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Old 02-22-2007
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Ok, so this is simpler than I first thought. This is way I should have done it the first time.

First of all, this is the situation. The distance the friend’s boat will travel is 5t (t = hours) NM on a course of 036 Deg T. You are 84NM from him at a bearing of 096 Deg T relative to his boat. In order to meet, you will travel a distance of 7t NM at some unknown heading.

It’s obvious that the angle between the track of your friend’s boat and the bearing from his boat to yours is 96 – 36 = 60 Deg
http://www.sailnet.com/photogallery/...ll=no&dthumbs=

Remember the law of sines? It’s handy. It says that for any triange the ratios of the length of each side to the angle opposite that side are equal.

If A is the angle between the heading you need and the bearing from your boat to your friends,

that means: sin60 / 7t = sinA / 5t

or sin –1 (5/7 * sin 60) = A

A = 38.2 Deg

Your heading then should be 96 + 38.2 + 180 = 314.2 Deg T

How long to get there?

We all remember that the sum of all the internal angles of a triangle = 180 so the remaining angle is 180 – 38.2 – 60 = 81.8 Deg

From the law of sines, 7t = 84 * sin 60 / sin 81.8

t =10.5 hours

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

Last edited by CapnHand; 02-22-2007 at 09:22 PM.
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Old 02-22-2007
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Pretty slick! And not a bad way to get thrown out of a cocktail party either!

Really nice job CapnHand.
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Old 02-22-2007
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Capt Hand,

Yeah, easy, but I believe confusing on several counts.

First, the friends boat (Vessel #1) is on a COURSE of 036T, not a BEARING. His speed is 5 knots. His distance from your boat (Vessel #2) is 84nm. The BEARING to/from the vessels is ambiguous, the way it was presented. Normally, bearings are taken FROM your vessel, but the wording in the original is ambiguous, as in your solution. Either way, the answers provided do not agree with the computed values. Your boatspeed is 7 knots.

Let's assume that the bearing of 096T was, indeed, from Vessel 1 to Vessel 2. The reciprocal bearing, from your vessel to your friend's vessel, is then 276T.

If we then set up the problem in a conventional way, it looks like this:

Vessel #1 (friends boat): Bearing 276T Distant 84nm Course 036T Speed 5 knots

Vessel #2 (your boat): speed 7 knots

Required: course, time, and distance to intercept

According to the calculator (both of the aforementioned calculators agree):

Course to intercept: 314.21 T
Time to intercept: 10 hours 30 mins
Distance to intercept: 73.5 nm

Bill

Last edited by btrayfors; 02-22-2007 at 10:55 AM.
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Old 02-22-2007
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Quote:
Originally Posted by CapnHand
Ok, so this is simpler than I first thought. This is way I should have done it the first time.

First of all, this is the situation. The distance the friend’s boat will travel is 5t (t = hours) NM on a bearing of 036 Deg T. You are 84NM from him at a bearing of 096 Deg T relative to his boat. In order to meet, you will travel a distance of 7t NM at some unknown heading.

It’s obvious that the angle between the track of your friend’s boat and the bearing from his boat to yours is 96 – 36 = 60 Deg
http://www.sailnet.com/photogallery/...ll=no&dthumbs=

Remember the law of sines? It’s handy. It says that for any triange the ratios of the length of each side to the angle opposite that side are equal.

If A is the angle between the heading you need and the bearing from your boat to your friends,

that means: sin60 / 7t = sinA / 5t

or sin –1 (5/7 * sin 60) = A

A = 38.2 Deg

Your heading then should be 96 + 38.2 + 180 = 314.2 Deg T

How long to get there?

We all remember that the sum of all the internal angles of a triangle = 180 so the remaining angle is 180 – 38.2 – 60 = 81.8 Deg

From the law of sines, 7t = 84 * sin 60 / sin 81.8

t =10.5 hours
While I applaud your attempt to answer the question, I see it as seriously flawed.

Given that your distance from his boat is 84NM to start with, and his is moving away from you, and you only move at 7 knots... I don't see how the answer could be 10.5 hours.

Draw the vectors... In 10.5 hours you have gone only 73.5nm on a heading of 73.8˚T. His position has now become 52.5nm @ 36˚T + the original 84NM @ 96˚T in the same 10.5 hours—119NM from your original position, or 45.5NM further along the same heading.

If you assume you were at coordinate (0,0) and have moved to (70,20.5)... boat B's position has gone from (83, -8.5) to (114, 33). You're still nowhere near them... Your solution assumes an interior angle of 60 degrees, but it has an interior angle of 120 degrees (96 -> 36). However, you are on a direct vector to their current position.

Sailingdog

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Telstar 28
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Last edited by sailingdog; 02-22-2007 at 11:11 AM.
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Old 02-22-2007
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dog,

I think his time to get there (10.5 hours) is correct, that is, if you assume that the ambiguous bearing info given is actually the bearing FROM the friend's boat TO your boat, as calculated in my response.

This confusion would not have existed if the original post had been unambiguous and/or phrased in the conventional way (i.e., FROM the maneuvering vessel TO the target vessel).

Bill
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Old 02-22-2007
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True... the wording is somewhat unusual...and makes a huge difference in the calculations.

Sailingdog

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Telstar 28
New England

You know what the first rule of sailing is? ...Love. You can learn all the math in the 'verse, but you take
a boat to the sea you don't love, she'll shake you off just as sure as the turning of the worlds. Love keeps
her going when she oughta fall down, tells you she's hurting 'fore she keens. Makes her a home.

—Cpt. Mal Reynolds, Serenity (edited)

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Old 02-22-2007
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On a paper chart, plot your friend's DR track out past the time you want to meet, then set a compass to the distance you can make during that same period. Set the compass point on your current position, and strike an arc through your friend's DR track; this is point where and when the meeting will take place, if you steer the course between your current position and there. If the arc does not intersect your friends track, one of your assumptions is unrealistic (speed, etc.). You can adjust for set and drift as you near each other. That's the way I did it thirty years ago, in the Navy, on the bridge.
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Old 02-22-2007
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tmhdgpth,

That's a useful technique, but doesn't respond directly to the problem as stated.

This is because there is no fixed time (time desired) for meeting. Rather, the desired meeting time is the EARLIEST time possible, given the relative positions of the vessels, the (fixed) course of one of them, and their respective speeds.

It's a classic intersect problem used in war gaming and targeting, as well as air and maritime navigation. Maneuvering Boards were the classic way to solve the problem, and a few brave souls used trig, but now calculators and computers make it easy.

Regards,

Bill
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I'm impressed with all the effort that has gone into solving this. I apologize for the ambiguity in my original post. I meant the relative bearing TO the other boat (friend)was 096T, but did not make that clear. The actual USCG question uses different courses, I was looking for the method, not an answer, and if I understood trig better (at all?) I think you guys have done it. The testing is multiple choice with four possible answers given. The answers only vary by 2-3 degrees each so you have to be right on. In the actual test nothing except a simple calculator, plotter, parallel rule, dividers and a pencil are allowed. By the way, the CG has a data bank of 14,000 total questions, (Navigation General, Rules of the Road, Deck General, Navigation problems) and I am working my way through all of them(lot's of duplicates). Thanks again for all the help and ideas. If I come across something else that stumps me I'll post it. Now, back to how many round turns to put on a bitt before figure eights using nylon line:-)) John

John
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Old 02-22-2007
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The more I read, the easier the maneuvering board solution seems to be.
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