The Windward Turn Theory

[Wizofoz]

Well, velocity is in relation to a frame of reference. If we are talking groundspeed, then relative to the ground.

The magnitude of velocity is the rate of change of position-m/s, km/hr, furlongs/ fortnight- it is how much your positionn has changed in the reference frame and how long the change took. I don't see how that can have a negative value.

[Wizofoz]

How do you quote on an Android??

Vesbot, you didn't deal with the guy standung sideways- is his groundspeed positive or negative?

[Vessbot]

It's embarrassing that the people on the correct side are chasing their tails on definitions and reference frames. Brercrow must be (justifiably) having a good laugh.

Yes, if you consider the aircraft a point object (like a ball) without a heading, then groundspeedvelocity can only have a direction and positive magnitude. But it's slightly more complicated than that, in that an aircraft has a set of body axes and everything is normally calculated relative to them as the direction component of the reference frame.

Again, before anyone got lost in the weeds of this thread, everyone would agree without a moment's hesitation of the trivial truth that in my scenario A, groundspeed is 90 initially and then 110. Not negative 110.

And again, imagine a dead calm day with no wind. Then, everything that is true about airspeed is true about groundspeed. If groundspeed is only positive, then airspeed is always positive, and therefore airplane should fly just the same in a tailslide as they do going forward. True or untrue?

[Vessbot]

How do you quote on an Android??

Vesbot, you didn't deal with the guy standung sideways- is his groundspeed positive or negative?

It's zero in relation to the body axes, but that definition starts being useless then. You can say it's 500 sideways, which is kinda clunky and not useful for much, but it is at least useful in figuring out how soon he'll get to the destination.

Now it's your turn again to deal with the Cessna flying at 50 knots into a 60 knot headwind. What's his groundspeed?

[Jet_Fan]

Well, velocity is in relation to a frame of reference. If we are talking groundspeed, then relative to the ground.

The magnitude of velocity is the rate of change of position-m/s, km/hr, furlongs/ fortnight- it is how much your positionn has changed in the reference frame and how long the change took. I don't see how that can have a negative value.

Right, helicopter flying into a headwind such that it has a groundspeed of -2kts but at the same time an airspeed of 5kts . Pilot of helicopter uses yaw pedal to change heading by 180. What's the helicopter's groundspeed when it's on its new heading?

[Wizofoz]

Vesbot,

Yes we are just talking semantics.

The magnitude of the ground velocity of the Cessna is 10 knots, its direction being 180 degrees to its heading.

Its position over the grond is changing by 10 nautical miles every hour, and there is no such thing as a negative nautical mile.

I would maintain speed is a scalar quantity, and therefore has no direction. By invoking a positive or negative you are now actually defining a reference frame, and we are talking velocity, not speed.

I think our passengers actually demonstrate this- how can you change speed without accelerating?

[Wizofoz]

Jetfan

7kts.

I don't understand your point.

[Jet_Fan]

Jetfan

7kts.

I don't understand your point.

It's flipped around and gone from -2kts groundspeed to +7kts GS. Or are you saying GS is never negative in relation to the motion of an aircraft?

[Wizofoz]

I understood what you meant, sure.

lets say he pedals throgh 90 degrees- now whats jis groundspeed?

[Wizofoz]

Point being, I would say his ground speed has gone from 2kts to 7 knots. His ground velocity frim 2cknots North to 7 knots South (or whatever).

[Jet_Fan]

Point being, I would say his ground speed has gone from 2kts to 7 knots. His ground velocity frim 2cknots North to 7 knots South (or whatever).

A cessna flying into a 60kt headwind at 50kts, what's its GS?

[Capt Pit Bull]

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.

The reference to the Earth is purely one of convenience.

Because we measure wind velocity relative to the Earth!

e.g. if you are heading 270 and you fly through a gradient from 270/10 kts to 270/30 kts then we have a +20 knot effect.

The same as if you went from 090/5 to 270/15.

It really doesn't matter what the absolute values are relative to the Earth, only what the change is.

But, if you wanted to calculate it based on reported wind values,we measure wind relative to a stationary point on the ground. This might give the impression that the relationship to the ground is fundamentally important rather than being a book keeping convenience.

[Capt Pit Bull]

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.

This, and everything else resting upon it, is total bollocks.

Groundspeed is a Scalar. An objects ground vector is comprised of its ground speed and its track.

When an aircraft is heading directly into a wind that exceeds the TAS then it's track is the reciprocal of the heading.

If you try and call that a negative ground speed you are taking account of the same thing twice.

[Vessbot]

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.

This is true. But why does it matter?

In both cases airspeed is positive.

This is true too.

In the first case the groundspeed reverses direction and in the second case the groundspeed is in the same direction.

Also true...

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.

Hold the phone! Why does that justify taking the sum? That makes no sense. The whole premise is that we are comparing the difference between the two groundspeeds (i.., before and after) which, by definition of the mathematical operation, is performed by subtraction. In no universe is addition meaningful to any of this. There is a difference in one of the initial conditions (the airmass is so fast compared to the airspeed that the groundspeed starts out negative) but this does not justify changing the manner in which we compare the initial and final condition. Groundspeed is airspeed minus headwind (equivalent to airspeed plus tailwind) always.

[Capt Pit Bull]

If groundspeed is only positive, then airspeed is always positive, and therefore airplane should fly just the same in a tailslide as they do going forward. True or untrue?

Well, AOA will be cracking on for 180 degrees. That might have a small part to play.....

+ve airspeed and Alpha(plus Beta for 3d) gives us a system that covers all the bases for where the air is coming from.

[Vessbot]

This, and everything else resting upon it, is total bollocks.

Groundspeed is a Scalar. An objects ground vector is comprised of its ground speed and its track.

When an aircraft is heading directly into a wind that exceeds the TAS then it's track is the reciprocal of the heading.

If you try and call that a negative ground speed you are taking account of the same thing twice.

No it's not, if you simply define groundspeed as the component of velocity of the ground relative to the airplane's X axis, signed as positive moving nose-to-tail.

And don't act as if this is esoteric or meaningless, as we, conventionally, do exactly that with airspeed as I've pointed out a few times by now.

[Vessbot]

Well, AOA will be cracking on for 180 degrees. That might have a small part to play.....

+ve airspeed and Alpha(plus Beta for 3d) gives us a system that covers all the bases for where the air is coming from.

Or you can say that the airspeed is negative. I think communication is goddamn impossible sometimes. One time my boss was trying to tell me there's no such thing as negative lift. Definitions and accounting is like 99.9% of most problems.

[Jet_Fan]

This, and everything else resting upon it, is total bollocks.

Groundspeed is a Scalar. An objects ground vector is comprised of its ground speed and its track.

When an aircraft is heading directly into a wind that exceeds the TAS then it's track is the reciprocal of the heading.

If you try and call that a negative ground speed you are taking account of the same thing twice.

...is just another way of saying its GS has become negative.

[Vessbot]

Jet Fan, I think it's time me and you resign from this part of the project, and think up a new and different counterexample to demonstrate that initial and final ground velocity is irrelevant to accelerations that are measured and enacted only by reference to the uniform airmass... :ugh::{

[Jet_Fan]

Jet Fan, I think it's time me and you resign from this part of the project, and think up a new and different counterexample to demonstrate that initial and final ground velocity is irrelevant to accelerations that are measured and enacted only by reference to the uniform airmass... :ugh::{

If you walk down to the tail of an airliner your 'track is [now] the reciprocal of the [your] heading' though. ;)

[Winemaker]

Now it's your turn again to deal with the Cessna flying at 50 knots into a 60 knot headwind. What's his groundspeed?
His ground speed is 10 knots. His ground velocity is 10 knots in the direction of the wind. Speed and velocity are two different things. A good definition of velocity is from Wikipedia, which I use for its clarity:

The velocity of an object is the rate of change of its position with respect to a frame of reference, and is a function of time. Velocity is equivalent to a specification of an object's speed and direction of motion (e.g. 60 km/h to the north)

Please note that velocity must have a frame of reference and that's what's leading to all this BS. Define your frame of reference then do all the vector sums from that frame. You can't mix both the 'air' reference and the 'ground' reference. Use one or the other. There is no energy generated when a plane turns constant circles in a constantly moving air stream. The velocity of the reference air stream over the ground doesn't make one whit of difference to the aircraft.

I'm a dumb winemaker and even I understand this.

[Jet_Fan]

His ground speed is 10 knots. His ground velocity is 10 knots in the direction of the wind. Speed and velocity are two different things. A good definition of velocity is from Wikipedia, which I use for its clarity:



Please note that velocity must have a frame of reference and that's what's leading to all this BS. Define your frame of reference then do all the vector sums from that frame. You can't mix both the 'air' reference and the 'ground' reference. Use one or the other. There is no energy generated when a plane turns constant circles in a constantly moving air stream. The velocity of the reference air stream over the ground doesn't make one whit of difference to the aircraft.

I'm a dumb winemaker and even I understand this.

As you love wiki so much, let's see what it has to say about groundspeed:

Ground speed can be determined by the vector sum (https://en.wikipedia.org/wiki/Vector_sum) of the aircraft's true airspeed (https://en.wikipedia.org/wiki/True_airspeed) and the current wind speed and direction; a headwind subtracts from the ground speed, while a tailwind (https://en.wikipedia.org/wiki/Tailwind) adds to it. Winds at other angles to the heading will have components of either headwind or tailwind as well as a crosswind (https://en.wikipedia.org/wiki/Crosswind) component.

What is 50 - 60?

[Vessbot]

What is 50 - 60?

Apparently groundspeed = | airspeed - headwind |

Ever since the 3rd grade I always wondered when in the world you'd ever use the | | symbol to strip out a negative, and I've finally found it!

[Jet_Fan]

Apparently groundspeed = | airspeed - headwind |

Ever since the 3rd grade I always wondered when in the world you'd ever use the | | symbol to strip out a negative, and I've finally found it!


This might help some in here:

https://www.youtube.com/watch?v=u99LjJ32qOo

[Winemaker]

Thanks Vessbot, you said it for me.

Race Fan, I didn't say I loved Wiki, I said their definition was clear.

[Capt Pit Bull]

...is just another way of saying its GS has become negative.

NO. It really isn't. It might look like it, but if you just choose to invert the sign of something on a whim, expect difficulties if you apply that thought process to a navigation or reference system.

[Vessbot]

NO. It really isn't. It might look like it, but if you just choose to invert the sign of something on a whim, expect difficulties if you apply that thought process to a navigation or reference system.

It's not a whim, it's perfectly in accordance with the simple principle that groundspeed is airspeed minus headwind component.

[Jet_Fan]

Thanks Vessbot, you said it for me.

Race Fan, I didn't say I loved Wiki, I said their definition was clear.

Said what for you?

[Wizofoz]

We are all on the same page, but just someth8ng for the poitive_negative guys to mull- you are defining positive ground speed as being in the direction the aircaft is ponting- well, if there is any cross wind, the aircraft isn' t tracking that way- so is a teack 30 degrees off the nose " moslty positove "

answering the question on a 50 knot airspeed in a 60 knot headwind- how far will the aircaft be from its starting position after one hour?

[bluepost22]

Can I propose a gedanken (thought experiment)?

A Cessna is flying into a 10 kt headwind and has ground speed 90 kts. Instantaneously the headwind increases to 20 kts. How long does it take for the Cessna's ground speed to attain 80 kts?

I've always wanted to know.

[Jet_Fan]

We are all on the same page, but just someth8ng for the poitive_negative guys to mull- you are defining positive ground speed as being in the direction the aircaft is ponting- well, if there is any cross wind, the aircraft isn' t tracking that way- so is a teack 30 degrees off the nose " moslty positove "

answering the question on a 50 knot airspeed in a 60 knot headwind- how far will the aircaft be from its starting position after one hour?

The idea that GS can’t fall into a negative value is really quite stupid. No need to report GS as mainly positive as its regarded as positive unless stated otherwise.

[Winemaker]

Said what for you?

Absolute value.

Speed is distance over time. Velocity is distance over time with a direction component, a vector. The ground speed of this slow flying Cessna is 10 knots; it may not be making any progress towards its intended target, but its speed over the ground is 10 knots. Its velocity vector is 10 knots in the direction of the wind. There's no 'negative' speed, there's speed in the direction opposite to the intended travel. If you want to plot this on an X -Y axis then you could certainly make the direction it ends up traveling as -X, the speed (magnitude) is still 10 knots, the direction is negative.

[Jet_Fan]

Can I propose a gedanken (thought experiment)?

A Cessna is flying into a 10 kt headwind and has ground speed 90 kts. Instantaneously the headwind increases to 20 kts. How long does it take for the Cessna's ground speed to attain 80 kts?

I've always wanted to know.

How long is a bit of string?

[Vessbot]

If you want to plot this on an X -Y axis then you could certainly make the direction it ends up traveling as -X, the speed (magnitude) is still 10 knots, the direction is negative.

Aka the speed is negative 10

[Jet_Fan]

Absolute value.

Speed is distance over time. Velocity is distance over time with a direction component, a vector. The ground speed of this slow flying Cessna is 10 knots; it may not be making any progress towards its intended target, but its speed over the ground is 10 knots. Its velocity vector is 10 knots in the direction of the wind. There's no 'negative' speed, there's speed in the direction opposite to the intended travel. If you want to plot this on an X -Y axis then you could certainly make the direction it ends up traveling as -X, the speed (magnitude) is still 10 knots, the direction is negative.

GS isn’t calculated as the absolute value of airspeed - headwind. If you think it is then you should edit that wiki page accordingly.

[A Squared]

Can I propose a gedanken (thought experiment)?

A Cessna is flying into a 10 kt headwind and has ground speed 90 kts. Instantaneously the headwind increases to 20 kts. How long does it take for the Cessna's ground speed to attain 80 kts?

I've always wanted to know.

You could measure that pretty easily. Start with the the the airplane stabilized in a 100 knot cruise, dive the airplane until the speed is above 110 knots, level out, start timing when the airspeed decreases through 110 knots, and stop timing when the airplane stabilizes at 100 knots again. That's how long it takes.

[Winemaker]

Aka the speed is negative 10

As I have said, I'm not a pilot, I'm a winemaker with an interest in aviation. I think we are confusing aviation terms with physics terms. I am thinking in physics terms; speed is a quantity, a magnitude, and combined with the direction of travel generates a vector. The length of the vector is the speed, the way the vector points is the direction. The vector sum of the wind speed and the slowly retreating Cessna is the resultant that defines the aircraft vector. In this case the speed is 10 knots and the direction is along the negative X axis. The speed is not negative, the resultant of the vectors is -10X after one hour. The plane has traveled 10 nautical miles in a direction opposite to where the pilot wanted to go.

When you are discussing ground speed I assume you are thinking in aviation terms, but this whole thread is not about aviation terms, it's about somehow magically gaining energy when turning windward in a steady wind environment. This seems to me to indicate that we need to treat this as a physics problem, not an interpretation of 'ground speed' in the aviation sense.

So call the Cessna's ground speed -10 knots/hr if you wish, and I'll call it 10 knots/hr at angle Θ, but I think we mean the same thing.

[Wizofoz]

The idea that GS can’t fall into a negative value is really quite stupid. No need to report GS as mainly positive as its regarded as positive unless stated otherwise.

So, if you have a groundspeed of negative thirty knots for an hour is your displacement negative thirty miles from your origin?

[Wizofoz]

Can I propose a gedanken (thought experiment)?

A Cessna is flying into a 10 kt headwind and has ground speed 90 kts. Instantaneously the headwind increases to 20 kts. How long does it take for the Cessna's ground speed to attain 80 kts?

I've always wanted to know.

Well, the math is that drag is now more than thrust, so it will cause acceleration in the "tailwards" direction.

This would be the same as flying at 100 kts, then quickly reducing throttle to the thrust you know will sustain 90 knots.

From experience I would say a matter of 5-10ish seconds.

[Wizofoz]

As you love wiki so much, let's see what it has to say about groundspeed:



What is 50 - 60?

ET TU Wiki-

https://en.wikipedia.org/wiki/Speed

in everyday use and in kinematics (https://en.wikipedia.org/wiki/Kinematics), the speed of an object is the magnitude (https://en.wikipedia.org/wiki/Magnitude_(mathematics)) of its velocity (https://en.wikipedia.org/wiki/Velocity) (the rate of change (https://en.wikipedia.org/wiki/Time_derivative) of its position (https://en.wikipedia.org/wiki/Position_(vector))); it is thus a scalar (https://en.wikipedia.org/wiki/Scalar_(physics)) quantity.

A position can't have a negative rate of change.

[Vessbot]

So, if you have a groundspeed of negative thirty knots for an hour is your displacement negative thirty miles from your origin?

Yes!! When defined as positive one way, negative the other. You know, like that nonsensical useless system called longitude.

[Wizofoz]

Yes!! When defined as positive one way, negative the other. You know, like that nonsensical useless system called longitude.

Thus, you have added direction to magnitude and are talking velocity not speed.

The distance between two paces can't be negative thirty miles.

[Vessbot]

Thus, you have added direction to magnitude and are talking velocity not speed.
OK, and?

text filler

[Vessbot]

OK,

Scenario B2 (really fast running person inside airliner. He runs at 501 knots and can turn around on a dime.)
final groundspeed minus initial groundspeed
(airspeed + tailwind) - (airspeed - headwind)
or, if you like:
(walk + tailwind) - (walk - headwind)
(501+500)-(501-500)
1001-(1)
1000

Scenario B3 (really really fast running person inside airliner. He runs at 1000 knots and can turn around on a dime.)
final groundspeed minus initial groundspeed
(airspeed + tailwind) - (airspeed - headwind)
or, if you like:
(walk + tailwind) - (walk - headwind)
(1000+500)-(1000-500)
1500-(500)
1000

Holy schnike, it turns out that if you substitute any number for the airspeed, the groundspeed difference is 1000, which is twice the wind speed when going from upwind to downwind.

Unless that number is less than the wind speed, then it becomes a special case for some reason that wraps people around the axle over scalars and quantum chromodynamics and whatever else.

[Wizofoz]

OK, and?

text filler

We were talking about ground speed, not ground velocity.

[Vessbot]

We were talking about ground speed, not ground velocity.

Get real. You know the two are collquially interchangable, and you know exactly what I meant. The point isn't just flying by you, you're actively dodging it. Look one post up.

[Wizofoz]

Get real. You know the two are collquially interchangable, and you know exactly what I meant. The point isn't just flying by you, you're actively dodging it. Look one post up.

No, speed and velocity are completely different things and are useful; in different circumstances.- and if your whole argument has been that they are the same, you have been sadly mistaken.

Your one post up is just silly- turning your back does not change your velocity.

[Vessbot]

No, speed and velocity are completely different things and are useful; in different circumstances.- and if your whole argument has been that they are the same, you have been sadly mistaken.

Your one post up is just silly- turning your back does not change your velocity.

Do you think there's an error in the math? If so, what is it? First you weren't happy that there were negative groundspeeds, so now that they're all positive, then what?
Do you apply that same criticism to scenario A? If not, why not? What's the difference?

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

[Wizofoz]

Do you think there's an error in the math? If so, what is it?
Do you apply that same criticism to scenario A? If not, why not? What's the difference?

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

I don't dispute the math- I dispute you are making any pertinent point.

[Vessbot]

To make sure you haven't lost the plot -- what do you think my (attempted) point is?

[Wizofoz]

To make sure you haven't lost the plot -- what do you think my (attempted) point is?

I assume justifying your position that the way you are facing defines "negative" speed.

[Winemaker]

Okay, let's take off in the Cessna and fly at 50 knots. We manage to get aloft, our flight plan is to fly at 360° for one hour. The the wind picks up; all of a sudden it's blowing 60 knots from 90°. Fortunately we've managed to account for the sudden wind and know our starting point. The speed of our little plane is 50 knots and we chug along. We are flying north at 50 knots and we have no headwind; so our ground speed is 50 knots (air speed - wind speed). After one hour, where are we? Well, we've flown 50 nautical miles north, but we've also been blown 60 nautical miles west. Damn, where are we? From our starting point we're about 78 nautical miles away at a bearing of 320°. So (and please correct me if I'm wrong) we are about 78 nautical miles from our starting point, but west and north of where we want to be. So what's our ground speed? Air speed - wind speed = 50 knots/hr. But we're 78 nautical miles from our point of departure! Damn! Our ground speed is 78 knots/hour at a heading of 320°. Direction matters because we are dealing with velocity, not speed. There is no negative speed.

[George Glass]

Yikes! This the scariest thread on this forum. The fact that some professional(???) Pilots might entertain this drivel for a minute is frightening. My employer has recently abandoned Year 12 ( ie A level or equivalent)Physics and Maths as a prerequisite for interview for employment."Human Factors" and gender issues are apparently far more important. Lord save us.The "profession" of Pilot is in big trouble if this becomes the norm.Time for a major re-think and institution of formal training pathways and truly professional qualification standards.

[Wizofoz]

Yikes! This the scariest thread on this forum. The fact that some professional(???) Pilots might entertain this drivel for a minute is frightening. My employer has recently abandoned Year 12 ( ie A level or equivalent)Physics and Maths as a prerequisite for interview for employment."Human Factors" and gender issues are apparently far more important. Lord save us.The "profession" of Pilot is in big trouble if this becomes the norm.Time for a major re-think and institution of formal training pathways and truly professional qualification standards.

Just to be clear, whose drivel are you objecting to?

[Jet_Fan]

ET TU Wiki-

https://en.wikipedia.org/wiki/Speed

in everyday use and in kinematics (https://en.wikipedia.org/wiki/Kinematics), the speed of an object is the magnitude (https://en.wikipedia.org/wiki/Magnitude_(mathematics)) of its velocity (https://en.wikipedia.org/wiki/Velocity) (the rate of change (https://en.wikipedia.org/wiki/Time_derivative) of its position (https://en.wikipedia.org/wiki/Position_(vector))); it is thus a scalar (https://en.wikipedia.org/wiki/Scalar_(physics)) quantity.

A position can't have a negative rate of change.

According to NASA: https://www.grc.nasa.gov/www/k-12/airplane/move2.html

If the plane took off to the West it would have a 20 mph tail wind (wind at your back). This gives a negative wind speed. At lift off, the airspeed is still 100 mph, the wind speed is -20 mph and the ground speed will now be 120 mph.

and:

Comparing this example with the ground reference, (https://www.grc.nasa.gov/www/k-12/airplane/move.html) we see that the magnitudes of all the velocities are the same, but the sign of the wind speed has changed with the reference velocity direction. For a ground reference, we chose a positive wind velocity to be in the same direction as the aircraft (towards the nose). For an aircraft reference, we choose a positive wind velocity to be towards the tail.

There is no negative speed.
​​​​​​​

[Wizofoz]

According to NASA: https://www.grc.nasa.gov/www/k-12/airplane/move2.html



and:




​​​​​​​

but the sign of the wind speed has changed with the reference velocity direction.

They are talking velocity, not speed.

Try it this way- if you drive your car in reverse at -30KPH for an hour, how far are you from your origin?

That figure is a MAGNITITUDE (which is what scalar quantities like speed reference)- they do not have positive or negative signs. if you are 30K from where you were an hour ago, your average SPEED has been 30K/PH.

If you want to reference an origin and call a direction from it negative, you can, but this is velocity, not speed.

[Jet_Fan]

but the sign of the wind speed has changed with the reference velocity direction.

They are talking velocity, not speed.

Try it this way- if you drive your car in reverse at -30KPH for an hour, how far are you from your origin?

That figure is a MAGNITITUDE (which is what scalar quantities like speed reference)- they do not have positive or negative signs. if you are 30K from where you were an hour ago, your average SPEED has been 30K/PH.

If you want to reference an origin and call a direction from it negative, you can, but this is velocity, not speed.

NASA are very clear in their example 'the wind speed is -20 mph' in reference to the plane. A high enough positive sign wind can produce a negative sign GS in reference to a plane. There is no absolute value rule in the calculation for GS.

[Wizofoz]

NASA are very clear in their example 'the wind speed is -20 mph' in reference to the plane. A high enough positive sign wind can produce a negative sign GS in reference to a plane. There is no absolute value rule in the calculation for GS.

I understand they used the term "speed", but they did so referencing a direction- that makes it a velocity.

Please answer my question regarding traveling at what you call a negative speed for a period of time- the displacement you have from your origin is a magnitude without sign or direction- that is what speed measures.

[Wizofoz]

Thing is, we agree on the actual physics, this is totally a matter of semantics- but it DOES lead to confusion when people don't understand the difference between a scalar and a vector quantity- it is part of our Down-Windian friends problem.

[Brercrow]

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 your reasoning is flawed Look at this diagramhttps://cimg6.ibsrv.net/gimg/www.gmforum.com-vbulletin/600x400/groundspeed_conundrum_508b880072f1aa89141b6405ee342eae848d32 60.jpg
Clearly when the airspeed Va is greater than the windspeed W (a) when the aircraft makes a 180, the ground-velocity Vg reverses direction and the difference in groundspeed is twice the wind-speed (in your example 20 correct)
However, when the airspeed is less than the windspeed (b) after the 180 the groundspeed continues in the same direction and the difference in groundspeed is twice the air-speed (In your example 2 not 1000)

[Brercrow]

Wizofoz thinks that wind is the same as no wind because all inertial frames of reference (IFR) are equally valid and he can measure a velocity relative to any IFR

Discuss

[Wizofoz]

Wizofoz thinks that wind is the same as no wind because all inertial frames of reference (IFR) are equally valid and he can measure a velocity relative to any IFR

Discuss

Wizofoz thinks this because Newton proved it.

Brercrow says this is not so, yet simultaneously says he understands Newton.

Explain to Brercrow his mistakes.

[Jet_Fan]

Wizofoz thinks this because Newton proved it.

Brercrow says this is not so, yet simultaneously says he understands Newton.

Explain to Brercrow his mistakes.

Brercrow's thinking is just hopelessly flawed. He's like a flat earth believer. He's a tragic figure.

[hans brinker]

Can we please stop talking about negative winds/speed/feelings and go back to making fun of brercrow?

[Peter H]

Brercrow: Wizofoz thinks that wind is the same as no wind because all inertial frames of reference (IFR) are equally valid and he can measure a velocity relative to any IFR
Discuss

[To be picky. To transform a velocity measure in one IFR to that which would be measured in another IFR you need to factor in the relative velocities of the two IFRs.]

If Wizofoz is wrong it looks like Einstein's theory of relativity must be as well.

https://en.wikipedia.org/wiki/Inertial_frame_of_reference (https://en.wikipedia.org/wiki/Inertial_frame_of_reference)
Measurements in one inertial frame can be converted to measurements in another by a simple transformation (the Galilean transformation (https://en.wikipedia.org/wiki/Galilean_transformation) in Newtonian physics and
the Lorentz transformation (https://en.wikipedia.org/wiki/Lorentz_transformation) in special relativity).

Without exceptional evidence to the contrary I'll side with Wizofoz on this one (but see PS).

Regards, Peter

PS
I'd been wondering if one way of making the discussions more productive would be to try and devise a "critical experiment" which would provide real concrete evidence
rather than opinions and potentially flawed calculations.

I'm thinking of something that could be measured with a GPS receiver on a light plane. You'd probably also need a ground-based windsock and anemometer mounted
sufficiently high to be out of near-ground wind-shear and turbulence.

https://en.wikipedia.org/wiki/Experimentum_crucis
In the sciences (https://en.wikipedia.org/wiki/Science), an experimentum crucis (English (https://en.wikipedia.org/wiki/English_language): crucial experiment or critical experiment) is an experiment (https://en.wikipedia.org/wiki/Experiment) capable of decisively determining whether or not a
particular hypothesis (https://en.wikipedia.org/wiki/Hypothesis) or theory (https://en.wikipedia.org/wiki/Theory) is superior to all other hypotheses or theories whose acceptance is currently widespread in the scientific community.

[DaveReidUK]

Clearly when the airspeed Va is greater than the windspeed W (a) when the aircraft makes a 180, the ground-velocity Vg reverses direction and the difference in groundspeed is twice the wind-speed
However, when the airspeed is less than the windspeed (b) after the 180 the groundspeed continues in the same direction and the difference in groundspeed is twice the air-speed

That's utter nonsense.

It would only work if you could mix velocities and speeds in the same calculation, i.e. ignore the direction component of some terms but not of others. You can't do that in any known universe.

[A Squared]

Wizofoz thinks that wind is the same as no wind because all inertial frames of reference (IFR) are equally valid and he can measure a velocity relative to any IFR

Discuss

Wizofoz thinks this and so does anyone else with a functional grasp of physics. This is established physics. Any problem in newtonian physics will derive the same result in any un-accelerated (inertial) frame of reference. The arithmetic may be more complicated in some frames than others. As PeterH pointed out, you have to include the relative velocities between frames of reference, but in the end, if you do the arithmetic correctly, you get the same result. *THIS* is why people keep telling you over and over that you don't understand newtonian physics. Because you don't. It is abundantly clear to anyone with an understanding of physics that you don't grasp this fundamental concept. It's just the icing on the cake that you actually put it into words and mock Wizofoz, for "believing" something that is a fundamental tenet of physics.

[Brercrow]

Brercrow: Wizofoz thinks that wind is the same as no wind because all inertial frames of reference (IFR) are equally valid and he can measure a velocity relative to any IFR
Discuss

[To be picky. To transform a velocity measure in one IFR to that which would be measured in another IFR you need to factor in the relative velocities of the two IFRs.]

If Wizofoz is wrong it looks like Einstein's theory of relativity must be as well.

https://en.wikipedia.org/wiki/Inertial_frame_of_reference (https://en.wikipedia.org/wiki/Inertial_frame_of_reference)
Measurements in one inertial frame can be converted to measurements in another by a simple transformation (the Galilean transformation (https://en.wikipedia.org/wiki/Galilean_transformation) in Newtonian physics and
the Lorentz transformation (https://en.wikipedia.org/wiki/Lorentz_transformation) in special relativity).

Without exceptional evidence to the contrary I'll side with Wizofoz on this one (but see PS).

Regards, Peter

PS
I'd been wondering if one way of making the discussions more productive would be to try and devise a "critical experiment" which would provide real concrete evidence
rather than opinions and potentially flawed calculations.

I'm thinking of something that could be measured with a GPS receiver on a light plane. You'd probably also need a ground-based windsock and anemometer mounted
sufficiently high to be out of near-ground wind-shear and turbulence.

https://en.wikipedia.org/wiki/Experimentum_crucis
In the sciences (https://en.wikipedia.org/wiki/Science), an experimentum crucis (English (https://en.wikipedia.org/wiki/English_language): crucial experiment or critical experiment) is an experiment (https://en.wikipedia.org/wiki/Experiment) capable of decisively determining whether or not a
particular hypothesis (https://en.wikipedia.org/wiki/Hypothesis) or theory (https://en.wikipedia.org/wiki/Theory) is superior to all other hypotheses or theories whose acceptance is currently widespread in the scientific community.

My dictionary defines the wind as 'motion of the air.' Personally, if I want to know the wind , I would use an anemometer set up on the ground.
What would you do? Read Principia Mathematica - in latin?

[Brercrow]

Wizofoz thinks this and so does anyone else with a functional grasp of physics. This is established physics. Any problem in newtonian physics will derive the same result in any un-accelerated (inertial) frame of reference. The arithmetic may be more complicated in some frames than others. As PeterH pointed out, you have to include the relative velocities between frames of reference, but in the end, if you do the arithmetic correctly, you get the same result. *THIS* is why people keep telling you over and over that you don't understand newtonian physics. Because you don't. It is abundantly clear to anyone with an understanding of physics that you don't grasp this fundamental concept. It's just the icing on the cake that you actually put it into words and mock Wizofoz, for "believing" something that is a fundamental tenet of physics.

It may be established physics.
It is not practical meteorology

[Brercrow]

That's utter nonsense.

It would only work if you could mix velocities and speeds in the same calculation, i.e. ignore the direction component of some terms but not of others. You can't do that in any known universe.

Seriously? Whats wrong with the diagram?

[Brercrow]

Can we please stop talking about negative winds/speed/feelings and go back to making fun of brercrow?

Shame on you

[Brercrow]

Brercrow's thinking is just hopelessly flawed. He's like a flat earth believer. He's a tragic figure.

Jet Fan

You think turn and turd are synonymous.
What does that say about your flying?

[Brercrow]

Wizofoz thinks this because Newton proved it.

Brercrow says this is not so, yet simultaneously says he understands Newton.

Explain to Brercrow his mistakes.

I think that was Galileo

[Wizofoz]

It may be established physics.
It is not practical meteorology

So, the "discussion" doesn't seem to be going you way now, does it...….

Thank you for admitting the fact that in one IRF the air is still (actually, in an INFINATE number, but I digress),

So, that is established physics. Can you dynamically soar instill air?

[Wizofoz]

I think that was Galileo

Yes, "Galilleo's ship" is often sited, but Newton expressed it best.

If you know this, how come you can't figure out it invalidates your claim?

[Wizofoz]

Seriously? Whats wrong with the diagram?

You use airspeeds and groundspeed in the same diagram-that is, vectors from different IRFs. The velocity of an aircraft doesn't "consist of" airspeed components and groundspeed components, it's velocity can be expressed as one or the other (or, indeed, be expressed reference any OTHER IRF). You try and take velocity from one frame and claim it effects the aircraft in another.

You can't do that.

[A Squared]

Wizofoz thinks that wind is the same as no wind because all inertial frames of reference (IFR) are equally valid and he can measure a velocity relative to any IFR

Discuss

then:

It may be established physics.
It is not practical meteorology


Your responses are getting less and less rational as time goes on. This *is* physics, the idea that it has anything more than a passing connection to meteorology is just absurd. If we were discussing a boat in the river would you insist it was hydrology, not physics?

[Jet_Fan]

Jet Fan

You think turn and turd are synonymous.
What does that say about your flying?
Actually, I think your theory is a turd, a stinking one.

[A Squared]

You use airspeeds and groundspeed in the same diagram-that is, vectors from different IRFs. The velocity of an aircraft doesn't "consist of" airspeed components and groundspeed components, it's velocity can be expressed as one or the other (or, indeed, be expressed reference any OTHER IRF). You try and take velocity from one frame and claim it effects the aircraft in another.

You can't do that.

You should probably realize that when the guy with whom you're discussing the physics of a turning airplane says that the principles of physics don't apply because it's "meteorology", any chance of in intelligent conversation has completely vanished.

Jet Fan called it when he said:

Brercrow's thinking is just hopelessly flawed. He's like a flat earth believer. He's a tragic figure.

Exactly, you may as well beat your head against the wall debating with a Flat Earther or a Chemtrail believer. All these type possess a defect in their reasoning. It doesn't matter how rationally you explain his errors, there will always be some irrational defense mechanism which kicks in to avoid seeing reality. In this case, Brercrow has pretty conclusively demonstrated that he's a member of that club by claiming that the principles of physics don't apply to a discussion of the motion of an airplane.

[Jet_Fan]

I never expected the guy to turn up. I just imagined we'd all agree the theory was rubbish. I obviously never expected others to rally to his cause.

His not exactly 100% behind his own theory either. There's zero verifiable information about Colin Taylor on his website, no picture, no nothing. He may not even exist and this could all be a massive troll. Maybe he saw something nasty in the woodshed, involving a naked man wearing a big watch who had a small....

[DaveReidUK]

Seriously? Whats wrong with the diagram?

I've already told you what's wrong with it. :ugh:

It would only work if you could mix velocities and speeds in the same calculation, i.e. ignore the direction component of some terms but not of others. You can't do that in any known universe.

[A Squared]

You can't do that in any known universe.

You can in the universe where the earth is flat, and airliners spray population control chemicals on the masses.

[Goldenrivett]

Brercrow (https://www.pprune.org/members/477529-brercrow)
Seriously? Whats wrong with the diagram?

In diagram A) you say VG = VA +/- W: Therefore difference = 2W.
In diagram B) you say VG = W +/- VA : Therefore difference = 2VA.

In A you use the aircraft’s velocity as a reference, in B you use the Wind vector as a reference.

In the link you posted #201 The Downwind Turn, “Ft” is Zero. It does not exist. There is no FWD / AFT acceleration on the aircraft. There is only a centripetal acceleration towards the centre of the turn in the moving air mass (i.e. relative to the aircraft wing). The ground velocity changes by the vector addition of wind + aircraft's instantaneous heading and airspeed.

If you don't understand the difference then I can't help you.

[Wizofoz]

No fair!!!!

Pprune towers has taken our chew-toy away!

Who are we going to point at and laugh now??