Originally Posted by
WideScreen
No paper show to you, though I do have a PhD in physics......
Maybe consider the following situation. We all know, that to physically "destroy an airplane" against a solid surface, we need to start with a sufficient amount of Potential + Kinetic energy (so, either height or speed, or a combination of that).
Assume, have an airplane (stall speed 35Kts) 1 mm above the runway, have a headwind of 40 Kts (Alaska, can happen) and an airspeed V of 40 Kts. Engine powered to keep that airspeed. Your ground speed will be zero. Your potential energy (relative to the runway) is very small and can be ignored.
Do the thought-experiment to snap turnoff the headwind as well as the engine. Of course, the airplane will immediately settle, with minimal mechanical movement, since it was not moving compared to the runway, even, the wheels will not start turning.
Will the airplane bleed off a Kinetic energy of 0 (IE effectively nothing happens) ?
Very flawed experiment/example. To "snap turn it off", you would have to decelerate those air particles instantly, and that is impossible.
Or will it bleed off a Kinetic energy of 0.5 * m * V(airspeed) ^2 and effectively destroy the airplane ? Given common sense tells us, the airplane will not be destroyed, the presumed amount of Kinetic energy has to go "somewhere". When not knowing where Energy goes, it is usually turned into heat. So, will the airplane (or the ground, or the -now no longer moving- air) heat up ?
The reverse of this experiment: Airplane on the runway. Have a headwind of 40 Kts, use the engine to keep the airplane steady on (!) the runway, wheels not moving. Your airspeed will show 40 Kts. To avoid, the airplane climbs above the runway, hold the elevator down, to stay on the ground. What is the Kinetic energy of the airplane ?
That depends on your frame of reference
Do the same, by holding the brakes (instead of the engine). What airspeed will be shown ? What is the Kinetic energy of the airplane ?
Now, return to the first experiment. The airplane not 1 mm above the runway, though on the ground, with non-compressed struts. What is the Kinetic energy of the airplane ?
The same, though now with struts compressed. What is the Kinetic energy of the airplane ?
Next step. No headwind. No ground speed. With increasing headwind, the airspeed will go live, with the airplane stationary using brakes. What will be the Kinetic energy of the airplane ? What will happen when we snap-turn-off the headwind ? Assuming the airspeed will be the measure for the Kinetic energy, where will the Kinetic energy go, when the headwind turns off ?
Gives me the feeling to argue with "The Truth"........
So, I will try to not make this personal and have omitted references to your PhD.
First of all. Kinetic energy is not absolute, but depends on your frame of reference. As I sit here in a chair typing I am not moving, so my kinetic energy (1/2 M x V^2) is 0, using my house as a reference. But that "V" is not an absolute value. If I use the center of the earth as my reference I am spinning at 1250kmph (40 degree latitude). If you were standing on the sun, you would have to add 100000kmph to that. and the sun itself spins around in the Milky Way, and that flies away from the center of the universe where the big bang took place. It is all relative. Same but easier to understand is potential energy. The height in there is normally defined, because you can go below the surface of the earth, and most understand it is from the starting point to the stopping point that counts. The same goes for unaccelerated motion, you have to define the reference. And there is a slope here, because my motion around the earth & sun are circular, and thus not un-accelerated, but close enough for flying.
If you sit in a swing ride, you make a circular motion around its central axis, and your seat gets swung out. you will feel a vertical acceleration but no lateral or horizontal acceleration once the ride reaches it operating speed using your seat as your horizontal reference. Now imagine you put that ride on a truck and drive at the same speed as you were spinning. Once the truck and swing-ride reach that speed, you will feel the excact same thing as when the ride was stationary and spinning, because of the unaccelerated motion of the truck (aside from the curvature of the earth). For a person standing on the side of the road, it will appear you make a motion where you decelerate to come to a stop on his side of the road, accelerate to twice his speed on the other side, and come to a stop on his side again and over and over until you throw up. Your kinetic energy will be a constant to the truck driver, but go from 0 to 4 times the value the truck driver sees to the pedestrian every rotation.
To get back to flying:
(the following is based on steady wind, no shear, disregards the curvature of the earth, and the rotation of the earth itself, and using the air as your reference grid)
While taking off you move in reference to the earth, and thus are affected by the wind. That is why in GA airplanes you hold controls into the wind, and have to use rudder to stay on the centerline (and correct for prop spin). As you get rotate there is a period where your frame of reference becomes the air. Once that has happened, the ground doesn't matter. Your plane only senses the air particles around you. What the ground does underneath does not matter. If you make a coordinated turn at a constant airspeed and bank angle, your kinetic energy will stay constant. Your acceleration will be perpendicular to you, so if you look at your energy as a vector it will change direction. Depending on the bank angle your load factor will change, but the only acceleration is in the vertical plane, with reference to the aircraft vertical.
If there are wind changes, like when descending, or turning out of the jetstream, there will be a period to reach a new equilibrium but that is not what we are discussing here.
Finally, getting back to ground reference based maneuvers. That doesn't mean the plane flies ground based, but that the pilot uses the ground for reference to adjust speed and turn rate to get to a certain point on the ground.
If you turn to final with a tailwind (but why would you?), the turn will be more than 90 degrees, as opposed to the more expected less than 90 degrees. Similar for the tailwind on base, your turn to final will, if un-anticipated have to be made at a higher rate. If both happen at the same time, sometimes accidents happen, that is why it is considered dangerous. The pilot using the ground as a reference, but the plane flying in reference to the air is what is the problem, never the plane flying in reference to the ground.