Levelling loops
Reciprocal levelling across a river crossing
A 1,360 ft river crossing is levelled reciprocally from both banks; the mean of the two differences removes the systematic error, and the half-difference is compared with the predicted curvature and refraction.
Given
- Benchmarks A and B on opposite banks of a river, 1,360 ft apart. Elevation of A = 596.482 ft (US survey feet); B is to be determined. Readings to 0.001 ft.
- Set-up on the A bank, instrument close to A: reading on the near rod at A = 3.472. Three readings on the far rod at B: 8.256, 8.261, 8.260.
- Set-up on the B bank, instrument close to B: reading on the near rod at B = 9.318. Three readings on the far rod at A: 4.614, 4.619, 4.618.
- Combined curvature and refraction correction, in feet, is 0.0206 times the square of the distance expressed in thousands of feet.
- Both observations were made within the same half hour, with the instrument moved directly between banks.
Required
- The difference in elevation from A to B from each bank.
- The accepted difference in elevation and the elevation of B.
- The magnitude of the systematic error removed, compared against the predicted curvature and refraction.
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
Why the crossing cannot be levelled ordinarily
Balanced sights are the standard defence against a collimation error and against curvature and refraction, and a river crossing makes them impossible: the instrument cannot be set in the middle of the water, so one sight is always short and the other is always a quarter of a mile.
Reciprocal levelling recovers the cancellation by symmetry in time instead of symmetry in space. The crossing is observed from both banks. From the A bank the long sight falls on B; from the B bank the long sight falls on A. The systematic error therefore enters the two computed differences with the same magnitude but opposite sign, and the mean of the two is free of it.
Mean the repeated far readings
The far rod is read three times at each set-up. At 1,360 ft the rod graduations subtend a small angle and the target shimmers, so repetition is not ceremony; it is the only way to get a far reading to 0.001 ft. The spread of the three readings is a useful field statistic in its own right, and here it is only 0.005 ft at each bank, which is good for the distance.
far readings on B: (8.256 + 8.261 + 8.260) / 3 = 24.777 / 3 = 8.259
far readings on A: (4.614 + 4.619 + 4.618) / 3 = 13.851 / 3 = 4.617
spread on B = 8.261 - 8.256 = 0.005 ft
spread on A = 4.619 - 4.614 = 0.005 ftDifference from each bank
For a single set-up the difference in elevation from one point to another is the reading on the first point minus the reading on the second. The sign convention takes care of itself if that rule is applied literally: whichever rod reads higher is standing on the lower ground.
The two banks disagree by 0.086 ft, which is far too large to be random at this distance. That disagreement is the systematic error made visible, and it is exactly what the procedure was designed to expose.
from the A bank: 3.472 - 8.259 = -4.787 ft
from the B bank: 4.617 - 9.318 = -4.701 ft| Set-up | Reading on A (ft) | Reading on B (ft) | dH A to B (ft) |
|---|---|---|---|
| Instrument on the A bank | 3.472 (near) | 8.259 (far, mean of 3) | -4.787 |
| Instrument on the B bank | 4.617 (far, mean of 3) | 9.318 (near) | -4.701 |
| Mean | - | - | -4.744 |
| Half difference | - | - | -0.043 |
The mean and the elevation of B
The accepted difference is the simple mean of the two, because the systematic error is equal and opposite in the pair. No weighting is appropriate; the two set-ups are of the same quality and the symmetry is what does the work.
B is 4.744 ft below A.
dH = (-4.787 + -4.701) / 2 = -9.488 / 2 = -4.744 ft
elev B = 596.482 - 4.744 = 591.738 ftAccount for the half difference
Half the difference between the two results is the size of the systematic error affecting each long sight, here 0.043 ft. That figure should not simply be discarded, because it can be predicted: curvature and refraction over 1,360 ft account for most of it, and whatever remains is the instrument's collimation error over the same distance.
The residual works out at 0.005 ft over 1,360 ft, which is 0.0004 ft per 100 ft of sight, or well under two seconds of arc. That is a well-adjusted instrument, and the agreement between the predicted and the observed systematic error confirms that nothing else, such as a blundered near reading, is hiding in the crossing.
half difference = (-4.787 - (-4.701)) / 2 = -0.043 ft, magnitude 0.043 ft
M = 1360 / 1000 = 1.36 ; M squared = 1.8496
predicted c + r = 0.0206 x 1.8496 = 0.038 ft
residual collimation = 0.043 - 0.038 = 0.005 ft over 1360 ft
= 0.0004 ft per 100 ft of sightField practice that makes the method work
Reciprocal levelling only cancels what is genuinely the same at both banks. Refraction over water changes through the day as the air and water temperatures diverge, so a crossing observed from one bank in the early morning and from the other in the afternoon will not cancel; the two set-ups must be made close together in time, as they were here.
For work of high order the crossing is normally observed several times, on different days and in both directions, and the results are meaned. A single reciprocal pair, as here, is appropriate for ordinary control work and is defensible provided the half difference is examined rather than ignored.
Answer
- Difference from the A bank = -4.787 ft; from the B bank = -4.701 ft
- Accepted difference in elevation, A to B = -4.744 ft
- Elevation of BM B = 591.738 ft
- Systematic error removed by the reciprocal pair = 0.043 ft on each long sight
Check
The two independent results straddle the mean symmetrically: -4.787 and -4.701 are each 0.043 ft from -4.744, as they must be if the systematic error is equal and opposite.
Predicted curvature and refraction over 1,360 ft is 0.0206 x 1.36 squared = 0.038 ft, which accounts for 88 percent of the observed 0.043 ft half difference. The 0.005 ft residual corresponds to a collimation error under 2 seconds of arc, consistent with a level in good adjustment.
Correcting the A-bank observation directly for curvature and refraction gives 3.472 - (8.259 - 0.038) = -4.749 ft, within 0.005 ft of the reciprocal mean, an independent route to the same answer.
More levelling loops
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- 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.