Compass Surveying

Problem 1:

The following bearings are taken on a closed compass traverse. Assume the observed bearing of line AB to be correct.
a. Compute the error in closure.
b. Compute the corrected bearing of the lines.

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Problem 2:

In the deflection angle traverse with a transit survey data below, find the corrected bearing of the lines. Bearing of line AB is N51º37'W.

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Problem 3:

An engineer's notebook gives the observed magnetic bearings of the following traverse. Compute the corrected bearings of the lines.

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Problem 4:

Traverse Bearings

Line	Forward Bearing	Backward Bearing
AB	N 30°30′ W	S 32°15′ E
BC	S 80°45′ W	N 82°45′ E
CD	S 53°00′ W	N 50°15′ E
DE	S 13°00′ W	N 11°30′ E
EA	N 66°30′ E	S 69°30′ W

Solution:
Compute the Central Angle
A = 180° − 30°30′ − 69°30′ = 80°00′
B = 32°15′ + 80°45′ = 113°00′
C = 180° − 82°45′ + 53°00′ = 150°15′
D = 180° − 50°15′ + 13°00′ = 142°45′
E = 66°30′ − 11°30′ = 55°00′

Sum of Interior Angles = (5 − 2)180° = 540°
Σ angles = 80° + 113° + 150°15′ + 142°45′ + 55° = 541°
Error = 541° − 540° = 1° (too big --> subtract the error on each side)
Correction = 1º / 5 = 12′

Adjusted Interior Angles

A′ = 80°00′ − 12′ = 79°48′
B′ = 113°00′ − 12′ = 112°48′
C′ = 150°15′ − 12′ = 150°03′
D′ = 142°45′ − 12′ = 142°33′
E′ = 55°00′ − 12′ = 54°48′

Forward and Back Azimuths

Line	Forward Azimuth	Backward Azimuth	Difference
AB	149°30′	327°45′	178°15′
BC	80°45′	262°45′	182°00′
CD	53°00′	230°15′	177°15′
DE	13°00′	191°30′	178°30′
EA	246°30′	69°30′	177°00′

In choosing the best line, we select the value nearest to 180º and take the difference. Reference line = DE (nearest to 180°, difference = 1°30′ from 180º-178º30'=1º30'). We treat this as the "best line", but since the forward and backward bearings of DE are not intially equal, we add one half of the difference to the azimuth of line DE. Here, we "add" since the nearest difference, 178º30', is lesser than 180º. In the same way, we must subtract half of the difference if the nearest difference is more than 180º.

Therefore,
Half of 1°30′ = 0°45′.
Corrected Azimuth of DE = 13° + 0°45′ = 13°45′.

Adjusted Bearings and Azimuths

Line	Bearing	Azimuth
DE	S 13°45′ W	13°45′
EA	13°45′ + 54°48′ = N 68°33′ E	248°33′
AB	180° − 68°33′ − 79°48′ = N 31°39′ W	148°21′
BC	112°48′ − 31°39′ = S 81°09′ W	81°09′
CD	150°03′ + 81°09′ − 180° = S 51°12′ W	51°12′

Problem 5:

Given the following deflection angles of a closed traverse, compute the bearing of all the lines if the bearing of AB is S40ºE.

The corrected deflection angles are shown below. By analyzing the observed values, it will be identified that the sum of the left deflection angles must be reduced and the sum of the right deflection angles must be increased.

In other words, the summation of the left deflection angles is larger than the summation of the right deflection angles, and since the total angle exceeds 360º, we need to subtract the correction for the larger summation (left) and add the correction to the lesser summation (right).