Down wind faster than the wind

[humber]

If anyone is going to discuss Kinetic Energy^WP here, I suggest they get on the same page and pay particular attention to "Frame of Reference".

And how does that affect two adjacent objects? Those same calculations can be made from any reference. Just don't mix one with another. Don't try and compare the kinetic energy of a rock on the Earth with one on Pluto, and all will be well. You could do, but should they collide, perhaps that will happen when Pluto hits the earth, but then they will be in the same reference.

 

[Marcus]

Do you expect that if you pick up a cart at 10kph, it will have no kinetic energy? Yet, that is obviously so, it shows no reaction at all. It is not travel ling any faster than it appears to be.

When you pick up the cart from the treadmill, you are both moving at the same speed, 18 1/2 miles per second. There is no 10kph difference. If you want to include the earth's orbital velocity as well as rotational, it would be a different number. Kinetic energy is entirely dependent on your reference frame.

 

[Dan O.]

I certainly hope that humber is just a troll. If not, the mental impact when the truth finally hits him will probably destroy him.


ETA: for the benefit of those that can't follow links, in my previous post I wanted to specifically point out...

Thus the kinetic energy of a system is lowest with respect to center of momentum reference frames, i.e., frames of reference in which the center of mass is stationary. In any other frame of reference there is additional kinetic energy corresponding to the total mass moving at the speed of the center of mass.

 

[H'ethetheth]

Tell me: From a frame of reference travelling with the air (or standing still with it), what is the kinetic energy of the cart in both cases?

In the real world 1/2mv2, where V is relative to the ground, in both cases, because the treadmill is also in the real world. It is going as fast
as it appears to be.

I ask you what the kinetic energy is relative to the air, and you tell me what it is relative to the ground. I didn't ask this question for no reason; it is a key to understanding the problem.
I will stress again: Kinetic energy is not absolute:

Relative to the air, the cart has zero kinetic energy.
Relative to the road, the cart has 1/2mV² of kinetic energy.

This is true for both treadmill and road.

Only the ground is relevant.

No. The difference between air and ground velocity is the only thing that's relevant.

It is common to the roadside and treadmill side observer.

No, it is common to a roadside oserver and one standing on top of the moving belt.

You could do it from the belt, but that will not mean any effective change in any parameter. If you find a real moving road then perhaps it may be useful to know from that vantage point, but why consider it?

Because it can be used to demonstrate that this cart will accelerate when it is travelling at wind speed in the comfort of ones own gym.

Velocity in that last case cannot be used to calculate power for example.
It means nothing without knowing the force as well. Low friction in this case, means low force, and it is clearly low

There is a word game here. Equivalency means for example, that if you view on object from another, they remain in the same relative locations. Also equivalency means, that unless velocities are very high, all objects are unchanged relative to each other by their velocities. There are no Einstein effects. Two parked cars may have a lot of KE from traveling around the sun, but they, like all objects on this planet have it too, so the relaative difference is zero. That is our reference frame. There is nothing to be gained from making the observation that one can be viewed equally from the other. So what?
This has nothing to do with the problems of the treadmill. It is a distraction.

It is not a distraction, it just enables one to demonstrate the cart's abilities in a convenient way for youtubers to look at.

Imagine a cart in a 20kph wind but on a belt. When the beltspeed is zero, that is like a fixed road. The only way that the cart can be at 0V relative to a ground observer, is if the belt and wind are equal and opposite. That is, the belt is going back as in the treadmill, at 10kph...

That should be 20kph, I guess?

...Two opposing velocities mean that the cart is stationary.
This is also true for the observer in the car. In the real world, that is the only case, save for no motion at all, where these two views are the same

How is it that the treadmill can model a situation, that of being at windspeed, but stationary to a ground observer and the driver?
There is no relative velocity between you and the driver. When that happens it is not possible to be moving. The model is wrong

The equivalent to a roadside observer is one moving with the belt! Not on the ground!

Only the boundary layer is modeled, and this is insufficient to couple the cart to the belt in a useful way. in fact, it tends to lift the wheels away, which is why the cart appears to float. What should happen, is the cart should go with the belt, dragged back by the missing wind. The cart would need to drive against that to achieve windspeed.
The entire model is wrong, because there is no real world equivalent that it could model.
You are seeing an artifact that results in the balance I have mentioned, that lends the illusion that the cart is at windspeed. Without that, you would not give it a second thought.

Now don't start about boundary layers. Not after having discarded them as subtleties before; a sensible observation I heartily agree with.

It is just this simple:

(1) Reference frame fixed to ground/belt: Cart and air move at 10 kph. Ground does not move.
(2) Reference frame fixed to air/cart: Cart and air do not move. Ground moves at -10 kph.

This is true for both belt and ground, do you agree?

 

[ThinAirDesigns]

The problem is that you can't show the cart going faster than the wind more than a short distance, the length of the treadmill. A test showing the cart going faster than the wind a substantial distance is need to rule out other explanations.

Ok, I'm game -- give me some "other explanations" for:

A: a device that will run indefinitely on a treadmill at *exactly* the speed of the wind while climbing a ~4.5degree incline.

B: Will run indefintely on a level treadmill at *exactly* the speed of the wind while generating enough thrust to tensioning a string held from behind.

C: Can be dragged down to below windspeed and will speed itself up to above windspeed as many times as one cares to drag it down.

I'm not talking about people who don't know physics here -- I'm asking for physics explanations/hyposthesis that fit what we DO know about the cart and it behaviors which could lead to our results, but not DDWFTTW.

I'm not trying to be pissy -- it truly would be a fun conversation to have. Certainly a lot more fun than what humber has presented lately.

Fire away.

JB

 

[spork]

give me some "other explanations" for:

A: a device that will run indefinitely on a treadmill at *exactly* the speed of the wind while climbing a ~4.5degree incline.

B: Will run indefintely on a level treadmill at *exactly* the speed of the wind while generating enough thrust to tensioning a string held from behind.

C: Can be dragged down to below windspeed and will speed itself up to above windspeed as many times as one cares to drag it down.

Hopping. Seriously dude - are you new here!?

 

[RossFW]

New to the forum and won't wade through ALL of this!

Just had a question regarding the "Aircraft on a treadmill" concept. Is someone trying to say an aircraft moving relative to a treadmill, but not moving relative to the earths surface or the air around it (assuming nil wind) will still fly?

I'm actually a pilot, and that is not correct.

(Appologies if I'm "Straw Manning" something that was not meant!!)

 

[Christian Klippel]

...(more corectley a Turbine, as it is being moved by the air, rather than the other way around)....

Sorry to ask, but you have read the thread, did you?

It has been said times and times again that the propeller in this cart works as a propeller, not a turbine. It rotates in the other direction than it should if it would work as a turbine/windmill.

 

[Christian Klippel]

New to the forum and won't wade through ALL of this!

Just had a question regarding the "Aircraft on a treadmill" concept. Is someone trying to say an aircraft moving relative to a treadmill, but not moving relative to the earths surface or the air around it (assuming nil wind) will still fly?

I'm actually a pilot, and that is not correct.

(Appologies if I'm "Straw Manning" something that was not meant!!)

Just saw that you edited your post in the meantime.

No, this is not about an airplane/aircraft on a treadmill. No one said that an actual aircraft with only the treadmill moving below it would fly.

This is about an vehicle that interfaces to the air _and_ to the ground to operate. Take either away and it won't work anymore. And because it interfaces the "ground", for this cart a test on a treadmill is valid.

Greetings,

Chris

 

[humber]

When you pick up the cart from the treadmill, you are both moving at the same speed, 18 1/2 miles per second. There is no 10kph difference. If you want to include the earth's orbital velocity as well as rotational, it would be a different number. Kinetic energy is entirely dependent on your reference frame.

OK. Let's get this old saw out of the way. Are you aware of how many devices from the late 19th century claimed for this effect? Even Tesla had a go, but he was not so honest, anyway. The idea dead and buried. It's a dinosaur. Now apart from that, it would be of no consequence in this case.
Your car may have huge KE from orbiting the sun. Try an insurance claim upon the basis. Now if Dan_O wants to pay extra cover for this, please do, because premiums are increasing, so that helps everybody else. A fool and his money etc.

Now to answer the question. Two cars in the street, both wizzing around the sun. The only thing that matters, is not that, but the difference between them, which is zero. There, solved.

A cart on treadmill has virtually no difference in velocity between one sitting on the nearby ground. Conclusion. No difference. No need to measure the sun. That has all been taken care of, because all things are equally affected, AND they share a common history. Nature keeps pretty tight books, so we can be sure that all credit and debts have long been paid. Everything is the same. All views are the same. Get over it.

 

[John Freestone]

Surely it shouldn't be too hard to build a bespoke windtunnel in miniature (for those with access to a suitable size of workshop and a few common materials). A piece of clear plastic tube somewhere in the region of 10 or 20 cm in diameter could form the tunnel, or one with a rectangular cross section could be fashioned from sheets of acrylic. A suitably small version of the cart could then be run through it, preferably on some kind of track to keep things straight (train track from model shop would be fine). Then all that's needed is a fan to blow air through the tunnel.

Perhaps the most difficult part of this would be suitable methods of data collection. It's difficult to photograph or video correctly, but maybe a set of trip switches at measured points along the track could be triggered by the cart passing over them, and hence its speed found. They could close to give pulses to a computer. This actually makes clear plastic unnecessary, of course, but clear might be useful for other tests, and might be best for measuring the velocity of the air photographically, using suspended particles, bubbles, etc..

 

[humber]

New to the forum and won't wade through ALL of this!

Just had a question regarding the "Aircraft on a treadmill" concept. Is someone trying to say an aircraft moving relative to a treadmill, but not moving relative to the earths surface or the air around it (assuming nil wind) will still fly?

I'm actually a pilot, and that is not correct.

(Appologies if I'm "Straw Manning" something that was not meant!!)

Strawmen dine here.
No, they are saying that the cart on the treadmill is moving as if travelling at windspeed. That is, the belt moving back at 2/ms, is like a road passing underneath at that speed.
If the belt is then said to be the velocity of a downwind, then the cart is traveling at that same speed.

If this happens, it is written that the differential in velocity between the cart and the wind will be zero.
Windspeed - Cartspeed = 0
Windspeed - Beltspeed = 0.
Conclusion: The cart is exactly as if traveling with the wind.

 

[humber]

Surely it shouldn't be too hard to build a bespoke windtunnel in miniature (for those with access to a suitable size of workshop and a few common materials). A piece of clear plastic tube somewhere in the region of 10 or 20 cm in diameter could form the tunnel, or one with a rectangular cross section could be fashioned from sheets of acrylic. A suitably small version of the cart could then be run through it, preferably on some kind of track to keep things straight (train track from model shop would be fine). Then all that's needed is a fan to blow air through the tunnel.

Perhaps the most difficult part of this would be suitable methods of data collection. It's difficult to photograph or video correctly, but maybe a set of trip switches at measured points along the track could be triggered by the cart passing over them, and hence its speed found. They could close to give pulses to a computer. This actually makes clear plastic unnecessary, of course, but clear might be useful for other tests, and might be best for measuring the velocity of the air photographically, using suspended particles, bubbles, etc..

Yes, John. Some difficulties, but nothing insurmountable. Only the motive is missing. I wonder why someone with an idea that they so vehemently support, would let it rust for lack of attention?
They will never test this device under scrutiny.

 

[Christian Klippel]

If the belt is then said to be the velocity of a downwind, then the cart is traveling at that same speed.

Do not try to twist the meanings of what was said and claimed.

What was said, and more than once:
A cart on a treadmill in still air, the cart standing still with respect to the ground the treadmill stands on, with a belt moving at speed x is the same as the cart moving at speed x, in a downwind of the same speed x, on a non-moving ground.

If this happens, it is written that the differential in velocity between the cart and the wind will be zero.

Yes, if there is wind with speed x, and the cart moves at the same speed x, then there is 0 difference between the speed of the cart and the wind. But then, the ground (or belt) is moving at -x relative to the cart, or the wind.

Windspeed - Cartspeed = 0
Windspeed - Beltspeed = 0.
Conclusion: The cart is exactly as if traveling with the wind.

If you assume that the cart is on the street at a day with no wind, or the cart is on a treadmill whose belt doesn't move, in a room with 0 air speed, then yes. But then, nothing would move at all. As soon as there is wind at speed x, or a ground moving at speed x, then there is a difference between windspeed and groundspeed. So, Windspeed (assumed 0) - Beltspeed (assumed non-zero) = -Beltspeed. And logically, if the cart moves at windspeed, then Windspeed - Cartspeed = 0.

Is it really that hard? You do see what is wrong with what you just wrote, do you?

 

[humber]

Do not try to twist the meanings of what was said and claimed.

Close enough. I didn't want to make him laugh too hard.

What was said, and more than once:
A cart on a treadmill in still air, the cart standing still with respect to the ground the treadmill stands on, with a belt moving at speed x is the same as the cart moving at speed x, in a downwind of the same speed x, on a non-moving ground.

Yes, that is what I said, so add that to the pile.

Yes, if there is wind with speed x, and the cart moves at the same speed x, then there is 0 difference between the speed of the cart and

There, you agree, you just forgotted it.

the wind. But then, the ground (or belt) is moving at -x relative to the cart, or the wind.

If you assume that the cart is on the street at a day with no wind, or the cart is on a treadmill whose belt doesn't move, in a room with 0 air speed, then yes. But then, nothing would move at all. As soon as there is wind at speed x, or a ground moving at speed x, then there is a difference between windspeed and groundspeed. So, Windspeed (assumed 0) - Beltspeed (assumed non-zero) = -Beltspeed. And logically, if the cart moves at windspeed, then Windspeed - Cartspeed = 0.

Which is a long way of saying the same thing. Read what you think.

Is it really that hard? You do see what is wrong with what you just wrote, do you?

When I wake up, it often is.
Wrong ? Right? Isn't it all relative?

Fixed it.

 

[spork]

Is it really that hard? You do see what is wrong with what you just wrote, do you?

humber has dedicated 47 pages to not understanding this thing - and to convoluting anything posted by anyone. Do you really have to ask this?


Frankly, I'm willing to stipulate that he has no chance of ever understanding even the most basic principles of physics, if that will get him to stop trying to prove the point.

 

[Christian Klippel]

humber has dedicated 47 pages to not understanding this thing - and to convoluting anything posted by anyone. Do you really have to ask this?


Frankly, I'm willing to stipulate that he has no chance of ever understanding even the most basic principles of physics, if that will get him to stop trying to prove the point.

Agreed. But you must admit that he does quite well in word-twisting.

 

[Christian Klippel]

Which is a long way of saying the same thing. Read what you think.

Dude, do you even remember what you wrote?

You said:

No, they are saying that the cart on the treadmill is moving as if travelling at windspeed. That is, the belt moving back at 2/ms, is like a road passing underneath at that speed.
If the belt is then said to be the velocity of a downwind, then the cart is traveling at that same speed.

If this happens, it is written that the differential in velocity between the cart and the wind will be zero.
Windspeed - Cartspeed = 0
Windspeed - Beltspeed = 0.
Conclusion: The cart is exactly as if traveling with the wind.

If the Belt is moving at 2m/s and there is no wind in the room, then Windspeed - Beltspeed = -2m/s. But you claim it is 0. Aren't you able to do basic math?

Either you are completely dumb, or you are deliberately trolling around. Either way, i'm done with you. You just don't want to understand.

Have fun talking to yourself soon.

 

[humber]

humber has dedicated 47 pages to not understanding this thing - and to convoluting anything posted by anyone. Do you really have to ask this?


Frankly, I'm willing to stipulate that he has no chance of ever understanding even the most basic principles of physics, if that will get him to stop trying to prove the point.

Keep changing the deckchairs. Goodman's, Bauer's and your cart, all exploit momentum. The treadmill is not only a sham, but an unecessary one.
When you can store energy, and nobody is counting, windspeed over an impressive enough distance is easily done.
If you actually knew what you were doing, and replaced you 'intuitions' with 'insight' (not the same thing BTW) you could have fully exploited that idea, and have already won the bet.

FYI.
That long drive shaft acts as a torsion spring. When the propeller is turned by the wind from the front, some energy is lost as it retards the vehicle, but some is stored in that spring and returned. Not a lot but everything counts. Add to that the momentum of the propeller, and you have a nice little energy pump going. Perhaps it may even resonate.

Inappropriate wind measurements, taken on face value rather than long term calculation of the actual energy consumed, would have allowed you stake a claim. Too late.

 

[humber]

Dude, do you even remember what you wrote?

You said:



If the Belt is moving at 2m/s and there is no wind in the room, then Windspeed - Beltspeed = -2m/s. But you claim it is 0. Aren't you able to do basic math?

Either you are completely dumb, or you are deliberately trolling around. Either way, i'm done with you. You just don't want to understand.

Have fun talking to yourself soon.

Volgens mij, niet.