Traverse Computations MLS

Example Traverse B 110° 05' 00" C 111° 25' 30" 65° 33' 30" A 72° 54' 30" D Balancing Angles Surveyors usually mea

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Example Traverse B 110° 05' 00"

C 111° 25' 30"

65° 33' 30"

A

72° 54' 30"

D

Balancing Angles Surveyors usually measure interior angles of a polygon, know the bearing or azimuth of one of its sides and need to calculate the azimuths or bearings of all the sides of the traverse. This is a very common problem that all students must be capable of solving. As such, consider the following example of data and try to solve for the azimuths of all the sides:

Equal Corrections • Issue

• Distribute closure

Station

Measured Angle

Correction

Adjusted Angle

A

65° 33' 30"

00"

65° 33' 30"

B

110° 05' 00"

30"

110° 05' 30"

C

111° 25' 30"

30"

111° 26' 00"

D

72° 54' 30"

30"

72° 55' 00"

359° 58' 30"

1' 30"

360° 00' 00"

SUM

Example Traverse: Balanced Angles B 110° 05' 30"

C 111° 26' 00"

65° 33' 30"

A

72° 55' 00"

D

Directions of Lines • Use adjusted angles to compute line direction

• Azimuths or bearings?

BEARING Vary from 0 to 90 degrees

From North and South Clockwise or counter clockwise Two letters + numerical value AZIMUTH Vary from 0 to 360 degrees

From North or south Clockwise or counter clockwise Numerical value

Bearing

Azimuth

Azimuths are called forward or back azimuths to represent, for example, the azimuths of AB and BA (see figure). The back azimuth differs from the forward azimuth by 360º on a grid map.

Computing Directions B 110° 05' 30"

C

A

Compute Directions Clockwise Around Traverse B 110° 05' 30"

C Assume or find azimuth of AB 30° 15' 30" Compute azimuth of BA (azimuth of AB  180°) 210° 15' 30"

A

Subtract angle at B 210° 15' 30" - 110° 05' 30" = 100° 10' 00"

Directions of Traverse Lines Station

Line

Angle

AB

B

Azimuth

Bearing

30° 15' 30"

N 30° 15' 30" E

100° 10' 00"

S 79° 50' 00" E

168° 44' 00"

S 11° 16' 00" E

275° 49' 00"

N 84° 11' 00" W

30° 15' 30"

N 30° 15' 30" E

110° 05' 30" BC

C

111° 26' 00" CD

D

72° 55' 00" DA

A

65° 33' 30"

AB

EXERCISES 1. If the azimuth of a line is in the second quadrant, then: a. Bearing = S (180 - azimuth) E b. Bearing = S (azimuth + 180) E c. Bearing = S (azimuth - 180) W d. Bearing = S (180 + azimuth) W 2. The azimuth of AB is 231o19’, and the angle at B is 134o42’ to the left. Find the bearings

of BA and CB. 3. The azimuth of BC is 98o24’, and the angle at C is 92o35’ to the left. Find the azimuths of

CB, CD and DC.

Directions of Traverse Lines B

C

A D

Departures and Latitudes • Departures (deps)

• Latitudes (lats)

Computing Departures & Latitudes • Compute by:

Dep = L sin 

Lat = L cos 

• Where:

 = azimuth

L = length of line

North (+Y) 

Lat. FG = L cos

G L

F

East (+X) Dep. FG = L sin

Example Traverse B

C

A D

Example Traverse Departure = Length x sin (azimuth) e.g. Dep. AB = 30.141 m x sin(30° 15' 30")

Latitude = Length x cos (azimuth) e.g. Lat. AB = 30.141 m x cos (30° 15' 30") Line

Distance

Azimuth

Departure

Latitude

AB

30.141 m

30° 15' 30"

15.1880 m

26.0347 m

BC

38.576 m

100° 10' 00"

37.9703 m

-6.8091 m

CD

25.605 m

168° 44' 00"

5.0026 m

-25.1116 m

DA

58.437 m

275° 49' 00"

-58.1361 m

5.9223 m

0.0248 m

0.0363 m

Sum

152.759 m

Closure in Deps and Lats • For mathematically closed traverse

• Geometrically closed traverse

• Closure: difference between known/computed position – Linear error of closure (LEC) – Relative error of closure (REC)

Linear Error of Closure dep = 0.0248 m

A' lat = 0.0363 m 0.044 m LEC  (dep)2  (lat)2

A

e.g. LEC  (0.0248)2  (0.0363)2  0.044 m

LEC 0.044 m 1 1 REC     perimeter 152.759m 3471.8 3470

Traverse Adjustment • Goal • Some methods – Arbitrary method – Compass (Bowditch) rule

– Least squares adjustment

Compass Rule Adjustment • Application • Works for traverses with limited number of lines

lat correction line length  lat closure traverse length dep correction line length  dep closure traverse length

Compass Rule • Proportion is rearranged for computational efficiency lat closure lat correction   line length traverse length dep correction 

dep closure  line length traverse length

Compass Rule – Balance Departures Departure correction   dep.corr 

departure closure  length perimeter

- (0.0248m) 152.759m

 length - 0.00016235 length

Corrected departure = Departure + Departure correction e.g. Corrected departure AB = 15.1880 + (-0.0049) = 15.183 m Line

Distance

Departure

Departure Correction

Corrected Departure

AB

30.141 m

15.1880 m

-0.0049 m

15.183 m

BC

38.576 m

37.9703 m

-0.0063 m

37.964 m

CD

25.605 m

5.0026 m

-0.0042 m

4.998 m

DA

58.437 m -58.1361 m

-0.0095 m

-58.145 m

-0.0248 m

0.000 m

Sum 152.759 m

0.0248 m

Compass Rule – Balance Latitudes Latitude correction  lat.corr 

latitude closure  length perimeter

- (0.0363) 152.759m

 length - 0.00023763 length

Corrected latitude = Latitude + Latitude correction e.g. Corrected latitude AB = 26.0347 + (-0.0071) = 26.028 m Line

Distance

Latitude

Latitude Correction

Corrected Latitude

AB

30.141 m

26.0347 m

-0.0071 m

26.028 m

BC

38.576 m

-6.8091 m

-0.0092 m

-6.818 m

CD

25.605 m -25.1116 m

-0.0061 m

-25.118 m

DA

58.437 m

5.9223 m

-0.0139 m

5.908 m

Sum 152.759 m

0.0363 m

-0.0363 m

0.000 m

Calculate Coordinates XB  X A  Departure AB  10000  15.183  10015.183m YB  YA  Latitude AB  5000  26.027  5026.027m

XC  XB  Departure BC  10015.183 37.964  10053.147m YC  YC  Latitude BC  5026.027 6.818  5019.209m Station A

Line Departure

Latitude

AB

26.027 m

15.183 m

B

BC

37.964 m

A

5026.027 m

10053.147 m

5019.209 m

10058.146 m

4994.092 m

10000.000 m

5000.000 m

4.998 m -25.118 m

D DA

10015.183 m

-6.818 m

C CD

X Coordinate Y Coordinate 10000.000 m 5000.000 m

-58.146 m

5.908 m

Adjusted Azimuths and Lengths North (+Y) 

Lat. FG

G

L

F L = length of line FG L  dep2  lat 2

 = azimuth of line FG  dep  α  tan    lat  1

Dep. FG

East (+X)

Calculating Azimuths and Lengths L = length of line JK North (+Y)

L  dep2  lat 2

 = azimuth of line JK

Dep. JK

 dep  α  tan    lat  1

J L K



East (+X) Lat. JK

Compute Adjusted Azimuths/Lengths Station

Line

Distance

Azimuth

A AB

30.132 m

A

10015.183 m

5026.027 m

10053.147 m

5019.209 m

10058.146 m

4994.092 m

10000.000 m

5000.000 m

25.610 m 168° 44' 40"

D DA

5000.000 m

38.571 m 100° 10' 55"

C CD

10000.000 m 30° 15' 25"

B BC

X Coordinate Y Coordinate

58.445 m 275° 48' 10"