Secret Aerodynamics of B2 Stealth

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 February 6, 2010, 14:46 #21 Administrator     Peter Jones Join Date: Jan 2009 Posts: 684 Rep Power: 10 The leading edge shape is indeed novel, but not something you would want in an airfoil which must operate well in a range of different incidence angles. The rest of the airfoil is not very novel I think. It is difficult to create an airfoil shape that is novel. Most ideas have been tried. What cruise Ma number is the airfoil designed for? Have you done any kind of assesment or simulations involving transition modeling of your airfoil?

February 6, 2010, 18:44
#22
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bearcat
Join Date: Jul 2009
Location: Ohio
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Quote:
 Originally Posted by dijinj The lift in airfoil is generated by pressure difference; the lower pressure in upper side of wing is due to higher velocity of air. I am only tryin to make vertical component of velocity zero at trailing edge. the total velocity at upper side is higher in my design too.
1. You don't fully understand the physics. Vertical velocity won't be zero at trailing edge if lift is generated. Consider helicopter's rotor as a wing.

2. Your idea doesn't work. Don't waster time.

 February 6, 2010, 20:29 #23 Member   Diji N J Join Date: Jan 2010 Posts: 37 Blog Entries: 1 Rep Power: 8 Indeed lift is generated but pressure difference is at front end; check pressure contours of cycloid wing. helicopter rotor and propellers are designed in such a way that it pull air downward or back ward respectively. similar to a turbine blade. Here is the data from an inviscid CFD analysis of both cycloid wing and NACA 4412 airfoil cycloid New improved equal spacing (Angle of Attack 0 degree) ------------ iter continuity x-velocity y-velocity cl cd time/iter 584 1.0582e-06 1.9681e-09 1.4305e-09 3.3978e-01 5.9761e-04 0:02:03 416 585 1.0061e-06 1.9298e-09 1.4192e-09 3.3978e-01 5.9760e-04 0:03:01 415 586 1.0351e-06 1.8866e-09 1.4134e-09 3.3978e-01 5.9760e-04 0:02:25 414 ! 587 solution is converged 587 9.9777e-07 1.8463e-09 1.4049e-09 3.3978e-01 5.9759e-04 0:01:55 413 lift to drag ratio = 0.33978 : 0.00059759 = 568.583811 ************************************************** * aerofoil (Angle of Attack 0 degree) -------- ter continuity x-velocity y-velocity cl cd time/iter ! 419 solution is converged 419 9.8097e-07 1.2270e-08 3.7687e-09 6.4636e-01 1.0542e-03 0:02:48 581 lift to drag ratio = 0.64636 :0.0010542 = 613.12843 ************************************************** **

February 6, 2010, 20:49
#24
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Diji N J
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Quote:
 Originally Posted by bearcat 1. You don't fully understand the physics. Vertical velocity won't be zero at trailing edge if lift is generated. Consider helicopter's rotor as a wing. 2. Your idea doesn't work. Don't waster time.
What is your Idea on how lift is generated in airfoils

 February 7, 2010, 02:55 #25 Member   Skeptic Join Date: Mar 2009 Posts: 67 Rep Power: 9 If the leading and trailing edges are cusped, it has been done before, e.g. "A boundary integral equations approach for the study of the subsonic compressible flow past a cusped airfoil", Adrian Carabineanu.

 February 7, 2010, 04:36 #26 Member   Diji N J Join Date: Jan 2010 Posts: 37 Blog Entries: 1 Rep Power: 8 Actually both Tangents at trailing edge is horizontal but cusp is more general term cycloid is only a member of that group. can you suggest a link of the article "A boundary integral equations approach for the study of the subsonic compressible flow past a cusped airfoil", Adrian Carabineanu. so that I can Cross check whether cycloid is included in article. and especially horizontal velocity vectors are mentioned.

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