Tolerance for closed-loop Leveling

Why does the tolerance for leveling has the form \(C = k \sqrt D\). Specifically why \(\sqrt D\) ?

In Surveying, leveling refers to figuring out the elevation of points relative to a benchmark point. In closed loop traverse we start at a point, follow a traverse (a sequence of points) and loop back to the starting point. And during closed loop leveling, the task is to figure out the elevation of the points in the traverse.

The acceptable tolerance of errors is \(C = k \sqrt D\), and our total error \(E\) when closing the loop should be less than the tolerance \(C\). Here, \(k\) is some constant.

Let's trick the system!

Instead of doing leveling of the whole distance \(D\) where we get total error \(E\). Lets divide it into \(N\) equal parts of distance \(D/N\) and then do closed-loop leveling in those segments.

For each segment, we have to go forward \(D/N\) distance and close the loop by coming back \(D/N\) distance, so the total distance covered would be \(2D/N\) and since the error of distance \(D\) was \(E\), we would have measurement error of \(2E/N\) for \(2D/N\) distance, and acceptable tolerance of \(k \sqrt {2D / N}\).

So what?

Although the error reduced by factor of \(N/2\) the tolerance is reduced by just \(\sqrt{N/2}\) and thus, we have higher tolerance than the expected error. This essentially means, we can get within the tolerance with worse instrument or technique if we keep divide the big loop into small ones.

Right?

The hidden issue:

There can be systematic as well as random errors during the measurement. Systematic error are handeled by calibration and proper technique:

1. equal forward sight distance and backward sight distance.
2. proper calibration of the instrument to fix collimation error
3. taking 3 point measurement (top, middle, bottom) and checking for tolerance in each measurement

And thus the error tolerance is for random errors. Such errors are independent, and thus the sum of the error doesn't increase proportionally with distance. These error sometimes add up while sometimes they cancle each other. Thus if a 10 km loop has error \(E\), a 100km loop doesn't have error \(10E\). Conversely, if the 100km loop has error \(E\), a 10 km loop doesn't have error \(0.1 E\).

For independent random errors, the total error is the square root of the sum of the squares of the individual errors (\(E_{T}=\sqrt{\sum E_{i}^{2}}\)). So, if we make \(N\) measurements the total error is proportional to \(\sqrt N\) times the individual errors. Or in other words, the individual errors are not \(E/N\) but rather \(E/{\sqrt N}\).

Since the number of measurements is proportional to the total distance, the total error becomes proportional to the square root of the distance.

Thus, in the trick we tried to perform, we were wrong to assume that the error accumulates linearly, i.e. If \(E\) was total error in \(D\) distance, the error for \(2D/N\) distance would be \(E / \sqrt {2/ N}\) instead of \(2E/N\) and this matches the formula of acceptable tolerance is.