Levelling loops
Two-peg test worked to a collimation correction
A two-peg test on a level is reduced to a collimation error per unit sight length, expressed in seconds of arc, and then used to correct a pair of unbalanced sights and recover a turning-point elevation.
Given
- Two firm pegs A and B set 300.0 ft apart on nearly level ground. All readings in feet to 0.001 ft.
- Position 1, instrument set midway between the pegs: rod reading on A = 5.284, rod reading on B = 4.117.
- Position 2, instrument set on the line 5.0 ft outside peg A, so that the sight to A is 5.0 ft and the sight to B is 305.0 ft: rod reading on A = 4.652, rod reading on B = 3.462.
- The instrument is later used on a job where the sights could not be balanced: a backsight of 5.216 was read on BM-14, elevation 742.186 ft, at a sight length of 250 ft, and a foresight of 8.940 was read on TP-3 at a sight length of 480 ft.
- Elevations are carried to 0.001 ft.
Required
- The true difference in elevation between the pegs.
- The collimation error of the instrument, expressed per 100 ft of sight and in seconds of arc, with its sense.
- The corrected elevation of TP-3.
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
Position 1 gives the truth
With the instrument exactly midway between the pegs, both sight lengths are equal. A line of sight that is not truly horizontal is then in error by the same amount on both readings, and that error cancels out of the difference. So the midpoint difference is the true difference in elevation, whatever the state of the instrument.
That cancellation is the entire logic of the two-peg test, and it is the same logic that makes balanced backsights and foresights the standard field procedure. The test does not repair the instrument; it measures how much damage the instrument does when sights are not balanced.
dH(A to B) = reading on A - reading on B
dH(true) = 5.284 - 4.117 = +1.167 ft (B is 1.167 ft higher than A)Position 2 exposes the error
From just outside peg A the two sights are wildly unbalanced: 5.0 ft to the near rod and 305.0 ft to the far rod. Any tilt in the line of sight now affects the far reading sixty-one times as much as the near one, so the difference computed from this setup is contaminated by very nearly the full effect of the tilt over 300 ft.
The apparent difference is 1.190 ft, which is 0.023 ft larger than the truth.
dH(apparent) = 4.652 - 3.462 = +1.190 ft
discrepancy = 1.190 - 1.167 = +0.023 ft over a 300.0 ft difference in sight lengthCollimation error and its sense
Define the collimation error e as the amount by which a reading is too large per foot of sight length, so that a true reading equals the observed reading minus e times the distance. Writing the difference from position 2 in those terms and setting it equal to the true difference solves for e directly.
The value comes out negative, which means the readings are too small at long range: the line of sight droops below horizontal. The far rod at 305 ft was read 0.023 ft low, which made B appear higher than it is and inflated the apparent difference. A drooping line of sight is the expected symptom when a level's compensator is sticking or the bubble is out of adjustment upward.
dH(true) = dH(apparent) + e x (dB - dA)
1.167 = 1.190 + e x (305.0 - 5.0)
e = -0.023 / 300.0 = -0.00007667 ft per ft of sight
per 100 ft of sight: -0.0077 ft
as an angle: arctan(0.00007667) = 0°00′15.8″, the line of sight below horizontalCorrect the unbalanced job readings
Every reading taken with this instrument is too small by 0.00007667 ft for each foot of sight. Correcting a reading means adding that amount back, in proportion to how far the rod was. The backsight at 250 ft picks up 0.019 ft and the foresight at 480 ft picks up 0.037 ft, and because the sights are unbalanced the two corrections do not cancel.
correction to a reading = -e x d = +0.00007667 x d
BS: +0.00007667 x 250 = +0.019 -> 5.235
FS: +0.00007667 x 480 = +0.037 -> 8.977
shortcut on the difference: e x (dFS - dBS) = -0.00007667 x 230 = -0.018 ft| Reading | Sight length (ft) | Observed (ft) | Correction (ft) | Corrected (ft) |
|---|---|---|---|---|
| BS on BM-14 | 250 | 5.216 | +0.019 | 5.235 |
| FS on TP-3 | 480 | 8.940 | +0.037 | 8.977 |
| Difference | 230 unbalanced | -3.724 | -0.018 | -3.742 |
Corrected elevation of TP-3
The corrected difference in elevation is the corrected backsight less the corrected foresight, applied to the held elevation of BM-14. The uncorrected figure would have been 738.462 ft, so ignoring the collimation error on this one unbalanced setup would have cost 0.018 ft.
That is a large error for a single setup. On a line of twenty such setups all leaning the same way it would accumulate to more than a third of a foot, because a collimation error is systematic and does not average out with repetition.
dH = 5.235 - 8.977 = -3.742 ft
TP-3 = 742.186 - 3.742 = 738.444 ft
uncorrected would have been 742.186 - 3.724 = 738.462 ft, high by 0.018 ftWhat to do with the instrument
Whether 16″ of collimation error warrants adjustment depends on the work. For construction staking with sights routinely balanced to within 20 or 30 ft, the residual effect is under 0.003 ft per setup and the instrument can stay in service. For a level line run to a published order of accuracy, or for any work where obstacles force unbalanced sights, the instrument should be adjusted and the test repeated.
Either way the test result should be recorded with a date in the instrument's log, because a collimation error that is drifting steadily larger over successive tests is a sign of mechanical trouble that a single test cannot reveal.
Answer
- True difference in elevation, A to B = +1.167 ft (B higher)
- Collimation error = -0.0077 ft per 100 ft of sight, that is 0°00′16″ with the line of sight below horizontal
- Corrected readings: backsight 5.235, foresight 8.977
- Elevation of TP-3 = 738.444 ft
Check
Apply the collimation error to the position-2 readings and re-derive the true difference: 4.652 + 0.00007667 x 5.0 = 4.6524 and 3.462 + 0.00007667 x 305.0 = 3.4854, giving 4.6524 - 3.4854 = 1.167 ft, which reproduces the midpoint result exactly.
Balanced-sight check: if the job sights had both been 365 ft, the corrections would have been +0.028 ft each and would have cancelled in the difference, confirming that a collimation error is harmless only when the sights are balanced.
The shortcut correction on the difference, e times the difference in sight lengths, gives -0.018 ft and matches the result of correcting the two readings individually.
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.