Georgs, My estimate of 12 engine hp per 100 sq. ft. of projected area to hold position in a 20 kt. wind is based on the following calculations. Wind drag on a boat is a function of the projected area at right angles to the wind, the square of the wind speed, the density of the air, and the dimensionless coefficient of drag which depends on the shape of the boat. Drag coefficients have been determined from wind tunnel tests. Some representative drag coefficients are: Open parachute (or efficient spinnaker) = 2.0 Hollow hemisphere, concave to wind = 1.7 Flat rectangular plate = 1.28 Wires, cylinders, masts = 1.0 Cargo ship, wind dead ahead = .95 Fishing trawler, wind dead ahead = .9 to 1.05, depending on superstructure, outriggers, etc. Streamlined passenger vessel = .70 Recreational trawler = .70 to 1.0, depending on superstructure, masts, outriggers, etc. Sphere = .47 Hollow hemisphere, convex to wind = .38 Modern automobile = .26 to .35 Airplane = .09 Using Area in sq. ft., wind Velocity in knots, and the U.S. Standard Atmosphere for air density, the equation for Drag in lbs. can be written as: Drag = .00339 x Coefficient of Drag x Knots^2 x Area In a 20 kt. wind, a boat with an area of 100 sq. ft. at right angles to the wind with a drag coefficient of 1.0 will have 135.6 lbs. of wind pressure on its surface. To determine the amount of power required to hold position, let's calculate how much power it would take to move that 100 sq. ft. area into still air at a speed of 20 kts. or 33.77 feet per second. The required power comes to 4580 ft. lbs. per second or 8.33 hp. But that's power applied directly directly to the boat hull with 100% efficiency. In practice, most boat propellers operate at 50% efficiency or less. Ignoring power transmission losses, the boat's engine would have to generate 16.66 hp to hold position in a 20 kt. wind. Since the power required varies directly as the coefficient of drag, a streamlined boat might hold position with only about 12 hp. The force of the wind increases as the square of its speed. The pressure of a 40 kt. wind is 4 times stronger than that of a 20 kt. wind. Thus holding each 100 sq. ft. area in position in an 40 kt. wind might require the boat's engine to produce between 52 to 75 hp. A typical hurricane condition 80 kt. wind produces 4 times the pressure of a 40 kt. wind. To hold position in this strength wind would require an engine output of between 186 to 267 hp. for each 100 sq. ft. of projected area. Since the Nordhavn 40 has approximately 200 sq. ft. of area and is not particularly streamlined, the estimated engine power required to hold position in a 20 kt. wind might be 25 hp. and in an 80 kt. wind might be nearly 400 hp. You would have to run this latter test yourself. Clearly the prudent strategy in extreme wind conditions is not to buck the wind but to run before the wind streaming a drogue to avoid broaching. Larry Z