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Old 09-07-2009, 08:58   #1
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Wind : Tide Ratio

I was told once about the amount of wind it required to counteract a knot of current/tide. But I am suffering CRS and can not remember the numbers. Does anyone know these ratios?
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Old 09-07-2009, 09:36   #2
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I'm not sure what you're asking. Is it how much wind it takes to just break even sailing against current? If so, I don't see how there could be a single answer, every boat requires a different amount of wind to achieve a certain speed.


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Old 09-07-2009, 10:30   #3
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Originally Posted by cal40john View Post
I'm not sure what you're asking. Is it how much wind it takes to just break even sailing against current? If so, I don't see how there could be a single answer, every boat requires a different amount of wind to achieve a certain speed.


John
I am certainly aware of the zillion factors that effect the boats movement, size of boat, windage, etc etc. But the numbers I was looking for me was a general rule.
An example would be: the wind is pushing me away from a dock, while the tide is pushing me towards the dock. So how much wind does it take to nullify the tidal effect?
The person who gave these numbers said he learned the numbers while being a helmsman in the CG
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Old 09-07-2009, 10:36   #4
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I would think that depended on windage, and hull, or hulls. I can't imagine there being a formula without having a HUGE variable?........i2f
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Old 09-07-2009, 10:40   #5
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A very general rule of thumb...

15 kts wind : 1 kt current

...was what we went by in the Navy (destroyers).
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Old 09-07-2009, 10:42   #6
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For rough estimates, the French Navy teaches that one knot of current or tide is equivalent to about 30 knots of wind for warships. Of course, the exact ratio for a boat depends on the hull shape and top hamper, and also on the directions of wind and current. IME, for a sailing boat, the ratio would be closer to one knot of current to 20 knots of wind.

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Old 09-07-2009, 12:10   #7
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Fluid* dynamics can get pretty complicated, and though I don’t know how, the forces (kinetic energy content) could certainly be resolved mathematically.

*Wind is moving air & currents are moving water - both are fluids, though air is compressible & water is not. The energy content of a moving fluid varies as the square of it's velocity - so we appear (to me) to be shooting at a moving target. I think the ratio will be different at differing base-line velocities.

The equation would have to include the relative densities (854:1) of water (±1025 kilogram per cubic metre) & air (±1.2 kg/m3), the effective projected areas (above & below water), the square of their respective velocities, some factor for compressibility/not ...

Where’s all of our scientists, mathemeticians, & engineers?
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Old 09-07-2009, 12:38   #8
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Fluid* dynamics can get pretty complicated, and though I don’t know how, the forces (kinetic energy content) could certainly be resolved mathematically.

*Wind is moving air & currents are moving water - both are fluids, though air is compressible & water is not. The energy content of a moving fluid varies as the square of it's velocity - so we appear (to me) to be shooting at a moving target. I think the ratio will be different at differing base-line velocities.

The equation would have to include the relative densities (854:1) of water (±1025 kilogram per cubic metre) & air (±1.2 kg/m3), the effective projected areas (above & below water), the square of their respective velocities, some factor for compressibility/not ...

Where’s all of our scientists, mathemeticians, & engineers?
You're also going to need the co-efficient of drag for each.

John
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Old 09-07-2009, 13:52   #9
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John:
The co-efficient of drag for each may be accounted for in EFFECTIVE projected area, but if not, I refer you to " ... "
As I indicated, I don't really know, and call on someone with a better understanding of fluid dynamics.
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Old 09-07-2009, 14:52   #10
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Gord, I think you have it and it and pretty much rules rules of thumb. Military ships sort of fall into a similar category based on the basics so it works for them pretty well. Recreational boats being lighter mass have more variations.

A beach ball floating on the water vs a 10 knot breeze and 6 knot current can have almost any current you can imagine and be blown where the wind goes. In the extreme example comes the answer. It all depends on how much of the above and below the water is exposed vs the mass of the boat. Submarines are not effected by the wind.

Newtonian physics says Force = Mass X Acceleration In this case we subtract the drag for each effective fluid as noted.
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Old 09-07-2009, 16:40   #11
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Well a fat old full keeler will lie to the current a hell of a lot longer than a Bene 393; but any Cat will already have swung ages ago.
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Old 09-07-2009, 18:56   #12
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Quote:
Originally Posted by GeoPowers View Post
15 kts wind : 1 kt current

...was what we went by in the Navy (destroyers).
that is the one he quoted, as he was on a CG cutter
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Old 10-07-2009, 14:43   #13
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Quote:
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Where’s all of our scientists, mathemeticians, & engineers?
Gord, I'm here .
In my first post, I wanted to keep away of technicalities but you seem to be interested in them. Then, I have tried hard to write a post explaining this subject with more detail but it is too much work for today. I give up for tonight and I go to bed.

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Old 10-07-2009, 15:09   #14
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Gordmay, If I remember correctly from my pilot training compressibility effects can be ignored up to speeds of about 200 MPH, or at least airspeed indicators don't have to be corrected for it until you exceed that speed. I would guess that we're talking about speeds less than that here.
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Old 10-07-2009, 16:18   #15
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Thanks Alain & Bill.
If 1 Kt current = 15 Kts wind
What 2 Kts current = ?
Remembering that force varies as velocity squared.
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