The Windward Turn Theory
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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.
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.
These other guys dont get it
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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.
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.
Brercrow, you did a VERY rough spreadsheet with 10 degree increments (refer section 6 of his Post), and then claimed a VERY SMALL change in airspeed as a conclusion. Didn’t that ring any alarm bells for you? Especially considering that your conclusions actually negated Newtonian physics, and if true would have earned you a Nobel Prize.
Last edited by Derfred; 4th Aug 2018 at 13:16.
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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).
You are right Fc gives the curvature of the air path and Fc * cos d gives the curvature of the ground track. But the ground track is a throw-away item because we are not interested in the ground track. What we are discussing is the effect on airspeed.
Forces exist in all frames of reference and for an aircraft the effect of tangential and centripetal forces are different in each frame. The forces described are acting on the aircraft not on the ground. Hence a component Ft = Fc .sin d causes a tangential acceleration of the ground -velocity. That then causes a rate of change of component K which is almost exactly balanced by the opposite rate of change of component H. The difference between the two gives the slight acceleration of airspeed.
Didn't you read the whole thing? I suspect not!
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.
My work complies with Newtonian physics but obviously there are several levels of complexity in this problem.
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.
I have tried different resolutions. The end result is the same but the curves are smoother with fine resolution.
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.
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).
You are right Fc gives the curvature of the air path and Fc * cos d gives the curvature of the ground track. But the ground track is a throw-away item because we are not interested in the ground track. What we are discussing is the effect on airspeed.
Forces exist in all frames of reference and for an aircraft the effect of tangential and centripetal forces are different in each frame. The forces described are acting on the aircraft not on the ground. Hence a component Ft = Fc .sin d causes a tangential acceleration of the ground -velocity. That then causes a rate of change of component K which is almost exactly balanced by the opposite rate of change of component H. The difference between the two gives the slight acceleration of airspeed.
Didn't you read the whole thing? I suspect not!
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.
My work complies with Newtonian physics but obviously there are several levels of complexity in this problem.
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.
I have tried different resolutions. The end result is the same but the curves are smoother with fine resolution.
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.
In the second example the math is wrong (500+1) - (500-1) = 2 not 1000
I refer you to post #116 for an example of what happens at jet/Jetstream speeds
But again you are right. My work is partly analytical and partly numerical and that is because albatross dynamic soaring depends on the particular values of wind speed and angles of bank an climb
So far as the downwind turn is concerned the effect is greatest when the drift angle is very large
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Vessbot has it.
Brercrow, you did a VERY rough spreadsheet with 10 degree increments (refer section 6 of his Post), and then claimed a VERY SMALL change in airspeed as a conclusion. Didn’t that ring any alarm bells for you? Especially considering that your conclusions actually negated Newtonian physics, and if true would have earned you a Nobel Prize.
I love the way you guys imply that anything you don't understand must be wrong and a violation of Newtons Laws
It never occurs to you that there might be more complexity in this problem than first appears.
The simple fact is that my work is based on Newtons laws of motion, the triangle of velocities and the aerodynamic forces acting on the aircraft plus gravity
If you want to understand it then YOU have to put in the effort as I have
And I have tried different resolutions in the spead sheet and I get the same result
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final groundspeed minus initial groundspeed
(airspeed + tailwind) - (airspeed - headwind)
(1+500)-(1-500)
501-(-499)
1000
If you don't think that's right, compare it with the ealier Cessna example, where the self-created windshear is 20 knots:
final groundspeed minus initial groundspeed
(airspeed + tailwind) - (airspeed - headwind)
(100+10)-(100-10)
110-90
20
For both examples, the final and initial groundspeeds are states of equilibrium. But the 1000 knot and 20 knot wind shears? Those are happening during the turn, just as you say.
I refer you to post #116 for an example of what happens at jet/Jetstream speeds
The situation that post 116 is a reply to, is a nonstarter. First, it's on an arrival procedure, where your track is anchored to the ground. So it does not apply to our discussion, which is premised on the motions being isolated to the airmass. Second, it starts with a tailwind and turns away from it (equivalent to turning into a headwind), so any purported effect should be an increase in performance and not a decrease! You note the same thing and dismiss the nose down effect as PIO overshoot of an initial nose up.
Last edited by Vessbot; 4th Aug 2018 at 14:56.
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The Downwind Turn
He has a whole website of it, which has been linked to multiple times in the thread:
The Downwind Turn
The Downwind Turn
There, that sounds better.
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In still air, angle d would be zero and therefore Fc sin d would be zero, and therefore Ft would be zero.
Vessbot, et al, you are never going to convert a 'believer' no matter how rational you are. Belief isn't arguable, it's belief. The premise is on its face false; frames of reference are ignored, logical physics arguments are twisted. I really think there is no point feeding this troll as this is where he gets his energy.
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The simple fact is that my work is based on Newtons laws of motion, the triangle of velocities and the aerodynamic forces acting on the aircraft plus gravity
1. Given that you claim to respect Newton’s laws of motion, then you must agree that a calculation performed from the inertial frame of reference of the airmass, an aircraft with a steady angle of bank with steady thrust to overcome drag, will travel in a perfect circle at a steady speed. True or False?
2. You must also agree that any other inertial frame of reference is defined as one that is moving at a constant velocity with respect to the first one (i.e. it is not being acted upon by an external force, so is not accelerating). True or False?
3. You must also agree that Newton’s laws of motion are respected regardless of which inertial frame of reference is being used. True or False?
If you cannot answer True to all three questions, then you are not respecting Newton’s laws of motion, and we cannot possibly move on to what may be right or wrong with your “work” (which you claim is based on Newton’s laws of motion).
If you can answer True to all three, then I am prepared to study your work further, notwithstanding that you haven’t provided a mathematical proof, you have merely programmed a “simulation”.
If you are indeed correct, in order to win that Nobel Prize you will need your work to be peer reviewed, and those peer reviews will hold a lot of weight if verified to be true by your biggest sceptics.
Regards, Fred
Last edited by Derfred; 5th Aug 2018 at 12:46.
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When the wind is changing from one state of equilibrium to another, the air is accelerated and the air-velocity will change by a small amount (as in turbulence) causing changes to the aerodynamic forces.
Last edited by A Squared; 5th Aug 2018 at 17:55.
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Please tell me, does this person work for a well known Canadian simulator company? If it's the same person then, yes he is
a total know all, loads of hours military and civil heavy, but a total Luddite.
a total know all, loads of hours military and civil heavy, but a total Luddite.
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Vessbot, et al, you are never going to convert a 'believer' no matter how rational you are. Belief isn't arguable, it's belief. The premise is on its face false; frames of reference are ignored, logical physics arguments are twisted. I really think there is no point feeding this troll as this is where he gets his energy.
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Let's back up a step and consider in broad terms what he is claiming: That a soaring bird can extract a net energy gain by maneuvering within a uniform moving air mass (ie: not crossing wind gradients as in real dynamic soaring) This is as fundamentally at odds with the basic tenets of physics as is the idea that a ball could bounce higher and higher off an unaccelerated surface with each successive bounce, or perhaps more similarly, a bicyclist, without pedaling, can go faster and faster by repeatedly coasting up and down a hill. That, in a nutshell is what his theory proposes. Yet, he makes absolutely no attempt to understand why this is either impossible, or a refutation of basic principles of physics accepted for centuries. He just keeps reiterating that his theory " is based on the laws of Newton".
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If I think I am right - Why should I admit I am wrong? In fact the original version of my website was based on calculations of energy and was quite unsatisfactory. I listened to my critics and I changed the emphasis to force acceleration and rate of change of momentum Which is F= ma.
Derfred - Do I understand Newtons Laws? Sure I do
Why don't you ask me whether I understand algebra trigonometry aerodynamics etc? Not so sure about those are you?
A Squared - My theory demonstrates how albatross dynamic soaring extracts momentum and energy from the wind. It has nothing to do with the examples you gave. And you are demonstrating your flawed verbal reasoning
My theory is an alternative to the accepted Rayleigh cycle ie wind-gradient dynamic soaring. Wind gradient effects are real but are not sufficient to explain albatross dynamic soaring.
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The Downwind Turn
Derfred - Do I understand Newtons Laws? Sure I do
Why don't you ask me whether I understand algebra trigonometry aerodynamics etc? Not so sure about those are you?
A Squared - My theory demonstrates how albatross dynamic soaring extracts momentum and energy from the wind. It has nothing to do with the examples you gave. And you are demonstrating your flawed verbal reasoning
My theory is an alternative to the accepted Rayleigh cycle ie wind-gradient dynamic soaring. Wind gradient effects are real but are not sufficient to explain albatross dynamic soaring.
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OK, let's talk about what's happening DURING the turn then. Do you notice anything unusual during the turn as you turn through the thousand knot self-created shear within a second? I don't. And my math is right, I just rechecked it:
final groundspeed minus initial groundspeed
(airspeed + tailwind) - (airspeed - headwind)
(1+500)-(1-500)
501-(-499)
1000
No No No You are talking about a person in an aircraft. The person walks toward the tail at 1 kt turns around and walks towards the nose at 1kt. But relative to the ground he is still moving with the aircraft
final groundspeed minus initial groundspeed
(airspeed +tailwind walk) - (airspeed - headwind walk)
(+500+1)-(+500-1)
=501-(499)
=2
If you don't think that's right, compare it with the ealier Cessna example, where the self-created windshear is 20 knots:
final groundspeed minus initial groundspeed
(airspeed + tailwind) - (airspeed - headwind)
(100+10)-(100-10)
110-90
20
OK that's 20 kts (say 10m/s) in say 1 minute at Rate 1 That's 10/60 m/s^2 = 0.17 m/s^2
That's 0.17/9.81 G = 0 .017G longitudinal g loading. You are not going to notice that are you?
Its as I say in my website: the effects are very small and easily missed
For both examples, the final and initial groundspeeds are states of equilibrium. But the 1000 knot and 20 knot wind shears? Those are happening during the turn, just as you say.
The second example is correct but they are both states of equilibrium
The situation that post 116 is a reply to, is a nonstarter. First, it's on an arrival procedure, where your track is anchored to the ground. So it does not apply to our discussion, which is premised on the motions being isolated to the airmass. Second, it starts with a tailwind and turns away from it (equivalent to turning into a headwind), so any purported effect should be an increase in performance and not a decrease! You note the same thing and dismiss the nose down effect as PIO overshoot of an initial nose up.
final groundspeed minus initial groundspeed
(airspeed + tailwind) - (airspeed - headwind)
(1+500)-(1-500)
501-(-499)
1000
No No No You are talking about a person in an aircraft. The person walks toward the tail at 1 kt turns around and walks towards the nose at 1kt. But relative to the ground he is still moving with the aircraft
final groundspeed minus initial groundspeed
(airspeed +
(+500+1)-(+500-1)
=501-(499)
=2
final groundspeed minus initial groundspeed
(airspeed + tailwind) - (airspeed - headwind)
(100+10)-(100-10)
110-90
20
OK that's 20 kts (say 10m/s) in say 1 minute at Rate 1 That's 10/60 m/s^2 = 0.17 m/s^2
That's 0.17/9.81 G = 0 .017G longitudinal g loading. You are not going to notice that are you?
Its as I say in my website: the effects are very small and easily missed
For both examples, the final and initial groundspeeds are states of equilibrium. But the 1000 knot and 20 knot wind shears? Those are happening during the turn, just as you say.
The second example is correct but they are both states of equilibrium
The situation that post 116 is a reply to, is a nonstarter. First, it's on an arrival procedure, where your track is anchored to the ground. So it does not apply to our discussion, which is premised on the motions being isolated to the airmass. Second, it starts with a tailwind and turns away from it (equivalent to turning into a headwind), so any purported effect should be an increase in performance and not a decrease! You note the same thing and dismiss the nose down effect as PIO overshoot of an initial nose up.
The Downwind Turn