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Recently, Peter Duniho posted:
"Neil Gould" wrote in message I wasn't claiming that it does. It's just that the amount of lift after stall isn't sufficient to be relevant. That's a false claim. As the lift drops off in a continuous manner, there is a region "after stall" where the lift coefficient is just as high as usable regions "before stall". You may equivocate on whether a pilot can maintain the airplane at the angle-of-attack required to obtain that "after stall" coefficient of lift. But the fact remains that the lift is theoretically obtainable. As long as you don't want a lift coefficient very close to the maximum lift coefficient for the wing, it may not even be that hard to obtain the desired coefficient. Again, the _context_ is a response to Matt & Jose's claim of controlling a typical SEL to "greaser full-stall landings". I was agreeing with George that this is probably not what they were experiencing, and Todd's explanation regarding the high pitch angle typical of stall speeds is in agreement with this, albeit for other reasons. So, in context, how is your theoretically available lift relevant? When one refers to the "angle of attack" (and, yes, I know that "AOA" is the acronym), one is definitely referring to motion having both direction and magnitude. No, they are not. The angle-of-attack is a specific angle, measured between the wing's chord and the relative wind. In fact, a motionless airplane can still have an angle-of-attack, just as long as there is some wind. If there is wind, there is motion, direction and magnitude relative to the wing, ergo, a vector. If there is no wind, there is no "attack", and that angle then describes something entirely different. [...] However, it would have been better stated if I had said "... relative direction of _the wing's_ travel...", even though the typical SEL's wing pitch isn't drastically different from the rest of the aircraft. ;-) The angle of incidence (which is what you appear to be talking about now...that is, the angle between the wing chord and the longitudinal axis of the airplane) is yet again something else entirely different from angle-of-attack. Two different things are being described. In context (the direction of travel), the difference between the AOA and the angle of incidence is not "drastically different". [...] The confusion here is not between the airplane's pitch angle and the wing's angle-of-attack. It's your insistence on calling the angle-of-attack a vector, when it's a scalar (and, it appears, your confusion between "relative wind" and "angle-of-attack"). To be a scalar, it would have to lack motion, ergo no "attack". [...] I responded to that. In the context of landing, if one flies slowly enough to stall, one can stall "flat" relative to the ground because the decrease in forward "relative wind" increases the AOA. That is what my remark addresses. Your claim is incorrect. As long as the airplane is flying just above the ground, the relative wind is parallel to the ground. No change in the angle-of-attack will occur from any decrease in speed, not directly. My claim is that if the aircraft is flying parallel to the ground just before touch-down, it isn't stalled. [...] It is simply impossible to do what you suggest one might do. If one "flies slowly enough to stall", the angle-of-attack is at the stalling angle-of-attack, period. And all I'm saying is that this is independent of the pitch angle relative to the ground. [...] What WILL happen is that as the aircraft slows, the pitch angle of the aircraft will need to be increased, so as to continually increase the angle-of-attack of the wing. We are describing the same phenomena from two perspectives. In the context of my usage, if one maintains the pitch angle as the aircraft slows, the AOA will continually increase (normally, the pitch angle changes as the aircraft slows). But, again, the context of what happens during landing; one is maintaining a safe pitch angle as the aircraft slows, not necessarily increasing the pitch angle to insure a stall. [...] In *either* case, the airplane will touch the runway before the wing stalls, assuming a safe landing. In fact, I stated that the risk is something quite different from a safe landing. Neil |
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"Neil Gould" wrote in message
... [...] So, in context, how is your theoretically available lift relevant? It is relevant only to your false claim that a stalled wing provides no lift. Had you not made that false claim, I would have had no reason to bring up that element of the discussion. [...] To be a scalar, it would have to lack motion, ergo no "attack". Wrong. "To be a scalar" it needs to be a single value. And it is. Angle-of-attack is just an angle. A single value. Every angle requires two reference lines in order to define that angle. That doesn't change the fact that the angle itself is a single value, without any direction component. Likewise, the fact that two reference lines (one defined by a direction of travel) are used to define angle-of-attack DOES NOT MAKE ANGLE-OF-ATTACK ITSELF A VECTOR. It's still just an angle. [...] I responded to that. In the context of landing, if one flies slowly enough to stall, one can stall "flat" relative to the ground because the decrease in forward "relative wind" increases the AOA. That is what my remark addresses. Your claim is incorrect. As long as the airplane is flying just above the ground, the relative wind is parallel to the ground. No change in the angle-of-attack will occur from any decrease in speed, not directly. My claim is that if the aircraft is flying parallel to the ground just before touch-down, it isn't stalled. That's a new claim. Your previous claim (quoted above) was that you COULD stall while flying parallel to the ground. That is, one could "stall 'flat'". In any case, other than the issue with the geometry of the airplane, there is absolutely no justification in claiming that flight parallel to the ground precludes a stall. [...] It is simply impossible to do what you suggest one might do. If one "flies slowly enough to stall", the angle-of-attack is at the stalling angle-of-attack, period. And all I'm saying is that this is independent of the pitch angle relative to the ground. It is NOT independent of the pitch angle relative to the ground if the airplane is being flown in a flight path parallel to the ground. [...] What WILL happen is that as the aircraft slows, the pitch angle of the aircraft will need to be increased, so as to continually increase the angle-of-attack of the wing. We are describing the same phenomena from two perspectives. I am fairly certain we're not. In the context of my usage, if one maintains the pitch angle as the aircraft slows, the AOA will continually increase (normally, the pitch angle changes as the aircraft slows). You cannot "maintain the pitch angle as the aircraft slows" without touching the runway. If the aircraft slows and the pitch angle is not changed, lift is reduced and the airplane will descend onto the runway. I'll say it again: the scenario you propose is an impossibility. Pete |
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Recently, Peter Duniho posted:
"Neil Gould" wrote in message ... [...] So, in context, how is your theoretically available lift relevant? It is relevant only to your false claim that a stalled wing provides no lift. Had you not made that false claim, I would have had no reason to bring up that element of the discussion. The only falsity here is your claim that I made such a statement. So, apparently you have no reason to bring up that element in the discussion. What I *did* claim is: "You're holding the aircraft at just above the stall speed. When you're stalled, you're falling, not flying." This makes *no* reference to the amount of lift that a stalled wing provides beyond it being inadequate to support flight, which should be obvious to one who pays so much attention to details as you. [...] To be a scalar, it would have to lack motion, ergo no "attack". Wrong. "To be a scalar" it needs to be a single value. And it is. Angle-of-attack is just an angle. A single value. Wrong. A scalar can not contain elements of direction by definition. Ergo, AOA has no meaning as a scalar. Every angle requires two reference lines in order to define that angle. That doesn't change the fact that the angle itself is a single value, without any direction component. Likewise, the fact that two reference lines (one defined by a direction of travel) are used to define angle-of-attack DOES NOT MAKE ANGLE-OF-ATTACK ITSELF A VECTOR. It's still just an angle. Wrong, it's AOA only has meaning when referenced by a direction and motion. Without that direction and motion, THERE IS NO ANGLE OF ATTACK. If you still don't understand this, what is the AOA when the aircraft is parked in the hangar (with the doors closed and no fans running, if you insist on picking nits)? ;-) Finally, when you apply a direction to a scalar, what does it become? ;-) [...] I responded to that. In the context of landing, if one flies slowly enough to stall, one can stall "flat" relative to the ground because the decrease in forward "relative wind" increases the AOA. That is what my remark addresses. Your claim is incorrect. As long as the airplane is flying just above the ground, the relative wind is parallel to the ground. No change in the angle-of-attack will occur from any decrease in speed, not directly. My claim is that if the aircraft is flying parallel to the ground just before touch-down, it isn't stalled. That's a new claim. Wrong, that's my original claim: Jose: " Then what am I doing when I practice stalls at altitude, holding the aircraft at the stall buffet?" I wrote: "You're holding the aircraft at just above the stall speed." Your previous claim (quoted above) was that you COULD stall while flying parallel to the ground. That is, one could "stall 'flat'". Again, I made no such statement. My above claim is actually a clarification of my original statement for your benefit. My "stall flat" comment refers to pitch angle relative to the ground (as was stated in the same paragraph). I made no claim about travelling parallel to the runway. Indeed, one would *not* be going parallel to the ground in a flat stall. In any case, other than the issue with the geometry of the airplane, there is absolutely no justification in claiming that flight parallel to the ground precludes a stall. I made no such claim as this, either. [...] It is simply impossible to do what you suggest one might do. If one "flies slowly enough to stall", the angle-of-attack is at the stalling angle-of-attack, period. And all I'm saying is that this is independent of the pitch angle relative to the ground. It is NOT independent of the pitch angle relative to the ground if the airplane is being flown in a flight path parallel to the ground. Again, "a flight path parallel to the ground" is *your* assumption, and not a claim of mine. A flight path parallel to the ground, but with a high AOA is Todd's assumption, and again not a claim of mine. I have no problem with these scenarios, they just have nothing to do with what I've written. [...] What WILL happen is that as the aircraft slows, the pitch angle of the aircraft will need to be increased, so as to continually increase the angle-of-attack of the wing. We are describing the same phenomena from two perspectives. I am fairly certain we're not. I am fairly certain that you are confused about this. In the context of my usage, if one maintains the pitch angle as the aircraft slows, the AOA will continually increase (normally, the pitch angle changes as the aircraft slows). You cannot "maintain the pitch angle as the aircraft slows" without touching the runway. If the aircraft slows and the pitch angle is not changed, lift is reduced and the airplane will descend onto the runway. That is NOT incongrous with what I've written. The aircraft "descends" because the AOA increases as the aircraft slows and the pitch angle is maintained. Do you *really* disagree with this? I'll say it again: the scenario you propose is an impossibility. And, I'll say it again, you are confused about what I've written. You've repeatedly attributed claims to me that I never made, and set up straw men that have nothing to do with the issue at hand. Neil |
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"Neil Gould" wrote in message
.. . This makes *no* reference to the amount of lift that a stalled wing provides beyond it being inadequate to support flight, Interestingly enough, a stall occurs at the critical angle of attack, which is the AOA at which the coefficient of lift is the *maximum possible*. Just past the critical angle of attack (that is, further into the stall), the lift coefficient is no longer maximal, but is still well above what it is in ordinary cruise flight. What *does* happen just past the critical AOA--that is, just into the stall--is *not* that there's insufficient lift to support the plane's weight, but rather that there's a loss of *vertical damping*. John Denker (a physicist and a pilot) has a nice explanation he http://www.av8n.com/how/htm/vdamp.ht...rtical-damping. Wrong. "To be a scalar" it needs to be a single value. And it is. Angle-of-attack is just an angle. A single value. Wrong. A scalar can not contain elements of direction by definition. Ergo, AOA has no meaning as a scalar. No, an angle is unquestionably a scalar, not a vector. Check any introductory math text. If an angle were a vector, then a symbol representing an angle would be set in boldface; but it is not. You're right that an angle is defined by reference to vectors, but so is (for example) the *dot product* of two vectors (yet the dot product is a scalar); or so is the *magnitude* of a vector (but the magnitude is a scalar). So being defined by reference to vectors does not preclude a quantity from being scalar. --Gary |
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Recently, Gary Drescher posted:
"Neil Gould" wrote in message .. . This makes *no* reference to the amount of lift that a stalled wing provides beyond it being inadequate to support flight, Interestingly enough, a stall occurs at the critical angle of attack, which is the AOA at which the coefficient of lift is the *maximum possible*. Just past the critical angle of attack (that is, further into the stall), the lift coefficient is no longer maximal, but is still well above what it is in ordinary cruise flight. What *does* happen just past the critical AOA--that is, just into the stall--is *not* that there's insufficient lift to support the plane's weight, but rather that there's a loss of *vertical damping*. John Denker (a physicist and a pilot) has a nice explanation he http://www.av8n.com/how/htm/vdamp.ht...rtical-damping. Thanks for that! I'm not sure that resolves the issue of trying to land with these parameters. Wrong. "To be a scalar" it needs to be a single value. And it is. Angle-of-attack is just an angle. A single value. Wrong. A scalar can not contain elements of direction by definition. Ergo, AOA has no meaning as a scalar. No, an angle is unquestionably a scalar, not a vector. Check any introductory math text. If an angle were a vector, then a symbol representing an angle would be set in boldface; but it is not. Mathematically, an "angle" by itself *is* a scalar, and I'm not arguing otherwise. I'm saying that "Angle Of Attack" requires direction to have meaning. Without direction, there is no AOA. You're right that an angle is defined by reference to vectors, but so is (for example) the *dot product* of two vectors (yet the dot product is a scalar); or so is the *magnitude* of a vector (but the magnitude is a scalar). So being defined by reference to vectors does not preclude a quantity from being scalar. Well, OK. Then, how do you determine the AOA when the aircraft is parked? If the component of direction is inseparable from the definition of AOA, how can it be a scalar? Neil |
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Mathematically, an "angle" by itself *is* a scalar, and I'm not arguing
otherwise. I'm saying that "Angle Of Attack" requires direction to have meaning. Without direction, there is no AOA. [...] Well, OK. Then, how do you determine the AOA when the aircraft is parked? If the component of direction is inseparable from the definition of AOA, how can it be a scalar? How long is it between 11 AM and 1:30 PM? The answer is 150 minutes. Now, is that 150 minutes AM or 150 minutes PM? In order to form the difference, you NEED to know whether the original times are AM or PM, but the result is a pure number of minutes. 11 AM is a =time of day=, 1:30 pm is a =time of day= but the difference is =not= a time of day, it is just a number (of minutes). In a similar vein, in order to form an angle (a pure number), you need to have not just one direction, but two. You need TWO quantites that have direction (they don't even have to have magnitude!). However, the result (the angle between them) has no direction (beyond the algebraic sign). When you say: I'm saying that "Angle Of Attack" requires direction to have meaning. all you're really saying is that you don't have an angle of attack if you don't have the requisite components (a relative wind, and a chord). But don't confuse the components with the result. Area is made up of length and width, but area is not in itself one dimesional. Cakes are made with raw eggs and flour, but I'm not likely to confuse the two any time soon. ![]() Jose -- "Never trust anything that can think for itself, if you can't see where it keeps its brain." (chapter 10 of book 3 - Harry Potter). for Email, make the obvious change in the address. |
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Recently, Jose posted:
Mathematically, an "angle" by itself *is* a scalar, and I'm not arguing otherwise. I'm saying that "Angle Of Attack" requires direction to have meaning. Without direction, there is no AOA. [...] Well, OK. Then, how do you determine the AOA when the aircraft is parked? If the component of direction is inseparable from the definition of AOA, how can it be a scalar? [...] When you say: I'm saying that "Angle Of Attack" requires direction to have meaning. all you're really saying is that you don't have an angle of attack if you don't have the requisite components (a relative wind, and a chord). I'm not sure that I follow your analogies, here, Jose. But, it may be a good idea for you to look up the definition of "scalar". It *can not* include a directional component. Conversely, AOA can not exist without one. Neil |
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"Neil Gould" wrote
Mathematically, an "angle" by itself *is* a scalar, and I'm not arguing otherwise. I'm saying that "Angle Of Attack" requires direction to have meaning. Without direction, there is no AOA. Neil, give us an example of AOA having a "direction". Well, OK. Then, how do you determine the AOA when the aircraft is parked? When parked with no wind, there is no relative wind and therefore NO AOA. Bob Moore |
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Recently, Bob Moore posted:
"Neil Gould" wrote Mathematically, an "angle" by itself *is* a scalar, and I'm not arguing otherwise. I'm saying that "Angle Of Attack" requires direction to have meaning. Without direction, there is no AOA. Neil, give us an example of AOA having a "direction". Well, OK. Then, how do you determine the AOA when the aircraft is parked? When parked with no wind, there is no relative wind and therefore NO AOA. Stated another way, AOA doesn't exist *without* a directional component. Neil |
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