Disagree------
C G of the earth should be the reference point. Level flight is circular motion because of the shape of the earth. Thrust and drag determine the airspeed.
When turning downwind the drag tries to decrease--but thrust will accelerate the aircraft
relative to the Great Circle around the earth if the time span is long enough. If the time span is to short---there will be a loss of airspeed.
In a rate one turn (3 degrees / sec) it takes one minute to turn 180 degrees. In the example above the aircraft would accelerate 100 mph in one minute relative to earth. If the ball in the turn and bank is centered--the g force is going to be on the seat of your pants. You are already pulling about 1.15 G's ---my old body would not be able pick up the small difference in G force.
In comment 2 (aerodynamically the airplane sees no difference). IMHO the aircraft experiences changes in the drag which causes acceleration if turning downwind and negative
acceleration in an upwind turn relative to the earth.

I stand by my points. Physics? Absolutely, but only if you get your points of reference right! Acceleration can mean an increase of airspeed, but it more correctly should be understood as a change in inertia. This can be a positive or negative change, and can be in any direction, not just forward. An aircraft can accelerate ground speed while decreasing airspeed or vice versa in certain circumstances. Or accelerate/decelerate up/down, or laterally (usually in a turn). What the airplane "feels" (indicated airspeed) and what WE feel are two different things sometimes. This question has everything to do with point of reference- and there are two: Airspeed is relative to air - whether the air itself is moving or still. Seat of the pants perceptions generally derived from inertia.

But the question at hand- seat of the pants perceptions- are pegged onto significant changes in inertia, not necessarily the airspeed indicator. At least apart from wind noise. This feeling of acceleration from turning into or away from the wind (not the G force acceleration because of the turn, OK?) will SELDOM ever be noticed. But could it potentially in more extreme situations? YES! The reality will be the reality regardless of what any of us think. This is a bit difficult to empirically test unless we have strong winds aloft but mild enough winds to safely get airborne. But it is easy to test on a river with a decent current with a small boat. Or even an airport "moving sidewalk" conveyor as I mentioned before. And RCH, drag has everything to do with airspeed & nothing to do with upwind/downwind, right?

To AF6IT
Hope this will help answer your question------

An airplane is cruising at 120 knots (2 degrees of the Great Circle per hour), flying straight and level (circular motion). It is equipped with standard flight instruments and an Inertial Navigation System (accelerometers are the sensors for this unit). There is no wind. The airspeed and the INS indicated speed in the gravitational field of the earth are the same. Some time later we notice during a thirty second time period the indicated INS speed has increased to almost 150 knots (2.5 degrees/hr). The airspeed has remained constant. The airplane has flown into a 30 knot tailwind. The airplane has accelerated, proven by the accelerometers in the INS. Thrust and drag determine the airspeed. Thrust and drag act as a governor –seeking equilibrium –if the drag tries to decrease, thrust will accelerate the airplane relative to the Great Circle so the airspeed remains almost constant.

It is a bit head-ache inducing but the truth is that in the case of manoeuvring within a uniform parcel of air, irrespective of whether it is moving relative to the ground or not, your turn does not involve a longitudinal acceleration. The proof is in the IAS though - if anyone is doubting it, check this next chance you get. Turn into and away from the wind and watch the ASI.

I think the confusion here (and I have had it too) is probably more likely to hit people with sailing experience. In that case you are operating at the interface of the water and wind but your vessel is moving through the water. Turning into or away from the the wind has an immediate and noticeable effect.

If you turn low enough, in something with some wingspan, you discover that the wind does matter to the aircraft. That's because, if you're low and have span, in a turn into wind the lower wingtip is down in the wind shear and has a lower airspeed than the fuselage does, while the higher wingtip is higher up and has a greater airspeed. The combination results in the aircraft trying to roll into the turn. Turning downwind you get the opposite effect: the aircraft wants to roll out of the turn.

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.

Maddog, in the unlikely event that you happen to see this, would you drop me a line at gabrieljb AT gmail dot com?

While you are correct about the airplane and pilot feeling the same in a turn with or without wind, the explanation you give is incorrect. I'd like to discuss it with you.