Solution: Traverse Adjustment
Problem (2)
These are the answers, computations are left to the reader.
| Unadjusted | Adjusted | |||||
| Line | Direction | Length (ft) | Latitude | Departure | Latitude | Departure |
| A-B | 226°54'23" | 639.15 | -436.662 | -466.732 | -436.631 | -466.761 |
| B-C | 00°00'00" | 523.69 | +523.690 | 0.000 | +523.715 | -0.023 |
| C-D | 235°02'29" | 409.54 | -234.660 | -335.645 | -234.640 | -335.663 |
| D-A | 79°35'02" | 815.93 | +147.517 | +802.484 | +147.556 | +802.447 |
| sums: | 2388.31 | -0.115 | +0.107 | 0.000 | 0.000 | |
(a) Compute linear closure and precision
LC = 0.157
Prec = 1/15,200
(b) Adjust the latitudes and departures using the Compass Rule
See above table for answers and math checks
(c) Compute the adjusted lengths and directions
| Adjusted | ||
| Line | Direction | Length |
| A-B | 226°54'36.6" | 639.150 |
| B-C | 359°59'50.8" | 523.715 |
| C-D | 235°02'42.5" | 409.544 |
| D-A | 79°34'50.5" | 815.901 |