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Gillespie
15th Feb 2005, 16:30
Hi,

Given; Distance 'A' to 'B' is 90nm, a fix is obtained 60nm along and 4nm to the right of course. what heading alteration must be made to reach 'B'?

I can work this out using trig, but this is abviously far too clumsy to work in the air, what is the quickest way to calculate this. What is the '1 in 60' rule.

Thanks.

A251
15th Feb 2005, 16:42
1 in 60 rule

RxA=60xD

a being the angle
d being the opp
r being the adj



hope that helps

ted

Send Clowns
15th Feb 2005, 17:24
Which you will calculate puts you 4° off track (sorry, course in JAA speak), and requiring an 8° closing angle to recover track, for a total turn of 12° port to reach the turning point.

HandspringGuy
15th Feb 2005, 17:31
Gillespie,

In flight the easiest way is to use the inverted fraction rule
(this is based on the 1/60 rule but much easier to calculate in your head)

You are 60nm along a 90nm track -this is 1/3 of your track left to run
Invert the fraction and multiply it by your angle off (not distance)

4 x 3 /1 = 12 degrees left to correct

This works for any angle or distance along track

Lee Frost
15th Feb 2005, 19:18
Hi Gillespie,


Look for a copy of the air pilots manual vol3, Navigation. An excellent wee book which covers your question exhaustively (and much more). Well illustrated. ISBN 1-84037-140-4

Good luck,

LF