Transcript No Slide Title

Control
We have started toHorizontal
compute the
traverse station
coordinates from an anticlockwise
TRAVERSE polygon traverse.
We tested the sum of the observed angles against the
sum of the internal angles of the polygon, and if within an
acceptable value we distributed the misclosure.
We established the bearings of all the traverse lines
and then calculated the changes in Easting and Northing
for each line.
The algebraic sum of these changes should be ZERO.
The values give the error eE and the error eN.
The linear misclosure e =  (eE2 + eN2) was calculated and
used to find the FLM. If acceptable then distribute the errors.
By : a)
b)
Bowditch Method - proportional to line distances
Transit Method - proportional to DE and DN values
c)
Numerous other methods including Least Squares
Adjustments
a)
Bowditch Method - proportional to line distances
The eE and the eN have to be distributed
For any line IJ the adjustments are dE IJ and dN IJ
dE IJ = [ eE / SD ] x D IJ
dN IJ = [ eN / SD ] x D IJ
Applied with the opposite
sign to eE
Applied with the opposite
sign to eN
CO-ORDINATE DIFFERENCES
WHOLE
HORIZONTAL
CIRCLE
DISTANCE
CALCULATED
BEARING
q
D
DE
DN
00
00
00
638.57
306
12
51
1576.20
-1271.701
+931.227
195
54
06
3824.10
-1047.754
-3677.764
47
44
33
3133.72
+2319.361 +2107.313
3
9172.59 3
0.000
+638.570
-0.094
eE
-0.654
eN
e =  (eE2 + eN2) =  (0.0942 + 0.6542) = 0.661m
Fractional Linear Misclosure (FLM) = 1 in SD / e
= 1 in 9172.59 / 0.661 = 1 in 13500
Check 2
dE IJ = [ eE / SD ] x D IJ
eE = -0.094m
Applied with the opposite
sign to eE
SD = 9172.59 m
dE IJ = [+0.094 / 9172.59 ] x D IJ = +0.0000102479…... x D IJ
For line AB
Store this in the memory
dE AB = +0.0000102479…x D AB = +0.0000102479…x 638.57
For line BC
dE AB = +0.007m
dE BC = +0.0000102479…x D BC = +0.0000102479…x 1576.20
For line CD
dE CD = +0.039m
dE BC = +0.016m
For line DA
dE DA = +0.032m
CO-ORDINATE DIFFERENCES
CO-ORDINATES
CALCULATED
DE
ADJUSTMENTS
DN
dE
0.000
+638.570 +0.007
-1271.701
+931.227 +0.016
-1047.754 -3677.764 +0.039
+2319.361 +2107.313 +0.032
-0.094
-0.654
eE
eN
dN
ADJUSTED
DE
DN
E
N
S
T
A
T
I
O
N
dN IJ = [ eN / SD ] x D IJ
Applied with the opposite
sign to eN
eN = -0.654m
dN IJ = [+0.654 / 9172.59 ] x D IJ = +0.000071299…... x D IJ
Store this in the memory
dN AB = + 0.000071299… x D AB = + 0.000071299…x 638.57
dN AB = +0.046m
dN BC = +0.112m
dN CD = +0.273m
dN DA = +0.223m
CO-ORDINATE DIFFERENCES
CO-ORDINATES
CALCULATED
DE
ADJUSTMENTS
DN
dE
dN
ADJUSTED
DE
0.000
+638.570 +0.007 +0.046 +0.007
-1271.701
+931.227 +0.016 +0.112 -1271.685
DN
N
E
S
T
A
T
I
O
N
3000.00
4000.00 A
3000.01
4638.62 B
+931.339 1728.32
5569.96 C
+638.616
-1047.754 -3677.764 +0.039 +0.273 -1047.715 -3677.491
680.61
1892.46 D
+2319.361 +2107.313 +0.032 +0.223 +2319.393 +2107.536
3000.00
4000.00 A
-0.094
-0.654
eE
eN
S= 0
S= 0
Check 3