I asked Jim Michalak about my windage question. Here's his very prompt reply.
jim michalak wrote:
> Mark A. wrote:
> He gives the formula for wind force on any panel as .0034 x S x C x V x V in normal air at
> sea level.
> But because the theme is really the effect of hiking out, I get confused about whether
> pounds and foot pounds are equivalent here.
It is in pounds. it works out to about a pound per square foot in a
twelve knot wind. It would be four psf in a 24 knot wind.
> So ( with some uncertainty ), I calculate the force of a combined 10 kt wind on the Clam Skiff
> house's 25 sq. ft front end as 17 pounds. Since about 60 pounds of thrust or bollard pull
> equals 1 hp, then it takes barely 1/4 hp extra to power against that wind. A 15 kt breeze
> is still only 38.25 pounds, or a bit more than 1/2 hp.
Sounds about right. I originally planned for Harmonica to be an electric
boat but figured a normal trolling motor wouldn't push it into a 10 mph
wind. Same as your calculations.
> Figuring this way, it's hardly more dramatic if the boat is trying to go 20 kt dead
> against a 10 kt air: .0034 x 25 x 2 x 30 x 30 = 153 pounds, only 2 1/4 hp.
> This either demonstrates the reason why Philip Bolger stopped caring much about the
> windage on his boats; or that my own lead line's a little short.
I suppose in power needed it might not be a big deal but I do think it
can cause handling problems at times both because of the windage and
because of the raised cg. The side windage on say a 6' x 8' cabin in 24
knots would be about 190 pounds.