This formula calculates the drag force of the wind, for a given wind velocity, on a certain surface. Whether you chose to use this force to move your sailboat forward or to tear your sails is another issue though...
The metric version gives the result in Newtons, when you insert air density in kg/m3 (it is almost 1.2 under Standard Temperature and Pressure), wind velocity in m/sec, sail or surface area in m2.
The drag coefficient is the resistance of the unique shape of the object under wind effect. It defines how aerodynamic a certain shape or design is. If I recall
right, for most cars we drive, it is 0.25 - 0.30.
If you want to calculate the drag force, say, on a rectangular plate, where the wind is blowing directly (i.e. 90 degrees) onto it, you should be taking the drag coefficient as 1. This means the force of the wind is not lessened at all by the shape of the effected object, i.e. the rectangular plate.
I don't see why the drag coefficient should be taken 2 in the above post. 2 means the effective force is doubled. And why should it be doubled unless the object in effect also blows back with the same velocity and two forces collide ?
For winds effecting the surfaces (i.e. sail surface) at angles other than 90 degrees, you can multiply the wind velocity with the cosine of the angle of attack, to find the effective velocity.
As for the drag coefficient of the boats, I'm not sure if they are being wind-tunnel tested during design or production but that should be the place to get those figures.
If you use your boat's sail area in the formula, you will get the actual amounts of forces your sails/rig are subject to when the wind is blowing at 90 degrees to the boat and sails.
You can't, in my opinion find the force at other angle of attacks, because although the effective velocity at an angle to the surface can be calculated as referred above, the drag coefficient remains unknown at different angles. At 90 degrees we can take it as 1, but we can't know this at say 45 degrees. At 45 degrees the boat shape is different, the coefficient is slightly reduced, at 0 degrees it is almost like an arrow, thus further reducing the coefficient. However, imo, estimations will be within acceptable limits.
Also you will see that; when wind increases from 15 knots to 18 knots; which is a %20 increase; the effective force will increase %44. If the wind is doubled from 15 to 30 knots, the force would increase 4 times.
A 30 knot
wind blowing at 90 degrees on to a sail area of 26 m2, will be applying a total force of 3714 Newton, which is the equivalent of the force of the gravity on a 379 kg object on our lovely planet. (It means someone is pulling your sails/mast with 379 kg weight on the opposite direction the wind is blowing) On a 40 knot
blow, this force is 6605 Newton, roughly 673 kg.
All this from long back, so I might have some errors here and there. Please feel free to correct.