CyCrow
Scholar
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Split Thread The validity of classical physics (split from: DWFTTW)
- Thread starter Dan O.
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What's inconceivable to me is the notion that you've ever DONE anything. You've never even shown the follow-through to respond to any of our questions. You just keep telling us that you "could" easily build a model that advances on the treadmill (when we all know quite well you could not). You tell us you're coming back with the ultimate proof that the treadmill is false (whatever that means). You fail to even take a stance on my bold assertion that 2 + 2 = 4.
Built a pulse jet? I don't think so.
Dan O.
Banned
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Here is an exercise to demonstrate conservation of energy and frames of reference that any school kid can do once they've mastered arithmetic. We will calculate the energy released when 2 masses collide in an inelastic collision.
Assume 2 masses on a collision course (restricting to 1 dimension makes the math easier but you get the same results in 2 or more dimensions). For the example I will use 10 kg moving right at 5 m/s for mass A and 5kg moving right at 10 m/s for mass B. Obviously (except perhaps to humber), mass B is initially to the left of mass A if they are going to collide. We can compute the momentum (mass times velocity) for each of these masses and get 50 kg m/s to the right. The combined mass after the collision is 15 kg with conserved momentum of 100 kg m/s to the right which means it has a velocity of 6.67 m/s to the right. The KE of the initial masses (1/2 m v^2) are 125 and 250 joules and the KE of the combined mass is 333 joules which means that (125+250-333) 42 joules were lost in the collision (converted to heat, sound, spin or some other form).
Now, we can do the same calculation using a different frame of reference such as the initial velocity of mass A. In this new frame, Mass A has a momentum of 0 and KE of 0 while mass B is moving at 5 m/s to the right with a momentum of 25 kg m/s to the right and KE of 62.5 joules. After the collision, the 15 kg combined mass with combined momentum of 25 kg m/s to the right has a velocity of 1.67 m/s to the right giving it a KE of 20.8 joules. The lost KE is again (62.5 - 20.8) 41.7 joules.
You can repeat these calculations for any reference frame and will always get the same answer for the lost KE (except that humber will invariably get one of the vectors backwards and come up with a wrong answer).
Assume 2 masses on a collision course (restricting to 1 dimension makes the math easier but you get the same results in 2 or more dimensions). For the example I will use 10 kg moving right at 5 m/s for mass A and 5kg moving right at 10 m/s for mass B. Obviously (except perhaps to humber), mass B is initially to the left of mass A if they are going to collide. We can compute the momentum (mass times velocity) for each of these masses and get 50 kg m/s to the right. The combined mass after the collision is 15 kg with conserved momentum of 100 kg m/s to the right which means it has a velocity of 6.67 m/s to the right. The KE of the initial masses (1/2 m v^2) are 125 and 250 joules and the KE of the combined mass is 333 joules which means that (125+250-333) 42 joules were lost in the collision (converted to heat, sound, spin or some other form).
Now, we can do the same calculation using a different frame of reference such as the initial velocity of mass A. In this new frame, Mass A has a momentum of 0 and KE of 0 while mass B is moving at 5 m/s to the right with a momentum of 25 kg m/s to the right and KE of 62.5 joules. After the collision, the 15 kg combined mass with combined momentum of 25 kg m/s to the right has a velocity of 1.67 m/s to the right giving it a KE of 20.8 joules. The lost KE is again (62.5 - 20.8) 41.7 joules.
You can repeat these calculations for any reference frame and will always get the same answer for the lost KE (except that humber will invariably get one of the vectors backwards and come up with a wrong answer).
John Freestone
Graduate Poster
I think that may be part of the problem. It's interesting to me because it is a perceptual challenge that I often fail myself, and I have done it here (or particularly on the parent thread). If I am trying to follow something with that binary computation in it somewhere - from lines of computer code (I do a bit of programming) to checking how cogs go round, or doing my accounts - I can often get the complicated bits right, but have to think hard about the +or-. I often get in my own way and reverse things once too often. I did that when I first tried to describe what the land-based frame would look like if it had the same boundary - bits of the land on either side would be moving - er - and I should have said forward (with the cart, of course, like the room moves with the treadmill cart), but confused myself into saying backwards. I did it again with 'the train that would move backwards under humber' if he jumped, when of course I meant forwards.
But I've never known anyone to get things backwards quite so often as humber, and I'm fairly sure now that it's become another tool he uses to dodge issues. I wouldn't mind any amount of difficulty someone has, especially as these are not simple ideas, some of them (that's why it was posed as a brain-teaser and so many physicists get it wrong at first). It's just when it becomes downright arrogance, obfuscation and lying, it's a different problem. What that problem might be would be even more off topic, although I see two theories have been proposed.
CORed
Penultimate Amazing
Humber has demonstrated on numerous occasions that he does not understand the concept of frames of reference.
How many times?
AYou
Little by little you reveal what really drives you. I don't care if you believe me or not. Perhaps you can find the plans for it, and claim it as your own.
Little by little you reveal what really drives you. I don't care if you believe me or not. Perhaps you can find the plans for it, and claim it as your own.
First, this is so puerile. Allow me to hang you from your own respective petards. I will demonstrate by using your own misguided interpretation of "frames" that the cart cannot go DDWFTTW under any conditions.
First, let me say that although frames have been theorized, they are only a notional concept, and not detectable. You cannot put an inertial frame in an envelope and mail it to me any more than you could ship me a magnetic monopole or a Lorenz contraction. You must stop leaning on them so heavily, because the gluons that hold them together are not so strong.
First, consider the ill-fated treadmill. You claim that it is the same as the cart traveling DDWFTTW. But whatever works in one frame must work in all. I choose the frame where I am standing on the floor. There is no wind. How can the cart go faster than nothing? It is not even defined which direction it should go. This is clearly an unresolvable paradox, and it must be so in all frames.
Now the outdoors. First I will choose the frame. I choose the frame of windspeed. Again, there is no wind to the observer. What direction will the cart go? This is like saying that you are west of the north pole. It is nonsense. Thus by your own reasoning, DDWFTTW is impossible. I can similarly show that the treadmill itself does not exist. It is notional. Tell the wife that she can have it back.
First, let me say that although frames have been theorized, they are only a notional concept, and not detectable. You cannot put an inertial frame in an envelope and mail it to me any more than you could ship me a magnetic monopole or a Lorenz contraction. You must stop leaning on them so heavily, because the gluons that hold them together are not so strong.
First, consider the ill-fated treadmill. You claim that it is the same as the cart traveling DDWFTTW. But whatever works in one frame must work in all. I choose the frame where I am standing on the floor. There is no wind. How can the cart go faster than nothing? It is not even defined which direction it should go. This is clearly an unresolvable paradox, and it must be so in all frames.
Now the outdoors. First I will choose the frame. I choose the frame of windspeed. Again, there is no wind to the observer. What direction will the cart go? This is like saying that you are west of the north pole. It is nonsense. Thus by your own reasoning, DDWFTTW is impossible. I can similarly show that the treadmill itself does not exist. It is notional. Tell the wife that she can have it back.
Yes. That is what you would like others to think. Seems though, you are unable to actually defend your position.
Now all you have to do is:
(a) Show that it holds for the treadmiill. Ans: It does not.
(b) Show how it makes the cart go faster Ans: No difference at all.
Conclusion. Pimp my frame of reference.
Speaking of vectors, the translation from 0V w.r.t to backwards-with-the-belt, is wrong. Can you see why?
Woops! The observer will be carried forward by momentum. While airborne, the body will dissipate KE if it moves relative to the air in the caboose.
ETA:
You did not specify in which direction the wind was traveling .w.r.t the ground in the jet example.
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2 + 2 = 4 is only a notional concept. You can't put that in an envelope and mail it either. Is that why you "deny" it?
I knew that you could not let that 4+4 thing go. You had to let me know. It's not that I am psychic, but that you are ploddingly predictable.
No, Huh34 is closer to the answer than you are Dan_O. It is models a physically unrealizable situation. You should see that.
How does the force needed to drive the cart relate to gravity, and how do you know that is the necessary force if the coefficient of friction is indeterminate?
spacediver
Thinker
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humber, the air in the caboose is also moving wrt ground, so there'll be no dissipation.
If an oxygen mask were to be released from the ceiling of a fast moving plane, it will remain stationary so long as the plane maintains its velocity. Yet in your world, it should be dragged by the air molecules inside the cabin.
Do you understand your error here?
How could I!? Incidentally it's more of a "2 + 2" thing.
But really - give it a crack. You can phone a friend (oops - sorry, honest mistake).
CORed
Penultimate Amazing
Well, originally it was a boxcar, but boxcar or caboose, it's enclosed, so the air moves with it, so no KE is dissipated. Humber has to be trolling here, nobody is that dense.
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