Traverse adjustment
Closed traverse computed from deflection angles
A five-sided loop observed with deflection angles rather than interior angles. The closure condition becomes 360 degrees rather than (n − 2) times 180, and azimuths advance by simple addition.
Given
- A closed five-sided loop A-B-C-D-E-A observed by deflection angles, all turned to the right.
- Observed deflection angles: at B 57°02′00″ R, at C 66°16′30″ R, at D 84°42′15″ R, at E 58°18′50″ R, at A 93°40′40″ R.
- Distances: A-B 443.14, B-C 491.50, C-D 420.72, D-E 504.66, E-A 528.08 ft.
- Azimuth of A-B fixed at 28°18′00″ from a gyro observation.
- Station A held at N 6000.00, E 2000.00.
- Angles turned with a 20 second instrument; the working angular tolerance is 30 seconds times the square root of the number of angles.
- Specification 1:10,000.
Required
- The angular closure condition for deflection angles, and the misclosure.
- The adjusted deflection angles and the resulting azimuths.
- Latitudes, departures, misclosure, precision and adjusted coordinates.
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 deflection-angle closure condition
A deflection angle is turned from the prolongation of the back course to the forward course, positive to the right. Going once around a closed loop, the direction of travel must return to where it started, so the deflection angles have to sum to one full revolution.
For a simple closed figure traversed consistently in one direction, that condition is 360°00′00″ with right deflections counted positive and left deflections negative. It replaces the (n − 2) × 180° condition that interior angles satisfy, and it does not depend on the number of sides.
The observed sum exceeds 360 degrees by 15 seconds, which is well within the 30″√5 = 67 second tolerance, so the misclosure is spread equally.
required sum of right deflections = 360°00′00″
observed sum = 57°02′00″ + 66°16′30″ + 84°42′15″ + 58°18′50″ + 93°40′40″ = 360°00′15″
misclosure = +15″
tolerance = 30″√5 = 67″; 15″ passes
correction per angle = −15″ / 5 = −3″Azimuths by simple addition
This is where deflection angles earn their keep. Because the angle is measured from the prolongation of the back course, the forward azimuth is just the back azimuth plus the deflection, with no 180 degrees to add and subtract.
Right deflections add, left deflections subtract. Reduce into the range 0 to 360 as needed, and the chain must return to the starting azimuth after the last angle.
azimuth(next) = azimuth(previous) + deflection right
A-B = 28°18′00″ (given)
B-C = 28°18′00″ + 57°01′57″ = 85°19′57″
C-D = 85°19′57″ + 66°16′27″ = 151°36′24″
D-E = 151°36′24″ + 84°42′12″ = 236°18′36″
E-A = 236°18′36″ + 58°18′47″ = 294°37′23″
back to A-B = 294°37′23″ + 93°40′37″ = 388°18′00″ − 360° = 28°18′00″ (closes)| Station | Observed deflection | Correction | Adjusted deflection | Course | Azimuth |
|---|---|---|---|---|---|
| B | 57°02′00″ R | −3″ | 57°01′57″ R | B-C | 85°19′57″ |
| C | 66°16′30″ R | −3″ | 66°16′27″ R | C-D | 151°36′24″ |
| D | 84°42′15″ R | −3″ | 84°42′12″ R | D-E | 236°18′36″ |
| E | 58°18′50″ R | −3″ | 58°18′47″ R | E-A | 294°37′23″ |
| A | 93°40′40″ R | −3″ | 93°40′37″ R | A-B | 28°18′00″ |
| Sum | 360°00′15″ | −15″ | 360°00′00″ |
Latitudes, departures and misclosure
From here the computation is identical to any other closed traverse: resolve each course, sum the two columns, and take the resultant.
The precision of 1:15,037 meets the specification.
e = √(0.1489² + 0.0553²) = √(0.022171 + 0.003058) = 0.1588 ft
precision = 2388.10 / 0.1588 = 15,037, that is 1:15,037| Course | Azimuth | Distance (ft) | Latitude (ft) | Departure (ft) |
|---|---|---|---|---|
| A-B | 28°18′00″ | 443.14 | 390.175 | 210.087 |
| B-C | 85°19′57″ | 491.50 | 39.995 | 489.870 |
| C-D | 151°36′24″ | 420.72 | −370.109 | 200.062 |
| D-E | 236°18′36″ | 504.66 | −279.935 | −419.903 |
| E-A | 294°37′23″ | 528.08 | 220.023 | −480.061 |
| Sums | 2388.10 | 0.149 | 0.055 |
Adjustment and coordinates
Compass-rule corrections in proportion to length, accumulated from station A.
The traverse encloses 377,309 square feet, or 8.662 acres.
| Course | Lat correction (ft) | Dep correction (ft) | To station | Northing (ft) | Easting (ft) |
|---|---|---|---|---|---|
| A-B | −0.028 | −0.010 | B | 6390.15 | 2210.08 |
| B-C | −0.031 | −0.011 | C | 6430.11 | 2699.94 |
| C-D | −0.026 | −0.010 | D | 6059.98 | 2899.99 |
| D-E | −0.031 | −0.012 | E | 5780.01 | 2480.07 |
| E-A | −0.033 | −0.012 | A | 6000.00 | 2000.00 |
Deflection angles against interior angles
Deflection angles are natural for route surveys, where the traverse follows an alignment and the crew is always looking ahead down the line. The instrument is plunged, so the angle is turned from the prolongation directly.
They carry one hazard that interior angles do not. The direction, right or left, is part of the observation and it is easy to book wrong. A deflection recorded as right when it was left produces an error of twice the angle, which on this traverse would be well over a hundred degrees at station D and would be obvious. A small deflection booked with the wrong hand is far more dangerous, because it produces a modest error that may still close within a loose tolerance.
The relationship to interior angles is simple for a convex figure traversed in this direction: deflection = 180° minus the interior angle. At station B the interior angle would be 180° − 57°01′57″ = 122°58′03″.
deflection at B = 57°01′57″ R → interior angle at B = 180° − 57°01′57″ = 122°58′03″
sum of interior angles = (5 − 2) × 180° = 540°00′00″
sum of deflections = 5 × 180° − 540° = 360°00′00″, which is the condition used aboveAnswer
- Deflection angles must sum to 360°00′00″. The observed sum is 360°00′15″, a misclosure of +15″, corrected by −3″ at each of the five angles.
- Azimuths: A-B 28°18′00″, B-C 85°19′57″, C-D 151°36′24″, D-E 236°18′36″, E-A 294°37′23″.
- Latitude closure +0.149 ft, departure closure +0.055 ft, linear misclosure 0.159 ft over 2388.10 ft.
- Precision 1:15,037, which meets the 1:10,000 specification.
- Adjusted coordinates: A 6000.00 / 2000.00; B 6390.15 / 2210.08; C 6430.11 / 2699.94; D 6059.98 / 2899.99; E 5780.01 / 2480.07.
- Enclosed area 377,309 square feet, or 8.662 acres.
Check
The azimuth chain must close. Adding the last adjusted deflection to the azimuth of E-A gives 294°37′23″ + 93°40′37″ = 388°18′00″, which reduces to 28°18′00″, the fixed azimuth of A-B.
Convert the adjusted deflections to interior angles and test the other condition: 122°58′03″, 113°43′33″, 95°17′48″, 121°41′13″ and 86°19′23″ sum to 540°00′00″ = (5 − 2) × 180°. The two closure conditions are equivalent, and satisfying both confirms the angle arithmetic.
The adjusted latitude and departure columns sum to 0.000, and the traverse closes on station A at 6000.00 / 2000.00.
Inverse a coordinate pair independently: from A at 6000.00 / 2000.00 to B at 6390.15 / 2210.08 gives 443.11 ft on azimuth 28°18′02″, against an observed 443.14 ft on an adjusted azimuth of 28°18′00″. Both differences are the adjustment applied to that course, and both are small, as they must be on a traverse closing at 1:15,000.
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.
- 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.