AlexanderSahlin
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- Oct 7, 2014
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I used the same equation for lift-force as you used in Post 4. Then I assumed that the force from the sail was as large as the force from the foil in water, to set up the equation for equilibrium. Then I divided both sides of that equation by the the speed ^2 multiplied by the density of water/2.Yeah. I suppose that's true. I knew it would sail in equilibrium but wasn't sure where that equilibrium would be. Could you explain the equation you wrote. I think I understand it but I'm not sure. Why assuming cl=1. How did you get to 11m^2. Yeah sail area is around. 0.9 m^2.
I came to the sail-area 11 sqm. by the hypothetical assumption that the CL of the foil was 0.6 and that CL of the sail was 1.0.
In the real world you have the sail-area you have made, and the CL:s will adjust themselves so you get equilibrium.
My assumed CL ≈ 1 for the sail is a rough estimate from my experience of model-airplane wings. It can be lower, it can be a little higher depending on the aoa, but it is hard to get it higher than CL=2 for a single-element airfoil, especially at model scale. I recall that we had CLmax slightly above 2 in a windtunnel-measurement on an early version of the wing-sail for my paravane-speedsailer, but that was a very extreme high-lift wing-section, that depended on laminar boundary layer in the front part. The final wing-design for the paravane-speedsailer was a more robust one with slightly lower CLmax, that worked also for fully turbulent boundary layer.
However, my conclusion is that you can start sailing with the foil and sail you have built, and after you have made it work, you can try to reduce the foil-area to reach higher speed. Very best luck!