Mess with your minds: A wind powered craft going faster than a tail wind speed.

[IanB]

With 1:1 gear ratio and no slip is allowed the treadmill will be locked so there will be no movement.

Indeed. But if there is a generator to motor connection instead of a 1:1 belt, then there can be torque conversion and differential wheel speeds. In this case the system will not be locked.

The treadmill will try to push the vehicle back while the back wheel will try to move the vehicle forward so there will be equal but opposite power acting on the belt. In real world the M wheel will slip unless friction between wheel and red box is higher than treadmill power so assuming treadmill is not overpowered the vehicle will move backwards (right to left).

Which wheel slips depends on the coefficient of friction and the weight above the wheel. If there is a heavier weight on M wheel, then maybe the G wheel will slip and the M wheel will grip. Or maybe both wheels will slip a bit. In any given case, you cannot state what will happen without defining further details.

[bdunham7]

Now if you connect the wheel G and M with an ideal belt and gear ratio is 1:1

1:1 doesn't work as I wrote previously (correcting my own error) because those equations of motion that you didn't solve would result in a divide-by-zero error, infinite speed to the right and infinite force.

[PlainName]

the treadmill will be locked

What is locking it?  It would only be locked if the vehicle is prevented from moving.

We've agreed that without the belt the G wheel  is free to rotate and hence doesn't move the vehicle, so with the belt, the only thing working against the G wheel is the M wheel.

The only thing preventing the M wheel from turning is the vehicle, and since the vehicle is free to move the M wheel will move the vehicle.

[electrodacus]

the treadmill will be locked

What is locking it?  It would only be locked if the vehicle is prevented from moving.

We've agreed that without the belt the G wheel  is free to rotate and hence doesn't move the vehicle, so with the belt, the only thing working against the G wheel is the M wheel.

The only thing preventing the M wheel from turning is the vehicle, and since the vehicle is free to move the M wheel will move the vehicle.

Yes the vehicle will prevent from moving.  Think about this way the G wheel is breaking so treadmill pushes the vehicle from right to left and then the M wheel if it can have enough traction and ideal vehicle will push with equal force in the opposite direction so it is like a wedge.
What prevents the M wheel for turning the the G wheel not the vehicle as M wheel is connected to G wheel.

So in summary
a) unconnected wheels ideal vehicle as you mentioned will just free spin and vehicle will stay in place.
b) connected wheels with 1:1 gear ratio will still have the vehicle stand still but this time the treadmill will also stop if slip on wheels is not allowed
In real life there will be losses so power on G will be higher than power at M and so vehicle will move backwards.
c) any gear ratio and real vehicle will result in vehicle moving from right to left.

[PlainName]

the G wheel is breaking

OK, to be explicit the G wheel is being braked by the mass of the vehicle acting through the M wheel, yes? In other words, if the M wheel had no grip then the G wheel would again be free to rotate and the vehicle wouldn't move either way?

[electrodacus]

the G wheel is breaking

OK, to be explicit the G wheel is being braked by the mass of the vehicle acting through the M wheel, yes? In other words, if the M wheel had no grip then the G wheel would again be free to rotate and the vehicle wouldn't move either way?

In order to extract any power from G wheel you will need to essentially break so whatever energy you extract from there if it can go 100% to M wheel then that wheel will try to counter the breaking force. You can compare forces in this particular case as gear ratio is 1:1
So since breaking force on the G wheel equal and opposite direction to M wheel the vehicle can not move in any direction but in real world only part of the breaking force will get to M wheel and thus the vehicle will slide backwards assuming the M wheel is the one that slips.

[PlainName]

You can compare forces in this particular case as gear ratio is 1:1

Why does it have to be 1:1? Suppose it is 2:1 (or 1:2) - doesn't that make a difference?

assuming the M wheel is the one that slips

Why do we assume that? Let's instead assume the M wheel is grippier than the G wheel.

[electrodacus]

You can compare forces in this particular case as gear ratio is 1:1

Why does it have to be 1:1? Suppose it is 2:1 (or 1:2) - doesn't that make a difference?

assuming the M wheel is the one that slips

Why do we assume that? Let's instead assume the M wheel is grippier than the G wheel.

It can be any gear ratio the result will be the same as far as vehicle direction of travel. 
2:1 will mean G wheel turns 2x in the same time M will turn only once but since also the force on G is half of that on M (again ideal case) Power at M and G wheel will be the same just opposite direction so vehicle can not move as with no delta in power you can not change the vehicle kinetic energy.
0.5:1 (1:2) means M wheel can spin 2x faster than G wheel but then force at M is half that at G thus again Power applied by both wheels is equal and so no change in vehicle kinetic energy thus vehicle id not moved.

All of the above is based on ideal case in real case power at wheel M will always be slightly lower than at wheel G due to friction and so vehicle will move from right to left if M wheel slips else is G wheel slips then vehicle will stay where it is and energy balance will all be wasted as heat at the interface between the treadmill surface and the wheel.

So if wheel M has perfect grip vehicle remains stationary while energy delta will end up as heat at the G wheel interface with treadmill moving surface.

[bdunham7]

Is this all a Monty Python skit? 

[electrodacus]

Is this all a Monty Python skit? 

Just make a test.  The energy conservation will not care about your intuition.

[Kleinstein]

It can be any gear ratio the result will be the same as far as vehicle direction of travel. 
2:1 will mean G wheel turns 2x in the same time M will turn only once but since also the force on G is half of that on M (again ideal case) Power at M and G wheel will be the same just opposite direction so vehicle can not move as with no delta in power you can not change the vehicle kinetic energy.
0.5:1 (1:2) means M wheel can spin 2x faster than G wheel but then force at M is half that at G thus again Power applied by both wheels is equal and so no change in vehicle kinetic energy thus vehicle id not moved.

All of the above is based on ideal case in real case power at wheel M will always be slightly lower than at wheel G due to friction and so vehicle will move from right to left if M wheel slips else is G wheel slips then vehicle will stay where it is and energy balance will all be wasted as heat at the interface between the treadmill surface and the wheel.

So if wheel M has perfect grip vehicle remains stationary while energy delta will end up as heat at the G wheel interface with treadmill moving surface.

The power is the same (actually opposite sign) on both sides only in the ideal stationary case. In the ideal case there is no extra power needed to change the kinetic energy as the speeds are constant.   In the non ideal case the power on both side does not have to be the same. If needed to overcome friction or to accelerate the vehicle the power on both sides can be different (e.g. more power going in to accelecate, or more power coming out to slow down). The system has plenty of power (force exchanged through the vehicles times paltform speed)  - so there is no real arguing about not enough power to increase the kinetic energy. The vehicle could move and still produce extra power, e.g. for lights.

---------------------------

To calculate the stationary speed we don't even need to look at forces or power, just look at the velocities. The calculation was shown in a similar way before and is still correct:
The gear ratio (k) connects the speeds of the two wheels and thus the speed of the vehicle relative to the 2 platforms (V1 and V2). So as an equation V2 = k * V1.  The 2 plattforms move relative to each other at a speed V0. So V2 = V1 + V0. 

Do the math with the 2 equations, and one gets V1 = V0 / (k-1).   This gives a valid solution for all gear ratios except k = 1.  So the 1:1 gear ratio does not work as we have seen before.

With a suitable value for the gear ratio you can have V1 at nearly any speed you want: both directions ( sign of V1) and also faster or slower than the relative morement of the platforms. In a reall life situation very high speed ratios may not work well due to friction, but gear rations of -1, 0.5, 1.5 or 3 are not a problem. This would give you half and twice the relative speed of the platforms in both directions.

The equations show the the vehicle could move at the calculated speed to avoid slip or tearing the vehicle appart. The platforms will provide the necessary forces and power to make the vehicle move this way, as this is the path of least restance (only friction).

[PlainName]

Just make a test.

You posted a video of a test, which shows the exact opposite of what you say would happen. Why is that video not actually reality?

Come to that, have you tested your hypothesis? Where did you post the video?

[Alex Eisenhut]

"Braking", not "breaking". Please. I beg you. Make it stop.

[Brumby]

"Braking", not "breaking". Please. I beg you. Make it stop.

That grates on me also ... but that is the least of the transgressions.

[Brumby]

.... and for Pete's sake, can we please abandon these absurd, asinine, daft, foolish, futile, harebrained, idiotic, illogical, imbecilic, laughable, meaningless, mindless, pointless, puerile, ridiculous, senseless, silly and worthless discussions about two wheels and motor/generator function on the vehicle.  There is absolutely NO relevance between these discussions and the mechanism which operates between a wheel and a propeller.

The propeller is essential to the mechanism and ANY discussion which excludes it is simply a waste of time.

Sheesh   

[bdunham7]

.... and for Pete's sake, can we please abandon these absurd, asinine, daft, foolish, futile, harebrained, idiotic, illogical, imbecilic, laughable, meaningless, mindless, pointless, puerile, ridiculous, senseless, silly and worthless discussions about two wheels and motor/generator function on the vehicle.  There is absolutely NO relevance between these discussions and the mechanism which operates between a wheel and a propeller.

The propeller is essential to the mechanism and ANY discussion which excludes it is simply a waste of time.

Sheesh   

OK, so it is a Monty Python skit....I think we should all get together and make a video. 

The propeller, IMO, is a red herring.  Although it is a possible source for misunderstanding the machine, the basic misconceptions here go  way beyond propeller-related issues. Simplifying the the problem to take out some of the variables is usually good way to work past such errors.  See my reply #20.  I would have hoped that once it was understood that a vehicle driven by an overhead belt could be made to go either direction at an arbitrary speed, subbing the propeller and wind back in for the overhead wheel and belt would have made it all easier to understand.  Unfortunately this does not appear to have worked all that well.  But what can you do with someone who claims an advanced knowledge intuition about physics but fails to understand Archimedes, Newton, levers, gears and propellers and tries to applies conservation of energy principles to a system that is not closed?

[IanB]

That grates on me also ... but that is the least of the transgressions.

You're sure it doesn't "great" on you? 

[Brumby]

Don't YOU start!

(Yes, I'm sure)

[electrodacus]

The power is the same (actually opposite sign) on both sides only in the ideal stationary case. In the ideal case there is no extra power needed to change the kinetic energy as the speeds are constant.   In the non ideal case the power on both side does not have to be the same. If needed to overcome friction or to accelerate the vehicle the power on both sides can be different (e.g. more power going in to accelecate, or more power coming out to slow down). The system has plenty of power (force exchanged through the vehicles times paltform speed)  - so there is no real arguing about not enough power to increase the kinetic energy. The vehicle could move and still produce extra power, e.g. for lights.

---------------------------

To calculate the stationary speed we don't even need to look at forces or power, just look at the velocities. The calculation was shown in a similar way before and is still correct:
The gear ratio (k) connects the speeds of the two wheels and thus the speed of the vehicle relative to the 2 platforms (V1 and V2). So as an equation V2 = k * V1.  The 2 plattforms move relative to each other at a speed V0. So V2 = V1 + V0. 

Do the math with the 2 equations, and one gets V1 = V0 / (k-1).   This gives a valid solution for all gear ratios except k = 1.  So the 1:1 gear ratio does not work as we have seen before.

With a suitable value for the gear ratio you can have V1 at nearly any speed you want: both directions ( sign of V1) and also faster or slower than the relative morement of the platforms. In a reall life situation very high speed ratios may not work well due to friction, but gear rations of -1, 0.5, 1.5 or 3 are not a problem. This would give you half and twice the relative speed of the platforms in both directions.

The equations show the the vehicle could move at the calculated speed to avoid slip or tearing the vehicle appart. The platforms will provide the necessary forces and power to make the vehicle move this way, as this is the path of least restance (only friction).

If you remember I posted here a motor connected to a generator and it had also a light bulb (free energy generator or something like that was the text describing that) And what you say about the vehicle is exactly the same thing.

Obviously if vehicle is stationary in ideal case the kinetic energy will remain zero as vehicle is not moving relative to the ground (red box).
And also yes the power will not be equal at G and M in real world but power at M will always be lower than power at G meaning the vehicle can only move from right to left so the Kinetic energy vehicle will have will be in a certain direction.
We are talking about power available at the wheel not motor or generator power as in a motor or engine powered vehicle.
Say transmission efficiency from G wheel to M wheel is 90%
And say breaking power is 100W say that is 10N at 10m/s  then say 90W of this is available at M wheel (due to 10% loss) then you can select any gear ratio you want and say you want to increase force to 20N then speed will be 90W/20N = 4.5m/s
So vehicle will move backwards (right to left) and is only because of the power difference and there is no need to know the forces and speeds as they are irrelevant to know what direction the vehicle will move.
So in this particular example if you store energy for 1 second and then apply that stored energy for another second you can see best what happens.
first you will store the 100W from the generator and while doing so the vehicle moved backwards 10m since M wheel is not powered in this first second.
Then the stored energy 100Ws is used to M wheel but transfer and motor efficiency is just 90% so 90Ws are available.
The vehicle has higher force available but travels slower so with the energy it will be able to travel back 9m thus vehicle is now 1m to the left compared to when it started.
So if you ignore the storage that was just to understand the vehicle moves from right to left at 1m/s speed.
You can calculate this much easier just from the power delta.

As mentioned many time before. If you are not convinced by my correct calculations you just need to test. And I promise if you (any of you) can demonstrate vehicle moving from left to right I will pay you back for the experiment.

[electrodacus]

There is absolutely NO relevance between these discussions and the mechanism which operates between a wheel and a propeller.

The propeller is essential to the mechanism and ANY discussion which excludes it is simply a waste of time.

Sheesh   

Yes propeller is essential for energy storage in air.  Without energy storage this vehicle can not exceed wind speed. But energy storage is the key and that also means vehicle will travel above wind speed only for limited amount of time determined by the amount of stored energy and how low friction the vehicle is as the stored energy is mostly used to cover the losses plus just a bit of acceleration.

The wheel only vehicle is here to demonstrate that it will not work without energy storage as wheel only vehicle is the same as propeller used under water.
As you see plenty of people thing they can make this vehicle in my diagram move from left to right and once they can understand that is impossible they will look in a very different way at the propeller version.

[cgroen]

.
.
As mentioned many time before. If you are not convinced by my correct calculations you just need to test. And I promise if you (any of you) can demonstrate vehicle moving from left to right I will pay you back for the experiment.

The pole you are getting up on is getting very high, remember the distance to fall is also getting longer all the time.....

[Brumby]

OK, so it is a Monty Python skit....

We should be so lucky.

I think we should all get together and make a video.

Veritasium already did a vid... ..... ..... Oh - you mean that sort of video.   Hmmmm...

 
The propeller, IMO, is a red herring.

Can't say I agree with that.

Although it is a possible source for misunderstanding the machine,

I would change "possible" to "certain"

the basic misconceptions here go  way beyond propeller-related issues. Simplifying the the problem to take out some of the variables is usually good way to work past such errors.

I quite agree!

See my reply #20.  I would have hoped that once it was understood that a vehicle driven by an overhead belt could be made to go either direction at an arbitrary speed, subbing the propeller and wind back in for the overhead wheel and belt would have made it all easier to understand.  Unfortunately this does not appear to have worked all that well.

Personally, I found your variation more confusing - but that may be because you moved to a somewhat different framework.  Kudos for making the effort and being creative.

  But what can you do with someone who claims an advanced knowledge intuition about physics but fails to understand Archimedes, Newton, levers, gears and propellers and tries to applies conservation of energy principles to a system that is not closed?

Agree completely on this question - and I am struggling to come up with any suggestions to answer it.

[Brumby]

Yes propeller is essential for energy storage in air.

Serious question:  How do you store energy in air which is not contained in any way?

Your previous hand-waving and "intuition" has not been informative, let alone convincing.

[Alex Eisenhut]

And say breaking power is 100W say that is 10N at 10m/s

Give me a brake!

[bdunham7]

Yes propeller is essential for energy storage in air.

Can you cite any examples, formulas, equations, papers--any evidence at all--to support the theory that is possible to store non-negligible amounts of energy in uncontained air for any significant amount of time?  Or is this a heretofore unnoted phenomenon that you are positing to exist simply because it is the only explanation that suits your 'intuition' and resultant belief that conservation of energy some how rules out all the other explanations for how the vehicle works?