Navigation problem
Are there any navigators here? What is the procedure for plotting an intercept course, as in meeting another boat that is approaching from a different direction. For example, if my friend is on a steady course of 036T and travelling at 5K and I am 84NM away on a bearing of 096T and I make 7K, how do I plot a new course to intersect him so we arrive at the same spot at the same time?(assume a constant track and speed) Anyone? Thanks, John Davidson

I should begin by saying that I have never wondered this before. A relatively easy way occurs to me though assuming that you have and use paper charts. Plot your friend's position, and his course. Next mark of where you expect him to be in .5 hours, one hour, 1.5 hours etc, using a compass and your scale from the chart. Plot your position. Then use the compass to draw arc segments to show where you can be in half hour increments. At some point they should come very close to coinciding, and you can plot a course to that point.
Any better ways? 
I always got lost on the part about the "train leaves Boston traveling at 45mph and another train leaves Denver...." I knew I should have been paying more attention in school!!!

If only there were web enabled cell phones.....tests today must be a snap!

I was going to say that if you figure out where his track will put him in 7 hours, he'll have gone 35 miles. If you aim at that point for seven hours, you will have gone 49. 35 and 49 make the 84 miles you said separated you. However, you said your bearing from him was 96šT, so that's not right. Yotphix seems to have the right idea, but I'm thinking it will take something like 18 to 19 hours to get to see your friend based on the original distance and bearing.

You describe a maneuvering board problem. You use the maneuvering board sheet according to the directions printed on it.
It is a quick graphic method for calculating relative motion between two moving objects. (Your boat and your friend's.) It is an excellent way of calculating such things as the course and speed of another vessel or closest point of approach to it. Once the technique is learned, it can be used mentally for a rough but quick result when needed. Good stuff to know. Here are links to places that can supply a pad of them and a manual for using same: http://www.celestaire.com/catalog/products/5090M.html http://www.usskawishiwi.org/Crew/Spa...ingBoard.html 
John,
The maneuvering board solution, as others have mentioned, is the classic way to solve this problem. However, in these days of computers there are obviously programs which can do the calculations rapidly. I don't know what the best of these might be, but there is a little program which was designed for gaming (submarines, torpedos, etc.) which can solve such problems. While the calculations are in yards or meters, and there are some limits to distance, etc., with a little ingenuity you can find the solution. The program is called Intercalc 1.1 Just Google it any you'll find it. Bill 
It's a trig problem. In cartesian coordinates your buddies boat starts at x1=0, y1=0 and at any time will have x1=2.939t, y1=4.045t. Your boat starts off at x2=83.54, y2=8.78 and will move from there and at any time have a position of x2=7tsin(a)+83.54, y2=7tcos(a)8.78 where x and y have units of NM, t is in hours and a is the heading in degrees.
Solving this for when x1=x2 and y1=y2, you catch him after ~10.5 hours with your boat's heading of ~314 degrees. 
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?? 
Let's not even get into set and drift...although the sea will.

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