Consecutive Co-ordinates:

1

Consecutive Co-ordinates:The northing and easting of any point with reference to the preceding point are called the consecutive co-ordinates of that point.

Independent or Total Co-ordinates:The co-ordinates of any point with respect to a common origin are known as the independent co-ordinates or total co-ordinates of that point.

Horizontal Control

TRAVERSE

We have started to compute the traverse station coordinates from an anticlockwise polygon traverse.

We tested the sum of the observed angles against the sum of the internal angles of the polygon, and if within an acceptable value we distributed the misclosure.

We established the bearings of all the traverse lines and then calculated the changes in Easting and Northing for each line.

The algebraic sum of these changes should be ZERO. The values give the error eE

and the error eN.The linear misclosure e = (eE

2 + eN2) was calculated and used to find the FLM. If

acceptable then distribute the errors.

a) Bowditch Method - proportional to line distances

b) Transit Method - proportional to N valuesE and

c) Numerous other methods including Least Squares Adjustments

By : -

a) Bowditch Method - proportional to line distances

This rule, also termed as the compass rule, is used to balance the traverse when the angular and linear measurements are equally precise.

The eE and the eN have to be distributed

For any line IJ the adjustments are E IJ and N IJ

E IJ = [ eE / perimeter of traverse ] x D IJ Applied with the oppositesign to eE

N IJ = [ eN / perimeter of traverse ] x D IJ Applied with the oppositesign to eN

CO-ORDINATE DIFFERENCES

CALCULATED

WHOLE

CIRCLE

BEARING

HORIZONTAL

DISTANCE

D E N

00 00 00

306 12 51

195 54 06

47 44 33

638.57

1576.20

3824.10

3133.72

0.000 +638.570

-1271.701 +931.227

-1047.754 -3677.764

+2319.361 +2107.313

-0.094 -0.654eE eN

e = (eE2 + eN

2) = (0.0942 + 0.6542) = 0.661m

9172.59

Fractional Linear Misclosure (FLM) = 1 in SD / e

= 1 in 9172.59 / 0.661 = 1 in 13500Check 2

E IJ = [ eE / SD ] x D IJ Applied with the oppositesign to eE

eE = -0.094m SD = 9172.59 m

E IJ = [+0.094 / 9172.59 ] x D IJ = +0.0000102479…... x D IJ

Store this in the memoryFor line AB

E AB = +0.0000102479…x D AB = +0.0000102479…x 638.57

E AB = +0.007mFor line BC

E BC = +0.0000102479…x D BC = +0.0000102479…x 1576.20

E BC = +0.016mFor line CD

E CD = +0.039m For line DA E DA = +0.032m


CALCULATED ADJUSTMENTS ADJUSTEDCO-ORDINATES

E N E N E N E N

STATI

ON

0.000 +638.570

-1271.701 +931.227

-1047.754 -3677.764

+2319.361 +2107.313

-0.094 -0.654

eE eN

+0.007

+0.016

+0.039

+0.032

N IJ = [ eN / SD ] x D IJ Applied with the oppositesign to eN

eN = -0.654m

N IJ = [+0.654 / 9172.59 ] x D IJ = +0.000071299…... x D IJ

Store this in the memory

N AB = + 0.000071299… x D AB = + 0.000071299…x 638.57

N AB = +0.046m

N BC = +0.112m N CD = +0.273m

N DA = +0.223m



E N E N E N E N

STATI

ON

0.000 +638.570

-1271.701 +931.227

-1047.754 -3677.764

+2319.361 +2107.313

-0.094 -0.654

+0.046

+0.112

eE eN

+0.007

+0.016

+0.039

+0.032

+0.273

+0.223

+0.007

-1271.685

-1047.715

+2319.393

S= 0

+638.616

+931.339

-3677.491

+2107.536

S= 0

A

B

C

D

A

3000.00 4000.00

3000.01

1728.32

680.61

3000.00

4638.62

5569.96

1892.46

4000.00

Check 3

9

Transit Method

The eE and the eN have to be distributed

For any line IJ the adjustments are E IJ and N IJ

E IJ = [ eE / arithmetical sum of all eastings ] x easting of that side

N IJ = [ eN / arithmetical sum of all northings ] x northing of that side

Applied with the oppositesign to eE

Applied with the oppositesign to eN

The transit rule may be employed to balance the traverse when the angular measurements are more precise than the linear measurements.


CALCULATED

WHOLE

CIRCLE

BEARING

HORIZONTAL

DISTANCE

D E N

00 00 00

306 12 51

195 54 06

47 44 33

638.57

1576.20

3824.10

3133.72

0.000 +638.570

-1271.701 +931.227

-1047.754 -3677.764

+2319.361 +2107.313

-0.094 -0.654eE eN

e = (eE2 + eN

2) = (0.0942 + 0.6542) = 0.661m

9172.59


= 1 in 9172.59 / 0.661 = 1 in 13500Check 2

eE = -0.094m arithmetical sum of all eastings =4638.816

E IJ = [+0.094 / 4638.816 ] x easting of that side = +0.0000202637…... x D IJ

Store this in the memoryFor line AB

E AB = +0.0000202637…x E AB = +0.0000202637…x 0

E AB = +0 m For line BC

E BC = +0.0000202637…x E BC = +0.0000202637…x 1271.701

E BC = + 0.0257 mFor line CD

E CD = + 0.0212 m For line DA E DA = +0.047m

E IJ = [ eE / arithmetical sum of all easting ] x easting of that side

Applied with the oppositesign to eE



E N E N E N E N

STATI

ON

0.000 +638.570

-1271.701 +931.227

-1047.754 -3677.764

+2319.361 +2107.313

-0.094 -0.654

eE eN

+0

+0.0257

+0.0212

+0.047

eN = -0.654m

N IJ = [+0.654 / 7354.874 ] x northing of that side = +0.00008892…... x D IJ

Store this in the memory

N AB = + 0.00008892… x D AB = + 0.00008892…x 638.570

N AB =+0.0567m

N BC =

N CD =

N DA =

N IJ = [ eN / arithmetical sum of all northings ] x northing of that side

Applied with the oppositesign to eN


CALCULATED

WHOLE

CIRCLE

BEARING

HORIZONTAL

DISTANCE

D E N

00 00 00

306 12 51

195 54 06

47 44 33

638.57

1576.20

3824.10

3133.72

0.000 +638.570

-1271.701 +931.227

-1047.754 -3677.764

+2319.361 +2107.313

-0.094 -0.654eE eN

e = (eE2 + eN

2) = (0.0942 + 0.6542) = 0.661m

9172.59


= 1 in 9172.59 / 0.661 = 1 in 13500Check 2



E N E N E N E N

STATI

ON

0.000 +638.570

-1271.701 +931.227

-1047.754 -3677.764

+2319.361 +2107.313

-0.094 -0.654

eE eN

+0

+0.0257

+0.0212

+0.047

SURVEYING – I (CE- 128)

16

Local AttractionThe magnetic needle does not point towards the magnetic north under the influence of the external attractive forces. Such a disturbing influence is known as Local Attraction.Compass bearings cannot, therefore, be relied upon unless means are taken to detect the presence of local attraction and eliminate its effects.

To detect its presence, it is only necessary to observe the bearing of each line from both its ends. If the for and back bearings differ by 180o, there is no local attraction at either station, provided the compass is free from instrumental errors, and no observational errors are made. If the fore and back bearings differ by 180o nearly, the back bearing is increased or decreased by 180o to get the corresponding fore bearing, and the mean is taken between this and the observed fore bearing, e.g. the observed fore and back bearings are 90o0’ and 276o30’. The fore bearing calculated from the back bearing is= 96o30’. The mean of the observed and calculated fore bearings= 96o15’, which is the corrected bearing of the line.


17

The amount of local attraction is the same for each of the bearings observed at the affected station. It may, therefore, be remembered that the differences between the bearings of the lines observed at the station will give the correct values of the angles between the lines even though the station is affected by local attraction provided they are taken at the same time with the same instrument.

There are two methods of correcting the observed bearings of the lines.

(1)In the first method the true included angles at the affected stations are computed from the observed bearings. Commencing from the unaffected line and using these included angles, the correct bearings of the successive lines are computed as already explained.


18

Example:- Suppose, for example, the observed bearings of the lines AB, BC, CD, &DA are:

Line Fore Bearing Back Bearing

AB 46o10’ 226o10’

BC 119o20’ 298o40’

CD 169o30’ 351o10’

DA 280o20’ 99o20’


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(2) In the second method, which is in most common use, the included angles are not computed, but the amount and direction of error due to local attraction at each of the affected stations is found. Starting from a bearing unaffected by local attraction, the bearings of the successive lines are adjusted by applying the corrections to the observed bearings.

Example:-

Line Observed bearing

Line Observed Bearing

AB 44o40’ CD 30o40’

BA 225o20’ DC 212o2’

BC 96o20’ DE 320o12’

CB 274o18’ ED 140o12’


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Consecutive Co-ordinates: