Ex31-1 Calculate the missing data of the closed loop traverse ABCD:

Figure Example 31.1

Solution : For D ABD, let l and q be the length and bearing of the side BD. Now, from the condition of a closed loop traverse,

SL = 55° cos 60° + l cos q + 105° cos 310° = 0

Therefore, l cos q = - 949.927 m

Similarly, from SD = 0

or, 550 sin 60° + l sin q + 105° sin 310° = 0

Therefore, l sin q = 328.033 m

Therefore length of the traverse side, BD i.e.,

l = = 1004.971m

Or,

q = 180° - 19° 3' 3'.7 E = S 19° 3' 3.7" E

Bearing (WCB) of BD = 180° - 19° 3' 37" = 160° 56' 56".3

cos b = = 55° 13' 48"

Similarly, cos a = = 0.6947

a = 45° 59' 48"

cos g = = 0.1947

g = 78° 46' 24"

check = a + b + g = 45° 59' 48" + 55° 13' 48" + 78° 46' 24"

= 180° (O.K)

In the closed traverse BCD since it is being traversed clockwise the inlcluded angles will be exterior angles.

F.B of BC = F.B of DB + RDBC

= (180° + 160° 56' 56".3) + (360° - 45° 59' 48")

= 114° 57' 08".3

F.B of CD = F.B of BC + RBCD

= 114° 57' 08".3 + (360° - 55° 13' 48")

= 239° 43' 20".3

<< Back | Next >>