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MEASUREMENT OF HORIZONTAL DISTANCES
I. Taped Distance
SP.1 A steel tape with a coefficient of linear expansion of 0.0000116/ oC is known to be 50m long at 20 oC. The tape was used to measure a line which was found to be 532.28 meters long when the temperature was 35 oC. Determine the following: a) Temperature correction per tape length. b) Temperature correction for the measured line. c) Correct length of the line.
Solution:
a) Ct = KL (T-T (^) s ) = 0.0000116(50)(35 o-20 o) = + 0.0087m (correction per tape length due to temperature. The positive sign indicates that tape is too long) b) Ctโ = KL (T-T (^) s ) = 0.0000116(532.28)(35 o-20 o) = + 0.0926m (correction for the measured line due to temperature) Solution Check (Using ratio and proportion): ๐ถ๐ถ๐ก๐กโฒ 532.28 =^
50 ;^ ๐ถ๐ถ๐ก๐กโฒ^ =^
50 = +0.0926๐๐^ (๐๐โ๐๐๐๐๐๐๐๐)
c) Lโ = L ยฑ Ctโ = 532.28 + 0.0926 = 532.37m (correct length of measured line. The correction is added since the tape is too long.
SP.2 A steel tape known to be a standard length at 20 oC, is used in laying out a runaway 2,500 m long. If its coefficient of linear expansion is 0.0000116/oC, determine the temperature correction and the correct length to be laid out when the temperature is 42 oC.
Solution:
Ct = KL (T-T (^) s ) = 0.0000116(2500.00)(42 0 -20 0 ) = + 0.638m (Correction for the length to be laid out due to temperature. The positive sign indicates that tape is too long)
Lโ = L ยฑ Ct = 2500.00 โ 0.638 = 2499.36m (Correct length to be laid out. The correction is subtracted since the tape is too long)
SP 3. A heavy 50m tape having a cross-sectional area of 0.05sq.cm. has been standardized at a tension of 5.5 kg. If E = 2.1x10 6 kg/cm 2.^ Determine the elongation of the tape if a pull of 12kg is applied.
Solution:
SP.4 A 30m steel tape weighing 1.45 kg is of standard length under a pull of 5kg, supported for full length. The tape was used in measuring a line 938.55m long on smooth level ground under a steady pull of 10kg. Assuming E = 2.1x10 6 kg/cm 2 and the unit weight of steel to be 7.9x10 -3^ kg/cm 2 .Determine the following:
a) cross-sectional area of the tape b) Correction for increase in tension c) Correct length of the line measured.
Solution:
a) ๐ด๐ด = (^) ๐ฟ๐ฟ (๐๐๐๐๐๐๐ก๐ก๐๐ ๐๐๐๐๐๐๐๐โ๐ก๐ก) = 1 .45๐๐๐๐ 30๐๐๏ฟฝ 100๐๐๐๐ ๐๐ ๏ฟฝ๏ฟฝ 7 .9๐ฅ๐ฅ10๐๐๐๐โ3 3 ๐๐๐๐๏ฟฝ
= 0.06 ๐๐๐๐^2
b) ๐ถ๐ถ๐๐ = (๐๐๐๐๐ด๐ด๐ด๐ดโ๐๐ ๐ ๐ )๐ฟ๐ฟ= (^0). 06 (10โ5( 2 .0๐ฅ๐ฅ10)^306 ) = +0.00125๐๐ (Correction per tape length. The positive sign indicates that tape is too long)
๐ถ๐ถ๐๐ 938.55๐๐
c) Lโ = L ยฑ Cp =938.55 + 0.04 = 938.59m
SP5. A 50m steel tape weighing 0.035kg/m is constantly supported at mid โ length and at its end points, and is used to measure a line AB with a steady pull of 6.5kg. If the measured length of AB is 1268.256 m, Determine the following. a) Correction due to sag between supports and for the whole tape length. b) Total sag correction for the whole length measured. c) Correct length of line AB.
2 BA 53
3 AB 51
4 BA 53
5 AB 52
6 BA 53
Solution:
a) Determining Pace Factor L=45m length of line AB n 1 = 6 number of trials taken on line AB Sum 1 = (50+53+51+53+52+53) = 312 paces M 1 = Sum 1 / n 1 = 312/6 = 52 paces (mean number of paces to walk line AB PF = L/M 1 = 45m/52 paces = 0.865m/pace ( Pace factor of surveyor)
b) Determining Unknown Distance
n 2 = 6 number of trials taken on line CD Sum 2 = (771+770+768+770+772+769) = 4620 paces M 2 = Sum 2 / n 2 = 4620/6 = 770 paces mean number of paces to walk line CD PD = M 2 (PF) = 770 paces (0.865 m/pace) = 666.1m
C) Determining Relative Precision TD = 667.0 m taped distance PD = 666.1 m paced distance RP = (TD โ PD)/TD = (667.0 โ 666.1)/667 = 0.9/667. = 1/741 say 1/700 (relative precision of the measurement)
III. Distance by Tachymetry
SP1. A stadia rod held at a distant point B is sighted by an instrument set up A. The upper and lower stadia hair readings were observed as 1.3m and 0.9m, respectively. If the stadia interval factor (k) is 100, and the instrument constant (C) is zero, determine the length of line AB.
Soluton:
D = Ks+ C K = 100 S = 1.3-0.9 = 0. C = 0 D = 100(0.4) + 0 = 40m
SP2. A transit is set at point โPโ with elevation = 96m, H.I. = 1.2m, f/I = 100, (f+c) = 0.3. Find the elevation of points 1 and 2 and the distance to the points observed as per data below.
Points Observed Rod Intercept True Azimuth Vertical Angle
1 1.056 352.2 degree + 2 degree
on 1.2m
2 1.260 56.2667 degre - 4 degree on 1.65m
