From: billphill@earthlink.net (Badwater Bill) Newsgroups: rec.aviation.homebuilt Subject: Re: Wing profile Date: Tue, 10 Oct 2000 18:42:45 GMT If I recall correctly the bending theory for lift is much more simplistic than you fellows are making it out to be. If you take Cameron's flat plate with no differential in top or bottom surface, you do still get lift when you tilt the AOA up in a wind. What you are doing is bending the flow of the oncoming air downward. This simply means that you are deflecting mass downward. If you deflect mass in a certain unit of time, you get force from Newton's equation. F = ma = dp/dt where p is "momentum.mass times velocity" and remember it's a vector. So, you don't have to generate more or less mass to get lift, what you have to do is change it's direction only. Then you get force (lift). Of course in rocket engines the dp/dt translates to d(mv)/dt = mdv/dt + v dm/dt but since "v" is nozzle velocity and is constant, dv/dt = 0 you use dm/dt or mass flow rate times the nozzle velocity for your acceleration figure. Different problem all together. _________________ Let's just do a simple calculation on my RV-6. Let's assume I have 100 square feet of wing area or about 10 square meters. I'm going to have to do this in the metric system because it's ten times easier (pun intended here). The mass of my airplane is 600 kilograms at normal flying weight. Since gravitational acceleration on Earth is 9.8 meters/sec-sec I weigh about 5900 Newtons. That's the force I have to generate to keep me at a constant altitude. At 200 miles per hour I'm going about 90 meters per second and my wingspan is 8 meters. Let's assume I'm at sea level and the density of air is 1.2 kg/m^3. Another assumption is that I affect a cylinder of air as I fly through it that has a cross sectional diameter equal to my wing span which is 8 meters (radius is 4 meters). We know this is not exactly true but it's a good first order approximation. So, the total mass I effect each second is: Cross section = pi x r^2 = 3.14 x 4^2 = 3.14 x 16 = 50 square meters. Since I'm sweeping through 90 meters a second I'm effecting 90 meters/s x 50m^2 or about 4500 cubic meters of air each second. At 1.2 kg mass per cubic meter that's a mass equal to 5400 kilograms of air From Newton's equation F=dp/dt we can write F= d(mv)/dt = 5400kg x (Velocity change) in the air we are deflecting, divided by one second. So, we know we need 5900 Newtons of force and we can write the equation: 5900 Nt = 5400 kg x dv/dt Dividing each side of the equation by 5400 and solving for dv/dt we get about 1.1 m/s^2. Integrating over our one second interval gives us the number 1.1 m/s final air velocity of this horizontal column of air we've disturbed. What all this means is: If I am at sea level in my RV-6, the effect of me cruising along the beach at 10 feet and 200 mph is to effectively deflect the air down as I go by. If I were a helicopter I'd have to pump air down from above in the same fashion. An airplane is no different. It's an air pump. It pumps air down as it passes. I take what was relatively still air and throw 5400 kilograms (nearly 12,000 pounds of air) of it down each second at 1.1 meters per second which is about two and a half (2.5) miles per hour defection velocity. I think this is what is meant when people talk about bending the air down. It's a matter of what reference system you are in. If you are in the airplane then you are in a stationary inertial reference frame (no accelerations) and to you it looks like the wind is blowing by you. In this case, you take a 200 mph wind and defect it down about 2.5 miles per hour and you get the lift you need to stay afloat. Isn't that interesting that it's that small of a number. On the order of 1% of your horizontal speed. I find that interesting that it's such a small number. If you look at pressure differential on a wing to produce this pumping effect, it's a tiny number too. It's short and simple to calculate so I'll throw it in here for fun too. One hundred square feet of wing area supports 1500 pounds on my RV-6. That's 15 pounds/sq.-ft. Since each sq-ft is 144 sq inches, that's 15/144=0.104 pounds/sq.-inch (1.6 ounces). Since atmospheric pressure at STP is about 15 pounds/sq-inch, this represents a change in atmospheric pressure of 0.104/15= 0.0069 atmospheres (atm) or in percentage, 0.69%. So, the wing doesn't really have to do that much. All it has to do on average is change the pressure of the atmosphere less than 1% as it goes by to produce all the lift necessary for flight. I find that sort of amazing. BWB

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