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Traverse Calculations
Traversing Methods
• Angle and Distance (Total Station Surveys)
• Direction and Distance (Compass-tape
Surveys)
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Traversing Methods
• Angle and Distance
(Total Station
Surveys)
Traversing Methods
• Direction and Distance
(Compass-tape
Surveys)
3
Traversing Methods
• Direction and Distance
(Electronic
equipment)
Traverse Types
• Loop
– Geometrically closed
– Mathematically closed (checks possible)
• Connecting
– Geometrically open
– Mathematically closed (checks possible)
• Open
– Geomet. and Math. open (no check)
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Loop Traverse
• Geometrically closed
• Mathematically closed (checks possible)
Connecting Traverse
• Geometrically open
• Mathematically closed (checks possible)
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Open Traverse
Geometrically open
Mathematically open
Overview of Calculations
• Distance reduction
• Angle closure and angle adjustment
• Calculate direction for each side
• Calculate Latitude and Departure
• Calculate Linear error of closure (LEC) and
Relative error of closure (REC)
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Overview (cont.)
• Adjust latitudes and departures (traverse
adjustment)
• Compute coordinates (Northings (y) and
Eastings (x) )
• Compute Area
• Inverse Problem - Given the coordinates of
two points find, dir. and dist. between them
Distance Reduction
HD = SD⋅ Cos(α ) = SD⋅ Sin(ZA)
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Perform Angle Closure
• Closure = ( meas. or calc.) – (fixed or
known)
• Closure should meet standard of accuracy
required for traverse specification being
followed. See ALTA-ACSM specifications
for boundary traverses.
Angle Closure (cont.)
• If the angle closure fails to meet the
standard of accuracy, a gross blunder or an
unaccounted for systematic error is
suspected among the measurements.
• In either case the measurement with the
blunder or the error should be isolated and
replaced by remeasurement before
continuing with the calculations.
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Perform Angle Adjustment
• Starting point is correction per angle.
• CA= correction per angle = - Closure ÷ n
n = number of angles
• The CA should be an integer number. Use
integer arithmetic to calculate it.
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Angle Adjustment (cont.)
• Adjust for Horizon Closure or Calculate Mean Int.
• Mean Int. = Int. + CA
Angle Adjustment (cont.)
CA = - (+95”) ÷ 2 = - 48”
CA = - (+15”) ÷ 2 = - 07”
3 26° 53’ 35’ 333°06’15” 359°59’50” - 10” + 05”
2 83° 16’ 55” 276°43’20” 360°00’15” + 15” - 07”
1 69° 49’ 30” 290° 12’05” 360°01’35” + 95” - 48”
Sta Int. Ext. Sum Clos CA
Field Angles
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Angle Adjustment (cont.)
• Mean interior = Int. + CA
Sum = 179°59’10”
3 26° 53’ 35 + 05” 26° 53’40”
2 83° 16’ 55” -07” 83°16’ 48”
1 69° 49’ 30” - 48” 69°48’ 42”
Sta. field int. CA Mean int.
Angle Adjustment (cont.)
• Calc. Adj. Int. Angle or adjust for geometric sum
• Closure = 179° 59’ 10” – 180° 00’ 00” = - 50”
• CA = - (-50”) ÷ 3 (do division longhand, find
remainder and adjust that many (remainder) angles by
one second more ; prevents round off error)
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Angle Adjustment (cont.)
3 26° 53’40” + 16” 26° 53’56”
2 83°16’ 48” + 17” 83°17’ 05”
1 69°48’ 42” + 17” 69°48’ 59”
Sta. Mean int. CA Adj. int.
Adjusted Interior = Mean int. + CA
Check Sum = 180° 00’ 00”
Calculate the Azimuth of each
side of the traverse
Key points
•Use the adjusted or balanced angles
•Requires knowledge of the traverse configuration or the
direction of travel around the traverse. I.e., is the order of
stations clockwise or counter clockwise as you proceed
from station to station?
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Calculate the Azimuth of each
side of the traverse (cont.)
Key points cont.
•Requires knowledge of different horizontal angles - int.,
ext., deflection angle, angle-to-the-right, etc.
•Meridians are parallel
•Line intersecting parallel lines creates equal alternate
interior angles ( see Geometry Review)
Geometry Review
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Review (cont.)
Review (cont.)
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Calculate the Azimuth of each
side - using Adjusted Int. Angles
Find the Azimuth of side 2-3
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Find the Azimuth of side 3-1
Calculate azimuth - check 1 -2
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Calculate Latitude and Departure
• Latitude = lat = HD · Cos(β)= HD · Cos( Az)
• Departure = dep = HD · Sin (β) = HD· Sin (Az)
Sign Convention
Latitudes and Departures (cont.)
Connecting Σ lat = ΔN Σ dep = ΔE
Loop Σ lat = 0 Σ dep = 0
latitudes departures
Traverse Mathematical Condition
type
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Latitude and Departure (cont.)
Loop Traverse
Latitude and Departure (cont.)
Connecting Traverse
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Lat and Dep (cont.)
lat 104 919 cos 340 00 00 98 592
dep 104 919 sin 340 00 00 0 35 884
lat 217 643 cos 76 42 55 50 012
dep 217 643 sin 76 42 55 211819
= ⋅ ° = +
= ⋅ ° = −
= ⋅ ° = +
= ⋅ ° = +
. ( ' ") .
. ( ' " .
. ( ' ") .
. ( ' ") .
Example Calculation for lines 1-2 and 2-3
• Latitude = lat = HD · Cos(β)= HD · Cos( Az)
• Departure = dep = HD · Sin (β) = HD· Sin (Az)
Lat and Dep (cont.)
3-1 229°48’59” 230.222 - 148.548 -175.885
Totals 552.784 + 0.056 + 0.050
2-3 076°42’55” 217.643 50.012 211.819
1-2 340°00’00” 104.919 98.592 - 35.884
Line Az HD (ft.) Lat (ft.) Dep (ft.)
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Calculate L.E.C. and R.E.C.
Loop Traverse
Closure lat lat 0 lat
Closure dep dep 0 dep
L E C lat dep
R EC
L E C
HD
1
n
2 2
i
= − =
= − =
= +
= = Σ
Σ Σ
Σ Σ
. . (Σ ) (Σ )
. . .
. . .
L.E.C. and R.E.C. (cont.)
• The R.E.C. should meet the standard of accuracy
required for the traverse specification being followed.
See ALTA-ACSM specifications for boundary traverses
for an example.
• If the R.E.C. does not meet the standard of accuracy, a
gross blunder and/or an unaccounted for systematic error
among the measurements should be isolated and
corrected before continuing with further adjustments.
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L.E.C. and R.E.C. (cont.)
• If angle closure was satisfactory, a failure to meet the
standard of accuracy for the R.E.C. at this point, likely
points to a problem with distances.
L.E.C. and R.E.C. (cont.)
Loop Traverse
Closure lat lat 0 lat
Closure dep dep 0 dep
L E C lat dep
R E C
L E C
HD
1
n
2 2
i
= − =
= − =
= +
= = Σ
Σ Σ
Σ Σ
. . (Σ ) (Σ )
. . .
. . .
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L.E.C. and R.E.C. (cont.)
Connecting Traverse
Closure lat lat N E
Closure dep dep E E
L E C E E
R E C
L E C
HD
1
n
L
D
L
2
D
2
i
= − =
= − =
= +
= = Σ
Σ Δ
Σ Δ
. . ( ) ( )
. . .
. . .
L.E.C. and R.E.C. (cont.)
Loop Traverse
Closure lat lat 0 lat
Closure dep dep 0 dep
L E C lat dep
R E C
L E C
HD
2 2 2 2
i
= − = = +
= − = =+
= + = + =
= = = Σ
Σ Σ
Σ Σ
Σ Σ
0 056
0 050
0 056 0 050 0 075
0 075
552 784
1
7370
.
.
. . ( ) ( ) ( . ) ( . ) .
. . .
. . . .
.
Example Calculation
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Traverse Adjustment
• Adjustment Methods to remove random
errors
– Compass Rule*
– Transit Rule
– Crandall Method
– General Least Squares Method
Traverse Adjustment via
Compass Rule
C
lat
HD
HD correction a latitude
note:
lat
HD
Correction per ft
C
dep
HD
HD correction a departure
note:
dep
HD
Correction per ft
Li i
Di i
= − ⋅ =
− =
= − ⋅ =
− =
Σ
Σ
Σ
Σ
Σ
Σ
Σ
Σ
to
to
.
.
bal lat lat C
bal dep dep C
L
D
. . .
. . .
= +
= +
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Traverse Adjustment (cont.)
C
0 056
552 784
104 919 0 011
C
0 050
552 784
104 919 0 009
bal lat 98 592 0 011 98 581
bal dep 35 884 0 009 35 893
L
D
= − ⋅ = −
= − ⋅ = −
= + − =
= − + − = −
.
.
. .
.
.
. .
. . . ( . ) .
. . . ( . ) .
Example calculation for line 1-2
Traverse Adjustment (cont.)
sums +0.056 +0.050 -0.056 -0.050 0.000 0.000
3-1 -148.548 -175.885 -0.023 -0.021 -148.571 -175.906
2-3 +50.012 +211.819 -0.022 -0.020 +49.990 +211.799
1-2 +98.592 -35.884 -0.011 -0.009 +98.581 -35.893
Lat Dep
Bal.
Dep(ft.)
Bal.
Lat.(ft.)
Line Lat. (ft.) Dep. (ft.) Correction (ft.)
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Calculate Coordinates
N N ballat
E E baldep
i i1 i1i
i i1 i1i
= +
= +
− −
− −
.
.
,
,
N N bal lat 1000 000 98 581 1098 581
E E bal dep 1000 000 35 893 964 107
2 1 12
2 1 12
= + = + =
= + = + − =
−
−
. . . .
. . ( . ) .
Usually, the coordinates of the first point are assigned
arbitrary values so other coordinates will be positive.
Calculate Coordinates (cont.)
1 1000.000 1000.000
-148.571 -175.906
3 1148.571 1175.906
+49.990 +211.799
2 1098.581 964.107
+98.581 -35.893
1 1000.000 1000.000
Northing Easting
Bal. Dep. Coordinates (ft.)
(ft.)
Bal. Lat.
(ft.)
Sta
N N ballat
E E baldep
i i1 i1i
i i1 i1i
= +
= +
− −
− −
.
.
,
,
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Calculate Area by Coordinates
2 Area = E1·(Nn – N2) + E2·(N1 – N3) + . . . En·(Nn-1 – N1)
• where n = number of sides
• a term is written for each vertex or traverse station
• parenthetical term = (preceding N – following N)
Area by Coordinates (cont.)
For computational convenience the terms are written
in a vertical stack. An example for a 3-sided traverse
is as follows:
E1·(N3 – N2)
E2· (N1 – N3)
E3· (N2 – N1)
__________
Σ = 2 Area
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Area by Coordinates (cont.)
( ) ( )
( ) ( )
( ) ( )
1000.000 1148.571 1098.581 1000.000 49.990 49990.0000
964.107 1000.000 1148.571 964.107 148.571 143238.3411
1175.906 1098.581 1000.00 1175.906 98.581 115921.9894
⋅ − = ⋅ =
⋅ − = ⋅ − = −
⋅ − = ⋅ =
2 Area = 22673.6483 ft.2
Area = 22,673.6483 ft2 ÷ 2 = 11,336.8242 ft2
Area in acres = 11,336.8242 ft2 ÷ 43560 ft2/acre = 0.26 acres
Example calculation:
Alternative Coordinate Method
Determinate Method
Setup coordinate pairs to look
like fractions - eastings over
northings, and in sequence
around the traverse. NOTE:
the first coordinate pair is
repeated at the end.
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Determinate Method (cont.)
• Obtain cross products, “up” products are multiplied by a plus
one and “down” products are multiplied by a negative one.
• The algebraic sum of the cross products is twice the area.
Inverse Problem
Inverse Problem - Given the coordinates of two
points ( i and j ) find: the dir. and dist. between them.
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Inverse Problem
Determine
Quadrant by
Inspection
lat N N N
dep E E E
arctan
dep
lat
arctan
E
N
d dep lat E N
i j i j j i
i j i j j i
i j
i j
i j
i j
i j
i j i j
2
i j
2
i j
2
i j
2
. .
. .
.
.
.
.
.
. . . . .
= = −
= = −
= =
= + = +
Δ
Δ
Δ
Δ
Δ Δ
β
Inverse Problem (cont.)
lat N 1098 581 1000 000 98 581
dep E 964 107 1000 000 35 893
arctan
35 893
98 581
20 00 23
d 35 893 98 581 104 912
1 2 1 2
1 2 1 2
1 2
i j
2 2
. .
. .
.
.
. . .
. . .
.
.
' "
( . ) ( . ) .
= = − = +
= = − =−
=
−
+
= °
= − + + =
Δ
Δ
β
Sample Calculation for line 1-2
Note: The bearing quadrant is NW; therefore, the bearing
is: N 20°00’23”W or the Azimuth is 339°59’37”
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Inverse Problem (cont.)
3-1 229°48’55” 230.253
2-3 076°43’11” 217.619
1-2 339°59’37” 104.912
Line Corrected Az Corrected dist. (ft.)
Find the corrected Azimuth and Distance between
traverse stations by inversing.
Example problem: