BC Punmia SURVEYING Vol 1 - By EasyEngineering.net

Downloaded From : www.EasyEngineering.net 284 SURVEYING (3 a) I f lhe point P is outside lhe great _,.-·-·'\"7.1\"\\:·· •.•. circle, lhe position o f p' should be so chosen lhst lhe point e (got by lhe intersection of lhe two rays drawn to nearer points) is midway between lhe point p' and lhe ray to lhe most distsnt point (Fig. 11.19). ;• w(3 b) When P is outside lhe great triaogle i ,.but inside lhe great circle (say in one of lhe ,·\"-Great wsegments of great circle), lhe point p' must be , · clrole so chosen lhst lhe · ray to middle point may lie between p' and lhe point e which is lhe intersection wof lhe rays to lhe olher two extreme points (Fig. .ESpedal Cases ·~......... .. FIG. 11.20 The following are few rules for special cases: 1\\.20) a(4a) I f lhe positions of A, B, C aod P are such ~-P lies on or near lhe side sof AC of lhe great triaogle, lhe point p' must be so chosen lhst it is in between lhe ytwo parallel rays drawn to A aod C aod to lhe right (or to lhe same side of each of Enlhe rays) of each of lhe lhree rays to satisfy Rule 3 (Fig. 11.21). b c ~---~/ -/.p-••.••••••- ~--- _ _ _ _ _ _ _ __ _ _ J FIG. 11.21. FIG. 11.22 (4b) I f lhe point P (as in 4 a) lies on or near lhe prolonged line AC, lhe point p' must be chosen outside lhe parallel rays aod to lhe right of each · o f lhe lhree rays to satisfy bolh Rules 2 and 3 (Fig. 11.22). (4c) I f A, B and C happen to be in one straight line lhe great triaogle will be one straight line only and lhe great circle will be having abc as its arc lhe · radius of which is infinite. lhe middle point In such cases, lhe point p' must be so chosen lhst lhe rays drawn to is between lhe point p' aod lhe point e got by lhe intersection o f lhe rays to lhe extreme poiut (Fig. 11.23). (4d) I f lhe positions A, B, C aod P are such lhst P lies on lhe great circle, lhe point p' cannot be determined by lhree-point problem because lhree rays will intersect in one point even when lhe table is not at all oriented (Fig. 11.24). Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net 28S PLANE TABLB SURVEYING a' ;.· b .,,../ \" -,-c , ..--:-:.:;;- b,I·..........·~p2, '\\',, : ' Pt.c.:.-- I :'1 . . . . \\ '. II ....\"'..,.., I~.. \\ /,' \\ !,' \\ \"'.,., I C a.!\\.t..... .............. .........I'.\\.•,;.·~,.I'.I \\ \\~:;'~< ..................'·'P3' FIG. 11.23 FIG. i t . 2 4 11.9. TWO-POINT PROBLEM Statement. LoCIUion o f the posilion on the plan, of the station oicupitd by the plane table by means o f observlllions to two weU defined poinls whose positions have been previously plDtted on the plan. \" Let .us take two points A aod B, lhe plotted positions o f which are known. Let C be lhe point to be plotted. The whole problem is to. orient lhe table at C. eye judgment). Clamp lhe table. n (2) Keep lhe alidade at a and sight A. Draw lhe resector. Similarly, draw a resector gfrom b aod B to intersect lhe previous one in d. The position of d is lhus got, lhe degree o f accuracy o f which depends upon lhe approxintation lhst has been made in keeping iab parallel to AB. Transfer lhe point d to lhe ground aod drive a peg. n. (3) Keep lhe alidade at d and sight C. Draw lhe ray. Mark a point c, on lhe Proc:edure. Refer Fig. 11.25 ( I ) Choose ao auxiliary point D near C, to assist lhe orientation at C. Set lhe table at D in such a way lhst ab is approximately parallel to AB (eilher by compass or by eray by estimation to ·represent lhe distsnce DC. e(4) Shift the table to C, orient it (teutatively) by taking backsight to D aod centre .... rit with reference to c1• The orientation is, thus, t.he same as it was at D. iA ng.n~ B ' , ... ... ... ... ' -----;\\ ~ ~---:-----e--t-~~ '''' / / / / ......... / \\ \\' ... / / )< ..... ' '\\ .../ / ...... \\ 0 c; FIG. 11.25. TWO·POINf PROBLEM. Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net SURVEYING 286 (5) Keep the alidade pivoted at a and sight it 10 A, Draw the ray 10 intersecr with the previously drawn ray from D in c. Thus, c is the poinl representing the sration C, with reference ro the approximate orieoration made ar D. wline with ab' and ar a great disrance. Keeping the alidade along ab, rorare the rable till (6) Pivoting the alidade about c, sight B. Draw the ray ro iorersecr with the ray drawn from D to B in b'. Thus b' is the approximate represeoration of B with respecr P is bisected. Clamp the ·rable. The rable is thus correcdy oriented. · w(8) After having oriented.· ~/ lable as above, draw a resec10r from a 10 A and 10 the orieoration made at D. (7) The angle berween ab and ab' is the error in orientation and must be corrected wrable. for. In order that ab and ab' may coincide (or may become parallel) keep a pole P in It is 10 be noted here that unless the point P is chosen infinilely disr8nr, ab, and .ab' cannot be made parallel. Since the disrance of P from C is limited due 10 other considerations, Erwo-poinl problem does nor give much accurate resulls. ·,,/it the same time, more lBbour another from b to B, the inlersection· of which will give the position C occupied by the ais involved because the rable is also 10 be set on one more sration to assist the orienration. , ver/ s(I) Select an auxiliary point D y(making ba approximately parallel 10 BA). · EIf D is chosen in the line BA, orientation Alternative Solution of Two-point Problem (Fig. 11.26) n(2) With b as centre, sight B and .r near to B and orient the rable there by estimation r·~.·which a and b have been; previously plotted. Since the disraoce BD iS small, any small error in orientation will nor have appreciable can be done accurately. c draw a ray Bb. Measure the disraoce BD and plot the poinl d to the same scale 10 effect on the location of d' The dotte<J lines .i \\ I I I. show the first position of plane rable with \\ · i.I \\ \\· I .I .I I.j ' \\. I .I I .I '\\ I.I I.I.. .\\ \\ . approximare orienration. (3) Keep the alidade along da and rorare the rable .10 sight A, for orientation. Clamp the rable. The finn lines show the second position with correct orientation. (4) With d as centre, draw a ray '·, ,. ~B ( \\ towards C, the point to be acmally occupied --- ------·\"' by the plane rable. • .\\I . :I'I , ..\"'/ ' ___..,_...-,,.,.\\ .\\ (S) Shift the rable 10 C and orieoc ,\" .,..../ it by backsighting 10 D. D '· b w•• •• <:,::.. ..-.•...... (6) Draw a ray 10 A through a, in- ~~_-.---a---- j'~ d rersecting the ray de in c. Check the orienration FIG. 11.26. TWQ-POINT PROBLEM. Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineel8r7ing.net . PLANE TABLE SURVEYING by sighting B through b. The ray Bb should pass through c i f the orientation is correct. 11 shor#d be ·noled lhol the twa-point resection and three point reseaion give bath an orientatidiJ. as weU as fixing. 11.10. ERRORS IN PLANE TABLING The degree o f precision 10 be attained in plane rabling depends upon the character of the survey, the quality o f the instrument, the system adopted and upon the degree 10 which accuracy is deliberately sacrificed for speed. The various sources of errors may be classified as : This includes 1. Instrumental Errors : Errors doe to bad qoality of the instrument. all errors descn'bed for theodolite, i f telescopic alidade is used. 2. Errors or plotting. 3. Error due to manipulation and sighting. These include (a) Non-horizontality of board. (b) Defective sighting. (c) Defective orientation. (d) Movement of board berween sights. (e) Defective or inaccurate centring. n points are sighted. The plain alidade with open sight is much inferior 10 the telescopic·., r (a) Non-horiVJII/IIlily o f board · The effect of non-horizontality of board is more severe when the difference in elevation galidade in the definition of the line of sight. heiWeeri the poinls sighted is more. i(c) Defecli¥e orien/111ion (b) Defecli¥e sighting The accuracy of plane rable mapping depends largely upon the precision with which nOrieoration done with compass is uoceliable, as there is every possibility of local attraction. eshould be ealready plotted. r(d) Movemerrl of board between sights iDue to carelessness of the observer, the rable may be disturbed between any two nsights resulting in the disrorbaoce of orientation. To reduce the possibility of such movement, the clamp should be finuly applied. It is always advisable 10 check the orienration at the Erroneous orieoration contribute 10wards dis10rtion of the survey. This orieoration checked at as mu.ny sta~nrLc: as possible hy g!ghring distant prominent objects gend of the observation from a sration. ..(e) nIt is very essential 10 have a proper conception of the etby inaccurate centring, as it avoids uooecessary waste of time Inaccurate centring extent of error introduced in setting up the rable by repeated trials. Let p be the plotted position of P (Fig. 11.27), while the position of exact centring should have been p', so that linear error in centring is ~ e ~ pp' and the angular error in centring is APB - apb ~ (a + ~). Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net 288 SURVEYING {,~ '\\;~>· w \\,_j \"'·..., ,.,;~/ ·........... Ray due to __..:\".'r~,,\"·...... /'1~./ ' ww¥ ,,/ \\ .Aaydua to wrong centring.correct centring'\\ '·, // ...\\ ...\\·.............. ,.,.,...·':,/,.' (P) ,./ 1·/ ,.........· Ep a ·...... p' _,_,. b •.<..~.. ' \\ / :' 9(,,/ aFIG. 11.27. ERROR DUE TO WRONG;•cBNTIUNG. sDrop perpendiculars from p' to yThen ENow :~ nLet us find out the error in the plotting of a and b. Let us say that a • and b • are the positions of A and B for correct centting. Then the error in the position of A ap and bp at f and g respectively. p ' f =AP sii!. a. p'g =BP sii!. ~ \".(I) siiJ. a. = ~P'I \"a.. as a. is sDjllll and siiJ. ~ = ~•~ \" ~. as ~ is small. and B will be aa • and bb • respectively. aaa• =pa. =pa. ~~ and bb' =pb. ~ =pb. ~~ \".(2) Let p ' f= p'g = e metres and s = fractional scale (R.F.) Then and Heoq: pa =PA. s pb =PB. s from (2), aa• = s. PA. ~;, s. e metres and bb' = s. PB. ; B = s. e metres. Hence, we find that the displacement of the points is es metres. If we take 0.25 mm as the unit of precision in plotting, e. s = aa' =bb' =0.00025 metre e= 0.00-0. -25 metre \".(3) °·Thus, we have got an expression that the value of e should be less than 00025 metre. s ! Centting must be performed with care in large scale work. For a scale of 1 em = metre, s = 1/100. Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net 289 PLANE TABLE SURVEYING ~ e= 0j~ =0.025 m =2.5 em which shows I that for large scale work (such as I em to I metre), the maximum value of e = 2.5 em only and centring should be done carefully. Let us take the case of small scale work also, such as I em = 20 metres. s =20-l0··-0 e = ~·= = 0.5 m = 50 em. This shows that for such small scales, we can have the position of the ground points within the limits of the board. Example 11.1. In setting up the plone table at a station P the corresponding point on the pion . was not accurately centred above P. If the displacement of P was 30 em in a direction at right angles to the ray, how much on the pion would be the consequent displacement of a point from its true position. if. (i<) scale is : I em = 2 metres. (i) scale is : I em = 100 m Solution. Ca<r (<) n 1.. g Case (u) in.. Oil'= e. s metres .. s=-1- Scale : I em= 100 m 10,000 oo' =e. s = 30 x ~ em = 0.03 m.m (negligible) e11.11. ADVANTAGES AND DISADVANTAGES OF PLANETABLlNG eAdnntages .. s =-1- r(I) The plan is drawn by the out-door surveyor himself while the country is before200 ihis eyes, and therefore, there is no possibility of omitting the necessary measureruents. n(2) The surveyor can compare plotted work with the actual features of the area. g(3) Since the area is in view. contour and irregular objects may be represented accurately. (4) Direct measurements may be almost .entirely dispensed with, as the linear and .nangular dimensions are both to be obtaiped by graphial means. (5) Notes of measurements are seldom required and the possibility of mislal<es in etbooking is eliminated. Scale : I em == 2 m ; oo'= e s = :~ = 1.5 mm (large). (6) It is particularly useful in magoetic areas where compass may not be used. (7) I t is simple aod hence cbesper than the theodolite or any other type of survey. (8) It is most suitable for small scale maps. (9) No great skill is required to produce a satisfactory map and the work may be entrUSted to a subordinate. Downloaded From : www.EasyEngineering.net

\"'Downloaded From : www.EasyEngineering.net SURVEYING 290 Disadvantages recorded, it is a great inconvenienc.:e if different scale. work . Since ootes of measurements are not is required to be reproduced to some very accurate tabling is not in!ended for The plane (I) the map (2) . (3) (4) (5) w(6) wlost. It is essentially a ttopical instrumen!. and in wet climate. It is most inconvenien! in rainy season transport, likelihood Due to heavyness, iE is inconvenient to w.EasyEnAwtholissuoldscda2((b543Ii((bbls...e.e.lic)))i,ui ()c(DDDsaaEDRaas)eei)cuRx.asesscspedfcectueiuhAlDirsrndraasiieeaeitbbiscnistoeeietcpnrio,utolwe.hawscnblnpriaswe.riteeothtniehiieslvtt-hhfirpetlrsieyisaonlokb:iglnmenahleettdttechecovobrebfnapsiehtfunrssseuotr,t.ahlavbspgeaeeltmneteyhhmsdpoeaeolfpafias(anvnfiasipoaeidv)nkflprtadloeiolrotdltiIahwiccuniobashesbsuithaltneeeeisodrsgopmw,vsnooceaiieccavmnnLttcreohthtireeearooofithsgneihdwssemddosotsbldra4eesiion5ooetunfssffotu'emsopfmroUoaavlraRfyoteilenowyoun.ecau.alibtaen't·isestn;gttip.asanogJcbtgsfiaamlooenllentleheavhtseeehuiodtposarfvpdovbelseiDbalnrIeyOyntt.iite:nc:agbtnrlyteaa5occbl0vehipln0seerlag0so.rn.auerog_pybtSa.htahpebotWerhlwr(eeAhmam.ssMtettuhetahra.tetvIhoitr.ooedErnydosa.?r...)t Since there are. so many accessories, there is every of these being PROBLEMS 4' m6. What is two-point problem ? How is it solved ? Bessel's method (i1) Triangle of error m7e. lhWodb. at is three-point problem ? How is it solved by 8. H(WDao)ihswatitDndgeoausercisesrhipbtlehabenetelwhdteeiafebfnelienree'srtnhuetrsovedsceotyiuoorcnfco'emosarpniaeodrnfet'iinewnrgtiretohrprsscelachntaiineionn'tpsaulmbarnlveeeethyboitnydagbs lbiiannasgcpkaosp?iinpgtlhHietoidonfwgat.cocauprrel.ainctyehearynabdleeeli(xmsApui.ernMdvaie.teJyen.diEcnyg.)??. 9. (b) (c) surveyin1g0.. (a) Compare tha advantages aod disadvantages of plane !able surveying with those o f chain trial method giving rule(sb)whSitcahteyothurewe-ilplofionltlopwrobinleemsriminalipnlgantehetapbolisnitgiona od describe its solution by lhe point SOUght. (A.M. I.E.). of the Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net @] Calculation of Area 1smihnu2eer.cva1tEnea. nysreeGgOTsdlAt.inEhsBeheNaLTnEEohdauefRrne1tiAao2tthfs.Lo1ed.lalaesoprBtweerRripimmInrsTgoaqiInjSr.eyeHctatfetbtohVdlbeeNojerIoTcqgntSuasivacneOarosteFifsthi,etolShsarQwenizUdohoAfirnleResteluaaEarltriivntoMhepnywElmiaAnonbegrSeetkUr.t.wiRicsTEeTehuhtnoWeeniItdtTaush,erHneetieattrshMmtweoEoifonTfueRlnasIminCyttdshesetaEesaimQnurareUrseep.ImValsaqAen.noLetfIs!mNstuTheorteSvfreestyariarnecogatrj Square clwim Sq. polts OT· Square yards Square Jeer Square links Merrie AtrtS Pen:hes Equiva/e,us 1 Sq. mih n1 258.99 ha I -' - I 640 6,400 102.400 3.097.000 100,000 0.40467 ha I I 160 4.840 10 g1 43,560 4,356 10.000 404.67 m2 i I 404.67 cm2 I 16 484 neeNote. The sWldard of square measure is the Acre. I 30.25 272.25 625 25.29 m2 TABLE 12.1 (a) METRIC VNITS OF SQ-UARE MEASURE W1T1I BRITISH 9 20.7 0.836 m1 2.3 929 m2 riSqUllrt kilometre EQ- UIVALI!NTS n(tm'J gI .nI Sq. tentinutres British H<ctam AT< (a) Sq. marts (<m'l (m'J Equiva/e~ts (/r4) 10.000 - 100 1.000.000 100 10,000 - etI 0.3861 sq. mile I 2.4710 acres I 100 1,000.000 1076.4 sq. ft. I 1 10.000 10.764 sq. ft. I 0.155 sq. ft. J Note. The standard of square measure is the Are. (291) Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net SURVEYJNG r' 292 12.2. GENERAL METHODS OF DETERMINING AREAS The following are the general methods of calculating areas: w(c) By latitudes and departures : 1. By computaJions based directly on field measurements These include : w(il) By double parallel distance (D.P.D. method) (a) By dividing the area into a number of triangles (b) By offsets to base line w2. By computaJion based.· ~'n ''fneosurements scakd from . a map. (i) By double meridian distance (D.M.D. method) .12.3. AREAS COMPUTED BY SUB-DIVlSlON lNTO TRIANGLES EIn this method, the area is divided into a(d)By co-ordinates. number of triangles, and the area of each triangle ais calculated. The total area of the tract will then sbe equal to the sum of areas of individual triangles. yFig. 12.1 shows an area divided into several triangles. For field work, a transit may be set up at 0, and Ethe lengths and directions of each of the lines OA, nOB..... el!:. may be measured. The area of each triailgle 3. By meclwnical method : Usually by means of a planimeter. D c ~ can then be computed. In addition, the sides AB, BC.. ... etc. can also be measured and a check may be applied by calculating the area from the three known sides of a triangle. ThU&, if two sides and 8 one included angle o f a triangle is measured, the area of the triangle is given by AG. 12.1 where is =half perimeter = (a + b + c). The method is suitable only for work of small nature where the determination of the closing error of the figure is not imporrant, and hence the computation o f latitudes and departure is unnecessary. The accuracy of the field work, in such cases, may be determined by measuring the diagoual in the field and comparing its length to the computed length. @ A R E A S FROM OFFSETS TO A BASE LINE: OFFSETS AT REGULAR lNTERVALS · This method is, suitable for long narrow strips of land. The offsets are measured from the boundary .{o the base line or a survey line at regular intervals. The method can also ·j)e applied to a plotted plan from which the offsets to a line can be scaled· off. The area may be calculated by the following rules : Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net 293 CALCULATION OF AREA (z) Mid-ordinate rule ; (ii) Average ordinate rule (iiz) Trapezoidal rule ; (iv) Simpson's one-third rule. (1) MiD-ORDINATE RULE (Fig 12.1) ~ The method is used with the assumption that the boundaries be- tween the extrentities of the or!iinates (or offsets) are straigltt lines. The o, o, lo, ~,o, 4n base line is divided into a number '\\t '1 ·.· _ ~ • I·' 8 of divisions and nttihde-pooirndtisna~tefs . are ~i r measured at the each d->1 division, as illustrated in Fig: 12.2. ------~ L=nd The area is calculated by the AG. 12.2 formula Area = /!. = Average ordinate x Length of base = 0 , + 0 , + 0 , + ...... +On \"J-• \"'-~\\0, + Oz + O, + .... +On) d =d EO ... (12.3) n. o, 0 , ... = the ordinates at the ntid-points of each division where (2) AVE~g~2.3) n This rule also assumes that the boundaries between the extrentities of the ordinates are straight lines. The offsets are measured to each of the points of the divisions of the EO = sum of the mid-ordinates ; n = number of divisions gbase line. in =[ J(@ ...(The area is given by . L =length of base line = nd ; d =distance of each division eewhere /!. = Average ordinate x Lengt!!__of the b>§e ro.. 0, .... = ordinates at the end of each division. Oo+ 0, ...... + On]L = - L - E O 12.4) in(3) ~(1'1g.12.3) .•.. n + 1 (n+l) · Oo = ordinate at one end 'of fue base. On= or~t.e at the other end of the gThis rule is based on the as- base divided into n equal divisions sumption that the figures are trape- .zoids. The rule is more accurate than nthe previous two rules which are ap- - eproximate versions of the trapezoidal rule. •I• tReferring to Fig. 12.3, the area of thoe , +firsot, trapezoid is given by o, o, o, o, o, Io, .. - 34 n L=niCI--------->1 \\4---d 2 d----+1 1!.,=--2-d FIG. 12.3 Downloaded From : www.EasyEngineering.net

!'' Downloaded From : www.EasyEngineering.net SURVEYING 294 o,Similarly, the area o f the second trapezoid is given by u, z_- O,-+- d 2 Area of the last trapezoid (nth) is given by w~~w+;+or <1n- On- L +On d 2 ~· Hence the total area of the figure is given by .,- t . = .1, +<1, + ....... t., Oo+O, d + 0 -1 +- 0, + ..... + 0,_, +0, =--2- - 2 -d 2d wAdd the qyemre of the end offsets to the sum of the jnlermemqle nfl<ets. M!!!!!I!JY .... + o , . y ) : i } ... (12.5) requjrsd «tel!:_ .E(4) SIMPSON'S ONE-THIRD RULE a~hort lengths ---s---of boundary between the ordinates are parabolic arcs. This method is more useful when the bound- Equation (12.5) gives the ·trapezoidal rule wbich may be expressed as below : the I<Wll sum thus ob/llined by tKi common dUtance between the ordinates. io get the yary line departs considerably from the straight .G F -----------· Eline. E• • ----~ T .::-.>IC nThus, in Fig. 12.4, the area between the oL<:---------------·ifi·----------- 10, o, o, line AB and the curve DFC may be considered to be equal to the area of the trapezoid ABCD A 1a plus the area o f the segment betw<\\'n the parabolic arc DFC and the. corresponding chord DC. d d _ __.., Let 0 0 , 0 1 , 0 2 = any three consecutive the chord DG FIG. 12.4 ordinates in E ordinates taken at regular interval of d. to cut the Through F, draw a line EG parallel to and G. Area of trapeem.d ABCD = -Oo+- Oz · 2d . . . (1) 2 To calculate the area of the segment of the curve, we will utilize the property of the parabola that area o f a segment (such as DFC) is equal to two-third the area of the enclosing parallelogram (such as CDEG): 2 2([Thus, area of segment o, - -Oo-+O,l 2d,II DFC = 3 (FH x AB) = 3 , 2 ... (2) Adding ( I ) and (2), we get the required area (8..,) of first two intervals. Thus, 21[< 1 , , ,O=o-2+-O· 2z d + , . o ,O- -o2- +O2'ldl = d (0o + 40,+0 z) ... (3) 3 3 Similarly, the area of next two intervals (<11.4) is given by Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngine2e9r5ing.net CALCULATION OF AREA .1,,, ~3d (0, + 4 0 , + 0.) ... (4\\ ... (5) Area cif the last two intervals ( 8 , _ \" <1,) is given by o,. I . . = 3d ( 0 , . , + 40n-l + 0,) Adding all these to get the total area (<1), we get <1 = 3d [Oo + 40, + 20, + 4 0 , ...... + 20n-1 + 40, - I + 0,] ~ + 0,) + 4 (9.L+ a;!+ ... + o...aor + 2 (0, + D,t.+ ... 0 , - t l Q . . . (12.6) I t is clear that the rule is applicable only when the number of divisions of the area is even i.e., the total number o f ordinates is odd. I f there is an odd number of divisions (resulting in even number of ordinates), the area of the last division mllSt be calcnlated separately, and added to equation 12.6. Simpson's one third rule may be stated as follows : Tht area is equal to the sum o f 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 the common interval between them. n of ordinates. g ......E(ample 12.1. The follawing perpendicular offsets were taken at 10 metres imervals ifrom a survey line to an irregular boundary line : n3.2?. 5.60, 4.W. 6.65, 8.75, 6.20. 3.;15 4.2o. 5.6;5. Ca/culare the area enclosed between the survey line, the irregular bowulary line. eand 1he firsi aJUi /iJsl offsets, by tile app/icalion of (a) average ordinate rule. (b) trapezoidal ende, and (c) Simpson's rule. 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 smaller than those obtained by using the trapezoidal rule accordin as the curve of the bo~ndary is ''!!!\"~ or convex tow~ the base line. In dealing with irregularly shaped figures, the degreeof precision of either mpllod can be increased by increaSing the number rSolution. i(a) By average ordinate rule nFrom g.nHere et6 = 980 x 47.75 = 424.44 sq.metres = 4.2444 ares. equation 12.4 (a), we have <1 = _nL+_l EO m n = number of divisions = 8 ; n + I = number o f ordinates= 8 + I = 9 L = Length of base= 10 x 8 = 80 m l:O = 3.25 + 5.60 + 4.20 + 6.65 + 8.75 + 6.20 + 3.25 + 4.20 + 5.65 = 47.75 (b) By trapewidal nde \"j 1d . 12.5, .. = l( O-o-+2-O, + o, + o, + ... + o,-' From Eq. tJ. Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net 296 SURVEYING Here d = I0 m ; -Oo+2-0=, 3.25 + 5_.65 4.45 m 2 wFrom Eq. 12.6, Here m 01 + 0 , .... 0 , - 1= 5.60 + 4.20 + 6.65 + 8. 75 + 6.20 + 3.25 + 4.20 = 38.85 1!. = (4.45 + 38.85) 10-= 433 sq. metres = 4.33 ares. w4 (01 + o, + ..._.o;-1>;, 4 (5.\"~o + 6.65 + 6.20 + 9 0 > =.90.60 (c) By Simpson's nde ° w:. 1!. = d [(Oo + O,) + 4 (01 + o, + .... + o,_ 1) + 2 (0, + o...... + o,_ :i)) .E~pie 3 o,d = 10 m ; 0 0 + = 3.25 + 5.65 = 8.9 m 2 co,+ O.+ .... o,_2),;·z(4.:iO·+ 8.75 +3.25)= 32.40. ao. 2 .65, 3.80, 3:75, 4.'65, 3.60, 4.95, 5.85 m- . 1!. = 1 (8.9 ~ 90.60 + 32.40) = 439.67 sq. metres = 4.3967 ares. line to a curved 3 boundary A series o f offsets were taken from a chain sCompwe the area between the chain line, the cuJ;.;d boundary and the end- offsets ySolution.12.2. line al intervals of I~es in the following order. ' En(a) By average ordinate nde by (a) average ordinate rule, (b) trape2oidal rule, and (c) Simpson's nde. , From Eq. 12.4 (a), we have 1!. = _n!+: _ ! :W Hence n=7; n+l=S. L = nd = 7 x 15 = lOS m :W = 0 + 2.65 + 3.80 + 3.75 + 4.65 + 3.60 + 4.95 + 5.85 = 29.25 m 1!. = 105 x 29.25 = 383.91 sq. m = 3.8391 ares. 8 (b) By lrflpewidol rule lequaa0 on 12.5 From 1!.= (-Oo~-O+o0 I + O , + .... O,-t d From 2 Here d = .15 m ,, Oo + 0 , 0 + 5.85 = 2.925 m 2 2 o1 + o, + .... o . - 1 = 2.65 + 3.8o + 3.75 + 4.65 + 3.60 + 4.95 = 23.40 :. 1!. = (2.925 + 23.40) 15 = 394.87 sq. m = 3.9487 ares, (c) By Simpson's rule · equation 12.6, 1!. =~ [(Oo+ 0 , ) + 4 (01 + 0 3 + ... On-1) + 2 ( 0 , + 0 . + ... On-2)] Here, . -d = -I3S= 5 m. 3 It will he seen that the Simpson's rule is not directly applicable here since the number of ordinates (n) is even. However, the area between the first and seventh offsets may Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net 'J!T1 CALCULATION OF AREA he calculated by Simpson's rule, and the area enclosed between the seventh and last offseiS may he found by the trapezoidal rule. Thus,; ( 00 + 0 , ) = 0 + 4.95 = 4.95 o,4 (01+ + ... o.-1> = 4 (2.65 + 3.75 + 3.60) = 40 o.2 (0, + + ... 0 . - 2) = 2 (3.80 + 4.65) = 16.90 1!.' = 5\"(4.95 + 40 + 16.90) = 309.25 sq. m. 'iArea of the last trapezoid= (4.95 + 5.85) = 81.0 sq. m. Total area= 309.25 + 81.0 = 390.25 sq. m = 3.9025 ares. 12.5. OFFSETS AT IRREGULAR INTERVALS (a) First Method (Fig. 12.5) 3 4 5 In this method, the area of each trapezoid o, o, o, aiea.is calculated Separately and then added together to calculate the tbtal Thus, from Fig. IO, o, 12.5, 1 ! . =d2l ( 0 1 + 0 , ) +d2, (0 , + O ,) A B n Example 12.3. to-d,-4!+--d2 Ill <1,---oJ+-d,~ ,f gto an irregular boundary : AG. 12.5 + d, (0, + 0,) . . . (12.7) i Chainage nOffset 2 By method of co-ordinates : See § 12.7 (b) Second Method. The following perpendicular offsets were taken from a chain line eSolution. eArea 0 10 \\ 25 42 60 75m ', rinArea 15.5 26.2 J 31.8 25.6 29.0 31.5 Colculate the area berween the chain line, the boundary and the end offsets. g,·.-•• of tltirdoffustu:aP\"wid = 1!.,1 = 10-0 ' + 26.2) = 208.5 m2 .nArea of fourth (15.5 etArea of fifth .,, of second trapezoid= 1!., = 25'-10 (26.2 + 31.8) = 435 m' -- 20 trapezOl'd = 1!.3 = -4 22-- 25 (31.8 + 25.6) = 487.9 m2 trapezoid= 1!.4 60-42 (25.6 + 29.0) = 491.4 2 = - 2- m trapezot'd = . = 7 5 --6 0 (29.0 + 3LS) = 453.7 2 1!., - m 2 Total area=l!.=l!.l+l!.,+l!.,+l!.<+l!.s = 208.5 + 435 + 487.9 + 491.4 + 453.7 = 2076.5 m' = 20.765 ares. Downloaded From : www.EasyEngineering.net

• Downloaded From : www.EasyEngineering.net 298 / SURVEYING ~pie ;:7·· \\.t waTrhOCahefpfasediezrtCoSugsJioadeg(llameuc:lut)(ilmoarn)tue~le.1<th2-((eb.1)4).•..arS=Teiah.m,-8Ie,pS-_5bfso·eo.ltnlw-o'-s1ew31ei0n0rnu.g7leth-.pe~e1rs4p2u5e.r8nvdeiy~cu!bl~an-r4e~.,,·.oft.fhs9ee0.ts}he_d'gr,ee8.:0-at!a·n_.k-de.n(t17hf0.er9o081.m.e-.-...s1..:;a.~6-1·~2o(\\c{0hc.a':i:·el·.'·~.n~-\\41-)li4.nQ!.e-}-_-_(It~o) w(a) By Trapezoidal role wthe wwhere to 5th offset. There is another interval between The interval is constant .from first offset between 7th offset and lOth offset. The total 5th and 7th offset and. a third interval .&, =area of third seetion ; d, =interval for first Section = 15 Ed, =interval for second section= 10 m ; d3 =..imerval for third ._area t> can, therefore, be divided into three sections. d=&,+A,+&, 2£( aNow a,= area of first section ·; A2 =area of second section sy&, = ( m section = :10 m &1 = (En&3 = ( 7.60 + 10.6 + 8.5 + 10.7 + ';'). = 616.3 - 2. 12.8 15 m' 10.6+8.3 + 9.5 ) 189.5 ' .... 2 )110 = m 8.3 + 4.4 .) 2 0 = 4 1 3 m' 2 +7.9+6.4 a = 616.5 + 189.5 + 413 = 1219 m ' = 12.19 ares. (b) By Simpson's Rule o f ordinates, and therefore. ordinates (even number) : The first section and the second section have odd number applicable. The third section has 4 Simpson's rule is directly the first three ordinates only : the rule is applicable for &1 = ~5 [(7:·60 + 10.6) + 4 (8.5 + 12.8) + 2 (10.7)] = 624 m ' &2 = 1~ [(10.6 + 8.3) + 4 (9.5)] = 189.7 m' ;oa , = [(8.3 + 6.4) + 4 (7;9)1 ; ~ (6.4 + 4.4) = 308.6 + 108 = 416.6 m ' & = 6 2 4 + 1 8 9 . 7 + 4 1 6 . 6 = 1 2 3 0 . 3 m ' = 1 2 . 3 0 3 ares. 12.6. AREA BY DOUBLE MERIDIAN DISTANCES TtAarnhadivsreerdfmseeTerpeehtahniarsoctnueddrmeismestthhekeoornifddodiweaoiasnnucbhtailhseseliDtnmohen.eeMenroi.dfDmaias.otsnhsumetmdeoetitfrshdtatoeavndnet.ocreusTesspOoadaoscsrfeafoltctrhchuraeloclacoutuemlgilnhaptehtuesettdhin.eaagrreTemahthoecbesoyttmarratwphevueiaesstretsoemderf.leytaihsoscdttlaho,tesioetnhdnebtalralaoatvinfteucrdesthdeees.. Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineer2i9n9g.net CALCULATION OF AREA ~., ME~RIDIANcDISfANeCES of any point in a traverse is the distance o f that point to ihe reference meridian. measured at right angles to the meridian. 'lhe meridian distance of a survey line is defined as the meridian distance o f its mid-point. The meridian diiiance (abbreviated as M.D.) is also sometimes called as the longitude. Thus. in Fig 12.6, if the reference meridian is chosen through the most westerly station A, the meridian distance (represented by symbol m) o f the line AB will be equal to half its departure. The meridian distance o f the second line BC will be given by m•- . = m 1D2+, - +D2-1 Similarly, the meridian distanc~ of the FIG. 12.6 D,(D') D,D,. third line CD is given by m,=mz+T+ - T =mz+T-T n And, gine·tdidhtiseseptaadndneceupHAepracaeercsnrtouorcurfreierd.,ee.ain.nt,hgyIoenftploriaonuthpseltieephtilevyifpseionarrgebsecoiqtgevhutndehea,ienlfomgtrrhteouerlleieidant,mhiseateeenprrrpindomdliupiaesdsenrteraipdnhdaicaaairetsntlttuefamrnnedltcaihioeyseatnanbonddescfehe1ptas1hatueearoglrtadefuf1dirirtvebhseetae·soslipipfgfnaonreiteldhlcofeowewrdtioslillniwngtbeeheseltieiTnehmshqeeieuglanfdp.mlselupetsaoornifduhhitraaahenlel.ffby the meridian distance of the fourth (last) line DA is given m. = m3 + ( - ~' ) + ( - ~· ) = m,- ~'- ~· = ~· eAREA BY LATITUDES AND MERIDIAN DISTANCES ringtowhfuilslthebgeIentltiionnFnegi,eg.taornif1adn2tg.ht6hlee,eseliaanshneted-swi,ghetrtsJhatpeelizonifeubsmathssaee.retOridonarefnagwtslhenideeofrtrrooimaftnragepelaeeaczchihourmtsritaatrwntaigopillnelezibtuooemr tththrweeaiplrleezmfibeueermerindtchi(eaesnomdfeloiarsrittmdiatiuneaddcne)e, .of that line. Thus, line. narea of each triangle or trapezium = latiJude of the it is eor A1 =L1 x m, tThe latitude (L) will be taken positive if it line x meridian distauce o f the is a northing, and negative if a southing. the area o f the traverse ABCD is equal to the algebraic sum o f the In Fig. 12.6, areas of dDCc. CcbB, dDA and ABb. Thus, Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net 300 SURVEYING A =Area of dDCc +area of cCBb- area of dDA - area of ABb or A= L3 m3 + £, m, - L . m. .- L, m1 = r.Lm. DOUBLE MERIDIAN DISTANCE wThe double meridian distance of a line is eljUII] to the sum o f the meridian distances· o f the two extremitks. wThus, in Fig. 12.7, we, have : Double meridian distarice (r';Presented by symbol M) o f the first line AB is given by wM, =m of·Ac +in of B=O+D, =D, (It is to b e noted that the quantities L, m, and L 1 m1 bear negative sign since L, and L, of DA and AB are negative.) Similarly, if M, , M3 , M , are the double me- .ridian distances of the lines BC, CD and DA re- Espectively, we have M,=m· of B + m of C a=D 1 +(D1 +D2) =M1 +D1 +D1 s= D.M.D. of AB + Departure of yAB + Departure of BC E \"M, = m of C + m of D +1!:-- o,,--. 1 n= (D 1 + D,) + (D, + D,- D,) c =Mz+Dz-D3 = D.M.D. o f BC + Departure of BC + Departure of CD and M,=m of D + m of A = ( D , + D1 - D3) /4----D,----< o, + (D 1 + D , - D , - D4) = M , - D , - D4 FIG. !2.7 = D.M.D. of CD + Departure of CD .,. Depanure o i OA Hence, the rule for finding D.M.D. of any line may be stated as follows: \"The D.M.D. o f any line is equal to the D.M.D. o f the preceding line plus the tkpiuture o f the preceding line plus the deptuture o f the line iJself. \" I Due attention should be paid to the sign o f the departure. The D.M.D. o f the first ! line will evidently be equal to its departure. The double meridian distance of the last line ' is also equal to its departure, but this fact should be used simply as a check. AREA BY LATITUDES AND DOUBLE MERIDIAN DISTANCES Jo Fig. 12.7, the area of the traverse ABCD is given by Now, area A = area o f dDCc + area o f CcbB - area o f d.DA - area o f ABb = =!of dDCc ¥dD + cC) cd (M3) x L3 Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net ':~\" JOI CALCULATION OF AREA That is, area o f any triangle or trapezium = Half the product o f the latiblde o f the line and its meridian distance. Hence' A= j; [M, L, + M, £, - M , L . - M, L,] Thus, to find the area o f the traverse by D.M.D. method, the following steps are necessary Multiply D.M.D. \"Of each line by its latiblde. (I) Find the algebraic sum of these products. The required area will be half the sum. (2) (3) AREA FROM DEPARTURES AND TOTAL LATITUDES 0)] From Fig. 12.8, the area (A) o f ABCD is given by A = area of ABb + area of BbcC + area of dcCD + area of DdA I f L 1'. L,', L,' are the total latirudes · o f the ends o f the lines, we get A = ! [(D,)(O- L,') + (D,.)(- L,' + L,') + ( - D,)(L,' + L,') + ( - D4) (L,' + = - Mu (D, + v,) + u <- v, + v,) + L.' (D, + v.)] n (I) Find the D Note. The negative sign to the area 4 '''/'''t.,' •!•• d g .... v.:..............................,........each station of traverse. has no significance. \"' Hence, to find the area by this method, ipartures of the two lines meeting at that c :he following steps are necessary : nstation. Li total hltitude (L') of e(3) Multiply the total latitude of each A- i b • er l o,Jof the departure (found in 2). (2) Find the algebraic sum of the de· L,':'•' iproducts gives th~ required area. •• ~station by the corresponding algebraic sum nAREA BY DOUBLE PARALLEL DISTANCES AND DEPARTURES ~---0, gA parallel distance of any line of a traverse is the perpendicular distance from the B .middle poim of that line to a reference line (chosen to pass through most southerly station) nat right angles to the meridian. The dcuble parallel distance (D.P.D.) of any line is the esum of the parallel distances of its ends. The principles of finding area by D.M. D. method tand D.P.D. method are identical. The rules derived above may be changed to get the (4) Half the algebraic sum o f these FIG. 12.8 corresponding rules for D.P.D. method, by substiruting D.P.D. for D.M.D. and 'departure' for 'latitude'. The method is employed as an independent method o f checking area cbmputed by D.M.D. method. Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net ~· 302 SURVEYING 12.7.~ Let (x,, y1), (X,, y,), (x,, y3) and (x4, y4) be :t~~~~-~\\\\\\\\\\\\\\\\'{ - .the co-ordinates of the stations A. B, C. D respectively, of a traverse ABCD. If A is the total area of the traverse, we have A = (Area aABb) + (Area bBCc) :n\\\\\\aB(X, y,) -(Area cCDd)- (Area dDAa) f w= [(y,- y,)(x, + x,) + (y, -_y,)(.i, + x3) w- (y, - y,)(x. + x,) ~ (y, - y,)(x, +x,)] w= ilYo(x,- X.) + y,(x,- x1) + y3(x4 - x,) + y4(x 1 - x3) ] .E,.In general, if we have n stations, we get a/ =} [y, (x, + x,) + y, (x3 - x,) + y, (x,- x,) + ... + y,(x, - x, -I)] syof __ FIG. 12.9 ·:,.. sr- ESide ... (12.7) \\, (in metres) : nAB --Eiample 12.5. The following toble gives the corrected latitudes and departures the sides o f a closed traverse ABCD · l.<Jiitude (-\\w N _l E / /08 4 BC IS 249 CD ~ 123 4 257 - - I ~ DA met/wd (iii) 0 Compute its area by (i) \"(Jk.:met!J.<Hj, (ii) D.M.D, Departures and total latitudes, (iv) Co-or!firwte metlwd. - Solution. (1) By meridian distances and latiJudes Area= l:(~) are arranged in the- point A is the most Calculate the meridian distance of 'each line. The calculations tabular limn ,below. By the inspection of the latitudes and departures, westerly station. AB is rakeo as the first line and DA as the last line-. line l.<Jiitude DejklltUr< ~Departure M.D. Area= mL (D) (m) (L) rtDJ AB + 108. +4 + - 2;-PI. 1' + 216 BC + 249 + 1928 CD + 15 +4 + 124.5.''\" 128.S - 3136S DA - 257 - 123. + 2 '-- 25S 0 0 - 128.S\" 128.S Sum - - 29221 Downloaded From : www.EasyEngineering.net

·\" Downloaded From : www.EasyEngine30e3r'ing.net CALCIJLATION OF AREA Tolal area= t. = l:mL = - 29221 m' \" !I Since the negative sign does not have any siiJnificance, The actual area= 29221 nl- = 2.9221 hectares. H - .,· tArea= l:mL (2) By D.M.D. mellrod : line l.<Jiitude Deptulure D. M.D. Area=mL (D) (m) (L)- AB + 108 +4 4 + 432 BC + IS + 249 257 + 3855 CD - 123 +4 510 DA 257 - 62,730 0 - 257 Sum 0 - S8,443 i· · Area= l:mL = 29221 m2 = 2.9221 - hectares. (3) By Deparlure and totoJ latiJudes : Let us first calculate the . total latitudes of ngin iJ:the with· A as the reference point, point, starting latitude of B = + 108 Thus, total latitude of C = + 108 + 15 = + 123 latitude of D = + 123 - 123 = 0 total latitude of A = 0 + 0 = 0 total (Total latitude x algebraic sum total e '1t/'AB The area·=erBC of adjoining departures) inCD lineg'DA l.<Jiitude /JtpiUIDn St4tion Totlll Algcbrokmm D.o,u.b.k .net.. (L) (D) of o4jololog + 108 +4 B + 108 +253 + 27,324 + 249 + 123 +253 + IS c + 31,119 +4 0 - 253 - 123 - 257 D 0 0 0 A - 253 Sum 0 S8.443 Area= ~58,443) m' = 29221 m2 = 2.9221 _hectares. (4) By Co-ortNnates : _For calculation of area by co-ordinates, it is customary to calculate the independent c<Hlrdinares of all the points. This can be done by uking the co-ordinates of A as ( + 100, + 100). The results are tabulated below : Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net 304 SURVEYING lndepttuknJ to-ortlbuiUJ rf\":v)'\\11M Slllllon I.IIJ/Ju4< .~ A North GJ &ut(x)_ (~-'''' 100,'. 100 . ' wBC AB + 108 +4 wwDA B 208- 104 + IS ' + 249 c 223. 353 CD - 123 - +4 I .I i EA = 2 (y,(x, -0'. - 257\\ D 100 I 357 A IQq I 100 . - . 223)) i as= (10800 + 12792- 38124- 43911) =- 29221 ~; Substituting the values of x and y in equation 12.7, we gef 1 .· x,l +y,(x, x1) ySince the negative sign does not bave significance,. the area = 2.9221 hectans.+ y,(x. x,)] x,) + y,(x, - - - E12.8. AREA COMPUTED FROM MAP MEASUREMENTS n(A) By sub-division of the area into geometric . fiiJIIl\"'S += [100(208 - 100) + 104(223 - 100) + 353(100 --}08) 357(100 - The area of the plan is sub-divided into common geometric figures, such as triangles, rectangles, squares, trapezoids etc. The length and latitude of eath such figure is scaled 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, ------------------------ are drawn on a tracing paper. The constant IM!tween the consecutive P'!'- (a) (b) allel lines represents some distance il in metres or links. Midway between FIG. 12.10 I each pair of lines there is drawn another pair of lines in a different colour Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net CALCULATION OF ARI!A 30l or dotted. The traCing is then placed on the drawing in such a way that lhe area is exactly enclosed between two o f the parallel lines. The figure is thus divided into a number of strips. As,luming that the strips are either trapezoids or triangles, the area of each is equal to the length of the mid~dinate multiplied by the constant breadth. The mid-ordinates o f the strips are represented by the length of the dotted lines intercepted within the maps. The total sum of these intercepted dotted lines is measured and multiplied by the constant breadth to get the required area. More accuracy will be obtained i f the strips are placed nearer. 12.9. AREA BY PLANIMETER A planimeter is an instrument which measures the area of plan of any sbape very accurately. There are two types of planintete1's: (I) Amsler Polar Planimeter, and (2) Roller Planimeter. The polar planimeter is most commonly used and is, therefore discussed here . Fig. 12.11 shows !lie essential parts o{ a polar planimeter. It consists of two arms hinged at a point .known as the pivot point. Ooe of the two arms carries an aucbor at The length of anchor arm is generally fixed, its end, and is known as the anchor arm. length of anchor arm is also provid~. The but in some of the planimeters a variable other arm carries a tracing point at its end, and is known as the tracing arm. The length of the tracing arm can be varied by means of a fixed screw and its corresponding slow 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 means of a wheel whose axis is kept parallel to the tracing arm. The wheel may either be placed between the hinge and the tracing point or is placed beyond the pivot point n away from the tracing point. The wheel carries a concentric drum which is divided into g100 divisions. A small vernier attached near the drum reads one-tenth of the drum division. ineering.netFIG. 12.11. AMSLER POLAR PLANIMETER. I . TRACING ARM 6. WHEEL 2. ,t.NCHOR ARM 1. GRADUATED DRUM 3. ANCHOR 8. DISC 4. TRAONG PO~ 9. MAGNIFIER 5. HINGE.. 10. ADJUSTING SCREW FOR I Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net 306 SURVEYING tTttoohheetthhtceeison,mtwhspahleeaetfenilx.derdTehvhuouinnlsudd,ureeoxednatshcnhseoafrorenatthdheteihnegwdhidsiersceulmoc,afanrefaonburdeeradtduhteiiolginstiehtdoasu-stdatoihnsecdkthnuasoncwrituosanttehbdetehienbngyuvmerarebnaeidsreuri.tooanfbInletthheaegdedtadiirmtiiisnoecgns, It is the anchor in position walong the cenrre line the zero of the disc has crossed the index. on three points - the wheel, the anchor poinr remains ijxed as the tracing point is moved ww.Ea(ttiThhnicfheeritueniaathbllleaomyTrureaeonaarardkrdoeaefilaornidlynsfg.disottohinltfelhsmettihhftaeiealglrc)ueworbamehooeureoesnilsfditatiostrthhyiseet.thnhoepepnfllca~sact;atleaa·tcdkeru;etl.itnanhi.tngeeasdnTiddpahenofecritnohhtthtmoer.earcTtitarnphhargoeeeciainnpftfgoion(ilianilfsplot.owtieihrsinine~ttahgndeoaikrwrnfeeogpaprtmmloaioceufsevxladeatlhdcaetr:logycuedlot)rs.cuoidkmv-eAewr·iistspheitoe.tiaanlalt¢Toreenhnaiges. clear from Fig. 12.11 that the planimeter rests poinr and the tracing point. Out of these three, while the wheel partly rolls and partly slides is perpendicular to the plane of the boundary. Since .the p)ane of the wheel only notrnal displacement - w h e n it of the tracing arm; ·the wheel measures aArea syEnwhere . . . (12.8) . (li) ~ M ( F - I ± 10 N + C) F ~ Final reading : I ~ Initial reading 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-clockwise direction. M ~ A multiplying constant, also sometimes known as the planimeter constant. It is equal to the area per revolution of the roller. C ~ Constant of the instrument which when multiplied by M. gives the area constant C is to be added only when the anchor of zero circle. The poim is inside the area. tracing point is to be moved in the clockwise direction only. It is to be noted that the to i''r'. The proof of the above formula is gi~en belo\\1.'. Proper sign mU:;t be given THEORY OF PLANIMETER Fig. 12.12 (a) shows the schematic diagram o f polar planimeter. where A , : Area to be measured, the anchor point being outside the area. L ~'Length of the tracing arm = Distance between the tracing point and the hinge. R ~ Length o f anchor arm = Distance between the pivot and the anchor point. a ~ Distance between the wheel and the pivot, the wheel being placed between the tracing point and pivot. w ~ Distance rolled by the roller in tracing the area. A,= Area swept by the tracing arm. Fig. 12.12 (b) shows the section of the perimeter of the area. Any such movement two simultaneous. motions : (I) translation o f the tracing arm of the arm is equivalent to (il) rotation of the tracing TP in parallel motion and arm about the pivot. F i g . 12:t2 Downloaded From : www.EasyEngineering.net

CALCULATION OF AREA Downloaded From : www.EasyEngine3e07ring.net ~ ~I .~ (c) shows ·the cjnnponents of the two motions separately. Thus, i f the tracing aim sweeps i a veiy Small arei dA,, such ·that dh is the movemeDI in parallel direction and d9 · is the t rotalion, we !Jave dA, = Ldh + -~ L'da 1 ·~ Since the recording wheel (W) is placed in plane perpendicular to that o f the tracing !'~1f arm, the wheel records only the movemeDI perpendicular to its. axis. I f dw is the distance rolled out by the wheel in sweeping the area tL!,, we get I~ . . . (1) dw= dh + ad6 or dh = d w - ad6 Substiruting the value o f dh in (1), we. get ~II idA, = L(dw- ad6) + L'd9 ... (2) I~ ...,___ L i+.!..ot ngineering.net(a) (c) (b) A, A (d) FIG. t2.t2. THEORY OF PLANIMETER. Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net 308 SURVI!YINO When the tracing point is moved along the boundary, the arm moves downwards along one side of. area and upwards along the other side. Heu:e, the net area A, swept wA0 =Lw-aLfda+iL' Ide by the tracing arm is equal to the area of the plan (Ao) IThus Ao= f dA_s=L l t t w - a L f dO + } L ' de ... (3) Now when wconstrained along anchor point but But f dw = total distance moved by the wbeel = w wfdO=O . ... (4) the anchor point is kept oulside the area,· the motion o f the pivot is .Eas ftbLhaeectkci.1rHc0loewiAlse1ov=feorArr,iagrdeiiinafUao.lthfeRpothaseanitncidohpnolarcnoapfmwtoephirneletntterisastchieonknegpeatntchriheenvosroiadlurepetiaoot.ihnnetHwea~irh)eCe;anek,epttthhtheeeinqptsruiiadvaceonitntigttypheopinoatdinrOetma=.oisv2ensbr..oaulognhgtthe the arc of a circle i.e., · i t never completes one revolution about that simply moves along the arc in upward and downward directions so Hence from (4), A0 = Lw ... (12.~) yAa = Area swept by the pivot. + EThen, the area A, = A, +Aa = IdA, + nR' = [L I dw - aL I de+ ~;L' I dO] n· =Lw- aL(2n) + ,j.L'(2it) + nR' =Lw + n(L'- 2aL + R') nR' · ... (12.9) Thus, equation 12.8 is to ·he used when the anchor point is oulside J.h('area while ... (3) equation 12.9 is to he used when the anchor point is kept inside the area. Now w = Total distance rolled by the wheel = 1tD n wbere D = Diameter o f t h e wheel n = Total change in the reading, due to the movement o f the tracing point along the periphery o f the area= F - 1± ION. Subsrii\"'Jting !he value f'f w ~n equation 1'2.9, we get the area A,. or !>. = LT<Dn + n ( L ' - 2aL + R');, Mn + n ( L ' - 2aL + R') ...(12.10 a) !I =Mr.+ MC = M(n +C)= M (F - H ION+ C) ... (12.10 b) I' I where M = The mtiltiplier = LxD = Length o f tracing arm x Circumference of the wheel C = Constant = n(L' - 2aL + R') M Thus; we get equation 12.10, which was given in the earlier stage. In the above equation C is to be added only if the anchor point is inside the area. ZERO CIRCLE The quantity MC = 1t (L' - 2aL + R') is known as the area o f the zero circle or correction the circle of correction is defined as the circle round the circumference circle. ~ zero circle or simply slide (without rototion) on point is moved, the wheel will tracing . arm of which if the tra~ing is possible when· the change in the reading. This the paper without any Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net 309 CALCULATION OF AREA is held in such a position relative to the anchor arm that the plane of the roller passes through the anchor point i.e., the line joining the anchor point and the wheel is at right angles to th~ line joining the tracing point and the wbeel. ~athol tradng point \",/ ~.--- 14/ L a-+! nW T A; A A (a) (t-~ FIG. 12.13 In Fig. 12.13 (a), the wbeel has been placed between the tracing point (1) and the pivot (P). Let R0 be the radius o f the zero circle. I f x is the perpendicular distance =(L' + t l - 2La + R ' - a') =(L2 - 2aL + R') n And area of the zero circle= n Rl = n(L'- 2aL + R ' ) o f the wheel W from anchor A, we get . r R0 = ( L - a ) ' + :i' = ( L - a ) ' + ( R ' - a') g In Fig. 12.13 (b), the wheel has been placed beyond the pivot. i Hence,nArea of the zero circle=1t(L2 +2aL+R') ... (12.11 a) eThus, the general expression for the area of the zero · circle ...(12.11 b) eArea of the zero circle=n(L2 ±2aL+R') rUse+ sign i f the wheel is beyond the pivot and -sign if iho Ro' = (L + a ) ' + ( R ' - tl) = L' + a ' + 2aL + R ' - t l = L' + 2aL + R' ... (12.11 c) irracing point and the pivot. ... (12.11 d) nTo find the area . of the zero circle practically, the tracing point is gthe perimeter of a figure, one with the anchor point oulside the figure, can be written as : .the ... (12.11) whtel is be<w\"\"a the n!>. = [Mn + n (L' ±2aL + 11')] = Mn' traversed along etn(L' ± 2aL + R') = M (n' - n) and then. with anchor point irui\\de it. same in both the cases, we get, from eqution 12.10 a Since the area swept is the ... (12.12) where n and n' are the two corresponding readings of the wheel. It is to be noted thoJ area o f the figure i s - greater than the area o f the. zero circle, n will be positive. if.. the smaller than the area of the zero if the area of the figure is while it will be negative circle. · Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net SURVEYING 310 MULTIPLIER CONSTANT (M) The multiplier constant or the planimeter comtant is equal to the number of units 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 length of the tracing arm is set to such a length that one revolution of the wheel corresponds to a whole number and convenient value of area. When the figure is drawn to a natural scale, and the area is desired in sq. incbes, the value of M is generally kept as eqlial to 10 sq. in of area. wFor any other setting of the tracing arm. the value of M can be determined by traversing the perimeter o f a·fi~. o f known· area (A), with anchor point outside the figure. wwThen It is to be noted that the value of M and C depends upon the length L which .is adjustable. The manufacturers, therefore, supply a table which gives the values 'of L Eand C for different convenient values of M.. /' ihii'The manufacturers always supply the values of M Known ilrea ~ ~ where n' = Change in the wheel readings aarm with the corresponding values of M and C. The following table is an extract fromn' n' / sthe values for a typical planinteter. yEnScak vernier setting on the .tracing .r Vemitr porilion Area .of one rel'Olution of ConsJJJnJ on lnJdng bar lht meMutr!nunl whtel (C) (M) SCIIle A.<lllal I: 1 I 33.44 ! 100 sq. em 100 sq. em I 23.521 1: l 21.58 I U.430 I ' 48 I U.97 I 10 sq. in. I 10 sq. in. 24.569 ! i 200 sq. ft. J 12.5 sq. in. I, 1 · '24 i 26 97 i SO sa fi I ~-125 $0 in Ii '24 569 I. I ' 50 21.66 0.4 ac;., 10.04 in. 26.676 : Thus, for full scale, value of M = 10 sq. in. in. f'.P.S. units, and for another setting of tracing bar. the value of M = 100 sq. em. Example 12.6. Calcula/e the area o f a figure from the following readings by a planimeter wiJh anchor poim outside the figure : Initial reading = 7.875, final reading = 3.086 ; M = 10 sq. in. The zero mark on the dial passed the fixed index mark twice in the clockwise direction. Solution. A = M(F - I ± ION+ C) Since anchor point is outside, C is not to be used in the formula, M = I O ; F=3.086 ; I=7.875 ; N = + 2 A= 10(3.086 -7.875 + 20) = 152.ll sq. in. Downloaded From : www.EasyEngineering.net

CALCULATION OP AREA Downloaded From : www.EasyEngineering.net 311 Example 12.7. Calculate the area o f a figure from the following readings recorded by the planimeter with the anchor point inside the figure. 'lnirial reading= 9.9I8 ; Final reading = 4.254 ; M = IOO sq. em : C = 23.52I It was observed thai the zero mark on the dial passed the index once in the ami-clockwise direction. Solution Theareais given by A = M ( F - I ± I O N + C ) Here M= 100 sq. em; I=9.918; F=4.254; C=23.521 and N = - 1 A= 100(4.254- 9.918- 10 + 23.521) = 785.7 sq. em. by a Example 12.8. The following readings were obtained when an area was measured planimeter the tracing arm being set to the nalural scale. The initial and final readings were 2.-268 and 4.582. 'J'fte zero o f disc passed the index mark once in the clockwise direction. The anchor poim was inside the figure with the value o f the constam C of the instrumem = 26.430. (a) Calculate the area o f the figure: (b) If the area o f the figure drawn be ·to a scale of 1 inch = 64 feet, find the area o f the figure. n . . A= 10(4.582- 2.268 + 10 + 26.430} = 387.44 sq. inches. r SoIuton. Since the tracing arm was set to the natural scale, the value of M = 10 sq.. inches. A =M ( F - I ± 10 N+ C) g The scale being I\"= 64 ft. Hence I sq. in. = 64 x 64 sq. ft. in:.Here F=4.582: I=2.268 ; N=+ I ; C=26.430 eExample 12.9. The perimeter of a figure is traversed clockwise with the anchor poim einside and with rhe tracing arm so ser rhal one revolution of rhe roUer measured 100 sq. em on the paper. The initial and final readings are 2.828 and 9.836. The zero mark rof the disc passed the fixed index mark twice in the reverse direction. The area of the Area of field 64 X 64 X 387.44 acres = 36.39 acres. 43560 izero circle is found to be 2352 sq. em. Find the area of the figure. nSolution. gThe area of the figure is given by .A = Mn + n (L' - 2aL + A2) nwhere etM = 100 sq. em. ... (12. 10 a) n ( L ' - 2aL +A')= Area of the zero circle= 2352 sq. em. n = F - I ± ION= 9.836 - 2.828 - 10 x 2 = - 12.992 .. Substiruting the values in Eq. 12.10 a, we get A= 100(- 12.992} + 2352 = - 1299.2 .+ 2352 = 1052.8 sq. em. Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net 312 SURVEYING Example 12.10. The foUowing observtJiions were mode with a planinreter. F.R. N Area I.R.. 8.286 0 w(I) The multiplier constant 2.326 5.220 +I (I) Known area of 60 sq. inches 8.286 both the cases Unknown area figure in area. w(1) The mulliplier constqilt (M)(2) the same A=M(F-1± !ON) with The anchor poinl was placed outside the setting o f the tracing ann. CalculoJe : wSubstiruting the values, we get The unknown and (2) Solution .E60 = M (8.286- 2.326 + 0), (2) The unknown area from .which M = _: = 10.027. sq. in. 50 asyEnbw11Wy.01eh1re8easn6q..p5EWtl.hTax2hiehna8neemi2cnmahpnezeatlesctheenrherodo,o1rna2m3tn.hp.1ac8peo1rh4aik.onp2trrleA.Toarc.hpwf=TieonathMihgsenfeo(t.Fplazldlower-aimwraacIoisel+n.d.bgpmoea1uiainsn0rtrssgeskeiNiadddd)eseoio,n=tfhgtthse1sthhee0.ete.wi0ns2teidahdn7remaieixatt(iela5o·mol.fbpni2agtaaea2rusnik0srndree-eed.odvf8nio,nc1t.tlw·hah·eu~· e8helt6ieoinn+rininneiadta1lhodien0aexif)n/ tg=achmasrle6oenaa9cdwrkw.k8ewfwhr0iienetasweseasqli5lc..dmem2;iirnee8ee.iaa6ancsdstuiiuanortfnhegneddes.s counter clockwise direction. · Find the area o f the zero . circle. Solution, Wilh the anchor point outside A= M ( F - I + 10 N) = 10 ( 1 . 0 8 6 - 5.286 + 10) =58 sq. inches. With the anchor point inside +A = M ( F - I ± 10 N+ C) ; Here A= 58 and N = - 2 5 8 = 10 (3.842 - 5.282 ~ 20 C) or 5.8 = ( - 21.440 +C) from which c = 5.8 + 21.440 = 27.240 hotorinfanecgtiheAnergreiEesvarxopi1amloum6oitn.if6poto.lnfezeemTtrho1h.oe2fe.T1tcwhhd2ieehri.scetdleameilns=TetcahiaesMnescuCoIerl.iefn=9nog2tg2fht7ehtewh.2meh4ow,e0fahenlettxhechelhae1on0frwrdt=rohapme2acoei7rinlne2tgath.b4efe0raoiornhmfmgsinqthgtp.bhe•leeaitncwizhse.eeidrenon1gb.e6ect8htiiwrseceel2eemtn2r. a..6ctihFneegimnd.hpioTntighhneeet and the diameter and the area o f Solution. (I) Area of one revolution of the measuring wheel = M = Length of tracing arm x Circumference of the wheel Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net ~ li CAI.C1JLATION OF AREA tracing point, minus sign 313 I'' = Lnd = 16.6 x n x 1.92 = 100 em'. will (il) ~ o f zero c i r c l e = \" (L' ± 2La + R') Since the· wheel is placed between the hinge and the he used with 2La. Hence, 1.68 + 22.6') = 2290 em'. by Area of zero circle= n ( L ' - 2a +R') = n(16.6'- 2 x 16.6 x recorded area o f a figure following readings Example 12.13. Calculall the poinl inside the from the the planimeter with the anchor figure : Initial reading= 2.286, jinai reading = 8.215 the counter-dackwise disc pnssed the indet mark twice in following observations The zero of the counting the instrument were not available, the direction. Since the constants of were also mode : the tracing poinl = 4.09 \" The distance the anchor point = 6. 28 \" The distance o f the hinge from o f the hinge from The perimeter o f the wheel = 2.5 \" The wheel was placed beyond the hinge at a distance o f = 1.22 \". n Area of zero circle= n(L' + 2La + R') = \" (4.09' + 2 x 4.09 x 1.22 + 6.28') = 208 Solution. (a) Ca/cul.ation o f inslnmumtol consllln/S M = Length of the tracing arm x its g Now circumference i = 10.225 (8.215 - 2.286 - 20) + 208 = - 143.88 + 208 = 64.12 ne -· PROBLEMS =4.09 X 2.5 = 10.225 in'. in. Inches. A= M (F - I ± 10 N +C)= M ( F - I ± 10 N) + MC sq. e1. sq. r2. ing.netbtarhoenODeuadifdnfssttfdihaeinrnsets·glcC3ytet.rawalaccaT:niushndlega11tt0e.0ah2a.rer36met8ha.lesaeTsottafhr1eeo0aS1afO5ztfhfe.s4ieiersntourbnmseqyaa.twurS1rkaya20isaml00or.dpfm2sscsteoahalnieesn'socdr(leIuMiSrs1dduOc8el=.ebd7p.y1a0bssa0eetdwsp2qle0a1t.e0hn6nee.i4mmitneht)ede.erTxc2whh52mae00iti.ahn8irnktilhtinieoaenl,ac3rne20etc02haih.end4oirntihngpeeogwci3unlaSo1tlsOac9okt.8u3w.t6bsiio2sdn8eenddat4hani01rel0d7}e(f.'Uc6itfs.iiauPonnrna.de)l. What is Simpson's rule ~ Derive an expression for it. a chain line 10 an irregular The following give the values in feet of the offsets taken from Cwttihhfaheeliicltesroealmtiaash4tdee.eindTgmsrtheahetewertaiennsrcaugoorrletrloidaeonrefgdaootfahfwresatecahaasteplreaol2acffni.io9gnaif8mug6rree1eat.cerrmtereavmnrogealucunto=dliarordnte2shpd0•lionaatmnrctehe2htOar1odersiec.nmlpgoocoiknxfwt 1i6so.e2oe6tsm0didirreewe,cvittoiaholnunot.tithoheFnersianndifncishuttohhrreee clockwise direction point outside. With was traversed and area of the fisure Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net 314 SURVEYING S. A figure is traversed clockwise with !he anchor point inside and with the ttacing arm tb.a1 0110 revolulion of !he roller measures 10 sq. incbes on !he paper. so set wwwpacTmsFaqlrorihoeeen.iceaanad:stkinuozw.o.trfeiheTT6ifn7rIsesA.os.h.!ehRhi,!eedhe1Da.Weem=f0wr.ieeuzzgbasiae3aetsuczrahTr.qrckrre0tooeii.hor0nbotefi9hegoicmsinesficr!co!ahhacmh;iabrte!rrelrkehncemsaeseFlcevaepzhor.eniooResvoroftdlrn>raei.fois=fseldb!:odchptyauh8pecfdoone.ilciiSbradufnpzlta4f!dopn:lteeisla7icaidre1srmnokpie0.cniwnpemesr1rcabetciib.0edeessesiartsespneeseT.e:br1adlhceaei6nnftpend4iiiavrs!d.h11s3siepente6s1lxswleyi4aoDtpwd..iinf3staleiihaqi1lxtsmsih!.eenchattedresrniit!ibqnahedttscecr.ehiininhanfdpdeitriag!nnsereehni.aaxcxeoncadhlnchrWieieonmmmrvpserga.blaaaesrrardipsr.etokkWoiovnuaTioigsrslnhsloloesauettnexttttmchioa:2oepeeurtein.eehs3tthrsaa4iiinnitoro3m!2dehd.feao0ees!!nh0thateaoeeh9neroandefrfdroearre!aenrefhdvt-vovdheoe9aleeeotllr.ereun8ffs3fsrrti.ieemSgig1ow!4uuhc9nidnd7roireteeiiirhrrnor?reregefeifeCcstissshgtptppiteiuohitooeetrrensnanccea.(vdtt.(iinUresvvUaco1rTr.eeThsPl.elllhIe1hoyyea'.de0)re)..r. .Ea2. 820.38 sq. yds. ANSWERS 3. 261 sq. em. ::.(~: . s4. 11.944 hectares yS. 119.69 sq. inches. E6. 116.6 sq. inches. n7. 119.7 sq. inches. Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net @] 'IIII Measurement of Volume 13.1. GENERAL adopted for measuring the volume. \"flleY. are There are rhree melhods generally (I) From cross-sections · (il) From spot levels (iii) From contours earth work while methods are commonly used for lhe calculation of capacities. the . The first two generally adopted for lhe calculation of reservoir third method is 13.2. MEASUREMENT FROM CROSS-SECTIONS into a series The total volume is divided measurement n are first calculated by standard formulae fundamental solids on which This is the most widely used method. depends upon the planes of cross-sections. The The spacing of lhe sections The area of of solids by the prism, wedge and prismoid. required in the measurement. of the gro.uod and the accuracy is based are lhe character gdeveloped below, and the volumes of i ~~the prismoids between successive· cross- ntrapezoidal formula or · by prismoidal lhe cross-section taken along lhe line e --- ..--------------rformula. er ---- -- ~be classed as sections are then calculated by either (a) (b) i(Figs. 13.1 a and 13.2) n(2) Two-level section, , .. ~' The various cross-sections may g(Fig. 13.1 b and 13.3) Cirtling (3) Side hill two-level section, / >( • \" FUiing (C) .(Fig. 13.1 c and 13.4) (I) Level section, n(4) etand (5) (d) ----------------------·· Three-level section, 13.5) (e) (Figs. 13.1 d and Multi-levei section. (Fig. 13.1 e and 13.6) FI!J. 13.1. (315) Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net 1'--.1l1i' II 316 SURVEYING II ii wtGeneral notations : li b = lhe constant formation (or sub-grade) widlh. h = lhe deplh o f cutting on lhe centre line. w, and w, = lhe side widlhs, or half breadlhs, i.e., lhe horizontal distances from lhe ,, centre to lhe intersection of lhe side slopes wilh original ground level. ,'. h, and h, = lhe side heights, i.e., lhe vertical distances from formation level to lhe wintersections of lhe slope wilh lhe original surface. n horizontal to 1 vertical = inclination of lhe side slopes. wm horizontal to 1 vertical = lhe transVerse slope of lhe original ground. A = lhe area of lhe cross-section w(1)~ (Fig. 13.2) .In this case lhe ground is level Etransversely. ai:. h,= h, = h lf---w c w_ 'i sw1 =wz=w=b2-+nh nh yEA=a+r~+nh ' ·; !h :: n~ •'' ~-----b------~ ... (13.1) ·' FIG. 13.2 (2) TWO-LEVEL SECTION (Fig. 13.3) wt 0 be lhe point on lhe ceotre line at which lhe two side slopes intersect. Hence BH:HO ::n:1 or HO=~ ~) ~) ~:}area DCEBA = t.DCO+ I> ECO- t . A B O = i l ( h+ Then w, + ( h + w,- r b ·, b1.} + 2n 2n J-lh1 r = 2 \\ (w, + w,) ... (13.2) The above formula has been derived in terms o f w, and w,, and does not contain lhe term m. The formula is, lherefore, equally applicable even i f DC and CE have different slopes, provided w1 and w2 are known. The formula can also be expressed in terms of h, and h, . Thus, Area DCEBA = I> D A R + t. EBH + I> DCH + I> ECH ... (13.3) b b } b }= ZIlf. 2 h, + 2 h, + hw, + hw, =2I { 2 (h 1 + h2) + h (w1 + w2) The above expression is independent of m and n. Let us now find lhe expression for . w11 w2, h1 and h2 in terms of b, h, m and n. Bl=nh, ... (1) Also BJ=HJ - H B = w,-2b ... (2) Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net MEASUREMENf OF VOLUME 317 Also, :. nh. = w•-zb ... (1) •:'' :''h1 w, =j (h, - h)m • l Substituting lhe value· of w, ... (11) in (,)';we. get b · · ....._ nh, = ( h , - h)m - 1 ,. ·1-4r! : or h1(m - n) = mh +b- lAlif::::b/2 i .__;.8./ ------ ~ J 2 - - ! ..- - w .1' ,w 1 :''' or h,= mm::1(i\\h. + . !... ) 2m Substituting lhe value of h, in (1), we get : V1 VV'I b b mn( b)w, =-2+nh, =2- +m- -- n- h +2-m- :''''' :' ... (13.4) I\",!•.•''•o'I'':' it can be FIG. 13.3. ... (13.5) .shown lhat 13.2 and simplifying, Proceeding in similar manner, ..h,= h2 in equation 13.3, ngin b!-Hn b)and ii!m+ii\\(- h _ l!._ ) 2m wz=-+ m+n h--- ... (13.6) 2 2m we get eer[ and w in equation ... (13.7) Substituting lhe values of w1 . 2 we get A r e a =m-' n- ( h +b- ) '-b-2 mz-nz 2n 4n Similarly, substituting lhe values of w,, w,, h, and in(Fig. 13.4) Area= n.( ~ ) ' +nf(bh + nh') ) . . . (13.8) ~~~ In this case, the ground slope crosses glhe formation level so lhat one portion .of lhe area is in cutting and lhe olher (3) SIDE HILL TWO-LEVEL SECTION nin filling. etAlso, Now, Bl=nh, BJ =.HJ- HB =w,-2-b nh, = w,- b ... (i) ... (ii) 2 w, w, = (h, - h ) in ~ But A Solving (i) and (iz) as before, we get FIG. 13.4 Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net SURVEYING :r 318 h · =mm--n·n- ( -2bm+ h ) ... (13.9) and w1 =b- +mm--nn- ( -2b-m+ h ) ... (13.10) wwwAlso 2 . . .(iil) . . . (iv) Let us now derive expressions for w, and h, .. /A= nh, /A= I H - A l i = w z - b / 2 Also .b .. - n. h 2 = W 2 -2- W,= (h + hz)m .or h z =mm--nn- ( -2bm- - h ) Ea(...!!....-£ £Hence =b- - mh nh, =(h + h,)m- b / 2 or h,(m- n) 2 ... (13.1 ~) syof w, = 2 + nhz = 2 +--m\"!!!-..n.. 2m h) ·!'-, . . . (13.12) expression (say). =H ..-' (b )' En~bm -+mh By inspection, it is clear that the expressions for w, and w, are similar ; also .r h1 and h2 are similar, except for =MAD= A, Now area of filling= &PBE =A, - h in place of +h. (say), And, area of cuiting A, =f(PB) (EI) £+mh){ m:n +h)}= i(m -n) ... (13.13) ·· J £-and Az=!z (AP) (JD) = l(!!.- mh]_{ _m_'-!!n._(.2..m.!!._-h)} mh ) ' ... (13.14) 2 ( m - n) z2 (4) THREE-LEVEL SECI'ION (Fig. 13.5) ground to one side and 1 in m, be !he Let 1 in m1 be the transverse slope o f !lie lhe cross-section. (Fig. 13.5). slope 10 the oilier side of ihe cenue line o f The expressions for w1, w2, h1 and c 1 inm, E h, can be derived in the similar way a; for case (2). Thus, ' ./ ' / l \\ \\t\\~- :. ,I I .!.)w, = ~m, n ( h + 2n ... (13.15) I ,I I I .' I I h+_!!._) ,I :I 2n :I:h _,.,·, . ' :ttl w,= m,n ( ... (13.16) I ,I I !I ,.,-',I iii;'=lit .., ·............ !I b)h,= : (h +w-1) = ~1 h + - ......'· ... : , / ______ J ml : m,-n 2m1 _,. ... (1_3.17) ---w,--- h, = (h -;w;;,)r\"\"'\"+' n (lh- b) FIG. 13.5. 2m,) ... (13.18) Downloaded From : www.EasyEngineering.net

MEASUR!!MENT OF VOLUME llDownloaded From : www.EasyEng3i1n9eering.net AHD + & BHE + & CDH + & CEH ~. ' h, X£)+ (h, X £ ) + hw, + =[ I The area H( hw,]ABECD = & ~ (h, Jz,)+ ~ (W, w,) ]. •. (13.19) l= \"' + + (5) MULTI-LEVEL SECTION (Fig. 13.6) tl. In the multi-level section the co- · i ordinate system prov_ides the mosi_ general •••••:' ~ - method of calculating the area. The cross- . •'••:• •' section notes provide with x and y co- B/._t___ ._:f ·'\" y t I' ordinateS for each vertex of the section, the origin being at the central point (H). [ 4 - -~- --W--[ 4-- W-1-b----W-1--~-~.- -+-i --~ The x co-ordinates are measured positive to the right and negative to the left o f H. Similarly, t h e y C<HJrdinates (i.e. the heights) are measured positive for cuts and negative for fills. In usual form, the notes are recorded as below: h , h, h0Hw,, H, FIG. 13.6 ;;;;- w, w, C<HJrdinates n -b/2r at extreme of formation points If the co-ordinateS are given proper sign and if the right), they appear and B are also included (one at extreme left and other g There are several methodsA algebraic sign is placed on the inappear as : as follows : ...&_ ...&_ h H, H, 0 -Wz -w., b/2 0 0 + w, + w, . + methods, the opposite The co-ordiantes then e -b/2+ to calculate the area. In one of the opposite side of each lower term. ersum isum !!. ~ ---l!J_ _ _o ___o__ ___!L __h_,_ nThus, we get -w1 + 0 + W1 - + W2 - +b/2- - w2 + tlw algebraic imlltip/ying each upper term by The algebraic the signs facing the upper term. area o f the cross-section. gA= f[h, The area can now be computed by· o f the two adjacem lower terms, using o f these productS will be double the -- - .neFor a numerical example, see Example 13.6. + O)+h (+w, '+- W,) + H, (0+ W,)+H,(- W, + b/2)] t13.3. h,( + w, ... (13.20) (+ b / 2 - w,) + THE PRISMOIDAL FORMULA obtained either bpyrispmriosmidsoidbaeltwfeoermn uslau.cceWsseiveshaclrlosfsi-rsset ctdioenrisvearaen expression for Th..e......volumes of the by trapezoidal formula or prismoidal formula. Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net 320 SURVEYING A prismoitl' is defined as a solid wlwse end faces lie in parallel planes and consist o f any two polygons, not necessarily of the same nwnber o f sides, the longitudinal faces being surface extended between the end planes. wperpendicular to the two end parallel /.11of triangles, parallelograms, or trapezium. The longitudinal faces take the form w .. ,,A, = area of cross-sO.:tion of· oneA., . 0, • Let d = length of !he prismoid measured ,,••• wA, = area of cross-section of the 9ther planes. /,~. , / ·'· ...........~~~~·-·-lf.~;:~l.::~. ::::~~:yAend plane. M = the ntid-area = the area of the ·\\.Eplane ntidway between, the end l~·::~;;a,J.:\\~<~~'· {'planes and parallel to them. ·, ----------;~-,:' -?' 2 / / / ),r,ii ··..end plane. ,/ ,,. , :r if In Fig. 13.7, let A, 8 1 c, D, be one end t:Ii splane and A2 82 c, D2 be another end plane yparallel to the previous one. Let P Q R S T \"i! . Erepresent a plane ntidway between the end . . li faces and parallel to them. Let Am be the , )'' narea of ntid-section. Select any point 0 in :,1 the plane of the mid-section and join it to i!l the vertices of both the end planes. The li .,.....\\\"·············~ .. / {; il II I\\ . 01 II \\\\ ' ! II I \\ II \\ :I II I ', ' I // \\ I '\\ ' '1 I ' !I 'I ' ' I '/ \\ ! ' \\ I \\· c, prismoid is thus divided into a number of FIG. 13.7 pyramids, having the apex at A and bases on end and side faces. The total volume of the prismoid will therefore be equal to the sum of the volume of the pyrantids. Volume o f pyramid OA,B,C1D, = f ( ~~ A1 = { A , d Volume of pyrantid - ' ,£, I - iOA,a,c,v, = A, d. To find the volumes of pyrantids on side faces, consider any pyrantid such as OA 1B1B,A2• Its volume=j<A 1B,B,.42) x h, where h =perpendicular distance of PT from 0. i j= j<d x P'I)h = d (26.0P1) = d ( M P 1 ) fSimilarly, volume of another pyratnid oc,D,D, on the side face = d (<I.OSR). :. Total volume of lateral (side) pyratnids = j d (PQRST) = j (Am) iHence, total volume of the pyrantid = A 1 d +~A, d + j A m d or V= d 1 +A,+ 4Am) ... (13.21) 6(A Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net MEASUREMENT OF VOLUME 321 1b1e1tewa eAeLn,e!Afi,ur,sstA·i n,,toh.w•r•e.e•.t.a..lsAceuncltaiostepnasct,heeditsvaotvluomaluemcoeonfswteaainlrltthbdeiws,toarnfkrcoembedtweeqaeupnaatniao. nnCu1om3nb.s2eidr1e,orifngsecthtieonsprihsamvoinidg_· = (2 tf) (A 1 + 4 A , + A,), 2d being the length of the prismoid. 6 volume of tjle second prumoid of leogth 2d will be Similarly, = 62 d (A, + 4 A, +A,), and volume of last prismoid of length 2d will be = 62 d (Ao-2 + 4 A n - i +An) Summing up, we get the total volume, ... (13.22) V= 3d [A,+ 4 A , + 2 A , + 4 A, ... 2An-2 + 4 An-I +A,] lor V = 3 [ ( A , + A , ) + 4 ( A , + A , .... An-J)+2(A 3 +A, .... A , . 2)] 13.4. aTphi.tsr.ea2istedddasl~esopuamrk.~nreOfrWy~:n a-arsiQ~tShrDepwe~lnuam'sre ~ebetewefvoenr et~h\"en!rMem.nuaimnibnergre. , it is necessary 10 rehave· ngareas. of sections, the end Snip must sections may be calculated by primoidal formula. i Y=di~+~ (AVERAGE END AREA METHOD) is the mean is given by neerinotetochhxoffeeaupclevdyctnlrooydaTlnbumheeaiaincisrdedpeosarob.iiosnstammfTitbnroabteehuyieiddeen.gbnpboetIriynendisqlaymustcahoacoremileeifdpaeteuotdhs(coheetafhaswveeopaiisrfntd,higsiptfmthyfhspeeroeuryaifemdrfpvnaicroicnidiilsetesuimndmbitsocseeoitdaamwhcailsaspceloluoefc)nrsfaaoeltcidrhctsmyhueelouoasvltfviaeena.odprcvlreueTeiamrssnhamatedgeicmsetuatchasaaoatelenrnncdedeae.aclaacwdrHuotcohelfooandwtrgetsrheeedoissecvliteadirkoone,nnnndmodltyhwsaitsheynaa;aimnnpadbwopesetlthhinneitocbchoddeheet,TRAPEZOIDALoftheend assumpdon that the mid-area This method is based on the the prumoid of Fig. 13.7 In the volume of that case, ... W gprismoidal correction. Let us now calculate .neareas tVolume between next two sections =~(A, +A,) a number of sections having the volume of earth work between A , A,, ..... An. spaced at a constant distanee d. . sections = id (A, +A',) Volume between first two Volume between ·last two sections =~(An. 1 +An) .. ... (13.23) ' .LUO....._. ............... Downloaded From : www.EasyEngineering.net

JDownloaded From : www.EasyEngineering.neti: SUR\\IEY!NG m i! '• ii wII 1aIAvlSer3o'ev,l.aua5ewlm.lw1fe'aATo,LNssyrewesomsHctaw,t'ussE.iulotaahscn~.t•ae's,ldc~.c'\"ahu2lOLlec'auaetIreltevalditeDteceA, r.bA,,,A1ytwh,.L;e~tte,ho.ot_hehO'iWpCtt\"(hrbplOi,eessrli+hhmRls~!omoclRonu-raihhoElOdrd2)ieaC.ds,alsaTb-leseIceftOoaocccntrsroJ.imurdo,!reb_rnucet(rltrcaCiaeotacifptnoeta)enrnAdtdfho'eti=sfrotrhohoemfet't!(>qhb~eyeratrs+hilcsesnames·irneovtods)oi.delwtuahclmheetefnoidricomfthfneuecrleaan.eatcn<edTeodhnbesebee!yccWteoinoterhendrenescaettainohorddeen wVolume by end area: rule is given by w. JAgain, EMid-area V=!!. nh')) = d[ bh + bh' + nh' + nb'' ... (1) 2 [h(b + n i t ) + h'(b + 2 2 2 2 aVolume by prismoidal formula is given by = -h +-h ' 2 the mid-areas ..V= :rh(b +nit)+ h'(b +nh') +4 ( h ; h') x (b + n(h; h') )]centreheight J yEor V=~ [3bh + 3bh' + 2n h2 +2 n h''+ 2 nhlt) = ( -h +2-h ' ) [ b + n ( h-+2-h ' ) ] nSubtracting =d [ -bh +b2-h' +n3h-' +n3h-'' +nh3-h' . . .(it) 2 (it) from (1), we get the prismoidal correction, c, = 6dn . - 2 . . . (13.24) (h h') The standard Similarly, the prismoidal correction for other sections can also be derived. expression for Cp are given below. · For !wo-ievel sectio11 ; /nCp = (w, - wt') (Wz- w2' ) ... (13.25) I For side hill two-level section : (~ + m'h')} mh)-1c, (cutting)= II I ~ (~n (w1 -' w,') { + ... (13.26) . . . (13.27) 1 mh)-@-Cp (filling)= ~n (w2 - w,') {(~- m'h')} For three-kvel section : . c, = d ( h - h')[(w1 + w . ) - (w,' + Wz')], ... (13.28) 12 I'I . 1th3e.6e. odTThHeseEcptiCroinsUmsRoaVirdeAalTinUanRpdEaratChlleeOl RtprRalapEneCezsoT.idIOaWlNhefonrm.thuelaecenwtreerelindeeriovfedcutotinngthoer assumption tha! an embankment I Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEnginee3r2i3ng.net MEASUREMENT OF VOLUM!! iwftSooer·rctehWveai'vrnieaodrpueasarisna!llloeepcflltaiocpnrnl,oas·sniseat-srse,eiscagtinicovdonemsnthmtebhnoeunlsoapwpgp.relaytctlintnitchgeseoemcqtoouerirvcecaacalltsecieonustnl,aatfetrhoeerathsCceoUrIvrVaeonacldtutuimroetnh.eenTfaohsrteoicsfuutarstvoehademathrredeeodepisxripssaermepcstpoisoliiidoneandsl formula. Level section : No correction is necessa<y since the area is sytrtrnetrical about (1) the centre line. section and three-level section : (i1) 1Wt>-level d b)C , =6-R( w ,2 - w ;')( h +2n- . . . . (13.29) I where R is the radius of the curve. curvature correction to the area (iii) For a two-level section, the . . . (13.30) = -Ae per um.t 1ength A e =the eccentricity, I.e., horizontal distance from the centre line to the where area - w,w, (w, + w,) ... (13.31) centrOid of the 3An The correction is positive if the centroid and the centre of the curvamre are to opposite side of the centre line while it is negative if the centroid and the centre the curvamre are to the same side of the ng Hwhere centre line. .- the (iv) For -side hill two-level section : of ine H~pieand . . . (13.32) 13.1. A Assuming 1he growui 10 ethe volume contained ·in a rin metres ·~.2. 3.7, ~3.8, Correction to area = ~ per unit length e= w, + ~- nh} for the larger area ... (13.33) e= Wz + ~ + nh) for the smaller area !<.(13.34) iSolndon. side slopes Q o 1. nFor a level cenrre ~;ne, ralculale railway embankment is 10 m wide with direcifon rrar.sve:se to rhe l 20 m intervals being be le'vel in a metres, the centre heights a length of 120 gSlope is 1f : I. Hence n = 1.5 4.0, 3.8, 2.8, 2.5. .netThe section, the area is given by A = (b + nh)h . areas at different sections will be as under : m2 A , = (10 + 1.5 x 2.2) 2.2 =29.26 m2 =; A , = (10 + 1.5 .x 3.7) 3.7 57.54 m2 m2 A , = (10 + 1.5 x 3.8) 3.8 =59.66 ; A , = (10 + 1.5 x 4.0) 4.0 = 64.00 m' ; A , = (10 + 1.5 x 2.8) 2.8 = 39.76 m' A , = (10 + 1.5 x 3.8) 3.8 = 59.66 aod A,.= (10 + 1.5 x 2.5) 2.5 = 34.37 m2 Volume by ·trapezoidal rule is given by Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net '.'I[':' SURVEYING li 322 . iJ 13.5. THE~PRISMOIDAL_ CORRECTIO~(Cp) · iL As stated earlier, the prismoidal correction is eqyal to the difference between the volumes as calculated by thO\" end-area formula and the rismoidal formula. Tlje correcnon ·i! area formula.It L Let us calculate the prismoidal correction for t!Je case when the end sections areil wlevel sections. Let A, w~. Wz, h1, h2, etc., refer to t~ss-section at one end and A', w1',WJ.', h,', h2' etc., to the crOss-section at lhe other end. IS always r. ve, 1. e. it should be subtracted from the volume c cu ated by the end wNow w nb'']V= t!_ [h(b + nh) + h'(b + nh')) = d[ bh + bh' + nh' + .Eil I!' a:! · A= h (b + nh) and A ' = h'(b + nh') Volume by end area rule is given by 22 2 2 ... (!) s]\\ Volume by prismoidal formula is given. by2 centre he1'gbt = -h +-h ' 2 Agam. , the yV=~[h(b + nh) +h'(b +nh') +4 ( h ~ h') x (b + n(h; h') )]!DI'd-area Enor V= ~ [3bh + 3bh' + 2n h' + ~}J·h'' + 2nhli) Mid-area = ( -h+2-h')[ b+n (h-+2-h')] .,, < ·!I'!I =d [ -bh+b-h' +nh-' +nh-'' +nh-h' ] ... (il) 22 3 3 3 Subtracting (il) from (1), we get the prismoidal correction, Cp = ~ (h - h')' ... (13.24) Similarly, the prismoidal correction for other sections can also be derived. The standard expression for c, are given below. · For .two-ievel section : c, = / . (w, - w,') ( w , - w,1 . . . (13.25) For side hill two-kvel section : j mh)-I ~ + m'h')} 1:Cp (cutting)= n (w 1 - w,') (~ + ... (13.26) ... (13.27) mh)-I (~- m'h')) nCp (filling)= 1: ( w , - w,') j(~- . . .(13.28) I For three-level section : . assumption that II c, = ~ (h - h')[(w, + w,) - (w,' + W>1]. an embanlanent ,[ il 13.6. THE CURVATURE CORRECTION !; The prismoidal and the trapezoidal formulae were derived on the the end sections are in parallel planes. Wben ,the centre line of cutting or l Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net 323 MEASUREMENT OF VOLUME iS ·cWYed in plan,· it is common practice to calculate the volume as i f the end sections were in parallel planes, and then apply the correction for curvamre. The standard expression for various sections are given below. In some cases, the correction for curvamre is applied to the areas o f cross-sections thus getting equiva/enl areas and then to use the prismoidal formula. (1) Level section : No correction is necessary since the area is symmetrical about the centre line. (il) Two-level section and three-revel section ' ... (13.29) d ' ' ( b)C, = 6R (w, - w;) h + 2n I wbere R is the radius of the curve. (iit) For a two-kvel section, the curvamre correction to the area =A-e per um'tl gent b . . .(13.30) A e =the eccentricity, i.e., horizontal distanCe from the centre line to the where centroid of the area +W1W2 (WI W2) ... (13.31) 3An n Correction to area = ~ per unit length The correction is positive if the centroid and the centre of the curvarure are to < the opposite side of the centre line 'while it is negative if the centroid and the centre gi Hwhere n Hand of the . curvamre are to the same side of the centre line. (W) For .side hill two-kvel section : . . . (13.32) ~piee13.1. A railway embanlcment is 10 m wide with side s l o p e s Q o I. eAssuming tile growui 10 b2 Ie11el in a direction fraiisrerse to th~ centre line, calcu.WlP. rthe volume contained ·in a length of 120 metres, the cemre heights at 20 m intervals being inin metres ';2.2. 3.7, ).8, 4.0, 3.8, 2.8, 2.5. Solution • gFor a level section, the area is given by A = (b + nh)h · e= w1 + ~- nh) for the larger area ... (13.33) e= w, + ~ + nh) for the smaller area .1<.(13.34) .Slope is Ii : I. Hence n = 1.5 nThe areas at different sections will be as under : eA = (10 + 1.5 x 2.2) 2.2 = 29.26 m2 ; A,= (10 + 1.5 .x 3.7) 3.7 = 57.54 m' t1 A , = (10 + 1.5 x 3.8) 3.8 = 59.66 m2 ; A• = (10 + 1.5 x 4.0) 4.0 = 64.00 m' ·A,= (10 + 1.5 X 3.8) 3.8 =59.66 m2 ; A , = (10 + 1.5 x 2.8) 2.8 =39.76 m' and A,.= (10 + 1.5 X 2.5) 2.5 = 34.37 2 m Volume by trapezoidal rule is given by Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net SURVEYING 324 V=,d [ -A , +2-A+, A , + A , + ... A , _ , ] ·.·. .r(= 2 0 t 29.26 + 34.37 + 57.54 + 59.66 + 64.00 + 59.66 + 39.76] 6258.9 -·' m °[ w= 2 29.26 + 4 (57.54 + 64.00 + 39.76) + 2 (59.66 + 59.66) + 34.37] = 6316.5 m'.2 = ~ple313.2. A railway embankment 400 m long is 12 m wide at the formation Volume by prismoidal rule is given by wlEvel and has the side slope} to 1. The ground levels at every 100 m along the cemre V= ~[A,+ 4(A2 + A , + A , + ... ) + 2(A, + A s + .... + Ao-2) +A,] wDistance .EThe formation lEvel at zero chainage is 207.00 and the embankment has a rising gradiem o f l .in lop. The ground is level across the cent[~ line. Ca/culaJe the voluine line are as under : aof earthwork. 0 100 200 300 400 sylevelR.L. 204.8 206.2 207.5 207.2 208.3 EnDlsutn<t · Solution. 100 the formation Since ihe embankment level is to have a rising gradient o f 1 in at every section can be easily calculated as tabulated below Ground Fonnatlon lewl Depth o fjiOing o. 204.8, 201.0' \"1 2.2 100 206.2 208.0 ..... 1.8 200 1.0\"/.S 209.0 l.S 300 207.2 210.0 2.8 400 208.3; 211.0 2.7 The area o f section is given by A = (b +~h) h = (12 + 2h) h \" A , = (12 + 2 ' 2.2) 2.2 ~ 36.08 m' , A2 = (12 + 2 • 1.8) 1.8 = 28.06 m' A , = (12 + 2 x 1.5) 1.5 = 22.50 m2 ; A, = (12 + 2 x 2.8) 2;8 = 49.28 m2 and A , = (12 + 2 x 2.7) 2.7 = 46.98 m2 Volume by ttapewida! rule is given by lV=d [ -A , +-A· +, A , +A,+ .. .A,_, 2 ~36 08 = 100 [ · 46 98 + 28.06 + 22:50 + 49.28] = 14,137 m' · Volume by prismoida! rule is given by V=~[ (A, + A , ) + 4(A, + A , ) + 2(A,)) V = 1 ~ [(36.08 + 49.98) + 4(28.06 + 49.28) + (2 x 22.50)) = 14,581 m'. Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net..·.·i 325 ·,i,,,l, MEASUREMENT OF VOLUME ~le 13.3. Find out the volume of earth work in a road cutting }32- metres !:1 long from the following data : · \"' l,i,'il The Jorrrla!;on width 10 metres ; side slopes 1 to 1; ;,erage depth of cutting along :ii the centre o f line 5 m ; slopes o f ground in cross-seCtion 10 to 1. Solution. and n, is given by equation 13.8 JIi The cross-sectional area,·- in terms o f m (2--\\a.et) II n ( !>.y + m2(bh + nh2) :! A= 2 • m2- nz Thus, Here n=l ; m = 1 0 ; h = 5 ; and b=10. :·11 !( J~~ + 102(10 X 5 + 1 X 5') = 7 6 m2 :.~ ,1 li A= 102 - 12 · • ~-I / V = A x L = 76 x 120 = 9120 cubic metres. d / E x a m p l e 13.4. A road embankment 10 m wide at the formation level. with side 'I slopes o f 2 to 1 and with an average height o f 5 m is constructed with an average gradient 1 in 40 from contour 220 metres to 280 metres. Find the volume o f earth work. Solution.n Area Difference in level between both the ends o f the road = contour 280 - contour 220 = 60 m Here, b = 10 m ; n = 2 m ; h = 5 m. g A = (10 + 2 x5) 5 = 100 sq. m. i. . Volume of embankment= Length x Area= 2400 x 100 = 2,40,000 cubic metres. Length o f the road = 60 x 40 = 2400 metres. nExample 13.5.~ of (b + nh) h the cross-section= ee50 metres apart. r 1.7 i1. ~ three level cross-sections at two sections The foUowing notes refer to n2.9 gThe widlh of cuning at the formation level is 12 m. Calculate the volume of cutting Sration Cross-section 2.8 4.6 .between the two stations. 7.7 0 nSolution. 10.6 3.7 6.9 - - -2. 8.9 0 12.9 4 eTbe area of a three-level cross-section, from Eq. 13.19, is given by tA= [ (w, + w,) + ~ (h, + h,)] At station I , b = 12 m ; h =2.8 m . w, = 10.6 m ; h, = 4.6 m ; w, = 7.7 m h, = 1.7 m Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net 'I'TT 326 SURVEYING A,= T2.8 (10.6 + 7.7) + 12 (4.6 + 1.7) = 44.52 m' 4 wVolume by trapezoidal formula is given by At station 2, b = 12 m ; h = 3. 7 m w, = 12.9 m : h, =6.9 m ; w, = 8.9 m ; h, =2.9 m w=}V (44.52 + 69. 73) x 50= 2856 cubic metres. A,= T3.7 (12.9 + 8.9) +·412 (6.9 + 2.9) = 69.73 m' To wmassiudm-i=ng, .Eb= 12 m : h let us first calculate the mid-area by calculate the volume by prismoidal rule, of those at the ends. Thus, for the w, h etc., as the average the quamiti.Saw, = we bave 7 syw1 - 2.8 + 3.7 - 3 . 2 5 m 2 310.6 + 12.9 2 EAm= ; -11.75; h , - 4.6 ; ; 6 . 9 - 5.75 nV 7.7 + 8.9 =8.3; h , - 1.7 + 2 . 9 -2. 3 2 2 5 (11.75 + 8.3) + (5. 75 + 2.30) = 56.73 m' = L ( A , + 4 A m + A , ) >=5' 60( 4 4 . 5 2 + 5 6 . 7 3 x4+69.73)=2843 , 6 m. Example 13.6, The following are the notes for a multi-level cross-section for a 10 m and the side slopes are 1 to 1. ColculaJe trhoead.croTshse-sewaiidotnhal oaf rethae. road bed is ~z5 s~9 5LBo172 ~1 8 Solution. the co-ordinates of the fonnarion I f thl! co-ordinates are given proper sign and if at extrell\\e right), they appear as points are also included (one at extreme left aod other follows : 2:2_ _ 2 ! _ _o_ Q_ ..l1_ 2.9 3.8 +5 -5 -7.5 -5 0 +7 + 10.8 Following the method explained in § 13.6 aod placing opposite algebraic sign on opposite side of each lower term, the notes appear as follows : the 3.8 5.2 5.8 0 0 2.5 2.9 +To.8 - +5- 0 +7 - ::-s+ =7.5+ ::-s+ the algebraic sum of The area can be tbecerommdsp,ouutubesldeingbtiy'liehmeaureslitagipnlsoyifnfgathcieneagcchrtohseusp-spueepcrptieotrne.rmtermb.y The algebraic sum of two adjacent lower these products will Thus, we get Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net 327 Ml!ASUREMBNI' OF VOLUME A =-} [2.5 (+ 5 - 5) + 2.9 (+ 7.5 + 0) + 3.8 (+ 5 + 7) + 5.2 (+ 0 + 10.8) + 5.8(- 7 + 5)] ur .=i;IO + 21.75 + 45.6 + 56.16- 11.6] =55.96 \"1Tritpatnheh3rhcaeic.ustt7teamshh.nr,etto,gahIVrtleniihenwszOaseoimtftdLhnoohdieituUsrsaetrvalpMtrenmtiphrCiacEntrenoieoctcogmhjlaFeilfonele!cbd!sreet.ee,.isdx.odlOaoctngnMhwaodeevsfopamltfStfahaiiePnoeteanhaenltOydes.lnoToruawowAirp,oneosLpgeshsrbeEtuke,rtmihVscgqcceioehEluo>mtlLanectrasehlvoSieinrnaesf.dtletssdofTidrioltihlbnifena[riFstsgedheeiciegavtdai.aorrietdntfsh1cigientro3lvheetgm.oee8dretyabphr(tresareoeic)sb)omma.geamrrofteTaovaeadbrerdneeeddiinfiafstdrwenaoeorldcksiwelotnalnaanotfnbnttwgheeutelenroeml.eatlhvboaaLeaerwbitregeliceteodhoorvntnufesrshsltetrosrpUqUabrsriDusuuezcsatrCsotufirenaoamentltcnseiateeed..sl base. us consider the ;!'I;''P•\"' abed of Fig. 13.8 (a). If h,, hb, h, and h , represent of excavation of the Let of the four comers, the volume right trUncated prisn\\ will ' the depth be given by V=(h.+h•+~+h•]xA ...(13.35) n the ~ 4• ., = averate height x the horizontal g -1( area of the rectangle. L 13.8 (b). If h., h• and h, are in = (average depth) ·x horizontal area of the triangle of Fig. Similarly, let us consider the triangle abc volume of the truncated triangular prism depths of excavation of the three corners, the e2 is given by e~ v = ( h , + h3• + h , l X A ... (13.36) rinI o ~~ 2 43 4/( ~I( ., \" EB g2 .net{a) {b) FlG. 13.8 Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net m SURVEYING ()( Volume or a group or rectaDgles o r squares ba.ntg the same area 'T Let us now consider a group of rectangles of the same area, arranged as shown iotatnohsnelryFteah,itrges.afeor)me.1re3erc.T8ethachentuia{ggasnlh,)ge.tslseinIsa(tsruejFwociihcginlo.limnasgb1me3aoa.t8nsteeett()onha, e)t,abwnycodeoaricrnsnehoescmrptaecenco(igtrnihnlodeeensiircgah(ttshhteusaedcitgh.ahsroeotanms wcetaohitmlelo'bmf)bf,oietgnhsueorutmeosheedebfiogyhuhealrsilgni hurmtemascrabtenaaeynrreugs)sl.teecismdome(sosmunoccanhes wLet wl: h,· = the sum of the heights used thrice. l: h . = the sum of the heights used fuur times. wA = horirontal area of the cross-section of one prism. l: h, .= the sum of the heighlli used once. l: hz = the sum o f the heighlli used twice. .EV A (II:k1 + 21:11, + 3l:h, + 41:/z.,) a aVolume or a gronp or triangles ba.ntg equal area syEnoshaaaosssefmigmtaperhtioatani/nIfny)ofcgt,ualterntiaahmsnetedbihosmeafgvesarosisouFnmusigtgen(hesd.deuercsqe1iehisus3vaa.elr8aenevsiaegt(rrthrbaieyimtat)a.)n.e,mbgsTlI)usen,hoc(sushmsustjhoo,ceimiuhsnnietindwncaugsaifclFsiaveeaiatge,titn.(tJgshsto):,ue1i.mtct3fhtcee.Th8osehracne(s(oesbarurmr)naec,e(taahirxnedidammah)ics,ceauhiaygmtasheotCdllbmOlneeiluen')Dw,medtirtbhilhslveeorihrimdbcefeeieieogdgfuhu(srtssietineuixmdtwcboheytiilsomlannaeacsubsemnecuab((omtsseuruursnbcscceee))hhd.r,r Then, . the total volume is given by ... (1 3.37) 4 [Fig. 13.8 (b)] Let l: h, = the sum o f height used once. l: hz = the sum of height used twice l: h, = the sum o f height used thrice. l: ha = t h e sum o f height used eight times. A = area o f each triangle. The total volurile o f the group is given by V = 3A {ll:h, + 21:11, + 3l:h, + 41:11. + 5l:h, + 6I:ko + ?l:h, + 8I:ka) ... (13.38) ~pieroad A reaangular p/or ABCD form:; rhe plane o f a pil excavated for work. 13.7. E is poinl o f inlerseaion of rhe diagonoJs: Calculale rhe volume o f lhe excavation in cubic merres from rhe following daUJ Poinl A BC D E 47.2 52.0 Originallevel 45.2' 49.8 51.2 40.8 42.5 Finallevel 38.~ 39.8 42.6 Lengrh o f AB = 50 m and BC = 80 ·m. Downloaded From : www.EasyEngineering.net

MBASUREMENT OF VOLuidB !I!Downloaded From : www.EasyEngin3e2e9ring.net il Solution. (Fig. 13.9) \" Area of each t r i a n g l e = ! x SOx 40 = 1000 sq. m. 11: vertices of each triangle and find the mean depth at each triangle. Thus, ttJiTake 1~ Depth o f cutting at A = 45.2 - 38.6 = 6.6 m • . / BDepth of cutting at = 4 9 . 8 - 3 9 . 8 = 10.0 m . , / \"•cDepth of cutting at = 51.2 - 42.6 = 8.6 m •,__.---- Depth o f cutting at D = 47.2 - 40.8 = 6.4 m I! Depth o f cutting at E = 52.0 - 42.5 = 9.5 m Now volume of any truncated triangular prism is given by v= (average height) x A = hA >~ rV1 = 8.7 x 1000 = 8700 m' For the triangular prism ABE 6.6 + 10 + 9.5 8.7 m h= _ 3 -- 50~v--,. ~,. .From the prism BCE. h - 10 + a ; + 9 '5 = 9.367 m :t ·:. ~8Gm· . fiG. 13.9. ~For ng v, = 8.167 x 1000 = 8167 m' v, = 9.367 x 1000 = 9367 m' prism CDE, in h the h= 8.6 + 6.4 + 9.5 -8.167 3 . e ~·~ ~ 7.5 x 1000 = 7500 m3 e.. Total volume= v, + V1 + v, + v, = 8700 + 9367 + 8167 + 7500 ~33,734 rif rAlternatively, the total volume may be obtained from equation 13.38. Thus, For the prism DAE. 6.4 + 6.6 + 9.5 = 7 .5 m 3 inV= 3A (ll:h, + 2l:hz + 3l:h, + .... 8l:h8) gHere . . . (13.38) .net(Since ll:h, = 0 m , =2 (6.6 + 10 + a.6 + 6.4) =63.2 height o f every outer corner is utilised in two triangles) 3l:h, , 5I:h, , 6l:h6 • ?l:h, and 8I:k, are each zero. 41:11. = 4(9.5) = 38 Substituting the values in equation 13.38, we get V = 1 ~ x (63.2 + 38) = 33733 m' Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net SURVEYING '1,,1\" i I 330 Example 13.8. An excavation is to be made for a reservoir 20 m long I2 m wide at the bo«om, having the side o f the exCill!arion slope at 2 horizontal to I vertical. Calculole the volume o f excavation if the depth i! 4 metres. The ground surface is level before excavation. Solution. Length of the reservoir at the top = L + 2nh = 20 + (2 x 2 x 4) = 36 m = B + 2nh = 12 + (2 x 2 x 4) = 28 m wWidth of the reservoir at the top wLength of the reservor.r at nu..d-he'tght = -W +2-% = 28 m wBm I i .A B E IE E -I l as\"' A' 8 ' y• Bm 12m•. j Enc· E I ·'•I• t!.l 20m 0 ! FIG. 13.10 i'' W1'dth of the reseiVm.r at m1.d-he1'ght = -12-+-28 = 20 m 2 Area of the bottom of the ·reservoir= 20 x 12 = 240 m' Area of the top of the reservoir = 36 x 28 = 1008 m' Area of the reservoir at mid height = 28 x W = 560 m' Since the areas ABCD and A'B'CD' are in parallel planes spaced :'i m part. prismoidal formula can he used. I :. V =~(A,+ 4 Am+ A,)=~ (240 + 4 x 560 + 1008) = 2325 ur. Example 13.9. Calculole the volume e f the excavation shown in Fig. 13.11. the I',j side slopes being 1 ~ horizonJal ro I vertical, und rhe original ground surface sloping aJ 1 in 10 in the direction o f the centre line o f the excavation. ' Solution. Since no two faces are parallel, the solid is not a prismoid and hence prismoidal formula will not he applicable. The total volume will he the sum of the vertical truncated prisms appearing in plan as ABCD, ABFE. DCGH, ADHE and BCGF. The depth h at the centre= 5 +-8 = 6.5 m The side widths - 2 w, and w, can he calculated from the formulae Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net MPASUREMENT OF VOLUME 7f 331 :\\1 IliIilil ~ II II' A II 14---30m D. ~ II' 20m B r-------~,G 11 \"\"'\"~m +5 Vj\"''\"\" ng Horiwntal breadth of the slope to the right of DC = 14.1 mjot---WtWz FIG. 13.11 ' inS.im.ilarly, 1.5(m:nw, = 2b + 30 ) mn ( 2bm ) = T30 + 10 x 6·5 + i x 10 = 15 + 14 1 h+ 10 - 1.5 · e:. Horizontal breadth of the slope to the left of BA ~ 6.52 m ePrism ABCD : w , =b- +mm-+n-n ( h -2b-m ) =32-0 + 10 x 1.5 ( 6 . 52 3x-01-0 ) =15+6.52. Area= 30 Y. 20 = 600 m'2 1 0 + 1.5 riAverage height= ~ (5 + 5 + 8 + 8) = 6.5 m nVolume= 600 x 6.5 = 3900 m3 gPrism ABFE .nAverage height=~ (0 + 0 + 5 + 5) = 2.5 m etVolume= 172.9 x 2.5 = 432.2 m3 Area= (20 + 6.52) 6.52_ = 172.9 m' Prism CDHG : Area= (20 + 14.1) 14.1 = 480.8 m Average height=~ (0 + 0 + 8 + 8) = 4 m Volume =-480.8 x 4 = 1923.2 m' Downloaded From : www.EasyEngineering.net