Traverse adjustment
Closed-loop traverse from field angles to compass-rule coordinates
A five-sided loop traverse is taken from raw interior angles and taped distances through angular closure, azimuths, latitudes and departures, misclosure and precision, to compass-rule adjusted coordinates.
Given
- A closed-loop traverse A-B-C-D-E-A run counter-clockwise, with the interior angles observed to the right at each station.
- Observed interior angles: A 94°44′00″, B 117°37′40″, C 103°06′05″, D 105°50′10″, E 118°42′20″.
- Observed horizontal distances: A-B 434.24 ft, B-C 398.95 ft, C-D 358.09 ft, D-E 450.30 ft, E-A 340.14 ft.
- The azimuth of course A-B is fixed at 114°30′00″ from a control line.
- Station A is held at N 5000.00, E 5000.00 on an assumed grid.
- Angles were turned with a 30 second instrument. The specification calls for 1:10,000 linear precision.
- Distances to 0.01 ft, angles to 1 second, coordinates to 0.01 ft. Latitudes and departures are carried to 0.001 ft and rounded only at the coordinates.
Required
- The angular misclosure and the adjusted angles.
- Azimuths and bearings of all five courses.
- Latitudes, departures, the linear misclosure and the precision.
- Compass-rule corrections and the adjusted coordinates of every station.
- Whether the traverse meets its 1:10,000 specification.
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
Angular closure
The interior angles of any closed polygon of n sides must sum to (n − 2) × 180°. For five sides that is 540°00′00″. The observed sum exceeds it by 15 seconds.
There is no basis for treating one angle as worse than another: they were all turned with the same instrument by the same crew in the same conditions. So the misclosure is distributed equally, 3 seconds off each angle.
Check the misclosure against a tolerance before adjusting it. A common working rule for a 30 second instrument is the least count times the square root of the number of angles, here 30″ × √5 = 67″. The observed 15 seconds is comfortably inside that, so there is no blunder to hunt.
required sum = (5 − 2) × 180° = 540°00′00″
observed sum = 540°00′15″
misclosure = +15″
correction per angle = −15″ / 5 = −3″
tolerance = 30″√5 = 67″; 15″ < 67″, accept| Station | Observed angle | Correction | Adjusted angle |
|---|---|---|---|
| A | 94°44′00″ | −3″ | 94°43′57″ |
| B | 117°37′40″ | −3″ | 117°37′37″ |
| C | 103°06′05″ | −3″ | 103°06′02″ |
| D | 105°50′10″ | −3″ | 105°50′07″ |
| E | 118°42′20″ | −3″ | 118°42′17″ |
| Sum | 540°00′15″ | −15″ | 540°00′00″ |
Azimuths around the traverse
Azimuths are carried from the fixed direction of A-B using the adjusted angles. At each station the forward azimuth is the back azimuth of the previous course plus the angle turned right, which is the same as the previous forward azimuth plus the angle plus 180 degrees, reduced into the range 0 to 360.
The computation must return to the starting azimuth after going all the way round. That closure is not an accident and it is not a second check on the angles: it is guaranteed by the fact that the angles were adjusted to sum correctly. What it does catch is an arithmetic slip in the running of the azimuths, which is why it is always worked.
azimuth(next) = azimuth(previous) + angle right at the common station + 180°
A-B = 114°30′00″ (given)
B-C = 114°30′00″ + 117°37′37″ + 180° = 52°07′37″
C-D = 52°07′37″ + 103°06′02″ + 180° = 335°13′39″
D-E = 335°13′39″ + 105°50′07″ + 180° = 261°03′46″
E-A = 261°03′46″ + 118°42′17″ + 180° = 199°46′03″
back to A-B = 199°46′03″ + 94°43′57″ + 180° = 114°30′00″ (closes)Latitudes and departures
For a course of length D on azimuth α, the latitude is D cos α and the departure is D sin α, with the azimuth measured clockwise from north. Latitude is the northing component, positive north; departure is the easting component, positive east.
On a closed loop both columns must sum to zero, because the traverse returns to where it started. What they actually sum to are the closures in latitude and departure, and their resultant is the linear misclosure.
| Course | Azimuth | Bearing | Distance (ft) | Latitude (ft) | Departure (ft) |
|---|---|---|---|---|---|
| A-B | 114°30′00″ | S 65°30′00″ E | 434.24 | −180.076 | 395.142 |
| B-C | 52°07′37″ | N 52°07′37″ E | 398.95 | 244.921 | 314.920 |
| C-D | 335°13′39″ | N 24°46′21″ W | 358.09 | 325.138 | −150.046 |
| D-E | 261°03′46″ | S 81°03′46″ W | 450.30 | −69.955 | −444.833 |
| E-A | 199°46′03″ | S 19°46′03″ W | 340.14 | −320.096 | −115.037 |
| Sums | 1981.72 | −0.069 | 0.147 |
Misclosure and precision
The linear misclosure is the resultant of the two column sums. Precision is expressed as the ratio of that misclosure to the total length of the traverse, written 1 to X with X rounded down, because claiming 1:10,000 for a traverse that computes 1:9,999.6 overstates the work.
At 1:12,235 this traverse meets the 1:10,000 specification. Note the direction of the misclosure, 295°10′, which is not close to the azimuth or back azimuth of any single course. That is what random error looks like; a closure line lying along one course would point at a distance blunder in it.
ΣLat = −0.0689 ft, ΣDep = +0.1466 ft (the column sums carried to four places)
e = √((−0.0689)² + (0.1466)²) = √(0.004747 + 0.021492) = √0.026239 = 0.1620 ft
ΣD = 1981.72 ft
precision = ΣD / e = 1981.72 / 0.16197 = 12,235
precision = 1:12,235, which meets 1:10,000
azimuth of the closure line = 295°10′Compass-rule adjustment
The compass, or Bowditch, rule distributes each closure in proportion to course length. It rests on the assumption that angles and distances are of comparable precision, which is the usual case for a traverse turned with a theodolite and measured with an EDM.
The correction to a latitude is the negative of the latitude closure times that course's share of the perimeter, and likewise for departures. The corrections must sum to the negative of the closures, which is the arithmetic check on the whole column.
Corrections are shown to 0.001 ft because at 0.01 ft several of them would round to the same value and the proportionality would be invisible.
correction to latitude of course i = −ΣLat × (Dᵢ / ΣD) = +0.069 × (Dᵢ / 1981.72)
correction to departure of course i = −ΣDep × (Dᵢ / ΣD) = −0.147 × (Dᵢ / 1981.72)| Course | Distance (ft) | Lat correction (ft) | Dep correction (ft) | Adjusted latitude (ft) | Adjusted departure (ft) |
|---|---|---|---|---|---|
| A-B | 434.24 | +0.015 | −0.032 | −180.061 | 395.109 |
| B-C | 398.95 | +0.014 | −0.030 | 244.935 | 314.891 |
| C-D | 358.09 | +0.012 | −0.027 | 325.151 | −150.072 |
| D-E | 450.30 | +0.016 | −0.033 | −69.939 | −444.866 |
| E-A | 340.14 | +0.012 | −0.025 | −320.085 | −115.062 |
| Sums | 1981.72 | +0.069 | −0.147 | 0.000 | 0.000 |
Coordinates
Coordinates accumulate from the held station: each northing is the previous northing plus the adjusted latitude of the connecting course, and each easting the previous easting plus the adjusted departure.
Running the last course from E must land exactly back on A, to the last digit carried. If it does not, the adjusted columns do not sum to zero and the adjustment arithmetic is wrong.
| Station | Northing (ft) | Easting (ft) |
|---|---|---|
| A | 5000.00 | 5000.00 |
| B | 4819.94 | 5395.11 |
| C | 5064.87 | 5710.00 |
| D | 5390.02 | 5559.93 |
| E | 5320.08 | 5115.06 |
| A (closing) | 5000.00 | 5000.00 |
Answer
- Angular misclosure +15″, corrected by −3″ at each of the five angles.
- Azimuths: A-B 114°30′00″, B-C 52°07′37″, C-D 335°13′39″, D-E 261°03′46″, E-A 199°46′03″.
- Latitude closure −0.069 ft, departure closure +0.147 ft, linear misclosure 0.162 ft over a 1981.72 ft perimeter.
- Precision 1:12,235, which meets the 1:10,000 specification.
- Adjusted coordinates: A 5000.00 / 5000.00; B 4819.94 / 5395.11; C 5064.87 / 5710.00; D 5390.02 / 5559.93; E 5320.08 / 5115.06.
- Area enclosed by the adjusted traverse: 264,208 sq ft, or 6.0654 acres.
Check
Re-inverse the adjusted coordinates and compare with the observed distances. From A to B: ΔN = −180.06, ΔE = 395.11, distance 434.20 ft against an observed 434.24 ft, a difference of 0.04 ft, which is the adjustment that course received. No course should differ from its observation by more than the corrections applied to it.
Sum the adjusted latitude and departure columns: both are 0.000 ft. This is the direct proof that the corrections were computed and applied correctly.
Run the closing course independently: from E at 5320.08 / 5115.06, adding the adjusted latitude −320.085 and departure −115.062 gives 5000.00 / 5000.00, which is station A held.
Recompute the azimuth of course E-A from the adjusted coordinates: atan2(−115.06, −320.08) = 199°46′20″, against the observed 199°46′03″ carried round from the angles. The 17 second difference is the rotation the compass rule applied to that course, and it should be small; a large one would mean the adjustment had been misapplied.
More traverse adjustment
All in this category- Transit-rule adjustment compared with the compass ruleThe same five-sided loop is readjusted by the transit rule, which distributes the closure in proportion to each latitude and departure rather than to course length, and the two sets of coordinates are compared.
- Closed-link traverse between two pairs of control monumentsA four-course traverse runs from one control pair to another. Azimuth closure comes from the two published control azimuths and position closure from the published coordinates of the far monument.
- A traverse that fails its precision specification, and what to do about itA four-sided traverse closes at 1:2,538 against a 1:10,000 requirement. The direction of the closure line points straight at the course carrying the blunder, which is remeasured and the traverse recomputed.