BC Punmia SURVEYING Vol 1 - By EasyEngineering.net

\"\"·f\"\"' Downloaded From : www.EasyEngineering.net 185 OMJTI'ED MEASUREMEN'rS :. Latirnde of EA = - l: L' = - 87.86 m and Departure EA = - l: D' = - 0.72 m. Since the latitude of EA is negative and departure is also negative it lies in the quadrant. The reduced bearing (a) of EA is given by SW tan a - Departure= 0.72 or a = o· 28' Latitude 87.86 Bearing of.EA = s o• 28' w = 180\" 28' Also Length of EA = ~ 87.86 = 87.85 cos 0° 28' cos a Example 8.2. A ci!Jsed traverse was conducted round an obstacle and the following observations were made. .Work out the missing quantities: AzimuJh Side Length (m) AB 500 98\" 3 0 ' BC 620 30\" 2 0 ' CD 468 298\" 3 0 ' DE ? 230\" 0 ' EA ? 150\" 1 0 ' ng u.. AB inBC CD eeSum Solution. f - The affected sides are adjacent. Fig. 8.3 shows the traverse ABCDE in which DA is closing line o f the polygon ABCD. The latitude and deparrure of the closing line DA can be calculated. The calculations are shown in the tabular form below LDIUude De1JQitUr< + - + 535.11 223.45 73.91 494.50 758.56 73.91 t L' = + 684.55 r.. Latitude of DA = - l: L' = - 684.55 313.11 ' I inSince both latitude and departure are negative, the line 807.61 1 411.29 411.29 gtan9=Q=~ I lY = + 396.32 .L 684.55 and Departure DA = - l : D' = - 396.32 nBearing of DA = S 30\" 4' W = 210• 4' DA is in third (i.e. SW) 9=30\"4' eLength of DA =I= L sec a= 684.55 sec 30\" 4' = 791.01 m quadrant, the reduced bearing (a) of DA is given by tFrom Fig. 8.3, LADE= a = 230\" 0 ' - 210\" 4' = 19\" 56' L DEA = ~ = 150\"10' - (230\" - 180\") = 100\" 10' LDAE = y = 210\" 4' - 150\" 10' = 59\" 54' (Check: \" + ~ + y = 19\" 5 6 ' + 100\" 10' + 5 9 \" 5 4 = 180\") From triangle ADE. using the sine rule, we get Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net :r 186 SURVEYING DE= DA sin py = 791 ·01 sin 59• S4' - 695.27 m sin sin 100• 10' sin a sin 1o9o°• 56' m EA = DA 's\"\"m0\"'\"\"/'i' = 791.01 sm. IO' Example 8.3. 1 - 273.99 Side AB wBC CD wDA A four sided traverse ABCD, has the following lengths and bearings: Length (m) Bearing 500 Roughly East Find the exact bearing o f the side AB. 245 17S 0 wSolution. 270° Not obtained 216 w• .Econtinous lines. The affected sides AB and CD of adraw a line AC' equal and parallel to BC. The A 8 2 s lcline CC' is thus equal and parallel to BA and Fig. S.6 shows the traverse ABCD in yclosing line of the traverse AC'D. The latitude Eand departure of C'D can be calculated as usual. nThe calculations are shown below : are not adjacent. To bring AB adjacent to CD, i 3 is adjacent to CD. The line C'D becomes t h e ' ; . t ' - l . - - - - - - - - - - - - - - ' ' 1 ; --- Closing line FIG. 8.6 Line Latitude Departure DA + 212.71 + 37.51 A C ' (BC) - 239.64 + 50.94 Sum - 26.93 + S8.45 L?~!t~Jo1f> nf r ' n - .J.. ?1\\. a~ ..,.nti Deparh!!'~ nf r n , - SI:St &5 The bearing (8) of C'D is given by tan 8 = Q. = 88.45 or 8 = 73• 4'. L 26.93 Bearing of C 'D = N 73• 4' W = 286• 56' Angle p = (270• - ISO•) - 73• 4' = J6• 56' Length of C'D = L sec 8 = 26.93 sec 73• 4' = 92.47 m From triangle CDC', we get cc· DC' or sin a = DccC.' sin p = 92.47 sin J6• 56' sinp=sina 500 .. a = 3 ° 55'. . . Bearing of BA = bearing of CC • = 210• - 3• 55 ' = 266• 5' .. Bearing of AB = 266• 5' - ISO• = 86• 5'. _,, Downloaded From : www.EasyEngineering.net

liDownloaded From : www.EasyEngineering.net OM!ITED MEASUREMENTS 187 ij Example 8.4. A straight tunnel is to be run between two points A and B , whose l' co-ordinates are given below : Co-ordinateS d Point :j NE '~ A 00 )j 256 Bc 3014 1398 ~; 1764 ·11' ~~ lt is desired to sink a shoft 01 D , the middle point o f AB, but it is impossible t~~.. ~ to measure along AB directly, so D is to be fixed from C, a third known point. Calculate : (a) The co-ordinates o f D. (b) The length and bearing o f CD. (c) The angle ACD, given thal the bearing o f AC is 38\" 24' E o f N. :I~ Solution. Fig. 8.7 shows the points A, B. C N and D. The co-ordinate axes have been chosen to pass 30141-- 8(3014, 256) .1 through point A. t ,:,~, (a) Since D is midway between A and B, its .''''' co-ordinates will be 1507 and 128. ~ (b) From Fig. 8.7, 1~~~-ti':------,. , C{1764, 1398) ~- Latitude of AD= 1507 ngand '. ! ineerinCD Departuse of AD= 128 ..''''' l Latitude of AC = 1764 ''''''' lrt Departure of AC = 1398 gJ E ; Latitude of DC= 1764- 1507= 257 l Departuse of DC = 1398 - 128 = 1270 ! Hence, Latitude of CD = - 257 1 Departure of CD = - 1270 FIG. 8.7. <!:W:~C: paii:~ing through ~ Since both latitude and departure are negative, line i ~ t:. :t.: ~bird quadn~~ •.•:ith r'?'~re~t t(' !J.o.. o:-\"'....n!\"rt~!12te I ! C. gLength of CD= 1(1270}' + (257}1 = 1295.7. The bearing (8) of CD is given by .(c) tan 8 = Q. = 1210 9 = 7S 0 34' 257 L n= 258• 34' - 38° 24' - ISO• = 49• 10'. :. Bearing of CD =S 7S• 34' W =zss• 34' eExample 8.5. A and B are two stalions of a location traverse, tin metres being L ACD = Bearing of C D - Bearing of Gil their total co-ordinaJes Total latitude Total Departure A 34,321 7,509 B 33,670 9,652 Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net 188 SURVEYING A straight reach o f railway is to run from C, roughly south o f A, to D, invisible from ·C and roughly north o f B, the offsets perpendicular to the railway being AC=l30 m and BD = 72 m. Calculate the bearing of CD. quadrant. Solution. (Fig. 8.8) Co-ordinates o f A, referred to B : Latitude= 34321 - 33670 = + 651 wDeparture= 7509 - 9652 = - 2143. Since latitude is positive and depanure is negative, line BA is in the NW wThe bearing (6) of BA is given by wtan .E. =Q = 2143 651 From F1g. 8.8, L or 9 = 73\" 6'. aOB+OA BD+AC 6 sor ru AG Length o f BA = ...f(651)2 + (2143)2 = 2238 m yma ~+72 OB BD :A OA = AC EnOA r u \" 1M 1M =1-30x 2238 = 1440 m 202 cos ~ = AC = 130 :e AO 1440 ~ ll\"43'.5 FIG. 8.8 p= 84\" 49'.5 a=84\"49'.5-73\"6' Bearing of CD = 90\" + a = 90\" + 11\" 43' .5 = 101\" 43' .5. Example 8.6. A and B are rwo of the stalions used in sening out construction lines of harbour works. The total latitude and departure of A, referred to the origin o f the system, U(c: te,:,pc:divety -r j-t~./ u u a - 3.31.2, wuitftu:n: u f B a r e + 713.0 und + S 0 / . 0 111 111urth lmltude and east deparrure being reckiJned as positive). A poinJ C is fixed IJy measuring Jrom --l''o A a distance o f 432 m on a bearing o f L 346\" 14: and from it a line CD, I I52 m in length is set out parallel to AB. It is required to check the position of D IJy a sight from B. Calculate the bearing of D from B. Solution. (Fig. 8.9) Latitude of A = + 542.7 Departure of A = - 331.2 A: e: Latitude of B = + 713 A ' FIG. 8.9 Departure :. Latitude of B = + 587.8 of B referred to Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net OMITfED MEASUREMENTS 189 = 713 - 542.7 = 170.3 Departure o f B referred to A line a lies in NE quadrant. The =587.8 - ( - 331.2) =919.0 Since both latitude and departure of B are positive, bearing (6) o f AB is given by tan 6 _Q_919.o - L - 170.3 6 = 79• 30' = beasing o f AB :. Bearing of AD= Beasing of AB = N 79\" 30' E Fig. 8.9 shows the traverse ABCD in which length and bearing o f the line BD are not known. Table below shows the calculations for the latitudes and departures o f the lines: Une Length Bearing Ultilutk - DeoNture AB + + 170.3 919.0 DC 1152 s 79° 30' w 211.8 1132.0 n Latitude of BD = - l.: L' = + 461.4 and Departure of BD = - J.: D' = + 110.2419.9 102.8 CA 432 S 13° 46' E 631.7 1021.8 gbeasing (a) being given by Sum 170.3 1132.0 in t __ r L' = - 461.4__ .- __'£ ! l = - 1 1 0 . 2 - - - eBeasing of BD = N 13\" 26' E. Since both latitude and departure of BD ase positive, it lies in NE quadrant, its e11nd P mm1 h, in nl'?\" ~rr-nf~hr fi!\"P rLine a n aD. = -1=10-.2- or a = l 3 ° 2 6 ' 461.4 L iAB nBC Example 8.7. For the following traverse, compute the length CD, so thoc A, D gCD .nSolution : Fig. 8.10 shows the traverse ABCD in which A, D Bearing Length in metres 1/0 83\" 12' 30\" 4 2 ' etcan be calculated as under : 165 346\" 6 ' DE 2/2 /6\" / 8 ' and E ase in the same line. Treating CA as the closing line o f the traverse ABC, its length and beasing line LaCicude Departure AB + 13.03 + 109.23 BC + 141.88 + 84.24 l.: L' = + 154.91 l.: D' = + 193.47 Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net SURVEYING 190 Latitude o f CA = - I: L' = - 154.81 Departure o f CA = - I: D ' = - 193.47 Since both latitude and departure are negative, it lies in SW quadrant, its bearing being given by tan a = Q = 193.47 L 154.91 a=S51'18'W I a wLength of CA = L sec a = 154.91 sec 51'18' I w= 247.92 m. I wNow, since A, D and E are in the same line, I bearing o f AD =bearing o f DE = 1 6 ' 18' I .From .:\\ ACD. a = (16' 18' + 180') -· (346' 6' - 180') = 30' 12' E~ = 346' 6 ' - (51 ° 18' + 180') = 114' 48' I I ay =51' 18'- 16' 18' = 35' I I I sCheck : a+ p + y = 30' 12' + 114' 48' + 35' = 180' I / . 11 yEnAgain, •,I_\",\"\\\"/ -x: ll' FIG. 8.10 ·~.r CD = _ c : i _ _ sin35° 282.70 m. sin y sin a sin 3 0 ' 12' - 247 ·92 _ siny CD - CA sin a Example 8.8. In a closed traverse ABCD, with the data given below, the bearings o f lines AB and CD are missing. 8• ~ • Line Length (m) Bearing AB /60.00m - • 2Born BC 280.00m 102'36' CD 120.00m DA 320.00m 270'00' Determine the missing bearings. 320m Solution : ------------------------- Here is a case in which the bearings -----E Closing o f two lines, not adjacent, are missing. line Method 1 : Semi-IIIUllytical soW/ion FIG. 8.11 Refer Fig. 8.11. The lines AB and draw CE parallel and equal to BA. Naturally, CD, wbose bearings a, and 61 are missing, the triangle ADE, the lengths and bearings of are not adjacent to u c h other. From C, AE will be parallel and equal to BC. In Downloaded From : www.EasyEngineering.net

OM11TI!Il MEASUREMENTS Downloaded From : www.EasyEngineering.net 191 AD and AE are known. Hence the lenith, and beaiing o f the closing line DE can be l· found. with the computations in the tabular form below DetHUturo I j: Line L<nRth_(m) &min• LIIJiJruk - 273.26 '. &< 280.0 282' 36' +320.00 320.0 + 61.08 + 46.74 ii AD 90'00' 0.00 -46.74 1,• I: + 61.08 i'' DE - 61.08 Hence line DE falls in the third quadrant. a = t a n - I QL = t a n - ' ( -- 4661.' 0748 ) = 3 7 ' 4. 2 4 .. Bearing o f DE= 180' + 37'.424 = 217'.424 = 217'26' l.eugth DE= -J D' + L' = -J (46.74)1 + (61.08)1 = 76.91 m Now for triangle CDE, s =-} (160 + 120 + 76.91) = 178.455 Nw cp1 2 = .-.; (178.455 - 7 6 . 9 1 ) ( 1 7 8 . 4 5 5 - 120) ;i 178.455 (178.455 - 160) !:] 0 tan ! Let us now calculate the bearings o f AB and CD. '\"i n Bearing of DE = 217'26' cp/2 = 53'.319 or cp = 106'.64 = 106'38' :·:i gBearing of AB = Bearing of EC = 37'26' + (360' - 45'56') = 351 '30' 45 , .94 = 45 , 56, /j Also p = sin _1 ( D C sin cp ) = sin_ 1 ( 120 sin 106' .64) = 160 i1 EC inAgain, i.! e.. . . Bearing o f ED= 2 1 7 ' 2 6 ' - 180' = 3 7 ' 26' erinor \\i •; gor .nAlso, \" or u etor f. Bearing o f DE= 217'26' Bearing o f DC= 217'26' + 106'38' = 324'04' 1( Bearing vi ;;r, \"\"'.:.:.;. i.M - ~sor..= 144o04· Melhod 2 : Allalytical solution Let us use suffixes I, 2, 3, 4 for lines AB, BC, CD and DA respectively. Since E D =0, /1 sin 81 + iJ sin 83 = - lz sin 82 - /4 sin 84 160 sin a , + 120 sin e , = - 280 sin 102'36' - 320 sin 270'00' 160 sin a , + 120 s i n e , = - 273.26 + 320 = 46.74 = P ...(!) I: L = o. :. 1, cos a, + 1, cos e, = - 1, cos e, - 1, cos e , 160 cos e, + 120 cos e , = - 280 cos 102' 36' - 320 cos 2 7 0 ' 00' ... (2) 160 cos e, + 120 cos e, = 61.08 + o.o = 61.08 = Q From (I) 160 s i n e , = 46.74 - 120 s i n e , . . .(3) From (2) 160 cos e , = 61.08 - 120 cos e , ... (4) Squaring and adding Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net •· 192 SURVEYING (160)' = (46.74)' + (61.08)' + ( 1 2 0 ) ' - 2 x 120 (46.74 sin a , + 61.08 cos a,) wBut (46.74)2 + (61.08)2 + (120)2 - (160}2 46.74 sin a , + 61.08 cos a , = - - 22.019 or 240 - 22.019 . cos (a, - a ) = - 1(46.74)' + ( 6 l . 0 8 j ' = - 0.28629 wa, = 106' 38' + 37' 26' = 144'04' wAlso, from (3), a, - a = 106'.64 = 106'38' lllll a=:~:~: or a = 37'.424 = 37'26' .or a, = 360' - 8'30' = 351'30' EExample 8.9. The following measuremems were made in a closed traverse aAB = 97.54 m; CD= 170.69 m; AD = 248.47 m a •=sm. . 1 [ 4 6 . 7 4 - 120 sin 144'04'] = - 8o 30-, 160 s= =LDAB yCalculate the nussrng measurements. ABCD: Solution : Taking lhe W. C. B. o f line • EAB = 90', lhe traverse is shown in Fig. 8.12. nrLet lhe angle ABC = 9 70'45' LADC 39'15' :•o and 1he lenglh BC be I. Sum o f interior angles = 360'. .. LBCD = 360'- (70'45' + 39'15' +a) =250'-a ~ Bearing o f AB = 9 0 ' ~\"b Bearing -~ bearmg of AD = 9 0 ' - 70'45' = 19'15' =oi i.JA. = 1~r1y T iou- ~~~-l.J\" Bearing o f D C = 1 9 9 ' 1 5 ' - 39'15' = 160'0' Bearing o f CD= 160' + 180' = 3 4 0 ' Bearing o f CB = 340' - (250' a) = 9 0 ' + 9 Bearing o f BC = (90' + 9) - 180' = 9 - 9 0 ' Bearing of BA = [(9 - 90') - 9] + 360' = 270' Bearing o f AB = 2 7 0 ' - 180' = 9 0 ' (check) Now, for the whole traverse, ' l:L=O and l : D = O 9' . . 97.54 _cos 9 0 ' + I cos (a - 9 0 ' ) A• 97.54m + 170.69 cos 340' + 248.47 cos 199' 15' = 0 re- ' I or · 0 + I sin 9 + 160.40 - 234.58 = 0 FIG. 8.12 Downloaded From : www.EasyEngineering.net

•·~ Downloaded From : www.EasyEngineering.net r OMITTED MEASUREMENI'S 193 ... (1) qr I sin 9 = 74.18 and 97.54 sin 9 0 ' + I sin ( a - 90') + 170.69 sin 340' + 248.47 sin 199' I S ' = 0 ...(2) or 97.54 - I co~ 9 - 58.38 - 81.92 = 0 or Ieos 9 = -42.76 From ( I ) and (2) I = -.Jr-(7-4-.l-,8)2,--+-(4-2.-76....,)' = 85.62 m Also, . - 74 18 = 119'58' 9 = tan_, · 42.76 LABC = 119' 58' LBCD = 2 5 0 ' - 9 = 250' - ( 119' 58') = 130'02' PROBLEMS ? 1. From a point C, it is required to set out a line CD parallel to a given line AB, such lllat C and D are not visible from A and B, and traversing is performed ABD is a right angle. as follows: Lensth in m BeaJlng LiM 360' 0' 290° 57' 352° 6' 263° 57' n 2. A closed traverse was conducted round an obstacle and the following observations wereBA51.7 gmade. BE 61.4. EF 39.3 r Compute the required length and bearing of CD.FC i BC WorknCDout' the missingeDE quantities : AlimuJh SidE Length In m ww ~ eI~ . ring.nsame 300 86° 23' 169° 23' 450 243° 54' EA 268 317° 30' and F may be in lhe 3. For the fol1owin!!: traverse, find the leugth of DE so that A. E straight line : Length in metres R.B. 200 S 84° 30' E LinE N 75'18' E ~ eare obtained BC 100 N 18' 45' E CD 80 tAB=2l9·m; BC=l10.5 m ; CD=245.75 m DE N 29' 45' E EF ISO N 64' 10' E 4. Two points A and D are conoected by a traverse survey ABCD and the following records Angle ABC= l i S ' 15' ; Angle BCD= ISO' 40'. Assuming that: AB is in meridian, determine : · (l} . The tatirode and departure of D relative to A. (2) 1 The, length AD. (3) · The angle BAD. Downloaded From : www.EasyEngineering.net

l 194 Downloaded From : www.EasyEngineering.net SURVEYING I 5. Fmd the co-o!dinates of the poim at which a line run from A on a bearing of 1'1 10' E ! will cut the given traverse, and find lh< length of this line. line Lo/ii1Jde DeJ>outure AB N s E w BC 1650 440 2875 1450 120 326 (J)3643 376 0 DE 0 wEA 6. Surface and A and B. The and 10451 N, wco-o!dinates of A and and 30415 W, 30624 W respectively. error in both wthose of A being as bearing aod distance underground traverses have been IUD between two mine shafts .counter clockwise direction, the length of CD and bearing 8560 N, :.4860 W of B as N ELine 10320 B given by the underground ttaverse are The surface traverse gave the co-o!dinates the before. Assuming rhe surface traverse ro be correct, find of the line AB, as given by the undergropnd traverse. aBC sCD in the 7. The following lengths and bearings were recorded in running a lheodolite traverse of DE having been omitted. R.B. yDE EA Length in m EnDeterntine the length of CD and the bearil!g of DE. AB 281.4 S 69° II' E 129.4 N21'49'E ? N 19° 34' W 144.5 ? 168.7 574°24' w ANSWERS I. 74.82 m : 180' m. 2. AB = 322.5 m : C D = 305.7 -.I 3. 66.5 m. 4. (1) 378.25 aod 383.0 lii) :>jH.J m (iii) 4 5 ' 21\" 5. 1991 N : 351 E : 2021. 6. o· 3 4 ' . 5 : 238. 7. 131 m : S 4 6 ' 9' W. Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net [0] Levelling 9.1. DEFINITIONS (Ref. Fig. 9.1) the Levelling. Levelling is a branch of surveying the ob~ct of which is : (1) to find elevations of given points with res,eect to a given or assumed daDJm and (~) to ~sh or assumed points at a given elevation or at different elevations with respect to a gjyep d e . The firSt operation is 'required to enable the works to be designed while the second operation is required in the setting out of all kinds of engineering works. Levelling deals with measurements in a vertical plane. Level Surface. A· level surface is defined as a curved surface which at each point at !he point. The surface of a still water is is perpendicular to the direction of gravity to the mean spheroidal surface of the earth a truly level surface. Any surface parallel n Horizontal Plane. surface at that point. It is, lherefore, a level surface. gineII· Level Line. A level line is a line lying in a level surface. It is, therefore, nonniu to the plumb line at all points. Horizontal plane through a point is a plane tangential to !he level is. therefore, perpendicular to the plumb line through the point. er :' , c.>~e~ \\\\(}!. ~ •••••_ ~fio(,.;.. \"'\"'!. ••.•• i' _..ng\\ .~.;·:::.\\ / n~. oatum =M.S~.l:0'/ i[:.s :~ et~\\ FIG. 9.1 (195) Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net ''fti 196 SURVEYING Horizontal Line. I t is slraight line tangential to the level line at a· point. It is also perpendicular to the 'plumb line. wabove or below sea level. It is often more convenient, however, to assume some other Vertical .Line. It is a line nonnal to the level line at a point. It is commonly considered to be the line defined by a plumb line. wElevation. The elevation of a point on or near the surface of the earth is its vertical Datum. Datum is any surface to which elevations are referred. The mean sea level affords a convenient datum world over, and elevations are commonly given as so mucb wwhiclr ·the two points lie. datum, specially i f only the relative elevations o f points are required. .EMeanpla'De. distance 'iiliOVe\" or below an arbittarily assumed level surface or datum. The difference in between two points is the vertical distance between the two level surfaces in elevation aof the tides. At any particular place it is derived by averaging the hourly tide heights sover a long period of 19 years. Vertical AniJe. Vertical angle is an angle between t\\vo intersecting lines in a vertical Deneb Mark. Bench Mark is a relatively permanent point o f reference whose elevation ywith respect to some assumed datum is known. It is used either as a starting point for Generally, one of these lines is horizontal. ·· \"' sea for all si~ges Sea Level. Mean sea level is the average height o£ the Elevelling or as a point upon which to close as a check. n9.2. METHODS OF LEVELLING (f' Three principal methods are used for determining difference in elevation. namely. barometric levelling, trigonometric levelling and spirit levelling. Barometric levelling. Barometric levelling makes use of the phenomenon that difference in elevation between two points is proportional to the difference in aonospheric pressu_res at these points. A barometer, therefore, jnay be used and the readings observed at different P9ints would yield a measure of the relative elevations of those points. · At a given point, the atmospheric pressure does not remain constant in the course -,.i.,.;.~ ,.:·.~ll ;·~ J.,.. . ..:.0o..~ ......... ~ . . . . . . . . . . . .~ . . . . . . . . ~............. ·~· .....,;;;.: ..... - • • .:.~··'-~) .!...~_¥:.;~-..:.i.~ l and is little used in surveying work except on reconnaissance Or exploratory surveys. Trigonometric Levelling (Indire<t levelling) : Trigonometric or Indirect levelling is the process of levelling in which the elevations o f points are compmed from the vertical angles and horizontal distances measured in the field, just as the length of. any side in any ttiangle can be computed from proper trigonomelric relations. In a modified form called stadia levelling, commonly used in mapping, both the difference in elevation and the horizontal distance between the points are directly computed from the measured vertical angles and staff readings. Spirit Levelling (Direct Levelling) : line It is that branch o f levelling in which the vertical distances with respect to a horizontal (perpendicular to the drrecnon of g~avity) .!)lay be us&! to determme the relative difference liielevation between two adjacent points. A horizontal plane o f sight tangent to lever surface a< any point is readily established by means . of a spirit level or a level vial. In spirit Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net . 11. LEVELUNG 197 I~lr levelling, a spirit level and a sighting device (telescope) are combined and vertical distances I are measured by observing on graduated rods placed on the points. The method is also I I t is the most precise method o f determining elevations and known as direct levelling. I the one most commonly used by engineers. [: 9.3. LEVELLING INSTRUMENTS ii The instruments commonly·· used in direct levelling are -~ ( I ) A level i l (2) A levelling staff. '·i' I. LEVEL :11- The purpose o f a level is to provide a horizontal line o f sight. Essentially, a level :J consists o f the following four parts : (a) A telescope to provide line of sight its centte o f run -~~ (b) A level tube to make the line of sight horizontal (c) A levelling head (lribrach and trivet stage) to bring the bubble in !I ,·,~, (d) A tripod to support the instrument. ·~ There are the 'following chief types o f levels : ;~ (r) DUMPY LEVEL The dumpy level originally designed by Gravatt, consists o f a telescope tube firmly •' .·It n secured in two collars fixed by adjusting screws to the stage carried by the vertical spindle. (I) Dumpy level (il) Wye (or Y) level :~ g 10 9(iv)Tilting level. :I· in r,' ' ' (iii) Reversible level I 2el=l ~ @• r-14II s-,. :'I.·i· e9 l ring.netJ. TELESCOPE :III' j!i5 - · -/'·X0\\ 3 :I: I 1 ·~ fll'll;\") f'~(fl?d7 .~ I: : : : !8 ·.~ I ~ c------', ~ I t 2 I FIG. 9.2. DUMPY LEVEL 7. FOOT SCREWS 2. EYE·PIECE 8. UPPER PARALLEL PlATE (TRIBRACH) 3. RAY SHADE 9. DIAPHRAGM ADJUSTING SCREWS 4. OBJECTIVE END 10. BUBBLE TUBE ADJUSTING SCREWS S. LONGITUDrNAL BUBBLE II. TRANSVERSE BUBBLE TUBE 6. FOCUSING SCREWS 12. FOOT PLATE (TRIVET STAGE). Downloaded From : www.EasyEngineering.net ~

Downloaded From : www.EasyEngineering.net SURVEYING 198 The modem form of dumpy level has the telescope tube and the vertical spindle cast in one piece and a long bubble rube is attached to the top of the telescope. This form is known as solid dumpy. . Fig. 9.2 shows the diagram.metic sketch o f a dumpy level. Fig. 9.3 shows the section of a dumpy level. Figs. 9.4 and 9.5 show the photographs of dumpy levels manufactured by M/s Wild Heerbmgg and M/s Fennel Kessel respectively. Fig. 9.6 shows a dumpy level by M/s W.F. Stanley & Co. The name 'dumpy level' originated from the fact that formerly this level was equipped with an. inverting eye-piece and hence was shorter than wWye level of the same magnifying power. However, modem forms generally have erecting eye-piece so that inverted image of the staff is visible in the field of view. wIn some of the instruments, a clamp screw is provided to control the movement of the spindle about the vertical axis. For small or precise movement, a stow motion screw w(or tangent screw) is also provided. The levelling head generally consists of two parallel plates with either three-font screws .or four-font screws. The upper plate is known as tribrach and .the lower plate is known Eas trivet which can be screwed on to a tripod. asyEn4 FIG. 9.3 SECTIONAL VIEW OF A DUMPY LEVEL. I TELESCOPE \\0 BUBBLE TUBE ADJUSTING SCREWS 2 EYE-PIECE 12 FOOT PlATE (TRIVET STAGE) J RAY SHADE 13 CLA~~p SCREW 4 OBJECTIVE END \\4 SLOW MOTION SCREW 5 LONGITUDINAL BUBBLE IS INNER CONE 6 FOCUSING SCREW 16 OUTER CONE 7 FOOT SCREWS 17 TRIPOD HEAD 8. UPPER PARAU.Fl. PLATE (TRI'BRACH) 18 TRIPOD. 9 DIAPHRAGM ADJUSTING SCREWS Downloaded From : www.EasyEngineering.net

LEVELUNG Downloaded From : www.EasyEngineering.net I99 The advantages o f the dumpy level over the Wye level are: (I) Simpler construction with fewer movable parts. (il) Fewer adjusbnents to be made. (iii) Longer life of the adjustments. (il) WYE LEVEL The essential difference between the dumpy level and the Wye level is that in the former case the telescope is fixed to the spindle while in the Wye level, the telescope is carried in two vertical 'Wye' supports. The Wye support consists of curved clips. I f the clips are raised, the telescope can be rotaied in the Wyes, or removed and turned end for end. When the clips are fastened, the telescope is held from turning about its axis by a lug on one of the clips. The bubble tube may be attached either to the telescope or to the stage carrying the wyes. In the former case, the bubble tube must be o f reversible type. Fig. 9.7 shows the essential fearures of Y-level. The levelling head may be sintilar to that of a dumpy level. In some cases, the instrument is fitted with a clamp and fine motion tangent screw for controlled movement in the horizontal plane. Fig. 9.8 shows the photograph o f a Wye level by Fennel Kessel. The Wye level bas an advantage over the dumpy level in the fact thai the adjustments ngineering.nI. TELESCOPE e2. EYE-PIECE longer t3. RAY SHADE can be tested with greater rapidity and ease. However, the adjustments do not have life and are disturbed more frequently due to large number of movable parts. A ~I I~ 7 115 15 FIG. 9.7. WYE LEVEL. 9. DIAPHRAGM ADJUSTING SCREWS SCREWS 10. BUBBLE TUBE ADJUSTING II. WYE. CLIP 4. OBJECTIVE END \\2. CUP HALF OPEN S. BUBBLE TUBE 13. CLAMP SCREWS 6. FOCUSING SCREW ·14. TANGENT SCREW 7. FOOT SCREW 15. TRIVET STAGE. 8. TRIBRACH Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net 200 SURVEYING (iii) REVERSIBLE LEVEL A reversible jevel combines the features o f both the dumpy level and the Wye level. The telescope is supported by two rigid sockets into which the telescope can be introduced from either end aod then fixed in position by a screw. The sockets are rigidly connected to the spindle through a stage. Once the telescope is pushed into the sockets and the screw is tightened, the level acts as a dumpy level. For testing and making the adjustnients, wthe screw is slackened and the telescope can be taken out and reversed end for end. The telescope can also be turned within the socket about the longitudinal axis. w(iv) ffiTJNG LEVEL In the case of a dumpy level and a Wye level, the line of sight is perpendicular wto the vertical axis. Once the instrument is levelled, the line of sight becomes horizontal and the vertical axis becomes truly vertical, provided the instrument is in adjustment. In the case o f tilting level, however, the line o f sight can be tilted slightly without \"tilting .Ethe vertical axis. Thus, the .line of sight and the vertical axis need not be exactly perpendicular to each other. This feature, therefore, helps in quick levelling. The instrument is levelled aroughly by the three-foot screws with respect eiiher to .the bubble tube or to a small circular bubble, thus making the vertical axis approximale/y vertical. While taking the sight sto a staff. the line of sight is made exactly horizontal by centring the bubble by means yEnof a fme pitched tilting screw which tilts the telescope with respect to the vertical axis. 42 \"3 FIG. 9.9 TILTING LEVEL. L TEL.ESCOPE 7. FOOT SCREWS 2. EYE-PIECE 8. TRIBRACH 3. RAY SHADE 9. DIAPHRAGM ADJUSflNG SCREWS 4. OSJECTIYE END 10. \\l. BUBBLE TUBE FIXING SCREWS s. LEY.EL TUBE 12. TILTIING SCREWS 13. SPRING LOADED PLUNGER 6. FOCUSING SCREWS TRJVET STAGE. Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net -::;- LBVELUNG 201 I t is, however, essential that the observer should have the view o f the bubble tube while sighting the staff. Fig. 9. 9 shows the essential features o f a tilting level. A tilting level is mainly designed for precise levelling work. It has the advantage that due to the tilting screw. levelling can be done much quicker. However, this advantage is not so apparem when it is required to take so many. readings from one instrument setting. Fig. 9.10 shows the photograph o f a tilting. level by M/s Vickers Instruments Ltd. 9.4. LEVELLING STAFF A levelling staff is a straight rectangular rod having graduations. the foot o f the staff representing zero reading. The purpose o f a level is to establish a horizontal line o f sight. The purpose o f the levelling staff is to determine the amount by which the station (i.e., foot o f the staff) is above or below the line o f sight. Levelling staves may be divided into two classes : (I) Self-reading staff, and (it) Target staff. A Self Reading Staff is the one which can be read directly by the instrument man through the telescope. A Target on the other hand, contains a moving target against which the reading is taken Staff, by staff man. n (c)(I) SELF-READING STAFF There are usually three forms of self-reading staff : gi ~7 n --2 ~v (c) Telescopic staff (Sopwith pattern). eer --- I 111~3 i-n --•:I I I -~;~ ::9 ~? :N g---------- .In§V I1 §_ - et~3 (a) Solid staff ; (b) Foldin_g staff ; Figs. 9.11 (a) and (b) show the patterns o f a solid staff in English units while and (d ) show that in metric unit. In the most common fonns, the smallest division ~!:J --:21 r-~ --;:=7v- -§1 - ---§3 \"- §I ~- -:91 I §9 ~a ~~ -=a ~ --=- --§7 --=- --E.V (a) (b) (c) (d) English Metric Hundredths. Fiftieths. Centimetres. Half-Centimetres. FIG. 9.11. (BY COURTESY OF MIS VICKERS INSTRUMENTS LTD.) Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net w. SURVEYING TifilsneuhntlelogoytfhsloteF0awxo.fi0gtffee1.rn4isd'b9fe6ot.gd.1\"xe,2noseoilrrstifadhl5oili5nsyw'gmsulmeimsnnaaut.gdoatelshlHoyt.hopoewfIwoniwfecthvetenhe1llretp4r,aast5lesftteaobrsmmnoo(xouedrsstotsaatfw5fafffvo,4emo'astd6rh)r\"eamhlnealatgevnhyoeigrndgetghhteha.i.nvaceToTlrthehhrfeeoreinesgpeetc1hoe4uotgedotrlrfiafeantd.slgcu1o0absplttoeiaiofxncefn,gfesttlhiehnsoouagpsrrattubor3mseso.,2mluidWsesumltirahdetmloeeslynp..s :f wof t ww.E.ctmooodandrvbkeeelndeTSineiihngnnievttncheesrltretoeeavoddef.clslaeitnortTlhrfgyh-erteehsaeitfatderveei·nletafergsdfotiamngrsge.srtsaiaddfomefunaae·arioetrse_k,f.-depadlttiahhi;cn;ee,eairnyeEssfttnooabgrfslelfaeit,cshehk(nesthautktooohnewtimhntSnetuhrgf.ebgrhoyemoriteghhrhaea·a'ttlc-blbythoeaevltdeheidsacvlodiesnopiedwews,en.hwiaonlalTlerhdFresnie.gua.mdtoinpb9ge.1rso2fo)a.fp,·tphTfeeehaseeert 1.5 m, 1.5 m and 2 m. 10 ft long having a binge at the middle folded about the hinge so that it becomes its Fig. 9.13 shows a folding staff usually length. Wheo not in .use, the rod can be asyEn,,(· r' i' FIG. 9.t2 TELI!SCOPIC STAFF FIG. 9.13 FOLDING STAFF FIG. 9.14 TARGET STAFF (BY COURTI!SY OF MIS VICKERS INSTRUMENTS LTD.) Downloaded From : www.EasyEngineering.net

i Downloaded From : www.EasyEngineering.net LEVELUNG 203 obosotldffaadfcfvfekieheitguwsrna.atrdhneueIendaaittrnhiotdShnitachstaeatniencddidansisctbeota,yrptuemtsaholeettfhenretasn,taoatwdftehfdhewiittseehrnietrtsdehapiaswamecndhedailreiksbnloldtahwioccefklaybtisownputghtanooctmelieelvs,estnhfhtooheowehtusretnomdtdhpraeeyfodiegtfvhuneasron.etbtSleaaaonppcmtppkheee.aatsirTprmsaheceiisnenhiwuntthn)hdledeeircneaffdtiiietetnhhllddges o f view, the red figure thus... indicating the whole feet. Folding Levelling Staff I n Metric Units coooofffmtstphhiareeifFsfesidgstiea.stftafw9ckh.oei1sap5b2tlne(t7maa5d)tyetmpshheoomorfwowuoaaignnthhedalyal41os8nelmgoaimcstuokmfdnionielngddraienlsdwgpesoevtorctiyicdtpipeevenewlaypltie.tihvetoceThuelelhstinegbawnafiocysthkltda.jifotnfhiTgne(htI.jeSjooiTinsn:htttae1faof7swf7sie9istdm-h1tehbj9ol6yisan1.tnt)ae.dfEdfaTtchihtshOiecgkmpesnittaeehadcsefeesrf in such a way that : (a) the staff may be folded to 2 m length. to (b) pieces may be detached from one another, wheo required, rigid the two easy handling and manipulation with one piece, and (c) facilitate when the two portions are locked together, the two pieces become and straight. o f 25-minute sensitivity is fitted at the back. The and correct the back bubble. A brass is screwed bas two folding baudles with spring acting locking ngnwmuimmth.eratFEhlieagc.ihnco9rmb.r1leea5stcprkeo()bn. )disiTnshgshueobwnddusiemvcitedihmreeadelstrdeien(tttaohnielusm20meo0refatlrdegirviasindsuiumomantaeisdor,eanlst.hceios·Entvtimhneiuracydokeunsedseisncthimorrofeeudtgrgehraaondluedutnagtttithhoheensissdtebacefffiiigm.nugerterd5ecased, A circular bubble, suitably to test staff bas fittings for a pluminet staff on to the bottom brass cap. The in(i1) TARGET STAFF device or an ordinary locking device. eering.nets7rsetobifishsioetnafxaedtaceffgrbk6tdfaarlsciaettsthfodsaotheFtoon.tfufdeoiflsagtdpihTtt.tseehmhteuthardese(ne9rai.tgnre.it7oe1felhsr.rato4tfeoatdfdnomt7ftihitfnsswehigmcrsfowotsatlsoaatwilhgapsmirmtdsereskrolgapreiadededsdoraootukibnfrw)aaettgtthionatselh·eowrotdsegalfewtabkeehctirnieatetunoedshgrrnpeutsegscph.dflet~tteoeaueht.serwWfdpelvtif,tedaeohlt·rnerhitbhentnPnadeyan:girgapnvegtatrihtehltnhlaroth.etsthgkiweengoenFgthPawsuiaatenolrlshblhirdarnltineshnleiaelrdstetnhianetd~ukahdoiifnrdinrsofenediogrtanpgarhfdbgseadeiiidsrrtitgnsanteahtehghrsagpechgtsldso.p.itae,dierfnvrtidotTeTngeaoxaagneshh.dbqdeetiboyuflna7eoeitbgupronhboe,tppteg.htfe.avpatetetdehkdehviFtreeanealnnctogrikwhnandplreaereiiatntrvteoihr,yrs,rtetehoilna-etofattvdhoghhmfeieefensurtanittnhgpghhntttiosphhaeeaeueoertrrd.gsr.gbnttaieeaaersnttrTTerlotdgogtcanhhhtewetiifheeeesssttf o f a foot. Relative Merits o f Self-Reading and Target Staffs (!) With the self-reading staff, readings can be taken quicker than with the target Staff. Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net II' '!!, 204 SURVEYING (ii) In !he case o f target staff, !he duties of a target staff-man are as important a\"s !hose o f !he leveller and demand !he services o f a trained man. In !he case o f a self-reading staff, on !he other hand, ordinary man can hold !he staff concentrating more on keeping !he staff in plumb. (iii) The reading with target staff can be taken with greater fineness. However, !he refinement is usually more apparent than real as the target man may not be directed accurateiy wto make !he line of sight bisect !he target. 9.5. THE SURVEYING TELESCOPE wThe optical principles of !he surveying telescope are based on !he fact that all parallel rays of light reaching a convex lens are bent when they leave it in such a manner that wthey intersect at a common point, called the . focus and that all the rays passing through another point called the optical centre pass through the geometrical centre o f lens without .bending. EThe surveyor's telescope is an adaptation of Kepler's telescope convex lenses ; the one nearest to the object is called the objectiVe athe eye is called the eyepiece. sThe object glass provides a real inverted image in front of the eye-piece which, yin rurn, magnifies the i..'Dage to produce an inverted virtual image. Fig. 9.16 shows the which employs two and the other near Eoptical diagram of such a telescope. n.,.. b.~ ''-~ -----(::::::-~~~,-A ~ a T ----- ....... ----- --- Eye b t'IU. ~.lo Ut'Ti(.AL DIAGRAM v r icLC.:,i...Vl'.C. The line of sight or line o f collimation is a line which passes through the optical centre o f !he objective and !he intersection of cross hoirs. The axis of the telescope is !he line which passes through the optical centres of objective and eye-piece. The cross-hairs are placed in front o f eye-piece and in the plane where the real inverted image is produced by the objective. Thus, the eye-piece magnifies the cross-hairs also. The distance from the objective of the image formed by it is connected with the \"distance o f the object by the relation ..; -I + -uI =f-I v where u = distance o f object from optical centre of objective v = distance of image from !he optical centre o f objective f = focal length o f the objective. Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net ' LEVELLING 20S The focal length o f an objective is constant. The establislnnent o f a telescopic line o f sight, therefore, involves the following two essential conditions : ( I ) The real image must be formed in front o f !he eye-piece. (2) The plane o f the image must coincide with that o f the cross-hairs. 'I Focusing. For quantitative measurements, it is essential that the image should always I be formed in the fixed plane in the telescope where the cross-hairs are situated. The 'I operation of forming or bringing the clear image o f the object in the plane of cross-hairs is known as focusing. Complete focusing involves two steps : !I . :: (l) Focusing the eye~piece. The eye-piece unit is moved in or out with respect to the cross-hairs so that the latter are clearly visible. By doing so, the cross-hairs are iii brought in the plane o f distinct vision which depends on eye-sight of a particular person. t..Ii (it) Focusing the objective. The purpose of focusing the objective is to bring !he image of object in the plane of cross-hairs which are clearly visible. Tbe focusing can n\"' i be done externally or internally. -n The telescope in which the focusing is done by the external movement of either ,,ut,,· external focusing telescope (Fig. 9.24) and the objective or eye-piece is known as an Hi one in which the focusing is done internally with a negative lens is known as inlemal n as described above. !tl: focusing telescope (Fig. 9.25). gWhether internal focusing or external focusing, a telescope consists of the following ll ~' Parallax. I f the image formed by objective lens is not in the same plane with cross-hairs, any movement of the eye is likely to cause an apparent movement of the image with !V1i respect to !he cross-hairs. This is called parallax. The parallax can be eliminated by focusing n(t) OBJECTIVEi(t) .i! essential parts :eIf simple (single) lenses are used, the telescope ewould have various optical defects, known as aberrations, Jl rwhich wouid resuit m curvarure, distortion, unwanted icolours and indistincmess of the image. In order to 'i neliminate these defects as much as possible, the objective 15 and eye-piece lens are made up of two or more simple · 1r glenses. The objective (Fig. 9.17) is invariably a compound .lens consisting of (a) the front double convex lens lr nmade of crown glass and (b) the back concave-convex ~''i~ lens made of flint glass, the two being cemented together ·,I,:I.~· etwith balsm at their common surface. Such compound •. Objective (il) Eye-piece (iil) Diaphragm (iv) Body and focusing device. f ·.·•..·-~:· FIG. 9.17. OBJEcriVE. !i lens is known as achromatic lens, and two serious m optical defects viz., sPherical aberration and chiomatic aberration are nearly eliminated. i,! (it) EYE-PIECE ~ In most of surveying telescopes, Ramsden eye-piece is used. It is composed of :-; plano-convex lenses of equal foc_allength (Fig. 9.18), the distance between them being two-thirds ·:! ·~ .~ ~Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net · 206 SURVEYING : j' o f the focal length of either. This eye-piece gives a flat field o f view. It is also known I>glass, appears still inverted to the observer. as positive or oon-erecting eye-piece for the inverted image, which is fonned by the object commonly used, is Huygen 's eye-piece. It is Eye composed o f two plano-convex lenses (Fig. 9.19), wthe distance between them being 'two-third of Aoother type o f eye-piece, thougb not the focal length o f ihe larier and twice the wfocal length of the smaiier. The ~lii:omatic air erration o f this combination is sli!Jhtiy less and spherical aberration is more than that ofRamsden's. w I>This is also a non-erecting eye-pie'Ce. FIG. 9.18. RAMSDEN EYE·PlECS .ereaing eye-pieces Einverted image of asyEnatTibohmrubhaetjaenegc(eetletr,i)hyveeecewf-topheraiiirenmcecdahcaentgiyhinesiegnndvoaceoftesle,vidlrteeeehsstcdehcideotoemphdreelaia,ggtditunne(isiirsafzealy)diintvoostgahbefneljeftepacxogdft.toweerr.fameTinrhAsi,lt·eeidnotadshnenieetysi'oe.iils-nnepavTnilceegherctitahresetdadvicornaeofinsnmtustanoaligtosgsetnessb-. ieenoroTeffbchltoeefiatsnotsgeurnreroostuneblfel-leteencbsrsaercieisucoslstlpiieanae(ngFtcmiihsgyteae. glsoenhi9smfi.ocf2iorae0ttgphed)eee.r,Eye Some telescopes are fitted with special which give a magnified but . the image formed by the FIG.,·9;19. HUYGEN'S EYE-PIECE l ; l .r- ~-;: A><l . I . li> ?I ;:~~- ~~ il' ?' J\"i'l'·' FIG- 9.'20 !:REC\"TINa EYE-?!!':C'E does not have to be erected but instead is formed by an achromatic eye-piece. For all precise work, the Ramsden eye-piece is to be preferred, as inverted images are not a great disadvantage and ~ a surveyor .very soon gets used to them. When the line o f sigbt ·is very much inclined -. to the horizontal, it becomes inconvenient for the eye to view througb ordinary eye-piece. In such a case, a diagonal eye-piece, such as shown in Fig. 9.21 is used. Diagonal eye-piece, generally o f Ramsden type, consists o f the two lenses and a reflecting prism or a ntirror fitted at an angle of 4 5 ' with the axis of the telescope. Such eye-piece is very much useful in astronomical observations. FIG. 9.21. DIAGONAL EYE·PlECE. Downloaded From : www.EasyEngineering.net

·~ Downloaded From : www.EasyEngineering.net l.I!VELUNG 11rl (iiz) DIAPHRAGM tatiasheneldheasopncrvooizepsTiorniehttntiieecotbaranclalChlryolfrauOe,nf6lIg!rtaSehbna·heyhdbeallicerdfsaro,oucisnrssdspl-eichasgsauailgptfliterlnsastetardJwnonm-thashtieiotecditoaahedlngeaiforvdliitfnesgsmtchaiornceevtadwoeltlemuesfbdtiehne(neiFrtteeicgdtaii.icnslaiuop9lbeeha.2e.lrsa2oogI)anf.mdpjs,umWoisgsotaishetdihdtbe,flmleatsch.nloeigginnebsdshtitlserytlump,moebeotfaonflttahst,hrivehnteshogerertiizchsroaceenlrtleditcawaulnlislynde. Diaphragm FIG. 9.22 DIAPHRAGM AND RECfiCULE. n- ocobrfefos.waos-fhhgiTvacliehharrsesys. (ahfAp)ian,liaresft(eebpwa)laortentiaynnpuudwimscuhaa(ilccwll)hyairreraaamrrneoeagfdeueemfsitlceeaodhnmfetseditnnhtorsfeifnlaeoedtvhfseeslvsife.llrkiront.miecsaInlcaonsacodnomdopenohiionnortsfsitzroutahumreteaelnbitlrsllou,inwsettnhrsaeteswprdehidtiiicecnhru,lFesibegcur.otvne9sm.i2saat3yss gineEB®@ffi®(a) @e@rin 0 0(~ (g)(b)(c)(d) (e) gFIG. 9.23. CROSS-HAIRS (h) (Q .netttshwuerovehmyoooTrirrhzeeotnohthoaorlsireziezocourontiastfalsl-thhhaaheiiarrs,isrtacfiafosflrleuidsusessdevraerdtitoniiacarclheoaamdilrapst,euthrtaieonlnlgyes.tadafiMbfsotoavansnetcdeastnetdhlebesyoctothwpseetorasdoviaanerretetiaqccaauhllaseloohdmaiiesreqtstaurnyiepc.npeasebdlbeewlotihwthe (iv) BODY AND FOCUSING DEVICE t9h.e24b) odToyhr eiasnfofcoiunrsmtienergndaldoeffvoitccweuosidntegupbeetnsedlesosncueoppocenapawb(hFleeigth.oefr9s.i2lti5d)iisn. gaInnaxeitaXhletleyrnewxaltiethrfnoinaclutshifneogcuosttihenleegrscboteypleesmc(eoFapinges.. Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net SURVEYING 208 t?oi \"-\" \"'~ I n w l : ' - ~'AG. 9.24. EXTERNAL FOCUSING TELESCOPE. ww.EasyEn·omoeeQfnyodejvre-,apeccmtiieckvcaeeernrg.tyceieaIonlnrlgfingiotshtthehewsecriortehebycwjeeaaec-s'pdetiisvpe,eitcnoetihroetehanlanetdoaivnbienajtlenchYcbeetreiv'dtoderuiatbotpiehsh,erthamsgeooomuutm,enthrtiealmldtefidoxtvehoeendfsocftruohiunscebiunesgfoionuccgatsuecrsrrrmiyeniwgronu.v.gbeemItnheaennstdocmirsotehsees-fchfaeeasycierteses-,dpiaethbnceydeII' ': I ·-':!'OJ' AG. 9.25. INTERNAL FOCUSING TELESCOPE. frorouufcbbueersaicwnhkIgonhlidacinihisnndgtcedparotninhnnaeeilobnelwfeonamicstnhuodsivisnheagdenmlpoettoexvoleteedfasrncndaoaaplloefsnrm,uogpilptlhbaeleeendmtdwboeeenbienatjnasedirc.dytteihFveeditgoh.dueaibna9lprd.eu2hb6reeacygoielmncl-aucprasirtvaeyrnecaidenteglsetahnrtteshheeeomkbpoeojrbpeiujnctentcctiievftpiidexvl.eeediTnubnhyadiasnemdrslsyebbatioohnnnnegs the focusing with a negative lens. · t:-;~~:~,,=!. --- -----l~~~C-ros-s·ha-lrs ----/~==~d~~ FIG. 9.26. In the absence of the negative lens B. the image will be formed at C. For the negative lens B, point C forms the vinual object the final image of which is at D (cross-hairs) For the lens A having f as the focal length rI JI. + J1,. ... ( ! ) For the lens B, having f ' as the focal length, Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net LI!VELIJNG 209 !-I' =( f-, -I d-) +(1--I d-) ... (2) a From the above two conjugate focal equations, the distance d can be koown for given vaiue ·of f, . When the object is at infinite distance, f, equals f and d will have its minimum value. AitshdevliatgenhlT(t(etaI2sh)g)ceieonTsRpaheiwdseovkeifasiognovIhtnnafetogrttaebealsrrlnaendfaalofkeelfinclnogtFgetcthdhoaetctehuoadenisfnidntpnteghaertenharaareTlrllueebtlblhfeueoesbcpcbuiolsamlspetieneindgiodsbtlueolbaevlbsoetselrferelittdrhahubeebdleeuetrexiotlnoteregsrcbngofealpoalecsdsu.fisosipcHcnolugaevs.cnieencrdTge,hodeafturherfceiooncmagubsspaiflanoafsgoncslucloessbiwlnoidogsxe:.f. telescope, is elimiuated. when transiting a thendnlite and pinion is less due to lesser movement of negative lens. (3) Wear on the rack Line of collimation is less likely to be affected by focusing. and (4) As the draw rube is not exposed to weather, oxidation is less likely to occur (5) telescope can be made practically dust and water proof. is almost the (6) ln making measurements by the· stadia methnd, an instrument constant n eolelfinmsthineas(ntc7edo)dmTfboahincenadeldnetphogleabatnjievecc.eotimvfTeophcuiaustanstdiineogxnntserlgaeanatsrieveqesuetirhlvveuaneslssenaistsimugfsoprecelfiaafulitleedrlo.eptnhtgiactnhalthpceaunrdpiosbsteaencbueeticlbiaseuetsdwe eetthone· power. focal length the objective g (8) Internal focusing also gives increase the iorder to reduce the aberrations of the nlens. e~ the optical desigrter an extra lens to work with, in e(I) The system and to increase the diameter of the objective r(2) The irepairs. Disadvantages of Internal Focusing Telescope The. principal disadvantages of an internal nOPTICAL DEFECTS OF A SINGLE LENS focusing telescope are as follows : gThe optical defects of a single lens are internal lens reduces the brilliancy of the image. for field cleaning and interior of the telescope is not so easily accessible .n(3) etbicfnryoodnmivsateirna(A(glc5eIetn)b)gseivisrSmlreispaCgnoahthguieoteprltrvnyhoicaasionat.tfuntlrAetethaaAbhrreeeeeyborrrefaonrdtrtbohiofatjoieeteniclaondtllnolet.inoscsrocltnolhltaneehncvaatnederiesdngvitndeihaegiostxltiaseaioenncnd(ctl6ltetyo)hnmfrsiomaetuthehaDgteagohveinsarietnatotywghrspaetiitoothorinfcunfoeltpcy.nlruitgirssTeshmph.tbhea(wuetFitrcrhiiaagceysla.nleslypu9astn.hrcu2earor7qotfle)aouu..caugelrhlfTesyo,dhrtmihrsteehiimfenrdgeaaedcgrftgaeeeayescdns.st (!) Spherical aberration (2) Chromatic aberration Coma (4) Astigmation Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net SURVEYING 2t0 or imperfection, arising from !he form of curvature · of !he Ieos is known as spherical aberralion or sometimes as axial spherical aberralion. The spherical aberration of a negative lens lends to neutralise that of a positive Ieos. Hence !he positive and negative elements of an achromatic Ieos can be so shaped as practically tci eliminale spherical aberration also. w(2) Chromatic Aberration. A,beam of while light wgreen, blue, indigo and violet. Since ihdocal lenglh wcolour of light, a beam of while light instesd of, (See Fig. 9.17 also). FIG. 9.27. SPHERICAL ABERRATION. is made up of seven colQurs~. oJange, yellow, .lens. is distribuled along !he axis in a series of Efocal points. The violet ray ·is refracled most and the red is refracted least (Fig. 9.28). This defect ais known as chromatic aberrtUion due to which of any single Ieos is different for 'each different A converging at a focus after passing mrough a singk syEnaactolaoobofbwentlmrrevureleraaeplrxaktseoiTtecodiwohvnlapaeeeesnnerdc.sheaoiilscsgmTieoimhshlbosieoniupunmafforaefleawtiixoidodcotpeineeneirpmn.notootIaseFgifftotipoeegsraciasbrwis9coneift.hwgori1criowr7ncaamshhltliesiyogcihdtnnhlh.oasiwtsTsopmsshoeaesn·iaisisemctcnuihuhdvctrorherooiansm!mllaihaZynaeanaOtctdijaohccincin!enhhenaverebeegoreodamrc·ftrihacavdrtot!eeioihncorm·eecnlrealamevnrtonieosicsfnqe-ceusao.asirnbiennvIetmetoregwrxeaainashttii,tivosIcecenhloohisoIne!oeshroifeoeds!ffheoporeoroaufefbsll,idielnrteerietpvsloaieddgtsgoeinsvlugIaieebbrsoelolslyeees.f ., ~·· FIG. 9.28. CHROMATIC ABERRATION. !he quality of telescope. a single lens, i.e. coma, astigmation, curvarure, distortion majority of surveyors. If further infoJ:II1lltion be required, The other possible defects in elementary standard text books on physics or etc.. ase of little importance to !he of the reference should be made to any optics. OPTICAL CHARACTERISTICS OF THE TELESCOPE afliTeemrlenhismeaSehg,iaendfrsfarT(p((ero12t3fehoem))i)edenifmiAADtoiasfdaobpcrepgneonjlhfhedesamirice;nnrotrdaiasiTamtebctilcipohsalsaoheoelnietsnrnip.os.daaommurssbpepaae.tDsoAtitruoecinrecApdplaavflolitcianitnbnohanhiygctrbnaeiohoet,lihsmarpnomreaso,niaaacswattcciielaotouielexssknrnrrlaani,e!csnoanbitygpwsietoicupsnsfl!ltaiaibonebksaeodasn·sefwdoctaneeiwthbcsolsnaetisenuanctserrnhcanivoncocaeafegpesl!ypseheisllneapoetaaghnshinnfscedaedoecrt1i!aipfhccihleecteeralechoen,ltwcsrmsscoopeiahosasmmobooptweilaebfclirotyetirsinnracaaaatdsibcttoeeiooasceopninpornne.remaaf.agnbostldAbiireoalsmiinxfntocyauli.esoltpdli.nmoooAoi.iwsnnpft.isohncdpuaogernvmu:ofdieidnspuiholtbceaiuoeinbnnnesldgne.. Downloaded From : www.EasyEngineering.net

LEVELLING Downloaded From : www.EasyEngineering.net 211 disOpiennvde(s4r's)uelplyloiunpmrlohinpeaotrmitoiaongnnaolirfyBitnorgigmphoatngwenesirfsi.caaTntihdoen!hilelaunndntuinmnaubtimeornbeaornrdobfrqiuglaheltnintsyeessso. fof!h!eheleimnsaegse. of 1elescope ,~. llluntinatinn I !i ..!bfP'sbhahyreeiooncuvat!fulihodsasecie(noaSbdln)evi filirMepoft!nurfhoagmaegpllhhaonigmgriifntomhiiaiofcefagniargnctatheaiilfe,omtiiincotaooa.angtbnin!ojMdiheisnfceiactttgihaioavsntapoeiitoef)tirnolcetosuoanuwtrrsbieeohlden!ti(hugnoiechd.ieeselhsodfetbotohjrbebefrlcoyhi!tclhserlaaiea!lthawinosealcineplylnobeogureabnlrothpujwtfeprceeoeote,iaffmnsa!l!aihnhgtgteohdeehoe,toedyawsrredmneehpid-gnaiepulclaneilhcerdeycssfseeoun.robuttbetThpearjeohnescncedct!uefshidrmeleahltwadeeaigeitlnllsoeri!iafhsfgiatcechipooavtpipteinieeeoygoaw)nes.,f J'I and wastes more time in focusing. ~!~.i !· : · I.! ;.; :~ I ii i is meant !he whole circulas asca seen at Iii\\'i view is not merely dependent upon lhe size I (6) Size of field. By field o f view The field of also increases as lhe magnification of the one time lhrough lhe telescope. I·I~ :f of reticule, but it lhe hole in ·the cross-hair [!I lelescope decreases. t,;l 9.6. TEMPORARY ADJUSTMENTS OF A LEVEL ··:! a!bwhneehdtwicihen(e2sEln)raUarcepIshDoemremsmnueta.ardvnepeeayPnsiatetnlst7ga11deoi0vjnruJesislerrbynlUinlnIeeDainsndetssnjlu.rtaUssnteITmDeeeeemddmnisptssroutsrwreabotnrteyeidnetygdap(deSbajsueenesdotmmfCapeadmhrdeeajspupaosterbonarnlrtyoenr1Sywt6tsa)h.leit:oonn( Ilt)haake1dienjfumgusnptmdooarbeammsseyesrnvataadtlijaousrnseetsmlattehwinooitnstshes, ··~il 1 ngine!oliaeshpngepsfr!iloxheaxeevs(1idme1e,l)satotsaSetnSonoleyedt.tt,tat!ihidhnhnejeagogunrsdtiuttzrueripcipdppn(obottdah)dt!lhhe,eablteLytSvheloleeevhmlvteleueilecnr.lnlgaiiTn~snnsrhgl!.preh(sU2e~ro·I)~ioDspmiueeneLnnnrsrdaltteert~vUlieeoiItsalhnDalsiereenaeodngtt,floalslwuahisenppoetestprlirnrpUcogorpxIoDnaiuvmvnepiendantetiw(eind3elicisyn)llhutwdEhibhte!eylhhseilmdieg(laihaenli)gtneasftmfitaioa!xnahndihdlnejlaugnosctrbdhfi!ir.nghecehepunTtatl-tiarrhh.niarbasellnlrrTaaUbdxtocurI.hiDbpabefoninliddxest The temporary adjusbnents for a level coosist of !he following : t~ j iii ~I eon !he tribrach. I riis I nof /iii by gscrews. :,,: dlelohvneee2ll.pinwlLagitelehvisele!ltlhvoienelgmshaeuldlcppee.petoAhnfedftsfevoreourtphtioacsvnacilnrewgawxhsileselvhateenrlrudlely!dhwevtirhteehertiacirnseaesflle.rUrlehTInrDheceeeentmlteoaavpneplnlhlreieonrxgoipmflsacatlreteeevlweylels,lvineaoglcsrc.utfhroTeauhteriensllleeprvvUueerIpllDlloiiennsngget .(a) Three Screw Head !h n1. Loose !he clasnp. 'I\\trn !he inslrUIDent until eis roughly pasallel to a line joining any two (such ..G t[Fig. 9.29 (a)]. the longitudinal axis of !he plate level ill as A and B) of !he levelling screws ,i·:·~]~; il a;e.ifnai,dcthhteUo2m'm.dierH!rehocetlumdinotnilulhneois!fheofermmtwbloyouvbesmbleolveemetlhilsainotgfc!ehstnehcterretahwlu.lesmftbbsIettthwmusemohevboneul!dehie[tshbeeelehrunrtFnooibwtgead.as' dn9sdt.h2teU9faircs(httha)eof]i.ln·hbgeu·erbr boolref each hand Iii' away from move !! will iii ~~ ill :~ ~~ II ,l ~~ :]j Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net 3. Turn the upper plate SURVEYING Q ;0through 90', i.e., until the axis212 on the level passes over the position of the third levelling screw C [Fig. 'f \\\\'.9.29 (b)]. ,. c ... C'. ww(i~>o I4. Tum this levelling screw 0·--- --- ----- __:·auntil the bubble is central. [Fig. 9.29 (a)] and repeat step w(2) tJU. the bubble is central. 5. Remrn the upper part through 90' to its original positjon c .7. Repeat steps (2) and (4) till the · bubble is central in both the positions. A B AB (b) Eaof (a) FIG. 9.29. LEVELLING-UP W m l TIIREE FOOT SCREWS. 6. ·Tum back again through 90' and repeat step (4). sNote. It is essemial to keep the same quarter circle for the changes in direCiion yand nor to swing through the remaining three quaners of a circle to the original position. 8. Now rotate the instrument· through 180' . The bubble should remain in the ceritre run, provided it is in correct adjustment. The vertical. ~ will then be truly verti¢al. its E(b) Four Screw Head nI. Tum the upper plate until the longirudi~. axis of the plate level is roughly parallel If not, it needs periTUlllenl adjustmenl. '··'· .· to the line joining two diagoually opposite screws such as D and B [Fig. 9.30 (a)]. 2. Bring the bubble central exactly in the same manner as described in step (2) above. 3. Turn the upper part through 90' until the spirit level axis is parallel to the other diagoually opposite screws such as A and C [Fig. 9.30 (b)]. 4. Centre the bubble as before. 5. Repeat the above steps tJU the o, ~--~ 8.two c bubble is central in both the positions. 6. Turn through 180' to check the Q '·,, ~-''·' permanent adjustment as for three screw ' ·,-..., , ,.' instrumeru. o· roA ,/:<....,._.~] / In modern instrument~. three-foot ,_,.,· ........ screw levelling bead is used in preference cf· ··-··o to a· four foot screw level!ing head. The (a) (b) three-screw arrangement is the better one, FIG. 9.30. t.EVELUNG-UP W m i FOUR-FOOT as three points of support are sufficient for stability and the introduction of an extra point of support leads to uneven wear on tlte Screws. On the other hand, a four-screw levelling bead is simpler and lighter as a three-screw head requires special casting called a tribrach. A three-screw instrument has also the important advantage of being more rapidly levelled. 3. Elimination of Parallax. Parallax is a condition arising when the :mage formed hy the objective is not in the plane of the cross-hairs. Unless parallax is eliminated, accurate Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net t.EVELL!NG 21~ sighting is impossible. Parallax can be eliminated in two steps : (I) by focusing the eye-piece for distinct vision of the cross-hairs, and (ir) by focusing the objective to bring the image of the object in. the plane of cross-hairs. (r) Focusing the eye-piece To focus the eye-piece for distinct vision of the cross-hairs. point the telescope towards the sky (or hold a sheet of white paper in front of the objective) and move eye-piece in or out till the cross-baris are seen sharp and distinct. In some telescopes. graduations are provided at the eye-piece so that one can always remember the particular graduation . position to suit his eyes. This may save much of time. (il) Focusing the objective The telescope is now directed towards the staff and the focusing screw in rurned till the image appears clear and sharp. The image so formed is in the plane of cross-hairs. 9.7. THEORY OF DIRECT LEVELLING (SPIRIT LEVELING) A level provides horizontal line of sight, i.e., a line tangential to :: level surface at the point where the instrument stands. The difference in elevation between two points is the vertical distance between two level lines. Strictly speaking, therefore, we must have a level line of sight and not a horizontal line of sight ; but the distinction between a level surface and a horizontal plane is ~ot an important one in plane surveying. n taken on these points. gdifference in elevation between widely separated points is thus obtained. Neglecting the curvature of earth and refraction, therefore, the theory of direct levelling in• (a) Differential Levelling. It is the method of direct levelling the object of which is solely to determine the difference in elevation of two points regardless of the horizontal epositions of the points with respect of each other. When the points are apart. it may ebe oecessary to set up the instruments serveral times. This type of levelling is also knowo is very simple. With a level set up at any place, the difference in elevation between any two points within proper lengths of sight is given by the difference between the rod readings By a succession of instrument stations and related readings. the SPECIAL METHODS OF SPIRIT LEVELLING ri• (b) Profile Levelling, It is the method of direct-levelling the object of which is to ndetermine the elevations of points at measured intervals along a given line in order to g• (c) Cross-Sectioning. Cross-sectioning or cross-levelling is the pr6cess of taking levels .on each side of a main line at right angles to that line, in order to determine a vertical ncross-section of the surface of the ground, or of underlying strata, or of both. as fly levelling. ·u e, (d) Reciprocal Levelling. It is the method of levelling in which the difference in ::~•.::U~u~1' televation between two points is accurately determined by two sets of reciprocal observations obtain a profile of the surface along that line. ]i!il when it is not possible to set up the level between the two points. :!HI (e) Precise Levelling. It is the levelling in which the degree of precision required is too great to be attained by ordinary methods, and in which, therefore. special. equipment ,.·'·'.un· ii·f '\" ' ! .;~ ~~ or special precautions or both are necessary to eliminate, as far as possible. all sources ·ll of error. ,.i,l, 'I d il :\\! Downloaded From : www.EasyEngineering.net ·:!

IDownloaded From : www.EasyEngineering.net SURVEYING I 214 i TERMS AND ABBREVIATIONS (!) Station. In IevelliDg, a station is that point where the level rod is held aDd not where level is set up. It is the point whose elevation is to be ascertaiDed or the wof known e/evaJion, to asce<tajn· the atnount by which the line of sight is above point that is to be established at a given elevation. (il) Height of Instrummt (H.I.) For any set up of the level, the height of instrument is the elevation of plane of sight (line of sight) with respect to the assumed datum. It does not mean the height of the telescope above the ground where the level s!aDds. wup from the point of known elevation · to the line of sight. It is also known as a p/Jls (iii) Back Sight (B.S.). Back sight is the sight taken on a rod held at a point wheight of the instrument. The object o f back sighting is, therefore, to ascertain the heightthat point and thus to obtain tjle height o f the instrument. Back sighting is equivalent to measuring :: .(iv) Fore Sight (F.S.). Fore sight is a sight taken on a rod held at a point' of Eunknown elevation, to ascertaiD the atnount by which the .point is sight as the back sight reading is. always added to the level of the datum to get the ameasuring down from the line of sight. It is als<i known as a minus sight as the fore of the plane of sight. ssight reading is always subtracted (except in speical cases of tunnel survey) from the height yof the instrument to get the elevation of the point. The object of fore sighting is, therjore, below the line of to ascertain the e/evaJion o f llle point. El' (v) TurniDg PoiDt (T.P.). TurniDg point or clumge point is a point on which both nmiDus sight and plus sight ate iaken on a line of direct levels. The minus· sight (fore sight and thus to obtain the elevation of the station. Fore sighting is equivalent to sight) is taken on the point in one set of instrument to ascertain the elevation of the point while the plus sight (back sight) is taken on the satne point in other set of the instrument to establish the new · height of the instrument. (vi) Intermediate Station (I.S.). Intermediate station is a point, intermediate between two turniDg points, on which only one· sight (miDus sight) is taken to determiDe the elevation of the station. STEPS IN LEVELLING (Fig. 9.31) : (a) to find by bow much atnount the line of to ascertain by bow much atnount the next point There are two steps in levelling sight is above the bench mark, aDd (b) is below or above the line of sight. I Line ofs~hl ELV. 213.176 I .. 1.836 2.324 B ELY. 211.340 B.M. ELV. 210.852 FIG. 9.31. Downloaded From : www.EasyEngineering.net

LEVELLING Downloaded From : www.EasyEngineering.net 2lS A level is· set up approximately ntidway between the bench mark (or a point o f known elevation) aDd the point, the elevation of which is to be ascertaiDed by direct levelling. A back sight ' i s taken on the rod held at the bench mark. TbeD . .. (1) H.l. = Elv. o f B.M. + B . S . TurniDg the telescope to briDg into view the rod held on point B, a foresight (minus sight) is taken. Then ... rn ~=ru-u For exatnple, if elevation of B.M. =210.852 m, B.S.= 2.324 m aDd F.S. = 1.836 m. Then H.l. = 210.852 + 2.324 = 213.176 m aDd Elv. o f B = 213.176-1.836 = 211.340 m. It is to be ooted that i f a back sight is rakeD on a bench mark located on the roof o f a tunnel or on the ceiliDg of a room with the instrument at a lower elevation, the back sight must be subtracted from the elevation to get the height of the instrument. Similarly, if a foresight is taken on a point higher than the instrument, the foresight must be added to the height of the instrument, to get the elevation of the point. 9.8. DIFFERENTIAL LEVELLING The operation of 1evelliDg to determiDe the elevation o f points at some distance apart is called differenriol levelling aDd is usually accomplished by direct levelling. When two poiDts. are at such a distance from each other that they cannot both be within range of the level ill the satne time, the difference in elevation is not found by single setting but the distance between the points is divided in two stages by turniDg points on which the staff is held aDd the difference of elevation o f each o f succeeding pair o f such turniDg n points is found by separille setting up of the level. g <D ine l I 1~I 24~024 !i! er\" iA 242.590 g I (240.000) ~ _ 2¥12 ngFIG. 9.32 .Referring to Fig. 9.32, A aDd B are the two points. The distance AB bas been~ T.P.2 ndivided into three parts by choosing two additional points on which staff readings (both eplus sight and miDus sight) have been taken. Points 1 and 2 thus se:ve as ruming points.T.P.1(240.490) tThe R.L. of point A is 240.00 m. The height of the first setting of the instrument (240.604) is therefore = 240.00 + 2.024 = 242.024. I f the followiDg.F.S. is 1.420. the R.L. of T.P. 1= 2 4 2 . 0 2 4 - 1.420 = 240.604 m. By a similar process of calculations, R.L. o f T.P. 2 =240.490 m aDd o f B =241.202 m. Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net 216 SURVEYING 9.9. HAND SIGNALS DURING OBSERVATIONS· When levelling is done at construction site located in busy, noisy areas, it becomes difficult for the instrument man to give instructions to the man balding the staff at the other end. through vocal sounds. In that case, the following hand signals are found to be useful (Table 9.1 and Fig. 9.33) TABLE 9.1. HAND SIGNALS wRefer SigiUll Message Fi• . 9.33 Movement of left Move to_ my left w(a} ann over 90° w(b) /.· I 1.E r 1(CIMovement of right ann over 90° ~ove ·'w,· my (d) right Movement of left (a) (b) (c) A,.,,, aAs' J\\ileiarm over 30° Move top/of (d) ~ AyEn(j} staff to my lefi ~ ~ Movement of I Move top of - (e) (Q I right ann over 30° staff to my right , Extension of ann Raise height peg· [ horizontally and or staff 1 moving hand ! I ! upwards Lower height .• ] Extension of ann peg or staff horizontally and moving hand downwards (g) Extension of both Establish lhe I: arms and slightly position . (g) (h) thrusling downwanls (h) IExtension of arms Rerum to me and placement of ! hand on top of i , head. FIG. 9.33. HAND SIGNALS. 9.10. BOOKING AND REDUCING LEVELS There are two methods of booking and reducing the elevation of points from the observed staff readings : (I) Collimation or Heig/a of Instrument method : (2) Rise and Fall method. (1) HEIGHT. OF lNSTR!lMENI METHOD I!{ this mehtod. the height of the instrument (H.l.) is calculated for each setting of the instrument by adding back sight (plus sight) to the elevation of the S.M. (First point). The elevation of reduced level of the turning point is then calculated by subtracting from H.l. the fore sight (minus sight). For the next setting of the instrument. the H.l. is obtained by ·adding the B.S taken on T.P. I to its R.L. The process continues till the R.L. of the last point ( a fore sight) is obtained by subtracting the. staff reading from height of the last setting of the instrument. If there are some intermediate points, the R.L. of those points is calculated by subtracting the intermediate sight (minus sight) from the height of the instrument for that setting. Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net LEVELLtNG 217 The following is the specimen page o f a level field book illustrating the method of booking staff readings and calculating reduced levels by beigbt of instrument method. Station B.S I.S. F.S. H. I. R.I.. I Remtuis I A 0.865 561.365 560.500. I o:M. on Gate B 1.025 2.105,. 56!).285' 559.26() c 1.580 558.705 Plalform D 2.230 1.865 560.650 558.420 E 2.355 2.835 560.270 \" 5S7.8t5 F 6.475 1.760 558.4t0 Chocked Ch<Ck 8.565 558.4t0 6.475 560.500 2.090 F~l 2.090 -- Arithmetic Check. The difference between the sum of back sights and the sum of fore sights sbould be equal to the difference between the last and the first R.L. Thus l:B.S. - 'EF.S. = Last R.L. - First R.L. (2) RISE AND FALL METHOD In rise and fall method, the height of instrument is not at all calculated . but the n difference of level between consecutive points is found by comparing the staff readings on the two points for the same setting o f the instrument. The difference between their gstaff readings indicates a rise or fall according as the staff reading at the pt>int is smaller ior greater than that at the preceding point. The figures for 'rise' and 'fall' worked out nthus for all the points give the vertical distance of each point above or below the preceding one, and if the level of any one point is known the level of the next will be obtained eby adding its rise or subtracting its fall, as the case may be. eThe following is the specimen page of a level field book illustrating the method rof booking staff readings and calculating reduced levels by rise and fall method : The method affords a check for the H.(. and R.L. of turning points but not for the intermediate points. .. inSlllli<lnB.S. I.S. F.S. Rise Fall i R.I.. i Remarks gI! 0.86~ . 1.580 • 2.tQS .nc 1.07.5 • l.24ll • 0.555 etE 2.355 560.500 B.M. on Ga[e 559.260 i ' 558.705 Platfonn D 2.236 .. 1,86~ I 0.285 558.420 2.815 0.605 557.8t5 'F 1.760 0.595 558.410 ; Check 6.475 8.565 0.595 2.685 I558.410 Checked 6.475 0.595 560.500 F~l 2.090 F~l 2.090 2.090 ~ Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net 218 SURVEYING Arltbmetli: Check. The difference between the sum of back sights and sum of fore sights should 1» equal to the difference !»tween the sum o f rise and the sum of fall and should also 1» equal to the difference between the R.I. o f last and first point. Thus, wand l:B.S. - U . S . = l : Rise - l : F a l l = Last R.L. - First R.L. a complete check on the intermediate sights also. The arithmetic check This provides the unlikely, but possible, case o f two more errors occurring in such would only fail in wbefore the reduced level calculatio~ \"are commenced. a manner as to balance each other. w.E..Tcliaesalhvllecerules)liqragiCutshmiieootrseemn.tdsahpnoHfadtdoororiwsfitaisoeanlnkvlteeemrrmom,oafereettdhhntioehuardeamtephbietTdsehii,wrgoghuhooltgtesfhsMosirsfemetatneiohdndoroisientdtograSuusvts.meadfieilraonaonTbutmdlhse,me,steihptmthherheeopoivlgsdemiah.dmiteiss!seHaolocfmaiweb~sostehfrtvurie~enulelmrs.tihuneceisnhsttitiaerrnbuccskllmeeeevtetieniintnlhntsgec,(pcaoaaclrcsshsesueu,clcaucoktnhlwilonihomnoaenstsaircetie!t(.fohdoonie.rrt I t is advisable that .on each page the rise and fall calculations shall be completed checked by comparing with· the' difference o f the back and fore sight column summations, asyEncti;hfoen0tsh.ti9ern8usEE8ctfrtxniUirotas;emmntrae2plntr.lhte0eewa9hd0o9aairn.vbk1gi;a,n.vge2wTp.hra8broese6efea4ilnfdteoailn;klmogelewnso/v.vi2enienlw6dgli2intahagsft;ptaeaaefr0ftgce.s.rt6htea0ioarf2ddff;i,nahg.es1s/lei.dx9vtw8he2oe/ rnabe;noadook1b.ebs0aie4egrn_nhv4dctehhd;carmsle2ucaa.uc6rdclk8aien4tsegsosivfthe:mle4y3e2Rt2r.w.2.e3Li2st8.h.84o.a;..fm. p.//.o.V6in0elt6.s Solution. iionnlnasts·rtthuthemereeFanBd.tSi.nS.wg. acscoionlusluhmtmihfntene.daFnA.adSfl.tsetorh,ceotrhtlehuifremodr,nef.,isrsixtAthtlehlreaafoondtuhdinretgrhei,gwrhesitaelhldvienrbnegetashdeiwnnagntieldslr,ednbthieneitsnheenttrrheeeeraaedddiBinn.ggSinss. Since the •i will be entered '~i will be entered column and the the I. S. column. of the -points may be calculated by rise and fall method as tabulated The reduced levels I.S. F.S. 1.606 FaD R.L. RelfiiiTts below · 2.864 432.384 B.M. 0.714 433.006 T.P.l Station B.S. Fall 433.624 2.228 1.380 432.850 · T.P. 2 I 1.640 T.P. 3 3.794 ~34.452 2 0.988 0.622 2.842 Cbockcd 3 2.090 D.6l8 0.952 433.on 431.432 4 1.262 1.602 432.384 431.432 s 0.602 1.982 2.842 · Fall 0.952 6 1.044 2.684 7 6.916 S.964 Check S.964 0.952 Downloaded From : www.EasyEngineering.net

LEVELLING Downloaded From : www.EasyEngineering.net 219 Example 9.2. It was required to ascertain the elevation o f two points P and Q and a line o f levels was run from P to Q. The levelling was then continued to a bench mark o f 83.SOOi the being as shown below. Obtain the R.L. o f P and Q. readings obtained B.S. I.S. F.S. R.L. Remarks i 1.622 I p I 1.874 0.354 I 2.032 1.780 .I 2.362 Q ,~,i .0.984 1.122 \"::!:. /.906 2.824 2.036 :!' ~ 83.500 B.M. \\i:j Solution• . ·i'j To find the R.J.s. o f P and Q, we will have to proceed from bottom to the top. '~·t='!il F. S. readings will have to be added to the R.L. o f the koown To find the H. I., therefore, !iii point and to find the R.L. o f the previous point, the B.S. will have to be subtracted from the so obtained H.l. as clearly shown in tbe table below · ~~ B.S l.S. F.S. H. I. R.L. Remaris 'ii,l n ' 'Q 2.362Station :·:':';J1 g 0.984 ' 1.874 83.198 p 1.622 84.820 '~1 0.354 86.340 84.466 I i 2.036 neCheck 2.032 1.780 86.592 84.560 I' 84.230 ! ! I 1.122 86.454 85.470 e I IRise i 1.906 2.824 85.539 83.630 ] ring.n·.0rmame.need3dttu8hrceo5eed;ndElteexav1rlaeen.mlv0dtlehi3pnle0lgae;losaofs/b9.t9o.a.t37h2vf.hefe5e·ofrni;ergTasrhdtac2eiod.n8pnigfe2ootsni5li.nltnoutwoC;ouiwafns3lalgcy.ts7uhl3eacs20lotoe0nlpi8;snite.neh/c2ge4u5.j(to.ij.rg.Jvi.enlrien5doi.nuurcgn;eeRda'd0udth.ial6nle~ete2g'vs5ofeaiulrstsw;cteooar2mfea.0mpnt0hadtaa5egnketehpn;ieoon·it3fnewltr.asia/vtslhatOblleyapvo;oerfillnies4t2evf.,i04eel8lad5mna.denbtdroTfeaohsl5ke:lI 83.500B.M. 8.418 8.116 83.500 I 8.116 83.198 0.302 0.302 Checked I Solution. etrwoenialdl ainbgeStuirncancaiennbgatchbkpeeosir5niegtahdmtainneagdtrnsedsthwethoerenseleylvta,aeskntaetnhnrdearodetniahndegairnegfwcooirnlwelt,iilnbluseiobxueatshlyfaorresbealaodcpiskniignghgsti.wgghiTrltol.huenbAdell,seovat,ehlesfthoemrceaanfx'isriimsbgtehumtrreeataadsdktiainelfgnyf calculated as shown in the tabular form below: Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net 220 SURVEYING Station B.S. /.S. ! F.S. Ri.se Fall R.L. : RtiiUIIb I 0.385 ,.. 2 1.030. 0.645 208.125 i 3 I 1.925 0.895 'tl 4 0.625 2.112S 0.900 207.480 I 3.730 0.905 : w1\"\"' s ' ' 206.585 i i[ ' ww·~ : I 0.955 6 I 2os.6ss I 7 2.005 1.380 8 I 1.010 i 3.110 4.685 I 1.105 204.780 i : 1.375 203.325 Check I Fall .-4.485 8.160 202.445 l ' 0.000 .9.-170 8.160 I 0.000 1.010 8..160 Fall I 19 61 .E1: Gradient of the line= ~O~~~ = ~ =I in 1?.~1 (falling). 201.340 199.965 208.125 199.965 asExample 9.4. The following 8.160 of the entries being illegible owing ycheck your res111Js. Rebook all the ... EStation I n2 figures were exJracted 1jiom a level field book. some to exposure to rain. Insert tile missing figures 'qnd figures fly tire 'rise' and 'fall\" method. ' B.S. J.S. F.S. I Ri.se I Fall R.L. : ReiiUIIb • 2.285 X ~· ' 232 .46() _j B.M. I I /.6SO 0.020 :i 3I i 2 .I05 IX 4 :X I.960 X 0.300 J 5 I.925 ' ! 2.050 X : •6 X I ' 232 .255 I B.M. 2 i I.690 0.340 i 7 X I 8 ! 2.865 X IX 9 ! I 2.100 ! i I I X i233.425 B.M. 3 Solution. can be calculated from the known rise. (r) The F.S. of station 2 is missing, but it will be lesser than the B.S. of station Since station 2 is higher than station 1, its F.S. I (higher the point, lesser the reading). Hence, F.S. of station 2 = 2 . 2 8 5 - 0.020 = 2.265 m ~ and tit) R.L. of station 2 = 232.460 + 0.02 = 232.480 m Fall of station 3 = 2.105 - 1.650 = 0.455 m 1,1 :. R.L. of station 3 = 232.480 - 0.455 = 232.025 m li (iit) B.S. of staion 4 can be calculated from the fact that the F.S. of station 5. having a fall of 0.300 m. is 1.925 m Thus, B.S. o f station 4 = I. 925 - 0. 300 = I. 625 m Downloaded From : www.EasyEngineering.net

lDownloaded From : www.EasyEngineering.net LEVELUNG 221 I Also, Rise of station 4 = 2 . 1 0 5 - 1.960 = 0.145 m !' and R.L. of station 4 = 232.025 + 0.145 = 232.170 m (iv) k.L. of station 5 = 232.170-0.300 = 231.870 m (v) From the known R.L. of stations 6 and 5, the rise of station 6 can be calculated Thus, Rise of station 6 = 232.255 - 231.870 = 0.385 i l.S. of station 6 = 2.050 - 0.385 = 1.665 I and (vi) F.S. of station 7 = 1.665 - 0.340 = 1.325 ;I (vir) R.L. of station 7 = 232.255 + 0.340 = 232.595 ·..~. ~ (viit) 14 Fall of station 8 = 2 . 1 0 0 - 1.690 = 0.410 1!-1 R.L. o f station 8 = 232.595 - Q.410 = 232.185 Since the elevation of station 9 is 233.425 m, it has a rise o f (233.425- 232.185) 1,~1I = 1.240 m. m. tabular form below .!A F.S. of station 8 = 2.865 - 1.240 = 1.625 the uI~'. I~' The above results and calculations are shown in n ' 4 1.625Station I B.s. I.S. F.S. Ri.se Fall R.L. Rtnuul<s ' 5 2.050I2.285 2.105 2.265 0.020 0.455 232.46() g6 2 1.650 232.480 I B.M. I I 3 232.025 i7 1.690 I ~~~ n8 2.865 I :!~ e9 1.960 0.145 232.170 I ,, 1.925 231.870 ; 0.300 L 232.255 i B.M. 2 er ICheck 1.665 0.385 1.325 0.340 232.595 intaken 2.100 I 0.410 232.185 1.625 1.240 233.425 B.M. 3 gReading on inverted staff on B. M. No. A.I •12.165 11.200 2.130 1.165 233.425 I ' 11.200 1165 Rise 232 46() l _ Checked_] i 0.965 0.965 - - 0.965 .Reading on peg P on grolllld : Rise nChange of instrumenl eReading on peg P on ground : ;;·as tReading on inverted staff on bottom of cornice B : Example 9.5. During a construction work, the bortom o f a R.C. Chhajja A as a temporary B.M. (R.L. 63.120). The following notes were recorded. 2. 232 l . 034 1. 328 4.124 Enrer the readings in a level book page and calculate tile R.L. of cornice B. Solution on an inverted staff and therefore it will have to be H.l. Sintilarly, the last reading was taken on an invened The first reading was taken B will be obtained by adding the F.S. reading to the subtracted from the R.L. to get the staff, and the R.L. of the cornice Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net ¥'1--.- 222 SURVEYING H.l. Use (-) sign for the B.S. o f A and F.S. o f B since both o f these have been taken in reverse directions than the normal ones. The calculations are shown in table the below: pPoinl B.S. I.S. F.S. R.I. R.L RetiUUb A - 2 .232 60 .888 63.120 wB 1.034 61.182 59.854 I Che<k Rise .I -4.124 65.306 wRise -3.090 65.306 Checked 1.328 63.120 ''' I 2 .186 -0.940 -3.090 + + 2.186 w9.11. BALANCING BACKSIGHTS AND FORESIGHTS When the difference in elevation between any .Eset-up by backsighting on one point and foresighting on o f line o f collimation and axis o f the bubble rube athe run) and also the error due to curvature and sIn Fig. 9.34, let ob- two points is delermined from a .single the other, the error due to non-parallelism (whe,i. the bubble is in the ceritre ·or refraction may be eliminated· if the yservations be made with a E$rlevel in which the line of ncollimation is inclined up- lengths o f two sights can be made equal. \"-----·\"'------ -------1 T . , wards by an amount a from horizontal, when the bubble is in the centre of its run, A o, D2· ojB the level being kept exactly midway between the two FIG. 9.34. BALANCING B.S. AND F.S. points A and B. The observed be eoual to backs1ght and foresight ar~ r. and X:. T h e ~orrect hacksiQ'ht on A will on B will be equal to XJ: - y2 where x1 - y1, where y1 = D 1 tan a_ . The correct foresight between A and B y, = D2 tan a . Hence the correct difference in level =~-~-~-~=~-~+~-~ =(x, - x,) + (D, tan a - D, tan a ) =(x, - x,) if D, =D, Thus, if backsight and foresight distances are balanced, the difference in eleva/ion between two taking difference o f the two readings and points can be directly calculated by no correction for the inc/inaJion of the line of sight is necessary. Fig 9.35 illustraleS how the error due to curvature can be eliminated by equalising horizontal line o f sight (and not backsight and foresight distances. Since the level provides The correct staff a level line), point B =hb . the staff reading at point A =h, and at a readings sbould have been H, and H• so thO! Ho = ho - ha' and H11 = hb - hb' The correct difference in elevation between A and B, therefore is given by Downloaded From : www.EasyEngineering.net

-. Downloaded From : www.EasyEngineering.net LEVE!.I.lNG 223 level lin, H level nne fhro• A (The eff&ct of refraction tlas not been shown) FIG. 9.35 H=&-fu=~-M-~-~=~-~-~-~ I f the horizontal distance AC and BC are not equal, true difference in elevation H cO!tllot be found unless ha' and ho are numerically found (see Art. 9.7). But if the distanceS AC and BC are balanced (i.e., made equal), ha' and ho would be equal and H will equal to ( h , - h•) . twO Thus, if the backsight and foresight distances are balanced, the elevation between points to the two points is equal to the difference betWeen the rod readings taken n In Fig. 9.36, it is required to set the level between two points T.P. I (turning and no correction for curva/Ure and refraction is necessary. BALANCING SIGHT ON A SLOPE ginjpstuuiogsritnhnittn)gjuaabsnptodoviberTnel.oPthTw.e.P2.tbh. oe2tLtoteomctapnthooeftfheltanehnveebleerxobldeeenstdwasebehdtleisnueropddheaflawtdr hAeoennnfoahurtehglehednotuouupnrghnhiiltnlhgeutpophtuoilbrilnnriitnntoggT.btPphro.iening2tli.ntTheT.ePho.elfinIels.eigvoheAltf When the points lie on a sloping ground, the level should be set off to one side far enough to equalise, as nearly as practicable the uphill and downhill sights. at A is nearly e20 m and 12 eto one side but at nearly the same elevation as at A, so rthat sights on T.P. 1 and T.P. in2 can still be taken and dis- tanceS o f T.P. 2 and T.P. g1 from B can be equal, the l .error due to non-adjustment nof collimation will be elimi- lnated. \" · eTo take a numerical ex- tample, let the level have line on the line between T.P. I and T.P. 2, the corresponding distance being m (say). On the contrary, if the level is set up at B, instead o f A. off :1 T.P.2 :0 1,..,.,.,., \"' \"' ~/ ........ § o , ~/ .................... ~ A«( --;.ElB 5 g> ......... \\ \\~ .g. ....... ...... \\ \\-''<.I.' .................4. z.......~......... \\ of sight inclined upwar,ds by :1 T.P.1 ,0_:: an amount 0.008 mettes in :0 Plan every 100 metres. When the FIG. 9.36. Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net SURVEYING 224 level is at A, lhe error in lhe rod readings will be Error in lhe levelling= 0.00064 m wAgain. if lhe level were at B. lhe emors in rod readings would be at T.P. 2 : 0.50 x 0.008 = 0.0040 m wT.P. I : 0.52 x 0.008 =.0.00416 m For T.P. 2 : 0.12 x 0.008 = 0.00096 m For T.P. 1 : 0.20 x 0.008 = 0.00160 m wThus. when lhe level is at B. 'lhe error in levelling is about {- lh of lhe error if .lhe level is set at A. By moving B farrher away, lhe error may be reduced until it approaches Ezero, as lhe lenglhs of lhe two sights from B become nearly equal. Error in levelling= ~.00016 m aA sar B, yIt is Ebeen obtained had t/rere been no collimation error. Example 9.6. A level set up an extended line BA in a position 70 metres from B reads 1.684 on a staff held Ill ~<'and 2.122 on a staff held and 100 metres from nSolution. Exact difference in elevation in\"B and A the bubble having been carefully brought to the centre o f its run before each reading. known thai the reduced levels of the tops o f the pegs A and II ·are 89.620 aild 89.222 respectively. Find (a) the collimation error, and (b) the readings 1:1at would have = 8 9 . 6 2 0 - 89.222 = 0.398 m, B being lower. As per observations, difference in elevation = 2 . 1 2 2 - !.684 = 0.438 m, B lower. This shows B to be lower lhan what it is. We know !hat. lower is lhe point, greater is lhe staff reading. Hence, lhe staff reading at B is greater lhan what it should be and lhus. lhe line of sight is inclined upwards, as shown in Fig. 9.37, by an amount 0.438 - 0.398 = 0.040 m in a distance of 30 m. Therefore tan a = 03~ = 0.0013333 We know lhat tan 60\" = 0.0002909 Hence by proporrion, a = 13333 X 60 seconds = 4' 34\" upwards. 2909 ,040 ~ 'Xf\"\"\"\"\"\"\"!'f\" False line of slaht Ill I ·-·-·-·-·-·-· ·-·-.1.-.-·-·-·-·-·-·-·-a· ~ B l lA 1 < - - 30m 70 m----~ FIG. 9.37 Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net LEVELLING !2S Exact reading i f !here were no collimation error. would be at A : ! . 6 8 j l - ( 0· : x 70·) = 1.684-0.093 = 1.591 m at B 2 . 1 2 2 - (0 · : X 100) = 2.122-0.133 = 1.989 m. So lhat lhe true difference .. in elevation = 1.989 - 1.591 = 0.398 m as given in lhe question. Example 9.7. A page o f a level book was defaced so that the only legible figures were (a) consecutive elllries in the column of reduced levels : 55.565 (B.M.) : 54.985 (f.P.) ; 55.170 ; 56.265 ; 53.670 ; 53.940: (f.P.); 52.180; 52.015: 5!.480 f f . P . l ; 53.145 : 54.065 (f.B.M.); (b) entries in the bocksighl column: 1.545: 2310: 0.105 : 3.360 in order from the top o f the page. Reconstruct the page as booked and check your work. Calculate the corrected level o f the T.B.M. if the instrumelll is known to have an elevared collimation error o f 60 \" and bock and foresight distance averaged 80 and 30 metres respectively. Solution. There are lhree turning points on which bolh back and foresights have been taken. The first sight is a backsight. The four backsight readings will. lherefore. '\"~;' be entered in order, one against lhe B.M. point and olher lhree against lhe lhree turning points. The last R.L. corresponds to T.B.M. on which a foresight is missing. All olher ;, sights will be J.S. and F.S. which are to be found. Knowing R.t.. and B.S. of any point. lhe F. S. of lhe point can very easily be calculated. The readings having ( x ) mark are j; n missing quantities which have been computed as shown in lhe tabular foim l\" g ,. ...<rMinn !: i I '\"I n2 1~ l e3 I R..<. ~ ; Rt ' Rcmatks ' 1.545 e l4 I ! 57.110 '' 55.565 ' B.M. r !5 I 2.310 !I 57.295 ' 54.985 T.P. ' I I i' x 2 .125 [ n6 0.105 X 2.125 [ f 55.170 g7 X 1.030 I 56.265 ! 53.670 .8 I I X 3.625 ' ' 53.940 ! n9 3.360 I 52.180 l 52.015 X 3.355 54.045 i T.P. ! et10 X 1.695; X 1.865 i ~ X 2.030 ' ' X 2 .565 54.840 51.480 T.P. 53.145 __j' I :II 54.065 T.B.M. ; X 0.775 I i ...! I 7.320 ~·~ 55.565 '' Check I 1.500 Fall I ~OtiS Chet:kc:{l 1.500 _l Fall ____1 ' '---- Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net SURVEYING 226 Due to collimation error each backsight staff reading is too great by an amount (80tan60\") metres. Also each change point F.S. reading is too great b y an amount (30 tan 60\") metre. Taking both errors together, it is as if F.S. readings were correct and B.S. too great by amount (50 taD 60'~ ·metres. As there are four set-ups, the total B.S. reading are great by an amount 4 x 50 tan 6 0 \" = 200 ·x 0.0002909 = 0.05818 ~ 0.058 metres. Now greater the B.S. readings, higher will be the H./. and. therefore, greater will be reduced levels calculated. The actual level of the T.B.M. will therefore, be = 54.065 - 0.058 = 54.007 m. w9.12. CURVATURE AND REFRACTION wFrom the definition of A (b) a level surface and a hori- '0 (a) wzontal line it is evident that a botizontalline departs from a level surface because o f .Ethe curvarure of the earth. Again, in the long sights, athe horizontal line of sight does nor remain straight but sit slightly bends downwards yhaving concavity towards ea Erth surface due to refraction. nIn Fig. 9.38 (a), AC is the horizontal line which FIG. 9.38. CURVATURE AND REFRACI10N. deflects upwards from the level line AB by an amount Be. AD is the acrual line of sight. Curvature. BC is the deparrure from the level line. Acrually the staff reading should have been taken at B where the level line cuts the staff, but since the level provides only the horizontal line of sight (in the absence of refraction), the staff reading is taken at the point C. Thus, the apparent staff reading is more and, therefore, the object appears to be lower than it really is. The correction for curvature is, rherefore, negan·ve as applied ro the staff reading, its numerical value being equal to the amount BC. In order to find the value BC, we have, from Fig. 9.38 (b). O C ' = OA' + A C ' , LCAO being 90' Let BC = Cr = correction for curvature AB = d ~ horizontal distance between A and B that of d AO = R = radius of earth in the same unit as (R + c<)' = R' + d ' or R1 + 2RCc+ C / = R2 + d 1 Cd2R + C,) = d 2 c, 'or , C, in comparsion to 2R) C, = _ d _ - _ !Jo 2Rd , (Neglecting 2R+ Downloaded From : www.EasyEngineering.net

LEVELLING Downloaded From : www.EasyEngineering.net 227 That is, to find the curvature correction, divide the square o f the length of sight by earth's diameter. Both d and R may be taken in the same units, when the answers will also be iij terms of that unit. The radius of the earth can be taken equal .to 6370 km. I f d i s to be in km, and R = 6 3 7 0 km,,C,-0.07849 tf metres]In the above expression. d i s to be substiruted in km, while c, will be in metres. Refraction : The effect of refraction is the same as if the line of sight was curved downward, or concave towards the earth's surface and hence the rod reading is decreased. Therefore, .the effect of refraction is to make the objects appear higher than they really are. The correction, as applied to staff readings, is positive. The refraction curve is irregular because o f varying atmospheric conditions, but for average conditions it is assumed to have a diameter about seven times that o f the earth. The correction of refraction, C, is therefore, given by C, = ~ ~ ( + ve) = 0.01121 d 2 metres, when d is in km. The combined orrecbon due to curvature and re action will be given by C =2d-R-7-1 2-dR' =76-2-Rd 2 (subtracn•ve) j:l = 0.06728 d' metres, being in km. \"!i:·: n = ·The corresponding values of the corrections in English units are I. gpistance to the visible horizon C, = j d ' = 0.667 d 2 feet] 'fd 2 :z !1 C, = 2 f d [, C ~ d' iIn Fig. 9.39, let P be the point of observation, its= 0·0952d is in miles and II nheight feet radius of earth = 3958 miles. d 1!1 · ~ 0.572 d 2 feet r: elevel line. If d is the distance to visible horizon, it is given eby A ')' ~.. ~ v r~I .· c 0 being equal to C and let A be the point on the id = horizon i.e., .a point where the tangent from P meets the = ng3.8553 -./C km, C ~eing in metres. .(Taking both curvature and refraction into account).0.06728 km n/Example 9.8. Find the correction for curvarure and for refraction for a distance o f (a) 1200 metres (b) 2.48 km. etSolution. FIG. 9.39. (a) Correction for curvature . = 0.07849 d 2 metres (where d is in km) = 0.07849 (1.2)2 ~ 0.113 m Correction for refraction = 'Ii C, = 0.016 m Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net m SURVEYING (b) Correction for curvature = 0.07849 (2.48)2 = 0.483 m Solution. = ~ C, = 0.069 m. correctlon for curvalure w(a) Combined Correction for refraction and refraction for distance ~pie 9.9. Find the combined o f (a) 3400 metres (bj ..1.29 km. .l? wCombined correction wand Q. a level was set upon the iine PQ, 60 metres from P and 1280 metres from Q. 'llle readings obtained on sta!f..~pt al P and Q were respectively 0.545 metre and 3.920 correction for curvarure and refraction = 0.06728 d ' m = 0.06728 (3.40)2 = 0.778 m. .m. Find the true difference in elevalion between P and Q. = 0.0672s (1.29)' = 0.112 m. ESolution. · aSince the distance of P from instrument is small, the _,correction for curvature etc. Example 9.10. 1n order to find .the difference in elevation between two poinrs P sCombined correction for Q yCorrect staff reading at Q E. . Difference in elevation between P and Q = 3.810-0.545 = 3.265 m, Q being lower. is negligible. n/ E x a m p l e 9.11. A Ught-house is visible jUSt above the horizon al a cenain station = 0.06728 (1.280)2 = 0.110 m (Subtractive) , =3.920-0.110=3.810 m ' al the sea level. The .mstance ~between the Stalion and the Ught-house is 50 fan. Find the height _of the light-house. r ... ~J ('d\"(~ Solution. UP The height ·of the light-house is given by r ._,/., C = 0.06728 d 2 metres= 0.06728 (50)' metres= 168.20 m I Example 9.12. An observer standing on the deck o f a ship just sees a light-house. .- The top o f the light-house is 42 metres above the sea level and the height o f the observer's eye is 6 metres above the sea level. Find the distance .A o f the observer from the Ught-house. '·---------------- 0 Solution. (Fig. 9.40) Let A be the position of the top of light-house '\\ i' i' and B be the position o f observer's eye. Let AB \\ i' be tangential to water surface at 0 . \\ i '\\ -rc.The distances d, and d, are given by d, = 3.8553 km \\\\ \\ \\ = 3.8553 -[42 = 24.985 km ' / \\\\ at!d d, = 3.8553 { 6 = 9.444 km \\.\\!/ . . Distance between A and B = d, + d, 0 = 24.985 + 9.444 = 34.429 lam FIG. 9.40. Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net • LEVEL~G 229 A x a m p t e 9.13. 'llle. observalion ray between two triangulation stations A and B just grazes the sea. If the heights o f A and B a r e 9,000 metres and 3,000 metres respectively, determine approxiqtalely the mstance AB (Diameter o f eanh 12,880 fan). Solution. In Fig. 9.40, let A and B be the two triangulation stations and let 0 be the ~~ of tangency on the horizon. )n_)ota Let A'A;, c, = 9000 metres= 9 km .& metres• = 3 km <7j \\ \"> \" B'B = C, = 3000 ? The distance d, is given by c1 d' =_.!._ 2R or d, = -./2iiC; in which d , R and C, are in same units d1 = .,;2 X 6440 X 9.0 = 340.48 km Similarly d2 = .,;2RC2 = .,;2 X 6440 X 3.0 = !96.58 km . . Distance AB = d, + d, = 340.48 + 196.58 = 537.06 Ian. Example 9.14. 1Wo pegs A and B are 150 metres apan. A level was set up n Determine (a) the angular error of the collimalion line in seconds, and (b) the length in the line AB produced and sights were taken to a staff held in tum on the pegs, the reading being 1.962 (A) and 1.276(8), after the bubble has been carefully brought to the cenrre o f its run in each case. The reduced level o f the tops o f the pegs A and B are ' gerror. Assume the radius of the eanh to be 6370 /an. iSolution. nObserved difference in elevation between A and B = 1.962 -1.276 = 0.686 m (A being lower) eThe difference in elevation = 121.324 - 120.684 = 0.640 m, A being lower. e-_,;; Hence, from the observations, A seems to be lower by an additionat amount= 0.686 - 0.640 = 0.046 m. known to be 120.684 and 121.324 m respectively. o f sight for which the error due to curvalure and refraction would be the same as collimation riSince B is nearer to the instruments than A, it is clear that the line of sight ninclined upwards ·by an amount 0.046 m in a length of 150 m. If a is the angular inclination (upwards) o f the line of sight with horiwntal, g.tan a = 0j~6 = 0.0003067 is netWe know that tan 60\" = 0.0002909 _3067x60 mm. utes _- 1, \"3. (upwards). a - \"'\"\"n Ln. For the second part of the problem, let the required line o f sight be L km. The combined correction for curvature and refraction would be ~ ~ (negative). The correction for collimation error in a length L will be L tan a . Equating the two, Downloaded From : www.EasyEngineering.net

/ Downloaded From : www.EasyEngineering.net SURVEYING 230 -6 -2LR' =L tan a =L (0.0003067) 7 L - 0.0003067 X 7 X 2 X 6370 = 4.557 km. 6 9.13. RECIPROCAL LEVELLING When it is necessary to carry levelling across a river. ravine or any obstacle requiring a long sight between two points so situated that no place for the level can be found wfrom which the lengths of foresight and backsigbt will be even approximately equal, special method i.e.. reciprocal levelling niust be used to. obtain accuracy and to eliminate the following: w(I) error in instrument adjusbnenl ; (2) combined effect of eanb's curvature aod the refraction of the abnosphere, and (3) variations in the average refraction. wLei A aod B be the points aod observations be made with a level, the line of sight of which is inclined upwards when the bubble is in the centre o f its run. The level .is set at a poim near A aod staff readings are taken on A and B with the bubble in Ethe centre of its run. Since B.M. A is very near to instrument, no error due to curvarure, refraction and collimation will be introduced in the staif¢readings at A. but there· will abe an error e in the staff reading on B. The level is then shifted to the other · Dank, on a point very near B.M. B, and the readings are taken on staff held at B and A. sSince B is very near, there will be no error due to the three factors in reading the staff, ybut the staff reading on A will have an error e. Let h, and hb be the corresponding EnHorizontal line Level line Line of River sight LeV9\\\\In8 ·- ___.....-·-·-·-·-·-·- -·- -·-·-·- -·-·-·- -·-- ....7 A Plan B (B.M.) (B.M.) FIG. 9.41. RECIPROCAL LEVELLING. Downloaded From : www.EasyEngineering.net

LEVELLING Downloaded From : www.EasyEngineering.net 231 staff readings on A aod B for the first set of the level and h ; and , . . be the reading> for the second set. .;taff From Fig. 9.41. it is evident that for the tirst set of the l•vcl. the corr•ct readings will be OnA:h,; OnB:h•-e ·e,. . T~ difference in elevation= H =- h0 - tilt-- Similarly for second set. the correct staff reading will be On A : h , ' - e ; On B : h> . . True difference in elevation =H =(h,' - e) - It> Taking the average o f the two nue differences in elevations. we get \"•12H =[ h , - (h•- e)+ ( h ; - e)- =( h , - hh) + l h o - \" ' ) H = t l ( h , - hb) + (lr,' - II>)J . The lrue difference in elevation, therefore, is equal 10 the mea11 of the two appearerrt differences in elevations, obtained by reciprocal observaJions. Example 9.15. The following notes refer to reciprocal levels take11 u'ith nne lel•e/: Jnst. at Staff readings on Remarks n Find (a) true R.L. of Q. (b) the combined correction for curvatrtre and refraction. and (c) the angular error in the collimation adjustment o f the inscrumem. g What will be the difference in answers of In) and (c) if ohsen•ed staff readin.~< iwere 2. 748 Oil P and 1.824 011 Q, the instrumcm beinx m P : and I. 606 nn P and n0. 928 on Q. the instrument being aJ Q. pQ p 1.824 2.748 Distance between Pand Q;JO!Om Q 0.928 /.606 R.L. of P;/26.386. eSolution. e(a) When the observations are [aken from P. Lhe apparent differcm:e in ckvaLion between P and Q = 2.748 - 1.824 ; 0.924 m. P being higher rWhen the observations are taken from Q. tht! apparent difference in dc\\·arion hcrwc..'t!n inP and Q = 1.606- 0.928 =0.678, P being higher. g.nand e(b) Combined correction for curvature and refraction t= 0.06728 d 2 = 0.06728 (1.010)' ~ 0.069 m Hence, the true difference in elevation nue = 0.924 + 0.678 =0.801 ru. P being higher m. 2 elevation o f Q = 126.386 - 0.801 ~ 125.585 (Q appears to be lower further hy 0.069 m due t<> !h'' · (c) When the level was at P. the apparent difference in clcvauon ~ !1.~1' n:. The difference in elevation = 0. 801 m Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net 232 SURVEYING Error in observation= 0.924 - 0.801 m = + 0.123 m This error consists of (1) error due to curvature and refraction (il) collimation error. Collimation error is said to be positive when the line of sight is so inclined that wit increases the staff reading at the farther point thereby making that point appear lower than what it is. Hence. the line of sight is inclined upward by an amount 0.054 m in Error due to curvature and refraction= + 0.069 m wa distance of 1010 metres.· Error due to collimation = 0.123-0.069 = + 0.054 m. wBut .a = 5\"3\"\"5\"x 6 011\" (upwards) ! Eas;i·; ' a =. 0.054 =0.0000535 tan 1010 tan 60· = o:0002909 y -(b) When the instrument is at P, the apparent difference in elevation between P If the staff readings are interchanged, then (a) True difference in R.L. between P and Q wiD: Jie 0.801 m (Q being higher) Eand Q = 0.924 m. Q being higher. nHence, Q appears to be higher by a funjler amounl of 0.924 - 0.801 = 0.123 m. R.L. of Q = 126.386 + 0.801 = 127.187 m. This error is due to (i) curvature and refraction, and (ii) faulty adjuslment of line , of collimation. Considering (1), the curvature and refraction tends to increase the staff reading at Q. thereby making Q appear lower than what it is by an amount 0.069 m (as already the point Q has been made 10 appear higher than I found out), but by actual observ'ations, it is clear that the line o f sight is inclined downwards what it is by an amount 0.123 m. Hence, ~ by an amount 0.123 + 0.069 = 0.192 m in a distance of 1010 m. I f a is the inclination of line of sight. we have tan a =~-~~ =0.000190 But tan 60\" = 0.0002909 a - 1920090x9 60 _ J9\" (downwards). Example 9.16. In levelling between two poi1Us A and B on opposile banks o f a river. I he level was sez up near .A. and the staff readings on A and B were I. 285 and 2.860 m respectively. The level was then moved and sel up near B and the respective readings 011 A and B were 0.860 and 2.220. Find the true difference of level between A and B. Solution. When the instrument is at A, AppeareD! difference in elevation between A and B = 2 . 8 6 0 - 1.285 = 1.575 m (A higher) Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net l.E'!I!LL!NG 233 When the instrument is Ill B, higher) Apparent diffemce in elevation between A and B 1 = 2.220 - 0.860 = 1.360 m (A True dz. fference m. elevao.on = 1.575 + 1.360 - 1.468 m (A higher) across a 2 river. Example 9.17. 1Wo poiflis A and B are I530 m apan wide The following reciprocal levels are taken with one level: Level at Readings on AB A 2.I65 3.8IO B 0.910 2.355 The error in the collimation adjustments o f the level is - 0.004 m in IOO m. Calculate the true dif!erelice o f level bezween A and B and the refraction. Solution. (I) True difference in level between A and B n where = (3.810- 2.165) + (2.355- 0.910) -1.545 m. - (il) Error 2 . . When gi j:c, =correction due to collimation= 0 due to curvature= 0.07849 d 2 metres= 0.07849 (1.53)2 = 0.184 m the level is at A, corrected staff reading on B = 3 . 8 1 0 - ( C , - C,) + c, n. . Corrected staff reading on B = 3.810- (0.184- C,) + 0.0612 = 3.6872 + C, e. . True difference in level between A and B = (3.6872 + C,- 2.165) = (1.5222 + C,) Cc = correction due to curvature = 0.184 m C, = correction due to refraction eBut it is equal to 1.545 m. r1.5222 + c, = 1.545 x 1530 = 0.0612 m ior C, = 1.545 - 1.5222 =0.0228 £!. 0.023 m. n9.14. PROFILE LEVElLING (LONGITUDINAL SECTIONING) gProfile levelling is the process of detemtining the elevations of pouits at short measured .intervals along a fixed line such as the centre line of a railway, highway. canal or sewer. nThe fixed line may be a single straight line or it may be composed of a succession of etlnngitadinal sectioning. straight liues or of a series of straight lines connected by curves. It is also known as By means of such sections the engineer is enabled 10 study the relationship between the existing ground surface and the levels of the proposed construction in the direction of its length. The profile is usually plotted on specially prepared profile paper. on which the vertical scale is much larger than the horizontal. and on this profile. various studies relating to the fixing of grades and !he estimating of costs are made. Field ProcedUFe : Profile levelling, like differential levelling. requires the establishment of turning points on which both back and foresights are taken. In addition. any number Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net SURVEYING 234 o f intermediate sights may be obtairied on points along the line from each set up o f the instrument (Fig. 9.42). In fact, points on the profile line are. merely intermediate stations. I t is generally best to set up the ·level to one side of the profile line to avoid too short wof the ground is regular or-' ~dually curving, levels are taken on points at equal distances sights on the points near the ilstrument. For each set up, intermediale sights should be lllk£n afte; the foresight on the next turning stalion hos been taken. The level is then wof slope occur, the points shoul<!. 'be· chosen nearer. For purose of checking and future set up in an advanced position and a backsight is taken on that turning point. The position o f the intermediate points on the profile are simultaneously located by chaini:J.g along the p~ofile and noting their distances from the point of commencemen1. When the vertical profile wField notes for profile levelling. are commonly kept in the standard form shown in apart and generally at int!:rvals of a .chain length. On irregular ground where abrupt changes .method as computations are easier by that method. The distances of the points on, the Eprofile are also recorded. The values shown in the table ·are same as those illustrated in reference, temporary bench marks should be established along the section. r-<-ll-$- ---~- -~- $-- asC\\i C\\i !L, l-- ----- --A ---·- ------ -\" ~ ~I yg~ ~j ~B M. En:10.45 the table on next page. The method is almost the same as given for 'collimation height' Fig. 9.42. ,., \" N iFC\\i C\\i :~g ~~ :~\\)~Iillll ~ ~ 3 ,..:~ N~ <'!_ \"~!---'= ___ :1---~ -~ :rp \" .. I 12 9 10 56 T.P.2 (a) Section 4-...... 13 T.P.3 T.P.t·------ ••• , \" (b) Plan FIG. 9.42. PROFILE LEVEWNG. Plotting the Profile (Fig. 9.43) The horiwntal distances are plotted along the horizontal axis to some convenienl scale and the distances are also marked. The elevations are plotted along the vertical axis. Each ground point is thus plotted by the two CO-<>rdinstes (i.e., horizontal distance and vertical elevation). The various points so obtained are joined by straight lines, as shown in Fig. 9.43, where the readings of the above table are plotted. ·Generally, the horizontal scale is adopted as I em = 10 m ( o r I \" = 100 ft ). The vertical scale is not kept the same but is exaggerated so that the inequalities· o f the ground Downloaded From : www.EasyEngineering.net