Gerhard Fiedler wrote: > At first I'll put a fan in a box. In a box like a long hallway. No > doors. Just a fan blowing down the hallway. Then I take an elastic > foil, like a thin plastic foil, and cover the whole profile of the > hallway, perpendicular to the air stream, a few meters away from the > fan. We probably all agree that the foil will be stretched, due to > the force. Yes. > We explain this by thinking of air as tiny, elastic > particles, and of the > wind generated by the fan as an average movement of those particles -- > momentum, power and all that. Not so fast. Keep momentum and power separate. The lack of distinction is getting you into trouble later. > But the inquisitive mind then puts the foil a few hundred meters down > the hallway. (Remember, this is a thought experiment, and I don't > have a problem imagining an imaginary hallway a few hundred meters > long. All > flat, no doors, no change in profile.) I hope that we all can agree > that the foil will remain flat. No, I don't. Think of the limiting case where the crossection of the hallway is exactly the size of the fan. Instead of a hallway, think of a pipe with the fan blades just barely not touching the inside of the pipe as they spin around. In that case I hope you can agree that if both ends of the pipe are sealed, the fan will cause a pressure difference between the air on both sides of it. The entire pipe on the outflow side of the fan is at higher pressure than the entire pipe on the inflow side of the fan. The pressure on each side is constant, extending to the full length of that sid= e no matter how long it is. Let's say we hold this pipe fixed on the outside but can measure the latera= l force it exerts on the fixture. Hopefully you agree that it exerts no net force once equillibrium is reached regardless of how long the pipe is. The fan is exerting a force (thrust) in one direction on the pipe. That force is exactly countered by the force of the pressure difference on the ends of the pipe. In your case I think you intended the fan not to be sealed against the side= s of the hallway. This is the same as making the pipe diameter larger than the fan. Now the fan actually moves air to the outflow side, which then eventually returns to the inflow side thru the gap between the fan and the inside of the pipe. This does make the force ballance a bit more complicated, but it still has to ballance. As you make the pipe diameter larger, the pressure difference between the two ends goes down due to the high pressure air "leaking" back to the low pressure side around the outsid= e of the fan. Note though that the area of each end of the pipe is increased too. That means a smaller pressure difference is needed to ballance the thrust from the fan. I maintain that as before, the closed pipe will not exert any net force on the outside world once the airflow inside has reached equillibrium. Since there are no net forces, the sum of forces must be equal along any direction. To keep things simple, let's only look at the force ballance in the lateral direction (down the long axis of the pipe, also the direction the fan is pointing). There are only three forces that stuff on the inside of the pipe can exert on it in the lateral direction. There is the fan, the pressure difference at the ends times their area, and friction of air moving in the lateral direction by the inside of the pipe. All three of these have to sum to zer= o else the pipe would exert a force on its surroundings. Do you agree it doesn't exert any such force? In this case there is a force on the fan not so much due to a pressure difference but because it is imparting momentum per time (=3D force) on the air flowing thru it. For the same fan speed, this is actually a greater force than just resulting from pressure as in the previous case when the pipe was sealed around the fan. You can prove this to yourself by thinking what happens when you hold your hand over a vacuum cleaner hose. The motor speeds up, not slows down. This shows it's "harder" to maintain air flow than it is to maintain pressure. There is more backward push on the fan blades due to imparting momentum on the air passing thru it than there is due to any pressure difference the fan can cause. Anyway, the fan is still pushing the pipe in one direction from the inside. In the steady state case, this push is exactly countered by the total of th= e pressure difference on each end of the pipe times the pipe area, plus the friction forces of the air sliding against the inside of the pipe. There will still be a pressure difference but I think the problem is that it's a little less obvious to visualize. Note also that the required pressure goe= s down with the square of the pipe diameter. A short distance from the fan, your foil will perceive a push because it is stopping the momentum of the moving jet of air coming from the fan. This i= s a force due to momentum per time just like the fan's, except in the opposit= e direction. As you go further down the pipe, the jet spreads out and eventually stalls. This stalling is a momentum change, and exerts a force on the air further down the pipe, which causes a pressure increase in the air the remaining distance down the pipe. The pressure increase persists t= o the end of the pipe where it exerts a force on the pipe. > Since the wind is only an average movement, created > by the particles bumping into each other more in one direction than in > others, and this bumping isn't fully elastic, there's some kinetic > energy transformed into other energy (heat mostly). That slows the > average movement down. You are mixing energy and momentum arguments. The two can't just be exchanged. Both momentum and energy have to ballance separately. In this case, trying to figure out the energy ballance is a lot harder. Fortunately, the momentum ballance is easier to consider in this case. Let's stick to that. In any case, you can't use loss of kinetic energy as = a argument for momentum change. The units don't match, for one. > There's also statistics. Since the wind is the > effect of many "white noise" bumps, they go into all directions, and > spread out, and create those localized "momentum compensation loops". This is voodoo physics. > That is, the wind generated by the fan doesn't stream all the way down > to the foil several hundred meters away, bumps into it, then moves > laterally back, I agree. > All with Newton and the idea of air as almost elastic little > particles that move around arbitrarily to create pressure. (I'm not > sure it was Newton who first came up with this idea, but it is > necessary here.) You are supposing a micro-mechanism to explain a macro effect. Momentum still has to ballance regardless of what micro effect you think impedes motion, converts kinetic energy into heat, etc. > To keep the plane in the air, there is a balancing force needed, which > we assume stems from the (almost elastic) collisions with the air > particles on the lower side of the wing. That generates a downwind, > which makes perfect sense. I agree about the downwind part. Trying to assume the micro effect causing it will only confuse and possibly mislead. > However, considering our first experiment > with the foil, if the plane flies in a box that is high enough, say a > few kilometers, it's also very likely that nothing of that average > momentum of the air particles reaches the floor Oh yes it does. This is exactly the pigeon in the box problem. > the power is > partially transformed into heat, and the rest of the momentum is used > to create those local "flow backs". Again, you're mixing energy and momentum ballance. You can't convert one t= o the other to show ballance. Each has to ballance separately. In this case= , let's stick to just showing the momemtum (and thereby the force) ballance. > Without resulting average momentum at the bottom of the box, > there is no force, and hence, I'm convinced that > if you make the box big enough, what you'll have is a box that has the > same weight with and without a plane flying in it, Newton and > everything considered. No! Absolutely not! This is the crux of the matter. Instead of a plane passing thru, it might be easier to envision a hovering helicopter. The weight of that helicopter is ultimately borne by the ground somehow. That somehow is in the form of slightly elevated pressure over a area. The higher the helicopter, the smaller the pressure and larger the area. Perhaps it helps to draw the free body diagram around the earth instead of the helicopter. Earth with helicopter sitting on the ground is the obvious case. The earth is gravitaionally attracted to the helicopter, which pushe= s back on the earth thru the pressure it excerts on the ground. When the helicopter is hovering, the earth is still gravitionally attracted to the helicopter. Unless you think the earth is now accellerating towards the helicopter, some force must be pushing back on the earth to counter the gravitational attraction. That force is the atmosphere pressure increase over some area underneath the helicopter. Note that depending on how high the helicopter is, this force that is ultimately pressure on the ground comes from a mix of two causes. One is that the air pressure under the helicopter is slightly higher than ambient. This effect decreases as the helicopter is higher. The other is the force due to the momentum change of the downdraft as it hits the ground. This will be the dominant effect when the helicopter is "high". It will then also be spread over such a large area and the velocity change so little tha= t a human standing in the area below the helicopter would not notice with ordinary human senses. To put this in terms of my previous post, the helicopter imparts a momentum of some (Kg m/s) on the air. That same momentum is ultimately imparted on the ground, but the Kg to m/s tradeoff is very different. The ground sees many many more Kg, which are therefore moving with much much less m/s. Add to that the fact that this effect is spread over a much larger area on the ground than the helicopter's rotors sweep, and the helicopter doesn't have to be too high for a unaided human observer to not notice the effect. > Also, > while I agree with the basic assumption that the force stems from > (almost) elastic collisions with stochastically moving air particles, > it seems to me that the conditions around a wing are too complex to > be able to tell with > such simple "napkin calculations" what net effect you get (down, up, > none) with what size of box around a wing Again, you are supposing micro effects to explain a macro effect. Your explanation may or not be right, but that doesn't change the fact that the macro effect is there, in fact must be there. The napkin calculations I made were only about the macro effect. I specifically avoided talking abou= t the micro effects that explain it since these confuse more than illuminate. For the plane to stay up, it has to push down on air. This remains true regardless of whether you can imagine the correct micro mechanism for this or not. In fact if you can't, then your perceptions of the micro mechanism= s are wrong. > -- but the one thing I'm > quite certain of: if you make the box large enough, there is no > mechanical net effect (force) reaching the border of the box. Again, no! This would mean that the box with a pigeon in it would weigh less when the pigeon is in flight. > There > is probably some radiated heat, though, that reaches the border. Again, heat is a energy argument, which does not apply to a momentum or force argument. > Also, Newton didn't exactly say that any air particle needs to go > downwards to keep a plane in the air. As you correctly said, he > mumbled that forces need to be in balance, so /something/ needs to > balance the gravitational force -- but it doesn't necessarily have to > be air particles going downwards. What else is there? If you claim the plane is not pushing down on the air, you have to show something else it is pushing down on. Newton says it has to push down on something. > We assume that air pressure is created by > air particles moving stochastically around and bouncing off surfaces. For this argument, there is no need to assume anything about the micro structure of air. It only serves to obfuscate in this case. > So if > one would design a device that works the particles above a (flat) > surface so that the pressure above it is lower than the pressure below > it, we have a force that may balance the gravitational force. That is in fact what a wing does. > I'm not > sure it is possible to design such a device without a net downwards > movement of particles, but it seems possible. It's not, at least not if the thing you are pushing on can move (like the air can). Pushing something with a force over time IS imparting momentum o= n it. Force =3D mass x accelleration (see Newton) =3D mass x velocity / time (see definition of accelleration) =3D momentum / time (see definition of momentum) ******************************************************************** Embed Inc, Littleton Massachusetts, http://www.embedinc.com/products (978) 742-9014. Gold level PIC consultants since 2000. -- http://www.piclist.com PIC/SX FAQ & list archive View/change your membership options at http://mailman.mit.edu/mailman/listinfo/piclist .