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post #11 of 43 Old 02-21-2007
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Quote:
Originally Posted by camaraderie
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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post #12 of 43 Old 02-21-2007
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The maneuvering board is just a piece of paper with a polar plotting grid printed on it. For the problem you describe, all you need do is to put your vessel at the center and plot your friends range and bearing from you. That wil establish the initial relative positions of both boats.

Now, plot a vector representing the course and speed of your friend's boat, originating at the center of the plot. Draw a line parallel to the his relative bearing from you and long enough to intersect the circle that represents your vessel's speed. The intersections thus drawn will be at the courses that will give (1) best closing speed, and (2) best opening speed relative to your friend.

There are also other techniques that allow you, for example, to estimate course and speed of another vessel from its observed relative motion.

The principles are especially useful when using radar because of its ranging capability.
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post #13 of 43 Old 02-21-2007
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Good exerice but academic for sure given the effects of wind, set and drift of two sailboats moving slowly if at all. The calculation begins to have real-world relevance only for power boats and should be used when sailing only as a very rough estimate.
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post #14 of 43 Old 02-21-2007
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You can download a printable copy of a Maneuvering Board Plotting sheet from my website.

www.offsoundings.info/navfiles/dma5090.zip

All the best,
Robert Gainer
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post #15 of 43 Old 02-21-2007
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uh, you'll see him tuesday... or weds, or about an hour before your beer runs out.

We are not primarily on earth to see through one another, but to see one another through

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post #16 of 43 Old 02-21-2007
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This could be a CPA problem using real time vector (radar) Or using "Dutton's" and follow their directions on using the maneuvering board. Either way will solve your intercept problem.
Mark
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post #17 of 43 Old 02-21-2007
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CapnHand...yup that was the problem...180 degrees out of "phase". You answer makes perfect sense now. Thanks.

Funny thing is, there's probably a math forum out there somewhere with people discussing roller furlings.
The really funny thing is that with our penchant for thread drift (mea culpa too!) ....we'll end up talking about roller furling on this thread!!
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post #18 of 43 Old 02-21-2007 Thread Starter
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I have to agree with Camaraderie, it would have to be an Easterly heading. I think the trig approach is correct, if only I knew how to do it. I am home studying for my 100 ton cert. and have not been able to find an explanation of how to solve this in any of the huge stack of books and reference material I have. Thanks for all the ideas, this one makes set and drift problems look easy. John

John
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post #19 of 43 Old 02-21-2007 Thread Starter
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For some reason when I checked in I didn't get the last replies, CapnHand you must be a genius, your answer makes perfect sense from the reference you used. If you think of the trig solution I'd really appreciate you posting it. That question has made me question my sanity trying to "go it alone" for my captains license instead of taking a class. Thanks, John

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post #20 of 43 Old 02-21-2007
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Problem Solved, I Think

OK, John...

I think this will do what you want.

There's a shareware/freeware set of nav programs called EZ-NAV93.EXE which contains a number of useful navigation programs, including one called INTERCPT.EXE which will solve the problem you posed.

You can find EZ-NAV93.EXE at: http://ftp.linux.org.uk/pub/Navigation/

Just download it, save it to your computer in its own folder, and run it. It will automatically generate the 12 nav programs in the same folder, together with documentation. Run the INTERCPT.EXE program, and away you go.

By the way, the answer to your problem seems to be:

28 hours to intercept on a course of 57.8T and a distance of 196 nm.

This assumes: (1) that the program is accurate; and (2) no effect of current or leeway.

Cheers,

Bill
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