Bearing Question. Please help
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Bearing Question. Please help

[From: ] [author: ] [Date: 12-04-15] [Hit: ]
vcMH = Calc Heading & Mag of vd = 168.43990 , Heading =250.08635 deg.The distance from the starting point is 168.The heading back to the starting point is 250 deg.......
A ship travels 84 km on a bearing of 10 degrees, and then travels on a bearing of 100 degrees for 146 km. Find the distance of the trip from the starting point, to the nearest kilometer.

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= <84cos10 + 146cos100, 84sin10 + 146sin100>
|D| = √[(84cos10+146cos100)² + (84sin10+146sin100)²] ≈ 168 km

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When doing navigation type problems it is convienent to use a coordinate system with x North and y East.
and use heading (the angle from North that is measured positive toward the east)
a vector v of magnitude vm is then < vm cos(heading), vm sin(heading)>
bearing and heading are the same

vaMH = 84 km, Heading =10 deg.
va = Calc Vector from vaMH = < 82.72385, 14.58645> km

vbMH = 146 km, Heading =100 deg.
vb = Calc Vector from vbMH = < -25.35263, 143.78193> km
km
vc = va + vb = < 57.37122, 158.36838> km

start = < 0, 0>

vd = start - vc = < -57.37122, -158.36838>

vcMH = Calc Heading & Mag of vd = 168.43990 , Heading =250.08635 deg.
The distance of the trip is 84+146 = 230 km
The distance from the starting point is 168.4 km
The heading back to the starting point is 250 deg.
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