Delta v is of course 200 kts in both cases.

On a site inhabited by professional airmen (airpersons?) this is still going after 5 pages?

Why choose the surface of the earth as your frame? It is moving East at about 800 kph at my latitude, not to mention the 110,000 kph velocity of the earth around the sun.

If you do the calculations correctly for any frame of reference, the answer is the same, and the easiest one to use for an aircraft in flight is the air mass it is flying in. There are effects of gusts and wind shear, but these are irrelevant to this thread. :ugh::ugh:

Incidentally, I have looped a glider both into and down a strong wind to demonstrate to a confused pupil that there was no difference. Same entry airspeed, same 'G' pull, same airspeed over the top. (Drifting downwind while doing it and getting back to the airfield was a different challenge).

Don't forget we are moving at an average velocity of 828,000 km/hr around the galactic centre.
https://starchild.gsfc.nasa.gov/docs...uestion18.html

Thank goodness relativity made looping a glider and turning down wind so simple that we only have to consider IAS. Phew!

kinetic energy is not an absolute thing - it is relative to a reference frame. As the aerodynamic forces on the aircraft come from the airflow the energy that matters to me in a turn is the KE relative to the air. If it is a car we look at the forces impacting the cars motion come from the ground so the earth would be what mattered. If I was driving on a huge conveyer belt doing circles my speedometer would read constant and I would use the belt as my reference frame.

As for the loop in a glider I would happily do a loop in a glider with 0 knots groundspeed providing my entry airspeed was correct and it was a steady flow airstream.

I am a bit cautious of entering such august company, being only a former hang glider pilot. But two things:

1 In a constantly moving air mass my experience was that a turn in any direction felt pretty much the same regardless of windspeed. But a top of the hill landing in a 20kt wind meant a scary downwind 40kt over the ground at 200 feet - knowing I had to land on my feet. But when I turned into wind and landed my groundspeed would be zero and would feel like stepping off a curb. The feeling of acceleration was however just the same as in a nil wind landing.

2 When you are within ten or twenty feet of the ground the wind tends to reduce sharply as you descend. Birds can do clever things utilising the wind gradient. Hang glider pilots not so much. Get down to 20 feet with a tail wind and it is impossible to turn. Bank (say) left, and the left wing being close to the ground is seeing less of a tailwind relative to the right wing - which translates to more of a headwind when you factor in the movement of the glider. It experiences more lift than the right wing, you can't sustain the bank and are committed to landing downwind. If it is on heather you may well not be hurt, but if there is a Derbyshire stone wall in front of you then...

All in keeping with the laws of physics, don't be timid, you have a very good grasp of the situation.

I didn't ask what the change in airspeed was I asked what the change in velocity is. That wasn't an accident. Velocity is a vector quantity, and that's what kinetic energy is based on. This is the fundamental piece of knowledge you're missing which is preventing you from correctly understanding the physics here. Fitter2 has already given the correct answer, but I'll repeat is in a little more detail; The change in velocity in the turn is 200 knots. Going from 50 knots eastbound to 150 knots westbound is a net acceleration of 200 knots. going from 100 knots eastbound to 100 knots westbound is a net acceleration of 200 knots. The change in velocity is 200 knots relative to the frame of reference of the ground; the change in velocity is 200 knots relative to the air mass moving over the ground at 50 knots, and in the no wind condition the the ground based frame of reference and the frame of reference of the air are the same, so the change in velocity of the calm air airplane is 200 knots in either case. Bottom line is that in all frames of refrence, in both the wind and no wind case the change in velocity is identical. So, if the change in velocity is identical, then the change in kinetic energy must be identical.

This warrants a specific comment. This is true, as far as it goes, but from your statement it appears that you don't understand what an "inertial frame of reference" is. An inertial frame of reference is quite simply, an unaccelerated frame of reference, a frame of reference which isn't changing velocity. In this scenario, both the ground and the uniform air mass moving at a constant velocity over the ground are inertial frames of reference.

Some points to ponder:

What is acceleration? (the physics definition, not the non-technical "going faster" meaning.)

Is turning flight accelerated flight or is it unaccelerated flight?

If I'm walking down the side of the road, I can calculate my kinetic energy relative to the ground surface, or relative to the truck that's coming the other way. It's only the first that matters . . . unless I step in front of the truck.

and if you *do* step in front of the truck your change in kinetic energy in the frame of reference of the ground will be identical to your change in kinetic energy relative to the frame of reference of the truck.

But will it?

Say you are walking towards the truck at 5 Km/Hr relative to the ground then stop. Your change in KE relative to the ground is 0.5*m*V^2 say 25 units.
Relative to the frame of reference of the truck moving uniformly at say 50 Km/Hr over the ground towards you, it observed you had a relative velocity of 55 Km/Hr initially, then 50 Km/Hr after you had stopped walking towards the truck.
Change in KE according to the truck’s frame of reference = (55*55) - (50*50) = 3025 - 2500 = 525 units.

Goldenrivett -- you're too kind. In the scenario I was thinking of, the tape ends when my velocity equals that of the truck (which is probably a millisecond after the point where I'll wake up from a nightmare tonight . . .). So I think A Squared is right (assuming that my fateful step doesn't change my velocity parallel to the road, and ignoring the slight change in the velocity of the truck).

Yes, you are correct, the change in kinetic energy would differ with the different frame of reference, I guess I had momentum in mind. Which makes the last sentence in my next-to-last post incorrect. to wit: In the scenario with non-zero wind the change in velocity is still the same, independent of frame of reference, but the change in kinetic energy will differ. However, what *is* true, is that the change in kinetic energy in the frame of reference of the air mass is identical in both the scenario with wind and the calm air scenario. And if we must consider this from the aspect of kinetic energy, that frame of reference (the air mass) is the only relevant frame of reference, because it is the air mass which is performing work on the airplane, ie; the airplane is only accelerated from +100 knots to -100 knots by the force the air is exerting on the airplane.

This is true in the real world, but the downwind turn myth holds that even in a non-turbulent uniform air mass, you will lose more airspeed in a turn to downwind than you will performing the same rate turn in calm air. So, the abstraction of a uniform airmass moving relative to the ground with no relative movement or air within the parcel is correct to address the fallacy. wind gradients, shear and vertical movement only serve to muddy the waters.

1/ The downwind turn fallacy should have become extinct with the advent of GPS showing groundspeed. Easily compare actual airspeed with actual groundspeed. Actual numbers are hard to dispute, even for a Flat Earther.

2/ Our bird theorist needs to get in an aeroplane and fly in ground effect, over varying surfaces, including swells.

That is, if a dreaded downwind turn doesn't get the better of him first....