HUMBER...
These questions are intended seriously. They are being asked so that we can gain a better understanding of your interpretation of physics, and so that you can gain a better understanding of ours, so I'd appreciate if you'd answer them.
Yes, I know. I am not trying to "avoid the answer"
but to show what extremes you must go to, to get the result that you already expect.
There is nothing to understand in that way, that you seem to suggest Brian_M. You are "right".
What good can such an environment be? You can do nothing!. How could it be the basis of a useful idea? This is a pointless discussion in many ways, because the treadmill has nothing to do with frames at all. It should be just a game, to see if you can beat the objections, but in this case, one side simply removes all means until the only possible outcome is that you have nothing left. That would simply be time to got to bed, not the basis of a means of scientific inquiry. They are hypothetical ideas, that have been made real in the most literal of ways, to say that the treadmill could be a frame of reference.
Often, it's hard for us to distinguish between your serious and sarcastic answers, so I'm sure that many of us fail to understand exactly what you're trying to say in many cases.
Assuming the force he applies in jumping is exactly vertical, then yes, we are.
If you believe otherwise, can you tell us why you think so?
Can you explain in more detail why KE would be lost when he jumps? And how would this be different from what would happen if the train wasn't moving?
Detailed answers are never sarcasm. Sometimes, I do wonder what there is left to say. The treadmill is false, it is so. I am not interested in playing parlour games as a sideline. They are trivial. The solution is in any book on the topic. So what?
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Are you saying that that the person's horizontal velocity will change when he is no longer connected to the floor of the box-car?
Note: The air in the box car is traveling at the same speed as the train, and the train is traveling at a constant velocity with no acceleration.
Perhaps I am saying WTF does a hobo in a van have to do with anything?
You know, I have conceded this point before.
But I ask, what do you expect to happen? That's the odd thing. To say "Ah hah, this proves that you can't tell" rather begs the question at to who would think otherwise.
To make a case for whatever "ideas" come from this bizarre isolation ward, you first need to instill the idea the somehow "most people/science" would say the contrary, so you can "fix it" by calling upon faux relativity.
This "proof" is then used to support the treadmills notions; and they are notions, so that anything that is relative is "the same" in someway. You can't tell the real wind from "treadmill wind". In this case you can, but may be you could make a system where the effects were similar, and so would be a good model. The treadmill is full of exemptions to its own rules, to the point where "windspeed" means nothing.
The real road is connected to the ground, so the belt must also be the ground. The earth moving below the cart.
Stand on the treadmill belt so as to move with it. Take a rock from your pocket and throw it on the belt. Just like for the real road, it lands at your feet.
Now, take another rock so that it lands on the
floor. Where is that?
Windspeed, of course. Take a bus and do the same. Windspeed bus.
(No, it is not a 'boundary condition', if it were that would be for the belt, not the air). Treadmill primary rule:- "All things that are not going back with the belt are at windspeed[/U]". There are no intermediary states.
The belt is not the ground or earth, its the equivalent of a real moving road, set in the ground. That's where the rocks go. (Or the air is the ground..) When you realise that, the cart is going nowhere at all, figuratively or literally. It's an orange on a belt or the item slipping on the belt at the supermarket checkout.
Given two situations, where a cart begins at windspeed and released, please explain to me exactly how the forces involved differ...
Situation A: Cart on the road with the wind blowing East.
Relative to the cart, the air is still and the ground (road) is moving West.
Situation B: Cart on a treadmill in an enclosed room.
Relative to the cart, the air is still and the ground (belt) is moving West.
I honestly don't see any difference between these two situations.
What you are trying to say is that a person standing in moving wind, is the same as a person moving in still air?
That's a metaphysical and terminological nightmare. Never ask an expert
"What's the time?"
No, the question is, can you make something that emulates the first situation, so that you can do the work in the comfort of indoors?
The treadmill, should it work, is a test platform, not a "frame"
If you look at Drela's cart proposal, there are two drawings. That is the correct use of "relative velocites". It's convenient to look at it that way, that's all. You can't take that idea, and make it real in the sense of the treadmill.
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You still haven't answered the question about why you don't think a balloon travelling at windspeed would continue to travel at windspeed.
A drawing for that.
And of course, could you answer the skateboarder question for me sometime?
OK
