The Windward Turn Theory

[Jet_Fan]

ttp://www.dynamic-soaring-for-birds.co.uk/html/windward_turn_theory.html#Windwardturntheory

Utter rubbish!

[oceancrosser]

Windward Turn Theory (http://www.dynamic-soaring-for-birds.co.uk/html/windward_turn_theory.html#Windwardturntheory)

Utter rubbish!

There, I fixed that for you. No comment on the "theory" yet though.

[FE Hoppy]

Why is it utter rubbish?
dynamic soaring on the leeward side of ridges is very popular in the model gliding world. It's principally the same thing.
https://www.youtube.com/watch?v=hFPJ6DUAY10

[Jet_Fan]

Why is it utter rubbish?
dynamic soaring on the leeward side of ridges is very popular in the model gliding world. It's principally the same thing.


He has a new theory. Have a read.

[Jet_Fan]

There, I fixed that for you. No comment on the "theory" yet though.

Thanks. The board won't like me put up links yet.

Looking forward to reading your opinion on this.

The Leeward Turn

The albatross gains momentum in the leeward turn using a component of aerodynamic force to act as a propulsive force. This component provides the acceleration which is seen as an increase in ground-speed rather than airspeed. Thus it gains horizontal momentum and kinetic energy without losing potential energy other than a small drag loss during the turn reversals. This propulsive force is a component of the horizontal resultant which, in turn, is the vector sum of the horizontal component of lift and the drag force.

According to the author, this effect is what produces lift and not the wind gradient.

[Chu Chu]

If you write "Dynamic Soaring for Birds," I guess the target audience is birdbrains.

I think the description of loss and gain of kinetic energy may actually be correct, when viewed from a ground-based frame of reference. But the kinetic energy determined from that frame of reference is wholly irrelevant -- until it's time to land, anyway.

[Wizofoz]

I've actually had significant communications with the author- he tried pushing this barrow on some physics sites I hang around on.

It was explained to him ad nauseum that you cannot gain lift by turning in a constant wind, but he was unreachable by logic and facts.

He's also a "downwind turn" proponent, which tells you all you need to know.

Most worrying thing is he is also still an active instructor......

[abgd]

It was explained to him ad nauseum that you cannot gain lift by turning in a constant wind, but he was unreachable by logic and facts.


The point of dynamic soaring is that the wind isn't constant. Over the sea, there's more of it the higher you go, and there are also gusts, and the ability to slope soar over the crests of waves. I have the flu and my brain is currently too fuggy to contemplate his exact theory though.

[Wizofoz]

The point of dynamic soaring is that the wind isn't constant. Over the sea, there's more of it the higher you go, and there are also gusts, and the ability to slope soar over the crests of waves. I have the flu and my brain is currently too fuggy to contemplate his exact theory though.

He is postulating a reverse of the "Downwind-turn"- he thinks by gaining groundspeed downwind, then turning into-wind, you gain airspeed and can climb.

It's barking, but he won't be told.

[jcomm]

Just out of curiosity:

https://www.youtube.com/watch?v=xlPI0AOj5MI

[Wizofoz]

Dynamic Soaring is completely real, just not the way the gut in the OP believes it to be.

[jimjim1]

Presentation about dynamic soaring by the present speed record holder.

https://www.youtube.com/watch?v=nv7-YM4wno8
about an hour long

I have been aware of apparently ludicrous speed claims on youtube for years and have often wondered if they were actually true.

The speaker in the video, Spencer Lisenby, seems convincing.

519mph and running into transonic effects.

The next glider will be swept wing.

I wonder if there is a theoretical speed limit?

[A Squared]

Dynamic Soaring depends on crossing wind gradients. This guy's theory is about "dynamic soaring" without crossing wind gradients.

Second paragraph from the top.

In the windward turn, the albatross maintains height and loses momentum due to the unbalanced drag force. The loss of momentum is seen as a loss of ground-speed rather than a loss of airspeed. Airspeed is constant because the tendency to lose airspeed due to drag is balanced by the tendency to gain airspeed from the increasing headwind components, whilst turning relative to the wind.

As someone else remarked, the opposite of the "downwind turn". If you start with the belief that you can gain airspeed by turning from crosswind to upwind, you know that everything which follows is nonsense.

[Jet_Fan]

Dynamic Soaring depends on crossing wind gradients. This guy's theory is about "dynamic soaring" without crossing wind gradients.

Second paragraph from the top.



As someone else remarked, the opposite of the "downwind turn". If you start with the belief that you can gain airspeed by turning from crosswind to upwind, you know that everything which follows is nonsense.

Exactly. I don't know why some people insist this is the case. It's almost like a mental illness.

[Chu Chu]

Of course, we're overlooking the real problem with the theory -- it doesn't account for the much larger kinetic energy effect of east and west turns. :}

[Chris Scott]

Quote (my emphasis):
"In a leeward turn, aerodynamic forces combined with a large angle of bank and a large angle of drift provide a propulsive force enabling acquisition of ground momentum and ground kinetic energy without gaining airspeed or losing potential energy. During the climbing part of the wing-over, extra potential energy is gained due to the propulsive force."


What "propulsive force"? I may be missing something (don't answer that!) but, on the face of it, that is gobbledygook. Pity, because the sight of an albatross sustaining flight - mainly in ground-effect - with no apparent form of propulsion is awe-inspiring.

Cannot see how - assuming the wind is horizontal and constant at any given height above the sea - energy can be harvested to maintain flight-sustaining airspeed simply by manoeuvring. Sounds like "perpetual-motion" to me.

[scifi]

Same as Windmills, it is Perpetual Motion..... Until the wind stops.
.

[Basil]

I don't understand and me 'ead 'urts! :confused:

I always thought seabirds were just ridge soaring on waves.

[Chris Scott]

Same as Windmills, it is Perpetual Motion..... Until the wind stops.
.

Presume you are being frivolous? In case not: windmills can harvest energy from the wind because they are anchored to terra-firma Flying machines are not...

[Jet_Fan]

Quote (my emphasis):
"In a leeward turn, aerodynamic forces combined with a large angle of bank and a large angle of drift provide a propulsive force enabling acquisition of ground momentum and ground kinetic energy without gaining airspeed or losing potential energy. During the climbing part of the wing-over, extra potential energy is gained due to the propulsive force."


What "propulsive force"? I may be missing something (don't answer that!) but, on the face of it, that is gobbledygook. Pity, because the sight of an albatross sustaining flight - mainly in ground-effect - with no apparent form of propulsion is awe-inspiring.

Cannot see how - assuming the wind is horizontal and constant at any given height above the sea - energy can be harvested to maintain flight-sustaining airspeed simply by manoeuvring. Sounds like "perpetual-motion" to me.

Yeah, he thinks he’s found an all new force that everyone else has missed.

[Brercrow]

Yeah, he thinks he’s found an all new force that everyone else has missed.

Nothing new here Its the same force that makes your groundspeed increase when you turn downwind and decrease when you turn upwind

[autoflight]

Its obviously Coriolis, So all you northerners please be careful when you do your downwind turns down under

[Wizofoz]

Nothing new here Its the same force that makes your groundspeed increase when you turn downwind and decrease when you turn upwind

No, no it isn't.....

[Jet_Fan]

Nothing new here Its the same force that makes your groundspeed increase when you turn downwind and decrease when you turn upwind

No, not at all where he is coming from. In his rather daft opinion, a headwind increases the speed of the relative wind (airspeed). This is a newbie/layman's mistake, I hear this spoken and see it written all the time, it's a pet hate of mine. He's taken the fact that a takeoff into wind is useful and turned it into a gain in lift whenever you point into wind in free flight. He's an idiot.

The CAA and FAA should do more to dispel this BS. Apparently this guy is/was an instructor with CAA certification. How many more jokers like him out there then?

[Wizofoz]

How many more jokers like him out there then?

I was a keen Aerobatic plot back in the day. I had an "interesting" conversation with no less than the chief pilot of a major airline, also an Aerobat, who insisted you would get a longer vertical line if you pulled into it downwind, as you had more kinetic energy.

Attempts at explaining frames of reference went no where.

And before anyone pipes in, vertical lines were not supposed to be corrected for wind, the judges were supposed to allow for the fact that you would drift.

[Jet_Fan]

I was a keen Aerobatic plot back in the day. I had an "interesting" conversation with no less than the chief pilot of a major airline, also an Aerobat, who insisted you would get a longer vertical line if you pulled into it downwind, as you had more kinetic energy.

Attempts at explaining frames of reference went no where.

And before anyone pipes in, vertical lines were not supposed to be corrected for wind, the judges were supposed to allow for the fact that you would drift.

I'm not entirely surprised that it goes that far up having met so many with the same stupid gene. There should be no hiding place for such people, certainly isn't in the military, but the CAA and FAA aren't interested. I spoke to some retired CAA guy who claimed the whole debate was a 'matter of opinion'.

[A Squared]

The CAA and FAA should do more to dispel this BS.

That's gonna be tough. No dealings with the CAA but I would bet long odds that I could easily find FAA inspectors who believe in the downwind turn myth.

[Basil]

Just keeping it local, speeds approximate and in knots:
At the Equator: 970kn
Earth orbital speed around Sun: 58,000kn
Sun orbital speed around galaxy: 447,000kn (Really?)
Less local:
Galaxy through space: 1,166,310kn

So, when you make a 180 in your 50kn glider you change from going forwards at over a million knots to backwards at the same speed.
Better lock your hold on those flight controls! ;)

p.s. Foregoing a bit of Shiraz induced reductio ad absurdum - did you notice?

[Wizofoz]

That's gonna be tough. No dealings with the CAA but I would bet long odds that I could easily find FAA inspectors who believe in the downwind turn myth.

Richard Collins, editor of the US Flying magazine was a believer, writing articles about it.

Another of his mantras was that if a modification meant an engine delivered more power, it MUST increase fuel burn.

"Efficiency" was apparently a foreign concept to him.

[Jet_Fan]

Richard Collins, editor of the US Flying magazine was a believer, writing articles about it.

Another of his mantras was that if a modification meant an engine delivered more power, it MUST increase fuel burn.

"Efficiency" was apparently a foreign concept to him.

It's the equivalent of flat Earth and Apollo hoax.

[Vessbot]

I was a keen Aerobatic plot back in the day. I had an "interesting" conversation with no less than the chief pilot of a major airline, also an Aerobat, who insisted you would get a longer vertical line if you pulled into it downwind, as you had more kinetic energy.

Attempts at explaining frames of reference went no where.

And before anyone pipes in, vertical lines were not supposed to be corrected for wind, the judges were supposed to allow for the fact that you would drift.

If the vertical line was established by bouncing the airplane off a trampoline mounted to the ground at a 45 degree angle... :ugh:

[A Squared]

Richard Collins, editor of the US Flying magazine was a believer, writing articles about it.

Another of his mantras was that if a modification meant an engine delivered more power, it MUST increase fuel burn.

"Efficiency" was apparently a foreign concept to him.

Interesting, I did not know that about Collins.

[Wizofoz]

Interesting, I did not know that about Collins.

Actually, after a little googling, I think I defamed Collins- it was J McClellan, also a former Flying editor, who wrote an article full of all the non-science associated with the downwind turn myth.

[DaveReidUK]

Actually, after a little googling, I think I defamed Collins- it was J McClellan, also a former Flying editor, who wrote an article full of all the non-science associated with the downwind turn myth.

I know a good lawyer ...

[Wizofoz]

I know a good lawyer ...

There are lots of good lawyers- but do you know a good one who is also alive?

[16024]

In the early days I was struggling with S&L in an out-of-trim Cherokee, and was discouraged from "playing" with the rudder trim on the basis that the tendency for it to keep flying one wing down was because of the crosswind...

[Vessbot]

In the early days I was struggling with S&L in an out-of-trim Cherokee, and was discouraged from "playing" with the rudder trim on the basis that the tendency for it to keep flying one wing down was because of the crosswind...

In recent days in an airliner, (which was a bit crooked) I was approaching to land with some aileron trim to the left, to hold a neutral force. The other pilot pointed out that we're expecting a right crosswind and I'm trimmed against the crosswind. I pointed out that it's trimmed neutral right now (and physically let go of the yoke to show him) and expected him to realize his brainfart. No, he persisted in trying to convince me that "I'm just making my job harder" being trimmed against the crosswind. I wish I was kidding.

[Gauges and Dials]

Someone should ask him:

"Is there any conceivable air data instrument that you could stick onto an aircraft (or a bird) that would tell you what said aircraft or bird was doing relative to the ground?"

"Is there any conceivable air data instrument that you could stick onto an aircraft (or a bird) that would tell you what the wind was doing?"

It's astounding the difficulty people have with the concept of frames of reference.

[Gauges and Dials]

In recent days in an airliner, (which was a bit crooked) I was approaching to land with some aileron trim to the left, to hold a neutral force. The other pilot pointed out that we're expecting a right crosswind and I'm trimmed against the crosswind. I pointed out that it's trimmed neutral right now (and physically let go of the yoke to show him) and expected him to realize his brainfart. No, he persisted in trying to convince me that "I'm just making my job harder" being trimmed against the crosswind. I wish I was kidding.

Take out a pencil and re-label the current position of the aileron trim indicator, "Neutral," wink at the other pilot, and say, "Thanks for catching that."

[Gauges and Dials]

Neither here nor there, but I love birds:

Aren't albatrosses effectively doing wave soaring: positioning themselves in the updraft part of the standing waves that are created when a constant wind blows over an irregular surface? With the irregular surface being the ocean? Constant speed wind striking the face of an ocean wave is very much like constant speed wind striking a mountain ridge: enormous updraft starting from the windward face and extending up and to leeward -- monster downsmashing rotor immediately to leeward of the wave crest / mountain peak. Albatross knows enough to stay in the updraft.

[john_tullamarine]

Aren't albatrosses effectively doing wave soaring

I wouldn't have thought so. The surface irregularities associated with the (comparatively) minor ocean wave profile would be unlikely to produce much perturbation in the wind trajectories.

However, even if you have an absolutely smooth surface for miles and miles around, there will be a near surface wind profile similar to the boundary layer near the surface of an aircraft in flight.

The typical atmospheric boundary layer model is the one-seventh relationship, commonly used in certification and FT work.

There are a few threads speaking to this topic .. for instance, a typical example (https://www.pprune.org/tech-log/10456-737-wind-increment-vref.html?highlight=737+Wind+increment+to+Vref) was looking at speed additives.

Your albatross might be gliding downwind toward the ocean's surface, then wheeling around into wind and climbing into the shear to gain some energy, then wheeling around downwind and so on .. ? The penalty for the bird, of course, is that the gross flightpath inevitably is downwind.

For the ridge/wave soaring folk, one can remain more or less in the same area or, indeed, near motionless ..

[Basil]

Vessbot, Perhaps the rudder was out and the slip indicator was wrong. ;)

Flew half my first B744 sim session without the slip indicator. Finally asked if it had one and instructor pointed to tiny white rectangle :O

[Wizofoz]

Aren't albatrosses effectively doing wave soaring

I wouldn't have thought so. The surface irregularities associated with the (comparatively) minor ocean wave profile would be unlikely to produce much perturbation in the wind trajectories.

However, even if you have an absolutely smooth surface for miles and miles around, there will be a near surface wind profile similar to the boundary layer near the surface of an aircraft in flight.

The typical atmospheric boundary layer model is the one-seventh relationship, commonly used in certification and FT work.

There are a few threads speaking to this topic .. for instance, a typical example (https://www.pprune.org/tech-log/10456-737-wind-increment-vref.html?highlight=737+Wind+increment+to+Vref) was looking at speed additives.

Your albatross might be gliding downwind toward the ocean's surface, then wheeling around into wind and climbing into the shear to gain some energy, then wheeling around downwind and so on .. ? The penalty for the bird, of course, is that the gross flightpath inevitably is downwind.

For the ridge/wave soaring folk, one can remain more or less in the same area or, indeed, near motionless ..

I believe it's been shown that Albatrosses can dynamicly soar and make progress UP wind.

[Gauges and Dials]

Albatrosses stay fairly close to the ocean surface. Speaking as a sailor, I can assure you that what the wind is doing at the top of a 20m (60 ft) mast is often very different from what it is doing at deck level. With wave heights on the open ocean being 5m and up (they don't usually feel so big because they are not steep) it wouldn't surprise me a bit to have a lot of vertical movement of air in the space from sea level to, say, 20 or 30 meters up.

[A Squared]

Albatrosses and the like may be able to gain advantage by exploiting wind gradients and/or air deflected upward by waves. I wouldn't dispute that for a moment.

I am however, fairly confident that they cannot achieve a net energy gain relative to their basic gliding efficiency by turning within a homogeneous body of moving air. As nearly as I can tell, this is what it being claimed by the article linked in the first post.

[Wizofoz]

I am however, fairly confident that they cannot achieve a net energy gain relative to their basic gliding efficiency by turning within a homogeneous body of moving air. As nearly as I can tell, this is what it being claimed by the article linked in the first post.

That is what is being claimed and you are right it isn't possible.

[Jet_Fan]

Albatrosses and the like may be able to gain advantage by exploiting wind gradients and/or air deflected upward by waves. I wouldn't dispute that for a moment.

I am however, fairly confident that they cannot achieve a net energy gain relative to their basic gliding efficiency by turning within a homogeneous body of moving air. As nearly as I can tell, this is what it being claimed by the article linked in the first post.

Yep that’s the false claim.

[john_tullamarine]

I believe it's been shown that Albatrosses can dynamically soar and make progress UP wind.

Such shouldn't be too difficult in suitable conditions with a reasonably strong gradient. A bit similar to tacking in sailing vessels .. but, probably, not an effective way to make progress ? Then, again, I guess that the bird has plenty of time on its hands, so long as the necessary fish for eating keep coming along ..

it wouldn't surprise me a bit to have a lot of vertical movement of air in the space from sea level to, say, 20 or 30 meters up.

Plenty of evidence to support that thought. However, I suspect that the bird will be doing better from the more predictable gradient profile than the more random turbulent mixing ?

[megan]

One study on how they do it.

https://publish.wm.edu/cgi/viewcontent.cgi?referer=https://www.google.com.au/&httpsredir=1&article=1100&context=reports

Spent hour upon hour watching these birds from the ships helideck, failing that, going to the bow and passing the time watching dolphins. Never ever observed the birds attempt to get airborne on no wind days, they'd just paddle out of your way. Only times ever saw them above sea level was heading offshore (flying) and a bird was soaring along in a dry front at 1,500 feet (Golden Beach JT), the other was popping out of a cloud at 500 feet to be confronted by a bird. I assume he was soaring in some localised disturbance that wasn't obvious to us.

[john_tullamarine]

Did a bit of net trolling to see what papers might come to light.

The following story (http://m-selig.ae.illinois.edu/pubs/SukumarSelig-2010-AIAA-2010-4953-DS-OpenFields.pdf) sounds to be reasonably persuasive to me and, probably, is a useful read in conjunction with Megan's link. Don't fuss too much with the mathematics .. that's just there to confuse those who haven't been mathematically anointed with this and that.

To my simple, dumb, engineer's mind, but with a pilot's overlay .. it is a matter of playing cyclically and skilfully with overshoot and undershoot shear within the surface boundary layer. A bit of skill (which the albatross has from however long its genetic lineage might be) and one can run downwind, tack across and into wind according to one's penchant at the time.

Megan, being a flingwing player of very considerable maritime experience would be expected to have a pretty intimate appreciation of what the bird folks might be doing out there over the water ... I note that his link is to an old paper .. which I will contemplate over coffee this evening for interest.

the other was popping out of a cloud at 500 feet to be confronted by a bird.

The occasional benefits of VFR become more evident .. last time I had such an experience was in a glider tug (45 or so years ago) on descent (sort of maintaining VMC, I guess) .. don't you just love gliders skimming along at the cloud base ? How we didn't kill our collective selves as young idiots often causes me no end of wonderment ..

[megan]

In our operation JT VFR stood for Very Frequently Resisted.

[Gauges and Dials]

don't you just love gliders skimming along at the cloud base ? How we didn't kill our collective selves as young idiots often causes me no end of wonderment ..

In the days before gyroscopes, I've heard that our grandparents were taught, "If you find yourself in the clouds, enter a spin. You'll lose altitude (not to mention your lunch) predictably without covering a lot of real estate or bending the airplane; if you break clear with enough altitude you can recover, and if you don't you were probably going to die anyhow."

I can't decide if the appropriate comment is, "that's bold, not old" or "Canvas airships and iron pilots."

[Gauges and Dials]

A bit similar to tacking in sailing vessels


Similar, but with an important distinction: in either case you need an inertial frame of reference that is decoupled from the moving air. A sailboat achieves this by sticking a big fin (keel or centerboard) into the water; conceivably an albatross achieves this via physical inertia while transiting the transition between layers of air that are moving in different directions relative to each other.

Ultimately their aint no free lunch, though. unless you continue to see a sharp gradient between wind velocity vectors across a very short distance, you're out of harvestable energy. I'll bet (being way to lazy to figure this out) that the albatross is fully at one with the "new" wind by the time it is 50 meters or so into it, and has run out of inertia imparted by the "old" wind by then.

All this aside, let's pause for a moment to contemplate how spectacularly beautiful the albatross is, eh?

[Pugilistic Animus]

Yes I heard about the 'spin out of the cloud method" from the autobiography of an airmail pilot...back then way before 14CFR 91.3 the Post Office decided when you can fly but that all changed with the Avigation Act of 1926

[DavidN]

Just want to start with, I'm not a pilot, just an average joe with an engineering background so I welcome any criticisms and explaining you feel like (theres no hard feelings on the internet right). I've been reading a lot about downwind turn myth stuff and just wanted to put out how i see it and see if ive got it right. Also want to pre-empt; im not including drag forces and other stuff thats relevant for real world applications here, this is just about clearing up the concepts that have been thrown around.

- I dont agree with the author of the OP link,
- there is no change in the magnitude ground speed of the aircraft when you turn into/out of wind (assuming no motor, no drag, no friction, ignoring increase in speed due to gravity or conversion of gravitational potential energy into KE); I think the big problem is that people forget about a direction change does not equal change in magnitude.
- upon changing headwind/tailwind you will change your "true airspeed" (as far as i know true airspeed is defined as velocity of airmass relative to the aircraft).

- DS relies on wind shear gradient,
- all DS flight patterns revolve around climbing in headwind and descending in tailwind
Ok so here im sort of guessing;
DS takes advantage of
- flying into headwind which assists in generating lift due to higher true airspeed flowing over the wings (assume you turn before you stall)
- flying into tailwind does impart a little kinetic energy to the aircraft (might be one of these or a combination or maybe something else)
- lower true airspeed results in less drag forces so you get higher efficiency in converting potential gravitational energy into KE
- I assume that as the heading of the aircraft changes (during the turn) some portion of the headwind is being used as a propulsive force and not a lift gaining force (AoA and
pitch/yaw will make this relative but you get the idea).

Its the relatively high frequency of loops over short 'straight legs' which allows to DS work in the real world as 'straight' legs dont net you any increase in energy while drag will still occur. Said another way is; its the turns which could theoretically(?) add energy coupled with the wind gradient which allows for more efficient energy conversion between height and momentum which DS operates on. Side note; unless there is negative true airspeed (from trailing edge to leading edge) there can be no propulsive force exerted by the wind and lets just ignore this scenario.

By all means let me know if I've used jargon incorrectly, forgot something or theres a flaw in my logic. I'm just here to learn from people more knowledge and experience than me.

Cheers guys

[itsnotthatbloodyhard]

- there is no change in the magnitude ground speed of the aircraft when you turn into/out of wind (assuming no motor, no drag, no friction, ignoring increase in speed due to gravity or conversion of gravitational potential energy into KE); I think the big problem is that people forget about a direction change does not equal change in magnitude.
- upon changing headwind/tailwind you will change your "true airspeed" (as far as i know true airspeed is defined as velocity of airmass relative to the aircraft).

David, unless I’m misreading this somehow (and I apologise if I am), it looks suspiciously like a rewording of the downwind turn myth. If you keep all your other assumptions, but change ‘no motor’ to ‘motor set at stable setting for a constant turn’, do your conclusions change at all?

Supposing you fly an ordinary powered aircraft at altitude in a constant 30 degree banked turn, in a constant airmass. You don’t know what the wind velocity is, and don’t care. The aircraft will go round and round in circles. The indicated airspeed (and true airspeed) won’t change at all. The magnitude of the groundspeed will be constantly changing, as long as the wind velocity is greater than zero. The position of your orbits will be moving across the ground due to the wind, but the aircraft doesn’t know this.

Most pilots would agree with all the above, yet a few of those same pilots (some with 20000+ hrs) still believe that their indicated airspeed will increase as they turn onto final into a strong wind. I’m not sure how the aircraft now knows it’s turning towards a runway, and they never seem able to explain this.

[pineteam]

A friend of mine gave me a simple example to understand it: Imagine you are on walking on a transportation belt moving at 30 km/h. No matter if you turn or not, your speed reference to the belt will be exactly the same as long as the belt keeps a steady 30 km/h speed.

[A Squared]

- there is no change in the magnitude ground speed of the aircraft when you turn into/out of wind

The magnitude of your groundspeed certainly *does* change turning into or out of the wind.

[FlightDetent]

It's inertia, INTBH. Those 20k+ pilots must have flown commercially, i.e. heavier planes, where the effect can be seen. Although like you say the craft is moving within a practically uniform parcel of air, Newton's laws still apply. The inertial system is the spheres, so the physics is there.

I have no claim how pronounced or measurable the effect is on those beauties with low wing loading, but our aircraft do show. Level flight, fixed thrust, steady heading with 90° cross-wind: there is a difference in which way you'd turn.

[itsnotthatbloodyhard]

It's inertia, INTBH. Those 20k+ pilots must have flown commercially, i.e. heavier planes, where the effect can be seen. Although like you say the craft is moving within a practically uniform parcel of air, Newton's laws still apply. The inertial system is the spheres, so the physics is there.

I have no claim how pronounced or measurable the effect is on those beauties with low wing loading, but our aircraft do show. Level flight, fixed thrust, steady heading with 90° cross-wind: there is a difference in which way you'd turn.

So the downwind turn myth isn’t a myth after all? Oh dear.

I fly the same heavy commercial jets (and previously, things with much higher wing loading again), and contend that the effect can’t be seen, because it doesnt exist.. When you’re flying a holding pattern up in a jetstream, what happens to your IAS - a.) alternating stall and overspeed warnings, or b.) nothing ?

[FlightDetent]

I cannot comment on your eye-sight, and have no desire to enter that dispute you are having above. It's just physics, especially knowing the "frame of reference" of kinetic energy and vectors. And no, my claim does not justify any of those "create energy out of thin air" theories, which a consider just bags of hot air.

Towards the example you gave above: I only held in a jet-stream once, and while the effects were not as extreme as your hyperbolic choice of words, in a gist you are correct. But that's because the FMS was trying to fly a fixed pattern over the ground. So that should not be relevant here. Or is it?

[hans brinker]

It's inertia, INTBH. Those 20k+ pilots must have flown commercially, i.e. heavier planes, where the effect can be seen. Although like you say the craft is moving within a practically uniform parcel of air, Newton's laws still apply. The inertial system is the spheres, so the physics is there.

I have no claim how pronounced or measurable the effect is on those beauties with low wing loading, but our aircraft do show. Level flight, fixed thrust, steady heading with 90° cross-wind: there is a difference in which way you'd turn.
I think your problem is that you can’t see air, and can’t stop seeing ground. If you are flying at 60kts, with a 60kts crosswind, a turn into the wind gets you a ground speed of 0kts (don’t worry, you won’t stall), if you then turn towards tailwind it will give you a 120kts ground speed (don’t worry, you won’t over speed). While all this happens, your airspeed and tas stay the same. How I try to explain it to new pilots: As soon as you are out of ground effect the wind doesn’t impact HOW the airplane flies, the wind is just something that moves the ground underneath the air. If you are above a layer of clouds you have no idea what the wind on the ground is, neither does your plane. The only thing the wind does, it moves your destination, so you have to take it into account for navigation.
The part where people in the pattern get into trouble is by not anticipating the wind and correcting their pattern for it. People spin/stall turning base/final because the picture is different with strong tailwind than calm air and if you don’t start your turn early you will have to increase bank angle, you are faster across the ground so you slow down......
with a tailwind on base, keep your pattern wide, and look at your INDICATED speed.

[itsnotthatbloodyhard]

FlightDetent, I’m sorry, but when it comes to IAS, the frame of reference that matters is the air mass you’re flying in. It doesn’t matter whether you fly fixed holding patterns or shifting orbits, the effect of a constant wind on IAS is the same - nothing (except that to fly a fixed racetrack you’ll sometimes have to tighten or relax the turn to allow for the wind, with a consequent change in induced drag).

If what you believe about wind and inertia were true, it would also change a whole lot about close-in air combat. For the last hundred years, fighter pilots have been doing their utmost to maximise their turn performance and energy state, in an effort to gain every tiny advantage possible. Yet never the slightest thought for the supposed effects of turning in or out of the wind. Perhaps as well as bewaring the Hun in the sun, they should have been worried about the Hun attacking from downwind? :)

Hans has it right.

[A Squared]

So the downwind turn myth isn’t a myth after all? Oh dear.

I fly the same heavy commercial jets (and previously, things with much higher wing loading again), and contend that the effect can’t be seen, because it doesnt exist.. When you’re flying a holding pattern up in a jetstream, what happens to your IAS - a.) alternating stall and overspeed warnings, or b.) nothing ?




c) Same thing which happens holding in still air, which is that the IAS bleeds off during the turn then increases during the leg, then bleeds off during the turn. lather, rinse repeat. In my airplane (> 100,000 lb, not equipped with autothrottles) this is easily noticeable.

[A Squared]

It's inertia, INTBH. Those 20k+ pilots must have flown commercially, i.e. heavier planes, where the effect can be seen. Although like you say the craft is moving within a practically uniform parcel of air, Newton's laws still apply. The inertial system is the spheres, so the physics is there.

I have no claim how pronounced or measurable the effect is on those beauties with low wing loading, but our aircraft do show. Level flight, fixed thrust, steady heading with 90° cross-wind: there is a difference in which way you'd turn.

Lets start with basics. Assume a 100 kt airplane flying east in still air with 100 kt groundspeed (obviously). The airplane turns to a west heading. groundspeed is now 100 kt West

What is the velocity change from the east heading to the west heading?

Now, same airplane flying East in a 50 kt wind out of the East (direct headwind) at 50 kt GS. The airplane turns to a west heading. Groundspeed is now 150 kts.

What is the velocity change in this example?

[FlightDetent]

A^2, do you really insist that flying a fixed pattern over the ground, there is no difference between still air and flowing air conditions?

[jack11111]

The "downwind turn danger myth" is as difficult to beat out of believers as the "sky god myth".

Fly a constant bank turn in a moving air mass, no change in power, no change in attitude equals no change in airspeed. However, you will drift over the landscape.

Don't believe me, try it.

[A Squared]

A^2, do you really insist that flying a fixed pattern over the ground, there is no difference between still air and flowing air conditions?

No, I did not say that. Flying a hold is not a fixed pattern over the ground. It is two standard rate turns, one of them initiated at a fixed point, connected with straight legs. That is not the same as flying a fixed track over the ground. In fact, your track over the ground will be different with a significant wind than in still air. How exactly they are different depends on the direction and strength of the wind. However, for a wind aligned with your holding track, the airspeed indication in your cockpit will be identical to the airspeed indication for the same hold in still air. .


I'm waiting for your answer to my questions above. To wit:

Lets start with basics. Assume a 100 kt airplane flying east in still air with 100 kt groundspeed (obviously). The airplane turns to a west heading. groundspeed is now 100 kt West

What is the velocity change from the east heading to the west heading?

Now, same airplane flying East in a 50 kt wind out of the East (direct headwind) at 50 kt GS. The airplane turns to a west heading. Groundspeed is now 150 kts.

What is the velocity change in this example?

[jonkster]

If the downwind myth were true you would notice some weird effects inside trains and cars and aircraft. Inside a train/car/aircraft the occupants and all inside it are carried along with the train/car/aircraft (much the same as an aircraft travels with the air in a constant wind).

If you were to throw a paper plane that circled inside the carriage when a train was waiting at the platform it might circle nicely. If the myth were true and the train was travelling at speed, the paper plane would now lose or gain airspeed as it turned towards the back or front of the train in its orbits (because it would have 'inertia' relative to the earth that needs to be 'made up' or 'lost' and so the plane would climb and descend as it gained and lost airspeed).

Similarly if you walked about a train's or aircraft's aisle in a circle, you would stumble about as you changed direction as you 'gained' and 'lost' inertia depending on the direction you were walking. Assuming the train or aircraft is travelling at a constant speed and direction, that is not what happens. Even at high speeds, you can comfortably move about the cabin without fighting inertia due to your 'inertia' relative to the earth.

Aircraft fly relative to the air around them, not the earth, same as passengers in constantly moving vehicles measure their practical (ie what it feels like) inertia relative to the vehicle not to the earth.

[Goldenrivett]

Hi FlightDetent,
Newton's laws still apply. The inertial system is the spheres, so the physics is there.

I agree, Newton's laws still apply. But you have to be careful to consider what the momentum and energy is relative to.
If the downwind myth were true, then doing a loop in a glider starting into a headwind which matched the flying speed (so ground speed = 0) would be impossible.
What kinetic energy (when ground speed = 0) is being converted into the height gain?
Would you gain more height if you start the loop downwind?

[eckhard]

Further to Jonkster’s post above, I humbly re-submit an earlier post of mine:

“You wake up, not sure where you are. You see that you are in a small, windowless room. Apart from your chair, the room is empty.

On the floor, you find a small model aeroplane with a battery-powered motor driving counter-rotating propellers. A simple instruction sheet tells you that when the motor is switched on and the 'plane is hand-launched, the controls are fixed so that it will fly perfect circles until a timer turns the motor off after 30 seconds.

You launch the model and watch, pleased and impressed, as it flies circles around you.

After the flight has ended, you notice that there is actually a window blind in the wall behind you. You open the blind and are amazed to find that you are in fact inside a carriage being pulled by a train along a perfectly straight, smooth track at 60mph.

You consider the fact that the model aircraft was flying in a parcel of air which was moving across the ground at 60mph. To an observer not on the train, it was, in effect, flying in a 60mph wind. This is a speed which is about five times greater than its own flight speed, yet the model did not seem to be affected at all! It did not, for example, exhibit any signs of 'stalling' or losing height when it was turning 'downwind'.

You come to the conclusion that the model was simply unaware of its location and speed relative to the tracks, as were you until you looked out of the window.”

[Wizofoz]

Further to Jonkster’s post above, I humbly re-submit an earlier post of mine:

“You wake up, not sure where you are. You see that you are in a small, windowless room. Apart from your chair, the room is empty.

On the floor, you find a small model aeroplane with a battery-powered motor driving counter-rotating propellers. A simple instruction sheet tells you that when the motor is switched on and the 'plane is hand-launched, the controls are fixed so that it will fly perfect circles until a timer turns the motor off after 30 seconds.

You launch the model and watch, pleased and impressed, as it flies circles around you.

After the flight has ended, you notice that there is actually a window blind in the wall behind you. You open the blind and are amazed to find that you are in fact inside a carriage being pulled by a train along a perfectly straight, smooth track at 60mph.

You consider the fact that the model aircraft was flying in a parcel of air which was moving across the ground at 60mph. To an observer not on the train, it was, in effect, flying in a 60mph wind. This is a speed which is about five times greater than its own flight speed, yet the model did not seem to be affected at all! It did not, for example, exhibit any signs of 'stalling' or losing height when it was turning 'downwind'.

You come to the conclusion that the model was simply unaware of its location and speed relative to the tracks, as were you until you looked out of the window.”

Beautifully put.

[ShyTorque]

Further to Jonkster’s post above, I humbly re-submit an earlier post of mine:

“You wake up, not sure where you are. You see that you are in a small, windowless room. Apart from your chair, the room is empty.

On the floor, you find a small model aeroplane with a battery-powered motor driving counter-rotating propellers. A simple instruction sheet tells you that when the motor is switched on and the 'plane is hand-launched, the controls are fixed so that it will fly perfect circles until a timer turns the motor off after 30 seconds.

You launch the model and watch, pleased and impressed, as it flies circles around you.

After the flight has ended, you notice that there is actually a window blind in the wall behind you. You open the blind and are amazed to find that you are in fact inside a carriage being pulled by a train along a perfectly straight, smooth track at 60mph.

You consider the fact that the model aircraft was flying in a parcel of air which was moving across the ground at 60mph. To an observer not on the train, it was, in effect, flying in a 60mph wind. This is a speed which is about five times greater than its own flight speed, yet the model did not seem to be affected at all! It did not, for example, exhibit any signs of 'stalling' or losing height when it was turning 'downwind'.

You come to the conclusion that the model was simply unaware of its location and speed relative to the tracks, as were you until you looked out of the window.”


The railway carriage effect, although a very good analogy, doesn't take into account gusts, which occur in the "real" outside world. To provide a few strong gusts, briefly open and shut the front door of the train and then see what happens.

A small, lightweight model aircraft has very low inertia and will easily be blown "downwind" during gusts, but see only relatively small variations in IAS.
A very large aircraft, such as an airliner, will see more variations in IAS but tend to follow the original flight path more closely.
Hence windshear being more of a concern on the approach in a very large aircraft.

The energy "found" during the previous video is being harvested as lift from updraughting airflow and then being converted to airspeed as the aircraft is descended on the "still air" side of the hill.

[Gadget freak]

Try throwing a ball into the air while driving along in a car. Just after throwing it up have the driver turn sharply. What does the ball do, stay within the cars frame of reference, moving with the air mass? Now try doing the same with a sheet of paper, something with less inertia and more air resistance. Same result?

If you leave a loose article in the cockpit during turbulence does it stay fixed to the aircraft’s plane of reference, moving with the air mass in the cockpit? Why do we think that that an inertial/gravitational frame of reference works differently laterally to vertically?

I think you are all just arguing about the relative balance of inertial to aerodynamic forces. All sort of depends how much inertia you have and how much lift/drag you have.

[itsnotthatbloodyhard]

Try throwing a ball into the air while driving along in a car. Just after throwing it up have the driver turn sharply. What does the ball do, stay within the cars frame of reference, moving with the air mass? Now try doing the same with a sheet of paper, something with less inertia and more air resistance. Same result?

That’s no longer the same discussion. We’re talking about the behaviour of aircraft in a constant airmass. Using car analogies is fine, but once you start swerving around in the car, it’s no longer analogous to a constant airmass.


Why do we think that that an inertial/gravitational frame of reference works differently laterally to vertically?

We don’t.


I think you are all just arguing about the relative balance of inertial to aerodynamic forces. All sort of depends how much inertia you have and how much lift/drag you have.

Er, really?
Eckhard’s post was about the most simple and elegant explanation of all this that you’re likely to read, and should’ve put an end to the discussion. Yet here we are.

[jonkster]

Try throwing a ball into the air while driving along in a car. Just after throwing it up have the driver turn sharply. What does the ball do, stay within the cars frame of reference, moving with the air mass?

It will move sideways (relative to the car). Once not in contact with anything in the car, the ball is not accelerated by the car as the car accelerates sideways in the turn and will maintain its straight line motion (relative to the earth) as it is not being accelerated by any force (other than minutely by air resistance).

In your analogy it appears the ball would be the aircraft and car and the "air" contained in the car would be the wind, is that correct?

If so, in that analogy, by turning the car rapidly, you are causing a sudden change in the direction of the "wind", not a change in direction of the "aircraft".

Of course if the wind direction or magnitude changes rapidly there will be IAS changes and that is seen when flying in turbulence or with windshear.



Perhaps we need to define the myth we are arguing against or the phenomena we are arguing for.

If we are talking about turbulence, varying wind speeds, windshear situations etc then of course, IAS will vary as we encounter changing wind velocities.

If the wind is steady and we perform a constant rate turn in a wind, the myth I will argue against, is that the airspeed will vary as the aircraft changes direction from flying the into wind direction to the downwind direction. In a constant wind, it won't.

I spent many hours in my early flying years low to the ground circling around stock in many different wind conditions from dead still to steady flow to strong and gusty days with lots of thermals. Always I found the aircraft flew relative to the air, any variations in airspeed were due to turbulent flow and didn't limit themselves to any particular part of the orbit.

Early on I had to learn to resist the very strong urge to over bank or over rudder the aircraft in strong steady winds due to the illusion the aircraft was skidding or slipping when orbiting low level in a wind. There be dragons.

[Gadget freak]

That’s no longer the same discussion. We’re talking about the behaviour of aircraft in a constant airmass. Using car analogies is fine, but once you start swerving around in the car, it’s no longer analogous to a constant airmass.




We don’t.




Er, really?
Eckhard’s post was about the most simple and elegant explanation of all this that you’re likely to read, and should’ve put an end to the discussion. Yet here we are.

I humbly suggest you have a think about the basic forces that are keeping you attached to the planet as it carves it's way through the universe.

Eckhard's post doesn't really explain the basic physics because the train isn't accelerating.
Try doing the same as the train slows, accelerates or goes round a corner. is the aircraft now fixed to the moving air mass?

[Gadget freak]

Jonkster,
I'm not trying to argue for or against any myths or theories, just trying to get the basic physics straight.

An aircraft going downwind at 120kts airspeed, into a 60 kt headwind with 180 kts groundspeed tailwind has a lot more kinetic energy and inertia than one going at 120 kts airspeed into a 60 kt headwind with only 60 kts groundspeed. Compare the braking distances if you try and land off those conditions. The question is so what? When turning downwind that extra kinetic energy has to come from somewhere and over the course of a downwind turn the aerodynamic forces provide that acceleration without drama.
However, someone above suggested trying a loop in a glider into a strong headwind starting with zero knots groundspeed. Try that and let us know how you get on?

[A Squared]

Eckhard's post doesn't really explain the basic physics because the train isn't accelerating.

Neither is the wind. The Downwind theory myth is that an airplane in a steady, un-accelerated wind, (no gust, no turbulence, no wind shear) turning downwind will lose airspeed. A railroad train traveling at a constant velocity on a straight track is perfectly analogous to a steady, gust-less, shear-less wind.

[A Squared]

Jonkster,
I'm not trying to argue for or against any myths or theories, just trying to get the basic physics straight.

An aircraft going downwind at 120kts airspeed, into a 60 kt headwind with 180 kts groundspeed tailwind has a lot more kinetic energy and inertia than one going at 120 kts airspeed into a 60 kt headwind with only 60 kts groundspeed.

Same two questions that Flight Detent refused to answer:

100 kt airplane, 50 knot wind out of the east. Airplane flies east into wind with 50 knot groundspeed, then turns 180 degrees and flies west with a 150 knot groundspeed.

What is the velocity change from upwind to downwind.

Same airplane, no wind. Flies East at 100 kt, turns 180 degrees flies west at 100 knots

What is the velocity change from eastbound to westbound?

[Fitter2]

Same two questions that Flight Detent refused to answer:

100 kt airplane, 50 knot wind out of the east. Airplane flies east into wind with 50 knot groundspeed, then turns 180 degrees and flies west with a 150 knot groundspeed.

What is the velocity change from upwind to downwind.

Same airplane, no wind. Flies East at 100 kt, turns 180 degrees flies west at 100 knots

What is the velocity change from eastbound to westbound?

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?

[Gadget freak]

Velocity can be measured relative to whatever frame of reference you chose. Flight dynamisists are used to working with “body axis” or “earth axis” references.

So the answer is that the air speeds (velocities relative to the bulk air mass) are constant while the ground speeds ( velocities relative to the earth) are not. Kinetic energy is relative to an inertial frame of reference and the V in 0.5*m*V^2 is velocity relative to the earth so the aircraft is being accelerated by the air mass that is moving relative to the ground.

[itsnotthatbloodyhard]

However, someone above suggested trying a loop in a glider into a strong headwind starting with zero knots groundspeed. Try that and let us know how you get on?

:eek:

As long as you enter the loop with whatever indicated airspeed your glider requires for a loop, you’ll be just fine. You’re surely not suggesting otherwise? :eek:

[Goldenrivett]

Velocity can be measured relative to whatever frame of reference you chose. Flight dynamisists are used to working with “body axis” or “earth axis” references.

So the answer is that the air speeds (velocities relative to the bulk air mass) are constant while the ground speeds ( velocities relative to the earth) are not.
Correct.

Kinetic energy is relative to an inertial frame of reference and the V in 0.5*m*V^2 is velocity relative to the earth so the aircraft is being accelerated by the air mass that is moving relative to the ground.
Not quite correct. The V^2 can be relative to whatever reference frame you want. In free flight the easiest reference frame is the air mass you are flying in. Hence it is possible to do vertical loops in a glider starting with zero ground speed (when head wind = flying speed).
If the air mass is moving uniformly across the surface of the earth - then there is no acceleration.

[Fitter2]

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).

[Goldenrivett]

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.
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/StarChild/questions/question18.html

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

[jonkster]

Jonkster,
I'm not trying to argue for or against any myths or theories, just trying to get the basic physics straight.

An aircraft going downwind at 120kts airspeed, into a 60 kt headwind with 180 kts groundspeed tailwind has a lot more kinetic energy and inertia than one going at 120 kts airspeed into a 60 kt headwind with only 60 kts groundspeed. Compare the braking distances if you try and land off those conditions. The question is so what? When turning downwind that extra kinetic energy has to come from somewhere and over the course of a downwind turn the aerodynamic forces provide that acceleration without drama.
However, someone above suggested trying a loop in a glider into a strong headwind starting with zero knots groundspeed. Try that and let us know how you get on?
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.

[911slf]

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...

[Wizofoz]

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.

[itsnotthatbloodyhard]

And thanks for a very interesting perspective that most of us would never see.

[A Squared]

So the answer is that the air speeds (velocities relative to the bulk air mass) are constant while the ground speeds ( velocities relative to the earth) are not. Kinetic energy is relative to an inertial frame of reference and the V in 0.5*m*V^2 is velocity relative to the earth so the aircraft is being accelerated by the air mass that is moving relative to the ground. 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.

Kinetic energy is relative to an inertial frame of reference ... 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?

[Chu Chu]

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.

[A Squared]

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.

[Goldenrivett]

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.

[eckhard]

Any individual body (thing) has an “infinite” number of different values of kinetic energy and gravitational potential energy, assuming that there are an “infinite” number of other bodies in the universe.

I am sitting at rest at a fixed table. My k.e. with respect to the table and the earth is zero. My g.p.e with respect to the chair-seat is zero.

The earth is moving towards a “stationary” planet at a speed of 20km per second, several parsecs distant. My k.e. with respect to the planet is enormous. My g.p.e with respect to the planet is also huge, as I would have the opportunity to accelerate to a very high speed before I hit it, should I be unlucky enough to fall towards it, without any other outside influence acting on me.

It’s helpful to ignore all these other possible frames of reference when discussing movement within a parcel of air.

Of course, the real atmosphere is untidy and non-uniform, especially close to the ground. Accelerations and turbulence take place, but we must try to keep a clear mind.

[Chu Chu]

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).

[A Squared]

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.

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: So, if the change in velocity is identical, then the change in kinetic energy must be identical. 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.

[A Squared]

Of course, the real atmosphere is untidy and non-uniform, especially close to the ground. Accelerations and turbulence take place, but we must try to keep a clear mind.

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.

[eckhard]

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.
I agree! Thank you for stating it so clearly.

[currawong]

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....

[itsnotthatbloodyhard]

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.



Sadly, I think this thread demonstrates that it’ll never become extinct. :(

[Bergerie1]

911slf,

You are indeed right about the clever use by birds of the wind gradient with altitude. I have spent hours in the southern ocean watching albatross using this method and also using ridge-lift over huge waves. Birds know more about flying than we ever will!!

https://www.researchgate.net/publication/320597531_Albatross-Like_Utilization_of_Wind_Gradient_for_Unpowered_Flight_of_Fi xed-Wing_Aircraft/fulltext/5a25fb04aca2727dd880f345/320597531_Albatross-Like_Utilization_of_Wind_Gradient_for_Unpowered_Flight_of_Fi xed-Wing_Aircraft.pdf?origin=publication_detail

[hans brinker]

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....

Sadly, I see the GPS made it worse, they make the downwind turn and see the groundspeed increase, and think they have increased momentum...............

[Wizofoz]

Our friend is doubling down.

He has now directly championed the downwind turn myth.

The Downwind Turn (http://www.dynamic-soaring-for-birds.co.uk/html/the_downwind_turn.html)

[zzuf]

Been thinking that it is probably possible to construct a simple analogy for this, though I haven't put any real thought into it.
It involves the idea that, in aerodynamics, it doesn't matter if the aircraft is moving through the air or the air is moving past the aircraft.
Anyway, it involves a large stationary cube of air with a solid floor (the ground) and a solid ceiling.
The floor is moving from (say) left to right and the ceiling is moving from right to left, both at the same speed WRT the air cube.. This being the equivalent of equal speed winds blowing in opposite directions.
Place an aircraft in the stationary air cube, in a constant turn at a fixed bank angle and CAS.
So according to the turning myth, with reference to both the floor and the ceiling, the aircraft is turning both upwind and downwind simultaneously, and according to the theory of the myth it would be both accelerating and decelerating simultaneously.
This cannot possibly be correct.
But as I said I have put no real thought into this.

[currawong]

"The Downwind Turn".

The man is clearly a fool. If the downwind turn was so deadly I would be dead many thousands of times over.

And to some it might be deadly, if susceptible to disorientation.

I would like to see his take on loading it to a knife edge, right to the region of reversed command. Lose a knot and you are going down. Any turn other than into wind would bring you unstuck, if his theory is correct.

[Wizofoz]

Been thinking that it is probably possible to construct a simple analogy for this, though I haven't put any real thought into it.
It involves the idea that, in aerodynamics, it doesn't matter if the aircraft is moving through the air or the air is moving past the aircraft.
Anyway, it involves a large stationary cube of air with a solid floor (the ground) and a solid ceiling.
The floor is moving from (say) left to right and the ceiling is moving from right to left, both at the same speed WRT the air cube.. This being the equivalent of equal speed winds blowing in opposite directions.
Place an aircraft in the stationary air cube, in a constant turn at a fixed bank angle and CAS.
So according to the turning myth, with reference to both the floor and the ceiling, the aircraft is turning both upwind and downwind simultaneously, and according to the theory of the myth it would be both accelerating and decelerating simultaneously.
This cannot possibly be correct.
But as I said I have put no real thought into this.

Yes, similar thought experiments have been put to down-wingians, such as "you are in an empty oil-tanker which is moving at X-knots. Thus the air inside is moving relative to the earth a x-knots. Will a model aeroplane turning downwind inside the tank suffer the "downwind turn" phenomenon.

Tha response is usually "yes.....errr.....no....errr......that's not relevent" followed by a quick subject change.

[Jet_Fan]

I see this thread has been reignited by a non flying theorist and then another non flying theorist stepped up. Then, after being soundly beaten they run away. Did they learn anything?

[autoflight]

the earth is not flat
aircraft is actually fly through air even if the air itself is moving
Newton’s Laws of Motion apply
the rotational speed of the earth at the equator,
the speed of light
the universe is expanding
I believe in the above, but only two apply. A tank full of air in a truck turning a corner with a drone inside or throwing a ball in the air in a speeding train and a few other things previously claimed do not apply.

In the early days of windshear understanding, we used a technique of increasing approach speed with the expectation of loss of IAS due to known loss of headwind closer to touchdown. Loss of headwind and then increasing tailwind is an approach that many experienced pilots would tend to avoid. My modern airliner might even warn of the the danger with a wind display or a warning. Classic horizontal windshear on approach is flight between areas with changing winds with a particularly bad case being loss of the entire headwind and increasing tailwind, with urgent action required to survive.

If the above can be reasonably accepted, we are almost there.

So what has windshear got to do with a simple downwind turn? In both cases the aircraft is flying into a reducing headwind and in the case of the downwind turn, we have set the circumstances to be eventually an equal tailwind.

In the downwind turn we are deliberately flying from an area with a headwind to an area with a tailwind. We are effectively creating our own windshear.

On approach, windshear might develop over maybe 5 - 10 seconds with urgent action required, but our downwind turn might take a minute. Depending on the actual wind for our downwind turn, little or no pilot response may be required. If the wind was 50K, and the aircraft was already low and slow, the loss of IAS would be more noticeable.

Fortunately for aviation, turns would normally be around rate 1. The minute taken to more gently progress from a 50K headwind to a 50K tailwind gives the aircraft auto flight systems the opportunity to recover some of the IAS loss. If our 200K IAS aircraft could make that same 180 degree downwind turn in 30 seconds, recovery of lost IAS would take more effort.

My downwind turn is simply a 180 degree turn from a headwind to a tailwind. It could be at any altitude and does not necessarily refer to a light aircraft turning from take-off heading to downwind in a circuit pattern.

[A Squared]

.

So what has windshear got to do with a simple downwind turn? In both cases the aircraft is flying into a reducing headwind and in the case of the downwind turn, we have set the circumstances to be eventually an equal tailwind.

In the downwind turn we are deliberately flying from an area with a headwind to an area with a tailwind. We are effectively creating our own windshear.
.

No you are not. In the real winshear examples you are crossing wind gradients. In a turn to downwind in a constant wind you are not crossing any wind gradient. A wind gradinent is the fundamental element in windshear and it doesn’t exist in a turn in constant wind. To be more specific, if you are flying into a 20 knot wind out of the east and you turn west, you have not crossed any wind gradient, because gradient means change and the wind hasn’t changed one iota. The wind is still blowing the same 20 knots in the same direction, out of the east.

So, no, turning downwind is nothing like windshear

[Vessbot]

So what has windshear got to do with a simple downwind turn? In both cases the aircraft is flying into a reducing headwind and in the case of the downwind turn, we have set the circumstances to be eventually an equal tailwind.

In the downwind turn we are deliberately flying from an area with a headwind to an area with a tailwind. We are effectively creating our own windshear.


No, we're not. A Squared beat me to it. In wind shear, you're transitioning from one air mass to another, in a small enough timespan that the airplane's inertia temporarily preserves groundspeed, and thereby changing your airspeed. If given enough time after the transition to reach steady state, the plane would stabilize at the original airspeed.

In a turn within the same airmass, there is nothing affecting the airspeed other than drag (the same way that drag would affect your airspeed if the airmass were not moving, i.e., zero wind). Rate of turn does not matter for the ability to recover lost airspeed, because there is no airspeed loss to recover. If it was a function of a gradual enough rate of turn, as you claim, that implies that there is some rate of turn/wind combo for which the "graduality" safety effect would not work, and there would be a significant airspeed loss. How about when you are walking upwind at 1 knot at knots in an airliner, toward the tail, with a groundspeed of negative 499 knots. You turn around to face the nose (downwind) in one second, and keep walking at 1 knot airspeed, but now your groundspeed is 501 knots. You just walked through a one thousand knot self-created windshear in one second, according to your theory, which predicts that there should be some effect. But there is none. When should this effect start kicking in? Remember that your theory also has to account for the "groundspeeds" around the center of the Earth, around the Sun, around the center of the galaxy, etc.

[Winemaker]

I'm a winemaker, not a pilot, but have a huge interest in aviation and really enjoy this site. Reading this thread I have only one thought: If you're flying in a constant wind it doesn't matter how you turn, upwind or downwind, as you are moving in the frame of reference of the wind. The wind velocity can be 10 knots or 150 knots relative to the ground, it just doesn't matter. You are flying in a volume of air and the airplane is only reacting to forces generated by its movement though that volume. Ignore the ground speed, that's not what generates lift. Turn left, turn right, turn downwind, turn upwind, it's all good as you are effectively turning in still air as you are swept along over the ground. You are going to move across the ground pretty fast flying downwind in a 150 knot breeze and you might be standing still in a 150 knot headwind, but it just doesn't matter. Thanks for all the great comments on these threads, it's pretty educational for me.

[currawong]

How about the boat crossing the river, for those that prefer to communicate in grunts.

Turn 180 degrees. Downstream. Rate one.Turn takes up lots of room as getting washed downstream.

Turn 180 degrees. Upstream. Rate one. Turn takes up less room as getting washed downstream.

Boat speed and rate of turn relative to water is constant.

Boat speed relative to bank varies due current.

Closest I can get to using monosyllables.

[itsnotthatbloodyhard]

Might as well give up, folks, there’s no point. It’s like a sort of religious zealotry. Doesn’t matter how simply you explain it or how much you appeal to logic and reason, you’re fighting an unshakable belief (with occasional abuse and misunderstanding of terms like ‘frame of reference’ and ‘Newton’ thrown in for a bit of pseudoscientific gravitas). Nice try though.

[Meikleour]

Most people seem to agree that airspeed fluctuations due to windshear are the result of the aircraft having an inertial groundspeed which causes the changes to be apparent as it experiences differing air masses and takes a finite time to regain its momentum ......................now, when flying in a stable but moving air mass where does the effect of aircraft momentum go? (why is it always ignored?)

I have observed many times descending into NRT with a tailwind of 200kts (not uncommon at certain times of year) with the autopilot in IAS hold and smooth wind conditions. The arrival called for a 90 degree heading change and whilst the IAS remained constant the rate of descent increased dramatically once the turn was complete (ie. much lower groundspeed) before stabilising back to the normal rate. This behaviour always seemed at odds with the theory.

[Wizofoz]

Most people seem to agree that airspeed fluctuations due to windshear are the result of the aircraft having an inertial groundspeed which causes the changes to be apparent as it experiences differing air masses and takes a finite time to regain its momentum ......................now, when flying in a stable but moving air mass where does the effect of aircraft momentum go? (why is it always ignored?)


It's not ignored, just often misunderstood.

An aircraft turns by tilting it's lift vector. Lift is proportional to airspeed, not ground speed. Momentum is relative to velocity which is relative to a frame of reference. It is not exclusively proportional to ground speed.

[Vessbot]

......................now, when flying in a stable but moving air mass where does the effect of aircraft momentum go? (why is it always ignored?)


Can you describe how the effect of momentum should be taken into account in what should happen to the human body when turning around inside a moving airliner and experiencing a near-instantaneous thousand knot reversal in groundspeed?

[Brercrow]

Most people seem to agree that airspeed fluctuations due to windshear are the result of the aircraft having an inertial groundspeed which causes the changes to be apparent as it experiences differing air masses and takes a finite time to regain its momentum ......................now, when flying in a stable but moving air mass where does the effect of aircraft momentum go? (why is it always ignored?)

I have observed many times descending into NRT with a tailwind of 200kts (not uncommon at certain times of year) with the autopilot in IAS hold and smooth wind conditions. The arrival called for a 90 degree heading change and whilst the IAS remained constant the rate of descent increased dramatically once the turn was complete (ie. much lower groundspeed) before stabilising back to the normal rate. This behaviour always seemed at odds with the theory.

This is explained as follows: The aircraft turns due to a horizontal component of the lift force. Lift force is the equal and opposite effect of momentum given to the air. In a turn, part of that momentum is horizontal. As the aircraft turns from downwind to crosswind it loses momentum as it loses groundspeed. This acceleration is caused by the aero-forces acting on the aircraft modified by the angle of drift. The air gains horizontal momentum and part of that is in the same direction as the wind. So aircraft loses momentum and wind gains momentum.
At the same time, due to inertia, the airspeed is affected by the reducing tailwind component and there is a slight tendency for airspeed to increase. (Same as an increasing headwind)
If the aircraft is descending in IAS hold and constant thrust, the effect is that the auto pilot will hold the airspeed and the rate of descent will decrease slightly but only during the turn.
When the aircraft rolls out on the crosswind heading, the thrust is insufficient to maintain the reduced rate of descent and the IAS hold pitches the nose down to maintain the airspeed. The rate of descent increases rapidly but settles back once equilibrium is restored at the original rate of descent.

[Goldenrivett]

experiencing a near-instantaneous thousand knot reversal in groundspeed
Errrr.....by my maths, that is going from 499 kts to 501 kts ground speed i.e 2 kts change in ground speed with respect to both earth and to the aircraft.

What momentum is being ignored?

[Jet_Fan]

Errrr.....by my maths, that is going from 499 kts to 501 kts ground speed i.e 2 kts change in ground speed with respect to both earth and to the aircraft.

What momentum is being ignored?

You turn to face the other direction so your ground speed becomes negative INSTANTLY!

[Meikleour]

Wizofoz: I agree with you however I think you will find that I never claimed lift was proportional to groundspeed! For the windshear analogy to apply the momentum must be related to the groundspeed ie. relative to the earth. Why is it therefore that when operating in the "bubble of air" concept then reference to the earth is discounted? Likewise can you help me understand the often observed phenomenon that I cited.

[Jet_Fan]

Wizofoz: I agree with you however I think you will find that I never claimed lift was proportional to groundspeed! For the windshear analogy to apply the momentum must be related to the groundspeed ie. relative to the earth. Why is it therefore that when operating in the "bubble of air" concept then reference to the earth is discounted? Likewise can you help me understand the often observed phenomenon that I cited.


While free flying, the relationship with the ground is purely a navigational one and has nothing to do with the aerodynamics of the plane. Can you really not grasp this?

[Meikleour]

Vessbot:Turning around inside an airliner does not change your momentum with respect to the earth. It is still related to the vector of the aircraft. The only change you could make is to run from nose to tail down the aisle and then you would only reduce it by your running speed times your mass! A tiny reduction compared to the aircraft. This is the same fallacy about being in a runaway lift(elevator) where the occupants are asked to jump upwards just before impact with the bottom to save thenselves! Won't work.

Jet Fan: I think you are being ironic!!

[Jet_Fan]

Vessbot:Turning around inside an airliner does not change your momentum with respect to the earth. It is still related to the vector of the aircraft. The only change you could make is to run from nose to tail down the aisle and then you would only reduce it by your running speed times your mass! A tiny reduction compared to the aircraft. This is the same fallacy about being in a runaway lift(elevator) where the occupants are asked to jump upwards just before impact with the bottom to save thenselves! Won't work.

Jet Fan: I think you are being ironic!!
m

the downwind turn myth Is for stupid people to believe in. Turning into a headwind does not translate into an increase in airspeed derived in any way from the headwind. For exactly the same reason, when I turn in the aisle and face the other direction, thus instantly changing my groundspeed to a negative, I feel no change in momentum.

[D120A]

Winemaker has it exactly right. If you don't believe him/her, or me, ask a friendly mathematician to construct a position vector from a fixed point on the earth to an aircraft performing a constant rate turn in a steadily moving airmass, and differentiate it twice to determine the acceleration of the aircraft. It is always towards the centre of the circle of the constant turn, a point which itself is moving with uniform speed over the ground. To demonstrate it in flight, choose a windy day and ask an instructor (better still, a test pilot) to fly an accurate rate one 360 degree turn under the hood while you look out of the window. The acceleration will feel fine, bank angle constant and ball in the middle, but at low level your track over the ground will scare you stiff and illustrate why trying to turn near the ground in such conditions using visual references is so dangerous.

[Jet_Fan]

Winemaker has it exactly right. If you don't believe him/her, or me, ask a friendly mathematician to construct a position vector from a fixed point on the earth to an aircraft performing a constant rate turn in a steadily moving airmass, and differentiate it twice to determine the acceleration of the aircraft. It is always towards the centre of the circle of the constant turn, a point which itself is moving with uniform speed over the ground. To demonstrate it in flight, choose a windy day and ask an instructor (better still, a test pilot) to fly an accurate rate one 360 degree turn under the hood while you look out of the window. The acceleration will feel fine, bank angle constant and ball in the middle, but at low level your track over the ground will scare you stiff and illustrate why trying to turn near the ground in such conditions using visual references is so dangerous.

yep,

exactly that

[Vessbot]

Errrr.....by my maths, that is going from 499 kts to 501 kts ground speed i.e 2 kts change in ground speed with respect to both earth and to the aircraft.

What momentum is being ignored?
From negative 499 to positive 501, that's a thousand knot increase. The airspeed is 1 knot before and after, unchanged.

And *I* don't think that there is anything ignored about the momentum, but the windward turn theorists do (like the one I quoted in my reply) and I'm challenging then to apply their reasoning to that situation.

[Vessbot]

Vessbot:Turning around inside an airliner does not change your momentum with respect to the earth. It is still related to the vector of the aircraft. The only change you could make is to run from nose to tail down the aisle and then you would only reduce it by your running speed times your mass! A tiny reduction compared to the aircraft.


Yes, and is this any different (and why?) from turning around inside a uniformly moving airmass?)

[Derfred]

I always think the easiest way to explain the argument is to use an extreme example.

An aircraft doing 50 KIAS in a constant 5000 KT airmass conducts a rate 1 orbit. What happens to it, compared to the same aircraft conducting the same orbit in nil wind?

Who votes for crash and burn?

The correct answer is: nothing. The aircraft doesn’t know it’s in a 5000 KT airmass. It behaves according to Newtonian physics from the frame of reference of the airmass.

[Goldenrivett]

Hi Vessbot,
From negative 499 to positive 501, that's a thousand knot increase. The airspeed is 1 knot before and after, unchanged.

And *I* don't think that there is anything ignored about the momentum, but the windward turn theorists do (like the one I quoted in my reply) and I'm challenging then to apply their reasoning to that situation.

I am definitely not a windward turn theorist. My reply was to your confused post #109
How about when you are walking upwind at 1 knot at knots in an airliner, toward the tail, with a groundspeed of negative 499 knots. You turn around to face the nose (downwind) in one second, and keep walking at 1 knot airspeed, but now your groundspeed is 501 knots. You just walked through a one thousand knot self-created windshear in one second, according to your theory, which predicts that there should be some effect.

You haven’t walked through a “”1000 KT self created wind shear”, you have only changed your groundspeed by 2 kts. The fact that you turned around and re-label which is +ve and -ve destroyed your inertial reference frame.

[Jet_Fan]

Hi Vessbot,


I am definitely not a windward turn theorist. My reply was to your confused post #109


You haven’t walked through a “”1000 KT self created wind shear”, you have only changed your groundspeed by 2 kts. The fact that you turned around and re-label which is +ve and -ve destroyed your inertial reference frame.

jesus wept, we know! It’s according to the downwind myth idiots that changes in groundspeed somehow affect airspeed through some mad principal that they can’t explain.

[Vessbot]

Hi Vessbot,


I am definitely not a windward turn theorist. My reply was to your confused post #109


You haven’t walked through a “”1000 KT self created wind shear”, you have only changed your groundspeed by 2 kts. The fact that you turned around and re-label which is +ve and -ve destroyed your inertial reference frame.
This thread is moving fast and I'm at work so I can't get detailed until later, but yes this is exactly a 1000 knot self-created wind shear according to the premises of the windward turn theorists. Check it again.

[Wizofoz]

Wizofoz: I agree with you however I think you will find that I never claimed lift was proportional to groundspeed! For the windshear analogy to apply the momentum must be related to the groundspeed ie. relative to the earth. Why is it therefore that when operating in the "bubble of air" concept then reference to the earth is discounted? Likewise can you help me understand the often observed phenomenon that I cited.

No, you do not need to calculate groundspeed to understand winds hear- because the aircraft is not interacting with the ground, so it's momentum relative to it is irrelevant. Windshear happens when the aircraft's momentum relative to the AIR changes.


Say you are turning on a still air day- but on the ground there is a passing truck- you could calculate your momentum relative to IT- it is as relevant a momentum figure as that of the earth, but it doesn't change the dynamics of the situation does it?

As to the increased rate of descent you cite, did it come with an increase in airspeed also? I have often observed older FMCs get caught out but changes in wind and do some radical stuff to maintain programed path.

[Wizofoz]

This is explained as follows: The aircraft turns due to a horizontal component of the lift force. Lift force is the equal and opposite effect of momentum given to the air. In a turn, part of that momentum is horizontal. As the aircraft turns from downwind to crosswind it loses momentum as it loses groundspeed. This acceleration is caused by the aero-forces acting on the aircraft modified by the angle of drift. The air gains horizontal momentum and part of that is in the same direction as the wind. So aircraft loses momentum and wind gains momentum.
At the same time, due to inertia, the airspeed is affected by the reducing tailwind component and there is a slight tendency for airspeed to increase. (Same as an increasing headwind)
If the aircraft is descending in IAS hold and constant thrust, the effect is that the auto pilot will hold the airspeed and the rate of descent will decrease slightly but only during the turn.
When the aircraft rolls out on the crosswind heading, the thrust is insufficient to maintain the reduced rate of descent and the IAS hold pitches the nose down to maintain the airspeed. The rate of descent increases rapidly but settles back once equilibrium is restored at the original rate of descent.

In case anyone was wondering, pretty sure this is the author of the website. Same ridiculous toss.

[Vessbot]

Here's a typical flight example of the windward turn that should suffer an airspeed loss according to the windward turn theorists:

Vehicle making the turn: Cessna 172
Wind (uniform airmass over the ground): 10 knots, East to West
Initial heading: East
Initial airspeed: 100 knots
Initial groundspeed: 90 knots (airspeed minus headwind, 100-10)
Groundspeed after 180 degree turn: 110 knots (airspeed plus tailwind, 100+10)
Groundspeed change, aka "self created wind shear:" +20 knots (final groundspeed minus initial groundspeed, 110-90)

And my equivalent example:

Vehicle making the turn: Human body inside an airliner
Wind (uniform airmass inside the airliner): 500 knots, East to West
Initial heading: East
Initial airspeed: 1 knot
Initial groundspeed: negative 499 knots (airspeed minus headwind, 1-500)
Groundspeed after 180 degree turn: 501 knots (airspeed plus tailwind, 1+500)
Groundspeed change, aka "self created wind shear:" +1000 knots (final groundspeed minus initial groundspeed, 501 minus negative 499)

The windward turn theorists say that in the first example there is a tendancy to lose airspeed due to the headwind loss/tailwind gain, but in most situations it's minor enough not to notice since the turn is so slow comapared to the shear amount (20 knots over 1 minute) that the momentum is gradually changed to stay caught up with the airspeed.

If this logic is true, it implies that there could be an example, if we tweak the numbers enough, where the momentum doesn't have a chance to gradually change, and the result would be a noticable airspeed loss. So how about it, if 20 knots over a minute is not enough, how about 1000 knots over a second? Has anyone ever noticed any effects of this while walking up and down the isle of an airliner? Maybe it's still not enough for the effect to rise above the noise floor. Do we have to tweak the numbers further to make the airmass be the inside of a Concorde? Or an Apollo command module on translunar coast?

Of course not, this is bogus and Newtonian relativity holds true, i.e., if you close the window shades all physics occur as if it's sitting still. Or moving uniformly in any direction at any speed.

[Vessbot]

Well things went dead quickly... surely I couldn't have had the last word, could I? :ooh:

[Wizofoz]

Well things went dead quickly... surely I couldn't have had the last word, could I? :ooh:


No, THIS is the last word......

[A Squared]

No, THIS is the last word......

Are you sure?

[Wizofoz]

Are you sure?

Yes, that was definitely the last post.

[A Squared]

Yes, that was definitely the last post.

I'm not so certain.

[Vessbot]

Has the thread turned downwind and lost airspeed?

[Brercrow]

I want to make it clear that nowhere in my website do I say that groundspeed affects an aircraft in flight.

Read the link at post #102 and you will find a rational explanation of the effect of the wind during turning flight.

Stall spin crash burn - beware the downwind turn. But it is mostly pilot error.

[A Squared]

I want to make it clear that nowhere in my website do I say that groundspeed affects an aircraft in flight.

Read the link at post #102 and you will find a rational explanation of the effect of the wind during turning flight.

No, you will not, you will find complete nonsense which ignores physics, couched in pseudo-scientific terms not understood by the author.

I'll ask you the same question that the other Myth-ers couldn't handle. (either refused to answer or answered incorrectly)

2 identical airplanes with a cruise airspeed of 100 knots. Both are flying east. Airplane A is in still air. Airplane B is flying into a 50 knot wind blowing out of the east. Both turn 180 degrees and fly west. What is the change in velocity from before the turn to after the turn for each plane?

[Wizofoz]

I want to make it clear that nowhere in my website do I say that groundspeed affects an aircraft in flight.

Read the link at post #102 and you will find a rational explanation of the effect of the wind during turning flight.

Stall spin crash burn - beware the downwind turn. But it is mostly pilot error.

You say wind affects the dynamics of an aircraft in flight- it's the same thing and equally wrong.

[donpizmeov]

You are a patient man Wiz.

[Wizofoz]

You are a patient man Wiz.

I always remember that Mark Twain quote-

“Never argue with stupid people, they will drag you down to their level and then beat you with experience.”

Then I go and do it anyway...…..

[Jet_Fan]

Nothing new here Its the same force that makes your groundspeed increase when you turn downwind and decrease when you turn upwind

Which force is that then?

[Brercrow]

No, you will not, you will find complete nonsense which ignores physics, couched in pseudo-scientific terms not understood by the author.

I'll ask you the same question that the other Myth-ers couldn't handle. (either refused to answer or answered incorrectly)

2 identical airplanes with a cruise airspeed of 100 knots. Both are flying east. Airplane A is in still air. Airplane B is flying into a 50 knot wind blowing out of the east. Both turn 180 degrees and fly west. What is the change in velocity from before the turn to after the turn for each plane?

A plausible question which is irrelevant. You are comparing two states of equilibrium. You are not thinking about what happens during the turn

[Brercrow]

Which force is that then?

Seriously? You started this thread and you have not read the website. If you had then you would not have to ask that question.

(I would show you now but Pprune does not allow me attachments)

[Wizofoz]

A plausible question which is irrelevant. You are comparing two states of equilibrium. You are not thinking about what happens during the turn

During the turn the lift vector, which is identical whether the wind blows or not, causes a turn of constant rate and constant (air) radius.

[A Squared]

A plausible question which is irrelevant. You are comparing two states of equilibrium. You are not thinking about what happens during the turn

Nope, not irrelevant at all. It is the very crux of the "downwind turn" fallacy and the fact that you are trying to dismiss it as irrelevant just illustrates that you don't have the fundamental grasp of physics required to understand why you are mistaken.

[jonkster]

I want to make it clear that nowhere in my website do I say that groundspeed affects an aircraft in flight.

Read the link at post #102 and you will find a rational explanation of the effect of the wind during turning flight.

Stall spin crash burn - beware the downwind turn. But it is mostly pilot error.
I have looked at that site.
You state (when discussing an aircraft, in a steady wind, doing a constant rate turn)

1.3 If you think, at this point, that the wind is irrelevant to an aircraft in flight, remember that the spiral ground track is the vector sum of aircraft velocity relative to the air and wind-velocity relative to the ground. The acceleration of the aircraft is caused by forces acting on the aircraft and is seen as acceleration components of air-velocity and ground-velocity. In this context, the acceleration of the aircraft is affected by the wind.


I don't get what you are saying and it seems to be the crux of your argument.

1. You can only get an acceleration by applying a net force.

The only force, acting on the aircraft, "caused" by the earth, is gravity.

Yet you seem to argue that somehow the earth acts to accelerate the aircraft laterally. How? What force is this, where does it come from and how does it act on the aircraft?

2. That spiral path would appear exactly the same for any circling object if it was viewed from a constantly moving point of reference. No additional force is required to make the spiral path. (eg one of those conical pendulums dropping sand on paper - pull the paper at a constant rate and you get the very same spiral.) Not getting where any additional force is needed to get that pattern.

[Brercrow]

I have looked at that site.
You state (when discussing an aircraft, in a steady wind, doing a constant rate turn)


I don't get what you are saying and it seems to be the crux of your argument.

1. You can only get an acceleration by applying a net force.

The only force, acting on the aircraft, "caused" by the earth, is gravity.

Yet you seem to argue that somehow the earth acts to accelerate the aircraft laterally. How? What force is this, where does it come from and how does it act on the aircraft?

2. That spiral path would appear exactly the same for any circling object if it was viewed from a constantly moving point of reference. No additional force is required to make the spiral path. (eg one of those conical pendulums dropping sand on paper - pull the paper at a constant rate and you get the very same spiral.) Not getting where any additional force is needed to get that pattern.

I am not saying that the Earth accelerates the aircraft laterally. You have just made that up. The aircraft is accelerated by the aerodynamic forces acting on it. Read the link again.The whole thing! There are diagrams!

[Jet_Fan]

I am not saying that the Earth accelerates the aircraft laterally. You have just made that up. The aircraft is accelerated by the aerodynamic forces acting on it. Read the link again.The whole thing! There are diagrams!

I’m still mystified by your assertions because none of it makes sense, especially when you are asked to distill it all down, whereupon you appear unable.

Try again talking us all through an upwind turn, explaining what’s happening to all the forces as you go.

I await your reply with great interest.

[jonkster]

I am not saying that the Earth accelerates the aircraft laterally. You have just made that up. The aircraft is accelerated by the aerodynamic forces acting on it. Read the link again.The whole thing! There are diagrams!

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?

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?

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).

Why does drag change?

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.

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?

[Ian W]

I hesitate to get involved in this argument as the positions are so entrenched. However, a little observational input may assist.
I was working as local/tower controller at Valley in Anglesey. It was a quiet Sunday and as was quite common there the wind was gale force although steady and not gusting as it came off the Irish Sea. The surface wind was almost due West and probably 50Kts at least (it was a mbmmble time ago).
A Stampe biplane wished to depart back to England to the East and 'taxied' if that is the right word for driving straight into wind onto the grass at close to full throttle with members of the visiting aircraft flight on both wings and fuselage near the tail plane. The pilot asked for takeoff and waved off the VAF crew and the tail of the Stampe lifted up and then the aircraft effectively took off close to vertically into wind. The aircraft climbed to around 200 ft straight up, then turned left steeply cross wind then downwind onto East. As it did so it lost a significant amount of height as it also gained a significant amount of speed. The last call to me in the tower from the pilot was: "I won't be coming back!".

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. A more gentle turn may have masked the effect but it would still have been there.
I think it is a case of relative speeds: wind speed to aircraft speed. A military fighter at 400 kts + in a relatively low wind may not notice the effect. An albatross at 20kts may notice the inertial effect of a turn in a wind of equal velocity.

[Jet_Fan]

I hesitate to get involved in this argument as the positions are so entrenched. However, a little observational input may assist.
I was working as local/tower controller at Valley in Anglesey. It was a quiet Sunday and as was quite common there the wind was gale force although steady and not gusting as it came off the Irish Sea. The surface wind was almost due West and probably 50Kts at least (it was a mbmmble time ago).
A Stampe biplane wished to depart back to England to the East and 'taxied' if that is the right word for driving straight into wind onto the grass at close to full throttle with members of the visiting aircraft flight on both wings and fuselage near the tail plane. The pilot asked for takeoff and waved off the VAF crew and the tail of the Stampe lifted up and then the aircraft effectively took off close to vertically into wind. The aircraft climbed to around 200 ft straight up, then turned left steeply cross wind then downwind onto East. As it did so it lost a significant amount of height as it also gained a significant amount of speed. The last call to me in the tower from the pilot was: "I won't be coming back!".

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. A more gentle turn may have masked the effect but it would still have been there.
I think it is a case of relative speeds: wind speed to aircraft speed. A military fighter at 400 kts + in a relatively low wind may not notice the effect. An albatross at 20kts may notice the inertial effect of a turn in a wind of equal velocity.

As has already been stated, gravity was the only force being exerted on that aircraft by the Earth. As the pilot turned he would have felt some g-force but as he turned downwind he wouldn't have been pushed into the back of his seat as his groundspeed went from 0 to 120 due to anything other than centrifugal force, which would feel the same as on a calm day.. He could have lost height for any number of reasons but it will always look more dramatic going downwind to a ground based observer.

[Derfred]

I suspect the answer lies in your own narrative:

turned left steeply

A steep turn at 200 feet in that kind of wind doesn’t sound wise to me. Not because of the wind strength, and not because he was turning downwind, but because a 50kt wind at 200 feet is NEVER a steady air mass, and is thus irrelevant to this discussion.

Or maybe the pilot was a subscriber to the “downwind turn myth” and lowered the nose because he was scared of losing airspeed :p:p:p

[jonkster]

May also have succumbed to the (potentially deadly) skid/slip illusion. I see it sometimes with students (and sometimes licenced pilots) flying circuits in strong winds, they try and fly their turns using reference to the ground, they try and make things look the way they normally experience their path over the ground by over ruddering or over banking whilst using the same nose attitudes they normally do.

Both actions will reduce climb performance and increase descent rates. Can also set you up for stall/spin at low level :(

It is a powerful illusion when low to the ground at low airspeeds in strong wind.

[Goldenrivett]

Hi 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. A more gentle turn may have masked the effect but it would still have been there.

Relative to you, the aircraft had to accelerate from 0 to 120 kts (change of 120 kts). Relative to the aircraft it had to accelerate from 60 kts heading North say to 60 kts heading South (i.e. 120 kts change in air speed vector - i.e. the same vector value)

Your estimation of distance is made using the apparent size of a known object (angle subtended at you eye). Your estimation of height is made using the angle between the object and the horizon. As the aircraft appeared to accelerate unusually rapidly down wind, it's distance from you increased unexpectedly rapidly and the angle between the horizon, the aircraft and your eyeball reduced rapidly giving you the illusion of a height loss.

The illusion is similar when observing a "slow" jumbo compared to a faster 737 on the approach when both are flying at say 160 kts. The jumbo appears slower because it appears closer and yet the change the angular position is the same as the "further away" 737.

[Brercrow]

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

[Brercrow]

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]

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]

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]

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]

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]

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]

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]

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

:ok:

Nonessential characters added to make the software happy.

[Ian W]

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]

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.



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/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.

[Wizofoz]

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]

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]

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]

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]

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.

[Brercrow]

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.

You are on the right lines talking about acceleration. Particularly that tangential acceleration is different to centripetal acceleration for an aircraft

These other guys dont get it

[Wizofoz]

You are on the right lines talking about acceleration. Particularly that tangential acceleration is different to centripetal acceleration for an aircraft

These other guys dont get it

Nor, apparently, did Isaac Newton.

[Derfred]

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.


Vessbot has it.

Brercrow, you did a VERY rough spreadsheet with 10 degree increments (refer section 6 of his Post (http://www.dynamic-soaring-for-birds.co.uk/html/the_downwind_turn.html)), 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.

[Brercrow]

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/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.

You give two examples in #133 Both compare two states of equilibrium and do not consider what is happening DURING the turn
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

[Brercrow]

Vessbot has it.

Brercrow, you did a VERY rough spreadsheet with 10 degree increments (refer section 6 of his Post (http://www.dynamic-soaring-for-birds.co.uk/html/the_downwind_turn.html)), 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 cannot tell you how many people have said I should get 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

[Vessbot]

You give two examples in #133 Both compare two states of equilibrium and do not consider what is happening DURING the turn
In the second example the math is wrong (500+1) - (500-1) = 2 not 1000


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

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.

[Vessbot]

To those who are absolutely certain that ‘windward turn theory’ is fact, I say, show those who are nonbelievers the math.

Otherwise this thread will just be going round and round in circles!
He has a whole website of it, which has been linked to multiple times in the thread:

The Downwind Turn (http://www.dynamic-soaring-for-birds.co.uk/html/the_downwind_turn.html)

[Jet_Fan]

He has a whole website of it, which has been linked to multiple times in the thread:

The Downwind Turn (http://www.dynamic-soaring-for-birds.co.uk/html/the_downwind_turn.html)

The Downwind Turd

There, that sounds better.

[Vessbot]

I think I asked the wrong question. Perhaps it should have been; please explain how you arrived at figures 1b and 1c and validate the vectors you claim to prove your theory.

Oh, and what about providing another ‘plan view’ for still air conditions for comparison.


Everything in black in those figures is correct. The overlaid blue triangles, though, are nonsense for the reason I explained earlier in post 174.

In still air, angle d would be zero and therefore Fc sin d would be zero, and therefore Ft would be zero.

[Winemaker]

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.

[Derfred]

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

Brercrow,

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

[gemma10]

Well having read the whole of this thread from start to finish this afternoon, I shaal not fly my Grumman spamcan in any wind from now on!

[A Squared]

Brercrow,

...

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? ...






From his page on his "windward turn" theory (which is agreed by all to the simply the obverse of the downwind turn myth) :
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.

In a uniform, constant velocity wind, the air *is* in equilibrium, it doesn't change to another state of equilibrium, and it is not accelerated, so the author fails in understanding the basic premise of his "theory". Alternately, the theory depends on flying within an accelerated air mass, which is a completely different scenario, and it was never disputed by anyone that energy gains can be achieved by crossing wind gradients. In either case, anything which follows is necessarily nonsense.

[one dot right]

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.

[Vessbot]

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.

I don't have as pessimistic of an outlook of it as you do. I think he is actually engaging in honest argumentation.

[A Squared]

I don't have as pessimistic of an outlook of it as you do. I think he is actually engaging in honest argumentation.

I think you are badly mistaken, Vessbot. Can you point to anyplace, either here or on other forums where he has shown even the slightest indication that he is willing to understand how he might be mistaken? I certainly can't.

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".

[Brercrow]

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.
Home (http://www.dynamic-soaring-for-birds.co.uk/home.html)
The Downwind Turn (http://www.dynamic-soaring-for-birds.co.uk/html/the_downwind_turn.html)

[Wizofoz]

Derfred[/b] - Do I understand Newtons Laws? Sure I do



OHHHH no you don't.

[Brercrow]

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.

It does not matter whether it is an FMS turn or a HDG turn. It is a turn with a reducing tailwind component ( same effect as an increasing headwind component). The increase in performance should be a reduction of the rate of descent during the turn This was not reported by the author who only reported the increase of rate of descent after wings level. This occurred because the Airspeed hold and fixed thrust setting could not sustain the reduced rate of descent after wings level and yes it was an autopilot PIO Remember that at jet speeds, even small changes of pitch attitude make big changes in rate of climb or descent

The Downwind Turn (http://www.dynamic-soaring-for-birds.co.uk/html/the_downwind_turn.html)

[Brercrow]

OHHHH no you don't.

I have had a long correspondence with Wizofoz going back years

Denial is not a counter-argument.

[Wizofoz]

I have had a long correspondence with Wizofoz going back years

Denial is not a counter-argument.

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

Are all frames of reference equally valid?

[Ian W]

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,

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.

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
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

[Vessbot]

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

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.

[Jet_Fan]

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

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

[A Squared]

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

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.

[Brercrow]

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

[Vessbot]

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?

[Brercrow]

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.

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

[A Squared]

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.

[Jet_Fan]

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.

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.

[A Squared]

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.



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

[Jet_Fan]

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

A steady wind has no aerodynamic effect on the aircraft.

Ask any physicist.

[Vessbot]

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

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.

[Jet_Fan]

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

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.

[PDR1]

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

[Jet_Fan]

Utter drivel. That would imply a 2hz rotation rate results in a 1000kn/sec linear acceleration - roughly 2,000m/sec^2 or 200G. Why is the person not squashed to a pulp by this acceleration?

PDR

Why do you think?

[Vessbot]

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


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

[Jet_Fan]

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

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

[A Squared]

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.


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.

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.

[Vessbot]

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).

[A Squared]

The airlner is flying WEST at 500 knots.

OK, fine, the airliner is flying west, so west is the positive direction of our frame of reference. The velocity of the westbound airliner is +500 knots. The man is walking walking in the east direction , so that's -1 knot. a positive 500 knots, plus a negative 1 knot equals positive 499 knots, the man is still moving west at 499 knots. it doesn't become negative because of the direction he is facing.

[A Squared]

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

"groundspeed" by definition, is relative to the ground. His velocity in his frame of reference is zero, because it's impossible for him to move relative to himself.

[Jet_Fan]

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.

He is moving backwards. If I reverse my car at 30mph, what is my car's groundspeed? Clue: the speedo won't be reading 30mph.

[Vessbot]

OK, fine, the airliner is flying west, so west is the positive direction of our frame of reference. The velocity of the westbound airliner is +500 knots. The man is walking walking in the east direction , so that's -1 knot. a positive 500 knots, plus a negative 1 knot equals positive 499 knots, the man is still moving west at 499 knots. it doesn't become negative because of the direction he is facing.

Oh Christ. You're anchoring the +/- to the geographical direction. I was leaving them anchored to the body. Yes I can belabor the point and rework everything to keep to a geographical frame, but the whole purpose of the analogy was to relate it to what we all intuitively understand from flying bug smashers at 100 knots in 10 knot winds (which is where a positive sign on groundspeed always means that ground is moving opposite the direction the nose is pointing). Look at scenario A!

[Jet_Fan]

"groundspeed" by definition, is relative to the ground. His velocity in his frame of reference is zero, because it's impossible for him to move relative to himself.

That's his speed relative to the airframe. When I move backwards relative to the ground, my groundspeed has a minus in front of it.

[A Squared]

Oh Christ. You're anchoring the +/- to the geographical direction.

Well, yeah, what else are you going to measure ground speed relative to? Again, groundspeed is inherently relative to (in the frame of reference of) the ground.

You can't make any meaningful analysis when the axes of your reference system move whenever a person looks in a different direction.

[Vessbot]

Well, yeah, what else are you going to measure ground speed relative to? Again, groundspeed is inherently relative to (in the frame of reference of) the ground.

You can't make any meaningful analysis when the axes of your reference system move whenever a person looks in a different direction.

Sure you can. If your last sentence is true, then it is not a meaningful analysis that in Scenario A the Cessna went from 90 to 110 groundspeed when turning East to West. Did you look at scenario A? Do you think that's meaningful or meaningless?

[Vessbot]

Hell we do the same thing about airspeed too, and necessarily so. The positive sign is always applied when air is coming from the direction of the nose, and we take that axis with us wherever we point the nose. If we didn't, then an airplane should fly equally well in a tailslide as in normal flight, and I can tell you from personal experience that it certainly doesn't.

[A Squared]

Sure you can. If your last sentence is true, then it is not a meaningful analysis that in Scenario A the Cessna went from 90 to 110 groundspeed when turning East to West. Did you look at scenario A? Do you think that's meaningful or meaningless?

From a physics standpoint, it is completely meaningless. You are making the identical mistake that Ian was making. In your Scenario A, From the frame of reference of the ground, the airplane went from a velocity +110 knots to a velocity of -90 for a total change in velocity of 200 knots.

[A Squared]

Hell we do the same thing about airspeed too, and necessarily so. The positive sign is always applied when air is coming from the direction of the nose, and we take that axis with us wherever we point the nose. If we didn't, then an airplane should fly equally well in a tailslide as in normal flight, and I can tell you from personal experience that it certainly doesn't.

Look, it's pretty clear that you simply are unable to grasp the concept of Velocity, how it is different than "speed", why it is imperative that velocity is used in physics and not speed, and why a constant, unchanging frame of reference is necessary to make any kind of a meaningful analysis. I apologize for having wasted your time trying to explain these things and I will stop now.

[Vessbot]

From a physics standpoint, it is completely meaningless. You are making the identical mistake that Ian was making. In your Scenario A, From the frame of reference of the ground, the airplane went from a velocity +110 knots to a velocity of -90 for a total change in velocity of 200 knots.

Say there's no wind (it's one of those perfect mornings) and you're flying around. You fly East for a while at 100 knots airspeed, and turn around and fly West. By that same premise, your airspeed is now -100 knots too, for a total change of 200 knots. Hmmmmm?

(if there's no wind then the airmass frame and the ground frame are the same, so you should just be able to do a find-and-replace of "ground" with "air" to your post and everything still hold true.)

No, it's perfectly valid, for certain contexts, to rotate the axes as needed.

[Brercrow]

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?

The difference between these two examples is that in one the windspeed is less than the airspeed and in the other the windspeed is greater than the airspeed. In both cases airspeed is positive. In the first case the groundspeed reverses direction and in the second case the groundspeed is in the same direction. That means that in the first case you are entitled to take the difference between the groundspeeds but in the second case you have to take the sum. In other words along with common sense there is a minus sign missing.
I don't really mind where you stick it.

I should add that this is because calculation of groundspeed is a triangle of velocities problem envolving the cosine of the wind-angle. Less than 90 degrees is positive Greater than 90 degrees is negative

(1+500)+(1-500)
=501 +(-499)
=2

[Vessbot]

Look, it's pretty clear that you simply are unable to grasp the concept of Velocity, how it is different than "speed", why it is imperative that velocity is used in physics and not speed, and why a constant, unchanging frame of reference is necessary to make any kind of a meaningful analysis. I apologize for having wasted your time trying to explain these things and I will stop now.

Quit being a martyr. You know it's perfectly reasonable, given the standard set of assumptions, to say that your groundspeed is 90 knots one way, then 110 the other way. In fact, before this thread, you would not have even blinked at saying exactly that. You know it.

[Jet_Fan]

From a physics standpoint, it is completely meaningless. You are making the identical mistake that Ian was making. In your Scenario A, From the frame of reference of the ground, the airplane went from a velocity +110 knots to a velocity of -90 for a total change in velocity of 200 knots.

It's not even a mistake, let alone the same mistake. The whole point is to explain to people like Ian, that what he saw was different to what the plane actually experienced.

Now, a ball can't travel backwards, but people and planes can so + and - is a way of expressing that relative movement.

[Wizofoz]

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.

No-it doesn't matter which way you are facing, it matters which way you have nominated as positive and negative in the reference frame you are using.

Acceleration is absolute- if his velocity changed by 1000kts in a fraction of a second, he'd be crushed by the Gs.

[Jet_Fan]

No-it doesn't matter which way you are facing, it matters which way you have nominated as positive and negative in the reference frame you are using.

Acceleration is absolute- if his velocity changed by 1000kts in a fraction of a second, he'd be crushed by the Gs.

We are looking at is groundspeed not his velocity. His groundspeed can be + or -.

[Wizofoz]

He is moving backwards. If I reverse my car at 30mph, what is my car's groundspeed? Clue: the speedo won't be reading 30mph.

The magnitude of your ground VELOCITY is 30 MPH

[Wizofoz]

We are looking at is groundspeed not his velocity. His groundspeed can be + or -.

What does a negative ground speed look like?

[Wizofoz]

We are looking at is groundspeed not his velocity. His groundspeed can be + or -.

So if two people are standing in the isle facing each other, one is going positive and one negative? Does one therefore not get to the destination?

What is the ground speed of someone facing sideways?

[Jet_Fan]

What does a negative ground speed look like?

Land a parachute with a negative groundspeed and have someone video you. Whether groundspeed it + or - does matter sometimes.

[Wizofoz]

Land a parachute with a negative groundspeed and have someone video you. Whether groundspeed it + or - does matter sometimes.

It's still a pretty imprecise term. What would your answer to the "sideways"case be?

[Jet_Fan]

So if two people are standing in the isle facing each other, one is going positive and one negative? Does one therefore not get to the destination?

What is the ground speed of someone facing sideways?

Yep

No, they both get to the destination, unless one it on the wrong flight.

Same as a ball that's facing sideways.

Next question.

[Wizofoz]

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

Not in the frame where the air is still.

[Jet_Fan]

It's still a pretty imprecise term. What would your answer to the "sideways"case be?

If you move sideways, are you moving forwards or backwards?

[Wizofoz]

Thats the point-neither.

so do you not HAVE a groundspeed?

[Jet_Fan]

Thats the point-neither.

so do you not HAVE a groundspeed?

The groundspeed is still there, it didn't go anywhere. A chopper can have a - groundspeed and a + airspeed at the same time, while a ball just has groundspeed.

[Wizofoz]

Yes, but can you have a groundspeed which is neither poitive nor negative?

A good starting point would be a definition of speed- I always just assumed it was the magnitude of velocity. You seem to have a different take on it.

[Vessbot]

What does a negative ground speed look like?

It looks like you're moving backwards. It's what happens when you fly at 50 knots airspeed into a 60 knot headwind. Groundspeed is -10 knots.

So if two people are standing in the isle facing each other, one is going positive and one negative? Does one therefore not get to the destination?

Fortunately for the guy with the negative groundspeed, the destination is behind him.

[Jet_Fan]

Yes, but can you have a groundspeed which is neither poitive nor negative?

A good starting point would be a definition of speed- I always just assumed it was the magnitude of velocity. You seem to have a different take on it.

Speed in relation to what?

[Jet_Fan]

It looks like you're moving backwards. It's what happens when you fly at 50 knots airspeed into a 60 knot headwind. Groundspeed is -10 knots.



Fortunately for the guy with the negative groundspeed, the destination is behind him.

Yes, good one. :)