RE: All about Gamma (or k if you want)
I think your pressure cooker analogy for Cv is essentially correct, but the open pot analogy for Cp is not quite correct. To determine Cp, you would need to heat air in a pot with a piston-like lid. If you kept the same amount of weight on the lid, allowing the lid to be displaced upward as the air is heated, you could measure the amount of energy required to raise the temperature of the air at constant pressure..
Posted on: 6/1/2012 12:12 PM by Author "Shoe" in the forum "Aerodynamics"
http://www.rcuniverse.com/forum/fb.asp?m=11102746

RE: Forward swept wing will it turn like a normal three channel?
I think what you want to know is: in what direction will this wing want to roll in response to sideslip? Forward sweep will normally make a wing roll in the opposite direction that dihedral (or polyhedral) will. The question is: which geometric feature will overpower the other? I think this is pretty easy to answer... hold the wing into the wind and give it some sideslip. Which direction does it want to roll? Is the magnitude of the roll tendency comparable to what you would feel for other designs? If the wing ends up high-mounted, that will also tend to add some roll tendency in the same direction as the polyhedral.
Posted on: 4/30/2012 8:38 AM by Author "Shoe" in the forum "Aerodynamics"
http://www.rcuniverse.com/forum/fb.asp?m=11062515

RE: Maximum altitude??
Definitely consider Hofstadter's law in this project: Hofstadter's Law: it always takes longer than you expect, even when you take into account Hofstadter's Law. You can make an aviation corollary to this law by replacing "takes longer" with "costs more". Seriously, I think a manned option might be cheaper, quicker, and more flexible in terms of accessible airspace (not saying this just because I make a living flying airplanes). I do suspect the manned option might be more administratively burdonesome (getting your payload approved for wing mounting) and less fun.
Posted on: 4/3/2012 4:44 AM by Author "Shoe" in the forum "Aerodynamics"
http://www.rcuniverse.com/forum/fb.asp?m=11026302

RE: Maximum altitude??
Brandon, Some "back of the envelope" calculations... Let's say you end up with a gross weight of 20 lb., a wing area of 2,200 sq in., and an Aspect Ratio of 13. Unless you do something really strange with the planform (or twist), I would expect your lift distribution to be pretty close to elliptical (especially with an aspect ratio of 13). For an elliptical lift distribution, the lift coefficient is approximately: CL = 2 * pi * Aspect_Ratio * AoA / (Aspect_Ratio + 2). For your configuration, I would expect your CL to be about 0.09 per degree. AoA at seal level would be roughly: AoA = 4,100 / TAS^2 (where AoA is in degrees, and TAS is in knots) So, for example, if your speed at sea level was 25 kt, your level-flight AoA would be about 6.5 degrees. At 35 kt, your AoA would be 3.3 degrees. I would recommend a wing incidence that puts the fusealge "into the wind" at your cruise airspeed (i.e. about 3 degrees if you plan to cruise at 35 knots). If your cruise L/D ends up being about 25:1 (achievable with some care), your drag would be just under a pound. The power required for level flight would be about P = 1.8 * TAS (where P is in Watts, and TAS is in knots). At 35 kt, your power consumption would be about 65 Watts. For a 10S setup, this would mean a current of 1.75 Amps. With a battery capacity of 12,000 mah (and 100% motor and propulsive efficiency) this would give a cruise endurance of 6.8 hours. For a 10S setup, the battery charge used during a 100% efficient climb (without drag) would be: Ch = 0.204 * Alt (where Ch is in mah, and Alt is in feet). For a climb to 15,000 ft, this would equate to a charge of 3,062 mah. If you were to assume a more realistic net efficiency (prop and motor) of 65% and try to account for the aerodynamic drag during the climb, I think the climb energy would be more like: Ch = 0.37 * Alt (again Ch in mah, and Alt in feet). So a climb to 15,000 ft would probably consume more like 5,500 mah. Putting everything together: Endurance = (12,000 - 0.37*Alt)/2,700 As an example a mission at 15,000 ft would have an endurance of about 2.4 hours. This assumes you're willing to use all of the battery capacity and you don't need any power for descent or landing. I would definitely check my math. Hope this is useful.
Posted on: 4/2/2012 6:22 PM by Author "Shoe" in the forum "Aerodynamics"
http://www.rcuniverse.com/forum/fb.asp?m=11025833

RE: Cambridge scientist debunks flying myth
[quote]ORIGINAL: Tim Green when something changes the momentum of the air, that something will react with a force equal to that momentum change and in the opposite direction. [/quote] Agreed (as long as by "momentum change" you mean "rate of momentum change"). My issue is you have been arguing the converse of this... that when something (e.g. air) exerts a force on another object it will experience a rate of change of its momentum. This is true ONLY if that something experiences an UNBALANCED force. You illustrated a few posts back that a RC helicopter hovering over a scale results in the air experiencing no unbalanced force (upward force exerted by the scale on the air is equal and opposite the downward force exerted by the helicopter on the air). If you don't conclude from this that the air is experiencing no net rate of vertical momentum change, then you truly do not understand Newton's Second Law. You have stated over and over again that a lifting wing imparts net vertical momentum to the air, yet you have provided nothing to show that a wing imparts more downward than upward momentum (other than a flawed argument that Newton's Laws require this to be the case).
Posted on: 2/29/2012 12:10 PM by Author "Shoe" in the forum "Aerodynamics"
http://www.rcuniverse.com/forum/fb.asp?m=10981774

RE: Cambridge scientist debunks flying myth
[quote]ORIGINAL: beepee For me, the train analogy does not work. You are not transferring linear momentum to the train cars in the circle, you are transferring rotational momentum. It is still momentum. Bedford [/quote] Fair enough, but several here have invoked Newton's Second Law to suggest a required balance between the rate of change of the air's linear vertical momentum and the lift force acting on the wing. The point I am trying to make is that balance doesn't exist (it's definitely not required by Newton's Laws). Newton would certainly make a distinction between linear and angular momentum.
Posted on: 2/29/2012 1:52 AM by Author "Shoe" in the forum "Aerodynamics"
http://www.rcuniverse.com/forum/fb.asp?m=10981084

RE: Cambridge scientist debunks flying myth
Apologies for the lengthy post, but I have a (fairly) simple “thought experiment� that might help convey my concerns about the wing momentum transfer discussion. Imagine you are standing next to a long, straight, level stretch of railroad track (frictionless of course), with a string of boxcars on it extending for a mile in either direction. If you were to exert a force of magnitude F on one of the boxcars, and the direction of the force was to the right as you faced the track, then Newton’s Second Law says the string of boxcars would accelerate to your right at a rate of F/m (where m is the total mass of the string of boxcars). Newton’s Third Law says you would experience a “reaction� force to your left of magnitude F. Now imagine that you took the track directly underneath the string of boxcars and bent it into a circle with a circumference of 2 miles (so you’d be able to connect the first and last boxcars). If you were to again exert a force of magnitude F (to your right) on a boxcar, the result would appear very much the same. The boxcars in front of you would accelerate to the right at a rate of F/m, and you would experience a reaction force to your left of magnitude F. There's a difference though... In the case of the straight track, the center of mass of the string of boxcars accelerates to your right at a rate of F/m (the string experiences net momentum change). In the case of the circular track, the boxcars in front of you accelerate to your right at a rate of F/m, but their counterparts on the opposite side of the circle accelerate to your left at a rate of F/m. The center of mass of the circular string of boxcars remains at the center of the circle (no net momentum change). The harder you push to the right, the harder the track pushes the string of boxcars to the left to keep them on the track. This example shows it’s possible to achieve a reaction force by pushing against an object without transferring net momentum to it (even though you have put the object in motion). All that is required is that you cause something else push on that object with equal magnitude in the opposite direction. If you make a mental leap from boxcars to “bits� of air, which of the above setups better represents a lifting wing? Before going there, I think it's easiest to start out considering a RC helicopter hovering in ground effect over a scale. As pointed out a few posts back, the scale will show a reading equal to the helicopter's weight when the helicopter is sitting on it, and a reading very nearly equal to the helicopter's weight when the helicopter is hovering over it. This shows that the force exerted by the scale on the air is (very nearly) equal and opposite the force exerted by the helicopter rotor on the air. In theory, if you made the scale's "footprint" big enough and the hover low enough, the forces would balance exactly. For a rotor in ground effect, the air experiences no unbalanced force, and therefore its net momentum doesn't change. I think the circular train track setup is an excellent analogy to the helicopter in ground effect (where the force you exert on the boxcar is analogous to the force exerted by the helicopter rotor on the air, and the force exerted by the circular track on the boxcars is analogous to the force exerted by the scale on the air). A rotor in ground effect imparts no NET momentum to the air. Any downward momentum created in one location is offset instantaneously by upward momentum created elsewhere (if Newton is to be believed). What happens as the helicopter moves out of ground effect? If it moves completely out of ground effect, it makes sense that the straight track setup becomes the better analogy. Far away from the ground, if the downward force exerted by the rotor on the air is no longer balanced by an upward force exerted by the ground (or scale) on the air, then the air can experience a net rate of vertical momentum change. How about a wing in steady, level flight? In ground effect, I think it again makes sense that the circular track is the more accurate analogy (the ground inhibits any net momentum transfer). How about as you move the wing out of ground effect? It's tempting to conclude that the straight track becomes the better analogy, just as it did for the rotor, but this is not necessarily the case. If you account for all of the air set in motion, it turns out that a lifting wing (unlike a rotor out of ground effect) adds just as much upward as downward momentum to the air (if the aspect ratio of your control volume is constrained by the ground). It has been suggested that a balance between upward and downward momentum transfer would mean the wing couldn’t produce lift. The circular track example above illustrates how it is possible have a reaction force without net momentum transfer. The bottom line is that both cases (the straight and the circular track) represent POSSIBLE ways to model the underlying physics of a wing flying out of ground effect (both are entirely consistent with Newton’s Laws). To know which one better represents what’s really going on, you have to actually sum up all the momentum transfer. Those who have done the summation have concluded that the circular track analogy better represents how a wing pushes against the air to generate lift. Those who suggest that a wing works by transferring downward momentum to the air need to qualify that statement by noting that, in a global sense, the air’s net downward momentum doesn’t change. Upward momentum transfer is just as much a feature of wing-generated lift as downward transfer. Wings do not generate force the same way rockets do (rockets use pure, unadulterated NET momentum exchange), and they are different from rotors and props in subtle, but important ways.
Posted on: 2/28/2012 2:11 AM by Author "Shoe" in the forum "Aerodynamics"
http://www.rcuniverse.com/forum/fb.asp?m=10979593

RE: Cambridge scientist debunks flying myth
A bit of a nit pick, but lift is traditionally defined as the force component acting perpendicular to a wing's direction of motion. Drag is traditionally defined as the force component acting opposite a wing's direction of motion. Only in the case where a wing's motion is a straight horizontal line do these definitions match up with the vertical and horizontal components you suggest. With the traditional definitions, the lift can't consume any work or energy because there is no displacement of the wing in the direction of the lift. As an analogy, if you were to grab a 10kg object from a 1m high shelf, carry it across a room and place it on another 1m high shelf, the upward force you exerted on the object while carrying it would not have done any work (the object's total energy is unchanged). It's the drag force that consumes all the energy. I'm OK with the statement that a wing must accelerate the air around it to fly. You can accelerate the air without adding net momentum as long as you don't accelerate all the air in the same direction. Yes, you ultimately heat the air, but I don't think that's a feature of the process that really contributes to an explanation of how lift is generated.
Posted on: 2/19/2012 10:05 AM by Author "Shoe" in the forum "Aerodynamics"
http://www.rcuniverse.com/forum/fb.asp?m=10965998

RE: Mythbusters
[quote]ORIGINAL: combatpigg How about model deltas and flying wings need special airfoils with built in reflex to fly at their optimum speed and efficiency...? [/quote] I'm pretty sure that for static stability (CG in front of the neutral point), an UNSWEPT flying wing needs to have some effective reflex in order to trim for level flight. Once you go to a delta, or add enough sweep, the connection between stability and need for reflex goes away (there's a reason you don't see many unswept flying wings). I suppose this is one of those myths where something that applies to a special case (usnwept flying wings) gets communicated as a general truth.
Posted on: 2/17/2012 9:34 PM by Author "Shoe" in the forum "Aerodynamics"
http://www.rcuniverse.com/forum/fb.asp?m=10964084

RE: Cambridge scientist debunks flying myth
Eventually, when the flight is over and the hangar doors are closed, all the disturbances made by the airplane's flight will have been dissipated by viscosity. All you're then left with is the heating. It's important to note that a wing's lift force is defined as being perpendicular to the wing's direction of travel. This means that the lift force, by definition, does no work (work results from displacement in the direction of a force). If there's any drag acting on the wing (and there usually is), then to maintain steady flight, you must provide power equal to: drag x speed. All the energy you are providing eventually ends up as heat, but in the near term, some of it goes into the kinetic energy of the trailing vortex system (causing induced drag), some of it goes almost directly to heat (through viscous shearing), and some of it goes into the turbulent motion of the air (viscosity dissipates this pretty quickly). There are other places the energy can go in the near term once you get transonic. A wing generating lift without vertical momentum transfer will still require work/ energy/ power to maintain steady flight as long as there is drag acting on the wing.
Posted on: 2/17/2012 10:50 AM by Author "Shoe" in the forum "Aerodynamics"
http://www.rcuniverse.com/forum/fb.asp?m=10963324

RE: Cambridge scientist debunks flying myth
Interesting to see the conclusion you draw from the RC helicopter hovering over a scale. I think we'd agree the rotor is pushing down on the air with a force equal to the weight of the helicopter. The reading on the scale shows it is pushing up on the air with a force almost exactly equal to the weight of the helicopter. If the rotor and the scale are the only two things pushing up or down on the air, then the air is not experiencing an unbalanced force. If the air is not experiencing an unbalanced force then Newton's 2nd Law requires it's rate of change of momentum to be zero. Counterintuitive, but the rotor in ground effect is adding as much upward as downward momentum to the air.
Posted on: 2/17/2012 9:19 AM by Author "Shoe" in the forum "Aerodynamics"
http://www.rcuniverse.com/forum/fb.asp?m=10963210

RE: Cambridge scientist debunks flying myth
The concept of a wing creating lift without momentum exchange doesn't really appeal to me either. The idea that lift is the result of a wing pushing down and imparting downward momentum to the air makes for a very tidy and intuitive explanation. It just makes sense from a Newtonian point of view. There's only one itty bitty problem with this explanation. If it's accurate, then you should be able to add up all the momentum being added to the air and show that the rate of change is equal and opposite the lift force. This addition requires some care, but it's certainly something you can do. When you do this, you are faced with the rather inconvenient result that, for the case of a wing in steady flight, there isn't any momentum exchange going on. Sure the air is deflected downward behind the wing (rather vigorously), but the air is also deflected upward outboard of the wingtips (not as vigorously, but there's a lot more air set in motion outboard of the wingtips than between them). I'm not aware of any way to determine the relative magnitudes of the upward and downward contributions short of actually adding them up and comparing them. Again, when you do this you find that they are the same... no net momentum transfer. Is this result in some way inconsistent with Newton's Laws? No. Again, as long as there is no unbalanced force being exerted on the air, then there is no need for its momentum to change. The same analysis that allows you to add up the momentum allows you to find a wing's pressure footprint on the ground. The upward force from this pressure footprint sums up to equal the lift. Not a very satisfying or particularly intuitive result, but that's how a wing in steady flight does its business.
Posted on: 2/17/2012 8:14 AM by Author "Shoe" in the forum "Aerodynamics"
http://www.rcuniverse.com/forum/fb.asp?m=10963119

RE: Cambridge scientist debunks flying myth
[quote]ORIGINAL: Lnewqban Then,.........is the atmosphere heavier when an airplane flies in it? [/quote] The weight of the atmosphere isn't changed, but the pressure it exerts on the ground is increased (assuming the density of the airplane is greater than that of the air). This is similar to when a boat is set afloat on the ocean. Because a boat displaces water, it increases the depth of the ocean VERY slightly. This extra depth means that the ocean pushes down on the earth with slightly more pressure. All the additional pressure times area adds up to the weight of the boat. For an airplane in ground effect, the pressure "footprint" is fairly compact. For an airplane hundreds of spans above the ground, the pressure footprint is enormous (and the amount of pressure increase correspondingly VERY small).
Posted on: 2/16/2012 12:48 PM by Author "Shoe" in the forum "Aerodynamics"
http://www.rcuniverse.com/forum/fb.asp?m=10961906

RE: Cambridge scientist debunks flying myth
I can't stop the "nonsense" you refer to because it just happens to be reality. You are basing your description of how wings work on the physics of props/rotors/fans/leaf blowers. There are subtle but very important differences between wings and spinning things that have compact wake structures. Unfortunately, you have to open your mind to be able to appreciate this. If you can't see that wings also deflect air upward (and there's a WHOLE lot more air outboard of a wing than between the wingtips), then you clearly haven't spent enough time studying the airflow around a wing. Please stop your nonsense about transfer of net downward momentum to the air until you can SHOW that it exists. Just saying it exists doesn't make it so. Several posters have provided a rigorous accounting that shows your position is flawed. You have provided essentially nothing beyond "this appeals to my intuition". Sorry, but that doesn't cut it, your intuition has led you astray with respect to wings.
Posted on: 2/16/2012 5:18 AM by Author "Shoe" in the forum "Aerodynamics"
http://www.rcuniverse.com/forum/fb.asp?m=10961376

RE: Cambridge scientist debunks flying myth
[quote]ORIGINAL: Tim Green Air cannot support anything that's heavier than it. [/quote] A vehicle that has the same density as air still has weight. In order for that vehicle to be held airborne by the air, the air must provide an upward force equal to the vehicle's weight. I think everyone would agree that there is no momentum transfer when the air provides an upward force of this kind (you don't feel upwash or downwash around a hot-air balloon... yes RMH, flatulent passengers could provide such a thing). At first, this might appear to be a violation of Newton's Laws. The air pushing up on the balloon means that the balloon must be pushing down on the air (Newton's 3rd law). If the balloon is pushing down on the air, then doesn't Newton's 2nd law require the air's momentum to change? The answer is obviously "no". The reason the air's momentum doesn't change is because the downward force the balloon exerts on the air is balanced by an upward force that the ground exerts on the air. Newton's 2nd law only requires an object to experience a rate of momentum change when there is an UNBALANCED force on it. What does this have to do with airplanes? Although the mechanism is different (dynamic lift vs. buoyancy), the same fundamental principle still applies. If the air doesn't experience an unbalanced force, then it won't experience a rate of momentum change. Anyone suggesting the air MUST experience a change in its vertical momentum in order to generate a lift force is misapplying Newton's Laws. Is it POSSIBLE for air to experience a change in momentum associated with an aerodynamic force? Yes, but it is certainly not REQUIRED by Newton’s Laws. [quote]ORIGINAL: Tim Green Airplanes lift in response to the large amounts of air they accelerate in a downward direction. [/quote] Possibly. In addition to the large amounts of air that airplanes accelerate downward, airplanes also accelerate large amounts of air upward (there are plenty of photos showing smoke going downward behind a wing, and going upward outboard of the wing tips at the same time). As far as Newton is concerned, the air going upward is JUST as important as the air going downward. In order to show that airplanes lift in response to the large amounts of air they accelerate in a downward direction, you have to show that the difference between the rate of downward momentum transfer and the rate of upward momentum transfer is equal to the lift. Tim, you have not shown this, you have simply stated it to be the case. Those who have done a disciplined accounting (Drela, Lissaman, Betz, Kroo among others) have shown that the above statement is misleading at best. Under almost all conditions, the downward force exerted by a wing on the air is balanced by the upward force exerted by the ground on the air, and there is no net momentum transfer. If your statement is true, then it must be possible to provide an accounting for the total momentum (other than "FEEEAAL the Physics" ) that shows a direct balance between lift and the rate of net vertical momentum transfer. You say no math is required, and I would welcome an approach that doesn’t include math, but simply ignoring all the air that is being deflected upward isn’t a valid approach. If you can't provide such an accounting, then you're simply stating a belief that makes sense to you. I will point out that this belief isn’t consistent with the reality of the situation. I used to believe in a balance between wing lift and downward momentum transfer. It wasn’t until I added up the momentum of all the downward going air and subtracted the momentum of all the upward going air that I realized this balance doesn’t exist. Still skeptical, I calculated the pressure footprint of a wing on the ground. Sure enough, under most conditions, the ground pushes up on the air with the same net force that the wing pushes down. Newton’s 2nd Law tells you this mean’s the wing isn’t changing the vertical momentum of the air. I suggest you take another look at Lissaman’s paper, it might change the way you look at airplanes.
Posted on: 2/14/2012 12:27 PM by Author "Shoe" in the forum "Aerodynamics"
http://www.rcuniverse.com/forum/fb.asp?m=10958812

RE: Mythbusters
[quote]ORIGINAL: rmh With enough thrust , airfoils are meaningless [/quote] With enough thrust, [b]and enough gas,[/b] airfoils are meaningless
Posted on: 2/11/2012 1:18 PM by Author "Shoe" in the forum "Aerodynamics"
http://www.rcuniverse.com/forum/fb.asp?m=10954461

RE: Cambridge scientist debunks flying myth
So Tim Green, it appears your finely tuned intuition grants you knowledge of very complicated processes, but prevents you from addressing a simple question that would give lesser people insight into your divine understanding. Can you answer the following? Is it possible for the air to support the weight of a vehicle without net momentum transfer to the air, or do Newton's 2nd and 3rd laws require the upward force on the vehicle to be accompanied by a downward rate of momentum transfer? Newton's 2nd Law: The net force on a particle is equal to the time rate of change of its momentum. Newton's 3rd Law: The forces exerted by two bodies on each other are always equal and are directed in opposite direction.
Posted on: 2/10/2012 12:32 AM by Author "Shoe" in the forum "Aerodynamics"
http://www.rcuniverse.com/forum/fb.asp?m=10952354

RE: Cambridge scientist debunks flying myth
Is it possible for the air to support the weight of a vehicle without net momentum transfer to the air, or do Newton's 2nd and 3rd laws require the upward force on the vehicle to be accompanied by a downward rate of momentum transfer? Newton's 2nd Law: The net force on a particle is equal to the time rate of change of its momentum. Newton's 3rd Law: The forces exerted by two bodies on each other are always equal and are directed in opposite direction.
Posted on: 2/9/2012 9:28 AM by Author "Shoe" in the forum "Aerodynamics"
http://www.rcuniverse.com/forum/fb.asp?m=10951365

RE: Cambridge scientist debunks flying myth
Is it possible for the air to support the weight of a vehicle without net momentum transfer to the air, or do Newton's 2nd and 3rd laws require the upward force on the vehicle to be accompanied by a downward rate of momentum transfer? Newton's 2nd Law: The net force on a particle is equal to the time rate of change of its momentum. Newton's 3rd Law: The forces exerted by two bodies on each other are always equal and are directed in opposite direction.
Posted on: 2/8/2012 1:23 PM by Author "Shoe" in the forum "Aerodynamics"
http://www.rcuniverse.com/forum/fb.asp?m=10950108

RE: Cambridge scientist debunks flying myth
BMatthews, I think irreverence for educated authorities/researchers/experts is appropriate in a forum like this. Ideas here are forced to stand on their own merits, and the ability of those who present them to make a convincing case without resort to jargon or appeal to consensus (I find few things more grating than "according to a NASA website..." ) . As long as minds remain open, this forum offers a unique opportunity to look at ideas from a wide variety of perspectives. Without peer review or other standard, you just need to be discriminating in what you take away. With this territory comes the reality that some won't be persuaded, no matter how airtight the argument (and determining who is stubborn and who is right usually comes down to what side of the argument you're on). It's interesting to see how a discussion develops when credentials are stripped away. Reminds me a bit of the caption from an old New Yorker cartoon... "On the internet, nobody knows you're a dog".
Posted on: 2/8/2012 11:45 AM by Author "Shoe" in the forum "Aerodynamics"
http://www.rcuniverse.com/forum/fb.asp?m=10949969

RE: Cambridge scientist debunks flying myth
[quote]ORIGINAL: Tim Green Think about that theory - it means you cannot move air and have a reaction to it. [/quote] If that's what you took away from the attached link, then you need to read it again... carefully. Drela in no way suggests: "you cannot move air and have a reaction to it". He is instead pointing out that it is quite possible for the air to exert a force on an object without absorbing any net momentum.
Posted on: 2/8/2012 8:35 AM by Author "Shoe" in the forum "Aerodynamics"
http://www.rcuniverse.com/forum/fb.asp?m=10949685

RE: Cambridge scientist debunks flying myth
This is going to get long winded, but I (naively) think I might be able explain the distinction(s) I have been trying to make. Consider three different systems that each generate the same amount of force (thrust/lift, call it what you want)... System 1 is a rocket engine that uses very dense propellant (imagine an extremely rapid fire machine gun with small, but dense bullets). System 2 is a propeller System 3 is a steadily translating wing Put each of these systems in a fairly large, but sealed chamber. Because the chamber is sealed, you can impart momentum to a portion of the air within the chamber, but in doing so you must impart equal and opposite momentum to another portion of air within the chamber. To see why this is the case, think about the center of gravity of the air within the chamber. If you were to impart any NET momentum to the air within the chamber, that would mean that you have put the air's center of gravity in motion (by definition). If the center of gravity were put in motion, air would have to leave the chamber (but it can't because it is sealed). With system 1, the rocket fires and it is easy to show that the thrust it generates must be equal and opposite the rate of change of the propellant's momentum (Newton's Second and Third laws). When the propellant eventually hits the wall/floor of the chamber, all of that momentum is taken back out. Take the walls away and the direct relationship between thrust and rate of momentum change is restored. With system 2, the propeller spins and it's very tempting to say that the propeller transfers momentum to the air at a rate equal and opposite its thrust. The case of the propeller is slightly different from the rocket in an important way: because of the continuous nature of the air in the chamber, the propeller never gets a chance to impart any net momentum to the air. For incompressible flow, the INSTANT the propeller starts to generate thrust, it creates a pressure field on the walls that induce flow within the chamber to keep the net momentum at zero (again the CG of the air within the chamber would translate if this were not the case). Take the walls away (completely) and the propeller will impart momentum to the air at a rate equal to its thrust. Newton breathes a sigh of relief. With system 3, the wing translates through the air, and just like the propellor, it cannot impart any momentum to the air in the chamber with the walls in place. The difference between the propellor and the wing comes into play when the walls are removed. Unlike the propellor (the flow around which has a fairly compact structure), the wing leaves an extended system of vorticity behind it (that for a steadily translating wing extends at least hundreds of spans behind the wing.. look up on a clear day and see the physics). The consequence of this is that even if you take the walls of the chamber away, the wing will still interact with the ground (a more scientific description is: the ground prevents you from enclosing the wing within a control volume that approaches infinitely tall and thin). The end result is that even with the walls of the chamber removed, the ground will always prevent momentum addition to the air. This is in no way a contradiction to Newton's laws. It simply means that the force exerted by the wing on the air is always balanced by the force exerted by the ground on the air (no rate of momentum change required). What I am trying to convey is: 1. Wings are different from propellers, and propellers are different from rockets in subtle ways. You need to be very careful trying to explain the behavior of one with an analogy to the other. 2. Wings certainly impart momentum to the air locally, but they CANNOT impart any net momentum. Realizing this you cannot invoke a balance between lift and momentum exchange (well you can invoke it, it just happens to be wrong).
Posted on: 2/7/2012 8:38 AM by Author "Shoe" in the forum "Aerodynamics"
http://www.rcuniverse.com/forum/fb.asp?m=10948074

RE: Cambridge scientist debunks flying myth
[quote]ORIGINAL: Tim Green [quote]ORIGINAL: Shoe [quote]ORIGINAL: Tim Green [quote]ORIGINAL: Shoe Yet you can't provide any accounting that shows that a wing imparts net vertical momentum to the air... curious. This sounds more like faith than Physics. [/quote] You can't feel the physics, apparently. Too bad you've never had a plane or or a fan in your hands. [/quote] I know
Posted on: 2/7/2012 12:47 AM by Author "Shoe" in the forum "Aerodynamics"
http://www.rcuniverse.com/forum/fb.asp?m=10947616

RE: Cambridge scientist debunks flying myth
[quote]ORIGINAL: Tim Green [quote]ORIGINAL: Shoe Yet you can't provide any accounting that shows that a wing imparts net vertical momentum to the air... curious. This sounds more like faith than Physics. [/quote] You can't feel the physics, apparently. Too bad you've never had a plane or or a fan in your hands. [/quote] I know this has been beaten to death, but try to stick
Posted on: 2/6/2012 8:28 AM by Author "Shoe" in the forum "Aerodynamics"
http://www.rcuniverse.com/forum/fb.asp?m=10946348

RE: Cambridge scientist debunks flying myth
Yet you can't provide any accounting that shows that a wing imparts net vertical momentum to the air... curious. This sounds more like faith than Physics.
Posted on: 2/5/2012 1:42 PM by Author "Shoe" in the forum "Aerodynamics"
http://www.rcuniverse.com/forum/fb.asp?m=10945393

RE: Cambridge scientist debunks flying myth
Not sure what "feeling the physics" buys you, but understanding the physics allowed him to design some pretty cool and successful aircraft, like this one... http://en.wikipedia.org/wiki/Gossamer_Condor
Posted on: 2/5/2012 6:40 AM by Author "Shoe" in the forum "Aerodynamics"
http://www.rcuniverse.com/forum/fb.asp?m=10944769

RE: Cambridge scientist debunks flying myth
The attached paper by Peter Lissaman tries to explain the origins of lift as simply as possible. It's lighthearted, but still a bit complex. This work is not at odds with the reference in the original post, but addresses the far more interesting case of "3D" lift I think he makes a pretty convincing case for the idea that the rate at which a wing imparts net downward momentum to the air is zero. link to the same file: http://www.redfish.com/friam/uploads/Lissaman_MeaningOfLift.pdf
Posted on: 2/4/2012 10:33 AM by Author "Shoe" in the forum "Aerodynamics"
http://www.rcuniverse.com/forum/fb.asp?m=10943576

RE: Cambridge scientist debunks flying myth
[quote]ORIGINAL: Tim Green Nawww - it's Newton. Stand behind a prop, and feel the physics. Newton states that for every action, there's a reaction. So logic dictates that since wings move air down while also accelerating that same air, then the wing reacts with an opposing force in the opposite direction - AKA lift. Stand under a helicopter, and feel the physics. No one can demonstrate lift through the miracle of pressure differentials only - not without also thrusting air down. The pressure differentials are just what happens when air moves anywhere. They are a red herring, provided by 60 years of disinformation in our textbooks. The new and correct thinking is what I stated above. And, if something's thrusting air down (hold onto your biggest plane with the props running full and feel the physics), then that something is reacting to it. You cannot deny Newton. [/quote] This is an apparently tidy explanation, but it neglects all the air that is being accelerated upward by a lifting wing (and to some extent a helicopter). Those who have gone through the rigor of summing up all of the momentum being imparted to the air (and to invoke Newton, you have to consider ALL the air being moved) have concluded that equality between the rate of momentum exchange and aerodynamic force does not exist. You cannot deny Newton, but you can misapply his laws.
Posted on: 2/4/2012 9:44 AM by Author "Shoe" in the forum "Aerodynamics"
http://www.rcuniverse.com/forum/fb.asp?m=10943508

RE: Cambridge scientist debunks flying myth
[quote]ORIGINAL: MTK Lift generated by pressure differential is a certain percent of the whole, but not 100%; Delta P is but one component of the total amount of lift required to lift any wing [/quote] This isn't correct. Unless you are considering electrical, magnetic, gravitational, or other forces "exerted at a distance" by the air on the wing, the net aerodynamic force acting on a wing must be EXACTLY equal to the sum of the surface stresses acting on the wing. A surface stress is the combination of pressure (acting perpendicular to an element of the wing surface) and shear (acting parallel to an element of the wing surface). This equality between net surface stresses and net aerodynamic force isn't some mysterious aerodynamic truth, it's basic Physics. Motion of the air away from the wing DOES affect the aerodynamic force acting on the wing, but only by influencing the stresses on the surface of the wing. The confusion usually arises because it is sometimes convenient (and accurate) to identify different mechanisms that contribute to the lift or drag on a wing. You can identify different mechanisms that contribute to the "lift generated by pressure differential", but you cannot (accurately) identify "lift generated by pressure differential" as a component of the total lift... it IS the total lift.
Posted on: 2/3/2012 6:00 AM by Author "Shoe" in the forum "Aerodynamics"
http://www.rcuniverse.com/forum/fb.asp?m=10941593

RE: Cambridge scientist debunks flying myth
[quote]ORIGINAL: Lnewqban higher lift means higher drag, always,...[/quote] Not always. If your airplane has multiple surfaces at different incidences, or a single surface with twist, it is quite possible for higher lift to mean less drag. [quote]ORIGINAL: Lnewqban ...and drag grows with the square of the velocity.[/quote] Only under specific conditions (like fixed AOA). In level flight, drag actually goes down with airspeed until you reach minimum drag. The basic equation for drag (drag = 1/2 * density * true_airspeed^2 * reference_area * drag_coefficient) can mislead you. In the case where you maintain Weight = Lift, the drag coefficient is also a function of airspeed, so drag is not simply proportional to the square of velocity.
Posted on: 2/3/2012 3:59 AM by Author "Shoe" in the forum "Aerodynamics"
http://www.rcuniverse.com/forum/fb.asp?m=10941494