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This really doesn't have to do with turning in wind. It is a general question of how the motion in one frame of reference compares with the motion in another frame of reference. Newton already showed, a few centuries ago, that the accelerations of a body relative to two different reference systems (or frame of references) that move about each other at constant speed is the same in both frames. The speed vectors are different, but the change of those speed vectors with time (definition of acceleration) is the same. I am really sorry if somebody thinks otherwise: It would be hard to convince Newton that he was wrong. I will neglect all considerations about the Earth not being an inertial frame of reference. Yes, an airplane flying "straight and level" is in fact following a path that is curved down to follow the curvature of the Earth, and so the motion is accelerated downwards and the lift needs to be a bit less than the weight (in an extreme case, the satellites manage to "fly straight and level" with zero lift). But has anyone bothered to calculate how much would that be in an airplane? Even in a fast and high flying one, it's as close to nothing as it gets. And then we have the Coriolis force (and other apparent forces that are the result of the Earth being a non-inertial rotating frame of reference). The Coriolis force makes, in the worst case, that if you want to fly a constant heading while overflying the North pole (the only heading there is South, by the way), you'll have to make a constant rate turn to keep with the ground rotating under you. That turn rate would be 360° in 24 hours, or in 1440 minutes. That is 720 times slower than your standard 2 minutes turn (nearly two order of magnitude slower). And in any event, the tangential speed of the surface of the Earth at the Equator is about 1666 km/h. A say 20, or even 80, kts speed won't make a great difference. So even if we want to take those "special" forces into account, they would affect almost exactly in the same way in the frame "Earth" or in the frame "wind". So I won't mess with these "special" forces. I'll take the surface of Earth as a flat inertial frame of reference (that is what Cessna and Boeing do to make their performance calculations too), and since any other frame that moves at constant speed about an inertial frame of reference is an inertial frame of reference too, a mass of air moving at constant speed about the surface of the Earth (aka steady wind) will be an inertial frame of reference too. Ok, so now that we have an inertial frame of reference, we can apply the Newton's laws of motion, in particular the second law: F = m * a (the net force F and the acceleration a are vectors, m is the mass of the airplane) Unless we bring Einstein aboard and become relativistic, we can assume that the mass is independent of the frame of reference (we are so much slower than the speed of light of 300,000 km/s that any relativistic effect is absolutely negligible). The force acting on a plane are... thrust, drag, weight and lift. These forces are also independent of the frame of reference. If the plane is in a level turn at low AoA, the thrust cancels the drag and the vertical component of the lift cancels the weight. Because the plane is banked, there will also be an horizontal component of the lift that is not cancelled with anything. Doing some simple trigonometry, we can see that since the lift vector L is tilted off the vertical the same than the fin (i.e. the bank angle) and the vertical component Lv needs to be equal to the wight to keep the turn level (neither climbing nor descending), the horizontal component of the lift Lh will be Lv * tan(bank angle). But since we've said the Lv has to be equal in magnitude to the wight, we have Lv = Weight = m * g (g is the acceleration due to the gravity, 9.81 m/s/s), and hence: Lh = m * g * tan(bank angle). Remember this is the only force that is not cancelled by some other force, and hence the net force F will be equal to Lh. Replacing in the second law of motion we get: F = Lh = m * g * tan(bank angle) = m * a, or a = g * tan(bank angle) I hope that everybody understands that the bank angle is the same measured about the surface of the Earth or about the wind, and the same goes for the acceleration due to the gravity g. Anybody that has that clear, should have no further doubt that g * tan(bank angle), and hence the acceleration a, IS INDEPENDENT OF THE FRAME OF REFERENCE (Earth or wind). Now, as many said here, an acceleration means a change in the speed vector, and that change can be either in magnitude, in direction, or both. What is the case in this case? Is this acceleration changing the magnitude of the speed, the direction of the speed, or both? Well, that DOES depends on the frame of reference. "Wait a minute", you'll say. "At the beginning you've almost quoted Newton saying that the change in the speed vector must be the same in both frames of reference". Yes, so? I wish I could make a very simple drawing here to show you, but since I can't you'll have to do it yourself. Don't worry. It's very easy and it'll take you like 30 seconds. So grab a piece of paper and a pen (or open your Microsoft Paint). Mark a point A. Starting in A, draw a vertical arrow of say some 5 cm (two inches). The pointy end of the arrow will be point B. Now draw a second, longer vertical arrow starting (say some 8 cm or 3 inches) from A, and call the pointy end C. If AB is the "initial speed vector" and "AC" is the final speed vector. Then "BC" is the "change in speed vector". In this case, the change is in magnitude but not in direction, you see? Now a couple of inches right of A mark another point A', and draw two arrows: A'B and A'C. If those arrows correspond to the initial and final speed in another frame of reference, then again "BC" is the "change in speed vector". As you can see now the speed affected both the magnitude and the direction of the speed vector, and yet, the change in speed vector is the same. However, IF the change in direction is the same in both frame of references, then the change in magnitude needs to be the same too. That doesn't mean that the speeds themselves are the same in the two frames, but the changes must be the same. As an example, draw the mirror image of the arrows that start in A'. We'll use this feature later. From a theoretical point of view, everybody should be convinced by now that the acceleration is the same measured about the ground or about a mass of air moving at constant speed about the ground (a steady wind). But I know that sometimes even when you rationally understand it, some counter-intuitive facts are hard to "buy", and someone can even be saying "ok, maybe the acceleration is the same, but why would that mean that the airspeed will not tend to change?" The answer to that question is that if it doesn't tend to change in no wind, then it won't tend to change in a steady wind either. But let's get practical and make an example. Not only that, but let's do a pretty extreme example. Someone said "maybe if the wind is very strong". Someone else said "maybe if the turn is too quick, like a canyon turn". So I'll tale my imaginary Piper Tomahawk, fly it at 100 KIAS, and make a test first with no wind (for reference) and then with a 90 kts wind. So here I am, no wind, flying at 100 KIAS (which is also 100 kts of ground speed because there is no wind), straight and level, heading North and then I do a 180° turn South. I do it pretty steep. Enough to turn the 180° in just 10 seconds (that's 18° per second, 6 times faster turn rate than the standard 2 minutes turn). Of course, to keep the altitude and speed during the turn I have to pull back and add power as needed. Now draw the arrows: Initial speed about the air: 100 kts North Final speed about the air: 100 kts South Change in speed about the air: 200 kts North to South in 10 seconds Initial speed about the ground: 100 kts North Final speed about the air: 100 kts South Change in speed about the air: 200 kts North to South Let's repeat the test, but now I'll have a 90 kts wind from the North. So here I am, heading North in my Piper Tomahawk at 100 KIAS (10 kts of ground speed due to the 90 kts head wind). Because there is no crosswind, the ground track is also fully North (there is no crab). Then I do the same turn as before: same bank angle, same pull-back and adding of throttle, which results in an equally steep and quick 18° per second turn, reaching a South heading in the same 10 seconds. Now I am heading South and tracking exactly South too, again because there is no crosswind (now I have a 90 kts tail wind). Because in both experiments one of the frame of references was the Earth, and the other two frames of reference (calm air and steady wind) were both moving at constant speed about the Earth (or not moving at all in the first case), the acceleration must have been the same in all of them: 200 kts North to South. Even further, because the change in DIRECTION was the same in all of them (initially North and finally South), the change in MAGNITUDE also MUST be the same. Will that hold? Let's see: Now draw the arrows: Initial speed about the air: 100 kts North Final speed about the air: 100 kts South Change in speed about the air: 200 kts North to South in 10 seconds Initial speed about the ground: 10 kts North Final speed about the air: 190 kts South Change in speed about the air: 200 kts North to South As you see, while I turned from a 90 kts headwind into a 90 kts tailwind in just 10 seconds, I didn't loose any single knot of airspeed nor there was the slightest tendency of that. As someone said, if you are in IMC and make an keep constant airspeed and constant bank turn, no matter how slow you are flying or how steep the turn, there is no way to tell if there is no wind, 20kts of wind, or 200kts of steady wind. If you stall you'll stall in any of them, and if you don't stall you won't stall in any of them. The turn will take the same (same turn rate) and the airspeed indicator will "say" nothing. You'll only note the wind when you look the FlightAware radar plot: With no wind you'll be making superimposed circles. With wind you'll be making "curls" like if you wrote eeeeeeeee in hand-script letters. A final example, just in case there is someone still in disbelief. Do you agree that the Physics of an airplane turning is the same regardless of the size of the airplane? Great, get a paper airplane. Practice to throw it in your living-room in two ways, one that it glides smoothly straight and another that it does a 180° turn. Bring it with you the next time you take a commercial flight. Now, a steady wind is a mass of air that is moving at constant speed about the ground, ok? The air INSIDE an airplane doesn't is moving with the airplane, otherwise we passengers would be all blown by a terrible wind. So if an airplane is flying say at 500 kts of ground speed, the mass of air inside the airplane is moving at 500kts about the ground too, ok? great. There we have a 500kts wind. Now get your paper airplane and throw it for a smooth straight glide in the direction of the airplane: The plane will be flying with at say 5 knots of airspeed, but with a tailwind of 500kts, it will be doing 505 knots of ground speed. Now do the same but throwing it back. Again the airspeed is 5 knots, but with a headwind of 500 kts it will be making a ground speed of minus 495 knots (yes, as viewed from the ground the paper planes moves tail first). Now throw it forward but in the way that it makes a 180° turn. But watch out!!! The paper plane will be launched back at some 500 kts. If you are hit at 500 kts you can result severely or fatally injured, even if that's with a paper plane!!! NOT!!! The paper plane will do the same than in your living-room. You cannot, by throwing a paper plane inside, tell whether the A380 you are in is parked on the ramp or crossing the Pacific Ocean at 500kts. With that I rest my case, and if anybody still has any doubt, may The Force be with you.