Wind : Tide Ratio

[AK_sailor]

Originally Posted by Pblais

.... 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.....

Mass of the boat is a factor, but I think the area exposed under the water to catch the current vs. above the water in the wind is most significant. Another extreme example:

River rafters in my area run into this situation frequently. Often a very strong afternoon wind blows upsteam. A river raft (even heavily loaded) tends to have a big area exposed above the water, but draws only a few inches below the water. In many cases the wind is strong enough that one needs to row hard downstream to make any progress at all. This has important safety implications if someone falls in. A swimmer has much less mass than the raft, but only a small area exposed above the water in the wind. Most of the person is below the water and pushed by the current. It can be impossible to rescue a swimmer in these conditions because they will be carried down stream much faster than the raft can travel even when being rowed hard.

Bottom line is that it is a contest between the area below the water being pushed by the current vs. the area above the water being pushed by the wind. Thus very different for each vessel. From experience one could no doubt come up with a rule of thumb for an individual boat or class of boats ("a Catalina 30 will do thus and thus..."), but I would be very skeptical of any general rule of thumb that claims to be even approximately correct for all boats.

[GeoPowers]

Originally Posted by GordMay

Thanks Alain & Bill.
If 1 Kt current = 15 Kts wind
What 2 Kts current = ?
Remembering that force varies as velocity squared.

2 kts current : 30 kts wind

It was a rule of thumb, barely empirical. Yet it worked for those that trained me, and hopefully those I trained. Really, the point was to respect the relationship of current/wind/set/drift, especially pierside. Besides, every engineer should know that experience in the field beats an equation any day of the week! And yes, stated from an engineer.

[GordMay]

I’ve stated the following as a set of assertions, but intend them to actually represent questions.
I’ll welcome any explanations invalidating or proving my hypotheses.

The Velocity-Squared Law states (something to the effect) that force on an object varies as the square of the fluid velocity.
Hence, doubling the velocity of a fluid (water or air), quadruples the force acting on a body immersed in that fluid.
I think that this represents an exponential (not linear) growth curve.

It seems to me, that the Force ratio between water power & wind power would have to be an exponential (or logarithmic ?) relationship - not linear.

Accordingly

IF we observe that:
Water @ 1 Kt = Wind @ 15 Kt

THEN I would extrapolate:
Water @ 2 Kt (1 x 2) = Wind @ 21.2 Kt (15 x 1.414*)

*Root 2 = 1.414

[Lodesman]

Originally Posted by AK_sailor

Mass of the boat is a factor, but I think the area exposed under the water to catch the current vs. above the water in the wind is most significant.

You can't treat current like wind. A boat whether moving or stationary moves with the water at the speed of the water. Wind will eventually move a boat, but no matter how long it blows, will only ever achieve a fraction of wind-speed. The size of this fraction depends on the windage, so will be less bow or stern-to than beam-on. Somewhere between 1/15th and 1/10th of windspeed is very general rule-of-thumb