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
Errors in Traverse Computations
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
Errors in Traverse Computations
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
EXPECTED TRAVERSE MISCLOSURE
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
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