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#1
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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 |
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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. |
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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. |
#4
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In article ,
cavelamb wrote: 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. LOL Sorry, caveman.... But conservation of momentum works well enough at the speeds at which aircraft operate. And Newton's laws tell us all we need to know. -- Alan Baker Vancouver, British Columbia http://gallery.me.com/alangbaker/100008/DSCF0162/web.jpg |
#5
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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 |
#6
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Alan Baker wrote:
In article , Jim Logajan wrote: 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. Nope! The "gas of a different density" inside does not cause the differential forces exerted by the air outside. Whether the balloon is filled with phlogiston or concrete, the difference between upward and downward forces exerted by the air outside depends on the height of the balloon. That's what my goofy 10,000 foot thick wing was, that would take off with no airspeed or power. A balloon. |
#7
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In article ,
Beryl wrote: Alan Baker wrote: In article , Jim Logajan wrote: 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. Nope! The "gas of a different density" inside does not cause the differential forces exerted by the air outside. Whether the balloon is filled with phlogiston or concrete, the difference between upward and downward forces exerted by the air outside depends on the height of the balloon. That's what my goofy 10,000 foot thick wing was, that would take off with no airspeed or power. A balloon. You're wrong. The "net downward force" includes the force of gravity. Which is lessened because a gas of lower density.. ....I can't believe I have to explain things so basic. -- Alan Baker Vancouver, British Columbia http://gallery.me.com/alangbaker/100008/DSCF0162/web.jpg |
#8
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Alan Baker wrote:
In article , Beryl wrote: Alan Baker wrote: In article , Jim Logajan wrote: 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. Nope! The "gas of a different density" inside does not cause the differential forces exerted by the air outside. Whether the balloon is filled with phlogiston or concrete, the difference between upward and downward forces exerted by the air outside depends on the height of the balloon. That's what my goofy 10,000 foot thick wing was, that would take off with no airspeed or power. A balloon. You're wrong. The "net downward force" includes the force of gravity. Which is lessened because a gas of lower density.. ...I can't believe I have to explain things so basic. I know, but you have to. Your first attempt didn't appear to include unsaid gravity. Gas that you described as "within the balloon" is now treated as "part of" the balloon. |
#9
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In article ,
Beryl wrote: Alan Baker wrote: In article , Beryl wrote: Alan Baker wrote: In article , Jim Logajan wrote: 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. Nope! The "gas of a different density" inside does not cause the differential forces exerted by the air outside. Whether the balloon is filled with phlogiston or concrete, the difference between upward and downward forces exerted by the air outside depends on the height of the balloon. That's what my goofy 10,000 foot thick wing was, that would take off with no airspeed or power. A balloon. You're wrong. The "net downward force" includes the force of gravity. Which is lessened because a gas of lower density.. ...I can't believe I have to explain things so basic. I know, but you have to. Your first attempt didn't appear to include unsaid gravity. Gas that you described as "within the balloon" is now treated as "part of" the balloon. Of course gas within the balloon is part of the ballon. Nothing I said ever implied otherwise. -- Alan Baker Vancouver, British Columbia http://gallery.me.com/alangbaker/100008/DSCF0162/web.jpg |
#10
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Alan Baker wrote:
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*. Please don't patronize when you've never studied a subject. A moment's research would have prevented you from posting something that incredibly ignorant. You might want to look up the "axioms of fluid dynamics" before you further add to your public embarrassment. 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. Shrug - your insults are lame and tiresome, but I will admit you're patronization is irritating. I do make mistakes about things I was taught, but I should point out to you that I did well enough in college physics to earn an undergraduate degree in the subject. Unlike you, I have had to solve a **** load of problems involving conservation of momentum to prove I understood the basics - including conservation of momentum in quantum mechanical systems. So far as I know, you HAVEN'T had to prove your mastery of the subject with ANYONE. (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". Well that probably explains your problem - you don't know how to set up the math properly, so you have no way to validate whether your "qualitative analysis" is correct. Ironically, all your posts contain the same violation of conservation of momentum - and yet you keep pointing to that concept as vindication. 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. You can't "focus" on one conservation law because the number of constraining equations has to equal the number of variables. Doing so simply leads to an infinite number of bogus results. 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. Sigh. This is a case where a little knowledge is a dangerous thing. It would take a book to explain the problem with your conceptual view of fluid dynamics. I don't have that sort of patience. 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 Of course if you had read my earlier post you'd see I'd ALREADY USED THAT EQUATION. But you obviously aren't familiar with the conventions used in fluid dynamics, so you probably had no clue what my "m_dot" meant or how I got the figures I did. 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. You math is correct and no one has denied there is a downwash (why you think otherwise continues to baffle me) yet your "net, now" comment violates conservation of momentum. Here's why: If we choose a reference frame so that at T=0 everything in the system is stationary with respect to that frame, we set the net momentum of the system to 0. Then, so long as the system remains closed, at all other times the conservation of momentum must yield 0. But according to your "qualitative analysis" the net vertical momentum P_net_z increases with time T, like so: P_net_z(T) = (100 kg/s)*T*V That's because the earth and the airplane maintain zero vertical momentums (P_earth_z(T) = 0, P_plane_z(T) = 0,) and there appears to be nothing in your conceptual view of the situation to correct that violation of conservation of momentum. I'm sorry if you don't get this, but it is very simple and absolutely irrefutable. It appears to salve your ego to ascribe assertions to me that I never made and then tell the world that those falsehoods prove I don't "get it." Probably because you've grown so much hubris and so little humility. |
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