TACHEOMETRY 3.1 INTRODUCTION ontal distance and elevations of Tachoometry is ch of surveying in which the horiz distance and elev : SSL teeter Awesome the operation of chaining and levelling. thus avoids chaining completely. Tacheometric s ng ts a rapid method as compared to the conventional surveying metho The instruments employed in tachcometry are the engineer's transit and the levelling rod or stadia rod, the theodolite and the subtense bar, the self-reducing theodolite and the levelling rod, the distance wedge and the horizontal distance rod, and the reduction tacheometer and the horizontal distance rod. Advantages of Tacheometry By observing staff intercepts, horizontal and vertical distance can be calculated, Tacheometric surveying is economical for route surveying such as roads, railways, canals.cte . It avoids the tedious and labories chaining process for the measurement of distances. 4. Itis useful for steep terrain and hilly areas where chaining is difficult. 5. Iti useful for recconaicces surveying for quick acquiring data, 1 2 3, Disadavntages of Tacheometry a; 1. Rough terrains such as, river valleys, sce a ‘ANCYs, steep slope: or swamps where chaining would be difficult,” "°K®® BtOUnds, stretches of water basinspulty in setting the instrument at more points can be avoided “Mri ideally suited for the preparation of contour maps, location surveys of roads, rattways, irrigation canals, reservoir projects and for filling the details in topographic survey where -Pegree of accuracy needed is not more. 4. Wis used for checking the already measured distances. $. Though it is not a accurate method, it is a quick and rapid method, hence economical 3.2 INSTRUMENTS USED IN TACHEOMETRY ‘The instruments used in tacheometric survey are 1. Tacheometer and 2. Ordinary levelling staff or a stadia rod. Tacheometer : A tacheometer is similar to an transit theodolite fitted with a stadia diaphragm( ic. a telescope _ provided with stadia hairs in addition to the regular cross-hairs. The stadia diaphragm has two onal horizontal cross hairs called stadia hairs. one above and the other below the central hair. i istics of tacheometer telescope of a tacheometer is usually longer than that used in a theodolite. magnifying power of telescope is high to make staff graduations clearly visible even at ort distances an ordinary levelling staff can be used. For long distances, a special large astadia rod is commonly used. The stadia rod is generally in one piece, 3 to 5 metres130. As it is inconvenient to ne stadia between the upper and lower staq, the ray along aia ‘ ache : increases with increase of distan endl and staff intercept S is kept fixeq shown in Fig-3.4 also fixed. The staff intercept varies with vice-versa. On the other hand, if stadia hairs are tachometer angle Bchanges with staff position as Fig. 3.4 : Movable hairs with changing teacheometer angle B with fixed intercept S 3.5 NON - STADIA SYSTEMS ___ This method of surveying is primarily based on principles of trigonometry and thus without stadia diaphragm are used. This system comprises of two methods: (i) Tangential method and (ii) Subtense bar method.135 = Line of sight inclined and staff held vertical case surface of ground is undulating staff readings are not possible with line of si of sight horizontal. ” ettreedings are to be taken with line of sight inclined upwards or downwards a 1) Line of sight inclined upwards B.M. A I = Fig. 3.9 : Line of sight inclined upwards Ais the tacheometer station B is the staff station. The tacheometer is set up at A. The vertical circle is set to zero and the staff reading on the BM. is taken. Then the telescope is raised till the three hairs cut the portion of the staff and the telescope is clamped. Then the vertical angle and three hair readings are noted. Let, S= staff intercept, @=angle of inclination of the line of sight (central hair) from horizontal, D = horizontal distance between the instrument and staff Station, V = vertical distance from the instrument axis to the central hair. h, = staff reading on B.M., h= central hair reading on the staff at B. Horizontal distance,| D = KS cos” 6+ C.cos ® Vertical comoonen eC sind 2 Now, R.L. of staff station B = R.L. of instrument axis + V—h where, R.L. of instrument axis = R.L. of B.M. + h, (or) RL. of instrument axis = R.L. of A + Height of instrument (if R.L. of 4 is known)Surveying . 140 3.11 WORKED EXAMPLES Se a an, ree Cross 1, A tacheometer has a dlapordt ie one and the distance from the object glas, f the object glass = eee ccseien ia is 120mm.Calculate tacheometric constan' Solution: Focal length,/ = 230mm, Distance from objective to the vertical axis, d= 120mm, Stadia interval (= 2x1,15 = 2.3mm f _ 230 an ae (f + d)= 230-4120 = 350mm Multiplying constant = 100 & Additive constant = 0.35m_ ‘The stadia readings with horizontal sight on a vertical staff held 50 m away from a tacheometer were 1.280 m and 1.880 m. The focal length of object glass was 249 mm. The distance between object glass & trunnion axis of the tacheometer was 145 mm, Calculate the stadia interval. Solution: Focal length, /= 240mm = 0.24m Distance from objective to the vertical axis, d= 145mm = 0.145m Horizontal distance, D = 50m, Staff intercept, S = 1.880 ~ 1.280 = 0,60m (fF +d)=0.24+0.145 =0.385m Using formula, d=(f pra) s0-(°2* oo +0585 144 “> = 50-0385 = 49.615 = 100 ? is 0.144 » 49.615 ~ 290%10" m= 2. 90m141 Seadia interval, (= 2.%0mm 50 and 80 metres were 4, Two distances accurately measured out and the staff Intercepts on the staff were 0.496m and 0.796m respectively calculate the tachometric constants. Solution: Horizontal Distance, pDe=KkS+C $0=K*0.496+C —(1) $0=K*0.796+C ---(2) Subtracting equation (1) from (2), {80 -50)= (0.796 - 0.496) =0.3K k=22-100 03 Substituting the value of K = 100 in equation (1), 50 =100x0.496+C 50=49.6+C C=50-49.6=0.4m 4. The following set of measurements taken from tacheometer. Calculate the tacheometric constants. Distance [Staff intercept (S) Solution: Considering first set of observations, D=KS+C 30=Kx0.295+C (1) 60=K*x0.595+C ---- (2) Subtracting equation (1) from (2), (60-30)= (0.595 - 0.295)K =0.3K 30 k-—- 03 oe Substituting the value of K = 100 in equation (1), 30=100x0.295+C 30=29.5+C C=30-29.5=0.5mobservations, emnaering cond wt of e pease 7 a 1ed = K < AONS + @ as + eek it - (1) from ( Saberacting equ 2 Q20- 100)- 1195 0.995) # 0.2K ‘Substituting the value of K = 100 in equation (1), 100= 100x0.995 + C 100=99.5+¢ C=100-99.5=0.5m ictiiatas ier The following observations lier of a tacheometer. The 2 ie taken Bo can baal at distances measured from the instrument ; were Stadia Readings " 60 120 180 Find the mean value of the constant given that the additive constant was 0.25M. Solution: Staft intercepts, 5,= 1.425 - 0.835 = 0.590 5,=2345 — 1.140 = 1.205 5,= 1.250 - 2.990 = 1.740 D,=K,S,+C 0.835, 1.425 60=K,x0.59+0.25 059K, = 60-0.25 = 59.75 59.75 K,=——— =101, Serre pe P=K,S, 0087 8+ Ccosg 120= K, x1.205xcos? 1°1540.25 051915i ee an 45=0.449K, +0.399 0.449K, = 45 ~0.399 = 44.601 44.601 _ 99 33 1 b) Observation-2 D, = K,S, cos’ 0+ Ccos0 90 = K, x0.90xcos? 2°15'+40.40x.c0s 2°15" 90 = 0.898K,, +0.399 0.898K, = 90-0.399 = 89.601 _ 89.601 0.898 c) Observation-3 = 99.78 K, D, =K,S, cos” 0+ Ccos0 135 = K, x1.35xcos? 0°40'+0.40xcos 0°40" 135 =1.349K, +0.399 1.349K, =135-0.399 = 134.601 99.33+99.78+99.78 I phe oe 7 99.63Surveying - Ci Ee CCS Solution: h,= Hat A= 1.350m Axial hair reading, A= 1.105m_ Vertical angle, @= +4°35', K = 100 & C= 0.00 Staff intercept, = 1.550 — 0.650 = 0.90m Distance between 4 and B, D=KScos* 0+ Ccos® D=100X0.90xcos? 4°35'+0 =89.425m Vertical Component, _100x0.90xsin(2x4°35') = z RL.of B=RL. of A+ h, +¥ —h=100.000+1.350 + 7.168 1.105 =107.413m 9. Determine the R.L. of a point P and its horizontal distance from the theodolite station from the following tacheometric observation made on a staff held vertically over P. v Csin® ae +0=7.168m 4 [ Vertical angle | Axial hair reading | Stadia hair readings | pitew | el ee] R.L. of instrument station = +37.120. Height of instrument axis = 1.450m, stadia constants are 100 and 0. Solution: Axial hair reading, h = 0.845m Vertical angle, @= +12°40'20" H.1=1.450m K=100&C=0 Staff intercept, S= 1.265 ~ 0.420 = 0.845m Horizontal distance, D = KS cos” 8+ Ccos® \ = 100x 0.84: 2190, ‘esti distance rom the instrument ais othe cental ha 124020+0= 20.43m _ KSsin2e 2 v +Coa@= 100% 0.845% sin(2 12%, 2 +0=18.00m RL of Instrument axis = RL of Instrument ‘stahon, RL of P=RL of Instrument axis + Vp *HI=31-120-5 1 =38.570+18,09._ 450 38.570m 0.845 = 55.815a eeerne ane at Take K= 100 and C= 0.40. 11. A Tachoometer was set a vertically held staff ee ed trees fae [are [sass] CT 361 1.650,2.515,3.380 ] RL. of BM. = 437.655 m. Calculate the distance between 4 & B and R.L. of B. Solution: 8, =2°18', 6, =8°36', h, = 3.550m, h,= 2.515m. Staff intercept, S, = 3.875 — 3.225 = 0.65m S,= 3.380 - 1,650 = 1.73m When instrument is at 4 and staff at BLM. i a 01 gr aD. cring = WMO ATE 9 Aaah RL. of Instrument axis at A =R.L. of BM.+h, + V, =437.655+3.55+ 2.62 =443.82m B.M. R.L = 437.655m Fig. 3.16tacheomelty 149 When instrument is at A and staff at B y= B282: .cxino, =! Tex axs36) E 5 +0.4sin 8°36'= 25.64m RL. of B= RL of inst. at A+ V, ~h, = 443,824 25.64 - 2.515 = 466.945m Distance between A & B D=KS, cos’ 8, + Ccos®, D=100X1.73x cos” 8°36'+0.4cos8°36'= 169.526m A Tacheometer was set up at a station P and the following readings were obtained on a vertically held staff. Inst. Staff Vertical | Hair Reading Remarks station | Point_| angle % BM. _|-4°22' | 1.050,1.103,1.156 | RL of BM = 1958.3 +10° 0" | 0.952,1.055,1.158 ‘The constants of the instrument were 100 and 0.1. Find the horizontal distance from P to Q and the reduced level of Q. Solution: 8, =4°22', @, =10°0', h,= 1.103m, h,= 1.055m. Staff intercept, S,= 1.156 — 1.050 = 0.106m S,= 1.158 — 0.952 = 0.206m RL = 1960.213mSinveyny. 4 * ve wt Pane walt at OM Whee gar Bo cane 100%0.106 8102 4°22") 4 0) poy A270 Hi ue ois 7 2 RL of Instrument axis at P= RL, of BM. 6h, 44, IIHT NO 6 O81 fey) 5 When instrument is at P and staff at 0 ¥,= KS, sin 20; +Csin 8, = ‘eone a ela #344n a 2 RL.of Q=RL.of inst at P+ V, hy =1960.219-+ 9,541,055 = 1961 44m Distance between P & Q D=KS, cos’, +C.c0s0, D=100x0.206% cos’ 10° +0.1c0810" = 20.077m 13. A tacheometer was set up at a station C and the following readin, Us Were wbrtwined om a vertically cae the distance CD and RL. of D, when constants Of the R.L. of B.M. : 750.000, Solution: 8,=5°20, 8, =8712', 4,= 1.800m, h,= 1.500m, Staff intercept, 5, = 2.450 ~ 1.150 = 1.30m 5,=2.250- 0.750 = 1.500m BM. RL = 750.000m Fig. 3.15a —<$£ @io@j$oo oq“ ‘wren instrument is at Cand staff at BM. 2 o 3™ xs” EE « cin, = 10081 SxHi0(25°20) "3 2 + 0.1 Sein $°20'= 12,045m RL of Instrument axis at C= RL of BM +h, +F, = 750.504 1.812.045 = 764.345 ‘When instrument is at C and staff at D KS,8in 20: | cing, = L0OX1.Sxsin(2x8"12') > sin@, = * +0,1Ssin 8°12 = 21, 196m RL.of D=R_L of inst. at C +V, — hy = 764.345 + 21.196 = 1,5 = 784.04 1m Distance between C & D D=KS, cos* @, +Ccos®, D=100X1.5xcos 8°12'+0.15.cos8°12'= 147.09m 14. The following notes refer to a line levelled tacheometrically with an anallatic tacheometer, the multiplying constant being 100. Inst. Vertical station | station | of axis | angle P B.M. 15 ~6" 12" 0,963,1.515,2.067 Hair Reading P Q 15 | +7°5' | 0.819,1.341,1.863 R 1.6 +12" 23" 1,860,2.445,3.030 RLL. of B.M. 460.50 m, Staff being held vertically. Compute the reduced levels of P, Q and R and the horizontal distances PQ and QR. Solution: @, =6%2', 0, =7°S', 0, =12°23', h,= 1.515m, h,= 1.341m, h,= 2.445m, H.L at P= 1.5m, Hat Q= 1.6m. Staff intercept, S,= 2.067 — 0.963 = 1.104m -863 — 0.819 = 1.044m 030 — 1.860 = 1.17m Instrument is at P and staff at B.M. KS,sin20, 104sin 2(6°12" a Gage ee tO sin 206 12) POAGiNA(6 12) | Gc me 2 _ RL of instrument axis = RL. of B.M. + h, +V, = 460.50 +1.515 + 11.85 = 473.865m of P=R_L. of instrument axis — H.1 at P = 473.865 —1.5 = 472.365mM. AL = 460,60m Fig, 3.19(a) Instrument is at P and staff at Q eet r= Hanna Cin = TS) 12 77m RL, of Om RL, of inst, axis at P+ V, ~h, = 473.865 + 12.77 -1.341 = 485.294m Instrument is at Q and staff at R Q ALL = 488,294m aaD, =KS, cos" 8, +Ceast, D, =100x4.95xcos* 32450 = 493 25S ‘When instrument is at B and staff is held at CFR t~<“—s~SCs When instrament is at P and staff ts held at 4 Silas a Fig. 3.22 (6) . ‘ 346" y= KS, 1 +Csin®, = Henne Sexes 36) 9 =9.20m RL.of P=RL.of A+h, —Vy — Hd = 319,076 + 1,95 - 9.2 = 1,87 = 310.256m 18. To determine the elevation of the first station A of a Tacheometric Survey, the following observations were made, the staff being held vertically. The instrument was fitted with an anallatic lens and the value of the constant was 100, Staff Readings ie a "| 1.332, 1.896, 2.460 +8°20" [0.780, 1.263, 1.746 1.746 6°24" | 1.158,1.617,2.076 R.L. of BM. = 158.205m. Calculate RL, of A, 8, =5°40', 0, =8°20', 6, = 6°24 A= 1.896m, h,= 1.263m, 4 HL at A= 1.380m. Z Staff intercept, S\= 2.460 — 1.332 = 1.128m S,= 1.746 — 0.780 = 0.966m 5,= 2.076 ~ 1.158 =0.918m = 1.617m, H.LatO= 1.440m,‘When staff is held at P.M. and instrument at 0 B.M. RL = 158.205m Fig, 3.23 (a) KS, sin 20, Tae «4 1, - i + Csind, = = 100%1-128xsin(25°40) . 9 14 og 2 RL. of instrument axis at O=R.L. of BM. +h, +V, =158.205 +1.896+11.08=171 181m When instrument is at O and staff at C.P. j 6 eet stoma 20°) +0 7, = Ss8072 4 csind, = =13.85m R.L. of C.P. = R.L. of instrument axis at O+V, —h, =171.181+13.85—1.263 =183.768m When instrument is at A and staff is held at C.P.pone + Cein @, = LOOX! 64 sin 2(5°67) 2 - +0~ 14.52 A B Fig. 3.24 (b) From fig. 2.30(b), ZAPB = 224° 134° =90° +. AB? = PA’ + PB? AB = 107.29? + 162.7? =194.89m Assuming R.L. of instrument axis = 100.000 RL. of A= RL. of instrument axis +V, —h, =100+19.94 ~1.915 =118.025m RL. of B=RL of instrument axis + V/, —h, = 100 +14.52-1.885 =112.635m Difference level between A & B=118.025-112.635 =5.39m Distance _ 194.89 Difference 5.39 Gradient of line AB = 1 in 36 (falling) 20. A tacheometer is used to obtain the difference of levels between two points P and Q. The instrument is set up at another point. R and the following observations were taken: Take K = 50 and C = 0.50 Gradient = =iois =D 3.500, 2.815, 2.130 1.870, 0.990, 0.110Surveying = ind Q and also determine AO Determine the horizontal distance, betwe R.L of P is 100.00 m. Solution: 0, = 6°30", 0, = 8°30", h, = 2.815m, h, = 0.990m. Staff intercept, S, = 3.500 ~ 2.130 = 1.370m S,= 1,870 ~ 0.110 = 1.760m P R.L = 100,00 pA Distance, RP = KS, cos? 0, +Ccos @, = 50x1.37 cos? 6°30! + 0.5cos 6°30'= 68.12m Distance, RO= KS, cos”@, +Ccos0, ~ 50x1.76cos*8°30' +0.5c0s8°30'= 86.57m Distance between, PQ = RP + RQ = 68.12+86,56 = 154.68m i 02m, y= AS15224. csing, SOW) sss 7 16m +0.5sin 8°30'= 12.93m in 28° iw Sa 20. +Csin@, = 50x1 Tos 48 30°) RLof O=RL.of P+h +¥,-V,—h, =100+2.815+7.76-12.93_ 9.99 - 96 655m Difference in level between P&0=100-96.655 =3.345m