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

Question

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 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 XXXXXXXXXXft. If the elevation of the occupied benchmark is XXXXXXXXXXft, what is the corrected benchmark elevation on the island? (Assume that the instrument does not correct for earth curvature and refraction.)

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Robert · Accepted Answer

Solution 
Corrected benchmark elevation difference can be calculated as
Where Δh represents corrected 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 = 0.

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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