An elevation must be established on a benchmark on an island that is XXXXXXXXXXft from the nearest benchmark on the lake's shore. The surveyor decides to use a total station that has a stated distance...

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An elevation must be established on a benchmark on an island that is 2536.98 ft from the nearest benchmark on the lake's shore. The surveyor decides to use a total station that has a stated distance measuring accuracy of ±(3mm + 3ppm), and a vertical compensator accurate to within +0.4". The height of instrument was 5.37 ft with an estimated error of +0.05 ft. The prism height was 6.00 ft with an estimated error of +0.02 ft. The single zenith angle is read as 87°05'32". The estimated errors in instrument and target centering are +0.003 ft. If the elevation of the occupied benchmark is 632.27 ft, what is the corrected benchmark elevation on the island? (Assume that the instrument does not correct for earth curvature and refraction.)


Answered Same DayDec 22, 2021

Answer To: An elevation must be established on a benchmark on an island that is XXXXXXXXXXft from the nearest...

Robert answered on Dec 22 2021
120 Votes
Solution
Corrected benchmark elevation difference can be calculated as
Where Δh represents correc
ted benchmark elevation difference
Now from the question we have
z = zenith angle = 87005’32”
S = distance between two levels = 2536.98ft
CR = Earth curvature and Refraction correction =...
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