adj comp lecture_3

7/24/2019 ADJ COMP lecture_3

1/28

ANALYSIS OF ERRORS IN INDIRECT

OBSERVATIONS


2/28

Analysis of indirect observations

LOPOVLaw of Propagation of Variance (Error)

)()( 2211 xxz xxz

T

xZ AA

1x

Z

2x

ZA=

x

12

2

1

XX

X

2

2

21

X

XX

Z2

Z

In matrix form:

2

2

2

2

1

1

2

XXZ

x

Z

x

Z


3/28

Matrix A:

Linear function, A = [c1 c2 cn]

Non-linear function

,

It can be shown that LOPOV;

where

1

1

2

1

1

........

x

Z

x

Z

x

Z

m

2

2

2

2

1

.......

x

Z

x

Z

x

Z

m

n

m

n

n

x

Z

x

Z

x

Z

......

2

1

22

2

2

2

1

1

2......

Xn

n

XXZx

Z

x

Z

x

Z

A =

021 XX


4/28

Derive the standard deviation of mean

Mean

Std deviation y1=y2= =yn = S

n

SS

x


5/28

Example

A B C

mm 012.0560.10 mm 015.0370.120

Find length AC (Z) and its standard deviation ( )

Z = x1 + x2

= 10.560 + 120.370= 130.93m

x1 x2

2

2

2

2

1

1

2

XXZ

x

Z

x

Z

019.0)015.0)(1()012.0)(1( 22 Z

mmLengthAC 019.093.130

Z


6/28

A B C

mm 012.0560.10 mm 015.0370.120

Find length AC and its standard deviation

1A 1

1

1TA

0

012.0 2

2015.0

0

=

019.0107.3 4 xAC

T

XZ AA

Z = x1 + x2

x1 x2

12

2

1

XX

X

2

2

21

X

XX

X

42 107.3 xZ Z

mmLengthAC 019.093.130


7/28

Sections

Length

(m)

Standard deviation

(m)

PQ 157.694 0.018

QR 381.978 0.024

RS 182.124 0.018

ST 227.325 0.024

A straight line PT was measured in sections and the measured length with its

standard deviation are shown in the following table. In matrix form, compute the

standard deviations of the length PT.

Z = x1 + x2 + x3 + x4

Answer = 949.121m +/- 0.042m


8/28

Z = x1 + x2 + x3 + x4 949.121

A(1x4)

1 1 1 1

Sigx(4x4)

0.000324 0 0 0

0 0.000576 0 0

0 0 0.000324 0

0 0 0 0.000576

At(4x1)

1

1

1

1

T

XZ AA

Asigx(1x4)

0.000324 0.000576 0.000324 0.000576

ASigxAt(1x1)

0.0018

std dev 0.042


9/28

mm 021.0163.37 R =

Find the area of a circle and its error

502.233)163.37(22

R

R

A

222 818.4338)163.37( mRA

22

904.4)021.0)(502.233( mA

2

2

RA

R

A

Answer = 4338.818 +/- 4.904 sq m


10/28

A = pi()*R

A(1x1) sigr(1x1)233.502 0.000441

At(1x1)

233.502

Asigr(1x1)

0.102974

AsigrAt(1x1)

24.04473

std dev 4.904


11/28

The elevation difference from points A to B is mm 012.0560.10

If the elevation of point A above mean sea level is mm 015.0370.120

(i) Calculate the elevation of point B above mean sea level

(ii) Calculate the standard deviation of elevation of point B

ABAB HH

Answer = 130.930 +/- 0.019m


12/28

Hb = Ha + dH = 130.93

A(1x2) sig(2x2)

1 1 0.000225 00 0.000144

At(2x1)

1

1

Asig(1x2)

0.000225 0.000144

AsigAt(1x1)

0.000369

std dev 0.019


13/28

As shown above, the distance between points B and C cannot be directly

measured as it was obstructed by the building. However, it was possible to

measure distances AB, AC and the horizontal angle A.

If the measured distance AB = 154.338m 0.063m, distance AC = 136.729m

0.063m and angle A = 80050 37 15, compute the distance BC and its

standard deviations.

Note that BC2= AC2 + AB22(AC)(AB)Cos A

Answer = 189.206 +/- 0.058m


14/28

L1 154.338 +/- 0.063 m

L2 136.729 0.063 m

a 80-50-37 80.84361 (deg) 15 (")

1.410987 (rad)

L3 189.206

rho 206265.000

A(1x3) sig(3x3)

0.701 0.593 110.111 0.003969 0 0

0 0.003969 0

0 0 5.28849E-09

At(3x1) Asig(1x3)

0.701 0.002781 0.002353 5.82318E-07

0.593

110.111 AsigAt(1x1) std dev

0.003 0.058


15/28

EC

B

A

D F

Back

Stn.

Inst. Stn. Fore

Stn.

Measured

angles

(0 )

Standard

deviation

( )

A B C 89-17-22 4.2

B C D 109-27-05 5.1

C D E 283-33-52 5.3

D E F 117-35-19 4.5

The clockwise angles of an open traverse ABCDEF were measured using aTotal Station. The measured angles and its standard deviations are as shown

in the table. Knowing the bearing of AB as 202016 37 and standard

deviation of 3.5, compute the bearing of BC, CD, DE and EF and their

standard deviations.


16/28

Y

X D

S

'0'0 010004

07.035.305

mmS

The slope distance (S) and vertical angle ( ) between points X and Y

were measured and their results are as shown;

i. Calculate the horizontal distance (D) between

points X and Y and its standard deviation in matrix

form.

ii. If the error in the horizontal distance (D) is

considered too large, determine whether the slope

distance or the vertical angle needs to be

measured more precisely.

sCosD


17/28

The distance (d) and bearing (b) of a traverse line PQ are as follows:

d = b =mm 02.087.456 "09"26'35230

In matrix form, calculate the coordinates of point Q and its standard

deviation given the coordinates of point P as

mm 05.000.100 Easting P =Northing P = mm 05.000.200

Answer:Easting Q = 282.838 0.054

Northing Q = 618.689 0.054


18/28

RANDOM ERRORS

IN ANGLE AND DISTANCE

OBSERVATIONS


19/28

SOURCES OF RANDOM ERRORS IN HORIZONTAL ANGLE

OBSERVATIONS

Instrument setup/centring and levelingTarget setup/centring

Circle reading

Target pointing

REDUCTION OF ERRORS

Errors in instrument leveling, reading and pointing can be reduced by

increasing the number of angle repetitions.

Errors in instrument and target setup can be reduced by increasing the sight

distances


20/28

READING ERRORS

Quality of the instruments optic

The smallest division of the circleOperator abilities

POINTING ERRORS

Quality of the instruments telescope optic

Target size

Operator abilities to bisect cross-wire on a target

Weather condition


21/28

TARGET CENTRING ERRORS

Environmental condition

Optical plummetQuality of the optics

Plumbob centring

Personal abilities

INSTRUMENT CENTRING ERRORS

Quality of instrument and optics

Optical plummet

Plumbob centring

Personal abilitiesQuality of tripod


22/28

I

B

F

D1

D2

D3

Angle measurement error:

Reading error ()

Pointing error ()

Target centring (rad)

Instrument centring (rad)

Note: x 206265 to convert (rad) to ()

r

p

t

i

2222

itpr

CosDDDDD 212

2

2

1

2

3 2


23/28

ner

r2

n

ep

p

2

ettDD

DD

21

2

2

2

1

eiiDD

D 221

3

2222

itpr

CosDDDDD 21

2

2

2

1

2

23


24/28

Example

An angle is measured four (4) times using a Total Station. The observer reading

error is1 and pointing error of 1.5. The observers centering error of the target

0.001m and centering error of instrument is 0.001m. The horizontal distance

from instrument to backsight target is 76.505m and foresight target 105.765m.

The measured angle is 65037 12.

Calculate the standard deviation in the measured angle.

ner

r2

n

ep

p

2

)(

21

2

2

2

1 pDD

DDett

)(221

3 pDD

Deii

2222

itpr

CosDDDDD 212

2

2

1

2

23

Note: p = 206265 to convert (rad) to ()


25/28

Error due to reading = 0.71

Error due to pointing = 1.06

Error due to target centring = 3.33

Error due to Instrument centering = 1.84 ; D3 = 101.780m

Combined error = 4.01


26/28

SOURCES OF RANDOM ERRORS IN ELECTRONIC DISTANCE

OBSERVATIONS

Instrumental errors : constant error (a) and scalar error (b) specified as

Target and instrument centring

)( ba ppm


27/28

Error in Electronic Distance Measurement (EDM)

2222 *bDatiD

Constant error (a)

Scalar error (b) in ppm

Instrument centring error

Target centring error

Distance measured (D)

i

t

610


28/28

Exercise

A Total Station with instrumental error (5mm + 10 ppm)

was used to measure a distance of 138.340m The

centering error of instrument is 0.001m and centering

error of reflector is 0.003m. Calculate the total error in

the measured distance.

mbDatiD 006.0* 2222

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