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"