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Philip Ouvry said-

>When instructing I would pose the following problem to my students.
>
>You are in a boat capable of 4 knots intending to cross a wide river.   Your
>destination is straight across the river.   In the centre of the river the
>current runs at 7 knots but close inshore on either side it reduces to 2
>knots.   What should be your strategy for crossing the river?
>
>My strategy was to cross the river at right angles until close inshore on
>the far side and then proceed against the current (2 knots) to reach the
>destination.   Is there any other strategy?
>
>Philip Ouvry

================

There's another strategy that's exactly equivalent, in that one could do
some, or all, the upstream part of the passage close to the near bank,
before the crossing, instead of close to the far bank, after it: but that's
only a trivial difference offering no saving.

Indeed, I've seen the little Kyle Rhea car ferry using just such a
strategy, when a spring tide was pouring through the narrows, far faster
than the ferry could counter. She would hug each bank, where the force of
the tide was broken by the rocks and boulders at the edge, to get some
distance uptide, making her dash across in-between. Presumably, that ferry
no longer exists, now that the road to Skye has been bridged.

Going back to Plilip's question, I think there's a real saving to be got in
the crossing phase. If, instead of heading at right angles (which I presume
was what Philip intended when he said "cross the river at right angles")
you can head just a bit upstream of that. If that angle is only small, then
the crossing distance and the crossing time are increased by a negligibly
small amount, but there's a significantly greater saving in reducing the
net downstream drift.

Presuming  the inshore 2-knot zone to be of negligible width, I find that
the optimum angle to head when setting off across the current, is something
like 26 degrees upstream from the right-angle heading that Philip proposed;
in which case there will be a saving of about 10% in the overall time taken
for the crossing.

I should have found this optimum-angle using calculus, but discovered that
mine has got a bit rusty, so used successive approximation to get the above
result, I'm rather ashamed to admit.

If Philip's students arrived at the same answer as mine, rather than
Philip's own result, I hope they didn't get marks deducted for doing so. If
I'm right, that is...

George.

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