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George Huxtable [mailto:george@HUXTABLE.U-NET.COM] said:
>
> I don't think Peter Smith's solution can possibly be correct: commonsense
> tells me so. We have two ships travelling in nearly the same direction,
and
> the second ship is slowly overtaking the first. In that case, the speed of
> the overtaking ship must be greater than that of the ship being overtaken.

Hmmm. I'm not sure from the above who's the "second ship" and who's
the "first ship". The target is ahead of us and to Port (bearing 322d).
The bearing is constant, so we are converging. Since the target is
ahead and the range is decreasing, we must be overtaking the target.

> ...
> If you add the vectors 12 knots at 150 and 1.45 knots at 142, using trig
or
> drawing, you end up with 13.44 knots at 149.1.

But that gives the target a greater speed than our own ship, even
though we are overtaking it.

Moreover, the target is on our Port bow (bearing 322d). For our
courses to converge (constant bearing), the target's course must
be a little to starboard (i.e., numerically GREATER) of ours. Thus,
if we're making 150d, he should be making a little greater then 150d.
If he's making 149.1d, we would be diverging.

Can you read an Excel spreadsheet? I have both the geographic and
relative vectors problems worked out trigonometrically. I'll send
it to you and see what you think.

 -- Peter

   

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