Levelling loops
Benchmark carried on two routes and reconciled by weighted mean
A new benchmark is levelled from two published control points over routes of different length. Test the discrepancy, weight each route inversely with its length, and compute the weighted mean elevation and its standard deviation.
Given
- A new benchmark BM-C is established and levelled independently from two published control points. Elevations in US survey feet, carried to 0.001 ft.
- Route 1, from BM-A: length 3.2 miles, giving BM-C = 1,214.372 ft.
- Route 2, from BM-B: length 5.6 miles, giving BM-C = 1,214.418 ft.
- The two published control points are themselves held fixed and are assumed error free for this computation.
- Tolerance coefficient for this order of work is C = 0.05 ft with the length in miles.
Required
- Whether the discrepancy between the two routes is acceptable.
- The weight of each route and the weighted mean elevation of BM-C.
- An estimate of the precision of the adopted elevation.
Work it through yourself before reading on — the solution below shows every step, so there is no way to skim it without giving the answer away.
Worked solution
Test the discrepancy first
Before averaging anything, find out whether the two routes agree well enough to be averaged at all. Taken together the two routes form a closed circuit from BM-A to BM-C to BM-B, of total length 8.8 miles, and the discrepancy between the two computed elevations is that circuit's misclosure.
If the discrepancy exceeded the tolerance the correct action would be to re-run the weaker route, not to average. A weighted mean of a good route and a blundered route is simply a contaminated answer wearing a statistical disguise.
discrepancy = 1214.418 - 1214.372 = 0.046 ft
combined length = 3.2 + 5.6 = 8.8 miles
allowable = 0.05 x sqrt(8.8) = 0.05 x 2.9665 = 0.148 ft
0.046 ft < 0.148 ft accept and combineWeights from route length
The variance of a levelled elevation difference grows in proportion to the distance run, so the weight, which is the reciprocal of the variance, is inversely proportional to length. The shorter route is the better observation and must count for more.
It counts for more in exact proportion to the length ratio, not to the square of it. Route 1 is 1.75 times shorter than route 2 and therefore carries 1.75 times the weight, giving it 63.6 percent of the say in the final answer.
w1 = 1 / 3.2 = 0.312500
w2 = 1 / 5.6 = 0.178571
sum of weights = 0.491071
normalised: w1 = 0.6364, w2 = 0.3636Weighted mean
The weighted mean is the sum of weight times value divided by the sum of weights. Working with the differences from a convenient base value, rather than with the full elevations, keeps the arithmetic short and keeps the significant figures where they belong.
offset = 0.008214 / 0.491071 = 0.016727 ft
weighted mean = 1214.372 + 0.016727 = 1214.38873
adopted elevation of BM-C = 1,214.389 ft| Route | Length (mi) | Elevation (ft) | Weight w | w x (elev - 1214.372) |
|---|---|---|---|---|
| 1, from BM-A | 3.2 | 1214.372 | 0.312500 | 0.000000 |
| 2, from BM-B | 5.6 | 1214.418 | 0.178571 | 0.008214 |
| Sums | 8.8 | - | 0.491071 | 0.008214 |
Residuals and precision
The residual of each route is its value less the adopted mean. The standard deviation of unit weight is the square root of the weighted sum of squared residuals divided by the degrees of freedom, and with two observations of one unknown there is exactly one degree of freedom.
The standard deviation of the weighted mean itself is that figure divided by the square root of the sum of the weights. It comes out at 0.022 ft, which is the honest one-sigma statement about the adopted elevation.
v1 = 1214.372 - 1214.38873 = -0.0167 ft
v2 = 1214.418 - 1214.38873 = +0.0293 ft
sum of w v squared = 0.3125 x 0.0167^2 + 0.178571 x 0.0293^2 = 0.0002405
sigma(unit weight) = sqrt(0.0002405 / 1) = 0.0155 ft
sigma(mean) = 0.0155 / sqrt(0.491071) = 0.022 ftSanity-check the weighting
The adopted value must lie between the two observations and must lie nearer the better one. It does: 1,214.389 is 0.017 ft from the short route and 0.029 ft from the long route, and the ratio of those residuals is 0.029 to 0.017, or 1.75 to 1, which is exactly the inverse of the weight ratio.
That relationship is a general property of a two-observation weighted mean and makes a fast field check. If the adopted value is not nearer the better observation, or the residual ratio does not invert the weight ratio, the weights were applied upside down.
|v2| / |v1| = 0.0293 / 0.0167 = 1.75
w1 / w2 = 0.312500 / 0.178571 = 1.75 (agrees)How to publish the result
The elevation is reported as 1,214.389 ft with a standard deviation of 0.022 ft, and the record should note that it rests on two routes of 3.2 and 5.6 miles that agreed within 0.046 ft. A user who later finds a third route disagreeing by 0.05 ft then has the information needed to judge the conflict rather than merely inherit an answer.
It is worth stating plainly that the 0.022 ft figure describes the internal consistency of these two routes only. It says nothing about the accuracy of the published elevations of BM-A and BM-B, which were held fixed here and which carry their own uncertainty on the network they came from.
Answer
- Discrepancy between the routes = 0.046 ft against an allowance of 0.148 ft; acceptable
- Weights: route 1 = 0.3125, route 2 = 0.1786 (63.6 and 36.4 percent)
- Adopted elevation of BM-C = 1,214.389 ft
- Standard deviation of the adopted elevation = 0.022 ft
Check
The adopted value lies between the two observations and nearer the shorter route, as it must.
The residuals -0.0167 and +0.0293 ft are in the ratio 1 to 1.75, the exact inverse of the weight ratio 1.75 to 1. This is the defining property of a two-observation weighted mean and confirms the weights were not inverted.
Recomputing the mean from the other base value: 1214.418 + 0.6364 x (1214.372 - 1214.418) = 1214.418 - 0.02927 = 1,214.3887 ft, reproducing the adopted elevation from the opposite direction.
Circuit check: the two routes together form a closed circuit of 8.8 miles whose misclosure is 0.046 ft, comfortably inside the 0.148 ft allowance, so the weighted mean is being applied to observations that genuinely belong together.
More levelling loops
All in this category- Differential level circuit reduced from field notesReduce a four-setup differential level run from BM-1 to BM-2 through three turning points, then prove the page with the sum of backsights against the sum of foresights.
- Closed level loop tested against a tolerance and adjustedA four-leg level loop returns to its starting benchmark 0.060 ft out. Test the misclosure against a C times root-M tolerance, then distribute it in proportion to leg length and compute adjusted elevations.
- Level net of three loops adjusted by inspectionThree benchmarks and an interior junction point are connected by six level lines forming three independent loops. Adjust the net by inspection, distributing each loop misclosure by line length and iterating until every loop closes.