ENGINEERING SURVEYING
(221 BE)
TRAVERSE
Sr Dr. Tan Liat Choon
Email: tanliatchoon@gmail.com
Mobile: 016-4975551
MEASUREMENT SEQUENCE
C
B
D
A
E
2
COMPUTATION SEQUENCE
1. Calculate angular misclose
2. Adjust angular misclose
3. Calculate adjusted bearings
4. Reduce distances for slope etc…
5. Compute (∆E, ∆N) for each traverse line
6. Calculate linear misclose
7. Calculate accuracy
8. Adjust linear misclose
3
CALCULATE INTERNAL ANGLES
Point
Foresight Backsight
Azimuth
Azimuth
Internal
Angle
Adjusted
Angle
A
21o
118o
97o
B
56o
205o
149o
C
168o
o
o
232
64
At each point :
D
232o
• oMeasure foresight
azimuth
352
120o
E
303o
Σ =(n-2)*180
Misclose
Adjustment
• Measure backsight azimuth
48
105o
• o Calculate internal
angle (back-fore)
For example, at B :
• Azimuth to C = 56o
• Azimuth to A = 205o
• Angle at B = 205o - 56o = 149o
4
CALCULATE ANGULAR MISCLOSE
Point
Foresight Backsight
Azimuth
Azimuth
Internal
Angle
A
21o
118o
97o
B
56o
205o
149o
C
168o
232o
64o
D
232o
352o
120o
E
303o
48o
105o
Σ =(n-2)*180
Adjusted
Angle
535o
Misclose
-5o
Adjustment
-1o
5
CALCULATE ADJUSTED ANGLES
Point
Foresight Backsight
Azimuth
Azimuth
Internal
Angle
Adjusted
Angle
A
21o
118o
97o
98o
B
56o
205o
149o
150o
C
168o
232o
64o
65o
D
232o
352o
120o
121o
E
303o
48o
105o
106o
535o
540o
Σ =(n-2)*180
Misclose
-5o
Adjustment
-1o
6
COMPUTE ADJUSTED AZIMUTHS
• Adopt a starting azimuth
• Then, working clockwise around the traverse :
Calculate reverse azimuth to backsight (forward azimuth
±180o)
Subtract (clockwise) internal adjusted angle
Gives azimuth of foresight
• For example (azimuth of line BC)
Adopt azimuth of AB
Reverse azimuth BA (=23o+180o)
Internal adjusted angle at B
Forward azimuth BC (=203o-150o)
23o
203o
150o
53o
7
COMPUTE ADJUSTED AZIMUTHS
C
B
150
o
Line
Forward
Azimuth
Reverse
Azimuth
Internal
Angle
AB
23o
203o
150o
BC
53o
D
CD
DE
EA
A
AB
E
8
COMPUTE ADJUSTED AZIMUTHS
C
65o
B
Line
Forward
Azimuth
Reverse
Azimuth
Internal
Angle
AB
23o
203o
150o
BC
53o
233o
65o
CD
168o
D
DE
EA
A
AB
E
9
COMPUTE ADJUSTED AZIMUTHS
C
B
121o
Line
Forward
Azimuth
Reverse
Azimuth
Internal
Angle
AB
23o
203o
150o
BC
53o
233o
65o
CD
168o
348o
121o
DE
227o
D
EA
A
AB
E
10
COMPUTE ADJUSTED AZIMUTHS
C
B
Line
Forward
Azimuth
Reverse
Azimuth
Internal
Angle
AB
23o
203o
150o
BC
53o
233o
65o
CD
168o
348o
121o
DE
227o
47o
106o
D
A
106o
E
EA
-59o
301o
AB
11
COMPUTE ADJUSTED AZIMUTHS
C
B
Line
Forward
Azimuth
Reverse
Azimuth
Internal
Angle
AB
23o
203o
150o
BC
53o
233o
65o
CD
168o
348o
121o
DE
227o
47o
106o
EA
301o
121o
98o
AB
23o (check)
D
A
98o
E
12
(∆
∆E,∆
∆N) FOR EACH LINE
• The rectangular components for each line are
computed from the polar coordinates (β,d)
∆E = d sin β
∆N = d cos β
• Note that these formulae apply regardless of
the quadrant so long as whole circle bearings
are used
13
VECTOR COMPONENTS
Line
Azimuth
Distance
∆E
∆N
AB
23o
77.19
30.16
71.05
BC
53o
99.92
79.80
60.13
CD
168o
60.63
12.61
-59.31
DE
227o
129.76
-94.90
-88.50
EA
301o
32.20
-27.60
16.58
(399.70)
(0.07)
(-0.05)
Σ
14
LINEAR MISCLOSE & ACCURACY
• Convert the rectangular misclose
components to polar coordinates
−1  ∆E 
β = tan 

 ∆N 
2
Beware of quadrant when
calculating β using tan-1
2
d = ∆E + ∆N
• Accuracy is given by
1 : (traverse length / linear misclose )
15
EXAMPLE
• Misclose (∆
∆E, ∆N)
(0.07, -0.05)
• Convert to polar (β
β,d)
β = -54.46o (2nd quadrant) = 125.53o
d = 0.09 m
• Accuracy
1:(399.70 / 0.09) = 1:4441
16
BOWDITCH ADJUSTMENT
• The adjustment to the easting component
of any traverse side is given by:
d∆E = ∆Emisc * side length/total perimeter
• The adjustment to the northing component
of any traverse side is given by:
d∆N = ∆Nmisc * side length/total perimeter
17
BOWDITCH ADJUSTMENT
• The adjustment to the easting component
of any traverse side is given by:
∆Eadj = ∆E +/- d∆E
• The adjustment to the northing component
of any traverse side is given by:
∆Nadj = ∆N +/- d∆N
18
EXAMPLE
•
•
•
•
•
•
•
•
East misclose
North misclose
Side AB
Side BC
Side CD
Side DE
Side EA
Total perimeter
0.07 m
–0.05 m
77.19 m
99.92 m
60.63 m
129.76 m
32.20 m
399.70 m
19
VECTOR COMPONENTS
(PRE-ADJUSTMENT)
Side
∆E
∆N
AB
30.16
71.05
BC
79.80
60.13
CD
12.61
-59.31
DE
-94.90
-88.50
EA
-27.60
16.58
Misc
(0.07)
(-0.05)
d∆
∆E
d∆
∆N
∆Eadj
∆Nadj
20
THE ADJUSTMENT COMPONENTS
Side
∆E
∆N
d∆
∆E
d∆
∆N
AB
30.16
71.05
0.014
-0.010
BC
79.80
60.13
0.016
-0.012
CD
12.61
-59.31
0.011
-0.008
DE
-94.90
-88.50
0.023
-0.016
EA
-27.60
16.58
0.006
-0.004
Misc
(0.07)
(-0.05)
(0.070)
(-0.050)
∆Eadj
∆Nadj
21
ADJUSTED VECTOR COMPONENTS
Side
∆E
∆N
d∆
∆E
d∆
∆N
∆Eadj
∆Nadj
AB
30.16
71.05
0.014
-0.010
30.146
71.060
BC
79.80
60.13
0.016
-0.012
79.784
60.142
CD
12.61
-59.31
0.011
-0.008
12.599
-59.302
DE
-94.90
-88.50
0.023
-0.016
-94.923
-88.484
EA
-27.60
16.58
0.006
-0.004
-27.606
16.584
Misc
(0.07)
(-0.05)
0.070
-0.050
(0.000)
(0.000)
22
COMPUTING FORWARD AND BACKWARD
BEARING WITH ONE FIXED LINE
C
N
B
N
A
23
COMPUTING FORWARD AND BACKWARD
BEARING WITH ONE FIXED LINE
C
B
B
E
N
F
A
24
COMPUTING FORWARD AND BACKWARD
BEARING WITH ONE FIXED LINE
Line
Forward Azimuth
Backward Azimuth
Internal Angle
AB
41° 35’ (Known)
221° 35’
129° 11’
BC
350° 46’
170° 46’
88° 35’
CD
259° 21’
79° 21’
132° 30’
DE
211° 51’
31° 51’
135° 42’
EF
167° 33’
347° 33’
118° 52’
FA
106° 25’
286° 25’
115° 10’
AB
41° 35’ (Check)
25
COMPUTING FORWARD AND BACKWARD
BEARING WITH ONE FIXED LINE
+
+
+
+
41° 35’ = AB
180° 00’
221° 35’ = BA
129° 11’
350° 46’ = BC
180° 00’
170° 46’ = CB
88° 35’
259° 21’ = CD
180° 00’
79° 21’ = DC
132° 30’
211° 51’ = DE
211° 51’ = DE
- 180° 00’
31° 51’ = ED
+ 135° 42’
167° 33’ = EF
+ 180° 00’
347° 33’ = FE
+ 118° 52’
466° 25’ - 360°
= 106° 25’ = FA
+ 180° 00’
286° 25’ = AF
+ 115° 10’
401° 35’ - 360°
= 41° 35’ = AB
When a computed azimuth exceeds 360°, the correct azimuth
26
is obtained by merely subtracting 360°
BALANCE INTERIOR ANGLES
Before the areas of a piece of land can be computed, it
is necessary to have a closed traverse (loop)
The interior angles of a closed traverse should total:
•
•
•
•
∑ = (n – 2) * 180
Total correction = ∑ -total angles in the traverse
Each line correction = Total correction / Number of sides
Where, n is the number of sides of the traverse
27
BALANCE INTERIOR ANGLES
Site
Measured Interior Angle
Correction
Adjusted Angle
AB
100°
45’
37”
- 2”
100°
45’
35”
BC
231°
23’
43”
- 2”
231°
23’
41”
CD
17°
12’
59”
- 2”
17°
12’
57”
DE
89°
03’
28”
- 2”
89°
03’
26”
EA
101°
34’
23”
- 2”
101°
34’
21”
540°
00’
10’
-10”
540°
00’
00
∑ = (n – 2) * 180 = (5 – 2) * 180 = 540°
Total correction = ∑ - Total angles of traverse = 540° - 540° 00’ 10’
= - 00° 00’ 10’
Each line correction = Total correction / Number of sides = - 00° 00’ 10’ / 5
= - 00° 00’ 02”
28
ERROR IN LATITUDE AND DEPARTURE
When latitudes are added together, the resulting error
is called the error in latitudes
The error resulting from adding departures together is
called the error in departures
Because of the errors in latitudes and departures, the
azimuth and distance have error, therefore, the traverse
will not close
29
MISCLOSURE IN LATITUDE AND
DEPARTURE
Because of errors in the observation traverse and
distances. The linear error of misclosure (e) represents
the distance from the actual location of point 1 to the
computed location of point 1
e = √ (Departure misclosure) 2 + (Latitude misclosure) 2
e = √ (∑E)2 + (∑N)2
30
MISCLOSURE IN LATITUDE AND
DEPARTURE
The relative precision of a traverse is expressed by a fraction that has the
linear misclosure as its numerator and the traverse perimeter of total
length as its denominator, or
Relative precision = linear misclosure / traverse length
Fractional Linear Misclosure (FLM) = 1 in D/e,
Where, D = total distance of survey
e = linear misclosure
Acceptable FLM values:
•
1 in 5000 for most engineering surveys
•
1 in 10000 for control for large projects
•
1 in 20000 for major works and monitoring for structural
deformation etc.
31
MISCLOSURE IN LATITUDE AND
DEPARTURE
Azimuth
Site
Degree Minute Second
Length (m)
Uncorrected
Departure
(∆E)
Uncorrected
Latitude
(∆N)
AB
00°
00’
00”
638.57
0.000
638.570
BC
306°
12’
51’
1576.10
- 1271.620
931.168
CD
195°
54’
06’
3824.10
- 1047.754
- 3677.764
DA
47°
44’
33”
3133.72
2319.361
2107.313
9172.49
∑E = - 0.013
∑N = - 0.713
e = √ (∑E)2 + (∑N)2
e = √ (-0.013)2 + (-0.713)2
e = 0.713 m
Fractional Linear Misclosure (FLM) = 1 in D/e
FLM = 1 in 9172.49/0.713
FLM = 1 in 12865
∆E = d sin β
∆N = d cos β
32
TRAVERSE ADJUSTMENT
There are two methods of
misclosure adjustment:
• Bowditch Adjustment
• Transit Adjustment
33
BOWDITCH ADJUSTMENT
The adjustment to easting component of any
traverse side is given by:
Eadj = Length of line (P12) * [Total departure
misclosure (∑Emisc) / total traverse length]
The adjustment to northing component of any traverse side
is given by:
Nadj = Length of line (P12) * [Total latitude
misclosure (∑ Nmisc)/ total traverse length]
34
BOWDITCH ADJUSTMENT
Site
Length (m)
Uncorrected
Departure
Uncorrected
Latitude
Adjustment
Departure
Adjustment
Latitude
AB
638.57
0.000
638.570
+ 0.001
+ 0.050
BC
1576.10
- 1271.620
931.168
+ 0.002
+ 0.123
CD
3824.10
- 1047.754
- 3677.764
+ 0.006
+ 0.297
DA
3133.72
2319.361
2107.313
+ 0.004
+ 0.243
9172.49
∑E = - 0.013
∑N = - 0.713
+ 0.013
+ 0.713
Adjustment departure AB = 638.57 * [0.013 / 9172.59]
= + 0.001
Adjustment latitude AB = 638.57 * [0.713 / 9172.59]
= + 0.050
Because of the uncorrected total departure/latitude are minus (-),
so, the adjustment must plus (+)
35
TRANSIT ADJUSTMENT
The adjustment to easting component of any
traverse side is given by:
Eadj = Total departure misclosure (∑Emisc) * length of line/
total traverse length
The adjustment to northing component of any traverse side is
given by:
Nadj = Total latitude misclosure (∑Nmisc) * length of line /
total traverse length
36
TRANSIT ADJUSTMENT
Site
Length (m)
Uncorrected
Departure
Uncorrected
Latitude
Adjustment
Departure
Adjustment
Latitude
AB
638.57
0.000
638.570
+ 0.001
+ 0.050
BC
1576.10
- 1271.620
931.168
+ 0.002
+ 0.123
CD
3824.10
- 1047.754
- 3677.764
+ 0.006
+ 0.297
DA
3133.72
2319.361
2107.313
+ 0.004
+ 0.243
9172.49
∑E = - 0.013
∑N = - 0.713
+ 0.013
+ 0.713
Adjustment departure AB = 0.013 * [638.57 / 9172.59]
= + 0.001
Adjustment latitude AB = 0.713 * [638.57 / 9172.59]
= + 0.050
Because of the uncorrected total departure/latitude are minus (-),
plus (+)
so, the adjustment must
37
TRANSIT ADJUSTMENT
Site
Uncorrected
Departure
Uncorrecte
d Latitude
Adjustment
Departure
Adjustment
Latitude
Corrected
Departure
Corrected
Latitude
AB
0.000
638.570
+ 0.001
+ 0.050
0.001
638.620
BC
- 1271.620
931.168
+ 0.002
+ 0.123
- 1271.618
931.291
CD
- 1047.754
- 3677.764
+ 0.006
+ 0.297
- 1047.748
- 3677.467
DA
2319.361
2107.313
+ 0.004
+ 0.243
2319.365
2107.556
∑E = - 0.013
∑N = - 0.713
+ 0.013
+ 0.713
0.000
0.000
38
T h a n k Yo u
&
Question And Answer
39