In a message dated 9/28/07 12:01:14 AM, Dan Green asks: > The standard canard for displacement hulls is that it does not take much > power > to get to hull speed in calm conditions but it takes more if you are > battling > head winds or seas. Has anyone ever seen figures on how much more it might > take? I realize that there are many variables such as windage of the hull > and > superstructure, amount of headwind, size of head seas and so forth, but > isn't > there some sort of approximation of how much horse power one should have to > overcome this adversity and be able to reach displacement hull speed into a > wind or sea? > 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 1.2, 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. The drag goes up as the square of the wind velocity. A conservative way to estimate frontal area is to multiply the beam by the height of the superstructure. An even simpler way is to multiply the beam by 3/4 of the beam. By this calculation, my Willard has about 100 sq. ft. of area. A Nordhavn 40 has 160 sq. ft. of area. A Nordhavn 47 has about 195 sq. ft. of area. A Nordhavn 72 has 330 sq. ft. of area. For each 100 sq. feet of area: 20 Kts = 136 Lbs. 40 Kts = 542 Lbs. 60 Kts = 1220 Lbs. 80 Kts = 2170 Lbs. 100 Kts = 3990 Lbs. 120 Kts = 4882 Lbs. 140 Kts = 6644 Lbs. 160 Kts = 8678 Lbs. 180 Kts = 10984 Lbs. It would take about 7.5 hp per 100 sq. ft. of area to move a boat slowly into a 20 kt. wind. Give a typical trawler propeller efficiency of 50%, the engine would have to develop at least 15 more horsepower than to move the boat at the same speed under calm conditions. The power requirements necessary to motor slowly onto the wind increase as the square of the wind speed. Larry Z ************************************** See what's new at http://www.aol.com