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#131
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visualisation of the lift distribution over a wing
In article ,
Beryl wrote: snip No downwash, no lift. No go learn something. Let's learn here. From you. Is that 90 degree turn *exactly* the same as a 180 degree turn that directs incoming air back in the opposite direction? Read this: "To determine [the angle represented by a greek letter in the original text], we observe that no downwash is generated when the wing generates no lift." I'm not disagreeing with that. I'll rephrase it, and say no circulation is generated. It is not even relevant. http://www.aoe.vt.edu/~cwoolsey/Cour...al/Aerodynamic Properties.pdf Read it over and over again until you get it. Get what? It's about wings and geometry. Find something about air moving through air. http://www.onemetre.net/Design/Downwash/Downwash.htm "The theory of downwash starts by noting that you only get downwash when you have lift.* No lift, no downwash." http://amasci.com/wing/airfoil.html ' The "Newton" explanation is wrong because downwash occurs BEHIND the wing, where it can have no effects? Downwash can't generate a lifting force? INCORRECT. Wrong, and silly as well! The above statement caught fire on the sci.physics newsgroup. Think for a moment: the exhaust from a rocket or a jet engine occurs BEHIND the engine. Does this mean that action/reaction does not apply to jets and rockets? Of course not. It's true that the exhaust stream doesn't directly push on the inner surface of a rocket engine. The lifting force in rockets is caused by acceleration of mass, and within the exhaust plume the mass is no longer accelerating. In rocket engines, the lifting force appears in the same place that the exhaust is given high velocity: where gases interact inside the engine. And with aircraft, the lifting force appears in the same place that the exhaust (the downwash) is given high downwards velocity. If a wing encounters some unmoving air, and the wing then throws the air downwards, the velocity of the air has been changed, and the wing will experience an upwards reaction force. At the same time, a downwash- flow is created. To calculate the lifting force of a rocket engine, we can look exclusively at the exhaust velocity and mass, but this doesn't mean that the rocket exhaust creates lift. It just means that the rocket exhaust is directly proportional to lift (since the exhaust velocity and the lifting force have a common origin.) The same is true with airplane wings and downwash. To have lift at high altitudes, we MUST have downwash, and if we double the downwash, we double the lifting force. But downwash doesn't cause lift, instead the wing's interaction with the air both creates a lifting force and gives the air a downwards velocity (by F=MA, don't you know!)' -- Alan Baker Vancouver, British Columbia http://gallery.me.com/alangbaker/100008/DSCF0162/web.jpg |
#132
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visualisation of the lift distribution over a wing
Alan Baker wrote:
Pressure waves can reach the ground, without the air in the column descending to the ground. I never said that the particular molecules that the aircraft touches are the ones that have to reach the ground. You said "The net flow is downward until it hits the ground and the momentum is transfer to the earth." And it is: the *net* flow. The molecules that "reach" the ground are the ones that were *already there* at ground level. Hmmmm...It's not too out of line to think of air particles at ambient temps as moving at the speed of sound. Its those air "molecules" that carry sound waves longitudinally. Air particles don't stay anywhere, in a manner of speaking... Brian W |
#133
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visualisation of the lift distribution over a wing
In article ,
brian whatcott wrote: Alan Baker wrote: Pressure waves can reach the ground, without the air in the column descending to the ground. I never said that the particular molecules that the aircraft touches are the ones that have to reach the ground. You said "The net flow is downward until it hits the ground and the momentum is transfer to the earth." And it is: the *net* flow. The molecules that "reach" the ground are the ones that were *already there* at ground level. Hmmmm...It's not too out of line to think of air particles at ambient temps as moving at the speed of sound. Actually, they move at considerably faster than the speed of sound... Its those air "molecules" that carry sound waves longitudinally. Air particles don't stay anywhere, in a manner of speaking... -- Alan Baker Vancouver, British Columbia http://gallery.me.com/alangbaker/100008/DSCF0162/web.jpg |
#134
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visualisation of the lift distribution over a wing
Alan Baker wrote:
Jim Logajan wrote: 3) Therefore if, say, the downwash is 1 kg/s at any given instant due to the wing, somewhere else in the fluid there must be an upwash at that same instant of 1 kg/s. Agree or disagree? Agree. At the surface of the Earth. The qualifier indicates to me that you don't agree with the statement as written. Unfortunately, I consider conservation of mass at all points in time in an incompressible fluid an essential element to understanding the behavior of downwash. If you don't, then I think any further debate between us is ended. (Just FYI, imagine a ~957 kg (Fg ~= 9379 N) helicopter dropped from a balloon from 3,000 m altitude (rho ~= 0.83 kg/m^3) and it's engine immediately started. After a small drop it levels out and maintains a downwash of air moving through its 6 m diameter disk (A ~= 28 m^2) at, say, 20 m/s. (So m_dot ~= 469 kg/s and hence Fe = Fg.) It would take ~150 s for that downwash to reach the ground if it maintained that speed. In the mean time, once the helicopter stopped descending, conservation of mass in an incompressible fluid seems to require an equal volume of air to have an upward vector of 20 m/s. So the surface of earth appears to be irrelevant for over two minutes.) |
#135
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visualisation of the lift distribution over a wing
Alan Baker wrote:
... First of all, the downward motion of the vortex clearly carries right out the bottom of the frame. Are you impaired? The airplane is approaching the camera. The camera is looking up at the airplane. The bottom of the frame contains the distant background. Objects farther than the airplane appear lower in the frame. If the camera was above the approaching airplane and looking down at it, distant objects would appear higher in the frame than the airplane. None of which refutes what I said. Oh, it was simply interesting to you that the vortex goes off into the distance, right out the bottom of the picture. |
#136
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visualisation of the lift distribution over a wing
In article ,
Jim Logajan wrote: Alan Baker wrote: Jim Logajan wrote: 3) Therefore if, say, the downwash is 1 kg/s at any given instant due to the wing, somewhere else in the fluid there must be an upwash at that same instant of 1 kg/s. Agree or disagree? Agree. At the surface of the Earth. The qualifier indicates to me that you don't agree with the statement as written. You're right. I shouldn't have stated it that way. Unfortunately, I consider conservation of mass at all points in time in an incompressible fluid an essential element to understanding the behavior of downwash. If you don't, then I think any further debate between us is ended. Look if you think that conservation of *mass* plays any role in this, you're missing out from the start. It's conservation of *momentum* that's in play here. The aircraft has a force exerted on it equal to its weight. That means that the aircraft must be exerting a force on the air in the opposite direction. That means that there is a constant change of momentum being done on the air by the aircraft. That means air *must* be moving down (net) after the aircraft has passed. (Just FYI, imagine a ~957 kg (Fg ~= 9379 N) helicopter dropped from a balloon from 3,000 m altitude (rho ~= 0.83 kg/m^3) and it's engine immediately started. After a small drop it levels out and maintains a downwash of air moving through its 6 m diameter disk (A ~= 28 m^2) at, say, 20 m/s. (So m_dot ~= 469 kg/s and hence Fe = Fg.) It would take ~150 s for that downwash to reach the ground if it maintained that speed. In the mean time, once the helicopter stopped descending, conservation of mass in an incompressible fluid seems to require an equal volume of air to have an upward vector of 20 m/s. So the surface of earth appears to be irrelevant for over two minutes.) Nope. The conservation of momentum says that there cannot be an equal amount of air moving upward at an equal speed. -- Alan Baker Vancouver, British Columbia http://gallery.me.com/alangbaker/100008/DSCF0162/web.jpg |
#137
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visualisation of the lift distribution over a wing
In article ,
Beryl wrote: Alan Baker wrote: ... First of all, the downward motion of the vortex clearly carries right out the bottom of the frame. Are you impaired? The airplane is approaching the camera. The camera is looking up at the airplane. The bottom of the frame contains the distant background. Objects farther than the airplane appear lower in the frame. If the camera was above the approaching airplane and looking down at it, distant objects would appear higher in the frame than the airplane. None of which refutes what I said. Oh, it was simply interesting to you that the vortex goes off into the distance, right out the bottom of the picture. No. That shows that the air continues to move downward far below the small portion of the vortex which is moving up. The net movement of the air after the plane's passing must be downward, because the plane is exerting a force on the air. Force is a change of momentum with respect to time. -- Alan Baker Vancouver, British Columbia http://gallery.me.com/alangbaker/100008/DSCF0162/web.jpg |
#138
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visualisation of the lift distribution over a wing
Alan Baker wrote:
Look if you think that conservation of *mass* plays any role in this, you're missing out from the start. It's conservation of *momentum* that's in play here. It appears you have never studied fluid dynamics (maybe elementary fluid statics?) and I doubt that you own any books on the subject. The aircraft has a force exerted on it equal to its weight. That means that the aircraft must be exerting a force on the air in the opposite direction. In other news, 1 + 1 = 2. That means that there is a constant change of momentum being done on the air by the aircraft. That means air *must* be moving down (net) after the aircraft has passed. You must have a devil of a time figuring out what keeps balloons afloat, what with no handy downward moving air! (Just FYI, imagine a ~957 kg (Fg ~= 9379 N) helicopter dropped from a balloon from 3,000 m altitude (rho ~= 0.83 kg/m^3) and it's engine immediately started. After a small drop it levels out and maintains a downwash of air moving through its 6 m diameter disk (A ~= 28 m^2) at, say, 20 m/s. (So m_dot ~= 469 kg/s and hence Fe = Fg.) It would take ~150 s for that downwash to reach the ground if it maintained that speed. In the mean time, once the helicopter stopped descending, conservation of mass in an incompressible fluid seems to require an equal volume of air to have an upward vector of 20 m/s. So the surface of earth appears to be irrelevant for over two minutes.) Nope. Dang - I try to use real numbers to establish a baseline example, and you manage to use a single word to demolish my attempts! Really helpful mathematical counter-example you produced - not. The conservation of momentum says that there cannot be an equal amount of air moving upward at an equal speed. I don't know what your problem is - maybe you are thinking this is a rocket problem where no external fluids are involved and you can't get your mind around the fact that THIS ISN'T A BLOODY ROCKET PROBLEM. Whatever the case, you seem to be fixated on applying one conservation law to one element in the entire system to the exclusion of everything else. Best of luck to you. |
#139
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visualisation of the lift distribution over a wing
Jim Logajan wrote:
Alan Baker wrote: Look if you think that conservation of *mass* plays any role in this, you're missing out from the start. It's conservation of *momentum* that's in play here. It appears you have never studied fluid dynamics (maybe elementary fluid statics?) and I doubt that you own any books on the subject. The aircraft has a force exerted on it equal to its weight. That means that the aircraft must be exerting a force on the air in the opposite direction. In other news, 1 + 1 = 2. That means that there is a constant change of momentum being done on the air by the aircraft. That means air *must* be moving down (net) after the aircraft has passed. You must have a devil of a time figuring out what keeps balloons afloat, what with no handy downward moving air! (Just FYI, imagine a ~957 kg (Fg ~= 9379 N) helicopter dropped from a balloon from 3,000 m altitude (rho ~= 0.83 kg/m^3) and it's engine immediately started. After a small drop it levels out and maintains a downwash of air moving through its 6 m diameter disk (A ~= 28 m^2) at, say, 20 m/s. (So m_dot ~= 469 kg/s and hence Fe = Fg.) It would take ~150 s for that downwash to reach the ground if it maintained that speed. In the mean time, once the helicopter stopped descending, conservation of mass in an incompressible fluid seems to require an equal volume of air to have an upward vector of 20 m/s. So the surface of earth appears to be irrelevant for over two minutes.) Nope. Dang - I try to use real numbers to establish a baseline example, and you manage to use a single word to demolish my attempts! Really helpful mathematical counter-example you produced - not. The conservation of momentum says that there cannot be an equal amount of air moving upward at an equal speed. I don't know what your problem is - maybe you are thinking this is a rocket problem where no external fluids are involved and you can't get your mind around the fact that THIS ISN'T A BLOODY ROCKET PROBLEM. Whatever the case, you seem to be fixated on applying one conservation law to one element in the entire system to the exclusion of everything else. Best of luck to you. Two dimensional Newtonian thinking in a three dimensional non-Newtonian world. |
#140
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visualisation of the lift distribution over a wing
In article ,
Jim Logajan wrote: Alan Baker wrote: Look if you think that conservation of *mass* plays any role in this, you're missing out from the start. It's conservation of *momentum* that's in play here. It appears you have never studied fluid dynamics (maybe elementary fluid statics?) and I doubt that you own any books on the subject. Sorry, lad, but conservation of mass is a principle that comes up mostly in *chemistry*. The aircraft has a force exerted on it equal to its weight. That means that the aircraft must be exerting a force on the air in the opposite direction. In other news, 1 + 1 = 2. What a pity then that you don't understand it. That means that there is a constant change of momentum being done on the air by the aircraft. That means air *must* be moving down (net) after the aircraft has passed. You must have a devil of a time figuring out what keeps balloons afloat, what with no handy downward moving air! I have no trouble figuring that out at all. A gas of a different density within the balloon causes the net upward force on the balloon exerted by the air outside the balloon to be greater than the net downward force on it. (Just FYI, imagine a ~957 kg (Fg ~= 9379 N) helicopter dropped from a balloon from 3,000 m altitude (rho ~= 0.83 kg/m^3) and it's engine immediately started. After a small drop it levels out and maintains a downwash of air moving through its 6 m diameter disk (A ~= 28 m^2) at, say, 20 m/s. (So m_dot ~= 469 kg/s and hence Fe = Fg.) It would take ~150 s for that downwash to reach the ground if it maintained that speed. In the mean time, once the helicopter stopped descending, conservation of mass in an incompressible fluid seems to require an equal volume of air to have an upward vector of 20 m/s. So the surface of earth appears to be irrelevant for over two minutes.) Nope. Dang - I try to use real numbers to establish a baseline example, and you manage to use a single word to demolish my attempts! Really helpful mathematical counter-example you produced - not. No math is necessary for this. Look up "qualitative analysis". The conservation of momentum says that there cannot be an equal amount of air moving upward at an equal speed. I don't know what your problem is - maybe you are thinking this is a rocket problem where no external fluids are involved and you can't get your mind around the fact that THIS ISN'T A BLOODY ROCKET PROBLEM. Whatever the case, you seem to be fixated on applying one conservation law to one element in the entire system to the exclusion of everything else. The law I'm focussed on is the one that counts. It doesn't matter whether the fluid is expelled from inside or whether it's an external fluid diverted down by the surfaces of the craft. In order for there to be a continuous force W equaling the weight of the craft acting on it, the craft must exert a force -W on the fluid. That -W means that there is a downward change of momentum in the fluid. Since the fluid is no accelerated indefinitely, there must be a continuous flow (mass per unit time M/t) of the fluid accelerated to a velocity V where the equation looks like: -W = M/t * V The velocity of the fluid will be: V = -W/(M/t) That is inescapable. If the craft weighs 9800N (newtons), and it moves 100kg of air every second, then the air must be moving downward (net, now!) at 98 m/s. I'm sorry if you don't get this, but it is very simple and absolutely irrefutable. -- Alan Baker Vancouver, British Columbia http://gallery.me.com/alangbaker/100008/DSCF0162/web.jpg |
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