Split Thread The validity of classical physics (split from: DWFTTW)

[H'ethetheth]

I did H'ethetheth, but you kept insisting that the treadmill and BMW/F1 are similar.

Yes. There is a treadmill under the F1 car, isn't there?

No. If there is wind, that is entirely different.

No. It's just one special case of the same model.

Remove the wind, and it's not just a case of less windspeed, but no model at all.

No. It's a fairly accurate model for a tailwind with the magnitude of the belt's speed.

The answer, as I wrote in the BMW/F1 case, is that the belt and wind are equal but opposite. So 100kph wind, means belt goes backwards at that same speed.

So the car sees ground coming by at 100 kph and wind at 100kph from the rear? Is that what you mean? That would be modelling a 200kph tailwind while travelling at 100 kph.

Do you think this is true?
A car in the driveway has a belt running under it, so that is passes under the chassis and so between the wheels. The car is not at beltspeed.
If you lift the car on to that belt, so that it spins the wheels, it is the same as if it were traveling at beltspeed.

Depends on the observer. An observer on the belt will tell you the car was travelling at belt speed all along. First it was flying by at wind speed, then it landed and kept going, except now the wheels are spinning.

You know, your own treadmill proves my point.

I doubt it.

If you stand on the running belt, you will see the cart moving away from you.

Yes

As I said, if the moment you step on the belt, you mentally expand the belt to an infinite plane, there is no motion between you and the belt,...

Depends on how you step onto the belt, but yes, generally.

...so there can be no motion between the cart and belt.

No. Stepping on the belt does not alter the relative velocities between cart and surface. You said so yourself above somewhere.

The reason is that the belt mimics only the surface of the road, so when the belt is expanded to be what it is said to approximate, a boundary free plane, you see only what is attached to that plane.

I'm not sure that that means anything, but I'll try to comment. The belt indeed only mimics the surface of the road. However, the fact that there is air above it that isn't travelling with the belt makes that air mimic a breeze at the (negative) speed of the belt.

There is only you, the cart, and the boundary air layer, and nothing else.

Yes.

There is no moving belt,...

Yes.

so there can be no motion between you and the cart or even the boundary layer.

No. To me the air seems to be moving.

You can move around, and there will be no up belt or down belt and the cart will remain motionless to that plane.

No. You missed the part where there is now wind my frame of reference, therefore the air has the capacity to blow the cart away from me.

This is part of your own idea about isolated frames of reference!

No. Try again.

When you step on the belt as you currently do, then you are effectively seeing other frames. You are looking "out of the window" to see a relative motion that is not strictly available from the belt.

What? With statements this muddled, I can't blame you for making mistakes.

Expansion of the belt, cuts off those views, and you see only what is in that frame. In "beltworld" all the KE's and velocites and so forth are once again normal and Newtonian.

Yes.

When correctly applied, you will see that from ALL frames, the cart is motionless.

No. This is where you jump from sensible statement into the world of papent nonsense. There is only one frame in which the cart is motionless. In other frames, such as all of them, the cart is, in fact, moving.

TAD et al, do not know what the F'k they are taking about.

Seeing how I'm definitely part of "et. al.", I'll assume you won't believe me when I say you're wrong.

 

[H'ethetheth]

Once again, the cart is not moving in any frame.

Okay, I'll repeat this one more time. This statement is very very wrong. It is infinitely wrong, in fact. There is an infinite number of frames in which the cart is moving. The fact that you can make this statement in earnest (as I still assume), means that you cannot possibly understand what a reference frame is.

 

[Christian Klippel]

There is an infinite number of frames in which the cart is moving.

Hello H'ethetheth,

just for the fun of it, i'll make that an infinite number of frames minus one, namely the cart's frame.

Btw, in another post from him he claimed that if someone steps onto the belt, the cart would _not_ move away from the person just stepped onto the belt. Nice example of how he either twists things around, tells someone to be wrong just to say that this someone is wrong, or that he silently corrects his errors without being able to admit being wrong in the first place.

Greetings,

Chris

 

[humber]

Yes. There is a treadmill under the F1 car, isn't there?

There's a belt at the supermarket.

No. It's just one special case of the same model.

The treadmill is an over-simplification of the same model

No. It's a fairly accurate model for a tailwind with the magnitude of the belt's speed. So the car sees ground coming by at 100 kph and wind at 100kph from the rear? Is that what you mean? That would be modelling a 200kph tailwind while travelling at 100 kph.

The road does not move, the car does. The fan simulates the car's motion through the still air, not the wind. FI races are not conducted in 300kph winds, nor do they depend upon the wind to take place. However, it makes no difference to the principle if it is tail or headwind.

Do you think this is true?
A car in the driveway has a belt running under it, so that is passes under the chassis and so between the wheels. The car is not at beltspeed.
If you lift the car on to that belt, so that it spins the wheels, it is the same as if it were traveling at beltspeed.

Depends on the observer. An observer on the belt will tell you the car was travelling at belt speed all along. First it was flying by at wind speed, then it landed and kept going, except now the wheels are spinning.

Take the BMW/FI. Lift the running car with a fork lift, so the wheels don't spin. Now what? Lower it a little so the wheels just make contact so as to spin. Now what?
Do things change for the belt observer? No.

Depends on how you step onto the belt, but yes, generally.
No. Stepping on the belt does not alter the relative velocities between cart and surface. You said so yourself above somewhere.

There is no motion between the cart and belt observer. Only when you include other frames can you ascribe motion. That is incorrect.
There is information leakage across frames.

I'm not sure that that means anything, but I'll try to comment. The belt indeed only mimics the surface of the road. However, the fact that there is air above it that isn't travelling with the belt makes that air mimic a breeze at the (negative) speed of the belt.

The belt is a partial and cyclic simulation of the road surface. The belt motor provides the power to move that simulated surface through the air just above the belt, so generating a boundary layer similar to the real thing.
The belt motor provides the necessary work for this, the bearings and other losses, but not motive power. For an F1 car to travel at 300kph in still wind, takes perhaps 600hp.

No. You missed the part where there is now wind my frame of reference, therefore the air has the capacity to blow the cart away from me.
No. Try again.

That is self-generated wind, derived by motion of the cart through still air.
There is no belt motion from this frame, because the belt cannot have motion w.r.t itself, and so no wind. Only the boundary that is attached to the belt is present, though this is notionally dimensionless.

Expansion of the belt, cuts off those views, and you see only what is in that frame. In "beltworld" all the KE's and velocites and so forth are once again normal and Newtonian.

No. This is where you jump from sensible statement into the world of papent nonsense. There is only one frame in which the cart is motionless. In other frames, such as all of them, the cart is, in fact, moving.

The cart is motionless w.r.t the treadmill observer. The correct model is that this observer is on the ground, as in the BMW/F1 observer.
That position is denied to allow the treadmill observer to be at 'windspeed'. This forces the condition that all frames must be internally consistent within that model. The treadmill 'windspeed' observer sees the cart as motionless, and exclusivity so. Forcing that state upon the belt observer, means that the cart is also at rest w.r.t that observer, and at a fixed distance determined by the initial displacement. That is consistent with all other frames, and no motion in any frame.

Seeing how I'm definitely part of "et. al.", I'll assume you won't believe me when I say you're wrong.

Belief is not relevant, H'ethetheth

 

[humber]

Okay, I'll repeat this one more time. This statement is very very wrong. It is infinitely wrong, in fact. There is an infinite number of frames in which the cart is moving. The fact that you can make this statement in earnest (as I still assume), means that you cannot possibly understand what a reference frame is.

In this case a frame is defined only by "velocity". I have made the point that this implies that any object anywhere traveling at the same speed as the cart, can be said to be in the same frame as the cart. The only supportable case would seem to be that the treadmill is an isolated system, giving specific views.

I hold that there is no frame in which the cart is in motion, including the belt.

This is also consistent with its zero KE status, and why it is not possible to measure it without moving the cart w.r.t the ground. To measure the KE, the cart must move out of the belt's frame.

 

[humber]

Btw, in another post from him he claimed that if someone steps onto the belt, the cart would _not_ move away from the person just stepped onto the belt. Nice example of how he either twists things around, tells someone to be wrong just to say that this someone is wrong, or that he silently corrects his errors without being able to admit being wrong in the first place.

Greetings,

Chris

I know what I said. Try and show me I am wrong about that.

 

[H'ethetheth]

Hello H'ethetheth,

just for the fun of it, i'll make that an infinite number of frames minus one, namely the cart's frame.

Btw, in another post from him he claimed that if someone steps onto the belt, the cart would _not_ move away from the person just stepped onto the belt. Nice example of how he either twists things around, tells someone to be wrong just to say that this someone is wrong, or that he silently corrects his errors without being able to admit being wrong in the first place.

Greetings,

Chris

In similar news, I realized I was slightly wrong before. There isn't just one frame in which the cart isn't moving. There is also an infinite number of frames that can be defined in which the cart is stationary. However, for every single one of these frames, there is an infinite number of frames in which it is not.

Greetings also,

Gijs

 

[John Freestone]

Does this remind anyone else of the ramblings of the HAL 9000 as it was going down?

More like The Titanic.

No. More like HAL. But let's not argue: your views compare quite well with that 'unsinkable' piece of rusting derelict.

We all feel like that from time to time. I have you tried new Relativit-ease?

"...Look Dave, I can see you're really upset about this. I honestly think you should sit down calmly, take a stress pill and think things over. I know I've made some very poor decisions recently, but I can give you my complete assurance that my work will be back to normal...

...Dave....Stop...."

He was managed off, yes.

You must be very pleased with yourself.

 

[H'ethetheth]

In this case a frame is defined only by "velocity".

No, a frame is defined as a vantage point. It has a position and a velocity. No more is needed. All forces observed in this problem in any inertial frame are identical.

I have made the point that this implies that any object anywhere traveling at the same speed as the cart, can be said to be in the same frame as the cart.

Correct.

The only supportable case would seem to be that the treadmill is an isolated system, giving specific views.

No, the treadmill need not be isolated. It just needs to be large enough and the air needs to be still enough.

I hold that there is no frame in which the cart is in motion, including the belt.

Yes, and I can only assure you that this is patent nonsense, even though I said I would repeat it only once more last time.

This is also consistent with its zero KE status, and why it is not possible to measure it without moving the cart w.r.t the ground.

To measure its KE, first we must define what we want to measure it relative to.

To measure the KE, the cart must move out of the belt's frame.

No. To measure the KE, we must move out of the cart's frame and into the belt's frame, or any other of the infinite possible reference frames.

 

[humber]

Here is the puzzle in full. The cart is a little different, but the principle is the same. The position is unstable and friction dependent. It hovers.


Will the propeller-driven Sledge Move?
A model propeller-driven sledge is set in motion by a propeller, is placed on a conveyer belt. What will the model's velocity be relative to the ground,.if the conveyer belt and the sledge simultaneously move in opposite directions i.e. will the sledge be at rest or will it move in either direction?

Solution
This problem is often solved in completely contradictory ways. Some people argue that the propeller-driven sledge will remain immobile, while others say it will move ahead.
Meanwhile the correct answer is the problem as given has no solution at all.

Let us consider two extreme cases. Suppose there is no friction between the conveyer belt and the sledge. Then the belt’s motion will not tell on the sledge’s motion, since the sledge is driven by the propeller. The vehicle will sort of hover over the road and the latter’s motion will not influence the motion of the sledge, just as the motion of the road cannot affect the motion of an aeroplane flying above it.

At the other extreme case, if there is a great deal of friction between the sledge and the conveyer such that the thrust produced by the propeller is insufficient to overcome the friction to the belt.
Then, of course, the vehicle will move in the same direction and with the same velocity as the conveyer belt.

There are as many different sledge velocities as there are intermediate cases. A special case is when the sledge’s position relative to the environment remains unchanged. i.e.. It will be motionless. This will occur when the propeller’s thrust equals the friction (the air resistance being negligible). However, this state would be unstable since even a slight push one way or the other, due to irregularities in the conveyor, will start the vehicle moving relative to the ground in the appropriate direction.

 

[humber]

No. More like HAL. But let's not argue: your views compare quite well with that 'unsinkable' piece of rusting derelict.

Busy re-arranging the deck chairs?

"...Look Dave, I can see you're really upset about this. I honestly think you should sit down calmly, take a stress pill and think things over. I know I've made some very poor decisions recently, but I can give you my complete assurance that my work will be back to normal...

...Dave....Stop...."

It was aimed at you. You ducked and missed.

You must be very pleased with yourself.

Pleasure yourself.

 

[John Freestone]

Zen and the Art of Argument-Recycling Maintenance

Here's an interesting conversation. Its origins are lost in the mists of time now, and who knows whether it will be picked up again. Michael_C may be having a rest or working. Brian M seems to have left it, or may still be deep in contemplation. The Zen Master Himself has been silent on the matter since it was put to him.

A car parked on a moving belt is exactly the same as a car parked on the road in the wind.

But you agree that the only apparent difference, is that on one case the wheels spin, and in another they do not?

In which case do the wheels spin?

Micheal_C, this is not Zen. I can simply ask you to nominate one case, of your choice, leaving the other to become the complement. That would still leave you to explain the difference. When you reject the absolute, the difference becomes that absolute by default.

No, this is not Zen: I asked you a simple question: in which case do the wheels spin? Please give a simple answer: is it the car parked on the belt, or the car parked on the road?

Why do you do that Micheal_C? All the information that you need it there.
You decide. On the floor on jacks, what ever.

Does putting the car on the belt make it move at windspeed, and what is the difference between any other form of car and wheel spinning combination?
No nonsense, please.

We must work out for ourselves why he is right. That is the way it has always been.

Humber - come on, answer the question. If the difference is that a car parked on one of those scenarios has its wheels spinning - no nonsense, please - in which scenario is it?

Or did 'No nonsense, please." just mean, "Don't press me on this issue, because I'm bluffing/have no idea what I'm talking about/am lying."?

 

[H'ethetheth]

Here is the puzzle in full. Will the propeller-driven Sledge Move?
...snip...
Solution
...snip...

I pretty much agree with this, except that it has very little to do with the cart. For the sledge, there is no coupling between speed and thrust, and the cart does not slip at all during normal operation.

 

[humber]

No, a frame is defined as a vantage point. It has a position and a velocity. No more is needed. All forces observed in this problem in any inertial frame are identical.

If that is the case, then the 'windspeed' observer's position is not unique. That implies that ground frame seen by that observer is also in that same frame. Only if each view is unique, can the treadmill's 'windspeed' claim be supported. Either way, there is no motion.

No, the treadmill need not be isolated. It just needs to be large enough and the air needs to be still enough.

The test must be specific to the cart, or there is no test. If the cart is not isolated, one may well choose any of the other objects within the same velocity frame. This would seem to be inappropriate and, well, 'unlikely'.
Only objects in contact with the belt surface are of interest. So perhaps selective, is a better description.

Yes, and I can only assure you that this is patent nonsense, even though I said I would repeat it only once more last time.

It is the point, and the crux of the claim. I deny that as valid.

To measure its KE, first we must define what we want to measure it relative to. No. To measure the KE, we must move out of the cart's frame and into the belt's frame, or any other of the infinite possible reference frames.

Yes, but not without changing the KE w.r.t the ground. The cart's frame is tenuous, because it would seem that it is both in the belt and windspeed frames. The "beltworld" isolation, reveals it from that perspective, in the same way as "treadmillside" isolates the windspeed perspective. Again zero motion and zero KE in each case.
The only energy source from which the KE could be obtained is the belt motor. The only viable coupling to the cart (the air is still w.r.t .the motor and so cannot transfer any useful level of energy), is via friction to the belt, leaving the belt as the only candidate.
Given the belt is a surface model, it seems unlikely that it could carry the KE of the cart. Yet there is no KE for the cart, relative to the ground.
The cart can obtain KE if pushed by the fork, suggesting that the KE actually resides in the cart, and is essentially zero. This too is in accord with the view from the belt as I describe it and with any possible ground frame.

Two questions remain, H'ethetheth

"Take the BMW/FI. Lift the running car with a fork lift, so the wheels don't spin. Now what? Lower it a little so the wheels just make contact so as to spin. Now what? Do things change for the belt observer? No."

and why my smoke test for the boundary flows is not adequate demonstration of failure of equivalence.?

 

[Christian Klippel]

Will the propeller-driven Sledge Move?

And a sledge has exactly what to do with a cart using wheels which drive the prop?

Do you recognize the fact that in the cart, the wheels drive the prop, whereas the skates of sledge do not and can not?

Do you know what it means to compare apples to oranges?

"Hey, i saw something with feathers, so it must be a duck!"

Your wannabe arguments are just ridiculous.

 

[Christian Klippel]

The test must be specific to the cart, or there is no test.

Mhhhh, i smell "fitting the experiment" here. Anyone else got that smell in his/her nose?

 

[RossFW]

This is also consistent with its zero KE status,

You know, I was actually going to try and explain to Humber, yet again, that KE was a relative value, but then I though to hell with it!!!

 

[humber]

I pretty much agree with this, except that it has very little to do with the cart. For the sledge, there is no coupling between speed and thrust, and the cart does not slip at all during normal operation.

There is no coupling in either case. The belt can only enforce the velocity of the wheel, but not the power. That is bound by the friction and the load. As is stated, it either hovers or goes back with the belt.
Therefore, the only matter of concern is how the cart stays in place in the hovering case, but not that it hovers.
The cart is different in that respect, but is explained by the fact that it does not move, rather than being in 'held' in place. If the propellor is driving the cart, the wheel must develop a contact patch behind the axle in order to drive it forward, but if that happens, it cannot of course,remain stationary w.r.t the ground (real motion w.r.t the belt surface). This is paradoxical, so the only solution is that drive is not developed.
It would seem to be the general case. No vehicle can move w.r.t the belt, if driven from the belt. Only if the power is external (a motor or wind driven prop) can controlled motion be sustained. The motor contrives to drive the vehicle to a state of minimum demand, because it cannot move it.

 

[RossFW]

I hold that there is no frame in which the cart is in motion, including the belt.

Guys, it really is time to call it quits.

He doesn't and WON'T understand something as simple as relative motion (E.G, apparently if I drive by the cart at 300KPH in a F1 car, the cart is still motionless from my frame of reference(Now watch him use this with a glib one liner indicating that that is the case and it's ME that doesn't understand the concept of frames of reference!!))

We really should save the bandwith and talk about something worthwhile!!

 

[humber]

And a sledge has exactly what to do with a cart using wheels which drive the prop?
Do you recognize the fact that in the cart, the wheels drive the prop, whereas the skates of sledge do not and can not?
Do you know what it means to compare apples to oranges?
"Hey, i saw something with feathers, so it must be a duck!"
Your wannabe arguments are just ridiculous.

The propeller is driven by the belt ,there is no wind. (Air resistance is in fact ignored). The sledge is used to overcome the wheel problem that bedevils the "plane on a treadmill" issue.
You do not understand the wheel. A wheel is a moving pad of friction that is replaced as the wheel turns. For a number of reasons, that makes them low friction, but that makes no difference to the problem.
The same goes for other classes of vehicle, as I am sure you appreciate.