Levelling loops

Level net of three loops adjusted by inspection

Three 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.

Analyze· about 35 minutes by hand· 6 steps

Given

  • Three benchmarks A, B and C and an interior junction point J, connected by six level lines. Elevations are in US survey feet and are carried to 0.001 ft.
  • Observed elevation differences and line lengths: A to B, +4.512 ft, 1.6 miles; B to C, -9.874 ft, 2.0 miles; C to A, +5.340 ft, 1.8 miles; A to J, +2.061 ft, 1.0 mile; B to J, -2.462 ft, 1.2 miles; C to J, +7.400 ft, 1.4 miles.
  • Elevation of BM A is held fixed at 4,182.615 ft. B, C and J are all to be determined.
  • The three independent loops are taken as Loop I (A to B to J to A), Loop II (B to C to J to B) and Loop III (C to A to J to C).

Required

  • The misclosure of each of the three loops.
  • A correction to each of the six lines that closes all three loops simultaneously.
  • Adjusted elevations of B, C and J.

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 six lines and three loops

The net has four points and six lines. Three of the lines are needed simply to reach the three unknown points from the fixed point A; the other three are redundant observations, and each redundancy shows up as a loop that ought to close and does not. Six lines minus four points plus one gives three independent loops, and adjusting the net means finding one set of corrections that closes all three at once.

Sign convention matters here more than anywhere else in levelling. Each line is stored in one direction, and when a loop traverses it the other way the stored value enters with a minus sign. Loop I runs A to B, then B to J, then J back to A, so it uses the A-to-B line forward, the B-to-J line forward, and the A-to-J line reversed.

loops = lines - points + 1 = 6 - 4 + 1 = 3 Loop I = (A-B) + (B-J) - (A-J) Loop II = (B-C) + (C-J) - (B-J) Loop III = (C-A) + (A-J) - (C-J)

Misclosure of each loop

Substituting the observed differences gives three small misclosures, all within a few hundredths of a foot. They are not independent of one another: the exterior triangle A to B to C to A is the sum of the three loops, so its misclosure must equal their sum. That relationship is the first check on the arithmetic, and it should be worked before any correction is computed.

Loop I = +4.512 + (-2.462) - (+2.061) = -0.011 ft Loop II = -9.874 + (+7.400) - (-2.462) = -0.012 ft Loop III = +5.340 + (+2.061) - (+7.400) = +0.001 ft exterior triangle A-B-C-A = +4.512 - 9.874 + 5.340 = -0.022 ft sum of the three loops = -0.011 - 0.012 + 0.001 = -0.022 ft (agrees)

Adjustment by inspection, pass by pass

Adjustment by inspection takes the loops one at a time. For the loop being treated, its current misclosure is distributed over its own lines in proportion to line length, exactly as for a single closed loop. Correcting one loop disturbs its neighbours, because the interior lines A-J, B-J and C-J are each shared by two loops, so the pass is repeated until nothing is left to distribute.

The corrections shrink by roughly a factor of three each pass here, so four passes reduce every loop misclosure below 0.001 ft. The table records the misclosure found in each loop at the moment it was treated, together with the running total of corrections applied to each line.

Successive passes: loop misclosure at the time of treatment (ft) and cumulative line corrections (ft)
PassLoop ILoop IILoop IIIA-BB-CC-AA-JB-JC-J
1-0.0110-0.0155-0.0066+0.0046+0.0067+0.0028-0.0013-0.0006+0.0025
2-0.0056-0.0040-0.0027+0.0070+0.0085+0.0040-0.0022+0.0002+0.0028
3-0.0017-0.0014-0.0009+0.0077+0.0091+0.0044-0.0024+0.0003+0.0030
4-0.0006-0.0005-0.0003+0.0079+0.0093+0.0045-0.0025+0.0004+0.0030
converged0.00000.00000.0000+0.0081+0.0094+0.0045-0.0025+0.0004+0.0030

Adjusted line values

Rounding the converged corrections to 0.001 ft and applying them to the observed differences gives the adjusted net. Rounding is not automatically safe in a net adjustment, so every loop is re-tested on the rounded values before anything is carried forward.

Loop I = +4.520 - 2.462 - 2.058 = 0.000 Loop II = -9.865 + 7.403 + 2.462 = 0.000 Loop III = +5.345 + 2.058 - 7.403 = 0.000 exterior = +4.520 - 9.865 + 5.345 = 0.000
Adjusted level net
LineLength (mi)Observed dH (ft)Correction (ft)Adjusted dH (ft)
A to B1.6+4.512+0.008+4.520
B to C2.0-9.874+0.009-9.865
C to A1.8+5.340+0.005+5.345
A to J1.0+2.061-0.003+2.058
B to J1.2-2.4620.000-2.462
C to J1.4+7.400+0.003+7.403

Adjusted elevations

With every loop closed, the elevation of each point is the same no matter which route is used to reach it. That is exactly the property the adjustment was performed to obtain, and it is also the strongest available check: compute each unknown by two different routes and require agreement to the last digit carried.

A = 4182.615 (held) B = 4182.615 + 4.520 = 4187.135 C = 4187.135 - 9.865 = 4177.270 J = 4182.615 + 2.058 = 4184.673

Second routes to every point

C is recomputed backwards along the C-to-A line, and J is recomputed from both B and C. All three alternative routes reproduce the elevations already found, so the adjusted net is internally consistent to 0.001 ft.

One practical note: this adjustment weights lines by length only. If some lines were run to a higher order of accuracy than others, the weights should reflect that instead, and a least-squares adjustment becomes the honest tool. Adjustment by inspection is appropriate when the lines are of comparable quality and the misclosures are small, which is the case here.

C via the C-A line: 4182.615 - 5.345 = 4177.270 (agrees) J via B: 4187.135 - 2.462 = 4184.673 (agrees) J via C: 4177.270 + 7.403 = 4184.673 (agrees)

Answer

  • Loop misclosures before adjustment: Loop I -0.011 ft, Loop II -0.012 ft, Loop III +0.001 ft
  • Corrections: A-B +0.008, B-C +0.009, C-A +0.005, A-J -0.003, B-J 0.000, C-J +0.003 ft
  • BM B = 4,187.135 ft
  • BM C = 4,177.270 ft
  • Junction J = 4,184.673 ft

Check

All three loops close at 0.000 ft on the rounded corrections, and so does the exterior triangle A-B-C-A.

C computed forward through B (4,177.270 ft) equals C computed backward along the C-to-A line (4,177.270 ft).

J computed from A, from B and from C all give 4,184.673 ft.

Consistency check on the misclosures: the exterior triangle misclosure of -0.022 ft equals the sum of the three loop misclosures, confirming that the three loops chosen are independent and together span the redundancy in the net.

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