4 On a cold day two surveyors measure a one mile distance 52

4.) On a cold day, two surveyors measure a one mile distance (5,280.00 feet). The chain they are using is known to be 99.97 feet. The elevation of Point A is 100.00 feet, the elevation of Point B 265.00 feet. Assume that it is a straight slope from point A to point B. At the time of the survey, the chain is recorded at 26°. What is the corrected distance from point A to point B? (30 points) What is the Zenith angle from A to B? (10 points)

Solution

There are 3 corrections to be applied in this problem to get the final length between the points A and B:

1) Correction for standard length

2) Correction for Standard Temperature

3) Correction for slope

1) Correction for standard length of chain

Corrected length due to incorrect tape length, C_L = l\' * (C/L) ;

where l\' = measured length = 5280 ft

C = Correction for chain length = (100-99.97)

L = standard length of chain, here it is 100\'

i.e., C_L = 5280 * ((100-99.97)/100) = 1.584 ft

(Here correction is negative as chain is too short)

2) Correction for temperature

Correction for temperature, C_T = LC(T_m - T_o)

where L = Length of tape or length of line measured.

C = coefficient of thermal expansion of tape = o.oooo116 /^oC

T_m = Temperature at the time of measurement = 26 ^oC

T_s = Standard temperature = 20 ^oC

C_T = 5280 * 0.0000116 * (26 -20) = 0.3674 ft

As T_m > T_o, C_T is positive.

3) Correction for Slope

C_SL = h²/(2L)

where L = measured length = 5280 ft

h = difference in elevation between points = 265 - 100 = 165 ft

i.e., C_SL = 100²/(2*5280) = 0.9469 ft

Slope correction is always negative.

Applying all the corrections,

Correct length of line = 5280 - 1.584 + 0.3624 - 0.9469 = 5277.8315 ft

To find zenith angle

Let, zenith angle be Z.

Slope correction, C_SL = L(1-cosZ)

i.e., 0.9469 = 5280(1-cosZ)

i.e., Z = 1^o4\'59\"