So you think that in cruise, with a steady wind from the left and no power changes, if you turn left your airspeed will increase, and if you turn right it will decrease?

Hans, Widescreen happens to be correct.

A change of vector is a shear. Period. If you change your track relative to a steady wind, you are introducing a shear through the change in vector, your inertial mass has to accelerate or decelerate. That shows up as an airspeed change. There is a lag in the change of GS for an instantaneous change, and then it is simply sematics.

If your aircraft has a finite inertial mass, and there is a rate of change, then, yes.

Now, you can go on line to a web page like this one: https://www.planeandpilotmag.com/article/downwind-turn/
and amongst the pretty pictures you will note a statement that is made that is the logical limit of the problem, and yet, the wrong conclusion is made from the limit case.
If indeed you turned 180 degrees from a 50 kt headwind to a 50 kt tailwind and you started at, say 50kts, you would indeed have a problem, you have just had a 100kt instantaneous shear occur, so how does the aircraft, which includes tin, plastic, protoplasm and gas, e.g., has mass, how does that go from the zero ground speed to 100kt ground speed in zero time? That would be an acceleration required of....:
(50m/sec)/0.5= 100m/sec,
which is a shade over 10g horizontal. In the same case, the C-172 or Piper Cub can accelerate at.... around 0.15g longitudinally. A lear gets about 0.2g with TFEs, a B773W gets about 0.18g... An F-18 gets just around 1.0g, an F-16 light gets about... 1.1g down in the weeds. roughly.

For most flight training, the ground reference case is not done pulling 2g in a steep turn, and with a 50kt wind, but if it did, you would still be looking at a modest shear rate... at 100kts CAS, you would take 9.8 seconds to do the 180, and meet a shear of 100 kts, so that is about 10kts/sec, which is going to need an inertial acceleration of 0.5g. That would actually exceed most aircraft out there. take the normal ground reference cases at a 25-degree bank, and that equates to rate 2, or just on 35 seconds for the same 180 turn, and that gives just on 3 kts shear per second, a relatively trivial amount, which you would see as a need for a bit of thrust change, as in... 0.15g longitudinal, well within a Lycoming or Conti's capability. And that is with a 50 kt wind... Most training is done in winds around 15-20kts for the same case, and the effect is perpetuating the position that planes. have no inertial effects. (15kts... 0.05g, etc... barely enough to set off an entry interface for the space shuttle).

Why this is a factor mainly at high altitude or with rapid wind shear is where there is a considerable rate of change, limited excess thrust, and high inertia. It is observable with 200kt jetstreams, or high vertical shear rates when climbing and descending at high rates of VS. It also is why losing a 30kt headwind on an approach will result in an instantaneous loss of airspeed, and a delay as the GS increases....

If you need more background, you can PM me, and I will forward some reading for you.

Newton has a lot to answer for ...

You chaps are confusing this humble helo driver, it used to occur that it was found necessary to hold in 70 kt winds occasionally, max endurance was 74 kt, turning downwind was no different than had there been zero wind. Close the eyes and there was no difference, visual experience was dramatic though.

"Give me leave, Sir, to insinuate that I cannot think it effectual for determining truth to examine the several ways by which phaenomena may be explained, unless where there can be a perfect enumeration of all those ways. You know that the proper method for inquiring after the property of things is to deduce them from experiments. And I told you that the theory that I proposed wa[s] evinced to me, not by inferring it is thus because not otherwise, that is, not by deducing it only from confutation of contrary suppositions, but by deriving it from experiments concluding positively and directly. The way therefore to examine it is by considering, whether the experiments which I proposed do prove those parts of the theory to which they are applied; or by prosecuting other experiments that the theory may suggest for its examination. And this I would have done in a due method...."

in "A Series of Quaeries Proposed by Mr. Isaac Newton to be determined by Experiments, positively and directly concluding his new Theory of Light and Colours, imparted to the Editor in a Letter of the said Mr. NEWTON'S, of July 8, 1672. No. 85 p. 5004. In The Philosophical Transactions of the Royal Society, London, from their commencement in 1665 to the year 1800 (abridged) Vol 1 p. 734

English was a second language to the pommies in 1672 apparently...

Isn't is true that "track" is ground referenced? How is a change of track (ground referenced) of any significance at all to the movement of the aircraft in a uniform airmass (airmass referenced).

Imagine swinging a weight attached to a rope around in around in a horizontal circle. It's swung at such a speed the angle of the rope is about 60 degrees from vertical. The ball experiences gravity and a put from the string. If you sat on the ball you would feel twice as heavy.
Let's assume we make the rope long enough that the ball moves at 60 miles per hour, and we swing the ball on a truck that moves at 60 miles per hour. If you stand on the truck the ball moves in a circle. To someone standing by the side of the road it .makes the same movement that would be made by a point on the wheel of the truck: ome to a stop, and leap forward, come to a stop and leap forward with dramatic changes in horizontal velocity and strong acceleration forces. The ball doesn't feel that, because it's frame of reference moves at that steady speed of 60mph. It still feels that same constant 2g pull from the rope.
Now change ball into airplane flying in circles, and truck into the air the plane is flying in.

quote from Airbus.

the only matter of concern to the aircraft is the instantaneous rate of change of a component to the steady inertial state of the aircraft. If you suddenly change your direction is the same situation to the wind suddenly changing its direction.

Is it relevant for most training cases? not really, the rate of change is modest, so the inertial accelerations are not noticeable.

For this thread on the F-50 short landing, this discussion is irrelevant, however, for a flight around jet streams, it is pertinent.

By definition, shear is determined to be a change in a component wind, and for low-level flight that is normally due to wind gust, turbulent boundary layer effects, gust lines or convective phenomena. That is as taught. At altitude and with extreme jet streams, a rate of change of the heading is a change of wind component, the aircraft has inertia and that has to change and takes a finite time to do so. The speed will alter, and recorded flight data shows that occurring. It is at that point that the assumption that the aircraft's inertial frame of reference is only to the airmass has an issue. For the climb and descent through a jet stream, the same issue arises. If the aircraft was bound inertially only to the airmass, a change in the wind would not change the IAS, it would only change the ground speed. That is not observed in the flight data.

For basic teaching we can ignore the fact that Newton wrote stuff in 1665, and assume that the plane is tied only to the airmass, and ignore the fact that it's motion is referenced to "the aether", not the universe/planet. However, if you look at chapter 1, THE KINEMATICS AND DYNAMICS OF AIRCRAFT MOTION in Stevens et al, 2015, you will note that the frame of reference isn't to cyclone Bob or Irma, it is to the earth, as is an INS, GPS (kind of). p. 3 of the section which is a riveting read, states:

INS, IRU's & ADIRU etc use accelerometer measurements added or subtracted to the platform alignment and original position georeferenced to the earth, and uses that to determine over time the velocity, and over time gives the distance shift which gives actual position. It doesn't need a wind input, it in fact gives that as a calculated output, from the component accelerations to the aircraft from wind components. The wings see IAS for stuff like V squares etc... the aircraft sees the sum of all fears accelerations.

As difficult as this may be to consider, the assumption that the aircraft miraculously is unbound to the earth's frame of reference would also suggest that as soon as the wheels came off the ground at the equator the aircraft would be doing 900 knots to the west, as apparently its inertial frame of reference is not to the earth. Most times that doesn't happen. For flying around the pattern you can disregard everything and just look out the window and enjoy the view, as the approximation that we are taught from day 1 suffices. That does not suffice at limit cases, where inertial changes are non-trivial.

The aircraft is immersed in an airmass that is tied to the planet and then to whatever reference you wish to take into consideration. The inertial moment of the aircraft is measurable to the earth as an inertial frame of reference. The airmass is tied to the external frame of reference itself.

Stevens B.L., Lewis, F.L., Johnson, E.N., (2015) Aircraft Control and Simulation: Dynamics, Controls Design, and Autonomous Systems (3rd edition), Wiley and Sons.

Why do you introduce a discussion of wind shear when you previously were arguing the case of steady wind and I specifically stated "uniform airmass"?

How about keeping it simple to start with.

Case 1. Aircraft makes continuous circles at constant altitude with no change in thrust. The aircraft is in a completely uniform airmass that has no motion relative to the surface of the earth. What happens to indicated airspeed during the turn?

Case 2. Aircraft makes continuous circles at constant altitude with no change in thrust. The aircraft is in a completely uniform airmass that is moving with contant speed and direction relative to the surface of the earth. What happens to indicated airspeed during the turn?

Good point. Ordinarily, would assume that the cases are identical, our experience and teachings suggest the cases are identical, we would not see an observable difference, as kinetic energy is irrelevant. The kinetic energy changes considerably but the IAS would essentially remain constant. for a wind of 20kts for a C-172 the difference in kinetic energy states is equivalent to a B747 in a 200Kt jet stream going from a tailwind to a headwind. For the B747 case, the IAS change that occurs is around 6kts, which is equivalent to around 1 kt for the C-172. 6kts is ~ 0.01M and wakes up the ATR.

Is there a transfer function? I have no idea. Is the B747 case observable at all times, it is repeatable and it suggests that the assumption that we have on changes of heading in an airmass is a useful approximation rather than an absolute truth. It has no relevance at low speeds and low inertia, it is observable and recordable on aircraft in a turn in the steady state jetstream where the only change is the track of the aircraft. It is additionally reversibile, going from a tailwind to a headwind results in the opposite effect. Do we read an IAS accurately enough on a C-172? no. The B747, B777 B787 A330/A380 etc, yes.

Before the first time I saw the aircraft behave in a steady state 200Kt + jetstream, in level flight with >90 degree turn, I did not believe it would make any difference. The most spectacular case is coming into southern Japan, but it is not the only case. Discount cases where the track is traversing the core of the jet in such a manner that it results in a shear, the observable curiosity is for the case where the wind is the same entering and exiting the turn, so wind shear is removed as a factor for the change in speed and subsequent thrust required. This happens to be memorable as getting to MCT limit thrust at high altitude and with decaying airspeed is attention-getting.

Clear as mud.

Yes.

Case 1 - Does the airspeed change, Yes or No?
Case 2 - Does the airspeed change, Yes or No?

hard to believe people don’t understand that.

1: No
2: imperceptibly, but yes, it does change

1: no
2: no
In both instances you are circling in an air mass moving up to 900 Kts, a few Kts more or less won't matter.

I would have expected a speed loss in a turn if power remains constant, due to the lift, the vector no longer being vertical, having to be increased by increase by an increase in AoA and resultant increase in drag. Never flown a big boy, just an expectation.

Both Case 1 and Case 2, as proposed by myself, specified continuous circles. The roll into the steady state turn is outside the condition being examined.

I understand your argument that the earth referenced kinetic energy of the aircraft will change as it circles in an airmass that is moving in that same frame of reference. However, I don't yet accept that this change in kinetic energy influences the airspeed.

The aircraft is assumed to be circling in a completely uniform volume of air. Suppose we place a gas balloon in that same volume of air and it carries an observer. The aircraft circles the balloon. The motion of the aircraft and its knetic energy can be referenced to the observer. (The frame of reference for kinetic energy can be whatever we define it to be. It does not have to be earth referenced). The observer will see the aircraft circling with constant angular velocity and constant altitude. In the observer's frame of reference the aircraft energy does not change. There is no reason for the aircraft indicated airspeed to change.

Why would the presence of the balloon and its observer change the way indicated airspeed behaves?

Wow, what a confusion.

Actually, when you "turn" through a wind field/direction, you get "blown" away. This blowing away, will take care, the wind will increase your kinetic energy. Though, this is not instantly. Have a heavy airplane with a lot of wind, and it may take enough time to get a stall surprise.

On a base leg, you will need to turn a little into the wind, to avoid, your base leg (ground track) is not perpendicular to the runway. When you start turning, your airplane starts to get caught by the wind and accelerated in the direction of the wind. Be heavy and turn fast, the wind does not have sufficient time to accelerate the airplane and the airspeed will drop. Have a light airplane, and the wind will accelerate the airplane within a few seconds.

For the example to make 360 turns in a steady wind, it is not defined, whether the circle should be ground based or relative to the (moving) air. With a ground based circle, you will have a constant kinetic energy, though juggling with the power (and varying airspeed) to make it a circle on the ground.

With a moving air circle, your airspeed will remain the same, though because your ground speed will continuously vary, the kinetic energy will continuously vary. To accomplish this, you will need to juggle with the power setting. The moment you turn "into the wind", you will need to bleed off ground speed, to avoid your airspeed going up (IE you will need a little less engine power at that moment). And the other way around.

Oh, and I fly and a often windy airport, where multiple 360s due to other traffic are pretty common. And flying a nice 360 ground based requires a constant power setting / trim juggle. Often, the 360s tend to become ovals.

Ehhhhhhh, maybe back to school ?