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#271
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lift, wings, and Bernuolli
In article ,
Jose wrote: F = m/t * v/t; the force is equal to the rate of mass per unit time, multiplied by the distance per unit time. I assume a typo: F = m/t * d/t (since v=d/t) Jose You assume correctly. g -- Alan Baker Vancouver, British Columbia "If you raise the ceiling 4 feet, move the fireplace from that wall to that wall, you'll still only get the full stereophonic effect if you sit in the bottom of that cupboard." |
#272
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lift, wings, and Bernuolli
"Jose" wrote Well, when an object passes through the air, does it not compress the air in front of it (and rarefy the air behind it)? This is how speakers work. Those are all pressure changes. Air is pressurized behind the speaker, just as well as the air in front of it. That is how bass reflex speakers work. -- Jim in NC |
#273
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The Impossibility of Flying Heavy Aircraft Without Training
"Immanuel Goldstein" The Impossibility of Flying Heavy Aircraft Without Training I would ask Nila Sagadevan to explain the video of Usama Bin Laden gloating about his accomplishments. Dallas |
#274
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The Impossibility of Flying Heavy Aircraft Without Training
"Immanuel Goldstein" What hijackers? http://news.bbc.co.uk/1/hi/world/middle_east/1559151.stm "Furthermore another article explains that the pilot who lives in Casablanca was named Walid al-Shri (not Waleed M. al-Shehri) and that much of the BBC information regarding "alive" hijackers was incorrect according to the same sources used by BBC." http://en.wikipedia.org/wiki/Waleed_al-Shehri Dallas |
#275
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The Impossibility of Flying Heavy Aircraft Without Training
"cjcampbell" Actually, he is not. Not in the US, anyway. There is no one by the name of Sagadevan currently holding a pilot certificate of any kind in the US Here he is: http://www.warpaintofthegods.com/wp/about.cfm Dallas |
#276
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lift, wings, and Bernuolli
Alan Baker wrote: In article . com, wrote: Jose wrote: The hovering spacecraft has zero horizontal and vertical momentum. It has weight, directed downwards. The engine accelerates mass downward producing an upward force equal in magnitude and opposite in direction to the weight of the spacecraft. This imparts an acceleration to the spacecraft equal in magnitude and opposite in direction from the local acceleration due to gravity. The flying wing has some horizontal momentum which is secondary here, How much? and zero vertical momentum. It also has weight, directed downwards. The wing accelerates mass downward (mass it finds in the air molecules) producing an upward force equal in magnitude and opposite in direction to the weight of the wing (and its presumably attached aircraft. It does so by finding air in front of it, flinging it downwards and forwards (which causes the air in front to try to get out of the way by rising). In the steady state, one can measure high pressure below and low pressure above, but this is just the macroscopic manifestation of the greater number of molecular collisions below, and the lesser number of collisions above. That's what pressure is - we have both agreed on this. The greater number of collisions below imparts an acceleration to the aircraft equal in magnitude and opposite in direction from the local acceleration due to gravity. I agree that lift is a force, exerted on the aircraft by the air, which in steady level flight is equal in magnitude and opposite in direction to the weight of the aircraft. Energy is 'pumped' into the air by the plane. There is no need for a net momentum exchange between the airplane and the air in order for energy to be exchanged or for forces to be applied. Indeed, in those last two paragraphs above, you make no mention of momentum. BTW, I was wrong to invoke conservation of momentum. Momentum is conserved in elastic collisions, like the collision between a cue ball and the eight ball. Momentum is not conserved in inelastic collisions, like the collision between a cue ball and a nerf ball. You are incorrect. Momentum is *always* conserved. How is momentum conserved when a cue ball hits a nerf ball? Roll the airplane into a 90 degree bank. The weight is now orthogonal to the lift. As teh airplane falls, it banks even though there is no Earth 'under' the belly. Why? Because the wings are exerting a force on the air and the air consequently experiences a change in momentum. Yes, both the airplane and the air experience a net change in momentum when the aircraft climbs, descends, or banks. In level flight at constant speed the aircraft has constant horzontal and zero vertical momentum. The air exerts a force on the wings. In level flight, this force is countered by an equal and opposite force exerted on the aircraft by the gravitational attraction of the earth. Without that countering force, the aircraft would accelerate upward. That's what an unbalanced force *does*. Yes, no question about weight being balanced by lift. But the wings also exert a force on the air (Newton, remember: for every force there is an equal and opposite, etc., etc.). That force is not countered by *anything*. Hence, the air is accelerated downward; a continuous stream of air receives an constant change in momentum. If the air has a net increase in downward momentum, how is momentum conserved. F = ma; that's the way we normally see it presented. This equation relates force, mass and acceleration. It assumes a constant force acting on a constant mass will produce a constant acceleration, and the mass will start moving faster and faster. But there is an equally valid presentation of that equation; one which is more useful for examining what happens with an aircraft moving through the air: F = md/t^2; force is equal to mass, times distance, divided by the time squared. If you keep velocity and time squared together, you get acceleration of course, but there's no rule that says you have to. In fact, the rules of equations say exactly the opposite: that an equation is equally valid regardless of the way you group multiplications and divisions. So: F = m/t * v/t; the force is equal to the rate of mass per unit time, multiplied by the distance per unit time. What that says is that if you change the velocity of a given mass flow (air) by a given velocity, then you will get a given force. Yes, Force is the time rate of change of momentum. In other words, an aircraft passing through the air will cause a portion of that air to be disturbed downward. Because the aircraft is moving forward a constant speed, it imparts a downward velocity to certain mass of air each unit of time. The air starts moving downward with a certain velocity. I don't deny that downflow occurs. The pont is that downflow is a consequence of lift, not the cause of lift, and it is balanced by upflow, (albeit a more diffuse flow) otherwise the upper atmosphere would run out of air. Once you understand this, you understand why induced drag is less at hight speeds than low. Go twice as fast, and you encounter twice as much air in any unit time, and thus only need to impart a velocity to it that is half as much. But because the kinetic energy involved is proportional to mass and proportional to the *square* of velocity. Twice as much mass doubles its contribution to energy lost, but half the velocity *quarters* its contribution; giving an overall kinetic energy lost to induced drag of half as much when going twice as fast. Interesting. -- FF |
#277
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lift, wings, and Bernuolli
How is momentum conserved when a cue ball hits a nerf ball?
The vector sum, before and after, is identical. The vectors themselves are different (kinetic energy is converted to heat and such) but looking at both balls, or even looking at a cue ball and a glue ball, the center of gravity moves with the same velocity before and after. If the air has a net increase in downward momentum, how is momentum conserved. ....by the air's eventual collision with the earth. Momentum is similarly conserved when an object merely falls. The momentum gained by the falling object is cancelled by the momentum acquired by the earth rising up to meet it. In the case of "mysterious phantom gravity" not associated with the earth, momentum is not conserved, it disappears into the phantom gravity. This is one of the reasons why phantom gravity is not experimentally supported. If you ignore the earth, you are in the same position. I don't deny that downflow occurs. The pont is that downflow is a consequence of lift, not the cause of lift, and it is balanced by upflow, (albeit a more diffuse flow) otherwise the upper atmosphere would run out of air. If there were no earth for the smooshed-together air to crowd up against, the upper atmosphere =would= run out of air. Jose -- Money: what you need when you run out of brains. for Email, make the obvious change in the address. |
#278
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lift, wings, and Bernuolli
Jose wrote: Do you agree that the net momentum transfered to the Earth by the air molecules is equal and opposite to the net momentum transferred to the wing by the air molecules? Yes. Do you agree, therefor that there is no net momentum transfered to the air? Overall, yes. Similarly, there is no net momentum transferred to the basketball when it is being used to support a (very fast) dribbler. But that is not to say that there is no momentum transfer. The basketball certainly moves around. I do agree that the net overall is zero. The air does not pile up permanently. Good. That was my point all along. There is no net momentum transfered to the air. There is a net transfer of energy to the air.. At which ponit the Earth throws the air molecule back up so that the net momemtum transferred to the air molecule is zero (averaged over the entire atmosphere) Yes. [it hits the wing on the way up] Which again transferes an equal and opposite momentum to the molecule which again is transferrred to the Earth leaving no net transfer of momentum to the air. Yes. Overall, there is no net (or "permanent") transfer of momentum to the air. The air is an intermediary, keeping the wing and the earth apart. There is certainly =energy= transfer to the air (mv^2/2), and there is a lot of momentum transfer =back=and=forth= with the air, but I will agree that the net is zero. The air is sort of a catalyst - ending up unchanged as it transfers momentum to the earth and then transfers it back from the earth to the wing. Yes, although we do not yet agree on the details of the mechanism. So.. after all that, I think we are in agreement - there is no =net= (permanent) vertical momentum transfer to the air, but there is locally momentum transferred to the air, which carries it to the earth and uses it to neutralize the momentum the earth has acquired being attracted to the plane, in doing so it acquires momentum in the opposite direction and transfers it to the wing, ending the cycle and leavint the air ready to act as momentum messenger again. No. Being attracted to something does not cause momentum. There must be relative motion for momentum. It carries momentum messages both ways, they (overall) cancel out, but do keep the earth and the wing separated. No, it is not momentum that keeps the aircraft from falling, it is lift. The lift is produced by a pressure difference through the wing. === In addition, the wing is throwing air forwards, due to its AOA and its own forward motion. (this acts as drag, counteracted by the engine). The air thrown forwards increases the pressure in front of the wing, that plus the air thrown down makes the air pressure in front of and below the wing higher, causing the air to rise in front of the wing. This rising air helps lift the wing; this is the source of induced drag. Some of the rising air spills around the wingtips, causing vortices. The vortices are not the cause of lift, they are an inescapable side effect of lift. Concur? No. -- FF |
#279
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lift, wings, and Bernuolli
Jose wrote: For the stationary fan if it were only _almost equal_ then you would eventually run out of air on one side of the fan. No, the pressure would build up on one side of the fan, and that pressure would push against the wall and against the other air that is being pushed by the fan. When the pressure on that side is sufficiently high, no more (net) air will be able to be smooshed together on that side, and the air will all be going around. If the air is ALL going around then the flow in one direction is equal to the flow going in the other direction, RIGHT? Not _almost equal_ but _exactly equal_, right? OK to be clear, by 'flow' I meant rate. While the fan is on there is a bit more air on one side than the other, but once equilibrium is achieved the flow rate in one direction equals the flow rate in the other direction. You have a closed loop. After equilibrium occurs the fan no longer puts any net momentum into the air mass. The momenta of the individual air molecules cancel. But a pressure difference will be maintained until the fan is turned off. Yes. The fan continues to do work. Consider your example of the person who 'hovers' by dribbling a basektball. His momentum is zero, the momentum of the Earth is zero and the momentum of the ball is constantly changing and reverses twice each dribble. The dribbler is pumping energy into the Earth yet there is no net exchange of momentum. I agree. Overall, no net change. Microscopically (at each impact) there is a momentum change. Inbetween dribbles, the earth and the dribbler experience momentum changes which each dribble then counteracts. The collison with the dribbler is inelastic. Energy is conserved, momentum is not. The dribbler changes the momentum of the basketbal without changing his momentum. That time rate of change of the basketball results in a force on the dribbler that is equal in magnitude and opposite in direction to his weight. Now look at the same situation with a "basketball transparant" earth, and an endless supply of basketballs being tossed at the dribbler (who is backed up against a frictionless wall, so for now we don't need to consider horizontal forces). But we do presume there is still gravity. The dribbler keeps on deflecting basketballs downwards, but they don't bounce back up - they pass through the earth. The dribbler (who admittedly is no longer really dribbling) is imparting momentum to basketballs, and once he stops doing that, he will himself experience a momentum change. He uses energy to impart momentum to the basketball without changing his own momentum Energy is conserved, momentum is not. Work is done. When he stops chucking the basketballs, gravititational potential energy will be converted to kinetic energy as he gains momentum by falling. Energy is conserved, momentum is not. This is in the reference frame of the Earth, of course. In his reference frame the earth falls toward him and if I am in freefall next to the dribbler he has no momentum with respect to me. In both cases, as far as the putative dribbler is concerned, he is throwing basketballs down. He imparts momentum to basketballs, and really doesn't care what happens to that momentum afterwards. Precisely. He does not need the earth beneath him any more than an airplane wing needs the Earth beneath it. -- FF |
#280
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lift, wings, and Bernuolli
Jose wrote: How is momentum conserved when a cue ball hits a nerf ball? The vector sum, before and after, is identical. The vectors themselves are different (kinetic energy is converted to heat and such) but looking at both balls, or even looking at a cue ball and a glue ball, the center of gravity moves with the same velocity before and after. Perhaps you are not familiar with nerf balls. Nerf balls are foam rubber. When a cue ball hits a nerf ball (sufficiently large) nerf ball it stops and the nerf ball just quivers a bit. The center of mas quits moving. The kinetic energy of the cue ball has been converted to heat. Energy is conserved, momentum is not. If the air has a net increase in downward momentum, how is momentum conserved. ...by the air's eventual collision with the earth. How is it conserved at the air/airplane collison? -- FF |
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