1. Atmospheric Correction

A distance of 826.39 feet was measure without including atmospheric corrections. If the temperature and pressure at measurement were 70°F and 28.5"Hg, what is the corrected distance?

From Equation (B-3)

c ex 1a

From Equation (B-4)

c ex 1b

2. Slope Reduction

A surveyor measures the distance between a section and quarter section corners and records a slope distance of 2677.36 ft with a zenith angle of 81°10'25" corrected for atmospheric conditions.

What error is introduced in the horizontal distance if refraction and curvature are not taken into account?

To solve this, compute the difference using Equations (A-2) and (B-5)

From Equation (A-2)

c ex 2a

From Equation (B-5)

c ex 2b 

The difference is 0.096 ft for a relative accuracy of 1/27,600.

While this is a pretty good relative accuracy, remember that this is just for an error introduced by ignoring refraction and curvature. It doesn't take into account other errors which can exist in the measurement.

3. Combined Random Errors

A horizontal distance of 985.37 feet is measured with a TSI having an MSA of ±(2mm + 3ppm). TSI centering error is estimated to be ±0.005 ft. The prism was pole mounted and hand held. Due to some wind the prism centering was assumed to be ±0.04 ft. Atmospheric conditions were accounted for at the time of measurement. What is the expected error in the distance?

These are all additive random errors since they affect parts of the line length. Therefore, we would use the Error of a Sum to determine the total error.

Recall that the Error of a Sum is: c ex 3a where Ei is a contributing error.

In this case we have four contributing errors: MSA constant, MSA proportional, TSI centering, and prism centering.

c ex 3b

Substituting into the Error of a Sum equation:

c ex 3c

This is an relative accuracy of 1/19,700. If a higher accuracy is needed then the first thing to consider is a more stable prism set up because that's the largest individual error.

What happens with shorter distances? Lets say a 156.53 ft long lot line was shot using the same equipment under the same conditions. What is the expected error in that distance?

The only contributing error which changes is the proportional part of the MSA; it becomes 0.000470 ft. Recomputing the Error of a Sum we get an error of ±0.04 ft for an accuracy of ~1/3800.

Notice that the absolute error didn't change much (±0.05' to ±0.04') but the relative error did (1/19,700 to 1/3800). That's because the biggest errors, prism and TSI setup, are considerably larger than those from the MSA specifications and they stay larger even for smaller distances.

How can we increase the relative accuracy? The largest contributing error is prism centering. If we mounted it on a tripod, we could achieve roughly the same centering error as the TSI's (it would be a liitle less than the TSI's becuse the prism tribrach usually has a fixed optical plummet). That single change raises the total and relative errors to ±0.0096' and 1/16,300, respectively.

For shorter distances, it is critical that the TSI and prism be centered as carefully as possible. Using a more accurate TSI with loose setups won't increase the overall accuracy. In those instances we've hit the limitations of the personnel not the equipment. 

Remember: Using accurate equipment inaccurately yields inaccurate results.