the following bearing were observed while traversing with compass and tape check the bearing for local attraction and correct the bearing by the method of including angle.

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written 6.6 years ago by

saxena_archit • 760

modified 6.5 years ago by

sanketshingote • 100

Line	F.B	B.B
AB	188 $^\circ$ 45'	7 $^\circ$ 45'
BC	118 $^\circ$ 15'	298 $^\circ$ 15'
CD	346 $^\circ$ 35'	166 $^\circ$ 30'
DE	337 $^\circ$ 05'	158 $^\circ$ 10'
EA	293 $^\circ$ 30'	113 $^\circ$ 00'

To find the correct including angle

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$\angle$ A = F.B of AB – B.B of EA = 188 $^\circ$ 45’ - 113 $^\circ$ 00’= 75 $^\circ$ 45’

$\angle$ B = F.B of BC – B.B of AB = 118 $^\circ$ 15’- 7 $^\circ$ 45’= 110 $^\circ$ 30’

$\angle$ C = F.B of CD – B.B of BC = 346 $^\circ$ 35’- 298 $^\circ$ 15’= 48 $^\circ$ 20’

$\angle$ D = F.B of DE – B.B of CD = 337 $^\circ$ 05’-166 $^\circ$ 30’= 170 $^\circ$ 35’

$\angle$ E = F.B of EA – B.B of DE = 293 $^\circ$ 30’-158 $^\circ$ 10’= 135 $^\circ$ 20’

Sum of all included angle = $\angle$ A + $\angle$ B + $\angle$ C + $\angle$ D + $\angle$ E = 75 $^\circ$ 45’ + 110 $^\circ$ 30’ + 48 $^\circ$ 20’ + 170 $^\circ$ 35’ + 135 $^\circ$ 20’ = 540 $^\circ$ 30’

Theoretical sum= (2n – 4) × 90 $^\circ$ = (2 × 5 – 4) × 90 $^\circ$ = 540 $^\circ$

Error = 540 $^\circ$ - 540 $^\circ$ 30’ = - 30’ (-ve error)

Error divide into five station = (- 30’ )/5 = - 6’ at each station

Find correct angle

$\angle$ A = 75 $^\circ$ 45’ - 6’ = 75 $^\circ$ 39’

$\angle$ B = 110 $^\circ$ 30’ - 6’ = 110 $^\circ$ 24’

$\angle$ C =48 $^\circ$ 20’ - 6’ = 48 $^\circ$ 14’

$\angle$ D = 170 $^\circ$ 35’ - 6’ = 170 $^\circ$ 29’

$\angle$ E = 135 $^\circ$ 20’ - 6’ = 135 $^\circ$ 14’

To find correct bearing

difference between F.B and B.B of BC is 180 $^\circ$

BB Of AB = F.B of BC – $\angle$ B (corrected angle) = 118 $^\circ$ 15’ – 110 $^\circ$ 24’ = 7 $^\circ$ 51’ F.B Of AB = 7 $^\circ$ 51’ + 180 $^\circ$ = 187 $^\circ$ 51’

BB Of EA = F.B of AB – $\angle$ A (corrected angle ) = 187 $^\circ$ 51’– 75 $^\circ$ 39’= 112 $^\circ$ 12’ F.B Of EA = 112 $^\circ$ 12’+ 180 $^\circ$ = 292 $^\circ$ 12’

BB Of DE = F.B of EA – $\angle$ E (corrected angle ) = 292 $^\circ$ 12’– 135 $^\circ$ 14’= 156 $^\circ$ 58’

F.B Of DE = 156 $^\circ$ 58’+ 180 $^\circ$ = 336 $^\circ$ 58’

BB Of CD = F.B of DE – $\angle$ D (corrected angle ) = 336 $^\circ$ 58’– 170 $^\circ$ 29’= 166 $^\circ$ 29’

F.B Of CD = 166 $^\circ$ 29’+ 180 $^\circ$ = 346 $^\circ$ 29’

BB Of BC = F.B of CD– $\angle$ C (corrected angle ) = 346 $^\circ$ 29’– 48 $^\circ$ 14’= 298 $^\circ$ 15’

Line	Incorrect		Incorrect included angle	-ve error	correct included error	Corrected
	F.B	B.B				F.B	B.B
AB	188 $^\circ$ 45'	7 $^\circ$ 45’	75 $^\circ$ 45’	6'	$\angle$ A = 75 $^\circ$ 39’	187 $^\circ$ 51’	7 $^\circ$ 51’
BC	118 $^\circ$ 15’	298 $^\circ$ 15’	110 $^\circ$ 30’	6'	$\angle$ B= 110 $^\circ$ 24’	110 $^\circ$ 24’	298 $^\circ$ 15’
CD	346 $^\circ$ 35’	166 $^\circ$ 30’	48 $^\circ$ 20’	6'	$\angle$ C=48 $^\circ$ 14’	48 $^\circ$ 14’	166 $^\circ$ 29’
DE	337 $^\circ$ 05’	158 $^\circ$ 10’	170 $^\circ$ 35’	6'	$\angle$ D= 170 $^\circ$ 29’	336 $^\circ$ 58’	156 $^\circ$ 58’
EA	293 $^\circ$ 30’	113 $^\circ$ 00’	135 $^\circ$ 20’	6'	$\angle$ E= 135 $^\circ$ 14’	292 $^\circ$ 12’	112 $^\circ$ 12’
			$\sum$ 540 $^\circ$ 30'	$\sum$ 30'	$\sum$ 540 $^\circ$

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