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| Aerodynamics What makes 'em fly? Designs & technology related to aerodynamics. |
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| | #25 (permalink) | ||||||||||||||||||
| Uber Contributer ![]() Join Date: Jan 2006 Location: Tucson, AZ, USA Age: 35
Posts: 169
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Although it is hard to discuss such subjects over a forum due to subtleties of the subject, I will throw in my 2 cents in response to keep the subject from getting too butchered. For symmetric airfoils (on which this thread started) the only sentence that is correct is the first one. The presence of curvature of the airfoil does not produce a large increase in lift. Like I said before (and I will keep saying it), for flat plates at moderate angles-of-attack (< ~11 deg) the lift curve slope is nearly the same as for most conventional, thin symmetric airfoils (yes, like the one on your airplane). Therefore, at angles < 11 deg, thin symmetric airfoils produce about the same lift as a flat plate. This is not intuitive, and grasping for the easiest answers will lead you quickly to a wrong understanding of what's going on. For thin symmetric airfoils (typically used on modern aviation), the curvature itself has little to do with the act of accelerating the air; a flat plate will accelerate the air just as well. The additonal curvature produces a very small acceleration compared to the presence of the airfoil itself. The curvature has to do with reducing drag, controlling separation characteristics (chordwise pressure distribution), controlling pitching moments, wave (compressibility) drag, and general lift and drag behavior of the airfoil at different angles and speeds. There's nothing in the physics that suggests that flat plates can't accelerate the air as quickly as symmetric airfoiled surfaces. The airfoil does a really good job of producing circulation, which is the fundamental basis for lift.Circulation is a measure of rotation of the airflow around the airfoil. The Kutta condition (that the flow must leave tangentially at the trailing edge) determines the amount of circulation for a particular airfoil. Lift is directly proportion to circulation, that is: Lift = circulation*density*airspeed*wingspan You produce more circulation, then you get more lift. Period. Any object (anything) will produce circulation in presence of a velocity and density field. Airfoils are good at producing large amounts of circulation (in the vertical direction) for minimum drag. By the way, since you;re probably wondering where I came from, I've got a BS degree in Mechanical and Aeronautical Engineering, and Masters Degree in aeroautical engineering. I have been working in aerodynamics for 8 years straight out of school; for NASA (Glenn and Ames Research Centers), and Military defense contractors. I deal with this stuff every day.
Last edited by teryn; 06-06-2007 at 11:28 PM. | ||||||||||||||||||
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| | #26 (permalink) |
| Bad-ass Super Contributer! ![]() Join Date: Jan 2006 Location: No Matter
Posts: 397
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I think Einstein plays into the whole lift problem here too (not literally, but sort of). We always think of air being accelrated over the airfoil from front to back. This is really only true in a wind tunnel. So what really happens? The plane is cutting through essentially stationary air. Lets look at a cambered airfoil flying at 0 AOA. The air passing underneath is hardly deflected downward at all. The air going over the top is accelerated vertically, then it falls in the vaccum behind the peak of the airfoil and maintains some of that momentum as it leaves the trailing edge of the airfoil as downwash. So what's the net effect? Well, Bernouli still applies. Bernouli says the total pressure of a gas can not change. The total pressure is the sum of the dynamic (moving) pressure and static (weight of the air) pressure. Therefore, since the air on top of the airfoil is moving vertically more than the air on the bottom, it gains in dynamic pressure. Its static pressure must therefore decrease. Whala... lift. Also, the air leaving the trailing edge is deflected down... Newton applies... change in momentum of air up so there must be an equal and opposite reaction... lift. Lift from 2 sources... noone has to be completely correct. Symmetrical airfoils... they only fly at a positive AOA. Some of the air from below the front of the chord-line is forced over the top of the airfoil due to the higher pressure and ram effect under the airfoil. Same as above happens. Same with a flat plate. Now... a vertical fin does produce 0 lift at 0 AOA. Now lets disturb the airplane in yaw... lets say slight yaw to the left. The tail now develops an AOA with the leading edge left. This produces lift which pulls the tail to the left, and inversely, the nose to the right, until it goes slightly too far to the right. Then the opposite happens. This happens contantly back and forth in reality, but VERY subtly. The efficiency of the system, mass, fuse length, etc... determines how good the damping is... or how small the movements are. Also... the drag of the tail behind the aerodynamic center adds to the stability of the tail in yaw. Makes the movements undetectable. This is all off the top of my head... especially the relativity stuff, but I don't think anyone can argue it.
__________________ Sicko |
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| | #27 (permalink) |
| Super Contributer ![]() Join Date: Jul 2006 Location: N. California
Posts: 140
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Thanks guys! A couple of great answers to a great question. If someone is initially overwhelmed by the answers, stop and read each one again. You will get it if you understand the fundamentals of aircraft flight. Fly right!
__________________ ______________________ FlyinTiger |
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| | #28 (permalink) | ||||||||||||||||||
| Old School Hucker ![]() Join Date: Jan 2006 Location: San Diego, CA Age: 40
Posts: 78
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Will airfoil make a difference in control rate? Sure it can. Will it make a difference on drag produced for a given deflection? Sure it can. I would guess that anyone who would tell you the shape of the fin/rudder doesn't matter just plain doesn't use the rudder except to steer the airplane on the ground. | ||||||||||||||||||
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| | #29 (permalink) |
| Flyin' Around ![]() Join Date: Aug 2007 Location: Cowes, Isle of Wight, UK
Posts: 1
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Surely all of this is Reynolds no dependent. A wise man once told me that a flat plate works really well up to Re=50,000 and that above 100,000 you really need at aerfoil shape to keep drag down. Because all my book learning was too long ago now, as an approximation Ive been using the quick formula Re = speed in meters per second * chord in centimeters * 680 If you have a plane with an average 4" chord on its rudder travelling at 25mph the Re no works out at just under 76000 so it probably doesnt matter whether you use a flat plate fin or a symemtrical section. However as the speeds go up or size increases through bigger scaled aircraft I honestly believe it pays to install a symmetrical section fin and tailplane for reduced drag and better performance. Richard |
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