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Old 23-02-2006, 23:24   #1
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Ocean Currents

I'm taking a class whose text book says that a boat’s progress against an ocean current will be lessoned by the exact same speed increment as the speed pickup when going with the current. This makes no sense to me considering the difference in the shapes of the typical bow and stern.

Does anyone know where I can get some authoritative information on this question?
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Old 23-02-2006, 23:51   #2
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I can't give you a reference, but I can give you the answer.

Have you ever climbed the stairs while riding up an escalator? You get to the top faster. But if you get on the up escalator and try to walk down, it takes longer to get to the top.

The same thing happens in the ocean current. Your boat is sitting on a moving surface (the water) and if you do nothing, it will be carried along at the speed and direction of the current.

If you use your engine to move the same direction as the current, it is just like walking up the moving up escalator. Your engine moves you forward at 5 knots while sitting on a surface that is moving 1 knot. Your speed relative to the ground is 6 knots, though your speed relative to the water is still only 5.

Now, suppose you run your engine to go 1 knot against the current. You move forward at 1 knot over the water, but the water is moving backward at 1 knot over the land. This is just like getting on an escalator and walking the opposite direction that it is moving, but at exactly the same speed. If you only look at the water (or the escalator stairs), you think you are moving, but if you look at a distant island (or something in the building) you see that you are actually standing still relative to other fixed objects.

But if you increase your speed to 5 knots, you will go forward 5 miles and backward 1 mile for every hour. The net result is that you are only moving at 4 knots relative to the ground.

Does that help?
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Old 24-02-2006, 02:14   #3
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The difference is not worth worrying about.....

Fritz,
I'm guessing your question is because you assume the tidal flow on your bow will have less impact on the boats speed as a reversed tidal flow on your stern?

Of course you are right - the profile of most yachts will see less resistance when the boat is moving into any flow as opposed to reversing into it - but the difference would be soooooooo minimal it is hardly worth trying to compute it into any calculation.

IMHO your lecturers / text is giving you the right base formula to use - so I'd stick with it.
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Old 24-02-2006, 03:46   #4
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Fritz:
You ask a good question, to which John gave a good answer.
The text is only referencing thwe mathematical vector relationship:
Either adding (angular) velocity when with the current, or subtracting the same angular velocity when against.
The funny thing about “exact” mathematical navigation/piloting calculations is that they ignore “unknowable” real-world factors, such as you mention, and others like steering accuracy etc.
As john indicated, don’t sweat it - you cannot steer a heading, nor estimate a current’s direction & velocity, with enough accuracy for the difference to matter.
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Old 25-02-2006, 16:52   #5
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Thanks

You all answered as I expected.

I didn't ask for any practical application -- ather, because one of my fellow students said it was simple physics, when it just didn't make sense to me.

Thanks,

Fritz
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Old 25-02-2006, 17:16   #6
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Hey Fritz.

Are you a "physics" teacher? Or a "mathematician?"

A friend of mine is pretty much into physics. And anything to do with science fiction as well.

Can you say "trekkie?"
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Old 26-02-2006, 02:21   #7
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Search "Vector Addition" - there are innumerable tutorials.
Your problem is a simple "head to tail" addition, verses subtraction. It becomes a little more complicated when the angles are other than 180 degrees apart (exactly opposite).

It is simple mathematics.

You have noticed (and introduced) one of the "practical" considerations that would make the simple mathematical model scientifically inaccurate.

It doesn't matter, because any calculation is only as accurate as it's least accurate component. In the real world, the actual differing "Force" (acting on the boat) between a "lifting" current and a "heading" current, will be much less than your estimation of it's exact angle and velocity (and even your boat's heading).

Ignore that which is unknowable - otherwise it will drive you crazy

HTH
Gord
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