PLATE 6-1

ERROR PROPAGATION IN ANGLE MEASUREMENTS

SOURCES OF ERRORS

1. Reading the circlepersonal value

2. Pointing on the targetpersonal valuedependent on instrument

3. Target setupreduced by increasing sight distance

4. Instrument setupreduced by increasing sight distance

5. Instrument mislevelinglargest affect when altitude of target high

AVE. FINAL READING INITIAL READINGn

r

f

n

20

n

2

PLATE 6-2

READING ERRORREPETITION METHOD

WHEN READINGS ARE TAKEN:1) read circle when zeroing circles

2) read circle after final pointing

FOR ANGLE TURNED n TIMES USinG BOTH FACES:

APPLYING ERROR IN SUM:

r

r 2

n

PLATE 6-3

READING ERROR REPETITION METHOD

If = = , then f 0 r

r± 1.5 2

6±0.4

PLATE 6-4

EXAMPLE

Assume = ±1.5" and angle is measured 6 times using ther

repetition method, what is the expected error in the angledue to the reading error:

AVE(RFS1 RBS1) (RFS2 RBS2) (RFSn RBSn )

n

PLATE 6-5

READING ERRORDIRECTIONAL METHOD

WHEN READINGS ARE TAKEN:1) read circle for each pointing

2) angle an indirect measurement,

= READING - READINGFS BS

For angle turned n times using both faces:

where R is a circle reading for the xxx sight.xxx

r

1n

2FS1r

2BS1r

2FS2r

2BS2r

2FSnr

2BSnr

r

n ×2 × 2r

nr 2

n

r FS1r BS1r FS2r BS2r

PLATE 6-6

APPLY ERROR IN SUM:

If

Then

r± 1.5 2

6±0.87

PLATE 6-7

READING ERRORDIRECTIONAL METHOD

EXAMPLE:Assume = ±1.5" and angle is measured 6 times usingr

the directional method, what is the expected error in theangle due to the reading error:

PLATE 6-8

ERRORS IN ANGLES DUE TO POINTING

Factors that affect pointing error size:a) optical qualities of telescope

b) target size

c) personal ability of observer to center on target

d) weather conditions (fog, heat waves, etc.)

p

2pBS1

2pFS1

2pBS2

2pFS2

2pBSn

2pFSn

n

p

p 2n

n

p2

n

pBS1pFS1

pBS2pFS2

pBSnpFSn

p

PLATE 6-9

ANALYSIS OF POINTING ERRORS

Every angle involves 2 pointings, thus for n anglemeasurements.

If

Then by applying the error in the sum,

p1.8 2

6

±1.0

PLATE 6-10

EXAMPLE

An angle measured 6 times by an observer whose abilityto point on a well-defined target is estimated to be ±1.8".What is the expected error in the angle due to pointing?

pr DIN 2

pr

2 DIN

n

PLATE 6-11

ESTIMATED POINTING AND READINGERROR WITH TOTAL STATIONS

DIN 18723 STANDARD:Based on a single direction measured with both faces

of the instrument.

Then for n measurements of an angle:

pr± 2×5

6

±4.1

PLATE 6-12

EXAMPLE

An angle measured 6 times by an observer with a totalstation having a DIN18273 value of ±5". What is theestimated error in the angle due to pointing and reading?

d

d

d

d

(b)

(a)

(c)

(d)

D

D

D

D

t

t

t

PLATE 6-13

ERRORS IN ANGLES DUE TOTARGET MISCENTERING

NOTE: This error is systematic for an individual setup,but will appear random over multiple setups.

td

D

PLATE 6-14

ERROR IN A SINGLE DIRECTION

where is in radians.t

t

d1

D1

2d2

D2

2

t

t

D1

2t

D2

2

t t1

D1

2 1D2

2

d1 d2 t

PLATE 6-15

ERROR IN AN ANGLE

IF , then:

t t

D 21 D 2

2

D1 D2

t

PLATE 6-16

ERROR IN AN ANGLE

where is in radians

NOTE: Since same error occurs on every pointing, itcannot be reduced in size by multiple pointings.

t0.02 2502 1502

250 × 150×

±32

PLATE 6-17

EXAMPLE

Hand held targets are centered over a station to within±0.02 ft. What is the error in the angle due to targetcentering if the backsight distance is 250 ft and theforesight distance is 150 ft?

where = 206,264.8"/radian

1

2 i

(b)

2

1

i

(c)(a)

1

2i

P

P

1

2

P1

P2P2

P1

PLATE 6-18

INSTRUMENT MISCENTERING

Facts:Error in every direction due to instrument

miscentering.

At 2 positions one of which is shown in (a),the errors cancel.

Although systematic for a particular setup, theyappear random with mulitple setups and multiplestations.

(P2 2) (P1 1)

(P2 P1) ( 2 1)

(P2 P1)

PLATE 6-19

ERROR IN AN INDIVIDUAL POINTING

PLATE 6-20

ANALYSIS OF ERROR CONTRIBUTION

1ihD1

y cos ( ) x sin ( )D1

yD2

y cos( ) x sin( )D1

D1 y D2 x sin( ) D2 y cos( )

D1 D2

PLATE 6-21

FROM SKETCHih = ip - qr

ih = iq cos( ) - sq sin( )

Let sq = x and iq = y, then

ih = y cos( ) - x sin( )

NOW

= y/D2 2

OR

x

D2 sin( )

D1 D2

y

D1 D2 cos( )

D1 D2

PLATE 6-22

ANALYSIS OF ERROR DUE TOINSTRUMENT MISCENTERING

Applying SLOPOV the parital derivative wrt x is

and wrt y is

x yi

2

2i

D 21 D 2

2 (cos2( ) sin2( )) 2D1D2cos( )

D 21 D 2

2

2i

2

PLATE 6-23

ANALYSIS OF ERROR DUE TOINSTRUMENT MISCENTERING

Since an error in x is as likely as the error in y, and sincethis represents a bivariate distribution:

Applying slopov, the error in the angle is:

i±

D3

D1D2 2i

i

PLATE 6-24

ANALYSIS OF ERROR DUE TOINSTRUMENT MISCENTERING

Rearranging, this can be simplified to:

Where is in radians

Mulitply by = 206,264.8"/radian to get value in seconds.

D3 1502 2502 2 (150) (250) Cos(50) 191.81

i0.005 191.81

(150) (250) 2±3.7

PLATE 6-25

EXAMPLE

What is the error in a 50° angle due to instrumentmiscentering, if the setup is within ±0.005 ft and the sightdistances are 150 and 250 ft?

So

f µd

I

D

S

P’P

v

PLATE 6-26

EFFECTS OF MISLEVELING

µ

PLATE 6-27

EFFECTS OF MISLEVELING

DERIVATION:SP = D tan(v)

PP’ = D

where D is the sighting distance and the angular error isin radians

PP’ = f µ (SP)d

where f is the fractional division of the bubble and µ isd

the sensitivity of the bubble

But PP' = f D tan(v)d

l±

(fd µ tan(vb))2 (fd µ tan(vf))2

n

PLATE 6-28

EFFECTS OF MISLEVELING

So the error in the direction due to misleveling is

= f µ tan(v)d

For an angle measured n times

l[0.5 (30) tan(1°30 45 )]2 [0.5 (30)tan(34°44 30 )]2/ 6

0.42 10.42/ 6

±4.2

PLATE 6-29

EXAMPLE

How much error in the horizontal angle can be expected ina sun shot, if the instrument has a bubble with a sensitivityof 30"/div and is leveled to within 0.5 divisions. Assume abacksight zenith angle of 91°30'45" and a foresight zenithangle to the sun of 55°15'30"? (Assume angle turned3DR.)

PLATE 6-30

ESTIMATED ERROR INA HORIZONTAL ANGLE

Assume a rectangular parcel, with sides of 251.75 and347.05 ft (2 acre parcel). Also assume that

= ± 5"DIN

n = 2

Method: Directional = ±0.005 fti

= ±0.01 ftt

What is the expected error in an angle? What is theexpected angular misclosure?

pr

2×5

2

±7.1

t0.01 251.752 347.052

(251.75) (347.05)

±10.1

PLATE 6-31

Solution: Solve each error component separately, and thensum these using error in sum.

Pointing and reading contribution:

Target miscentering:

D3 251.752 347.052 428.744 ft

i

0.005

2

428.744(251.75) (347.05)

±3.6

PLATE 6-32

Instrument miscentering:

7.12 10.12 3.62

±12.9

PLATE 6-33

Error in a single measured angle:

±13.4 4 ±26.8

95%12.705 × 26.8

2±241

PLATE 6-34

EXPECTED ERROR IN SUM OF ANGLES:

This is the value that you would expect a field crew toclose within 68% of the time.

If you were checking closure then use a 95% confidencelevel, or thus a t multiplier., n-1

Since each angle measured only twice, n - 1 = 1 and

t = 12.7050.025, 1

So