Thanks for the in depth analysis of wind drag. I calculated it for my boat, a Portsmouth Downeast '30 flybridge Cruiser. What this tells me is that even if I have adequate ground tackle to hold the boat in hurricane winds, I will have a problem with pilot house windows blowing in at those pressures. I think I will have to design and build boat "shutters" for those windows. Wind Drag Analysis 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. The answers to your specific questions are: The force is much greater on the beam than on the bow because of the greater area exposed to the wind. Wind forces have been measured innumerable times in wind tunnels and in actual hurricane conditions. That's how the coefficients of drag were determined. The results correspond to the figures I have given. The force increases as the square of the wind velocity. Wayward O'Malley Wind Pressure Estimates at High End of Scale Assumptions atmospheric pressure .0039 Drag coeficient 1.2 (takes into consideration radar arch, windshield) square footage at right angles to wind 190 (calculated by multiplying beam 10.6 times height 18) Formula is as follows =SUM(0.0039*1.2)*(75*75)*190 where 75 is wind force in knots 75 Knots of Wind5001.75Pressure on the part of the boat facing the wind 85 Knots of Wind6424.47 95 Knots of Wind8025.03 100 Knots of Wind8892 110 Knots of Wind10759.32 120 Knots of Wind12804.48 135 Knots of Wind16205.67 150 Knots of Wind20007 Wayward O'Malley Wind Pressure Estimates at High End of Scale Assumptions atmospheric pressure .0039 Drag coeficient 1.2 (takes into consideration radar arch, windshield)1 (does not count radar arch or windshield as important square footage at right angles to wind 150 (calculated by multiplying beam 10.6 times height 14) Formula is as follows =SUM(0.0039*1.0)*(75*75)*150 where 75 is wind force in knots 75 Knots of Wind3290.625Pressure on the part of the boat facing the wind 85 Knots of Wind4226.625 95 Knots of Wind5279.625 100 Knots of Wind5850 110 Knots of Wind7078.5 120 Knots of Wind8424 135 Knots of Wind10661.63 150 Knots of Wind13162.5