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Yeesh! Chill out, sailor, or take it to The Pit.
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I know this thread is pretty much dead (and without a final resolution) but I couldnt help myself from posting. For a long time I wanted to be a pilot. I am not claiming to be an expert, just some experience in the matter.
First a couple of things you guys arent taking into consideration. You are assuming the flat wing is 2 dimensional (ie it has a 1 dimensional -line- cross section). Almost any flat wing will have a height and the leading edge of this will cause B’s Principle. This would contribute to lift a little. Another thing that you arent considering is that even a flat wing has high and low pressure zones around it at any AOA other than 0 degrees (even at 0 if you want to nitpick). This is similar to the wake of moving boats. The high pressure zone forms under the wing (assuming + AOA) because of the air being deflected. The total lift of even a flat wing would still be the low pressure ‘suction’ + deflected air, with the ‘suction’ as the greater amount of force. Of course, the lift would be much greater with the typical aerodynamic wing design.
Have any of you seen those super gliders that they use other planes to launch? The wing design is very unusual. The wings are VERY narrow, only a few inches at the greatest width, and very long. The shape of the wing is pretty unique as well. it uses 2 convex curves, top and bottom. This wing utilises B’s P almost exclusively. They are extreemly efficient and only lose a few feet of altitude every few miles even when there are no updrafts. A glider made to take advantage of the rocket effect would probably fall like a rock.
This suction is a much stronger force than the ‘rocket’ effect of the downward traveling air.
Helicopters and props rely more on this suction and not the rocket effect to fly. BTW, a prop is similar in design to a wing, but with a slight twist to take advantage of this ‘rocket’ thrust effect. A prop gets its ‘thrust’ from both the rocket effect and B’s P. Also, NASA here by Cleveland designed and built a prop that took advantage of B’s P almost exclusivly. It is almost as powerfull as a jet engine, uses much less fuel, and is much quieter.
Have any of you been near (and I mean very close) to a helicopter taking off? It does not move enough air to knock you over, let alone flatten you to the ground.
Lets make some numbers to try to work this out. I dont know the formulas needed for this so this is very simplified. Will someone please work this out properly? Lets assume a 2 ton copter. Lets assume the diameter of the blades to be 10 feet. According to the ‘rocket’ effect, the copter would need more than 2 tons of downward force to lift off the ground. This is alot of force. Sure, some of it gets disipated by the atmosphere, but not very much when lifting off. A human would be crushed or tossed around from this if near. This would be enough to blow cars over from many feet away.
On a side note, whoever mentioned that a sail is flat and therefore the wind must be blowing against it (instead of around it) to produce movement: A sail is a flexible piece of material, that can take pretty much whatever shape it wants. If you look at a sail under a good wind, you can see that the wind shapes it into a big curve very similar to the top edge of a wing.
my conclusion: The downward thrust contributes very little to the lift of an airplane or copter, and only a negligable amount. That small amount may not even matter because of the loss of lift due to the turbulance it causes.
This all doesnt matter anyway; everyone knows that the best way to fly is to fall and forget to hit the ground…
-Fox
Challenge accepted.
OK, so you don’t know how to calculate it, but you know it would crush a human? Great answer.
The energy represented in flowing fluids is far greater than you can imagine. This is why hydroelectric power is so useful.
Let’s assume a perfectly flat rotor, taking specs from the Aeros:
Rotor chord: 8 in. (.2 meters)
Rotor diameter: 22 ft. (6.7 meters)
Rotor RPM: 450
Gross weight: 900 lbs. (410 kg)
Let’s further assume the simple case where the rotor blade encounters stationary air at the leading edge and accelerates it to directly down to the speed required for the air to not go through the blade.
Questions:
a) How fast is that exactly? This yields an approximation of how fast the air flows out of an imaginary tube of rotor diameter. This can be directly compared to standing in a wind of the same velocity.
b) Is that enough lift, or do we have to invoke some fancy schmancy pressure differential doo-wop?
The first question is pretty easy, if we ignore turbulence and such (that’s a big if, but so what.) For a perfectly flat blade, passing at an angle of attack theta at velocity v through the air, it accelerates the air to a speed of v tan(theta). So, for an angle of 20 degrees you get a downward velocity of air of .36v.
The tip of our Aeros rotor is moving at 450 * 22 * pi feet/min, or about 31000 ft/min, or about 353 mph. Thus the downward speed of air at the tip of the rotor is about 120 mph. Skydivers aren’t crushed when they jump at such speeds. But, better yet, the entirety of the rotor is moving slower than that speed by some amount. So, even if the average speed of air coming off the blades is 80 mph, it would dissipate in the atmosphere pretty rapidly. Your crushed human argument hasn’t a leg to stand on.
I don’t have time to answer question b) right now, and I refuse to speculate on the answer as I don’t like to say things like “blow cars over from many feet away” without actually having sound logic to back me up. Perhaps later, or perhaps not.
Thank you, I stand corrected. At the time (I was 8) it did not seem that the copter had nearly enough thrust to lift off. I was fighting to stay on my feet, but I figured it would need much more power than that. I asked my uncle (who is a veitnam copter pilot with a couple Dist. Fly. Crosses, and a very, very smart guy) and he said the copter got most of its lift from the B.s P.
Anyway, I dont even remember posting that, I was running on about 5 hours of sleep over the last 72 hours because of my job (hauling dead bodies).
-Fox
Actually, no. If your flat (let’s say it’s massless, too) wing is running at exactly zero degrees, then there is no reason why the buildup of air and thus the high pressure couldn’t happen on top of the wing. But it doesn’t happen that way. Your wing/plane/whatever has mass, so the force vector is angled downward. So the high pressure is going to be below ANYWAY, and with the wing moving forward, you get an effective angle of attack, since the plane of the wing is at 0 degrees, but the path of the unpowered wing will be down and forward.
Ok, let’s use your numbers.
10 foot rotor span
4000-lb vehicle
downward force > 4000 lbs for liftoff.
the rotor moves in a circle. radius 5 ft, or 60 in.
Rotor path has area of 60^2*pi, or about 11310 square inches.
4000 lb downforce spread over 8000 sq in (11310-3310 sq in, to account for the body of the vehicle, where air can’t blow through) makes 1/2 lb/sq-in above atmospheric pressure. That’s about 1mm on the barometer, or not very much. No crushed people, no blown over cars. Sorry.