

Mar 23, 2011, 10:45 PM  

Maybe you've accidentally discovered a new airfoil masquerading as an MH32. Better post up the coordinates
before it is lost. ian 
Latest blog entry: FAA makes clear intent to kill FPV via...


Mar 24, 2011, 06:32 PM  
Sydney
Joined Apr 2009
716 Posts

Steve, these should help.Note the RG14 appears to have an edge on the DS 19 at CL's >0.3.
Girsberger certainly designed some classic foils. The DS19 indicates however why it is so popular. A very fast and forgiving airfoil from these plots and would carry more weight. John PM Just discovered this on a reread.I Stumbled a bit here Steve wrt to the drag of the RG14. The RG14 has less drag(should have shown< instead of >) than the DS19 below approx 0.3CL  which may not mean much as Adam points out below. My guess would be that the DS19 would be the better overall choice. 
Mar 31, 2011, 05:21 PM  

Quote:
Flying the same weight and wing area plane at 300mph in a circle at a more human lap time of say 4s would require a CL of approx 0.14. If you inherited SL's flying skills and flew 3.3s laps, a CL of approx 0.17 would be required. The point I hope to illustrate is that, despite the tremendous 'G' forces, because of the high speed the actual CL or alpha required to generate the necessary lift force is surprisingly low. L= CL 1/2 p v^2 S. Because the value of v^2 is very large, in order to provide the required lift force the value of CL is low. Therefore, provided you can launch and land, an aerofoil with lower drag at low CL should be better. I worked it out thus: 300mph = 134 m/s. Circuit time = 4s, therefore circumference is 134 x 4 = 536m. radius = c/2pi so 536/2pi = 85m Centripetal acceleration = v^2 / r so 134^2 / 85 = 211m/s^2 Force = mass x acceleration (80oz = 2.27Kg) so F = 2.27 x 211 = 479N. This is close as dammit to the lift force required to fly the plane in the assumed circle. You can work out the total lift vector required to account for the vertical component to overcome gravity but you will find the difference it makes is only about 1N so can be considered negligible during these extremely hard turns. Rearranging the lift formula we get CL = L / .5 p v^2 S (480sq in = 0.31m^2) so CL = 479 / .5 x 1.225 x 134^2 x .31 = 0.14 If we be more realistic and use a racetrack pattern to simulate our DS path, this does require tighter radius turns. If we approximate a racetrack using straight edges, say 30m each, we end up with a 4s circuit requiring a CL of 0.16, still a low value. I hope this is of interest. Best regards Steve (another Steve, there are plenty of us in the DS world!) 

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