,/

,rn fhe '-'/Ocess, ?hpr

ffilrurd l

'to

l

Ith )

t'2; j*fi,,:f*:r*#ffiff,::rii

,r' where

-""*"" *1I;;Tiil'ifl""l'ffi,IlTi*31:.;,tr"Hr.d

?r ;:::""",::T,ll:*.,u " ^,,*

,,,',11.':, ".,." "-i#tlii;i*r*t'i'ffie2,14.2 uor.,o^,,

Th. distrr^-- --

."uor, for sro;sre*". rill._.*easured r;::*"poinr,. #::"#: r,o.i.orffi#. ,,.^^ .

il'*r', it .nu", ', r1e dishn":T'-il'J'#''r?i"oo..poraioe,;:X.ff,;1nffiffi;:

,---1D = {}l-;2, The srope correcrion, ? :::1 i .r, t- l,

'

=; - !L. - n1'" Fig. 2.2< c

:i-it -!l,rt'" -5 224

= !"_ ,; ;ii,i\;J,,:,,, _ ,

wftere z = dirrerenc "?r,l.l",ilrn"el';fi"c

;:"0',';;,.0.. ,"

;1*:;x*;{i;ffi1 - "ruerrerrns)

ff***fi*;#n***r"**

52 Surueying

co=P-ur!9 twhere Po = standard pull

P - pull applied during measurement

A = areaof cross-section of the tape (in cm2)

E = modulus of elasticity of taPe

= 2.1 x lOs N/mm2 for steel

= 1.54 x 10s N/mm2 for invar

I = measured length (in m)

Tension correction is positive, if the applied pull is more than the standard pull,

and negative, if the applied pull is less than the standard pull'

2.14.4 Correction for TemPerafure

Thetapelengthchangesduetochangesinthetemperaturewhiletakingthemeasurements. The teriperature correction C, which, therefore, needs to be made

is given by:

Ct=q(T^-TOLwhere 7* = mean temperature during measurement

Io = temperature of standardisation

a = coefficient of thermal expansion of material

= 0.0000035/oC for steel

= 0.0000001221"C fot invat

L = measured length (in m)

Thecorrectionispositive,ifthetemperatureduringmeasurementismorethanthe standard temperature, and negative, if the temperature during measurement is

less than the temperature at which the tape was standardized'

2.14.5 Sag Correction

when the tape is stretched between two points, it takes the form of catenary

(assumed to be a parabola). consequently, the measured length is more and the

correction is applied. The sag correction Cru is given by:

c*--@:!J =y*24P' 24P'

IftherearezequalSpanspertapelength,thecorrectionpertaPelengthisgivenby

co=nry# =vrl^:t:t'24P' 24P'

(wt)zt tft=-*,= -----;--24n"F 24n'P

Horizonht, &asa,sientwhere ra, = weigh,t of

_the tape per metre length

W = total weight of the tape - --"r

p = pull applied (in N)

,r_ = *u length of tape suspended betv= Iength ir m" op" = n/1 (in m)

veen two supports

Sag correction is always negative.

2.74.6 Reduction to Mean Sea LevelThe length of a line measured at an altitud e of hmetres above mean *"1,I"r."l Gt. ;.rfi alwaysmore as compared to- the length;;;.J on themean sea level (m.s.l.) surfaJe. ft " ,"lJrrirv ,rreducing distances to.o"'o.i"y* ui"'-to # ii:ry:-lii:HiH :li:The correction denoted O, i_ *;;; ;;l-,

c*=ftrthere R is the radius of the earth.The correction is always subtractive.

2.14.7 Combining CorrectionsIn actual practice,

ii:l "f the above correction, based on the length recorded,are combined bv addition. nut rtri"try -ri"rting,

"*"rr, ,n"r1o be combined:l,i::HffiTilil:'ta*'""h1Ji{'JJ;:'il;x?jt1,:l::i,,rrer.,owing-

Unit sag correction = cUnit slope correction = D

Unit temperature correction = cTlue length = Z

Recorded length _ /-,Then L=Ii (t + a) (t + b)(t + c)

;HHILj,.:#f ;ilff ,T#,Il";ff ;,:i;J:,-ffi Zl{"u?"***oL=Ll(l+a+b+c)

[*#:l " the correctio, ;t; {:":r'"i,1"t,1,*,n recorded,and combined

2.74.8 Normal TensionThe pull or tension whic.tr, when applied to a tape suspended in the air, equarisesthe correction due to pull and ,", ,ffir#j, no.rnut rension.

IIil-rl

RI

1

JFig.2.25

54 Surveying

For one tape length,

q=P=p r

and c*= Yl24P2

Since, Cp = C.u

(P-Po)l _Wzl--AE - ,4?

The value of P may be calculated by trial and error'

!::rd:.::..:.'.a,..,i::.i.:::.. A line was measured with a steel tape which was exactly 30 m$&6n''pte "'o at a temperature o{ 2o'c and a pull of 10 kg. The measured

length was 1650 m. The temperature during measurement was 3O'C and the pull applied

*"J ts kg. Find the true length of the line, if the cross-sectional area of the^tape was

0.025 cmZ. The coefficient ofLxpansion of the material of the tape per'C is 3'5 x 10-and modulus of elasticity ol the material of tape is 2'1 x 10o kg/cm''

Solution Correction to be applied are: (1) correction for temperature and (2)

correction for pull.

(1) Correction for temPerature:

'' =i!':;?',:'- 20) 1650

= 0.05775 m (+ve)

(2) Correction for pull:

c,=9;{l r(ls - 10)

0.025 x 2.1x 106

= 0.157101 m (+ve)

True length of the line = 1650 + 0.O5775 + 0.157101

= 1650.2'l'4831 m'

The downhill end ol a 30 m tape is held 90 cm too low' What is

the horizontal distance measured?

Solution

Correction for sloPe = hzlzL

= Q.il2Tz x 30)

= 0.0135 m

Hence, the horizontal distance = 30 - 0.0135

=29.9865 m

1650

tape ends.

Horizontal

^leasurement Ss

lJ'^? Il,iT:,;;i:ii,"Jf3"ilT:": ,L: :ro" unoer a purr ol;'200Nft rei,,,eishiJil;;EtTH",L,ff f [i"":?,lI:,#lg

Solution

correction for sag = w'lr=

z+F

= !(30),

x 1oo)/(24 x eoo)z)= 0.09375 m

Horizontal distance = 100 _ 0.09375

2.75boundaries.

"rfu.rr..^li rr6u. ur rerr of the chain Ii

types of offc.ro. /1r n- ' as.shown in Fig. 2.26 are,tvpes of offsers: frl p".p"nii"ui#'"ill"1lt;#6 areca,ed offsets. Th". ur" t*o(z) oblique offsets.

Perpendicular offset

B

FU. 2.26 Offsets

2.75.7 perpendicular Offsets

When the Iareral me:rine, rhe "'r;;;#::W:';:";r:::i::r,ri::::rs are ar right angres to rhe chajn

2.75.2 Ohligue OffsetsWhen the lateral measrine, the ffi;'#.J;:3':;;:;"W:.iil:"* nor at right angres ro the chainor perpendicutar offsets and ro i;;#il ;hese

are used ro .n.It tt" ,""r.*,:ners of buildings more

_2'1s.8 Limiting Length of offset

-:^r...16r t,ure accurately'

The offset should notwiil be *o;;;;":r

be loo long. otherw,r.l r.l" error produced by rong offsetsur' ,."r.u"y a"s*a]r,plPer' The limiting length depenil;;;;r;'factors,

such,r,rr'";;i;"'# #ii"""J.:fj:':s, maximum ".."' inl"yi,,g

"1, *, dire*ionEffict of emor d.ue t

rras oeen"lala'irr,lr'0" l? !:"",r"n Let p btr PD on, ln",i'i,,.T; ffiT,,.Tffi Xiil:T,,.j5.7

= 99.90A5 m.

fi Surueying

/ and a is the error rnade in laying the offsetpaper will be CP, in.t"ua';;7;'i":" r"'"'j::: ground' This, whea plotted oni'a."rr,g ,i; ;lr"#"d of CP Gie' 2'27)'.T,tri' ,r,. priri ilJiirpr*"a uyof ptoning is I cm = ,t,,lno

bv PrP, perpendicula. ,r;J;il; rclrr rhe scate

,o _lsina".tll]imir of accuracy in plorring =0.025 cm

fu$" = o'025

or l=O.O25Scoseca

-^Therefore, the limiting length ofoffset is.0.025 S cosecafbr an error ofainlayingthe offset and the disptace_ nI_nr^ol poinr perpendicutar .;;;;Irne Prp, on paper will be

PPr=1Qppr=tEF,p,

= "E(ttr)

=",lil sin aCorresponding displacernent on paper n

a,so :r;,:1",::: r..ig ?28

lt^jyi, of accuracy in ptorting is0.025 cm, then

tilttrl (1/S) = g.s25or t=(0.025/.,!T)rSalso .,1/ sin a/S = 0.025or t=e.025/o)Scoseca

Fig.2.27 Error in direction of offset

Error in directiort anct length afoffset

_ /_"T:,r"B

PrPz - Cpt - Cp, = 1- / cpl a

Combined effect of :r::r in direction and length Let p be a point fromwhere an orfset pc is raid..on ;;;',;;; ;; rn. in"urur"J rrrr,r.. .r rhe ofr.setcP, is / and the error in dir""ri"r;;';'ii$.\.rr.,.il;;;",iil

o""o.u"r,,;:;:i:"ff|t or the orfset and '"tr" "r orr"u;rg u" r cm = s ,,. ir,", rhe roral

Assuming angle pp,p, _ 96.The_ displacement due to angularerto\- Prp, = / sin a and it ufroriO Joequal to displacement drle to the linearerror, prp - l/r

also

4P

--tiFr l- ,/' lPtt /,,I /rt/r| /, I

I l',' I

ta/' I

t-.-/ i

_/_ 1 ,--fi,ain rine

c n- --- -B

2.15.4 Taking OffsetsHorizontal lleasurement Sl

t, ".f a point where rhe offset is ro beok:n ^(Fre

2.29). Ther"ua", iorll" ;";";end of the rape at the point p "rd;;;;l:Iow_er swings off the tape in an arc with p Tape

3,1[ lentre,-as shown in Fig.-izi.-it"3:::T11n_'Tging oj tape on trre crrain riJ \

2.15.5 Establishing perpendicular to a Chain UneThe most common m

:_+s *",r,"i:.ffi:J fi:Ti:dffiH:"Lffffi:1il ff Ji":;,lffi # Jl;c is the point from which a p".'p"ra"i"ri#,1'r" o" erected. erruurli'u poinr E onthe chain AB at 3 rinks, from i. ff;

"trin.rint, ,rrr?""ri'i"i,n r u, ,t"i:jl[ "T,fl[ Hil:;liP ti"'o' ti"',],.,, I'in 4 chain rink rrori i and with ccP is the desired ,d::fiil;".inrersecting

the previous -" ,n"rir".onsrrucred.

flffi:t:.::':,,^",r,Jr""i"i:i";:l#: - - V. - L _ .,,*[__chain rinedicutar from p. a, "n"ii"ill;Tff"; "______7way is also called swing offset. mg. z.2g Taking offsets

;*ChaintineB

A-

t-rg. 2.80 Establishing perpenclicular to a chain lineExamptc 2.9 An offset is laid 4" out fr;ffiil# ji ; rhl";rx: *imu{*{i#I;ir,;, ;m* d{;(i) tn rhe direction

(ii) ln the directon f:9,to the chain line'

Solution -.._ snEU,vr perpendicular to the chain line.

(i) Displacement of the point on ground parallel to chain line=/sina=8sin4o= 0.558 cmThe scale of plan is 6 m to I cm

Displacement of the point on plan = / stn a

=EIq = o'oe3 cm

58 Surueying

(ii) Displacement of the point on ground perpendicular to the chain rine

=/(l_cosa)=8(l_cos4o)= 0.01948 cm

Displacement of the point on plan - /(1 - c.os a)s

= o'019+8

6

= 0.003 cm

,',"fr,ry4 ;g,1,;j,,r1i{,11,.. Find.tl.re lim.iting length of an offser so that the disptacement of

raid 3. our rrom irs u," i,lJ"lii:ff,i:tSJ#"1,#',,"Jff:: ? ?'f. "'' rhe orrset was

Solution

I sin alS = 0.025or l=O.O25Scosec3o

= 0.025 x 10 x cosec 3o

= 4.78 m

-..gt . Fing the.maximum length of an otfset so that the displacement

sources or error oo". n:l :["31"t:3,I:;l"i#'"ll5, i:'*#,ffi fl?f, *,,,:Liiilof 1 in 30 and the scale used is i cm = 25 m.solution The dispracement of point on the ground from both sources of error

=arr=Et ^^_30..

The scale of plotting is 25 m = 1 cm

The displacemenr of point on paper - ;g+ "^rhen

*" = 0.025

or l=13.259m.Hence, the maximum length of offset should be 13.25g m.r., ]ff"iTj:"tiT:'J: g3s",[J:,"j,"jl,Ti:1f:i:"""J::,"Lil

which the offset shourd be measured so that the maximum displacement of point onpaper from both the sources of error be the same.Solution

I sin a=!or I sin 5" = llrOf r=COSSCS"

= 11.47

Hence, the offset must be measured with an accuracy of about r in rr.47 m.

Hoizontal Measurement 59

r;::t,r. 2.76

The chain survey work is recorded in a book known as field book. It is of200 mm x 120 mm size. The pages of the field book can have either a red lineor two blue lines 12.5 - 15 mm apart ruled down the middle of each page. Thefield work is commenced at the bottom of the page and worked upwards as shownin Fig. 2.31. Following details are recorded in the field book on the commence-ment of a chain line.

Fig.2.31

1. Name or number of the chain line.2. Name or number of the station.3. The symbol denoting the station mark.4. The direction of other survey lines at the stations.5. The initial chainage (generally zero) enclosed in the symbol.

?/aA To facilitate reference, no line is commenced on the same page that contains a finish ofanother Iine.

Booking the DataIn recording measurements as they are being taken in the field, the notekeepershould ensure that every measurement that should be recorded is taken andthat every measurement taken is recorded. Every linear measurement should berecorded in such a way that the last digit will indicate the degree of precisionwith which the measurement was made. The following points shiuld be observedwhile booking the data.

The measurement of distances consists of chaining and making offsets. Duringmeasurements, it is practically impossible to set out all the chain lines in astraightforward method because of a variety of obstructions to chaini'g andranging in the field. The difficulties can be overcome by running perpendicularand parallel lines or by running a few additional lines and measiring angles bysome instrument. The scope of the chapter limits the solution of the problemsinvolving only the essential equipments used in chain surveying. To find the bestand rapid solution, the surveyor should have a good knowledge- of geometric andtrigonometric principles.

The obstacies may be <tivided into trvo classes. Those which do not obstructthe ranging (view) like ponds, river:s and falr in the category of obstacres to tnea-surement. The others are those which we cannot see across, i.e. both the chainingand ranging are obstructed, e.g. houses, stacks, etc., and are known as obstacle.tto alignment only a few solutions have been discussed here and many more canbe developed by the surveyor himself, depending upon the fielct conditions andmethod resorted to.

N Surveying

1. Chainage is written in the central colurnn.2. chaiuage of the stations may be enclosed in a circle or ellipse.3. Objects are sketched along the sides of the chain line.4. Offsets are written close to the object.5. Chainage of corners of the object are also eiltered.6. when features like road, fence, lake, etc. cross the chain line, the chainage

of intersection is entered and the direction of the feature is sketched.7. oblique offsets are written along with dimension line in the direction of

the offset.

2.17

2.17.1 Obstacles to Measurement

First method (F;g. 2"32) Let ABCD be a chain lineobstructed by a pond. The problem consists in findingout the distance BC. Two ofTsets BE a:nd CF of equallength are made at B and C and chaining is done alongEF. The work is then continued from point C.

Seconrl method (Fig" Z.SS) Let DAB be a chainline obstructed by a river. Lay offAC of any convenientlength, perpendicular to the requirecl distance,4g and layoff DC perpendicular to BC. Then, AB = ,qdUO.Third method (Fig. Z.S4) Let AB be a chain lineobstrucred by a river. Assume a point 1 anywhere in linewith the required distance AB. Take a point Il in such away that HJ = HI and HK = I1B. Establish t in the iine

Di

icr------ r F

-

li:'-....-l-- I\--_\\ ;

\\.\\\ I

\)11lt t

.,t/,4/ / I

;Z)t/t/ i:1 IBl -----.-)E

I

AI

Fig.2.32

AH and at the same time in the line JI( produced. Then KL = AB.

Hoizontal Measurement 6l

Fis. 2.33 Fig. 2.34

\,H'r---7'G

\,/\,,"/,\/AF

Fig.2.ts6

2.17.2 Otrstacles to NignmentLet DE be a chain line obstructed by a house.

First mcthod (Fig. z.ss) Assume a point c arbitrarily. Make EC = cB and.DC = CA. Then AB = DE.

second, method, (Fig. 2.s6) Estabrish a point F at equal distances from D andE at any convenient distance. Make F/1 = rG. Th"n D:E = (HG x DF)/HF.

/fih', il'lllVv/ *,- E

t.., ,'

)'(0

ExawpLe Z..tS , A chain line ABC crosses a river at 90. as shown in Fig.2.S7.- B an!_C are two points located at the near anO tar Oants,respectively. AB = 57.79 m, BD =100 m and LABD = 90". The *11olu "ii"r"

magneticbearing (w.c.B.) of c and A taken al D are 30" and 120., respectir"rv. rino the widthof the river.

Solution

V/.C.B. of C at D = 30o

WC.B. of AatD=!2Oo

Fig.2..35

ZADC=120-30o=90"

62 SurueYing

Fig.2'37

Consider triangles BCD arrd BDA

ZCBD=ZABD=90"IBCD = ZBDA, arrd Z-BDC = IBAD

Hence, triangles BCD and BDA ate similar triangles'

BC _BDED_AN

or BC =%- =W = ti3.22 m

Hence, width of river = 173'22 m'

A big pond obstructs the chain line ab as shown in Fig' 2'38'

Frunpl,e 2.74 ., ; il; li*u, ,"""rred on the lett of the line ab for circumvent-

ing the obstacle. Th" b;;il';i'i *ut sor .m' Similarly' the line am. was measured on

the right of the line uO *"not" length was 1'100 m' Points m' b and / are in the same

straight line. Lengths "r'in" ri"r" oland bm are 502 m and 548 m, respectively. Find

the distance ab.

Solution Consider tr\angle alm

Let llma = a

cos d=

5O2m b 548m

Fig.2.3B

'(1100)2 + (1050)2 - (901;2=ffi= 0.6497

In triangle abm

(ab)z = (bm)z + (a*)2 - 2 (am) (bm) cos a

= (548)2 + (1100)2 - 2 x 1100 x 548 x 0'6497

or ab = 852-b6 m-

(am)z+(lm)2-@ltZ(am)(lm)

Horizontal Measurement 63

i..$,|#r,, 1t:t::l tiejef cros3ing. a lver at risht ansres cur ts banks;1;ffiffi;;ffiJ'":"fl*:i"1" *:yr'i l'n i t, i" ##Yff # ;tif?[following operation was carried out.A 60 m long line BE wasD and mid-noinr F nr ^"::::T::ghty

paraltel ro the river. Line CE was extendedp:=Tp lTll-T*i^,:"":_r^r-:L_"$ll;d"I rffi;.li,i,ji,T;,!lj.;i"-a'""ff,,i""1FG .= EF. Line DG was extenoeJt" .riii"

"r"vey line ABC at H. GH and HB were;";;;;and found to be 40 m and g0 ,, ,""p;;;;il"

Find the width of the river.Solution

Given BE =gQ m, BH = 80 m andHG =40 mGD=39=60m' HD=HG+GD

=40+60=100mConsider similar triangles CHD and CBE

CBICH =BE|HDor CBI(CB + BII) = BE\(HG + GD)i.e. CBI(CB + 80) = 6Ot(40 + 60) _ 0.6

CB =0.6 (CB + 80)CB (l - 0.6) = g6 1 6.6

CB=39x0.6/0.4

=4910.4 = 120 m

_6- ;ll""HT;_"":iTll'::1:,r:l an obstacre, a 200 m rons

"::, :i#1"" ::',::i l^?* :il;i ; ^

t, "

"i'J',I"#; il::!f, ?"?;:if,,e":y*"11";:f1^u^olg,:'li:,i"""'J"1iil:tri?"f,?""?f "'ffi 'f,Tfl ff

X;:'ffi jff ff iff :1""?:,"?;,^*^",Fi;#;ffi#;;,#.'j:lllffi .lI1,?,;L'$:.L:,IAlso determine the obstiucted length.Solution Refer to Fig. 2.40.From ABCD,

CD = gg sec 60"

= 200 x sec 60"

=400mFrom A,BCE, CE = BC sec 45o

= 200 x sec 45o

= 292.94 mand BC = BC tan 45o

200 x tan 45.

= 200 m.

Fig.2.39

fifuh[lf{ffi!::: i{ ifii! lir][fi;rri;l;:n s:

Fig.2.40

CE, are set our at angleJ irr io; ""JiirJ;,i, ;;

64 Suruaying

respectively. Determine the lengrths cD and cE so that.o and E may be on the prolon_gation of A8. lf the chainage ot a is 1OO m; find the chainage ot O. -

Solution From ABCD (Fig.2.4l)

cD =*3€06 =#s.= 173.205 m

From ABCE

ce=#" =#0. = 195'81 m

BP=lCDz-BC|z= [(t73.20r2 _ $50)2)yz=86.602 m

Chainage of D=chainage of B+BD= 100 + 86.602

= 186.602 m.

i:t; A- river is flowing from west to east. For determining fhe widthii:'dl!* ^, rha:n,^. &.r^ -^:-r-:ai,.:.1:i:::'*;1::,rr...ii:.Tiii:rl,r!t.Irfi!i.i Of the fiVef, tWO pgints A and B afe SeleCted On the S'Uthefn::ll.:1"1j:l_0p11?."9.18: 100 m. eginr {is west wards. rhe bearinss ar a tree con the northern bank are observed to be 40. ano sao",-iesp";tir"ry ;i;," A and B.Calculate the width of the river.

Solution In A,ABC (Fig.2.a2)

ICAB=90o-40o=50oICBA = 340'- 270" =70oZACB = tB}" - (ICAB + Z.CBA)

= l80o _ (50. + 70.) - 60.By sine rule

#fu=#=#fu

Fig.2.47

F---loom___+jFis.2.42

AC ='l4ii!Jq: - loo x 0.e396srn 60" 0.966

= 108.50 m

and BC = 44ii4,iq - 76.60srn 60" 0.966

= 88.455 mWidth of the river = AC sin 50o

= 108.50 sin 50o

= 83.116 mwidth of the river = BC sin 70o

= 88.455 x sin 70o

= 83.116 m.Width of the river = 83.116 m.

Hotizantal filtaeurement 6

2.1 Explain how a chain is tested and adjusted in the field.2.2 Bnefly describe the process of chaining.2.3 Describe the various methods of chaining on a slope along with their

advantages and disadvantages.

2.4 Describe the following with sketches:

(i) Line ranger (ii) Optical square

(iii) Prism square(iv) Clinometer2.5 Differentiate betwean the following terms:

(i) Base line and check line(ii) Main station and tie station(iii) Chainage and offset(iv) Cumulative and compensating errors

2.6 Explain the following terms: normal tension, hypotenusal allowance,cumulative error, and ranging.

2i Explain the various sources and nature of errors in chain survey.

2.8 Describe the various tape corrections with sketches.

2.9 The area of a plan of an old map plotted to a scale of 10 m to 1 cm mea-sures 100.2 cm2 as measured by a planimeter. The plan is found to haveshrunk so that line originally l0 cm long now measures 9i cm. Further,the 20 m chain used was 8 em too short. Find the true area of survey.

fAn* 105.6438 acresl2.10 An area actually measures 0.8094 hectares. How much will it measure

in m2 by a 30.48 m chain which was 20.32 cm too short at the start and60.96 cm too long at the end of the survey? lAns.7987.L5 mzl

2.ll The area of a piece of a land which had been surveyed with a chainwas calculated to be 9562 m2. Of this, 8935 m2 was the total area of thetriangles and 627 m2 was the area included between chain lines and theboundary. The 30 m chain used was found 0.05 m too long, and the 30m tape used for measuring offsets was found 0.03 m too short from theirnominal lengths. Calculate the correct area of the land.

[Ans. 9590.5 m2]2.12 A line was measured with a 30 m long steel tape, standardized at 15oC,

with a pull of 100 N. Find the correction per tape length, if the temperatureat the time of measurement was 20'C and the pull exerted was 160 N.Weight of I cm3 of steel

Weight of tape= 0.0786 N

=8NModulus of elasticity = 2.lO x 105 N/mm2Coefficient of expansion of tape/oC = 7.1 x l}a [Ans. 0.49 mm]

213 The distance between two points P and Q measured along a slope is250 m. Find the horizontal distance between P and Q, if(a) the angle of slope is 10",

(b) the slope is I in 4.5, and

66 Surueying

(c) the difference in elevation is 35 m.

fAns. 246.20 m.244.04 m.247.53 m]2.14 Calculate up to five decimar praces, the sag correction for a r00 m tape

weighing 13.0 N. It is used under a pul of 90 N and in four equal spansof 25 m each. [Azs. 0.00543i m]

2.15 To what precision would you measure the offsets, if the plan of the surveyis to be plorted to a scale of (i) 1 cm = 1m and (ii) 1 cm = 10 m.

/ [Ans.2.5 cm, 25 cm]

' rzed under no pull at 15'c. The tape has cross-sectional area of 3.75 mm2.If one-half of the line is measured at temperature of 20oC and the otherhalf at 26oC and the tape is stretched with apul of 22kg,find the cor-rected total length, given that the coefficient of expansion is tz x 10{ peroC weight of 1 cm3 of steel = 7.7504 g and E = 2.ll x t06 tgl"-r. '

[Ans. 2000.384 m]2.17 a-steeltape is 30 m rong between the end graduations at a temperature of

15"c when it is laid horizontally on the ground. Its sectional area equals0.065 cm2, total weighr is 15.g N and the coefficient of expansion being11.5 x 10-6 per oC. The tape is stretched on two supports 30 m apartand is also supported in the middle, the three supports being at the samelevel. calculate the actual length between tt e ena graduatiJns under thefollowing conditions: temperature = 25"C, pulr on tL" tup" = 100 N, andE = 2.It x 105 N/mm2. fAns. 29.991 m]

2.18 A base line measured with a steel tape gives an approximate length of1000 m. Compute the correct length of the base line at mean sea leverwhen the pull at the standardization equals 15 kg.,The applied pull is23 kg.^The cross-sectionar area of tape is 0.0645 cm2 and z = 2.ri x 106kg/cm2. Temperatures e and To aie 35.C and 15.C, respectively. Thedifference in level of the two ends of base line is 2 m. Radius of earth, R= 6400 km. Elevation of base line above mean sea level = 1000 m.

2.t9 Thethree bays of a base line were measured o, , uiSr'#if :l::#as 30.084, 29-973 and 25.233 m, under respective pults of z7 and 5 kg,temperatures of 12, 13o and 17'C and differences of level of supports of0.3' 0.7 and 0.7 m. If the tape was standardized on the flat at a temperatureof l5'c under a pull of 4.5 kg, what are the lengths of the bays? 30 m oftape is exactly 1 kg with steel at g300 kg/m2. The coefficient of expansionis 0.000011 per oc and the modulus of elasticity E=2.L x rOs i.i/mm2.

[Ans. 30.057 m, 29.94O m and 25.194 m)2.20 A steel tape was exactly 30 m long at 1g"c when supported throughout it,s

length under a pull of 8 kg. A line was measured with a tape under a pullof 12 kg and found to be 1602 m. The mean temperature o*ing measure-ment was 26'C. Assuming the tape was supported at every 30 m, carcurate*"rT:

.l:ngtr- of the line, given cross-sectional *u oi the tape = 0.04cm'. weight of 1 cubic cm is 0-0077 kg, coefficient of thermar expansion,{z = 0.000012 C, E = 2.1 x Lff kglcmz. fAns. t6OZ.22 ml

Hoizontal Measurement 67

2.21 A line measured with a steer tape which was exactry 30 m at a tempera-ture of 20'c and a pull of 10 kg. The measured length was 1650 m. Thetemperature during measurement was 30oc and a pull applied was 15 kg.Find the true length of the line, if the cross sectional area of the tape was0.025 cmz. The coefficient of thermal expansion of the material of thetape per "c is 3.5 x 106 and the modulus of elasticity of the material oftape is 2.1 x 106 kglcm2. [Ans. 27.776g m)

2'22 A steel tape of 30 m nominal length was suspended between two sup-ports to measure the rength of a line. The measured length on a slope bf4o 25' is 29.8605 m. The mean temperature during the ireasurement was15oc and the pull applied was r20 N. If the standard rength of the rapeis 30.008 m at 27"c at the standard pull of 50 N, calcuhG the correctedhorizontal length- Take the weight of tape = 0.16 N/m" it's cross sectionalarea = 2.75 mm2, coefficient o1tfr".rrrut expansion = 1.2 x 10-;;;;;;and E = Z.O5 x 106 N/mm2. [Ans. 29.]7002 ml

2.23 The slope distance between two stations A and B of elevations 1572.25 mand 4260.46 m, corrected for meteororogical conditions is 33449.215 m.Determine sea level distance, R = 6370 km. fAns. 33332.7g9 m)

2.3

2.1 In chain surveying, field work is limited to(a) linear measurements only(b) angular measurements only(c) both linear and angular measurements(d) none of the above

2.2 The accuracy of measurement in chain surveying, does not depend upon(a) length of the offset(b) scale of the plotting(c) importance of the features(d) general layout of the chain lines

Chain survey is well adopted for(a) small surveys in open ground(b) small surveys with ups and downs(c) large area with simple details(d) large area with crowded details

In a metric chain, number of links per metre can be(a) 2 (b) s (c) 8 (d) 10

Cross-staff is used for(a) setting out right angles (b) measuring horizontal angles(c) both (a) and (b) (d) measuring the bearing oi lin",

In chain surveying, perpendiculars to the chain line are set oui ty(a) a theodolite @) a prismatic compass(c) a clinometer (d) an optical ,qru."

2.4

,<

2.6

68 Surueying

2.7 Ranging is defined as(a) measuring the distance f,rom starting point(b) establishing intermediate points on a chain line(c) the distance between end points(d) a point on chain line

2.8 Chainage in chain survey means(a) the distance between end stations(b) the perpendicular distance of the object from the chain rine(c) the distance of the object along the chain line from the zero end ofthe chain(d) any disrance measured by chain in field

2.9 Main stations in chain survey are the points(a) lying in the area enclosed by survey lines(b) connected by main sorrey lines(c) on main survey lines to cover the rocal details(d) on main survey lines to check the accuracy of the survey work2.10 Oblique offsets are used to(a) locate broken boundaries(b) locate boundary lines of property(c) check the accuracy of tne plottea work in chain survey(d) plot the chainage

2'll which of the following methods results in higher accuracy for measuringhorizontal distance on rough grounds:(a) chaining @) raping(c) tacheometr

z'rz wn**""rr.rlrorizonrar am*""(1],i,io:[]nt, h,ls, it is berter tomeasure the distance by(a) stepping down slope @) stepping up slope(c) both of the-above (dj suspending tape in air2.13 The maximum tolerances in aZOrn

"ra 30 m chain are(a) t2 mm tg

(c) t5 .n- *, *- (b) t: mm +5 mm

2il pickrp rh";";;;Tatementls). (d) t8 mm +5 mm

(i) A metric chain is 30 m long(ii) A metric chain is 20 m lon!(a) (i) is correct (b) Gi) is correct(c) both (i) and (ii) are corecr iaj uotr, are wrong2'15 Prolongation of a chain rine across anobstruction in cha-in survey is done

(a) making angular observations(b) drawing perpendiculars with a chain(c) solution of triangle(d) ail of the above (e) onty (a) and (b)

Hotizontal Measuremenf 69

2.16 while measuring a line between two stations A and B intervened by araised ground,

(a) the vision gets obstructed(b) the chaining gets obsrructed(c) vision and chaining both get obstructed(d) none of the above

2.17 A well-conditioned triangle should not have angles more than(a) 30' (b) 120. (c) 45' (d) 60"

2.18 Offsets are

(a) short measurements from the chain line(b) ties or check lines which are perpendicular to the chain line(c) chain lines which go out of alignment(d) both (a) and (b)

2.19 The correction to be applied to each 30 m chain length along slope is(a) 30 (1 - sec a) m (b) 30 (sec a- 1) m

,/ (c) 30 (1 - cos a) m (d) 30 (cos a _ l) mYo A-tape of length / and weight w N/m is suspended at its end with a pull

of P newtons. The required sag correction is1a) twztzupz 1a1 Pw3tz+r, @) p*ztz4pz (d) lw2t24p

2.21 Compensating or accidental errors are proportional to(u) L,,, (b) L1t3 @) L (d) uL

2.22 Check lines (proof lines) in chain surveying are essentially required(a) to plot the chain line(b) to plot the offsets(c) to indicate the correctness of the survey work(d) to increase the efficiency of the surveyor

2.23 Perpendicularity of an offset may be judged by eye if the length of theoffset is less than(a)5m (b)10m (c)15m (d)20m

2.24 which of the following instruments is generally used for base linemeasurement?

(a) Chain @) Metallic tape(c) Steel tape (d) Invar tape

2.25 Invar tape is made of an alloy of(a) copper and steel (b) brass and nickel(c) brass and steel (d) steel and nickel

2.26 The length of a chain is measured from the(a) centre of one handle to the centre of other handle(b) outside of one handle to the outside of other handle(c) outside of one handle to the inside of other handle(d) inside of one handle to the inside of other handle

2.27 The angle of intersection of the two plain mirrors of the optical squareis

(a) 30' (b) 45. (c) 60' (d) 90'

7O Surueying

2.28 which of the folrowing angres can be set out with the help of Frenchcross-staff?

(a) 45. only @) 90. only(c) borh of the above (d) any angle

2.29 lt the length of a chain is found to be short on testing, it can be adjustedby

(a) straightening the links(b) rgmoving one or more small circurar rings and by placing bigger

rlngs(c) closing the joints of the rings, if opened(d) all of the above

2.30 lt the length of a chain is found to beadjusted by

(a) closing the opened joints of rings(b) reshaping elongated links(c) removing one or more circular rings(d) replacing the worn out rings(e) all of the above

too long on testing, it can be

2.31 The position of a point can be fixed accurately by(a) cross-staff @) optical square(c) clinometer (d) perpendicular offset

2.32 Figure 2.43 shows one of the brass tallies of a 30 m chain. Thedistance of this tally from the nearest end of the chain is r^\

)"/1)

Fig. 2.43

2.36 During chaining arong a straight rine with a 20 mchain, the leader of theparty has 4 arrows in his hand while the foilower has 6. Distance of thefollower from the starting point is

(a) 4 chains (b) 6 chains (c) g chains (d) lZ chains2'37 The main difference between an optical square and a prism square is

(a) the difference in the principle of working(b) that an optical square is more accurate than a prism square

(a) 5m(c) 15 m

(b) 10 m(d) 20 m

2'33 which of the following is an obstacle to chaining but not toranging?

(a) River @) Hiilock(c) Building (d) None of the above

2.34 A building is an obstacle to(a) chaining but not to ranging(b) ranging but not ro chaining(c) both chaining and ranging(d) neither chaining nor ranging

(c) Optical square (d) Cross-staff

2.35 which of the forowing is not used in measuring perpendicular offsets?(a) Line ranger (b) Tape

Hoizontal Measurement 71

(c) that no adjustment is required in a prism square since the anglebetween the reflecting surfaces cannot be changed

(d) all of the above

2.38 The allowable length of an offset depends upon the(a) degree of accuracy required(b) method of setting out the perpendicular and nature of ground(c) scale of plotting(d) all of the above

2.39 Normal tension is that pull which(a) is used at the time of standardizing the tape(b) neutralizes the effect due to sag

(c) makes the correction due to sag equal to zero(d) makes the correction due to pull equal to zero

2.40 The correction for sag is(a) always additive(b) always subtractive(c) always zero(d) sometimes additive and sometimes subtractive

2.41 The permissible error in chaining for measurement with a chain on roughor hilly ground is

(a) 1 in 100 (b) 1 in 250 (c) I in 500 (d) 1 in 1000

2.42 Corrcction for slope is given by

61 nznr g) htL (c) htzL (O zh2lL2.43 The required slope correction for a length of 30 m, along a gradicnt of 1

in 20 is(a) 3.75 cm (b) 0.375 cm (c) 27.5 cm (d) 2.75 cm

2.44 lf the length of a chain line along a slope of a is l, the required slopecorrection is

(a) 2l cot2 a/2 (b) 2l sinz al2 (c) I tanz al2 (d) I cosz atT2.45 Match the following

(I) Correction for standard length (A) C" = LCll

(II) Correction for tension (B) Cp =' ;{o t(III) Correction for temperature (C) C, = u (T^ - T)L

@) q" = \f tl24 Pz

(E)Cn=LhlR(IV) Sag correction

(V) Reduction to m.s.l.

(VI) Normal tension

where .L =measured length,C = correction applied, and

/ = nominal length of tape

(a) I-A, II-B, III-C, IV-D, V-E, VI-F(b) I-C, II_D, IIT_B, IV_A. V_F, VT_E

O.2O4w,lAEtrtr=:{P-Po

72 Surueying

(c) I-B, II-A, III-F, IV-C, V-E, VI-D(d) none of the above

2.46 For setting out an offset at an angle of 45' with a chain line, the instru-ment used is

(a) an optical square (b) an open cross-staff(c) a French cross-staff (d) a prism square

2.47 A pair of slots at right angles to each other are provided in(a) cross-staff (b) arrow (c) ranging rod (d) offset rod

2.48 The limiting length of an offset does not depend upon(a) accuracy of the work(b) method of setting out perpendicular(c) scale of plotring(d) the number of features to be surveyed

2.49 Pick up the correct statement.(a) The cost of making a horizontal measurement decreases with an

increase in the desired precision.(b) A base line may be measured with a precision of 1 in 106(c) Tie stations are generally located on the intersection of two main

survey lines(d) Base line is a line lying at the base of the area to be surveyed by

a chain2.50 Pick up the correct statement(s).

(i) offset is the distance from the foot of an object to the chain line(ii) Perpendicular offsets may have infinite length(a) only (i) is correct (b) only (ii) is correct(c) both (i) and (ii) are correct (d) none of the above is correct

2.51 Pick up the correct statement.(a) A revenue chain is 66 ft long.(b) Gunter devised the invar tape.(c) A tally is used to facilitate observation in an optical square.(d) A brass ring is provided at every metre length in a metric chain.

2.52 While testing a chain, a tension of 80 N is applied at the ends of a(a) 20 m chain (b) 30 m chain(c) both (a) and (b) (d) Gunter,s chain

2.53 Pick up the correct statement.(a) Invar is an alloy of sreel (36Vo) and fickel (64Vo)(b) A steel tape is soft and easily deforms as compared to invar tape.(c) Metallic tape is made by weaving linen with brass wires.(d) Steel tapes can be used comfortably in grounds with weeds and

vegetation.

2.54 Pick up the correct statement.(a) A ranging rod is provided with a stout open ring recessed hook.(b) An offset cannot be laid with a French cross-staff.

Horizontal Measurement 73

(c) Optical square and cross-staff are used for the same purpose.(d) clinometer is used to measure the directions of survey lines in chain

survey.

2.55 Pick up the correct statement(s).' (i) In the process of chaining, the leader inserts the arrows and the

follower picks them up.(ii) A leader follows the instructions of the follower.a) only (i) is correct. (b) only (ii) is correct.

(c) both (i) and (ii) are correct. (d) none of the above is correct.Pick up the incorrect statement.

(a) It is easy to measure distance down the slopes.(b) Incorrect plumbing, while measuring distances on slopes, is a cumu-

lative error-(c) Sag correction is a cumulative error.(d) Incorrect holding of chain at arrow is a compensating error.

2.57 Pick up the correct statement.(a) Sag correction may be positive or negative.(b) The limiting length of an offset is independent of the scale of

plotting.(c) Error due to laying of the direction of offset is negligible.(d) The slope correction is always subtractive.

2.58 Pick up the correct statement(s).(i) The length of the offsets in a chain survey is always limited to

reduce error in plotted work.(ii) Perpendicular offsets are used for filling in details.(a) only (i) is correct. (b) only (ii) is correct.(c) both (i) and (ii) are correct. (d) none of the above is correcr.

2.59 Pick up the correct statement(s).(i) Optical square is better than a prism square.(ii) In both the optical and prism squares, the principle of operation is

same.

(a) only (i) is correct. (b) only (ii) is correct.(c) both (i) and (ii) are correct. (d) none of the above is correct.

2.60 which of the following is the most precise instrument for measuringhorizontal distances.

(a) Chain (b) Tape (c) Tacheometer (d) Tellurometer2.61 Two mirrors are used for offsetting in

(a) cross-staff(c) miner's dial

(b) optical square(d) prismitic compass

2.62 What is the angle between two plane mirrors of an optical square?(a) 30" (b) 60' (c) 45"

2.63 The object of chain and cross-staff survey is to1. locate the boundaries of an area.

2. plot the figure to a scale.

(d) 90.

74 Surueying

3. find the area of the plot.

4. find the reduced levels of the plot.

Which of the above statements is/are correct?

(a) 1,2,3 and4 (b) 1,2, and 3 (c) 1 and 2 (d) 4 alone

2.64 Which of the following instruments have both horizon glass and indexglass?

1. Optical square

3. Box sextant

Select the correct answer using the codes given below:(a) 2,3, and 4 (b) 1, 3, and 4 (c) 1 and 3 only (d) 2 and 4 only

2.65 What is the slope correction for a length of 30.0 m along a gradient of 1

in 2O?

(a) 3.75 cm (c) 0.375 cm (c) 37.5 m (d) 0.0375 cm

2.66 A 30 m metric chain is found to be 10 cm too short throughout a measure-

ment. If the distance measured is recorded as 300 m, what is the actualdistance?

(a) 300.1 m (b) 301.0 m (c) 299.0 m (d) 310.0 m

2.67 A 20 m chain was found to be 10 cm too long after chaining a distance' of 2000 m. If was found to be 18 cm too long at the end of the day's

work after chaining a total distance of 4000 m. What is the true distance

if the chain was correct before the commencement of the day's work?

(a) 3962 m (b) 4019 m (C) 3981 m (d) 4038 m

Answers to Objectiae-type Questions

2.t (a) 2.2 (d) 2.3 (a) 2.4 (b) 2.5 (a) 2.6 (d)

2.7 (b) 2.8 (c) 2.9 (b) 2.10 (b) Z.rt (c) 2.t2 (a)

2.r3 (c) 2.14 (c) z.ts (d) 2.t6 (a) 2.t7 (b) 2.18 (d)

2.19 (c) 2.20 (a) 2.21 (a) 2.22 (c) 2.23 (c) 2.24 (d)

2.2s (d) 2.26 (b) 2.27 (b) 2.28 (c) 2.29 (a) 2.30 (e)

2.31 (d) 2.32 (a) 2.33 (a) 2.34 (c) 2.35 (a) 2.36 (b)

2.37 (c) 2.38 (d) 2.39 (b) 2.40 (b) 2.4t (b) 2.42 (a)

2.43 (a) 2.44 (b) 2.45 (a) 2.46 {c) 2.47 (d) 2.48 (d)

249 (b) 2.50 (a) 2.51 (d) 2.52 (c) 2.53 (c) 2.54 (c)

2.55 (c) 2.s6 (b) 2.s7 (d) 2.58 (c) 2.s9 (b) 2.60 (d)

2.61 (b) 2.62 (c) 2.63 (b) 2.64 (a) 2.6s (d) 2.66 (c)

2.67 (b)

2. Line ranger4. Pedometer