| Ray,
I have a simple graphic approach. First plot Boat A's course. Then
using dividers mark out concentric circles, with each circle equal to a
distance boat B can make good given boat B's average speed. Label each
concentric circle with the time needed to get there. Start the circles
at boat B's initial position. Eventully you will find at least 1 point
that will intersect boat A's line of position.
Now do the same for Boat A to see when boat A can get to that point.
This will be a trial and error method, but will give you a "good feel"
for the time and speed for the 2 boats.
Frank
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| Seems to me that the problem can be solved mathmatically. I'm too lazy
to create a cute line drawing but you are trying to solve a triangle
which looks something like
B
o
A C X
o o o
Boat A is maintaining a course A-X and boat B wishes to intercept at some
unknown point C along A's track. The trick is to solve the triangle ABC
to obtain angle ACB which is the amount above or below boat A's course
that B must sail to intercept at C. What's shown is an acute triangle -
it will also work with an obtuse triangle where B is starting out
"behind" A. At least it will if B is faster than A.
Since you know the course of boat A (A-X in this case) and you know the
relative positions of A and B you can therefore determine the angle BAC
either by plotting it or working out the coordinate transform. Now
consider the sides of the triangle A-C and B-C. Both boats will be
getting to the intercept point C at the same time and therefore the
distances they travel (the lengths of the sides A-C and B-C) will be
proportional to the speeds of the two boats.
Now, my geometery is pretty rusty (poor old Miss Noyes would never
forgive me for being unsure about this - "Someday you're going to be in a
boat trying to catch another boat and you'll wish then that you'd spent
more time studying and less time ... ") but I seem to remember that in a
plane triangle the ratio of the sine of each angle with the length of the
side opposite is the same for each of the pairs of angles and opposed
sides. A/sine(a) = B/sine(b) = C/sine(c).
Anyway, you know the angle BAC, and you know the ratio of the boat speeds
and you can thereby determine the angle ABC and subtracting angles BAC
and ABC from 180 degrees gives the course correction angle ACB. It works
out to be something like
Course(B) = Course(A) +/-(180 - BAC - ASIN(SIN(BAC)*Speed(A)/Speed(B)))
The triangle can't always be solved if A is faster than B.
I would guess that there exists a plotting device similar to the
"computer", used by aviators to work out wind drift problems, which would
give you an "analog" solution.
Of course, if you're actually ON the boat (and you can make sufficient
speed), it's much simpler. Just steer so as to approach the other boat
at a constant relative bearing - by and by you'll reach point C.
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