‘To{ind the distance AB by the swing ‘offset method, the head chainman tached the end of the tape to one end othe ne a5 al 8 and describes an arc ‘with center 8 and radius 100m. The ‘ear choinman stationed at lines nthe tend cl he tapa wth some distant object 250 and directs the setting of ins at poinsa and b where the endo he tape ‘tossed ine AO. A point C midway between a b les on the perpendicular 8B, Apinis setatC, andthe distances ‘BC and CA are measured fo obtain the neces aaa cpu he ng of AB, So4 ‘TAPE CORRECTION ‘Ang was determina fo bs 2885.25 m. when measured wit a 30m. sel tape suppored Sowa trot pial digste mean temperature of 35°C, Tape used ie of andar eng a 20 unde pal of 8 Fy: Groce-secSona area of ape Is 0.93 qian, Coefficient of thermal. expansion Is 0 0006118, Modus of easly of lpe i 2108 kgfom © Oetermine the err of the tape due to change n temperate, ® Cetormine the erat due to tension, ® Oeternine te comectod length ofthe ne. Solution: © Temp. corecton: GemT-T) = .0000116 (2395 2535-29 Gyeeaatsim © Tenslon conection: celal ae oo Gam Gp#-00089m D Comncted length: +L = 205 25 +0.4168 - 0.0389 b= 23986269, 25 ‘A 501 tape wa standardized ad was found (0 be 0.0082 m, 100 tong than the standard feng aan observed temperature of 56°C and 4 pull of 15 bls. The same tape waa used fo ‘measure a certain distance and wes recorded (o be 679.82 m. lang a an abserved temp. of {68°C ond a pull cf 15 kilos. Cooffclent of Feat expansion is 00000116 m7. © Detar sce onprt ® Doterrine te otal crecton. © Determine the ie length a the ine Solution: ‘Standard temperate: CreK (Ty Ty) Ly + 0.0042 = 0.000016 (58 -T;}50) + 000082 = 0.03364 0.00058 7} 7 = 5078 (snd temo) Total eerecton’ CreKIT-T)Ly Cy =0,000016 (68 - 50.76)50) Cr=00 (ape too ng) e s relearn = TOON G* CH, Toweorecon=opsiim 7 Ve The Mery © Tslengtn of ine (Carected hor. stance = 673.92 + 0.1348 Comected hor. stance = 674 085 m. ‘A-50 m, sted tape was stindardtzed and suppported throughout is whole length and foun io be 0.00206 m. nger at an served temperate of 318°C and a pul of 10 hos. This tape was used fo measure a Ene which ‘was found (0 be 662.702 m. al en average {temperature of 246°C using ihe same pul Use coefficient of enpanson of 00000116 m: er degree centigrade, (© Comput th standard temp, Compute the tl ep. coeeton. Conte corel Engh ah he Solution: ‘Standard lerperate: Cr=K(Tps TIL 0.00206 = 00000116(38- 7,X80) 31a © Toa correction: Cy=K(Ty- TIL 67 =00000116(246- 7827450) ¢7=000218 (ton sho) Teal caecons 02212 Teta cretion = 0.02823 m, Coected length of ine: Corected hottortal distance = 662702-002821 = 68207377 ABAD) 0.01 G13!TAPE CORRECTION S-5 ‘@ Whatls he true horizontal distance? Solution: © Actual length: Cr=K(T2- Tp ly Cr = 0.00001 16(5 - 20(30.005) Cr =- 0.00522 mm p= Filta AE Total correction = - 0.013 m ‘Actual length of tape during measurement 730.005-0013 '=29.992m. @ Total error: Too short by = 30 - 29.992 Too short by = 0.008 m Tel ors 7222008) Total ear = 0.126 m. (to be subtracted} ® True hortzontal distance: Horizontal Distance Comected incined distance = 47290- 0.126 2472714 oo 80°27 n= 472.774 (003) h=14183m (4.1837 Correcton fr sope * 3 era ray Correction far stop = 0.213m_ Comected horizontal distance = 472714 -0.213 = 472.561 m. ® What is the errorin area in sq.m.S-6 TAPE CORRECTION Solution: Solution: © Actual length: © Cross-sectional area: Cr=K(T2-T3) Ly A100) 1004786) _ 5 67 Net Cr = 0,000016(30 - 20}(30.002) 1000 : j Lo Cr=0.003481m (too tong) A=Q0M sam Wirg,? Ves) Actual length of tape during measurement a = 30.002 + 0.00348 ® Total correction: = 3000548 m. CrEK(Tg-TyL ® The area: Cr=7x 107 (20- sig Therefore the tape is 0.00548 m foo long Cr=0,00035 m. Forthe 144.95 m side: Total eror = 4 SSO 925, Pull creation: {P2-Pi)k True length = 144.95 + 0.026 Cp= TE True length = 444.976 m coe 10400 For the 113 m side: 0.034 (2) 10° _113{0.00548) | Cp=0.009m Raters et Comrection = 0.00085 + 0.000 Tae length = 113 +0.021 Cometon= 02085 m True fongth = 113.021 m True area = (144 976)113.021) No.of ape fengths = 208 = 4.396 \\ Tre area = 16,385.33 mi? Total comection = 43080 (0.00035) © Enrinaea: Total correction = 0.0403 m Eroneous area = (144.95(113) Erroneous area = 16,379.35 m2 Error in area = 16,385.33 - 16,379.35 Error in area = 6982 m? @ True length of baseline: True length of basefne = 430.60 + 0.0403 True length of Baseline = 430.6403 m ‘Abaseline was measured using a 100 m. tape’ which Is standardized at 15°C with a standard pull of 10:kg. | The recorded distance was ‘found out to be 430.60 meters. Al the ime af ‘measurement the temperature was 20°C: and. the pull exerted was 16 kg. The weight of one -cubic cm of steel is 7.86 gr. weight of tape Is 2.87 kg. E = 2x 108 kglom?, K= 7x 107 we. z ® eae a ®. Cobb taed arcen ‘D. Compute the true length of the basa line. ‘A’30 m. steel fape weighing’ 1.45 kg is of ‘standard length under a pull of 5 Kg supported for full tength.. The tape.was used In measuring a fine 938.55 m, long on a smooth tavel ground under a steady pat of 10 kg. a Assuming E = 2 x 10° kgfom’ nd the unit ‘weigh! of steel to ba 7.9 x 10° kgfom’, ©. Determine the cross-sectional area of the tape in cn’, @ Determine the correction for increase in tension, : @® Determine the correct.length ‘of the line5-7 TAPE CORRECTION Solution: © Cross-sectiona! area of tape: WEALYs 1.45 = A (9000)7.9) x 10 Ackil engineer used 1 30 m tape in measuring A= 0.061 cm? an Inclined distance, The measured length on ® Pull Correction: the siope was recorded to be 459.20 m long. (P-P)L- ‘The difference in elevaion between the Initial “ae point and the end point was found to be 125m. The 30m tape is of standard length at cp, ‘a temperature of 10°C and a pull of 50H. ‘P 0.061 (2x 10°) During measurement the temperature reading Cp=+0.00123 ooorza Was 1B ond We ape wna sipped ic = 9200123 (938 55) ‘ends with an applied pul of 75N. The cross-. Tetal Conecton = "30 sectoral area of he tape ls 650 mn? andthe Total Correction = + 0.038 m. modulus of elasticity is 200 GPa. The tape ® Correct length of line: Covrected length = 93855 + 0.038 Corrected length = 938,588 m. A sleel tape with @ coefficient of linear expansion of 0.00001 16 per degree centigrate Is knawn fo be 0 m. lang at 20°C, The tape was used fo measure a line which was found tobe $32.28 maters lori when the temperature ‘was 35°C. Determine the following: © Temperature comection per fape teng’h, ® Temperature correction for the measured (line. @ Corrected length of the ine. Solution: © Temperature correction per tape length: CreK(T- TL C= 0,0000116 (35 - 20,50) r= 0.0087 m. too long Temperature correction for the measured tne: Te > oat a S20 .00087 Total correction = + 0.0926 m. a Comected length = 5323726. 'm. ‘has a mass of 0.075 kgim. K = 0.000116" me. oO Pelee i a ctmem pebe @ deme th caret bsp. ® Determine the hocznntal distance. Solution: © Total correction per tape length: Cr=K(T2-Ty) by r= 0.000016 (15- 10)(30) Cr=+0.00174m Pull correction: Pa Paddy Op aE 5 "550 00% 10) p= +0,00058 m Sag Correction: c= He S* 24 c (0075x981? 0 “s 2475) Cs= 0.10827 m Total correction per tape length: 00174 + 0.00058 - 0.10827 0.10595 mS-8 TAPE CORRECTION ® Correction for slope: ® Tolal correction: Tat eorcton = ABAD oat 7) Te as! Total correction = + 1.622 Cr = 0.000116 (624.9932 - 20) Garect slope alstance = 459.20 1.622 Cy=+0.087 (too lag Carect slope alstance » 457.578 m i Cs*25 252 Os* 31457578) Cs= 0002 © Horizontal distance: Corected ‘A steel tape is 100 m. long at a temp. of 20°C gota pal ‘twas used to measure a distance of 624.95 m: at a temp. of 32°C with ‘an applied pull of 15 kg. during measurement with the taps supported at .both ends. Coefficient of thermal expansion Is 00000116 m’c and a modulus of elasticity of 2 x 108 kg/cm, Welght of tape ts 0.04 kg/m, anda «ross sectional area of 0.08 em? (© Compute the sag correction, ® Cares oe een: and temperature. ® Compute be comected engh ofthe fine by applying the combined corrections for Jension, sag and temperature. Solution: : @ ‘Seg compcfont ‘ = ch 1 = me % 3 . cx sm, (0.04)4 (24.95)3 _ C= age Total sag corection = 6(0.296) + 0.005 Total sag corection = 1.781 (too short}... ‘A Gil engineer used a 101m tape which is of standard length at 32°C in nieasuring a certain dlistanca’and found out that the fength of tape have different lengths at different tensions were applied as shown. K =<0,0000116 mic" Length ottape _Ter.sion Applied 99,988 m 10ig 90.962 m 4g 100.003m £20k ‘®. What tension must be eoplied fa the tape at a temp. of 32°C sp Frat it would be of Standard fength? @ What tension must be applied to the tape ‘at a temp, of 40.6°C 90! hat M would be of - ‘standard length? @ What tension mus! be g5plied to the tape ata temp. of 30°C so taal it would be of standard length? :TAPE CORRECTION Solution: @ Tension applied at 32". “ wt |, om aut x _ 0.008 6 oo x=4.36 kg, Tension applied = 18.36 kg . @ Tension applied at 40.6" ‘Temperature corection: Cr=K(T2-Ty)L Cr = 0.000116 (40.8 - 32) 100 C= 0.00998 Actual length of tape at40.6°C Tension 99.986 + 0.00998 = 99.99598 ‘99.992 + 0.00998 = 100.00198 100.003: + 0.00998 = 10001298 wmf anton UI, 100,00198 x _ 0.00402 4° 0.006 4= 2.68 kg Tension to be applied = 10+ 2.68 Tension to be applied = 12.68 kg ® Tension applied at 30°C: Temperature comection: =K(T2- ML 0000116 (30 - 32) 100 Cy=-0.00232 10kg 14g 201g 14) Actual length of tape at40.6°C Tension 99.986 - 0.00232 = 9998368 10kg 99.992 = 0.00232 = 99.98968 14kg 100.003 - 0.00232 = 100.00068 20kg jt ~(} x 0.01032 5 “O01T ¥=§63kg .+ Tension applied = 14 + 5.63 Tension appiied = 19.63 kg A.50 m. tape of standard length has a weight of 0.05 kg/m, with a cross sectional area of 0.04 sq.em_ It has a modulus of elasticity of 2.10 x 108 kg/em?, The tape is of standard length under a pull of 5.5 kg when supported throughout is length and a temp of 20°C. This tape was used to measure a distance between A and B and was recorded to be 458.65 m. long. At the lime of measurement the pull applied was 8 kg. with the tape supported only atts end points and the temperature observed ‘was 16°C, Assuming coefficient of linear ‘expansion of the tape is 0.0000116 mC. ‘© Compute the correction due to tha applied pull of 8 kg. @ Compute the comection due to weight of tape. ® Compute the true length of the measured fine AB due to the combined effects of tension, sag and temperature,(8 -594458.550 = oon 2.10 x108) 24m @ Number of paces for the new line: 893 = 893.5 + 891 +8955 No. of paces= Na of paces = 693.25 Distance of the new tine: Distance of new fine = 893.25 (0.691) Distance of new line = 617.236 m. 1s = + 1.832m. (always negative) ® Tre length of measured line AB: Gr=K(Tz-T)L Cr= 0.00001 16 (18 - 20/458.65) Cr=-0011m. Tolal correction = | baal 1832-0014 Total correction = - 1.6291m. True length AB= 458 65- 1.829 True length AB = 456.821 m. A line 100 im. long was paced by a surveyor’ four times'with the following data. 142, 145, 145.5 and 146,’ Then another line was paced for four limes again with the following results, 893,893.5, 891 and 805.5. * © Oetermine the pace factor, @® Determine number of paces for the new line. ®, Determine the distance of the new fine. Solution: ® Pace factor: No. of paces 3 Mb Ss 185 8 No. of paces = ‘Wass Pace factor =O = 0694 : m= This sides of a square lot having an area of 2.25 hectares were measured using a 100 m. tape that was 0.04 m. too short. Compute the error inthe area in sq.m. ‘The correct distance between two point is. 220.45 m._Using a 100 m. fape that is “x m. too long, the length to be laid on the ground should be 220.406 m. What Is the value of "x"? The distance from D tb E, as measured, is 165.2 m. If the 50 m. tape used Is 0.01 m. too short, what Is the correct distance in Solution: Aa22aDbeCtaES yi! ao Eorin area=225-22482 = ui, Enor in area = 0.0018 hectares Erorin area = 0,0018 x 10000.” Emo in area = 18 sq.m. Note: 1 hectare = 1000 sqm. Corrected distance: Corect dtance = 1652 922901) Conect nce = 168167