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COMPUTING MISSING
DATA IN A POLYGON
Robert Burtch
Surveying Engineering
Ferris State University
TRAVERSE WITH MISSING
ELEMENTS
= Traverse Station
= Distance of line = Line number
= Azimuth of line
BC
15 EA
A
SEA
S 1
E (^) S (^) DE DE
14 D
2
11
1 12
13 S^2
C
BC
11
S B
S
A
E
A
15
11
14 D'
C' 12
13
B'
TRAVERSE WITH MISSING
ELEMENTS
3 forms of missing data problem
- Missing distance and direction of same
line – trivial problem
- Missing 2 distances of 2 adjacent lines, or 2 directions of 2 adjacent lines, or
distance and direction of 2 adjacent lines
- 2 unknown distances or directions on 2
non-adjacent lines or unknown distance on one line and unknown direction on
another non-adjacent line
MISSING DATA ON 2
ADJACENT LINES
1
(^2 )
4
5
807.38' 92 41' 52"
171 20' 40"345.91' 340 27' 18"327.76'
o
o o
EXAMPLE ---MISSING DATA ON 2
ADJACENT LINES
4 1748.89 428.
327.76 340 O^ 27’ 18”
3 1858.54 120.
807.38 92 O^ 41’ 52”
2 1052.06 158.
345.91 171 O^ 20’ 40”
1 1000.00 500.00 Assumed
STADISTANCE AZIMUTH X Y COMMENTS
- Compute distance and azimuth of line
- Solve of missing elements in triangle
( ) ( ) ( ) ( )
( ) ( )
D X X Y Y
D
Az
X X Y Y
Az o
4 1 1 4
2 1 4
2 2 2
4 1
4 1
1 1 4 1 4
1
4 1
1000 00 1748 89 500 00 428 90
752 26
1000 00 1748 89 500 00 428 90
275 25 24
−
−
−
− −
−
= − + − = − + −
=
=
− −
=^
− −
=
.... . '
tan tan
.. ..
' "
Solve for interior angles
( ) [( ) ( )]
α
α
β
β
γ α β
γ
= − = −
=
= − = −
=
= − + = − +
=
− −
− −
Az Az
Az Az
o o
o
o o
o
o o o o
o
1 4 1 5
4 5 4 1
95 25 24 33 47 32
61 37 52
314 5118 275 25 24
39 25 54
180 180 61 37 52 39 25 54
78 56 14
' " ' "
' "
' " ' "
' "
' " ' "
' "
- Using sine law, distances can be
computed
D 4 − 5 D 1 5− D1 4−
sin α sin β sinγ
D
D
D
D
D
D
o
o
o
o
1 5
1 4
1 5
4 5
1 4
4 5
−
−
−
−
−
−
sin
sin
sin ' "
sin ' "
sin
sin
sin ' "
sin ' "
γ
β
γ
α
Use sine law to compute distances
sin 4441 ' 24 " sin 3707 ' 23 "
sin sin
D
D
sin 9811 ' 13 " sin 3707 ' 23 "
sin sin
D
D
o o
AB AC
o o
AB BC
β
γ
α
γ
Check results
0.00 0.
1
1034.52 2 O^ 21’ 43” +1033.64 +42.
5
602.53 245 O^ 11’ 32” -252.80 -546.
4
567.92 199 O^ 41’ 19” -534.72 -191.
3
735.04 145 O^ 14’ 20” -603.86 +419.
2
452.17 37 O^ 42’ 17” +357.74 +276.
1
STA DISTANCE AZIMUTH LATITUDE DEPARTURE
ANALYTICAL APPROACH
- Exploit fact that in closed loop traverse
that the sum of latitudes and
departures is zero
- Write 2 equations with 2 unknowns
- Equations solved simultaneously
- Unknowns in 3 categories
- 2 unknowns both lengths of traverse
- 1 length unknown and 1 direction of
another traverse course
- Direction of 2 lines unknowns
CATEGORY 1
- Write equations
- Rearrange 2
nd
equation
D Az D Az x
D Az D Az y
1 1 2 2
1 1 2 2
0
0
sin sin
cos cos
∆
∆
D
y D Az
Az
2
1 1
2
=
− ∆ − cos
cos
CATEGORY 1
Recognizing:
1 2 1 2
2 2 1
1 1 2 2 1 1 2 2
2
2 1 1 2 1 1
sinAzcosAz cosAzsinAz
ysinAz xcosAz D
DsinAzcosAz ysinAz DcosAzsinAz xcosAz 0
x 0 cosAz
ysinAz DcosAzsinAz DsinAz
−
∆ −∆
−∆ − +∆ =
+∆ =
∆ + −
sin( x − y )= sin x cos y −cos x siny
( 2 1 )
2 2 1 sinAz Az
ysinAz xcosAz D −
∆ −∆
CATEGORY 1 EXAMPLE
Given data:
Line Azimuth Distance
1 36
o 42’ 25” 468.38’
2 97
o 34’ 01” D 2
3 193
o 02’ 56” 723.00’
4 222
o 15’ 08” D 4
5 346
o 28’ 20” 967.30’
CATEGORY 1 EXAMPLE
Difference in azimuths from 2 to 4:
Az 4 - Az 2 = 124 o^ 41’ 07”
∆ ∆ ∆ ∆ ∆ ∆ x D Az D Az D Az
x
x
y D Az D Az D Az
y
y
o o o
o o o
= + +
= + +
= −
= + +
= + +
=
1 1 3 3 5 5
1 1 3 3 5 5
468 38 36 42 25 723 00 193 02 56 967 30 246 28 20
109 547
468 38 36 42 25 723 00 193 02 56 967 30 246 28 20
611635
sin sin sin
.. 'sin ' ". ' sin ' ". 'sin ' "
. '
cos cos cos
.. 'cos ' ". ' cos ' ". ' cos ' "
. '
CATEGORY 1 EXAMPLE
- The distance D 2 is
- Alternatively, can also compute D 2 as
( )
( )
- 75 '
sin 12441 ' 07 "
- 635 'sin 22215 ' 08 " 109. 547 'cos 22215 ' 08 " sinAz Az
ysinAz xcosAz D (^) o
o o
4 2
4 4 2
=
−−
−
∆ −∆
( ) ( ) ( ) ( )
D
y Az x Az
Az Az Az Az o o
o o o o
2
4 4
4 2 4 2
611 635 222 15 08 109 547 222 15 08
222 15 08 97 34 01 222 15 08 97 34 01
598 75
=
−
−
=
− −
−
=
∆ sin ∆cos
sin cos cos sin
. sin ' ". cos ' "
sin ' " cos ' " cos ' " sin ' "
. '
CATEGORY 2
- 4 can be factored out of quadratic
equation leading to this solution
where:
D
V V U
V V U
1
2
2
U x y D and
V xM yN
= + −
= +
∆ ∆
∆ ∆
2 2 2
2 ,
CATEGORY 2 EXAMPLE
Line Azimuth Distance
1 36
o 42’ 25” 468.38’
2 97
o 34’ 01” D 2
3 193
o 02’ 56” 723.00’
4 Az 4 719.
5 346
o 28’ 20” 967.30’
CATEGORY 2 EXAMPLE
- Solution
- Distance:
- Ignore second distance
( 109. 547 ) sin 9734 ' 01 "( 611. 635 ) cos 9734 ' 01 " 189. 13588
V xsinAz ycosAz
- 547 611. 635 719. 80 132 , 014. 1216
U x y D
o o
2 2
2 2 2
2 4
2 2
=− + = −
=∆ +∆
=− + − =−
=∆ +∆ −
75 'or 220. 48 '
13588 189. 13588 132 , 014. 1216
D V V U
2
2 2
= −
=−− ± − −−
=− ± −
CATEGORY 2 EXAMPLE
- Solve for azimuth of line 4
- NW or SW quadrant?
- Good drawing: 222 o^ 15’ 08”
Az
D M x D
o
o
4
1 2
4
1 598 75^ 97 34 01^ 109 547 719 80
42 15 08
=
+
=^
− − −
= −
− − sin sin
. sin ' ".
.
' "
∆
CATEGORY 2 EXAMPLE
SW QUADRANT RESULT
Travers e Adjus tm ent Program Azim uth Decim al Sta Dis t Deg Min Sec Degrees Departure Latitude 1 468.38 36 42 25 36.70694 279.96 375. 2 598.75 97 34 1 97.56694 593.54 -78. 3 723.00 193 2 56 193.04889 -163.24 -704. 4 719.80 222 15 8.7 222.25242 -483.99 -532. 5 967.30 346 28 20 346.47222 -226.27 940. 1 3477.23 0.00 0.
CATEGORY 2 EXAMPLE
NW QUADRANT RESULT
Travers e Adjus tm ent Program Azim uth Decim al Sta Dis t Deg Min Sec Degrees Departure Latitude 1 468.38 36 42 25 36.70694 279.96 375. 2 598.75 97 34 1 97.56694 593.54 -78. 3 723.00 193 2 56 193.04889 -163.24 -704. 4 719.80 317 44 51.3 317.74758 -483.99 532. 5 967.30 346 28 20 346.47222 -226.27 940. 1 3477.23 0.00 1065.
CATEGORY 2 ALTERNATIVE
- Multiply 1 st^ and 2 nd^ equation in slide 16
by -sin Az 1 and by cos Az 1 respectively
− − − =
D Az Az D Az Az x Az
D Az Az D Az Az y Az
1 1 1 2 2 1 1
1 1 2 2 1 1
0
1 0
sin cos cos sin sin
sin cos sin cos cos
∆
∆
( )
2
2 2
2 2 2 2 2 2
2
1
W 1 W
1 1
W W W W 1 cosAz
− ν
− ν± ν + −
+ν
− ν± ν − + ν − −ν
CATEGORY 3
- Square both equation on slide 16
- Add equations
D Az D Az x
D Az D Az y
1
2 2 1 2 2
2
1
2 2 1 2 2
2
sin sin
cos cos
( ) ( )
( ) ( )
D D Az x D Az y
D Az D Az x x D Az D Az y y
D D x Az y Az x y
1
2 2 2
2 2 2
2
2
2 2 2 2 2
2 2
2 2 2 2 2
2
2
2 2 2 2
2 2
2 2
2
= + + +
= + + + + +
= + + + +
sin cos
sin sin cos cos
sin cos
∆ ∆
∆ ∆ ∆ ∆
∆ ∆ ∆ ∆
CATEGORY 3 ALTERNATIVE
where
cosAz
y K x y x K
x y
x
1
2 2 2 2
2 2
=
− ± + −
+
∆ ∆ ∆ ∆
∆ ∆
∆
K
x y D D
xD
2 2 2
2 1
2
CATEGORY 3 EXAMPLE
o
Az 719.
4
o
2 Az 2 598.75’
1 36 o^ 42’ 25” 468.38’
Line Azimuth Distance
CATEGORY 3 EXAMPLE
( ) ( ) ( ) ( ) ( )( )
J
x y D D x D
=
− + − + − = −
∆ ∆ ∆
2 2 2
2 4
2
4
2 2 2 2
2
109 547 611635 598 75 719 80 2 109 547 719 80 3 460336
.... .. .
−
−
cos
x
x y
yJ x y xJ Az cos
1
2 2
2 2 22 1 4
Solution set of Az 2 = 97° 34’ 00.2” and Az 1 =
Travers e Calculation Program Azim uth Decim al Sta Dis t Deg Min Sec Degrees Departure Latitude 1 468.38 36 42 25 36.70694 279.961 375. 2 598.75 97 34 0.2 97.56672 593.536 -78. 3 723.00 193 2 56 193.04889 -163.241 -704. 4 719.80 222 15 7.6 222.25211 -483.989 -532. 5 967.30 346 28 20 346.47222 -226.268 940. 1 3477.23 0.000 0.
Solution set of Az 2 = 242° 07’ 28.8” and Az 1 =
Travers e Calculation Program Azim uth Decim al Sta Dis t Deg Min Sec Degrees Departure Latitude 1 468.38 36 42 25 36.70694 279.961 375. 2 598.75 242 7 28.8 242.12467 -529.275 -279. 3 723.00 193 2 56 193.04889 -163.241 -704. 4 719.80 117 26 21.4 117.43928 638.822 -331. 5 967.30 346 28 20 346.47222 -226.268 940. 1 3477.23 0.000 0.
