Ok here is the deal with the downwind turn. Its more about momentum then airspeed. I TOTALLY agree that the airplane doesn't feel the wind when its in the air. And I made that argument with an astrophysicist. His thought here is when you have a airplane with a lot of mass and turn from a headwind to a downwind it has to accelerate. e.g. You take off at 100 kts indicated and there is a 20 kt headwind thats 80 kts groundspeed and instantaneously turn downwind you will still have that 80kts momentum pushing you and would have to accelerate to 120 kts groundspeed in the opposite direction to maintain that 100 kts of indicated airspeed. Grant it the wind does the acceleration but it can take a period of time which is increased the faster you whip the airplane around. Now I know that this can't be done "instantaneously, but it still shows what forces are acting on the plane. Point is that this is more pronounced in a 747 then it would be in an R/C model because of mass and surface area.

To summarize: The radius of a constant altitude turn has nothing to do with aircraft mass or size. True airspeed and bank angle are the only inputs to the turn radius equation (r=V^2/11.26*tan(phi), V = TAS knots, phi=bank angle in degrees). The only time the relative motion of the earth has any direct bearing on an aircraft's performance or behaviour is when we want to know something about it's ground track (like speed and/or direction), or when the airplane is in contact with it. Either during takeoff or landing roll, or when making for example a crash landing (upwind please). The airplane acts on the airmass and only the airmass (not considering ground effect). The apparent difficulties with making an upwind or downwind turn with a model have nothing to do with aircraft performance and everything to do with pilot perception. When you or I make constant-altitude and bank turns IN an airplane, we are not interested in drift over the ground, only in making a circular path with respect to the airmass, and we can do that if we maintain constant altitude, airspeed and bank angle, whether the wind is calm or blowing steadily at 200 knots. Either case makes no difference whatsoever to an airplane's performance or the way the turn is performed, even one just lifting off, ignoring wind shear and gradient effects. There is no such thing as absolute velocity. When we want to make a constant radius turn, then bank angle is varied as groundspeed changes in order to maintain the desired radius.

I'm not sure I understand what you mean by, "80 kts momentum pushing you". The instant the wheels leave the ground, the ground's relative motion becomes irrelevant to the airplane, as far as it's performance is concerned. It's only relevant with respect the the pilot's desire to position the aircraft relative to the runway or a departure course, etc.

So-called 3D airplanes doing "walls", "waterfalls" and other maneuvers with very high pitch rates follow exactly the same path through the airmass whether performed upwind, downwind or crosswind, if of course they are performed with the same inputs under the same conditions, at the same entry speed, etc. They look very, very different, however when performed upwind and downwind because the airspeeds are so low that windspeed (and the resultant groundspeed differences) causes large variations in the paths relative to the "stationary" observers on the ground. This is true of foamies, 40% planes, full scale acro planes, fighers, and it would be true of 747s if they could turn tightly enough. Put another way, if an observer was watching the airplane perform one of these maneuvers while moving at exactly the same velocity as the wind the airplane was experiencing, then the airplane would scribe the same path relative to that observer as it would relative to a "stationary" observer watching the same maneuver in calm wind.

A bit wordy, I'm sure and sorry for that.

And I'm betting you weren't one of those guys. But spoken like a true airline pilot.

Its not about groundspeed. The airplane in moving in direction A. That momentum in turn is in that direction. If it turns instantaneously to that opposite direction B the momentum moving in direction A needs to be overcome and accelerated towards direction B.

OK. So you agree that an airplane doesn't try to descend in a turn from downwind and climb in a turn from upwind (because of the wind direction), like some people insist. That was my point. I understand what you are saying above.

Here’s a question about airplanes turning from upwind to downwind. Suppose you have a very efficient glider that can make a 180-degree turn maintaining 100 knots and lose only 10 ft of altitude (to eliminate ambiguity say that there in NO motion of the air relative to the ground and it starts 100 ft above sea-level). Next, suppose that the glider is going upwind into a 100-knot headwind. Relative to the ground, the glider would be hovering. As it makes a 180-degree turn, it will accelerate to 200 knots relative to the ground, again losing only 10 ft of altitude. My question is: in the no-wind case, the glider lost 10 ft worth of potential energy, but didn’t gain any kinetic energy. In the 100-knot headwind case, the glider again lost 10 ft worth of potential energy, but gained 200 knots worth of kinetic energy. If it’s true that the wind doesn’t make any difference to an airplane once it leaves the ground, then how did the wind give the plane all that extra kinetic energy?

If kinetic energy is the energy of motion, it is therefore relative in exactly the same fashion as motion itself. Saying the glider has 200 knots groundspeed is equivalent to saying it has so much kinetic energy relative to the surface of the earth. Does the glider at 100 knots airspeed and zero knots groundspeed have zero kinetic engergy? Relative to the earth, yes (not counting it's vertical speed). Relative to the airmass, no.

OK, to keep things simple and consistent, stick to "relative to the earth" reference frame. I'm not sure you answered how the wind added all the extra kinetic energy?

I'm not sure I understand your point. We both agree that the airplane's kinetic engergy relative to the surface of the earth is greater at a greater groundspeed. I never said anything different, in fact I stipulated it in an earlier post. But it does not have any effect on the airplane's vertical speed, in feet per minute, for example. Two gliders would take the same amount of time to reach the earth from the same starting geopotential whether the wind was calm or blowing at an arbitrary speed. The gradients relative to the earth would be different, but not the rates. And turns performed exactly the same way would have identical radii relative to the airmass whether the airmass was moving relative to the earth or not, assuming steady wind of course. It's those differing gradients and groundspeeds that give rc pilots trouble sometimes, not any direct effect on the airplane's performance.

My question was: In both cases, the glider's potential energy changed by the same amount. In the no-wind case, the glider's kinetic energy didn't change at all (in the earth-fixed reference frame). In the headwind case, the glider's kinteic changed by a lot (also in the earth-fixed reference frame). Where did all that extra kinetic energy come from? It had to come from somewhere.

I used to sweat bullets over the downwind turn issue... yes I have no life ....

But one day I started thinking about my free fight models and how they fly on the ragged edge of a stall to give the lowest sink rate. On a free flight model NOTHING matters but the sink rate.

Now some of these are light rubber models and some are heavier engine powered models but they all have this edge of stall minimum sink rate trim. So if the downwind turn thing matters you'd expect it on the windier days to stall or show signs of a near stall as it turns towards it's downwind leg. But this doesn't happen. They just merrily fly their circular path in fine form regardless of how tight or open the circles are. And I generally like a tighter pattern in order to stay in the thermals.

This same behaviour has also been noticed in my RC gliders where I commonly set in my thermal circles with the trims for longer flights and only nudge it here and there as it merrily circles and drifts downwind.

So to my mind I'm now happy that any down wind turn shenannigans are totally pilot induced and have nothing to do with the model

There's a component of total energy (sum of potential and kinetic) that is due to the glider's position in the moving airmass. So in a no-wind situation the glider has much less total energy than in the high-wind case. So as it turns downwind, it exchanges some of that total energy from potential (pointed upwind in a moving airmass) to kinetic (pointed downwind in a moving airmass). The change in total energy is the same as in no wind, it's just there was a lot more energy than you thought to start with because of the wind.

It comes from relative velocity, by definition. It only has a direct effect on the airplane when the airplane is in contact with the earth. Otherwise you would have to consider it's motion relative to the sun, moon, and other planets as well.

If I am walking at 3 mph toward a parked car and collide with it, the force of impact is calculable, based on the relative velocity between me and the car. Now if the car is moving away from me at 2 mph, and I collide with it while walking again at 3 mph, the force of impact will be calculated base on a relative velocity of 1 mph. My total energy hasn't changed, only my velocity relative to a car. Relative to another car moving toward me at the same time, my relative velocity is simultaneously different. My total energy of motion relative to the medium I am acting on in order to move hasn't changed (still moving 3 mph relative to the earth), but my kinetic energy relative to various cars is different for each car at any given time. Same with a plane. The plane is acting on the air in order to move, so motion of the earth beneath is not relevant unless in contact with it (or I want to plot a deduced reckoning course).

I think we probably agree, but we might just be trying to state the same thing in different ways. I think we all agree that a steady wind doesn't affect an airborne airplane's airspeed or vertical speed. And, of course if you hit the ground at a higher groundspeed, due to flying downwind, more damage will be done. That statement seems to sum up shoe's and my arguments.

Where does the big change in kinetic energy come from in the headwind case?

no wind: delta KE = 1/2 m(Vdownwind^2 - Vupwind^2) = 0

headwind: delta KE = 1/2 m(Vdownwind^2 - Vupwind^2) = 1/2 m Vdownwind^2 = Big Number

Both cases are observed from the same reference frame and both cases have the same change in potential energy. If energy is conserved, all that kinetic energy had to come from somewhere.

[>:][][][][][:-][:-][:-][&o][&:][:@]WOW[][:'(][:@][&:][:'(][&:][8D]

Course, if the airplane doesn't know that it is in a moving airstream, then it has no idea that it's flying upwind or downwind, nor that it's got lots more or less, kinetic energy. In full size, we poor pilots have no idea which way or how much the wind is blowing unless we get close to the ground. The energy is the same, wind or not.

Had to delete as it is really mindboggling and the answer was only partly correct.
New try:
Etot=mgh +mvv/2
As the glider has a mass and the inertial system for this is earth, the mass will be accelerated when turning from headwind to tailwind. Where is the energie coming from: To keep airspeed constant = mvv/2 (groundspeed rising)it only can be taken from the altitude =mgh. (glider) From Tailwind (now with higher mvv/2) into Headwind and keeping the airspeed constant (groundspeed decreasing) the stored energie will be now converted in altitude again. Assuming no drag, you should have the same altitude as the process has started, as the total energie was constant and only traded back and forth between height energie and kinetic energie.

More drastic:
Think about microbursts at approach, have already sent some airliners down. Why that: Rapid change from headwind into tailwind. If the aircraft is high enough (not touching the thrustlevers) it can trade altitude for speed to accellerate the given mass for the original airspeed. Out of the zone it now has a lower height energie but much more velocity energie and to get the airspeed back to the level started, speed can be traded for height again.

b.r.
Wolfgang

So if the extra kinetic energy comes from potential energy, then the glider would have to lose more altitude in the turn when turning downwind. That isn't consistent with the suggestion that the motion of the airmass wouldn't affect sink rate.

The aircraft does not "feel" this supposed acceleration.

The airborne glider is acting not directly against the earth, but rather on the atmosphere. Its reference frame for performance is the air.

Mass has no reference frame, first problem here.

Second, aircraft do feel the acceleration turning downwind... but it's no different than turning crosswind or any old turn at the same bank angle in still air, for that matter. It's kind of hard to see this with a model, but if you turn into a constant bank angle turn, you fly a helical course with the spacing of the turns being how far the wind carries you downwind (I've actually done this with a fullsize power plane and measured the spacing... it was a very windy day so it was nearly a mile per 2-minute turn... the math all worked out, of course). Flying in circles in still air is a special case of this.

If you are trying to counter my statement that the airplane does not feel the earlier referenced acceleration, then you misunderstand me. Your post seems to read as though you are trying to correct me, but I believe we agree perfectly. I'm a full-scale pilot too: a 4700 hour ATP. You make exactly the same points I've been trying to make, namely that airplanes in constant bank turns feel the same acceleration regardless of wind speed or direction (in steady wind). The exercise you described is an excellent way to illustrate this point, and also to measure wind velocity (your wind was a little less than 30 knots, though you didn't report the direction of drift).

The discussion so far has focused on the forces and accelerations that the glider "feels". There seems to be agreement that a glider would feel the same forces and therefore experience the same accelerations in both the no-wind and the 100 kt headwind turn. It's pretty clear that the glider gains a LOT more kinetic energy in the case where it is turning downwind when it starts with a 100 kt headwind. My question is where all that energy comes from. One explanation would be that the glider would need to lose quite a bit more altitude when turning downwind from a 100 kt headwind in order to convert potential into kinetic energy. Unfortunately, that explanation doesn't fit with the suggestion that the wind should have no effect on the "vertical performance" of the glider. So... where DOES all that kinetic energy come from?

Are you suggesting that this increase in kinetic energy occurs in an absolute sense? It does not. It's kinetic energy must be measured relative to another object. What about a collision with another glider? Head on, crossing, overtaking. The glider can have a large number of kinetic energy levels simultaneously, in fact all objects do. We isolate cases we are interested in and measure those, disregarding the rest. We care about the landing ground speed of the glider so we consider it's variation in kinetic energy relative to the earth based on wind, but we don't care about the glider's kinetic energy relative to the moon, or the Monorail at Disneyland, or the USS Ronald Reagan aircraft carrier, because the surface we want to land on is "at rest" relative to the earth, not those other objects. If landing on an aircraft carrier, the movement of the earth is not relevant (again purely from a kinetics viewpoint), only the movement of the carrier and the airmass.

Since kinetic energy is based on relative motion, it must be measured in that context. You say the glider has increased kinetic energy. How do you know it isn't the earth itself that has increased it's energy relative to the glider?. If you know anything about physics, you know the two concepts are indistinguishable from the standpoint of the energy that exists at the moment of contact. We arbitrarily declare the earth to be "fixed" and therefore it is the reference frame, but it is equally valid from a kinetics viewpoint to say the glider is "fixed" and is therefore the reference frame, or the aircraft carrier is "fixed" and is therefore the reference frame, even though it is moving relative to the earth in a certain direction to maximize headwind for the aircraft operating from it. You might as well be saying that the earth's kinetic energy is alternately increasing and decreasing as it's distance from the galactic center increases and decreases as it orbits the sun. You can see this concept quickly compounds itself as the number of object in the universe is considered, all moving in different directions relative to the earth.