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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)

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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 openMathematically 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 SD Sin ZA= ⋅ = ⋅( ) ( )α

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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 ÷ nn = 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”

+ 05”- 10”359°59’50”333°06’15”26° 53’ 35’3

- 07”+ 15”360°00’15”276°43’20”83° 16’ 55”2

- 48”+ 95”360°01’35”290° 12’05”69° 49’ 30”1CAClosSumExt.Int.Sta

Field Angles

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Angle Adjustment (cont.)

• Mean interior = Int. + CA

Sum = 179°59’10”

26° 53’40”+ 05”26° 53’ 35383°16’ 48”-07”83° 16’ 55”269°48’ 42”- 48”69° 49’ 30”1Mean int.CAfield int.Sta.

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.)

26° 53’56”+ 16”26° 53’40”383°17’ 05”+ 17”83°16’ 48”269°48’ 59”+ 17”69°48’ 42”1Adj. int.CAMean int.Sta.

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 ofstations 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.)

Σ dep = ∆EΣ lat = ∆NConnecting

Σ dep = 0Σ lat = 0Loop

departureslatitudes

Mathematical ConditionTraverse 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 592dep 104 919 sin 340 00 00 0 35 884lat 217 643 cos 76 42 55 50 012dep 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.)

-175.885- 148.548230.222229°48’59”3-1

+ 0.050+ 0.056552.784Totals

211.81950.012217.643076°42’55”2-3- 35.88498.592104.919340°00’00”1-2Dep (ft.)Lat (ft.)HD (ft.)AzLine

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Calculate L.E.C. and R.E.C.

Loop TraverseClosure lat lat 0 latClosure dep dep 0 dep

L E C lat dep

R E CL E C

HD1n

2 2

i

= − == − =

= +

= =∑

Σ ΣΣ 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 TraverseClosure lat lat 0 latClosure dep dep 0 dep

L E C lat dep

R E CL E C

HD1n

2 2

i

= − == − =

= +

= =∑

Σ ΣΣ Σ

Σ Σ. . ( ) ( )

. . .. . .

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L.E.C. and R.E.C. (cont.)

Connecting TraverseClosure lat lat N EClosure dep dep E E

L E C E E

R E CL E C

HD1n

L

D

L2

D2

i

= − == − =

= +

= =∑

Σ ∆Σ ∆

. . ( ) ( )

. . .. . .

L.E.C. and R.E.C. (cont.)

Loop TraverseClosure lat lat 0 latClosure dep dep 0 dep

L E C lat dep

R E CL E C

HD

2 2 2 2

i

= − = = += − = = +

= + = + =

= = =∑

Σ ΣΣ Σ

Σ Σ

0 0560 050

0 056 0 050 0 0750 075

552 7841

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

ClatHD

HD correction a latitude

note: latHD

Correction per ft

CdepHD

HD correction a departure

note: depHD

Correction per ft

Li i

Di i

= − ⋅ =

− =

= − ⋅ =

− =

∑∑∑∑∑∑∑∑

to

to

.

.

bal lat lat Cbal dep dep C

L

D

. . .

. . .= += +

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Traverse Adjustment (cont.)

C0 056

552 784104 919 0 011

C0 050

552 784104 919 0 009

bal lat 98 592 0 011 98 581bal dep 35 884 0 009 35 893

L

D

= − ⋅ = −

= − ⋅ = −

= + − == − + − = −

..

. .

..

. .

. . . ( . ) .

. . . ( . ) .

Example calculation for line 1-2

Traverse Adjustment (cont.)

0.0000.000-0.050-0.056+0.050+0.056sums

-175.906-148.571-0.021-0.023-175.885-148.5483-1

+211.799+49.990-0.020-0.022+211.819+50.0122-3

-35.893+98.581-0.009-0.011-35.884+98.5921-2

DepLat

Bal. Dep(ft.)

Bal. Lat.(ft.)

Correction (ft.)Dep. (ft.) Lat. (ft.)Line

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Calculate Coordinates

N N bal lat

E E bal depi i 1 i 1 i

i i 1 i 1 i

= +

= +− −

− −

.

.,

,

N N bal lat 1000 000 98 581 1098 581E E bal dep 1000 000 35 893 964 107

2 1 1 2

2 1 1 2

= + = + == + = + − =

−

−

. . . .. . ( . ) .

Usually, the coordinates of the first point are assignedarbitrary values so other coordinates will be positive.

Calculate Coordinates (cont.)

1000.0001000.0001-175.906-148.571

1175.9061148.5713+211.799+49.990

964.1071098.5812-35.893+98.581

1000.0001000.0001EastingNorthing

Coordinates (ft.)Bal. Dep. (ft.)

Bal. Lat. (ft.)

Sta

N N bal lat

E E bal depi i 1 i 1 i

i i 1 i 1 i

= +

= +− −

− −

.

.,

,

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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 writtenin a vertical stack. An example for a 3-sided traverseis 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 49990.0000964.107 1000.000 1148.571 964.107 148.571 143238.34111175.906 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 MethodDeterminate Method

Setup coordinate pairs to look like fractions - eastings overnorthings, and in sequence around the traverse. NOTE:the first coordinate pair isrepeated at the end.

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Determinate Method (cont.)• Obtain cross products, “up” products are multiplied by a plusone 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 twopoints ( i and j ) find: the dir. and dist. between them.

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Inverse Problem

Determine Quadrant by Inspection

lat N N Ndep E E E

arctandeplat

arctanEN

d dep lat E N

i j i j j i

i j i j j i

i ji j

i j

i j

i j

i j i j2

i j2

i j2

i j2

. .

. .

..

.

.

.

. . . . .

= = −

= = −

= =

= + = +

∆

∆

∆

∆

∆ ∆

β

Inverse Problem (cont.)

lat N 1098 581 1000 000 98 581dep E 964 107 1000 000 35 893

arctan35 89398 581

20 00 23

d 35 893 98 581 104 912

1 2 1 2

1 2 1 2

1 2

i j2 2

. .

. .

.

.

. . .. . .

.

.' "

( . ) ( . ) .

= = − = += = − = −

=−+

= °

= − + + =

∆∆

β

Sample Calculation for line 1-2

Note: The bearing quadrant is NW; therefore, the bearingis: N 20°00’23”W or the Azimuth is 339°59’37”

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Inverse Problem (cont.)

230.253229°48’55”3-1217.619076°43’11”2-3104.912339°59’37”1-2Corrected dist. (ft.)Corrected AzLine

Find the corrected Azimuth and Distance betweentraverse stations by inversing.

Example problem: