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Down wind faster than the wind

Yes, it is still at wind speed. The bar did not change the speed of the air in the room. The bar is also moving at wind speed. This is equivalent to the cart on the ground at wind speed being restrained by a bar that is being constrained to move at wind speed through some mechanism, such as being fixed to a belt that is moving at wind speed. The cart will behave the same in both situations.

No, it means that it is like it is next to the belt, but its wheels are driven by the belt. I expect to see the cart restrained by a bar in the forthcoming video #8.
Such a wonder will truly amaze the crowned heads of Europe.
 
Next slipping wheels reference puts you on fast scroll. I will take it that you are incapable of saying anything else.
Humber does not like being reminded of his mistakes . He's gonna put me on "fast scroll". I'm scared. That's okay, there's lot's more mistakes where that one came from. Like "It doesn't matter which direction the propeller points".
Or "Kinetic energy doesn't change with frame of reference". Or "The treadmill isn't a valid model".

So, you have no ideas why you think so, except that someone has told you?
So you say.
Lot's of ideas. None Humber is capable of understanding.
Want entertainment? Go play in the park.

No, thanks. Reading your nonsense is a lot more entertaining. Way funnier than most comedians.
 
I can do better - with that setup I can decrease the KE of one, leaving the other alone. Choose a frame in which the two objects are moving at 1 m/s in the opposite direction you're going to accelerate towards. Then, as you accelerate it, the one you act on will have its KE decrease to zero.

No again. My datum. I hold the doppler gun. I have one that I made, so I know its characteristics quite well. This is Cartworld City Airport, and one of the carts is leaving for a journey. Now try.

That definition makes no sense whatsoever. You cannot covert KE to heat without acting on the object with something. But depending on the characteristics of that something, you can generate arbitrary amounts of heat. For example, if the object rubs against the surface of a very massive object (like the earth), friction will eventually bring it to rest wrt that surface. But depending on the relative velocity of the surface to the object at the beginning, any amount of energy can be converted to heat.

I plan to crash the device in an inelastic collision. The calorimeter measures the heat so produced.
 
No, it means that it is like it is next to the belt, but its wheels are driven by the belt.

I'm not sure what you mean. If the cart is restrained by a bar constrained to move at wind speed, whether on the treadmill belt or on the road, the cart's performance will be the same. In either case you could say the wheels are driven by the surface, with or without the bar.
 
An inelastic collision with what? You still do not seem to grasp that KE is relative.

The bar is attached to the treadmill frame.
That would mean that the entire treadmill is at windspeed too, and so the room and planet.
The crash barrier has an oil filled buffer, as seen at the ends of a railway track.
That barrier is bolted to the ground 10m away from the stationary cart (object)
 
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humber, do you understand that a reference or inertial frame is not a special kind of picture frame?
 
That would mean that the entire treadmill is at windspeed too, and so the room an planet.

Correct, since we are talking about the wind speed of the air in the room. The surface of the belt is not at wind speed, of course.

The crash barrier has an oil filled buffer, as seem at the ends of a railway track.
That barrier is bolted to the ground 10m away from the stationary cart (object)
You seem to be discussing a scenario now that you have not introduced, and it is unclear what you mean. Keep in mind that if you are talking about the cart on the ground and a barrier bolted to the ground, the equivalent on the belt is a barrier that is bolted to the belt. If you are talking about the cart on the belt and a barrier bolted to the ground, the equivalent for the cart on the ground is a barrier that is moving at wind speed. In either case, any collisions between the cart and the barrier will produce the same amount of heat in both scenarios.
 
Correct, since we are talking about the wind speed of the air in the room. The surface of the belt is not at wind speed, of course.
If that's the case, and it does not matter how the cart stays on the belt, or if the belt is moving, then I may get the idea that someone is trying to get me to equate a spinning propellor with windspeed.

You seem to be discussing a scenario now that you have not introduced, and it is unclear what you mean. Keep in mind that if you are talking about the cart on the ground and a barrier bolted to the ground, the equivalent on the belt is a barrier that is bolted to the belt. If you are talking about the cart on the belt and a barrier bolted to the ground, the equivalent for the cart on the ground is a barrier that is moving at wind speed. In either case, any collisions between the cart and the barrier will produce the same amount of heat in both scenarios.[/QUOTE]
This refers to my question to Sol_Invictus.

ETA:
I have an idea for my own cart. It will stay on my belt, just like yours. I declare my cart to be at 'humberspeed'
 
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If that's the case, and it does not matter how the cart stays on the belt, or if the belt is moving, then I may get the idea that someone is trying to get me to equate a spinning propellor with windspeed.

If the belt of the treadmill is not moving, then the wind speed is zero. Wind speed for the cart on the treadmill is the speed of the air with respect to the surface of the belt, just as on the ground it is the speed of the air with respect to the surface of the road.

When you add connections to the ground for the cart on the treadmill as you did in your bar example, you must add the same connections to an immense object moving at wind speed for the cart on the road, otherwise you are not treating the two frames equivalently.

This refers to my question to Sol_Invictus.
OK, then, you have specified the frame of reference as the inertial frame in which the objects were initially (momentarily and approximately) stationary. That is arbitrary. KE of the first object is non-zero in that frame after acceleration, but not in all frames. KE is relative.
 
I have no need. I flatly deny that frames have any place at all in the role of the treadmill. It is a sham.

But have can you know this when you don't seem to understand what people mean when they talk about a reference frame?
 
If the belt of the treadmill is not moving, then the wind speed is zero. Wind speed for the cart on the treadmill is the speed of the air with respect to the surface of the belt, just as on the ground it is the speed of the air with respect to the surface of the road.
I don't know. It seems from the video that the profile above the belt is OK, but going the wrong way. It's a headwind. This seems to fall almost to zero at prop height, but then at that point, there should be a tailwind.

When you add connections to the ground for the cart on the treadmill as you did in your bar example, you must add the same connections to an immense object moving at wind speed for the cart on the road, otherwise you are not treating the two frames equivalently.
No, I think that immense object would be the ground. I think that the treadmill is the ground. Difficult to argue that it is then going at windspeed.
I can measure no change at all in wind or motion when I turn off the belt.
This seems to conform to my expectations.
OK, then, you have specified the frame of reference as the inertial frame in which the objects were initially (momentarily and approximately) stationary. That is arbitrary. KE of the first object is non-zero in that frame after acceleration, but not in all frames. KE is relative.

Not arbitrary. I have merely set the datum. I have confirmed that to be stable. I am indeed looking for a difference. It would seem to me that from both the 'frame' of the cart and that of the buffer, I should see the same result. No matter how relative, I should see at the work done returned to me.
 
But have can you know this when you don't seem to understand what people mean when they talk about a reference frame?

But I do understand what you mean by a frame of reference. That is why I know the treadmill isn't one. You should too.
Why velocity? You can make the streets of Manhattan your frame, or the top of the Empire State. Can you re-create one frame in another? How?
Yes.. but difficult..and then..... But a simple belt can do all that for velocity?
 
I don't know. It seems from the video that the profile above the belt is OK, but going the wrong way. It's a headwind. This seems to fall almost to zero at prop height, but then at that point, there should be a tailwind.
Surface effects are similar on the road.

No, I think that immense object would be the ground.
But that is obviously wrong. The ground has a different relation to the surface on which the cart travels in the two cases.

I think that the treadmill is the ground. Difficult to argue that it is then going at windspeed.
The speed of the air in the room is zero with respect to the body of the treadmill. That is what "wind speed" means.

I can measure no change at all in wind or motion when I turn off the belt.
This seems to conform to my expectations.
Again, wind is the speed of the air with respect to the belt surface. When you turn off the belt, that does change.

Not arbitrary. I have merely set the datum. I have confirmed that to be stable. I am indeed looking for a difference. It would seem to me that from both the 'frame' of the cart and that of the buffer, I should see the same result. No matter how relative, I should see at the work done returned to me.
That is correct. If the cart accelerates from zero on the treadmill and then crashes into a barrier fixed to the treadmill belt, the same amount of energy will be released as for a cart on the road accelerating from zero and crashing into a barrier fixed to the road.

Your mistake is that you keep wanting the earth to be equivalent in both frames, even though in one it is moving at wind speed and in the other it is moving at surface speed.
 
But I do understand what you mean by a frame of reference.
Followed by incoherent rambling demonstrating that he doesn't have a clue what a frame of reference is.
That is why I know the treadmill isn't one. You should too.
Why velocity? You can make the streets of Manhattan your frame, or the top of the Empire State. Can you re-create one frame in another? How?
Yes.. but difficult..and then..... But a simple belt can do all that for velocity?
 
Surface effects are similar on the road.
No, in the real world, the profile is the same, but going downwind. Also, the road does not generate that wind. It's a road, not a wind. The belt wind is anomalous.

But that is obviously wrong. The ground has a different relation to the surface on which the cart travels in the two cases.
If I take your view, yes, but it also works from mine. Both can't be right.

The speed of the air in the room is zero with respect to the body of the treadmill. That is what "wind speed" means.
I know that is what you say, but I can find no test that contradicts my interpretation. I can find no flaws in it, and it breaks none of Newton's laws, and I need never depart from that one frame, nor enforce that view upon a randomly chosen observer. To take your version, I must accept a lot of flaws, adopt an enforced view, and give Newton something to think about.

Again, wind is the speed of the air with respect to the belt surface. When you turn off the belt, that does change.
Only with respect to the belt, but at the belt surface level, that seems to be contradictory.

That is correct. If the cart accelerates from zero on the treadmill and then crashes into a barrier fixed to the treadmill belt, the same amount of energy will be released as for a cart on the road accelerating from zero and crashing into a barrier fixed to the road.
No, it seems not. Even a fork can halt the cart's progress. It goes backwards too. Therefore, I doubt that such a barrier would indicate the level of KE expected of a 6oz cart traveling at 10mph.

Your mistake is that you keep wanting the earth to be equivalent in both frames, even though in one it is moving at wind speed and in the other it is moving at surface speed.

I don't see that as a mistake.
 

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