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Chapter: Civil Surveying - Survey Adjustments

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Distribution of Error of The Field Measurement

Whenever observations are made in the field, it is always necessary to check for the closing error, if any.

DISTRIBUTION OF ERROR OF THE FIELD MEASUREMENT

 

Whenever observations are made in the field, it is always necessary to check for the closing error, if any. The closing error should be distributed to the observed quantities. For examples, the sum of the angles measured at a central angle should be 360 o , the error should be distributed to the observed angles after giving proper weight age to the observations. The following rules should be applied for the distribution of errors:

 

(i).The correction to be applied to an observation is inversely proportional to the weight of the observation.

 

(2) The correction to be applied to an observation is directly proportional to the square of the probable error.

 

(3)   In  case  of  line  of  levels,  the  correction  to  be  applied  is  proportional  to  the

length.

The following are the three angles ?, ? and y observed at a station P closing

 

the horizon, along with their probable errors of measurement. Determine their corrected values.

 

 

Solution.

? = 78 o 12? 12' 2'

 

? = 136 o 48? 30' 4'    

y = 144 o 59? 08' 5'    

Sum of the three angles  = 359 o 59? 50'

Discrepancy         = 10'

Hence each angle is to be increased, and the error of 10' is to be distributed in

 

proportion to the square of the probable error.

 

Let c1, c2 and c3 be the correction to be applied to the angles ?, ? and y respectively.

          c1 : c2 : c3 = (2) : (4) : (5) = 4 : 16 : 25    (1)

Also, c1 + c2 + c3 = 10'         (2)

From (1),    c2 = 16 /4 c1 = 4c1       

And            c3 = 25/4 c1        

Substituting these values of c2 and c3 in (2), we get c1 +      

          4c1 + 25/4 c1 = 10'      

or      c1 ( 1 + 4 + 25/4 )= 10'

                   c1 = 10 x 4/45 = 0'.89  

                   c2 = 4c1 = 3'.36  

And            c3  = 25 /4 c1  = 5'.55  

Check:   c1 + c2 + c3 = 0.'89 + 3'.56 + 5'.55 = 10'

 

Hence the corrected angles are

 

? =  78 o 12? 12' + 0'.89  = 78 o 12? 12'.89

 

? = 136 o 48? 30' + 3'.56 = 136 o 48? 33'.56

 

and                                y = 144 o 59? 08' + 5'.55 = 144 o 59? 13'.55

 

Sum  = 360 o 00? 00'+ 00

An angle A was measured by different persons and the following are the values

:                      

Angle             Number of measurements

65 o 30? 10'            2

65 o 29? 50'            3

65 o 30? 00'            3

65 o 30? 20'            4

65 o 30? 10'            3

Find the most probable value of the angle. Solution.

 

As stated earlier, the most probable value of an angle is equal to its weighted

 

arithmetic mean.

 

65 o 30? 10' x 2 = 131 o 00? 20'

 

65 o 29? 50' x 3 = 196 o 29? 30'

 

65 o 30? 00' x 3 = 196 o 30? 00'

 

65 o 30? 20' x 4 = 262 o 01? 20'

 

65 o 30? 10' x 3 = 196 o 30? 30'

 

Sum = 982 o 31? 40' ? weight = 2 + 3 + 3 + 4 + 3 = 15

Weighted arithmetic mean

 

= 982 o 31? 40'

 

= 65 o 30? 6'.67

Hence most probable value of the angle = 65 o 30? 6'.67

 

The telescope of a theodilite is fitted with stadia wires. It is required to find the most probable values of the constants C and K of tacheometer. The staff was kept vertical at three points in the field and with of sight horizontal the staff

 

intercepts observed was as follows.

Distance of staff     Staff intercept S(m)

from tacheometer D( m)

150     1.495

           

200     2.000

           

250     2.505

 

Solution:

 

The distance equation is

 

D = KS + C

 

The observation equations are

 

150 = 1.495 K + C

 

200 = 2.000 K + C

 

250 = 2.505 K + C

 

If K and C are the most probable values, then the error of observations are:

 

150 - 1.495 K - C

 

200 - 2.000 K - C

 

250 - 2.505 K - C

By the theory of least squares

(150 -1.495 K -C)2 +(200 - 2.000 K- C )2 +(250 - 505 K - C )2 = minimum---(i)

 

For normal equation in K,

 

Differentiating equation (i) w.r.t. K,

 

2(-1.495)(150 -1.495 K -C) +2(-20.)(200 - 2.000 K- C)

 

+2(-2.505)(250 - 505 K - C ) = 0     

208.41667 - 2.085 K - C = 0 --------  (2)

Normal equation in C   

Differentiating equation (i) w.r.t. C, 

2(-1.0)(150 -1.495 K -C) +2(-1.0)(200 - 2.000 K- C)

+2(-1.0)(250 - 505 K - C ) = 0         

200 - 2 K - C = 0          ---------  (3)

On solving Equations (2) and (3)     

K = 99.0196        

C = 1.9608

 

The distance equation is:

 

D = 99.0196 S + 1.9608





 

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