Saying you understand something when everything you say shows you don't is not an affirmation.

Are all frames of reference equally valid?

If you read the post from A-Squared where he suggested I go to a friendly university physicist - taking about 4 sentences - you will see why I said what I did.

I choose the ground as my frame of reference then. Aircraft stationary in my frame of reference and in 30 seconds accelerates to 120kts. Or alternately is traveling at 120Kts and in 30 seconds becomes stationary

Scenario A (Cessna)
final groundspeed minus initial groundspeed
(airspeed + tailwind) - (airspeed - headwind)
(100+10)-(100-10)
110-90
20

Scenario B (Person inside airliner)
final groundspeed minus initial groundspeed
(airspeed + tailwind) - (airspeed - headwind)
or, if you like:
(walk + tailwind) - (walk - headwind)
(1+500)-(1-500)
501-(-499)
1000

You're not understanding the analogy. I thought I was clear enough in being excessively verbose and redundant in all the details. The airmass inside the moving airliner (scenario B) represents the airmass that is the steady, uniform wind the Cessna is flying in (scenario A)

The airspeed of the person walking (1 knot in scenario B) represents the airspeed of the Cessna (100 knots in scenario A). It's the speed of the person moving relative to the airmass inside the airliner. Like the speed of the Cessna relative to the airmass it's flying in.

The "airspeed" of the airliner in scenario B (500 knots) is not analogous to the airspeed of the Cessna (it's only a coincidence in the use of the same word.) It is analogous to the wind speed in scenario A (10 knots). It is the speed of the airmass over the ground.

Your rearranging of the terms is incorrect. Please look up and read it again, while keeping straight what is analogous to what. Line by line, term by term, I really don't know how to make it more explicit. Examine it again.

All you're doing there is mixing frames of reference to get the answer YOU want, not an intelligent way to proceed.

And what you're not grasping is that going from stationary to 120 knots is the identical change in velocity as going from 60 knots east to 60 knots west, and it requires the identical force exerted on the airplane by the air for the identical period of time, and both scenarios require overcoming the identical amount of inertia.

Vessbot you cannot be serious. Even if it is a cessna flying at 1kt in a 500 kt wind there is still no 1000 kt change of groundspeed

Read #201 again

Yes, there will be, for the reason explained by the math I posted.

What's the groundspeed before the turn? -499
What's the groundspeed after the turn? 501

The difference is 1000.

I read your post again, and it still rearranges the terms in error. Look at scenario A, and see that obviously everything is right. This is common sense to a private pilot. And then look at how every element translates to scenario B. Have I made any mistakes in the translations?

No not an identical force.
0 to 120 requires a tangential force
Making a 180 requires a centripetal force
In purely Newtonian terms force is force but for an aircraft, tangential and centripetal forces have quite different effects because it envolves rotating forces.

Ask a mathematician

I'm speaking about the comparison between a 60 knot airplane in a 60 knot headwind doing a 180 degree turn, and the same airplane doing the same turn in still air.

Yeah, but he is choosing to go from 60kts (the airspeed) to 120kts downwing (the groundspeed) and claiming it's a much bigger change just because of the wind.

Ummm, if a person in an airliner with a velocity of +500 knots is walking toward the tail at 1 knot, his groundspeed is +499 knots, not -499 knots

A steady wind has no aerodynamic effect on the aircraft.

Ask any physicist.

He's moving backward. His (body) groundspeed could not be in the positive unless he's running toward the tail at greater than 500 knots.

What's a Cessna's groundspeed when flying at 50 knots airspeed into a 60 knot headwind? Come aawwn!

Again, we're talking about the speeds of the person not the airplane. The only way the airplane figures into this, is an enclosure of the 500 knot airmass.

no, it's -499 because he is facing the tail. If he stops it goes back to -500. If he turns to face the cockpit while stationary it goes from -500 to +500 almost instantly.

So you choose a ground-centric reference frame for the velocity vector, but a anthropomorphic reference frame for the maginitude?

You are out of your depth - stop digging.

PDR

Why do you think?

Do you apply the same criticism to a Cessna flying on an East heading in an East-to-West wind?

I'll tell you the same thing I told Brercrow: Look at scenario A, and see that everything checks out, as should be common sense to a private pilot. Then see how every element translates to scenario B. I'll even paste it right in here so you don't have to scroll up. Have I made any mistakes in the translations?

Scenario A (Cessna)
final groundspeed minus initial groundspeed
(airspeed + tailwind) - (airspeed - headwind)
(100+10)-(100-10)
110-90
20

Scenario B (Person inside airliner)
final groundspeed minus initial groundspeed
(airspeed + tailwind) - (airspeed - headwind)
or, if you like:
(walk + tailwind) - (walk - headwind)
(1+500)-(1-500)
501-(-499)
1000

It is simply HIS groundspeed measured from his frame of reference.

You're adopting a frame of reference which changes orientation with the person. The only way you can make sense of any of this is with a constant, non-accelerated frame of reference.

the airliner is flying East at 500 knots. the passenger is walking toward the tail at one knot, he is still moving East at 499 knots. He's not moving west merely because he's facing west.

While I obviously think Brecrow's theory is fallacy, I have to agree with him that the analogy the two if you are using here is just as flawed.

The airliner (= airmass container = wind) is flying WEST at 500 knots. The person is initially walking East (1 knot airspeed, -499 groundspeed) turning around to West (1 knot airspeed, 501 knots groundspeed).