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WingFlaps schrieb:
(the stall is now damn close -better hope there's no significant wind) .... Now we add in the energy losses from having to accelerate with the wind and to glide speed. Arrrgh! Not the old "turn into downwind" legend again! Better work out your understanding of physics before publicly reasoning about turns. There is nearly always somewhere flat to put the plane The operative word in this sentence is "nearly". |
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On Apr 25, 10:16*am, Stefan wrote:
WingFlaps schrieb: (the stall is now damn close -better hope there's no significant wind) ... Now we add in the energy losses from having to accelerate with the wind and to glide speed. Arrrgh! Not the old "turn into downwind" legend again! Better work out your understanding of physics before publicly reasoning about turns. Try reading the statement again, here it is: "Now we add in the energy losses from having to accelerate with the wind and to glide speed." Now perhaps you would like to revise some physics and try to critcise it for us? I await your stumbling analysis of my words with mild amusement. Cheers |
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WingFlaps schrieb:
Try reading the statement again, here it is: "Now we add in the energy losses from having to accelerate with the wind and to glide speed." Now perhaps you would like to revise some physics and try to critcise it for us? It's the "having to accelerate with the wind" part which is complete BS unless I completely misunderstand what you are trying to say. |
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On Apr 25, 10:48*am, Stefan wrote:
WingFlaps schrieb: Try reading the statement again, here it is: "Now we add in the energy losses from having to accelerate with the wind and to glide speed." Now perhaps you would like to revise some physics and try to critcise it for us? It's the "having to accelerate with the wind" part which is complete BS unless I completely misunderstand what you are trying to say. The latter I think. The imposed accelerations associated with the change in direction (from upwind to downwind) require control inputs that add drag and increased energy loss (from drag). To summarise your missed point, the pilot control inputs cost energy that is not factored into simple glide/time analysis. Cheers |
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#6
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Hold on while I try to correct some nits in what Bertie said and see
what happens ![]() Bertie the Bunyip wrote: Nope, the wind is going to help you in almost every way if you're turning back. [ many good points supporting this assertion, but... ] also, your best LD speed is going to occur at a lower airspeed, Well, technically, your best LD speed is related to angle of attack, and not the groundspeed, so that won't change. Your best glide speed certainly will be less... There is no inertia involved in making a downwind turn. None. Here's why I wonder about that. Let's suppose 65 KAS before and after a 180 turn from a 10 KT headwind. OK, before the turn, your groundspeed is 55KTS and after the turn your groundspeed is 75KTS. Your intertial frame of reference is tied to the g/speed, not the a/speed. So -- the kinetic energy of the aircraft and contents is about 33% higher (75/55)^2. That energy is only going to come from one place with no power -- trading in altitude (potential energy) for kinetic energy. |
#7
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tman inv@lid wrote in
: Hold on while I try to correct some nits in what Bertie said and see what happens ![]() Bertie the Bunyip wrote: Nope, the wind is going to help you in almost every way if you're turning back. [ many good points supporting this assertion, but... ] also, your best LD speed is going to occur at a lower airspeed, Well, technically, your best LD speed is related to angle of attack, and not the groundspeed, so that won't change. Your best glide speed certainly will be less... OK, not the LD, you're right. but your best glide speed is realted to your ground speed. To take an extreme example to illustrate this point, imagine that your published best glide is 70 and you're trying to glide into a 70 knot headwind. You're going nowhere. Increase your speed and you will make headway. In the tailwind scenario, your best glide distance over the ground will be better at a lower speed than published. the stronger the wind, the lower the speed required until you arive at your min sink speed. There is no inertia involved in making a downwind turn. None. Here's why I wonder about that. Let's suppose 65 KAS before and after a 180 turn from a 10 KT headwind. OK, before the turn, your groundspeed is 55KTS and after the turn your groundspeed is 75KTS. Your intertial frame of reference is tied to the g/speed, not the a/speed. So -- the kinetic energy of the aircraft and contents is about 33% higher (75/55)^2. That energy is only going to come from one place with no power -- trading in altitude (potential energy) for kinetic energy. Nope, your inertial frame of reference owes nothing to the ground whatsoever. None. Zero, nil zilch zippo. You ar entirely a creature of the air and owe nothing to the ground whatsoever ( except in the vertical, of course) The earth simply isn't that important in the bigger scheme of things! If it were, you'd have trouble making left or right hand tunrs in your car.... Bertie |
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On Apr 24, 6:41 pm, tman inv@lid wrote:
Hold on while I try to correct some nits in what Bertie said and see what happens ![]() Bertie the Bunyip wrote: There is no inertia involved in making a downwind turn. None. Here's why I wonder about that. Let's suppose 65 KAS before and after a 180 turn from a 10 KT headwind. OK, before the turn, your groundspeed is 55KTS and after the turn your groundspeed is 75KTS. Your intertial frame of reference is tied to the g/speed, not the a/speed. So -- the kinetic energy of the aircraft and contents is about 33% higher (75/55)^2. That energy is only going to come from one place with no power -- trading in altitude (potential energy) for kinetic energy. See the previous post. There's a change in kinetic energy, but very, very little. Many people make the mistake of thinking that the earth has an effect on the airplane. It does, but only vertically, by gravity. Gravity has no horizontal Component. Like a gyroscope, which is rigid with respect to space and cares not one bit about the earth, the airplane's mass, as it moves in the horizontal, is affected only by its relationship to space and the air it flies in. That isn't to say that the earth isn't going to get in the way a little harder. Landing downwind, as with landing into the wind, involves transferring the weight from the wings to the wheels, and downwind means much more groundspeed and maybe loss of control as the roll continues at higher speed while the flight controls feel a decreasing airspeed, or maybe the airplane will run out of runway. Bang. We do illusions created by drift turns with students, usually in a strong wind and at around 500 feet, so that they can see that the ball stays centered in the turn while they get the visual impression that the airplane is skidding or slipping on the downwind and upwind sides of the turn. The airspeed does not change. Not so's you could read it. If we put the student under the hood and make him fly on instruments while we do this, he can't tell us when he's turning into the wind or out of it. Can't feel anything, can't see any performance changes on the gauges. Dan |
#9
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On Apr 24, 6:41 pm, tman inv@lid wrote:
Here's why I wonder about that. Let's suppose 65 KAS before and after a 180 turn from a 10 KT headwind. OK, before the turn, your groundspeed is 55KTS and after the turn your groundspeed is 75KTS. Your intertial frame of reference is tied to the g/speed, not the a/speed. So -- the kinetic energy of the aircraft and contents is about 33% higher (75/55)^2. That energy is only going to come from one place with no power -- trading in altitude (potential energy) for kinetic energy. In your world, it's gonna be pretty hard for a sailplane to circle in a drifting thermal. In my world, it's not a problem. |
#10
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In rec.aviation.student tman inv@lid wrote:
Here's why I wonder about that. Let's suppose 65 KAS before and after a 180 turn from a 10 KT headwind. OK, before the turn, your groundspeed is 55KTS and after the turn your groundspeed is 75KTS. Your intertial frame of reference is tied to the g/speed, not the a/speed. So -- the kinetic energy of the aircraft and contents is about 33% higher (75/55)^2. That energy is only going to come from one place with no power -- trading in altitude (potential energy) for kinetic energy. This simply does not make any sense. Kinetic energy, like velocity, is a relative quantity. You cannot look at an object and say, "it has X joules of KE". You can only talk about KE relative to some frame of reference. Just like velocity. So forget about KE. It's in the same boat as velocity, so look at velocity. You make a turn and suddenly you gain a bunch of groundspeed. Where does the extra speed come from? It comes because you're maneuvering relative to a medium, the air, which is itself moving. Your KE relative to that medium is exactly the same as it was, so no energy has to come from anywhere. -- Michael Ash Rogue Amoeba Software |
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