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Old 02-21-2007
CapnHand
humble pie rat

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Quote:
CapnHand...I was trying to work that out myself on graph paper. Knew there had to be a trig solution but was at a loss! Are you sure of your answer? At 84 miles away...it would take 12 hours at 7 knots for "my" boat to reach my friends if it were stationary and in a position almost due East from me.

Since the friends boat is moving away from my boat on a course of 36 degrees T ...(Northeast)it does not make intuitive sense that I could close the gap in less time...especially with a heading of 314T or northwest.
It has been a long time since high school...where am I going wrong here?

I am thinking that your starting position for my boat is perhaps different than mine based on our reading of the problem. I am assuming my friends boat is on a bearing of 96T taken from my boat...perhaps you are reading it as a bearing of 96T taken from my friends boat??
The post read "I am 84NM away on a bearing of 096T" so I took both the distance and bearing to be relative to the friend's boat. If it stated that "HE is 84NM away on a bearing of 096T", then I'd have taken the distance and bearing relative to his boat.

So from the perspective I took, yes, I'm sure of my answer. Does my answer make more sense now? The fact that you disagreed indicates that you were thinking about it too and that's good.

Last time I was in high school was to pick up my son. This was a good mental excercise, jrd. Thanks for posting the question.

There's another way to figure this out using vector math instead of cartesian coordinates. It's simpler and more elegant, but I couldn't remember how to do it last night, so I just went ahead with the trig. It's coming back to me, I'll post it when I get it straight. If someone else knows, have a go.

Funny thing is, there's probably a math forum out there somewhere with people discussing roller furlings.

There are 10 kinds of people. Those who understand binary and those who don't.

Last edited by CapnHand; 02-21-2007 at 01:59 PM.
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