Error propagation
Angular closure of a seven-sided traverse against a k√n tolerance
Seven interior angles that sum 21 seconds over the geometric condition. Test the misclosure against the k√n tolerance a 10-second instrument justifies, then distribute it equally and prove the adjusted angles close.
Given
- A closed seven-sided traverse was observed with a 10″ total station, each interior angle turned by one direct and one reverse pointing and meaned.
- Observed interior angles, in station order:
- 1: 128°14′22″ 2: 96°37′48″ 3: 142°05′11″ 4: 118°49′36″
- 5: 154°22′07″ 6: 131°58′44″ 7: 127°52′33″
- The field specification allows an angular misclosure of k√n seconds, with k taken as the least count of the instrument, 10″, and n the number of angles.
- No angle is suspected of a blunder; all were turned with the same procedure and the same care.
Required
- The geometric condition the seven angles must satisfy, and the observed misclosure.
- Whether the misclosure meets the tolerance.
- The correction to each angle and the adjusted angles.
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
The geometric condition
The interior angles of a closed polygon of n sides must sum to (n − 2) × 180°. For seven sides that is 5 × 180° = 900°00′00″. This is a property of the figure, independent of how it was measured, and it is what makes a closed traverse self-checking in angle.
Convert every angle to seconds before summing. Degrees, minutes and seconds carry in different bases and adding them in mixed units is where the arithmetic goes wrong; converting to a single unit removes the problem entirely.
Had exterior angles been turned instead, the condition would be (n + 2) × 180° = 1620°00′00″. Confirm which was observed before testing anything — an exterior-angle set tested against the interior condition misses by exactly 720°, which at least announces itself.
required sum = (n − 2) × 180° = (7 − 2) × 180° = 900°00′00″ = 3,240,000″
for comparison, exterior angles would require (n + 2) × 180° = 1620°00′00″Sum the observed angles
Convert each angle to seconds and add. The total is 3,240,021″, or 900°00′21″, so the observed angles exceed the geometric requirement by 21 seconds.
The misclosure is small and positive. Positive means the angles as turned are collectively too large, so the corrections will all be negative.
| Angle | Observed | Seconds |
|---|---|---|
| 1 | 128°14′22″ | 461,662 |
| 2 | 96°37′48″ | 347,868 |
| 3 | 142°05′11″ | 511,511 |
| 4 | 118°49′36″ | 427,776 |
| 5 | 154°22′07″ | 555,727 |
| 6 | 131°58′44″ | 475,124 |
| 7 | 127°52′33″ | 460,353 |
| Σ observed | 900°00′21″ | 3,240,021 |
| Required | 900°00′00″ | 3,240,000 |
| Misclosure | +21″ | +21 |
Test against the k√n tolerance
The tolerance takes the same square-root form as every other accumulation of independent random error in this subject. Each of the n angles carries an independent error of roughly the instrument's least count, and the error of their sum grows as the square root of n, not in proportion to n.
With k = 10″ and n = 7 the allowance is 26.5″. The observed 21″ is inside it, so the misclosure is ordinary accumulated observing error and the traverse passes.
It is worth seeing how forgiving the alternatives are. A specification written at k = 20″ would allow 52.9″ and one at k = 30″ would allow 79.4″. The choice of k is what the specification is really about; the √n is fixed by the propagation. Note also that a linear tolerance of 7 × 10″ = 70″ — a common misremembering — would let through nearly three times the misclosure that the propagation justifies.
allowable = k √n = 10″ × √7 = 10″ × 2.6458 = 26.5″
observed misclosure = 21″ < 26.5″ PASS
at 79 % of the allowance
for comparison: 20″√7 = 52.9″; 30″√7 = 79.4″; the incorrect linear form 7 × 10″ = 70″Distribute the misclosure and adjust
Every angle was turned by the same instrument, with the same procedure, on the same kind of target. There is no basis for weighting one more than another, so the misclosure is distributed equally: each angle takes minus one seventh of 21″, which is exactly −3″.
The division coming out exact is convenient rather than typical. Where it does not divide evenly, apply the whole seconds equally and put the remaining seconds on the angles with the shortest sights, since those are the ones the instrument's centring error affects most. Never park the entire remainder on one angle to make the column add up.
Note that the correction of 3″ is below the instrument's own least count. The adjustment is not claiming to have found a 3″ error in each angle; it is enforcing the geometric condition so that azimuths carried round the traverse return to their starting value.
| Angle | Observed | Correction | Adjusted |
|---|---|---|---|
| 1 | 128°14′22″ | −3″ | 128°14′19″ |
| 2 | 96°37′48″ | −3″ | 96°37′45″ |
| 3 | 142°05′11″ | −3″ | 142°05′08″ |
| 4 | 118°49′36″ | −3″ | 118°49′33″ |
| 5 | 154°22′07″ | −3″ | 154°22′04″ |
| 6 | 131°58′44″ | −3″ | 131°58′41″ |
| 7 | 127°52′33″ | −3″ | 127°52′30″ |
| Σ | 900°00′21″ | −21″ | 900°00′00″ |
Answer
- The seven interior angles must sum to (7 − 2) × 180° = 900°00′00″.
- They sum to 900°00′21″, so the angular misclosure is +21″.
- The allowable misclosure is k√n = 10″ × √7 = 26.5″. At 21″ the traverse passes, using 79 per cent of its allowance.
- Each angle is corrected by −21″ / 7 = −3″ exactly, since all were observed with equal care and there is no basis for unequal weighting.
- Adjusted angles: 128°14′19″, 96°37′45″, 142°05′08″, 118°49′33″, 154°22′04″, 131°58′41″, 127°52′30″, summing to exactly 900°00′00″.
Check
Re-sum the adjusted angles in seconds: 461,659 + 347,865 + 511,508 + 427,773 + 555,724 + 475,121 + 460,350 = 3,240,000″, which is exactly 900°00′00″. The geometric condition is satisfied to the second.
Check the correction column independently: seven corrections of −3″ sum to −21″, exactly cancelling the +21″ misclosure. Any residual would mean the division left a remainder that was not accounted for.
Check the tolerance is being applied in the right form. If angular error accumulated linearly rather than in quadrature, seven 10″ angles would allow 70″ — 2.6 times what the square-root form permits. The distinction is not academic: a 60″ misclosure would pass the linear test and fail the correct one, and a traverse carrying 60″ of angular misclosure through seven courses puts the far end of a 500 ft course roughly 0.15 ft out of position.
Reasonableness of the misclosure itself: 21″ spread over seven angles is 3″ each, well within a 10″ instrument's resolution, and there is no single angle that would need to be grossly wrong to produce it. Had the misclosure been 21′ rather than 21″, the place to look would be a single angle mis-booked by 21 minutes, not a distribution.
More error propagation
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- Error in a taped distance measured in n tape lengths, against a single EDM shotThe same 648 ft line measured two ways. Series propagation over seven tape settings against one EDM observation with a constant-plus-ppm specification, and what the comparison says about when taping is still adequate.
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