Area & partitioning
Irregular boundary area by the trapezoidal rule and by Simpson's one-third rule
Nine offsets to a meandering creek bank at 25 ft intervals, reduced two ways. The two rules differ by 22.5 ft², and the reason they differ is the reason to prefer one of them.
Given
- A traverse line runs along the upland side of a parcel. Perpendicular offsets were taken from that line to the top of a creek bank at a constant interval of d = 25.00 ft.
- Nine offsets were taken, giving eight intervals: 12.4, 18.7, 24.2, 27.9, 29.5, 26.8, 21.3, 15.6, 10.2 ft, in order along the line.
- Offsets are recorded to 0.1 ft, which is as well as the top of bank can be identified.
- The strip between the traverse line and the bank is to be added to a parcel area computed by coordinates.
Required
- The strip area by the trapezoidal rule.
- The strip area by Simpson's one-third rule.
- Which value should be carried forward, and why they differ.
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
Confirm the offsets can take Simpson's rule at all
Simpson's one-third rule fits a parabola through each consecutive group of three offsets, so it consumes offsets two intervals at a time and needs an even number of intervals — an odd number of offsets. Nine offsets give eight intervals, which is even, so the rule applies to the whole strip.
It also requires the interval to be constant. Both rules do, in the form given here; unequal intervals mean falling back to summing individual trapezoids, one at a time, on their own spacings.
Had there been ten offsets and nine intervals, the usual practice is to apply Simpson to the first eight intervals and close the last one as a single trapezoid, and to say so on the sheet.
offsets n = 9, intervals = 8 (even) → Simpson's one-third rule applies
interval d = 25.00 ft, constantTrapezoidal rule
The trapezoidal rule joins the tops of consecutive offsets with straight lines. Collected, it is the interval times the sum of all the offsets with the two end offsets counted at half weight.
Every interior offset is shared between two trapezoids, which is where the half weight on the ends comes from; forgetting it and simply averaging all nine offsets over-counts the two ends and inflates the area.
A = d · [ (y₁ + y₉)/2 + y₂ + y₃ + y₄ + y₅ + y₆ + y₇ + y₈ ]
end terms: (12.4 + 10.2) / 2 = 22.6 / 2 = 11.30
interior: 18.7 + 24.2 + 27.9 + 29.5 + 26.8 + 21.3 + 15.6 = 164.00
A = 25.00 × ( 11.30 + 164.00 ) = 25.00 × 175.30 = 4382.5 ft²Simpson's one-third rule
Simpson's rule replaces the straight chords with parabolic arcs through successive triples of offsets. Collected, the ends carry weight 1, the odd-numbered interior offsets carry weight 4, and the even-numbered interior offsets carry weight 2, all over d/3.
Get the weights the right way round. The offsets that carry the weight of 4 are the ones at the apex of each parabola — the second, fourth, sixth and eighth in the list. Swapping the 4s and the 2s is easy to do and produces an answer of about the right size, which is what makes it dangerous.
| No. | Offset (ft) | Weight | Weighted offset (ft) |
|---|---|---|---|
| 1 | 12.4 | 1 | 12.4 |
| 2 | 18.7 | 4 | 74.8 |
| 3 | 24.2 | 2 | 48.4 |
| 4 | 27.9 | 4 | 111.6 |
| 5 | 29.5 | 2 | 59.0 |
| 6 | 26.8 | 4 | 107.2 |
| 7 | 21.3 | 2 | 42.6 |
| 8 | 15.6 | 4 | 62.4 |
| 9 | 10.2 | 1 | 10.2 |
| Σ | 528.6 |
Evaluate Simpson and compare
Multiply the weighted sum by one third of the interval. The result is 4405.0 ft², which is 22.5 ft² more than the trapezoidal figure — a difference of 0.51 percent.
Simpson's value is the larger, and it is larger for a structural reason rather than by accident. This bank is convex toward the traverse line over most of its run, and a straight chord between two offsets on a convex curve always falls inside the true boundary. The trapezoidal rule therefore under-reads a convex boundary and over-reads a concave one, every time; it is a biased estimator, not a noisy one.
Simpson's rule reproduces any boundary that is genuinely parabolic exactly, and errs only at the fourth power of the interval on anything else. For a smooth natural feature sampled at a constant interval it is the better estimate, and it costs nothing extra.
A = (d / 3) · [ y₁ + 4(y₂ + y₄ + y₆ + y₈) + 2(y₃ + y₅ + y₇) + y₉ ]
odd interior sum = 18.7 + 27.9 + 26.8 + 15.6 = 89.0 ×4 = 356.0
even interior sum = 24.2 + 29.5 + 21.3 = 75.0 ×2 = 150.0
ends = 12.4 + 10.2 = 22.6
Σ weighted = 356.0 + 150.0 + 22.6 = 528.6
A = (25.00 / 3) × 528.6 = 13,215.0 / 3 = 4405.0 ft²
difference = 4405.0 − 4382.5 = 22.5 ft² (0.51 %)What to carry forward
Carry 4405 ft², or 0.101 acre, into the parcel total, and say on the sheet which rule produced it. The difference between the two rules — 22.5 ft², or 0.0005 acre — is smaller than the uncertainty in identifying the top of bank in the first place, so the choice does not change the acreage reported to 0.001 acre. It is still worth making deliberately, because on a longer or more sharply curved reach the same bias grows quickly.
The honest limit on either figure is the offsets themselves. Nine offsets sample a meandering bank rather coarsely: anything that happens between two offsets is invisible to both rules, and no amount of refinement in the arithmetic recovers it. If the bank is the controlling boundary of the conveyance, the answer is more offsets, not a better rule.
4405.0 / 43,560 = 0.10112 ac → 0.101 ac
4382.5 / 43,560 = 0.10061 ac → 0.101 ac
both round to 0.101 ac at the reporting precisionAnswer
- Trapezoidal rule: 4382.5 ft² = 0.101 acre.
- Simpson's one-third rule: 4405.0 ft² = 0.101 acre.
- Simpson's value exceeds the trapezoidal value by 22.5 ft², or 0.51 percent, because the boundary is convex toward the traverse line and straight chords cut the corners off a convex curve.
- Carry 4405 ft² forward, and record on the computation sheet that Simpson's one-third rule was used with nine offsets at 25.00 ft.
Check
Apply Simpson's rule to the two halves separately and add. First half, offsets 1 to 5: (25.00/3)[12.4 + 4(18.7) + 2(24.2) + 4(27.9) + 29.5] = 2305.83 ft². Second half, offsets 5 to 9: (25.00/3)[29.5 + 4(26.8) + 2(21.3) + 4(15.6) + 10.2] = 2099.17 ft². The two sum to 4405.00 ft², matching the single application exactly.
Independent check by extrapolation. Recompute the trapezoidal rule on the coarse 50.00 ft interval, using only offsets 1, 3, 5, 7 and 9: 50.00 × [ (12.4 + 10.2)/2 + 24.2 + 29.5 + 21.3 ] = 50.00 × 86.30 = 4315.0 ft².
The trapezoidal error falls with the square of the interval, so halving the interval should remove three quarters of it: the refined estimate is T(d) + [T(d) − T(2d)]/3 = 4382.5 + (4382.5 − 4315.0)/3 = 4382.5 + 22.5 = 4405.0 ft². That reproduces Simpson's answer exactly — as it must, since the two are algebraically the same combination — and confirms the weighting table has no transposed 4 and 2.
Rough sanity check: the mean offset is about 20.7 ft over a run of 8 × 25.00 = 200.00 ft, giving roughly 4140 ft². The computed values sit a few percent above that, which is right for a strip that bulges in the middle.
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