c . L =length of base line = nd ; d =distance of each division
For field work, a transit may be set up at 0, and
the lengths and directions of each of the lines OA, (2) AVE~g~2.3)
OB ..... el!:. may be measured. The area of each triailgle This rule also assumes that the boundaries between the extrentities of the ordinates
~
can then be computed. In addition, the sides AB, are straight lines. The offsets are measured to each of the points of the divisions of the
BC.. ... etc. can also be measured and a check may
be applied by calculating the area from the three base line.
The area is given by /!. = Average ordinate x Lengt!!__of the b>§e
known sides of a triangle. ThU&, if two sides and
one included angle of a triangle is measured, the
8 =[ Oo+ 0, ...... + On]L =-L-EO J(@
.•.. n+1 (n+l) ·
...(12.4)
area of the triangle is given by AG. 12.1
where Oo = ordinate at one end 'of fue base.
On= or~t.e at the other end of the base divided into n equal divisions
o.. 0, .... = ordinates at the end of each division.
(3) ~(1'1g.12.3)
This rule is based on the as-
where s =half i
perimeter = (a + b + c). sumption that the figures are trape-
The method is suitable only for work of small nature where the determination of zoids. The rule is more accurate than
the closing error of the figure is not imporrant, and hence the computation of latitudes the previous two rules which are ap-
o, o, o, o, o,
I
and departure is unnecessary. The accuracy of the field work, in such cases, may be determined proximate versions of the trapezoidal o,
by measuring the diagoual in the field and comparing its length to the computed length. rule.
Referring to Fig. 12.3, the area
-
.. -
2 3 4 n
@AREAS FROM OFFSETS TO A BASE LINE: OFFSETS AT REGULAR lNTERVALS
· This method is, suitable for long narrow strips of land. The offsets are measured
of the first trapezoid is given by \4---d
o, + o, •I• d----+1
L=niCI--------->1
from the boundary .{o the base line or a survey line at regular intervals. The method 1!.,=--2-d
can also ·j)e applied to a plotted plan from which the offsets to a line can be scaled· FIG. 12.3
off. The area may be calculated by the following rules :
!''
295
294 SURVEYING CALCULATION OF AREA

... (4\
, _ O,-+-o,d
given by uz- d
.1,,, ~3 (0, + 40, + 0.)
Similarly, the area of the second trapezoid is 2
Area of the last trapezoid (nth) is given by Area cif the last two intervals (8, _ " <1,) is given by
... (5)
<1n-
On- L +On
d ~·
o,. I..= 3d (0,., + 40n-l + 0,)
2
Adding all these to get the total area (<1), we get
Hence the total area of the figure is given by
Oo+O, 0 1 +-0, 0,_, +0, d [Oo + 40, + 20, +40, ...... + 20n-1 + 40, -I+ 0,]
<1 = 3
t.= .1, +<1, + ....... t., =--2-d+--2-d+ ..... + 2 d
~ + 0,) + 4 (9.L+ a;!+ ... + o...a + 2 (0, + D,t.+ ... 0,-tlQ
~~+;+ .... +o,.y):i} or ... (12.6)
or ... (12.5)
It is clear that the rule is applicable only when the number of divisions of the
Equation (12.5) gives the ·trapezoidal rule wbich may be expressed as below : area is even i.e., the total number of ordinates is odd. If there is an odd number of
Add the qyemre of the end offsets to the sum of the jnlermemqle nfl<ets. M!!!!!I!JY divisions (resulting in even number of ordinates), the area of the last division mllSt be
the I<Wll sum thus ob/llined by tKi common dUtance between the ordinates. io get the calcnlated separately, and added to equation 12.6.

------
requjrsd «tel!:_ Simpson's one third rule may be stated as follows : Tht area is equal to the sum
G
(4) SIMPSON'S ONE-THIRD RULE . F -----------· of the two end ordinates plus our liiMs um intermediJJie orditwJes
+ twtce t e sum o t e odd. i rmediate ordinates the whole mu · · d bv one-third
~hort lengths E•• ----~ T .::-.>IC
of boundary between the ordinates are parabolic the common interval between them.
arcs. This method is more useful when the bound-
oL<:---------------·ifi·----------- .,- Comparison of Rules. The results obtained by the use of Simpson's rule are in
all cases the more accurate. The results obtained b using ·Simpson's rule ,are ter or
ary line departs considerably from the straight
smaller than those obtained by using the trapezoidal rule accordin as the curve of the
line. o, o,
10, bo~ndary is ''!!!"~ or convex tow~ the base line. In dealing with irregularly shaped
Thus, in Fig. 12.4, the area between the number
figures, the degreeof precision of either mpllod can be increased by increaSing the
line AB and the curve DFC may be considered
to be equal to the area of the trapezoid ABCD of ordinates.
plus the area of the segment betw<\'n the parabolic A
1
a ......E(ample 12.1. The follawing perpendicular offsets were taken at 10 metres imervals
arc DFC and the. corresponding chord DC. d d _ __.., from a survey line to an irregular boundary line :
Let 0 0 , 0 1 , 0 2 = any three consecutive 3.2?. 5.60, 4.W. 6.65, 8.75, 6.20. 3.;15 4.2o. 5.6;5.
FIG. 12.4
ordinates taken at regular interval of d. Ca/culare the area enclosed between the survey line, the irregular bowulary line.
and 1he firsi aJUi /iJsl offsets, by tile app/icalion of (a) average ordinate rule. (b) trapezoidal
Through F, draw a line EG parallel to the chord DG to cut the ordinates in E
and G. nde, and (c) Simpson's rule.
. Oo+Oz Solution.
Area of trapeemd ABCD = - - · 2d ... (1) (a) By average ordinate rule
2
To calculate the area of the segment of the curve, we will utilize the property of From equation 12.4 (a), we have <1 = _L_ EO
n+l
the parabola that area of a segment (such as DFC) is equal to two-third the area of the
enclosing parallelogram (such as CDEG): Here n = number of divisions = 8 ; n + I = number of ordinates= 8 + I = 9
2 2([ Oo+O,l 2d,II L= Length of base= 10 x 8 = 80 m
3 , o, - - 2-
Thus, area of segment DFC = (FH x AB) = ... (2)
3 l:O = 3.25 + 5.60 + 4.20 + 6.65 + 8.75 + 6.20 + 3.25 + 4.20 + 5.65 = 47.75 m
Adding (I) and (2), we get the required area (8..,) of first two intervals. Thus, 80 x 47.75 = 424.44 sq.metres = 4.2444 ares.
6 =9
Oo+Oz
<1,,,=- -·2d+ ,.o,-- 21[
Oo+O'll
- 2d = d (0o+40,+0z) ... (3)
2 3 2 3 (b) By trapewidal nde
Similarly, the area of next two intervals (<11.4) is given by . . ( Oo+O, "j
From Eq. 12.5, tJ. =
2
l--
- + o, + o, + ... + o,-' 1d
 'J!T1
CALCULATION OF AREA
296 SURVEYING
he calculated by Simpson's rule, and the area enclosed between the seventh and last offseiS
Oo+ 0, 3.25 + 5.65 may he found by the trapezoidal rule.
Here d =I0 m; --= _ 4.45 m
2 2 Thus,; (00 + 0,) = 0 + 4.95 = 4.95
01 + 0, .... 0,- 1 = 5.60 + 4.20 + 6.65 + 8. 75 + 6.20 + 3.25 + 4.20 = 38.85 m 4 (01 + o, + ... o.-1> = 4 (2.65 + 3.75 + 3.60) = 40
1!. = (4.45 + 38.85) 10-= 433 sq. metres = 4.33 ares.
2 (0, + o. + ... 0.- 2) = 2 (3.80 + 4.65) = 16.90
(c) By Simpson's nde
1!.' = 5"(4.95 + 40 + 16.90) = 309.25 sq. m.
From Eq. 12.6, 1!. =
d
[(Oo + O,) + 4 (01 + o, + .... + o,_ 1) + 2 (0, + o...... + o,_ :i)) 'i = 81.0
3 Area of the last trapezoid= (4.95 + 5.85) sq. m.
Here d = 10 m ; 0 0 + o, = 3.25 + 5.65 = 8.9 m Total area= 309.25 + 81.0 = 390.25 sq. m = 3.9025 ares.
4 (01 + o, + ..._.o;-1>;, 4 (5."~o + 6.65 + 6.20 + 90> =.90.60
2 co,+ O.+ .... o,_2),;·z(4.:iO·+ 8.75 +3.25)= 32.40. 12.5. OFFSETS AT IRREGULAR INTERVALS
5
:. 1!. =
1
3
°
(8.9 ~ 90.60 + 32.40) = 439.67 sq. metres = 4.3967 ares.
(a) First Method
In this method, the
(Fig. 12.5)
area of each trapezoid
3
4

~pie 12.2. A series of offsets were taken from a chain line to a curved boundary is calculated Separately and then added together
line al intervals of I ~es in the following order. ' to calculate the tbtal aiea. Thus, from Fig.
o, o, o, o,
o. 2 .65, 3.80, 3:75, 4.'65, 3.60, 4.95, 5.85 m- . 12.5, IO,
A B
Compwe the area between the chain line, the cuJ;.;d boundary and the end- offsets dl
1!.=2(01+
d, 0, +O,)
0 ,)+2(
to-d,-4!+--d 2 Ill <1,---oJ+-d,~
by (a) average ordinate rule, (b) trape2oidal rule, and (c) Simpson's nde.
,f
Solution. AG. 12.5
(a) By average ordinate nde
+ d, (0, + 0,) ... (12.7)
2
(b) Second Method. By method of co-ordinates : See § 12.7
From Eq. 12.4 (a), we have 1!. = _!:_! :W
n+ Example 12.3. The following perpendicular offsets were taken from a chain line
Hence n=7; n+l=S. to an irregular boundary : \
10 ', 25 42 60 75m
L = nd = 7 x 15 = lOS m Chainage 0
Offset 15.5 26.2 J 31.8 25.6 29.0 31.5
:W = 0 + 2.65 + 3.80 + 3.75 + 4.65 + 3.60 + 4.95 + 5.85 = 29.25 m
Colculate the area berween the chain line, the boundary and the end offsets.
105
1!. = x 29.25 = 383.91 sq. m = 3.8391 ares.
8 Solution.
2
10-0 '
(b) By lrflpewidol rule Area of fust u:aP"wid = 1!.,1 = .,, (15.5 + 26.2) 208.5 m
l
=

From equaaon 12.50

1!.= (Oo~Oo
- -+0I+O,+ .... O,-t d 25'-10
2 Area of second trapezoid= 1!., = -2- (26.2 + 31.8) = 435 m'
0

= , Oo + 0, 0 + 5.85 = 2.925 m
Here d .15 m,
2 2 ·-
,. •• of tltird trapezOl'd = 1!.3 = -42-- 25 (31.8 + 25.6) = 487.9 m2
2
o1 + o, + .... o.-1 = 2.65 + 3.8o + 3.75 + 4.65 + 3.60 + 4.95 = 23.40 60-42 2
:. 1!. = (2.925 + 23.40) 15 = 394.87 sq. m = 3.9487 ares, Area of fourth trapezoid= 1!.4 = - - (25.6 + 29.0) = 491.4 m
2
(c) By Simpson's rule · ' . 75-60 2
Area of fifth trapezotd = 1!., = - - (29.0 + 3LS) = 453.7 m
From equation 12.6, 1!. =~ [(Oo+ 0,) + 4 (01 + 0 3 + ... On-1) +2(0,+ 0.+ ... On-2)] 2
Total area=l!.=l!.l+l!., +l!.,+l!.<+l!.s
. d IS
Here, -=-=5 m. = 208.5 + 435 + 487.9 + 491.4 + 453.7
3 3
It will he seen that the Simpson's rule is not directly applicable here since the number = 2076.5 m' = 20.765 ares.
of ordinates (n) is even. However, the area between the first and seventh offsets may
• 299
SURVEYING CALCULATION OF AREA
298 /
.,
;:7·· \.
~

MERIDIAN DISfA NCES

~pi e 12.4•. The-following
1
-perpendicular offsets tw
~ ~·
re. taken from ..a chai··.·n line to
1
~ce of any point in
point to ihe
ahed ge: 1 -,_·
45 ~, 0 ' 80 . 100..-..1.120 . 14Q I a traverse is the distance of that
ChairuJge (m) =,IS . --30
.<().. .. at right angles to
n. measur ed
7.9 :; 6~{. ' 4.! __ reference meridia
Offsets (m) ~- . 85 10.7 12.8 !~-4.. .9.} _ ,
:!_.
hedge and the e ·o e -- (~) the meridian. 'lhe meridian distance of a survey
te the area between the survey bne, the
Calcula
-·- ( 8 -s- \ c: ~-\- line is defined as the meridian distance of its
Trapezoidal rule (b) Simpson's rule. mid-point. The meridian diiiance (abbreviated
~ ( ':l ~ ) .}-
Solution as M.D.) is also sometimes called as the longitude.
(a) By Trapezoidal role Thus. in Fig 12.6, if the reference meridian
is another interval between
The interval is constant .from first offset to 5th offset. There is chosen through the most westerly station A,
offset and lOth offset. The total
the 5th and 7th offset and. a third interval between 7th the meridian distance (represented by symbol
area t> can, therefore, be divided into three sections. 2£( m) of the line AB will be equal to half its
d=&,+ A,+& ,
._ departure. The meridian distance of the second
a,= area of first section ·; A2 =area of second section line BC will be given by
where
= 15 m D1
&, =area of third seetion ; d, =interval for first Section D,
al for third section = :10 m m-.=
• 2 -+-
m1+ 2
d, =interval for second section= 10 m ; d3 =..imerv
';'). -
Now &1 = (
7.60 + 10.6
+ 8.5 + 10.7 + 12.8 15 = 616.3 m' Similarly, the meridian distanc~ of the FIG. 12.6
2.
third line CD is given by
&, = (
10.6+ 8.3 )
+ 9.5 110 = 189.5 m
)
' .... D,(D')
m,=mz+T+ -T =mz +T-T
D,D,
2
is given by
8.3 + 4.4 . )
+7.9+ 6.4 20=41 3m
' And, the meridian distance of the fourth (last) line DA
&3 = (
m. = m3 + (- ~' ) + (- ~· ) = m,- ~'- ~· = ~·
2
a= 616.5 + 189.5 + 413 = 1219 m' = 12.19 ares.
as follows The meridian
(b) By Simpson's Rule Hence, the rule for the meridian distance may be stated
n distance of the preced ing line plus half
The first section and the second section have odd numbe
r of ordinates, and therefore. distance of any line is equal to the meridia ihelf.
line plus half the deparru re · of the line
Simpson's rule is directly applicable. The third section has 4 ordinates (even number) : the deparrure of the preceding
will be equal to half
the rule is applicable for the first three ordinates only
: · According to the above, the meridian distance of the first line
proper attenlion shauld be paid to the signs of the
its departure. In applying the rule,
&1 = ~5 [(7:·60 + 10.6) + 4 (8.5 + 12.8) + 2 (10.7)] = 624 m' depanures i.e., positive sign for eastern departure and 11ega1iv
e sign for westem depanure.

AREA BY LATITUDES AND MERIDIAN DISTANCES

&2 =
1
~ [(10.6 + 8.3) + 4 (9.5)] = 189.7 m' to the reference meridian,
In Fig. 12.6, east-west lines are drawn from each station
triangle or trapezium (so formed)
thus getting triangles and trapeziums. One side of each
a,= ;o [(8.3 + 6.4) + 4 (7;9)1 ; ~ (6.4 + 4.4)
will be one of the lines, the base of the triangle
or
m
trapezium will be the latitude
will be the meridian distance
of the line, and the heighJ of the triangle or trapeziu
= 308.6 + 108 = 416.6 m'
of that line. Thus,
&=62 4+189 .7+41 6.6 =1230 .3m'=1 2.303 ares. area of each triangle or trapezium = latiJude of the line x
meridian distauce of the line.
12.6. AREA BY DOUBLE MERIDIAN DISTANCES or A1 =L1 x m,
area of a closed traverse. g, and negative if it is
This method is the one most often usOd for computing the The latitude (L) will be taken positive if it is a northin
area by this method, the latitudes
This method is known as D.M.D. method. To calculate the a southing.
ted. The traverse is then balanced. to the algebraic sum of the
and departures of each line of the traverse are calcula In Fig. 12.6, the area of the traverse ABCD is equal
to pass through the most westerly station of the
A reference meridian is then assumed areas of dDCc. CcbB, dDA and ABb. Thus,
are computed.
traverse and the double meridian distances of the lines
 ':~" JOI
SURVEYING CALCULATION OF AREA
300
That is, area of any triangle or trapezium = Half the product of the latiblde of
A =Area of dDCc +area of cCBb- area of dDA - area of ABb
the line and its meridian distance.
or A= L3 m3 + £, m, -L. m. .- L, m1 = r.Lm.
Hence' A= j; [M, L, + M, £, -M,L.- M, L,]
(It is to be noted that the quantities L, m, and L 1 m1 bear negative sign since
Thus, to find the area of the traverse by D.M.D. method, the following steps are
L, and L, of DA and AB are negative.)
necessary
DOUBLE MERIDIAN DISTANCE (I) Multiply D.M.D. "Of each line by its latiblde.
The double meridian distance of a line is eljUII] to the sum of the meridian distances· (2) Find the algebraic sum of these products.
of the two extremitks. (3) The required area will be half the sum.
Thus, in Fig. 12.7, we, have :
AREA FROM DEPARTURES AND TOTAL LATITUDES
Double meridian distarice (r';Presented by symbol M) of the first line AB is given by
From Fig. 12.8, the area (A) of ABCD is given by
M, =m of·Ac +in of B=O+D, =D,
A = area of ABb + area of BbcC + area of dcCD + area of DdA
Similarly, if M, , M 3 , M, are the double me-
ridian distances of the lines BC, CD and DA re- If L 1'. L,', L,' are the total latirudes · of the ends of the lines, we get
spectively, we have +1!:-- o,,--. 1 A=! [(D,)(O- L,') + (D,.)(- L,' + L,') + (- D,)(L,' + L,') + (- D4) (L,' + 0)]
M,=m· of B+m of C
=D 1 +(D 1 +D2) =M1 +D 1 +D1
=- Mu (D, + v,) + u <- v, + v,) + L.' (D, + v.)]
D
= D.M.D. of AB + Departure of c Note. The negative sign to the area
has no significance. ''
AB + Departure of BC ''
M, = m of C + m of D
"' Hence, to find the area by this method,
:he following steps are necessary :
4 '
'
/t.,'
c
= (D 1 + D,) + (D, + D,- D,) (I) Find the total hltitude (L') of
•• Li
•
=Mz+Dz-D3 each station of traverse. ! ...........b,........
.A-... v.:i ...................
d •
= D.M.D. of BC + Departure of (2) Find the algebraic sum of the de· '•
partures of the two lines meeting at that '
BC + Departure of CD L,':
station. ••
and M,=m of D+m of A=(D,+D1 -D3) /4----D ,----< o,
~
+ (D 1 + D,-D, -D4) =M,-D, -D4
= D.M.D. of CD + Departure of CD
.,. Depanure oi OA
FIG. !2.7
(3) Multiply the total latitude of each
station by the corresponding algebraic sum
of the departure (found in 2) .
(4) Half the algebraic sum of these
l ~---0,
FIG. 12.8
B o,J
Hence, the rule for finding D.M.D. of any line may be stated as follows: "The products gives th~ required area.
D.M.D. of any line is equal to the D.M.D. of the preceding line plus the tkpiuture AREA BY DOUBLE PARALLEL DISTANCES AND DEPARTURES
of the preceding line plus the deptuture of the line iJself. "
I A parallel distance of any line of a traverse is the perpendicular distance from the
Due attention should be paid to the sign of the departure. The D.M.D. of the first middle poim of that line to a reference line (chosen to pass through most southerly station)
!' line will evidently be equal to its departure. The double meridian distance of the last line at right angles to the meridian. The dcuble parallel distance (D.P.D.) of any line is the
is also equal to its departure, but this fact should be used simply as a check. sum of the parallel distances of its ends. The principles of finding area by D.M. D. method
AREA BY LATITUDES AND DOUBLE MERIDIAN DISTANCES and D.P.D. method are identical. The rules derived above may be changed to get the
Jo Fig. 12.7, the area of the traverse ABCD is given by corresponding rules for D.P.D. method, by substiruting D.P.D. for D.M.D. and 'departure'
A = area of dDCc + area of CcbB - area of d.DA - area of ABb for 'latitude'. The method is employed as an independent method of checking area cbmputed

Now, area of dDCc =¥dD + cC) cd =! (M 3) x L3

by D.M.D. method.
 "
~· 303'
SURVEYING CALCIJLATION OF AREA
302
Tolal area= t. = l:mL =- 29221 m' "
12.7.~ icance, !I
Since the negative sign does not have any siiJnif
Let (x,, y1), (X,, y,), (x,, y3) and (x4 , y4) be
- .,· The actual area= 29221 nl- = 2.9221 hectar es. H
the co-ordinates of the stations A. B, C. D
respectively, of a traverse ABCD. If A is the
:t~~~~-~\\\\\\\\'{:n\\\a
- .B(X, y,)
(2) By D.M.D. mellrod : Area= t l:mL
total area of the traverse, we have Deptulure D. M.D. Area=mL
line l.<Jiitude (m)
(L)- (D)
A= (Area aABb) + (Area bBCc)
4 4 + 432
-(Area cCDd)- (Area dDAa) AB + 108 +

f
= [(y,- y,)(x, + x,) + (y, -_y,)(.i, + x3) BC + IS + 249 257 + 3855

- (y, - y,)(x. + x,) ~ (y, - y,)(x, + x,)] CD - 123 + 4 510 - 62,730

= ilYo(x,- X.) + y,(x, - x1) + y3(x4 - x,) DA 0 - 257 257 0

+ y4(x 1 - x3)] Sum - S8,443

__ FIG. 12.9
In general, if we have n stations, we get ,. 2
i
Area= l:mL = 29221 m = 2.9221 - hectares.
·:,.. ··
/ =} [y, (x, + x,) + y, (x3 - x,) + y, (x,- x,) + ... + y,(x, - x, -I)] ... (12.7)
(3) By Deparlure and totoJ latiJudes : Let
us first calculate the . total latitudes of
ted latitudes and departures (in metres) the point, starting with· A as the reference point,
--Eiample 12.5. The following toble gives the correc
of the sides of a closed traverse ABCD · Thus, total latitude of B = + 108
Side l.<Jiitude total latitude of C = + 108 + 15 = + 123
\,
N sr- _l E (-\ w total latitude of D = + 123 - 123 = 0
AB /08 : 4
/
total latitude of A = 0 + 0 = 0
BC
CD
IS
~ 123
249
4
The area·= iJ: (Total latitude x algebraic sum of adjoining depart
ures)
~

DA 0 257 -- I line l.<Jiitude /JtpiUIDn St4tion Totlll Algcbr okmm

.,..
Doubk
of o4jololog
Compute its area by (i) "(Jk.:met!J.<Hj, (ii) D.M.D
-
, met/wd (iii) Departures and total (L) (D)
'1t/'
latitudes, (iv) Co-or!firwte metlwd. + 108 + 4
AB
Solution. B + 108 +253 + 27,324

(1) By meridian distances and latiJudes BC + IS + 249

Area= l:(~) c + 123 +253 + 31,119
ed in the-
Calculate the merid ian distance of 'each line. The calculations are arrang CD - 123 + 4
es and departures, point A is the most - 253 0
tabular limn ,below. By the inspection of the latitud the last line
D 0
first line and DA as -.
westerly station. AB is rakeo as the 'DA 0 - 257
M.D. Area= mL A 0 - 253 0
line l.<Jiitude DejklltUr< ~Departure
(L) (D) (m) Sum S8.443
rtDJ
Area= ~58,443) m' = 29221 m = 2.9221 _hectares.
2
AB + 108. + 4 + - 2;-PI. 1' + 216
..
BC + 15 + 249 + 124.5.''" 128.S + 1928
by co-ordinates, it is customary to
- '-- (4) By Co-ortNnates : _For calculation of area
CD 123. + 4 + 2 25S - 3136S
points. This can be done by uking the
- 0 calculate the independent c<Hlrdinares of all the
DA 0 - 257 128.S " 128.S are tabulated below :
Sum - 29221 co-ordinates of A as ( + 100, + 100). The results
-
 SURVEYING 30l
304 CALCULATION OF ARI!A

lndepttuknJ to-ortlbuiUJ or dotted. The traCing is then placed on the drawing in such a way that lhe area is
11M I.IIJ/Ju4< .~
rf":v)'\ Slllllon
exactly enclosed between two of the parallel lines. The figure is thus divided into a number
(~-'''' North GJ &ut(x)_
of strips. As,luming that the strips are either trapezoids or triangles, the area of each is
A 100,'. 100 . ' equal to the length of the mid~dinate multiplied by the constant breadth. The mid-ordinates
AB + 108 + 4 of the strips are represented by the length of the dotted lines intercepted within the maps.
B 208- 104 The total sum of these intercepted dotted lines is measured and multiplied by the constant
BC + IS ' + 249 breadth to get the required area. More accuracy will be obtained if the strips are placed
c 223. 353 nearer.
I
CD - 123 - + 4
12.9. AREA BY PLANIMETER
100 I 357
D A planimeter is an instrument which measures the area of plan of any sbape very
0 ' . I
DA - 257\ accurately. There are two types of planintete1's: (I) Amsler Polar Planimeter, and (2) Roller
IQq .
. A
1
100 -
Planimeter. The polar planimeter is most commonly used and is, therefore discussed here .
Substituting the values of x and y in equation 12.7, we gef Fig. 12.11 shows !lie essential parts o{ a polar planimeter. It consists of two arms
I .·
A = (y,(x, - x,) + y,(x, - x1) + y,(x. - x,l + y,(x, - x,)]
hinged at a point .known as the pivot point. Ooe of the two arms carries an aucbor at
2 its end, and is known as the anchor arm. The length of anchor arm is generally fixed,
=i [100(208 - 100) + 104(223 - 100) + 353(100 --}08) +357(100 - 223)) but in some of the planimeters a variable length of anchor arm is also provid~. The
other arm carries a tracing point at its end, and is known as the tracing arm. The length
=i (10800 + 12792- 38124- 43911) =- 29221 ~; of the tracing arm can be varied by means of a fixed screw and its corresponding slow
Since the negative sign does not bave significance,. the area = 2.9221 hectans. motiou screw. The tracing point is moved along the boundary of the plan the area of
which is to be de1ermined. The normal displacetnent of the tracing arm is measured by
12.8. AREA COMPUTED FROM MAP MEASUREMENTS means of a wheel whose axis is kept parallel to the tracing arm. The wheel may either
(A) By sub-division of the area into geometric . fiiJIIl"'S be placed between the hinge and the tracing point or is placed beyond the pivot point
The area of the plan is sub-divided into common geometric figures, such as triangles, away from the tracing point. The wheel carries a concentric drum which is divided into
rectangles, squares, trapezoids etc. The length and latitude of eath such figure is scaled 100 divisions. A small vernier attached near the drum reads one-tenth of the drum division.
off from the map and the area is calculated by using the usual formulae.
(b) By sub-division into . square;; : Fig. 12.10 (a)
The method consists in drawing squares on a tracing paper each square representing
some definite num6er of square metres. The tracing paper is placed on the drawing and
the number of squares enclosed in the figure are calculated. The positions .of the fractioual
squares at the ·:1.!.!-\red tJlili~~· .: ..::~ ~t~t:~. 7!::-: :2L?.! ~E.:. ~f tll~ figw:e will then be
equal to the total number of squares ·
multiplied by the factor (i.e., sq. me-
tres) ••rresented by each square.
(c) By division Into trapezolli>;
Fig. 12.10 (b).
In this method, a number of --~-----------------·~-
·parallel lines, atconstantdistaru:e apart, FIG. 12.11. AMSLER POLAR PLANIMETER.
are drawn on a tracing paper. The
6. WHEEL
constant IM!tween the consecutive P'!'- I. TRACING ARM

allel lines represents some distance ------------------------ 2. ,t.NCHOR ARM

1. GRADUATED DRUM
8. DISC
il in metres or links. Midway between (a) (b) 3. ANCHOR
I 9. MAGNIFIER
each pair of lines there is drawn another 4. TRAONG PO~
pair of lines in a different colour FIG. 12.10 5. HINGE..
10. ADJUSTING SCREW FOR I
 307 ~
SURVEYING CALCULATION OF AREA ~I
.~
306 i
(c) shows ·thecjnnponents of the two motions separately. Thus, if the tracing aim sweeps
The complete revoluuons of the wheel are read on a disc acruated by
to the wheel. Thus, each reading is of four digits- the units being
a suitable gearing
read on the disc,
In addition
a veiy Small arei dA,, such ·that dh is the movemeDI in parallel direction
rotalion, we !Jave
and d9 · is the
t
1
ths on the vernier.
the tenths and hundredths on the drum, and the thousand ·~
utilised to know the number of the times dA, = Ldh + -~ L'da
to this, a fixed index near the disc can be of the tracing
the zero of the disc has crossed the index. Since the recording wheel (W) is placed in plane perpendicular to that
'~1f
axis. If dw is the distance
It is clear from Fig. 12.11 that the planimeter rests on three points
- the wheel, arm, the wheel records only the movemeDI perpendicular to its. !
poinr remains ijxed rolled out by the wheel in sweeping the area tL!,, we get
I~
the anchor poinr and the tracing point. Out of these three, the anchor
partly rolls and partly slides as the tracing point is moved dw= dh + ad6 or dh = dw- ad6 ... (1)
in position while the wheel
of the wheel is perpendi cular to the plane of the
along the boundary. Since .the p)ane Substiruting the value of dh in (1), we. get II

~
displacem ent -when it
cenrre line of the tracing arm; ·the wheel measures only notrnal i
dA, = L(dw- ad6) + L'd9 ... (2)
acrually rolls.
To find the area of the pl~;· the anchor point is either placed outside
the area ...,___ L i+.!..ot
(if the area is small) or it is placed inside the area
then marked on the boundary. of ate;. and the
initial reading of the wheel is then
the boundary till it comes to the
taken.
starting
The
tracing
tracing
point. The
(if

point
final.
the
point
is
area
kept
now
r~ading
is

of
large).
exactly
moved
the
:
drum
A
over·
clock-wis
is
point is

e
it. The
along
tal¢en.
I~
The area of the figure is then calculated from the following formula
(li) ~ M (F- I± 10 N + C) ... (12.8) .
Area
where F ~ Final reading : I ~ Initial reading (a)
N ~ The number of times the zero mark of the dial passes
the fixed index mark.
Use plus sign if the zero mark-df the dial passes the index mark in a
clockwise direction and minus sign when it passes in the anti-cloc kwise direction.
M ~ A multiplying constant, also sometimes known as the planimet
er constant.
It is equal to the area per revolutio n of the roller.
the area
C ~ Constant of the instrument which when multiplied by M. gives
of zero circle. The constant C is to be added only when the anchor
poim is inside the area. (b) (c)
e direction
It is to be noted that the tracing point is to be moved in the clockwis
formula is gi~en belo\1.'.
only. Proper sign mU:;t be given to i''r'. The proof of the above
THEORY OF PLANIMETER
Fig. 12.12 (a) shows the schematic diagram of polar planimeter. where
A,: Area to be measured, the anchor point being outside the area.
A
A,
and the hinge.
L ~'Length of the tracing arm = Distance between the tracing point
R ~ Length of anchor arm = Distance between the pivot and the anchor point.
between
a ~ Distance between the wheel and the pivot, the wheel being placed
the tracing point and pivot.
w ~ Distance rolled by the roller in tracing the area.
A,= Area swept by the tracing arm.
movement
Fig. 12.12 (b) shows the section of the perimeter of the area. Any such (d)
simultaneous. motions : (I) translatio n of the tracing arm
of the arm is equivalent to two
rotation of the tracing arm about the pivot. Fig. 12:t2 FIG. t2.t2. THEORY OF PLANIMETER.
TP in parallel motion and (il)
 309
SURVI!YINO CALCULATION OF AREA
308
of the roller passes
moves downwards is held in such a position relative to the anchor arm that the plane
When the tracing point is moved along the boundary, the arm through the anchor point i.e., the line joining the anchor point and the wheel is at right
the net area A, swept
along one side of. area and upwards along the other side. Heu:e, angles to th~ line joining the tracing point and the wbeel.
by the tracing arm is equal to the area of the plan (Ao)
~athol tradng point
Thus Ao= f dA_s=L lttw-aL f dO +}L' de I ... (3)

But f
dw = total distance moved by the wbeel = w ",/ ~.--- 14/ L a-+!
n W
Ide ... (4)
A0 =Lw-a Lfda+ iL' T
of the pivot is
Now when the anchor point is kept oulside the area,· the motion
completes one revolution about the A;
constrained along the arc of a circle i.e., ·it never
the arc in upward and downward directions so that
anchor point but simply moves along
fdO=O . A
A
... (12.~) (t-~
Hence from (4), A0 = Lw (a)
moves along
However, if the anchor point is kept inside the area, the pivot point FIG. 12.13
point is brought
the circle of radiU. R and completes one revolution when the tracing
back .10 ils original position after tracing the area. Hei)Ce the quantity dO = f
2n ..
In Fig. 12.13 (a), the wbeel has been placed between the tracing
point (1) and

A1 = Area of the plan when the anchor point ~ ; kept inside the
perpendicular distance
Let
area. the pivot (P). Let R0 be the radius of the zero circle. If x is the

Aa = Area swept by the pivot. of the wheel W from anchor A, we get

.r R0 =(L-a )'+ :i' =(L-a )'+ (R'- a')
Then, the area A, = A, + Aa = IdA, + nR' = [L I dw - aL I de+ ~;L' I
dO] nR' · +
... (12.11 a)
=(L' + tl- 2La + R'- a') =(L2 - 2aL + R')
· =Lw- aL(2n) + ,j.L'(2it) + nR' =Lw + n(L'- 2aL + R') ... (12.9)
... (12.11 b)
J.h('area while And area of the zero circle= n Rl = n(L'- 2aL + R')
Thus, equation 12.8 is to ·he used when the anchor point is oulside
the anchor point is kept inside the area. In Fig. 12.13 (b), the wheel has been placed beyond the pivot.
equation 12.9 is to he used when + R' ... (12.11 c)
Now w = Total distance rolled by the wheel = 1tD n ... (3) Hence, Ro' = (L +a)'+ (R'- tl) = L' +a'+ 2aL + R'- tl = L' + 2aL
2 ... (12.11 d)
wbere D = Diameter of the wheel Area of the zero circle=1t(L +2aL+ R')
point as :
Thus, the general expression for the area of the zero · circle can
written
n = Total change in the reading, due to the movement of the tracing be
along the periphery of the area= F- 1 ± ION. 2
Area of the zero circle=n (L ±2aL+ R')
... (12.11)
area A,.
Subsrii"'Jting !he value f'f w ~n equation 1'2.9, we get the Use+ sign if the wheel is beyond the pivot and -sign if iho whtel
is be<w""a the
!>. = LT<Dn + n(L'- 2aL + R');, Mn + n(L'- 2aL + R') ... (12.10 a)
or rracing point and the pivot.
!I is traversed along
I' =Mr.+ MC = M(n +C)= M (F -H ION+ C) ... (12.10 b) To find the area . of the zero circle practically, the tracing point
I one with the anchor point oulside the figure, and then. with
M = The mtiltiplier = LxD = Length of tracing arm x Circumference of the
wheel the perimeter of a figure,
where
the anchor point irui\de it.
C = Constant = n(L' - 2aL + R') eqution 12.10 a
Since the area swept is the same in both the cases, we get, from
M
In the above !>. = [Mn + n (L' ± 2aL + 11')] = Mn'
Thus; we get equation 12.10, which was given in the earlier stage. ... (12.12)
equation C is to be added only if the anchor point is inside the area. n(L' ± 2aL + R') = M (n' - n)

where n and n' are the two corresponding readings of the wheel.
It is to be noted thoJ
ZERO CIRCLE
area of the. zero circle,
or correction n will be positive. if. the area of the figure is- greater than the
The quantity MC = 1t (L' - 2aL + R') is known as the area of the zero circle area of the zero
the circle round the circumfe rence while it will be negative if the area of the figure is smaller than the
circle. ~ zero circle or the circle of correction is defined as
·
the wheel will simply slide (without rototion) on circle.
of which if the tra~ing point is moved,
in the reading. This is possible when· the tracing . arm
the paper without any change
 311
SURVEYING CALCULATION OP AREA
310

MULTIPLIER CONSTANT (M) Example 12.7. Calculate the area of a figure from the following readings recorded
by the planimeter with the anchor point inside the figure.
The multiplier constant or the planimeter comtant is equal to the number of units
'lnirial reading= 9.9I8 ; Final reading = 4.254 ; M = IOO sq. em : C = 23.52I
of area per revolution of the roller. Numerically, it is equal to LnD . Since the diameter
of the roller or wheel is a fixed quantity, the value of M depends on L. Thus, the It was observed thai the zero mark on the dial passed the index once in the ami-clockwise
length of the tracing arm is set to such a length that one revolution of the wheel corresponds direction.
to a whole number and convenient value of area. When the figure is drawn to a natural Solution
scale, and the area is desired in sq. incbes, the value of M is generally kept as eqlial Theareais given by A=M(F-I ±ION+C )
to 10 sq. in of area. Here M= 100 sq. em; I=9.918; F=4.254; C=23.521 and N=-1
For any other setting of the tracing arm. the value of M can be determined by A= 100(4.254- 9.918- 10 + 23.521) = 785.7 sq. em.
traversing the perimeter of a·fi~. of known· area (A), with anchor point outside the figure.
Then . /' Example 12.8. The following readings were obtained when an area was measured
by a planimeter the tracing arm being set to the nalural scale. The initial and final readings
M Known ilrea ~ ~ where n' = Change in the wheel readings
n' / n' were 2.-268 and 4.582. 'J'fte zero of disc passed the index mark once in the clockwise
direction. The anchor poim was inside the figure with the value of the constam C of
It is to be noted that the value of M and C depends upon the length L which
the instrumem = 26.430.
is adjustable. The manufacturers, therefore, supply a table which gives the values 'of L
(a) Calculate the area of the figure:
and C for different convenient values of M.
The manufacturers always supply the values of ihii' vernier setting on the .tracing (b) If the area of the figure drawn be ·to a scale of 1 inch = 64 feet, find the
arm with the corresponding values of M and C. The following table is an extract from area of the figure.
the values for a typical planinteter. .r SoIuton .
Since the tracing arm was set to the natural scale, the value of M = 10 sq .. inches.
Area .of one rel'Olution of
Vemitr porilion lht meMutr!nunl whtel ConsJJJnJ A =M (F-I± 10 N+ C)
Scak on lnJdng bar (M) (C)
Here F=4.582: I=2.268 ; N=+ I; C=26.430
SCIIle A.<lllal
.. A= 10(4.582- 2.268 + 10 + 26.430} = 387.44 sq. inches.
100 sq. em 23.521
I: 1

1:l
I
I
33.44

21.58
!
I
100 sq. em

10 sq. in. I 10 sq. in. I U.430

The scale being I"= 64 ft. Hence I sq. in. = 64 x 64 sq. ft.
64 X 64 X 387.44
:. Area of field acres = 36.39 acres.
U.97 i 200 sq. ft. J 12.5 sq. in. I 24.569 43560
I' 48 !
Example 12.9. The perimeter of a figure is traversed clockwise with the anchor poim
125 $0 in i '24 569 .
, 1 · '24 i 26 97 SO sa fi
inside and with rhe tracing arm so ser rhal one revolution of rhe roUer measured 100
I I ' 50 : 21.66 i 0.4 ac;., I 10.04 ~- in. I 26.676 I sq. em on the paper. The initial and final readings are 2.828 and 9.836. The zero mark
of the disc passed the fixed index mark twice in the reverse direction. The area of the
Thus, for full scale, value of M = 10 sq. in. in. f'.P.S. units, and for another setting zero circle is found to be 2352 sq. em. Find the area of the figure.
of tracing bar. the value of M = 100 sq. em. Solution.
Example 12.6. Calcula/e the area of a figure from the following readings by a planimeter The area of the figure is given by
2
wiJh anchor poim outside the figure : A = Mn + n (L' - 2aL + A ) ... (12. 10 a)
Initial reading = 7.875, final reading = 3.086 ; M = 10 sq. in.
where n(L'- 2aL +A')= Area of the zero circle= 2352 sq. em.
The zero mark on the dial passed the fixed index mark twice in the clockwise direction.
M = 100 sq. em.
Solution. A = M(F -I± ION+ C)
n = F- I± ION= 9.836 - 2.828 - 10 x 2 = - 12.992
Since anchor point is outside, C is not to be used in the formula,
.. Substiruting the values in Eq. 12.10 a, we get
M=IO; F=3.086 ; I=7.875 ; N=+2
A= 100(- 12.992} + 2352 =- 1299.2 .+ 2352 = 1052.8 sq. em.
A= 10(3.086 -7.875 + 20) = 152.ll sq. in.
 ~
313 li
312
SURVEYING CAI.C1JLATION OF AREA ''
Example 12.10. The foUowing observtJiions were mode with

(I)
(2)
Area
Known area of 60 sq. inches
Unkno wn area
I.R..
2.326
8.286
F.R.
8.286
a planinreter.

5.220 +I
N
0

the cases with the same

= Lnd = 16.6 x n x 1.92 = 100 em'.
(il) ~ of zero circl e=" (L' ± 2La + R')
Since the· wheel is placed between the hinge and the
he used with 2La. Hence,
tracing point, minus sign will

x 16.6 x 1.68 + 22.6') = 2290 em'.

I
The anchor poinl was placed outside the figure in both Area of zero circle= n(L'- 2a + R') = n(16. 6'- 2
setting of the tracing ann. CalculoJe : Example 12.13. Calculall the area of a figure
from the following readings recorded
unknown area. the figure :
(I) The multiplier constant and (2) The by the planimeter with the anchor poinl inside
Solution Initial reading= 2.286, jinai reading = 8.215
twice in the counter-dackwise
(1) The mulliplier constqilt (M) The zero of the counting disc pnssed the indet mark
not availa ble, the following observations
A=M (F-1 ± !ON) direction. Since the constants of the instrument were
Substiruting the values, we get were also mode :
= 4.09 "
The distance of the hinge from the tracing poinl
60 = M (8.28 6- 2.326 + 0), from .which M = _:0 = 10.027. sq. in. = 6. 28 "
5 The distance of the hinge from the anchor point
(2) The unknown area The perimeter of the wheel = 2.5 "
(5.22 0- 8,.~86 + 10) = 69.80 sq. in. distance of = 1.22 ".
A = M(F -I+ 10 N) = 10.027
. 1··· The wheel was placed beyond the hinge at a
red
Example 12.11. The following readings were obtained when an area was measu Solution.
by a planimeter, the tracing arm being so set
that one revolution of the wheel measufes
· (a) Ca/cul.ation of inslnmumtol consllln/S
10 sq. inches on paper. M = Length of the tracing arm x its circumferenc
e
final readings were 5. 286 and
When the anchor point was outside, the initial and =4.09 X 2.5 = 10.225 in'.
mark once in the clockwise direction.
1.1186. The zero mark of the. dial..passed the index
figure. the inilia/ and final ;eadings 2 x 4.09 x 1.22 + 6.28') = 208 sq. in.
When the anchor poinl was placed inside the same Area of zero circle= n(L' + 2La + R') =" (4.09' +
passed the index mark twice in the
were 5.282 and 3.842. The zero mark of the dial Now A= M (F -I± 10 N +C)= M (F- I± 10 N) + MC
counter clockwise direction. · = 64.12 sq. Inches.
. circle. = 10.225 (8.215 - 2.286 - 20) + 208 = - 143.88 + 208
Find the area of the zero
Solution,
Wilh the anchor point outside
-· PROBLEMS

sq. inches.
A= M (F- I+ 10 N) = 10 (1.08 6- 5.286 + 10) =58 1. What is Simpson's rule ~ Derive an expres
sion for it.
With the anchor point inside the offsets taken from a chain line 10 an irregular
2. The following give the values in feet of
N=- 2
A =M( F-I± 10 N+ C) ; Here A= 58 and boundsly :
ISO 200 250 3SO
300 400
58= 10 (3.842 - 5.282 ~ 20 C) + Distance
Offset
0
10.6
SO
15.4
100
20.2 18.7 16.4 20.8 19.3
22.4 17.6
l}' and
5.8 = (- 21.440 +C) from which c = 5.8 + 21.440
= 27.240 n the chain line, the inegul at bonnda
or ·Calculate the area in sq. yards included betwee (U.P.)
sq. in. the first and the last offset by Simpson's rule.
Area of zero circle= MC = 27.240 x 10 = 272.40 eter with the anchor point outside the fisure
between the tracing point and the 3. The area of a fisure was measored by a planim
Example 12.12. The length of the tracing arm diameter and the tracing arm set 10 the natural scale (M = 100
sq. em ). The initial reading was 8.628 and final
from the hinge is 22.6 em. The mark once in the clockwise direction.
hinge is 16.6 em. The distance of the anchor point and the reading was 1.238. The zero mark of the disc passed the index
being placed between the hinge
of the rim of the wheel is I. 92 em, the wheel area of Calcolate the area of the figure.
the hing• is 1.68 em. Find the revolutions in the clockwise direction
tracing point. The distance of the wheel from 4. The roller of a planimeter recorded a reading ofx 1.2606 em with the anchor point outside. With
wheel and area of the zero circle. while the measuring area of a rectangular plot 21
em
one revolution of the measuring and th• anchor point ootside , another fisure was traversed and
Solution. the same setting of the tracing arm ise direction. Find the area of the fisure
the reading recorded was 2.986 revolutions in thO .
clockw
(I) Area of one revolution of the measuring
wheel = M scale of 1 em = 20 metres
if it is drawn to a
= Length of tracing arm x Circumference
of the wheel
 II'II
314

S. A figure is traversed clockwise with !he anchor point

so set tb.a1 0110 revolulion of !he roller measure s 10 sq. incbes on !he paper.
SURVEYING

inside and with the ttacing arm @]

I.R.= 3.009 ; F.R.= 8.S47 respectively.
in !he reverse direction. The
The zero mark of !he disc bas passed !he index mark once area of !he figure? (U.I')
found 10 be 164.31 sq. inches. Wbat is the
area of !he zero circle is
. 6. Wbat is meant by zero cin:le 1 Describe !he various
methods of determi ning
on of the roller correspo
its
nds
area.
10 Measurement of Volume
The uacing arm of a planimeter is so set that one revolutioutside and then with the anchor
point
sq. in. A figure is traver>ed clockwise. first with !he anchor
are 2.342 and -9.319 respectively.
point inside. The observed differences in planimeter readings (U.P.)
Find the area of !he zero cirt:le.
7. Describe !he polal: pla!iimeter and explain its principle.
The perimeter of a figure is traversed 13.1. GENERAL
arm so set that one revolution of the roller the volume. "flleY. are
cloc:kwise, with the anchor point inside and with tracing readings are 3.009 and 8.S47 respectively.
The initial and final There are rhree melhods generally adopted for measuring
measures 10 sq. inches on the paper. ·
otx:e in !he reverse direCtion. The From cross-sections
The zero mark of !he disc bas passed !he fixed index markWllat is !he area of !he figure 1
(I)
area of !he zero circle is found 10 be 164.31 sq. inches. (il) From spot levels
(iii) From contours
tion of earth work while
ANSWERS . The first two methods are commonly used for lhe calcula
of reservo ir capacities.
::.(~: . the third method is generally adopted for lhe calculation
2. 820.38 sq. yds. 13.2. MEASUREMENT FROM CROSS-SECTIONS
divided into a series
3. 261 sq. em. This is the most widely used method. The total volume is
solids on which measurement
4. 11.944 hectares of solids by the planes of cross-sections. The fundamental
of lhe sections depends upon
S. 119.69 sq. inches. is based are the prism, wedge and prismoid. The spacing The area of
d in the measur ement.
6. 116.6 sq. inches. lhe character of the gro.uod and the accuracy require
7. 119.7 sq. inches. lhe cross-section taken along lhe line

~~
are first calculated by standard formulae
developed below, and the volumes of
the prismoids between successive· cross-
sections are then calculated by either (a) (b)
trapezoidal formula or · by prismoidal
formula. ---..--------------r
~
, .. ~'
The various cross-sections may
/ --
be classed as
(I) Level section,
>(
---- Cirtling
(Figs. 13.1 a and 13.2) •" FUiing
(2) Two-level section, (C) (d)

(Fig. 13.1 b and 13.3)

(3) Side hill two-level section, ----------------------··
(Fig. 13.1 c and 13.4)
(4) Three-level section,
(Figs. 13.1 d and 13.5) (e)
and (5) Multi-levei section.
(Fig. 13.1 e and 13.6) FI!J. 13.1.
(315)