3. Angles and Directions

a. Angles to Directions

Starting with a direction for one traverse line, directions of the others can be computed from the horizontal angles linking them. The process of addition or subtraction is dependent on the type of horizontal angle (interior, deflection, etc), turn direction (clockwise or counterclockwise), and direction type (bearing or azimuth).

(1) Examples - Bearing

Example 1

The bearing of line GQ is S 42°35' E. The angle right at Q from G to S is 112°40'. What is the bearing of the line QS?

Sketch:

;

Figure C-10

 

Add meridian at Q and label angles:

;

Figure C-11

 

At Q, the bearing to G is N 42°35' W.

Subtracting 42°35' from 112°40' gives the angle, β, from North to the East for line QS.

Bearing QS = N 70°05" E

Example 2

The bearing of line LT is N 35°25' W. The angle left at T from L to D is 41°12'. What is the bearing of the line TD?

Sketch:

 Figure C-12

 

Add meridian at T and label angles

 Figure C-13

 

At T the bearing TL is S 35°25' E.

The bearing angle, σ, is 35°25' + 41°12' = 76°37'

Bearing TD = S 76°37' E.

(2) Examples - Azimuth

Example 1

The azimuth of line WX is 258°13'. At X the deflection angle from W to L is 102°45' L. What is the azimuth of line XL?

Sketch:

Figure C-14

 

A deflection angle is measured from the extension of a line. The azimuth of the  extension is the same as that of the line. To compute the next azimuth, the deflection angle is added directly to the previous azimuth.

Because this is a left deflection angle, you would add a negative angle.    

Add meridian at X:

Figure C-15

 

Azimuth XL = 258°13' + (-102°45') = 155°28'

Example 2

The azimuth of line BP is 109°56'. The angle right at P from B to J is 144°06'. What is the azimuth of line PJ?

Sketch:

Figure C-16

 

At P, add the meridian and extend the line BP

Figure C-17

 

To get the azimuth of line PB: 109°56' + 180°00'= 289°56'.

Since 144°06' is to the right (+), add it to the azimuth of PB to compute azimuth of PJ: 144°06' + 289°56' = 434°02'

Figure C-18

 

Why is the azimuth greater than 360°? Because we've gone past North.

To normalize the azimuth, subtract 360°00': 434°02' - 360°00' = 74°02'

 

b. Directions to Angles

Given directions of two adjacent lines, it is a simple matter to determine the angle between the lines.

(1) Example - Bearings

The bearing of line HT is N 35°16' W , the bearing of line TB is N 72°54' E. What is the angle right at T from B to H?

Sketch:

Figure C-19

Label the back-direction at T and angle to be computed, δ.

Figure C-20

 

Based on the sketch, the desired angle is what’s left over after both bearing angles are subtracted from 180°00'.

δ = 180°00' - (72°54' + 35°16') = 71°50"

(2) Example - Azimuths

The azimuth of line MY is 106°12', the azimuth of line YF is 234°06'. What is the angle right at Y from F to M?

Sketch:

Figure C-21

 

Label the back-azimuth at Y and angle to be computed, ρ.

Figure C-22

 

ρ = 286°12' - 234°06' = 52°06'

 

c. Traverse Direction Computations

(1) Example - Bearings

Given the following traverse and horizontal angles:

Figure C-23
Bearing Traverse

 

Using a bearing of N 36°55' E for line AB, determine the bearings of the remaining lines clockwise around the traverse.

Many survey texts use a tabular approach to compute traverse line directions from angles. For the beginner this can be confusing and lead to erroneous directions. It helps to instead draw sketches in order to visualize the line relationships.

At point B:  

Figure C-24

Label the bearing angle, 36°55', from B to A.

Subtract it from 117°19'.

β = 117°19' - 36°55' = 80°24'Brg BC = S 80°24' E

 

At point C:

 

Figure C-25

Label the bearing angle, 80°24', from C to B.

Subtract it along with 87°42' from 180°00' tooo get bearing angle CD.

γ = 180°00' - (80°24' + 87°42') = 11°54'

Brg CD = S 11°54' W

 

At point D:

 

Figure C-26

Label the bearing angle, 11°54', from D to C.

Subtract it from 93°38' to obtain next bearing angle.

&eta= 93°38' - 11°54' = 81°44'Brg DA = N 81°44' W

 

The directions for all four traverse lines have been computed. Angles at B, C, and D have been used, but that at A has not. For a math check, use the Bearing of DA and the angle at A to compute the bearing we started with.

Figure C-27

Label the bearing angle, 81°44', from A to D.

Subtract it along with 61°21' from 180°00' to get next bearing angle.

α = 180°00' - (61°21' + 81°44') = 36°55

Brg AB = N 36°55' E  check

 

If our computed and initial bearings for AB don’t match it means one of two things:

        • there is a math error in our computations, or,
        • the interior angles weren’t balanced.

For this traverse the angles sum to 360°00' so there is no angular misclosure. If our math check had failed it would have been due to a math error in our computations.

Had the angles not been balanced and if there were no math errors, the math check would be off by the angular misclosure.

Summary:

Line Bearing
AB N 36°55' E
BC S 80°24' E
CD S 11°54' W
DA N 81°44' W

 

(2) Example - Azimuths

Given the following traverse and horizontal angles:

Figure C-28
Azimuth Traverse

 

Using a azimuth of 68°00' for line OP, determine the azimuths of the remaining lines counter-clockwise around the traverse. 

 

At point P:

 

Figure C-29

Line PQ is 92°48' right of Azimuth PO

Az PO is the back azimuth of Az OP

Az PQ = (Az OP + 180°00') + Angle PAz PQ

= (68°00' + 180°00') + 92°48' = 340°48'

 

At point Q:

 

Figure C-30

Line QR is 112°26' to the right from Az QP

Az QP is the back azimuth of Az PQ

Az QR = (Az PQ + 180°00') + Angle QAz QR

= (340°48'+180°00') + 112°26' = 633°14'

Normalize: Az QR = 633°14' - 360°00' = 273°14'

 

At point R:

 

Figure C-31

Line RO is 67°14' right from Az RQ

Az RQ is the back azimuth of Az QR

Az RO = (Az QR + 180°00') + Angle RAz RO

 = (273°14' + 180°00') + 67°14' = 520°28'

Normalize: Az RO = 520°28' - 360°00' = 160°28'

 

The directions for all four traverse lines have been computed. Angles at P, R, and R have been used, but not the angle at O. For a math check, use Azimuth RO and the angle at O to compute the original Az OP.

 

At point O:

 

c 25

Figure C-32

Line OP is 87°32' right of Az OR

Az OR is the back azimuth of Az RO

Az OP = (Az RO + 180°00') + Angle O

Az OP = (160°28' + 180°00') + 87°32' = 428°00'

Normalize: Az OP = 428°00' - 360°00' = 68°00' check!

 Summary:

Line Bearing
OP 68°00'
PQ 340°48'
QR 273°14'
RO 160°28'

 

There's a distinct pattern computing these azimuths:

    New Az = (Previous Az + 180°00') + Angle.

This is true for a loop traverse meeting these conditions:

        • Directions are counter-clockwise around the traverse, and,
        • Angles are interior to the right.

What if the directions are clockwise around the traverse and interior angles counter-clockwise?

    New Az = (Previous Az - 180°00') - Angle

Other patterns exist for clockwise travel with clockwise interior angles, clockwise exterior angles, counterclockwise exterior angles, etc. Rather than memorize the possible patterns, draw a sketch, and begin computing; the pattern will present itself after a few lines.