Traverse adjustment
Closed-link traverse between two pairs of control monuments
A 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.
Given
- A link traverse runs MON-15, 101, 102, 103, MON-22. It begins on the control pair MON-14 / MON-15 and closes on the control pair MON-22 / MON-23.
- Published coordinates: MON-14 N 12480.55 E 8210.33; MON-15 N 12905.12 E 8588.47; MON-22 N 13980.44 E 10255.60; MON-23 N 14320.00 E 10690.25. All US survey feet.
- Angles right, from the back station to the fore station: at MON-15 (back MON-14) 196°56′05″; at 101 180°24′45″; at 102 177°16′30″; at 103 178°21′20″; at MON-22 (fore MON-23) 177°20′10″.
- Distances: MON-15 to 101, 470.36 ft; 101 to 102, 524.83 ft; 102 to 103, 504.74 ft; 103 to MON-22, 484.96 ft.
- Specification: 1:15,000 linear, and 10 seconds times the square root of the number of angles for azimuth closure.
- Coordinates to 0.01 ft; latitudes and departures carried to 0.001 ft.
Required
- The two control azimuths, by inverse.
- The azimuth misclosure and the adjusted azimuths of the four traverse courses.
- The position misclosure at MON-22, the precision, and the compass-rule adjusted coordinates of 101, 102 and 103.
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
Control azimuths by inverse
A link traverse has no interior-angle condition to satisfy: it is not a closed polygon. Its angular check is instead the azimuth it delivers at the far end, compared with the azimuth the far control pair already defines.
So the first job is to inverse both control pairs. The plane inverse uses the surveying convention, azimuth = atan2(ΔE, ΔN), with northing as the first coordinate.
MON-14 to MON-15: ΔN = 424.57, ΔE = 378.14
distance = √(424.57² + 378.14²) = 568.55 ft
azimuth = atan2(378.14, 424.57) = 41°41′23″
MON-22 to MON-23: ΔN = 339.56, ΔE = 434.65
distance = √(339.56² + 434.65²) = 551.56 ft
azimuth = atan2(434.65, 339.56) = 52°00′08″Carry the azimuths and find the misclosure
Start from the known azimuth MON-14 to MON-15 and run the observed angles right through all five occupied stations. The fifth angle, turned at MON-22 to MON-23, delivers a computed value for the closing control azimuth.
The difference between the computed and published closing azimuths is the azimuth misclosure. It is distributed equally over the angles, one correction per angle, because every angle is equally likely to carry the error.
azimuth(next) = azimuth(previous) + angle right + 180°
MON-15 to 101 = 41°41′23″ + 196°56′05″ + 180° = 58°37′28″
101 to 102 = 58°37′28″ + 180°24′45″ + 180° = 59°02′13″
102 to 103 = 59°02′13″ + 177°16′30″ + 180° = 56°18′43″
103 to MON-22 = 56°18′43″ + 178°21′20″ + 180° = 54°40′03″
MON-22 to MON-23 = 54°40′03″ + 177°20′10″ + 180° = 52°00′13″
published = 52°00′08″; azimuth misclosure = +5″
tolerance = 10″√5 = 22″; 5″ passes
correction per angle = −5″ / 5 = −1″Adjusted azimuths
Applying −1 second to each angle shifts the first course by 1 second, the second by 2 seconds, and so on, because the corrections accumulate down the line. That cumulative pattern is what an equal angular adjustment always produces on a link traverse.
Re-running the chain with the adjusted angles must deliver exactly 52°00′08″ at the end, and it does.
| Station | Observed angle right | Adjusted angle | Course | Adjusted azimuth |
|---|---|---|---|---|
| MON-15 | 196°56′05″ | 196°56′04″ | MON-15 to 101 | 58°37′27″ |
| 101 | 180°24′45″ | 180°24′44″ | 101 to 102 | 59°02′11″ |
| 102 | 177°16′30″ | 177°16′29″ | 102 to 103 | 56°18′40″ |
| 103 | 178°21′20″ | 178°21′19″ | 103 to MON-22 | 54°39′59″ |
| MON-22 | 177°20′10″ | 177°20′09″ | MON-22 to MON-23 | 52°00′08″ |
Latitudes, departures and the position closure
On a loop traverse the latitude and departure columns must sum to zero. On a link traverse they must instead sum to the coordinate difference between the two terminal control stations. That difference is the condition the traverse has to satisfy.
Compute the required difference from the published coordinates first, then compare it with what the traverse produced.
required ΔN = 13980.44 − 12905.12 = 1075.32 ft
required ΔE = 10255.60 − 8588.47 = 1667.13 ft
closure in latitude = 1075.355 − 1075.320 = +0.035 ft
closure in departure = 1667.223 − 1667.130 = +0.093 ft
e = √(0.0351² + 0.0926²) = 0.0991 ft
precision = 1984.89 / 0.09905 = 20,038, that is 1:20,038, which meets 1:15,000| Course | Adjusted azimuth | Distance (ft) | Latitude (ft) | Departure (ft) |
|---|---|---|---|---|
| MON-15 to 101 | 58°37′27″ | 470.36 | 244.893 | 401.579 |
| 101 to 102 | 59°02′11″ | 524.83 | 270.022 | 450.039 |
| 102 to 103 | 56°18′40″ | 504.74 | 279.971 | 419.975 |
| 103 to MON-22 | 54°39′59″ | 484.96 | 280.470 | 395.630 |
| Sums | 1984.89 | 1075.355 | 1667.223 | |
| Required from control | 1075.320 | 1667.130 | ||
| Closure | 0.035 | 0.093 |
Compass-rule adjustment and coordinates
The compass rule works exactly as on a loop: each correction is the negative of the closure times that course's share of the total length. The only difference is that the closure was measured against the control rather than against zero.
Accumulate the adjusted latitudes and departures from the held coordinates of MON-15. The last course must land on the published coordinates of MON-22 exactly.
| Course | Lat correction (ft) | Dep correction (ft) | To station | Northing (ft) | Easting (ft) |
|---|---|---|---|---|---|
| MON-15 to 101 | −0.008 | −0.022 | 101 | 13150.00 | 8990.03 |
| 101 to 102 | −0.009 | −0.025 | 102 | 13420.02 | 9440.04 |
| 102 to 103 | −0.009 | −0.024 | 103 | 13699.98 | 9859.99 |
| 103 to MON-22 | −0.009 | −0.023 | MON-22 | 13980.44 | 10255.60 |
Answer
- Control azimuths: MON-14 to MON-15 = 41°41′23″ over 568.55 ft; MON-22 to MON-23 = 52°00′08″ over 551.56 ft.
- Azimuth misclosure +5″, adjusted by −1″ per angle. Adjusted course azimuths: 58°37′27″, 59°02′11″, 56°18′40″, 54°39′59″.
- Position closure: +0.035 ft in latitude and +0.093 ft in departure, a linear misclosure of 0.099 ft over 1984.89 ft.
- Precision 1:20,038, which meets the 1:15,000 specification.
- Adjusted coordinates: 101 at N 13150.00 E 8990.03; 102 at N 13420.02 E 9440.04; 103 at N 13699.98 E 9859.99.
Check
The adjusted traverse must land on the published far control exactly. Accumulating the four adjusted latitudes and departures from MON-15 gives N 13980.44 and E 10255.60, which are the published coordinates of MON-22 to the hundredth.
Re-running the azimuth chain with the adjusted angles delivers 52°00′08″ at MON-22 to MON-23, matching the azimuth inversed from the published coordinates of MON-22 and MON-23.
Independent inverse over the whole traverse: from MON-15 to MON-22 the published coordinates give √(1075.32² + 1667.13²) = 1983.84 ft on azimuth 57°10′39″. The traverse ran 1984.89 ft along a path that bows slightly off that straight line, so the traverse length must exceed the straight inverse. It does, by 1.05 ft.
A link traverse has no built-in closure. Both checks here are against external control, which is exactly the reason a link traverse must always terminate on a second known pair and never simply stop.
More traverse adjustment
All in this category- Closed-loop traverse from field angles to compass-rule coordinatesA 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.
- 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.
- 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.