ERROR PROPAGATION TRAVERSE

Departure

J- D + D

I

(a)

Departure

I

J

-

+

(b)

PLATE 7-1

ERROR PROPAGATIONTRAVERSE

Every course in traverse affected by:

distance measurement errors.

angular measurement errors causing

computed errors in azimuth.

PLATE 7-2

THE FUZZY TRAVERSETHE FUZZY TRAVERSE

Errors in distance and angle measurements result inincreasing uncertainty in computed coordinates at eachstation.

PLATE 7-3

PLATE 7-4

DERIVATION OF ERRORPROPAGATION FORMULA

LAT = D Cos( Az )

DEP = D Sin( Az )

Need to use G.L.O.P.O.V. since latitude and departure arefunctionally related by errors in distance and azimuth.

1 0 2 3

9 3 8 6

0 4 7 1

PLATE 7-5

SOLVING EQUATIONS WITH MATRICES

MATRIX: a set of numbers or symbols arranged in anarray of “m” rows and “n” columns.

E.G.

A = 34

1 0 2 3

9 3 8 6

0 4 7 1

a11 a12 a13 a14

a21 a22 a23 a24

a31 a32 a33 a34

PLATE 7-6

FUNDAMENTAL CONCEPTS

Elements designated by subscripted lower case letters.

E.G. For first row: a , a , a , a , 11 12 13 14

A =

a = 1 a = 0 a = 2 a = 311 12 13 14

a = 9 a = 3 a = 8 a = 621 22 23 24

a = 0 a = 4 a = 7 a = 131 32 33 34

A =

1 0 2 3

9 3 8 6

0 4 7 1

1 9 0

0 3 4

2 8 7

3 6 1

PLATE 7-7

MATRIX TRANSPOSE

Interchanging rows and columns, that is:a = aij ji

A =

then

A = T

A7 3 1

2 5 6B

1 5 6

4 2 3

A B7 1 3 5 1 6

2 4 5 2 6 3

PLATE 7-8

MATRIX ADDITION

Add matrices element by element

E.g. Let

then

A1 2 3

4 2 7B

4 8

6 2

5 3

31 21

63 57

PLATE 7-9

MATRIX MULTIPLICATION

Row is multiplied by column. Let

Then C = A × B2 2 32 3 2

c = 1×4 + 2×6 + 3×5 = 3111

c = 1×8 + 2×2 + 3×3 = 2112

c = 4×4 + 2×6 + 7×5 = 6321

c = 4×8 + 2×2 + 7×3 = 5722

or

C =

1 2 4

6 3 8

5 1 1

x

y

z

3

8

2

PLATE 7-10

REPRESENTING EQUATIONS USING MATRICES

Let1x + 2y - 4z = 36x - 3y + 8z = 85x + 1y + 1z = 2

This can be written in matrix form:

AX = Bwhere

2D 0

0 Az2

lat, dep

2lat lat,dep

lat,dep2dep

A

LatD

LatAz

DepD

DepAz

LatD

Cos(Az) LatAz

D Sin(Az)

DepD

Sin(Az) DepAz

D Cos(Az)

PLATE 7-11

APPLYING G.L.O.P.O.V.

= A Alat,depT

where

PLATE 7-12

EXAMPLE

A traverse course has a length of 456.87 ± 0.02 ft, andazimuth of 23° 35' 26" ± 9". What are the latitude anddeparture for the course, and the estimated errors in each?

Lat = 456.87 Cos( 23° 35' 26" )= 418.69 ft.

Dep = 456.87 Sin( 23° 35' 26" )= 182.84 ft.

lat, dep

Cos(Az) DSin(Az)

Sin(Az) DCos(Az)

0.022 0

0 92

Cos(Az) Sin(Az)

DSin(Az) DCos(Az)

lat, dep

0.9164 456.87 (0.4002)

0.4002 456.87 (0.9164)

0.0004 0

0 9 20.9164 0.4002

456.87 (0.4002) 456.87 (0.9164)

lat, dep

0.00039958 0.00000096

0.00000096 0.00039781

PLATE 7-13

THEIR ESTIMATED ERRORSBY G.L.O.P.O.V.

OR

OR

q11 0.00039958

q22 0.00039781

PLATE 7-14

ERROR ANALYSIS

q is the latitude's variance, 11

q is the departure's variance, and 22

q and q are their covariances.12 21

Thus estimated errors are:

= = ±0.020 ft. Lat

= = ±0.020 ft.Dep

PLATE 7-15

ESTIMATED ERRORSIN TRAVERSE

TWO COMPONENTS:1. Errors in azimuths

2. Errors in distances

Azc

2Azp

2

PLATE 7-16

ESTIMATED ERRORSIN COURSE AZIMUTHS

FORMULA:Az = Az + 180 + c p

whereAz is the azimuth of the course,c

Az is the azimuth of the previous coursep

USING S.L.O.P.O.V.

Azn

2Azinitial

n

i 2

2i

Azn

Azinitial

PLATE 7-17

FOR A TRAVERSE WITH AZIMUTHSCOMPUTED FROM ANGLES

where is the azimuth of the n’th course

is the azimuth of the initial course

3.183 3.52 3.12 3.62 3.12 3.92

±24.6

PLATE 7-18

(Each angle was measured with four repetitions.)Occ Std Distance S BS Inst FS Angle SA B 1435.67 0.020 E A B 110 24' 40" 3.5"B C 856.94 0.020 A B C 87 36' 14" 3.1"C D 1125.66 0.020 B C D 125 47' 27" 3.6"D E 1054.54 0.020 C D E 99 57' 02" 3.1"E A 756.35 0.020 D E A 116 14' 56" 3.9"

540° 00' 19"

EXAMPLE

Check angular misclosure at 95% level of confidence:

t = 3.1830.025,3

Since 19" is less than 24.6", the angular misclosure isacceptable.

3.12 3.62 ±4.8

4.82 3.12 ±5.7

5.72 3.92 ±6.9

PLATE 7-19

FROM TO AZIMUTH ESTIMATED ERROR

A B 0 00' 00" 0"B C 267 36' 14" ±3.1"C D 213 23' 41"

D E 133 20' 43"

E A 69 35' 39"

Errors in Traverse Computations

Cos(AzAB) DABSin(AzAB) 0 0 0 0 0 0 0 0

Sin(AzAB) DABCos(AzAB) 0 0 0 0 0 0 0 0

0 0 Cos(AzBC) DBCSin(AzBC) 0 0 0 0 0 0

0 0 Sin(AzBC) DBCCos(AzBC) 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 Cos(AzEA) DEASin(AzEA)

0 0 0 0 0 0 0 0 Sin(AzEA) DEACos(AzEA)

PLATE 7-20


Set up A matrix:alternate columns with partial derivatives for distanceand azimuthalternate rows with latitude and departure equationsfor each course.

2DAB

0 0 0 0 0 0 0 0 0

0AzAB

2

0 0 0 0 0 0 0 0

0 0 2DBC

0 0 0 0 0 0 0

0 0 0AzBC

2

0 0 0 0 0 0

0 0 0 0 2DCD

0 0 0 0 0

0 0 0 0 0AzCD

2

0 0 0 0

0 0 0 0 0 0 2DDE

0 0 0

0 0 0 0 0 0 0AzDE

2

0 0

0 0 0 0 0 0 0 0 2DEA

0

0 0 0 0 0 0 0 0 0AzEA

2

PLATE 7-21

Errors in Traverse Comps

Setup matrix:

0.00040 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0

0 0 0.00017 0.00002 0 0 0 0 0 0

0 0 0.00002 0.00040 0 0 0 0 0 0

0 0 0 0 0.00049 0.00050 0 0 0 0

0 0 0 0 0.00050 0.00060 0 0 0 0

0 0 0 0 0 0 0.00064 0.00062 0 0

0 0 0 0 0 0 0.00062 0.00061 0 0

0 0 0 0 0 0 0 0 0.00061 0.00034

0 0 0 0 0 0 0 0 0.00034 0.00043

PLATE 7-22


Substitute in appropriate values and compute:

= A Alat, dep T

LC ( LatAB LatBC LatEA )2 ( DepAB DepBC DepEA )2

LCLati

LatsLC

LCDepi

DepsLC

PLATE 7-23

EXPECTED MISCLOSUREIN A TRAVERSE

Computing the expected value for the linear misclosure ofa traverse.

The partial derivatives w.r.t. Lat and Dep for the i’thcourse are:

A LatsLC

DepsLC

LatsLC

DepsLC

LatsLC

DepsLC

PLATE 7-24

EXPECTED MISCLOSUREIN A TRAVERSE

Since the partial derivatives are course independent, the Amatrix structure is:

( 0.082)2 (0.022)2 0.085 ft.

PLATE 7-25

ACTUAL TRAVERSE MISCLOSURE

Course Latitude DepartureAB 1435.67 0BC -35.827 -856.191CD -939.811 -619.567DE -723.829 766.894EA 263.715 708.886

= -0.082 = 0.022

LC =

LC [ 0.9647 0.2588 0.9647 0.2588 0.9647 0.2588 ] Lat, Dep

0.9647

0.2588

0.9647

0.2588

0.9647

0.2588

0.9647

0.2588

0.9647

0.2588

[ 0.00229 ]

PLATE 7-26

EXPECTED TRAVERSE MISCLOSURE

From previously determined and G.L.O.P.O.V. Lat, Dep

0.00229 ±0.048 ft

PLATE 7-27

STATISTICAL CHECK OFTRAVERSE MISCLOSURE

Expected traverse misclosure:

= LC

PLATE 7-28

STATISTICAL CHECK OFTRAVERSE MISCLOSURE

At 95% level of confidence use a t multiplier of t =0.05, 3

2.35

Thus E = 2.35 × 0.04895

= ± 0.11 ft.

Actual traverse misclosure 0.085 ft. within 95%confidence interval.

1

A BC

D 2

PLATE 7-29

LINK TRAVERSE EXAMPLE

Distance ObservationsFrom To Distance S

1 A 1069.16 ±0.021A B 933.26 ±0.020 Control StationsB C 819.98 ±0.020 Station X YC D 1223.33 ±0.021 1 1248.00 3979.00D 2 1273.22 ±0.021 2 4873.00 3677.00

Angle Observations Azimuth ObservationsBS Inst FS Angle S From To Azimuth S1 A B 66 16' 35" ±4.9" 1 A 197 04' 47" ±4.3"A B C 205 16 '46" ±5.5" 2 D 264 19' 13" ±4.1"B C D 123 40' 19" ±5.1"C D 2 212 00' 55" ±4.6"

11.02 4.12 ±11.7

PLATE 7-30

ANGULAR MISCLOSURE

Course Azimuth1A 197 04' 47" ±4.3"AB 83 21' 22" ±6.5"BC 108 38' 08" ±8.5"CD 52 18' 27" ±9.9"D2 84 19' 22" ±11.0"

The actual difference in the azimuths:

84° 19' 22" - 84° 19' 13" = 9"

Using S.L.O.P.O.V., the estimated difference is:

Thus, there is no reason to suspect any problems in theangle measurements.

LC Dep2 Lat 2

( 0.128)2 ( 0.032)2

0.132

PLATE 7-31

CHECKING TRAVERSEMISCLOSURE

Course Latitude Departure1A -1022.007 -314.014AB 107.976 926.993BC -262.022 776.989CD 747.973 968.025D2 125.952 1266.975

= -302.128 = 3624.968

ACTUAL MISCLOSUREDep = Dep - (X - X ) = 3624.968 - 3625.00 2 1

= -0.032

Lat = Lat - (Y - Y ) = -302.128 - (-302.00) 2 1

= -0.128

A

Cos(Az1A) D1ASin(Az1A) 0 0 0 0 0 0 0 0

Sin(Az1A) D1ACos(Az1A) 0 0 0 0 0 0 0 0

0 0 Cos(AzAB) DABSin(AzAB) 0 0 0 0 0 0

0 0 Sin(AzAB) DABCos(AzAB) 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 Cos(AzD2) DD2Sin(AzD2)

0 0 0 0 0 0 0 0 Sin(AzD2) DD2Cos(AzD2)

PLATE 7-32

EXPECTED TRAVERSEMISCLOSURE

SETUP MATRICES

2D1A

0 0 0 0 0 0 0 0 0

0 Az1A

2

0 0 0 0 0 0 0 0

0 0 2DAB

0 0 0 0 0 0 0

0 0 0AzAB

2

0 0 0 0 0 0

0 0 0 0 2DBC

0 0 0 0 0

0 0 0 0 0 AzBC

2 2

0 0 0 0

0 0 0 0 0 0 2DCD

0 0 0

0 0 0 0 0 0 0AzCD

2

0 0

0 0 0 0 0 0 0 0 2DD2

0

0 0 0 0 0 0 0 0 0 AzD2

2

PLATE 7-33


lat, dep

0.000446 0.000263 0 0 0 0 0 0 0 0

0.000263 0.000492 0 0 0 0 0 0 0 0

0 0 0.000859 0.000145 0 0 0 0 0 0

0 0 0.000145 0.000406 0 0 0 0 0 0

0 0 0 0 0.00107 0.000467 0 0 0 0

0 0 0 0 0.000467 0.000476 0 0 0 0

0 0 0 0 0 0 0.002324 0.001881 0 0

0 0 0 0 0 0 0.001881 0.001565 0 0

0 0 0 0 0 0 0 0 0.004570 0.000497

0 0 0 0 0 0 0 0 0.000497 0.000482

A LatLC

DepLC

LatLC

DepLC

LatLC

DepLC

PLATE 7-34


Expected Traverse Misclosure:

= A A = [ 0.01000 ]LC lat, depT

= ± 0.10 ft.LC

E = t = 2.35 × 0.10 = 0.24 ft.95% 0.05,3 LC

ERROR PROPAGATION TRAVERSE

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