The Windward Turn Theory

[Brercrow]

Thanks Ian W that is a good example of the sort of thing I am talking about.

[Brercrow]

Originally Posted by jonkster

not trying to make stuff up, apologies if I misunderstand what you are saying but I am not getting what you are saying.

Your argument (as I read it) is that in a constant wind, an aircraft in a constant rate turn, will have (small) changes of airspeed as it moves from a headwind (relative to the ground) to a tailwind (relative to the ground). Is that correct?

Correct

Would you agree in this situation that the only forces that are directed laterally on the aircraft are thrust and drag and inclined lift, all are acting due to interaction with the air?
Correct. Except that the aircraft is accelerating tangentially in the direction of its ground speed. That acceleration can be accounted-for by a component of the centripetal force (horizontal component of inclined lift)

The angle of bank doesn't change so the lateral component of lift is constant.

We do not change thrust.

The aircraft can only generate an acceleration (which would be required to change the airspeed that you say happens) is if we change drag - that is the only force we have left (as far as I can see).

There is a rate of change of airspeed caused by the rate of change of the headwind/tailwind component as the aircraft turns. This is approximately equal and opposite to the rate of change of airspeed caused by the acceleration of groundspeed. Hence the rate of change of airspeed is approximately zero but NOT EXACTLY zero

Why does drag change?

It does not change apart from the effect of increased load-factor

The only way I can see that changing is because airspeed changes but that is assuming the effect we are trying to find the cause of - it would appear to me to be the airspeed changes because the drag changes and the drag changes because the airspeed changes therefore the airspeed changes... ?? ie we are assuming the effect we are trying to prove occurs.

I am not getting where an unbalanced force can come from that causes the (small) airspeed changes you claim are occur.

It is just that the effect of the changing head/tail wind components is slightly greater than the effect of the force components. The airspeed increases slightly in the windward turn and decreases slightly in the leeward turn

Now if the wind velocity varied (eg a gust or windshear in descent/climb etc) then that would cause a change in airspeed but not in a constant wind. What am I missing here?

Here you are saying that airspeed can change just because of changing head/tailwind components without a force being applied
Well that's the same as what I am saying

[Jet_Fan]

Originally Posted by Brercrow

Here you are saying that airspeed can change just because of changing head/tailwind components without a force being applied
Well that's the same as what I am saying

With all due respect, it’s not the same thing at all!

[Winemaker]

Put the airplane in a black box big enough for it to take off and fly circles. Accelerate the box to some speed in some direction. Fly the plane in circles. Record the path/speed of the plane. Now, Brercrow, calculate the speed and direction of the black box from variations in the path/speed of the circling plane; if what you say is true you should be able to do this. Bet you can't.

[CallmeJB]

For a fun cocktail cruise, I will often drive my boat up the river. It's a comfortable ride and the boat does well at about 3/4 throttle setting. Every now and then, I will turn around and go back down the river. I've never noticed a change in boat speed when I make this turn, nor have I had to increase throttle setting to maintain boat speed throughout the turn, no matter how fast the current is.

I do notice a large change in speed relative to the land on shore, but absolutely no change in driving characteristics relative to the water that the boat is driving in.

I've also never noticed water starting to come up over the transom as the boat slows down in this turn, or that the boat starts sliding backwards due to having a low inertia relative to the land on shore. Really, as I sit there with my cocktail, the only inertia that matters is the inertial relationship to the flowing water, even though someone sitting on the shore might see something entirely different.

[A Squared]

Originally Posted by Brercrow

Here you are saying that airspeed can change just because of changing head/tailwind components without a force being applied
Well that's the same as what I am saying

No, that absolutely is not the same as you are saying. He is talking about changing wind components when flying across wind velocity gradients, in other words, when the wind is changing relative to itself, a non constant, non uniform wind. You are talking about flying in a uniform, constant wind, where no wind gradients exist ...yet somehow through maneuvering the airplane causing a tailwind. That is something completely different than what he's talking about.

[Jet_Fan]

Originally Posted by A Squared

No, that absolutely is not the same as you are saying. He is talking about changing wind components when flying across wind velocity gradients, in other words, when the wind is changing relative to itself, a non constant, non uniform wind. You are talking about flying in a uniform, constant wind, where no wind gradients exist ...yet somehow through maneuvering the airplane causing a tailwind. That is something completely different than what he's talking about.

I think this rather exposes the fact that he really doesn’t understand the basics of aerodynamics. He talks about drift angle, which is obviously only in reference to the ground and then claims it’s having an aerodynamic effect. It’s like he can’t understand that there are two completely different frames of reference here, the ground frame of reference and that of the airmass.

[A Squared]

Originally Posted by Ian W

Yes I know all the arguments about being in a windfield so the ground speed doesn't matter - but I had the feeling that it did. The inertia of the aircraft that was close to stationary had to be overcome for the aircraft to depart at (guess) wind of 60 kts at 200ft plus aircraft eventual flying speed of say 60kts meant that the airframe had to accelerate from stationary to 120kts - agreed helped by the airmass it was in. That inertia has to be overcome regardless and that is due to the frame of reference to the Earth/ ground speed of the airframe. So an extreme example but I assure you that the aircraft lost a lot of height in the turn going past at my eye level in the tower.

Ian, Goldenrivet already addressed this but I wanted to add a little more detailed explanation, because this is an important concept. What you described is the very crux of why the downwind turn myth exists. The fact is, the situation you describe has no more inertia to overcome than the same airplane doing the same rate turn in still air. I know that it *seems* like it is different, because *WOW*, the airplane is going 120 knots at the end, but there is no more acceleration, and no more inertia to overcome when an airplane goes from 0 knots groundspeed in a 60 knot wind, to 120 knots downwind than there is when the same airplane goes from 60 knots East in still air to 60 knots West in still air. It *seems* like it's different, because often people think of the airplane in still air as not accelerating, because it's started out with a speed (air and ground) of 60 knots and it ended up with a speed of 60 knots. This is a fallacy, albeit an understandable one if physics isn't your strong point. The thing you have to understand is that from a physics standpoint the relevant quantity is not "speed" but "velocity", and velocity is a vector quantity. What that means is that velocity is a measure of your motion, and of the *direction* of your motion. If the direction of your motion changes, it's a change in your velocity, just like a change in your speed is a change in your velocity. So, going from 60 knots east to 60 knots west is very much a change in your velocity. This might be a little more clear if you consider two 60 knot airplanes flying past each other closely in opposite directions. When you think about that, it should be pretty obvious that the difference in their velocities is 120 knots. The same applies for an airplane flying 60 knots in one direction, then turning 180 degrees, and flying in the opposite direction 60 knots. Just like with the airplanes flying past each other the difference in velocity from before the 180 degree turn to after is 120 knots. Now going back to the airplane turning in the wind, it goes from 0 knots into the wind to 120 knots downwind, and obviously, the change in velocity is 120 knots. Which is identical to the a airplane turning 180 degrees in still air: 120 knots change in velocity. So, the change in velocity is identical. the airplanes use the same rate of turn, so the time for the velocity change is identical, so the acceleration is therefore identical, because acceleration is change in velocity divided by time. If the acceleration is identical, then too, the forces acting on the airplane must also be identical.

If you're really interested in understanding this, but you're not sure what I'm saying is correct, print this out and take it to the nearest university physics department and ask if you can have a few moments of an instructor's time to explain this. He or she will tell you that the same thing.

As far as what you witnessed, as others have said, there are a variety of explanations. It may be that the airplane did fly through a wind gradient, either a loss of windspeed (gust or shear)or a vertical (down) motion. Strong winds close to the ground are not nice and constant and uniform, even though they may seems so ot an observer.

[A Squared]

Originally Posted by Jet_Fan

I think this rather exposes the fact that he really doesn’t understand the basics of aerodynamics.

Well, re-exposes it anyway. He's already demonstrated a pretty spectacular lack of understanding of physics and aerodynamics .

[Jet_Fan]

Originally Posted by A Squared

Well, re-exposes it anyway. He's already demonstrated a pretty spectacular lack of understanding of physics and aerodynamics .

I meant to type ‘further’ rather than ‘rather’.

[A Squared]

Originally Posted by Jet_Fan

I meant to type ‘further’ rather than ‘rather’.

Nonessential characters added to make the software happy.

[Ian W]

Originally Posted by A Squared

Ian, Goldenrivet already addressed this but I wanted to add a little more detailed explanation, because this is an important concept. What you described is the very crux of why the downwind turn myth exists. The fact is, the situation you describe has no more inertia to overcome than the same airplane doing the same rate turn in still air. I know that it *seems* like it is different, because *WOW*, the airplane is going 120 knots at the end, but there is no more acceleration, and no more inertia to overcome when an airplane goes from 0 knots groundspeed in a 60 knot wind, to 120 knots downwind than there is when the same airplane goes from 60 knots East in still air to 60 knots West in still air. It *seems* like it's different, because often people think of the airplane in still air as not accelerating, because it's started out with a speed (air and ground) of 60 knots and it ended up with a speed of 60 knots. This is a fallacy, albeit an understandable one if physics isn't your strong point. The thing you have to understand is that from a physics standpoint the relevant quantity is not "speed" but "velocity", and velocity is a vector quantity. What that means is that velocity is a measure of your motion, and of the *direction* of your motion. If the direction of your motion changes, it's a change in your velocity, just like a change in your speed is a change in your velocity. So, going from 60 knots east to 60 knots west is very much a change in your velocity. This might be a little more clear if you consider two 60 knot airplanes flying past each other closely in opposite directions. When you think about that, it should be pretty obvious that the difference in their velocities is 120 knots. The same applies for an airplane flying 60 knots in one direction, then turning 180 degrees, and flying in the opposite direction 60 knots. Just like with the airplanes flying past each other the difference in velocity from before the 180 degree turn to after is 120 knots. Now going back to the airplane turning in the wind, it goes from 0 knots into the wind to 120 knots downwind, and obviously, the change in velocity is 120 knots. Which is identical to the a airplane turning 180 degrees in still air: 120 knots change in velocity. So, the change in velocity is identical. the airplanes use the same rate of turn, so the time for the velocity change is identical, so the acceleration is therefore identical, because acceleration is change in velocity divided by time. If the acceleration is identical, then too, the forces acting on the airplane must also be identical.

If you're really interested in understanding this, but you're not sure what I'm saying is correct, print this out and take it to the nearest university physics department and ask if you can have a few moments of an instructor's time to explain this. He or she will tell you that the same thing.

As far as what you witnessed, as others have said, there are a variety of explanations. It may be that the airplane did fly through a wind gradient, either a loss of windspeed (gust or shear)or a vertical (down) motion. Strong winds close to the ground are not nice and constant and uniform, even though they may seems so ot an observer.

A^2 I think you are getting carried away with the turns being air radius turns and forgetting your basic physics the the aircraft is actually ground referenced, for acceleration and deceleration.

1, Take a 3 ton weight that is stationary that then turns along a rate 2 turn radius and accelerates to 120kts in 30sec
2. Take a 3 ton weight that is traveling at 120kts that then turns along a rate 2 turn radius and decelerates to stationary in 30sec

Your argument is neither of these weights experience any acceleration. I say that is obviously false.

My understanding is that the force in the turn to hold the weights in the turn (centripetal force ꟺ) accelerates the weights as it is (as you say) a change in velocity that can only come with application of a force. The ground radius described by a slow moving (compared to the wind) aircraft in an air radius turn is uneven stretched by the wind velocity vector. In consequence the centripetal force ꟺ to accelerate the aircraft (to change its velocity vector ) is uneven. An aircraft with a tight ground radius first being accelerated more at first while the aircraft with a wide ground radius first is accelerated less at first. These ꟺ accelerations - changes in velocity vector - must be added to the ground speed accelerations (decelerations) These differences will only really be apparent when the wind velocity is close to or a large fraction of the aircraft velocity.

[A Squared]

Originally Posted by Ian W

A^2 I think you are getting carried away with the turns being air radius turns and forgetting your basic physics the the aircraft is actually ground referenced, for acceleration and deceleration. .

No, I am not forgetting my basic physics. Whether you use the frame of reference of the ground, or the air, the acceleration is the same. Acceleration is independent of frame of reference, assuming non-accelerated frames of reference, which (for our purposes) both the ground and a uniform air mass in constant motion both are.

Originally Posted by Ian W

1, Take a 3 ton weight that is stationary that then turns along a rate 2 turn radius and accelerates to 120kts in 30sec
2. Take a 3 ton weight that is traveling at 120kts that then turns along a rate 2 turn radius and decelerates to stationary in 30sec

Your argument is neither of these weights experience any acceleration.

No, my argument absolutely is *NOT* that neither experience any acceleration. The velocity of each changes, so by the definition of acceleration (change in velocity), they both experience an acceleration. I have no idea where you got that. What I did say, is that the same a weight accelerating from zero to 120 knots to the West experiences the identical change in velocity as the same weight accelerating from 60 knots east to 60 knots west. Both experience a change in velocity of 120 knots, regardless of frame of reference.


Yes, the radius of the turn relative to the ground will be different for the case with the wind. No, it does not make any difference because the ground is not accelerating the airplane. The only thing touching the airplane is the air, so the air is the only thing which can exert force on the airplane, so the only turn radius that matters is the turn radius of the airplane in the frame of reference of the air mass, which is identical for both the wind and zero wind cases.

[itsnotthatbloodyhard]

the the aircraft is actually ground referenced, for acceleration and deceleration.

No, it isn’t. The relevant accelerative forces - thrust, drag, and lift - are all ‘air referenced’, for want of a better term. (And as has already been pointed out, the change in velocity is the same, regardless of the frame of reference.)

If you want to reference the aircraft to something other than the airmass it’s flying in, what’s so special about the local surface of the earth? Why not the centre of the earth? Or the sun? Or a passing bus? And of course the earth’s rotation means the surface has a speed ranging from 900 kts at the equator down to zero at the poles, so maybe the downwind turn theorists need to factor this into their specious equations, with a cos latitude term.

[Vessbot]

I wish the tone in here didn't have to get nasty, though I understand (and experience myself) the frustration when the other side doesn't see things one's own way. Hopefully it can return to something more civil, where we can all simply address the other side's arguments.

I will reiterate the following: There is no preferred frame of reference, and as long as you do all the math right, you should be able to calculate any physics in any frame. It's just that lots of variables drop out and the calculations get a whole lot easier in some frames, namely the airmass frame when calculating things regarding aircraft. The danger comes when you mix frames within the same calculation without doing the necessary transformations.

And Brercrow, I'm afraid that's just what you've done. In the Downwind Turn page Figure 1b, you resolve a component Ft of force Fc (itself the horizontal component of lift, perpendicular to the fuselage) parallel to the ground track. You are correct that in the ground frame, that component exists and accelerates the ground velocity. However, my first hint of your trouble is that you don't likewise complete the resolution and take a component of Fc perpendicular to the ground track.

You have: Ft = Fc sin d (this is the force that accelerates the ground velocity forward when turning downwind)
You're missing: Fm (m stands for missing) = Fc cos d. (This is the weakening of the centripetal force, which accounts for the widening radius when turning downwind)

That would be the component of the horizontal component of lift that is centripetal to the curve, in the ground frame; not Fc. I'm not sure what the blue triangle is supposed to represent, but I am sure that it inappropriately mixes frames, as it connects Fc (which exists only in the airmass frame) and Ft (which exists only in the ground frame).

I've actually been revisiting your site for a few weeks, and this is the only hole I can poke in your math so far. The rest of the things you go on to do with your Ft seem to be valid. So why haven't I changed my mind? Maybe a small part of being an obstinate ass given to the natural human inclination to hang on to an opinion, but I would guess a larger part of lending more credibility to there being an undiscovered hole in your math, than that Newtonian relativity is wrong.

I do have a main suspicion, which is that instead of an analytical solution you have a numerical one based on an Excel spreadsheet, and that your resulting change in airspeed is an artifact of accumulations of rounding errors in the increments. I would please like to ask you to do 2 things:

1. I assume the resolution of your spreadsheet is 1 degree, therefore 360 increments. Can you rerun it with more increments, say half a degree and a quarter of a degree? If the airspeed change trends to zero with an increasing resolution, there's our problem.

2. With apologies to the huge number of counterexamples thrown at you in the thread, I'd like you to reread my post #133? https://www.pprune.org/tech-log/607454-windward-turn-theory.html#post10204374
and plug the constants of the second example into your spreadsheet, and see what answer it gives for an airspeed change; and decide if you've ever felt such a tendency while turning around in an airliner.

[Wizofoz]

Originally Posted by A Squared

Well, re-exposes it anyway. He's already demonstrated a pretty spectacular lack of understanding of physics and aerodynamics .

And has that breathtaking certainty of the truly ignorant.

He's been on about this for around ten years- popping up on forums and entering into email flame wars- each time retreating when he finally can't defend his BS, but neve EVER abandoning the core premise.

It's a case study in deciding you are right based on incomplete knowledge, then being impervious to evidence that shows you're wrong.

You KNOW you are right, so the counter arguments MUST be wrong- in this case a guy named Newton apparently made some fundamental errors..……

I suspect we've all been guilty of some degree of it somewhere, but most are able to be reasoned with on the end.

[Wizofoz]

Originally Posted by Ian W

A^2 I think you are getting carried away with the turns being air radius turns and forgetting your basic physics the the aircraft is actually ground referenced, for acceleration and deceleration.

1, Take a 3 ton weight that is stationary that then turns along a rate 2 turn radius and accelerates to 120kts in 30sec
2. Take a 3 ton weight that is traveling at 120kts that then turns along a rate 2 turn radius and decelerates to stationary in 30sec

Your argument is neither of these weights experience any acceleration. I say that is obviously false.

My understanding is that the force in the turn to hold the weights in the turn (centripetal force ꟺ) accelerates the weights as it is (as you say) a change in velocity that can only come with application of a force. The ground radius described by a slow moving (compared to the wind) aircraft in an air radius turn is uneven stretched by the wind velocity vector. In consequence the centripetal force ꟺ to accelerate the aircraft (to change its velocity vector ) is uneven. An aircraft with a tight ground radius first being accelerated more at first while the aircraft with a wide ground radius first is accelerated less at first. These ꟺ accelerations - changes in velocity vector - must be added to the ground speed accelerations (decelerations) These differences will only really be apparent when the wind velocity is close to or a large fraction of the aircraft velocity.

Ian,

You've made the same fundamental error- and you've been arrogant with it.

Before saying someone has forgotten basic physics, you'd best learn some yourself.

The earth is not a privileged frame of reference and no, no value HAS to be calculated in reference to it. All calculations of velocity, acceleration and therefore momentum and kinetic energy work in any frame of reference- as long as you STAY in one frame of reference.

Momentum is not related to groundspeed. It is related of whatever frame of reference you choose to calculate in.That is implicit in Newtons first law of motion,

[Wizofoz]

Originally Posted by Vessbot

I wish the tone in here didn't have to get nasty, though I understand (and experience myself) the frustration when the other side doesn't see things one's own way. Hopefully it can return to something more civil, where we can all simply address the other side's arguments.

I will reiterate the following: There is no preferred frame of reference, and as long as you do all the math right, you should be able to calculate any physics in any frame. It's just that lots of variables drop out and the calculations get a whole lot easier in some frames, namely the airmass frame when calculating things regarding aircraft. The danger comes when you mix frames within the same calculation without doing the necessary transformations.

And Brercrow, I'm afraid that's just what you've done. In the Downwind Turn page Figure 1b, you resolve a component Ft of force Fc (itself the horizontal component of lift, perpendicular to the fuselage) parallel to the ground track. You are correct that in the ground frame, that component exists and accelerates the ground velocity. However, my first hint of your trouble is that you don't likewise complete the resolution and take a component of Fc perpendicular to the ground track.

You have: Ft = Fc sin d (this is the force that accelerates the ground velocity forward when turning downwind)
You're missing: Fm (m stands for missing) = Fc cos d. (This is the weakening of the centripetal force, which accounts for the widening radius when turning downwind)

That would be the component of the horizontal component of lift that is centripetal to the curve, in the ground frame; not Fc. I'm not sure what the blue triangle is supposed to represent, but I am sure that it inappropriately mixes frames, as it connects Fc (which exists only in the airmass frame) and Ft (which exists only in the ground frame).

I've actually been revisiting your site for a few weeks, and this is the only hole I can poke in your math so far. The rest of the things you go on to do with your Ft seem to be valid. So why haven't I changed my mind? Maybe a small part of being an obstinate ass given to the natural human inclination to hang on to an opinion, but I would guess a larger part of lending more credibility to there being an undiscovered hole in your math, than that Newtonian relativity is wrong.

I do have a main suspicion, which is that instead of an analytical solution you have a numerical one based on an Excel spreadsheet, and that your resulting change in airspeed is an artifact of accumulations in the increments. I would please like to ask you to do 2 things:

1. I assume the resolution of your spreadsheet is 1 degree, therefore 360 increments. Can you rerun it with more increments, say half a degree and a quarter of a degree? If the airspeed change trends to zero with an increasing resolution, there's our problem.

2. With apologies to the huge number of counterexamples thrown at you in the thread, I'd like you to reread my post #133? https://www.pprune.org/showthread.php?p=10204374
and plug the constants of the second example into your spreadsheet, and see what answer it gives for an airspeed change; and decide if you've ever felt such a tendency while turning around in an airliner.

His error is that he uses vectors from different reference frames in the same vector diagram.

[Vessbot]

Originally Posted by Wizofoz

His error is that he uses vectors from different reference frames in the same vector diagram.

That is an error (as I pointed out in my own post that you quoted) but not the error, as Excel can't do math on diagrams. But you're welcome to look through his math more specifically, maybe you'll find an error there that I haven't yet.

[Wizofoz]

Originally Posted by Vessbot

That is an error (as I pointed out in my own post that you quoted) but not the error, as Excel can't do math on diagrams. But you're welcome to look through his math more specifically, maybe you'll find an error there that I haven't yet.

It's what invalidates his whole idea. An object has one velocity vector in any frame of reference. He tries to say energy calculated in the IFR where the earth is still can be calculated in the in where the air is still.