humber needs to meet LeeJerry. They are cut from the same cloth. The parallels are absolutely astonishing. And now they have another thing in common - welching on a bet.
http://forums.sailinganarchy.com/index.php?showtopic=82175
Split Thread The validity of classical physics (split from: DWFTTW)
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humber needs to meet LeeJerry. They are cut from the same cloth. The parallels are absolutely astonishing. And now they have another thing in common - welching on a bet.
http://forums.sailinganarchy.com/index.php?showtopic=82175
This is humber's response to being called out for not being a man of his word. Welching on a bet makes you a nothing. And that's just infinitessimally less than we thought you were.
To be fair I'm sure he's never been in a moving vehicle in which he could jump. I suspect he's been fully restrained, horizontal, and most likely sedated for transport. Better safe than sorry.
Sorry, RossFW, I wrote a more detailed reply, but it got lost when logging in.
I will shorten it.
Yes, details are important, but that is not my aim. For those who think I am discussing models, then you must play on your own. I don't know how many more times I need to tell you. The treadmill is the subject.
You think I don't understand your models. Good Grief. I put "if quick" deliberately. Why would I do otherwise? when I know the quick and easy cornflake packet answer of "on the same spot" would do. Your questions transmit the answer? How stupid do you think I am? The answers are always the same! Zero KE, zero V, up, down, no change...
The problem with the treadmill is fundamental. How do you know it works?
How do you compare it with the real world equivalent? You don't have to!
The outcome is the premise upon which it is built. You "can't do that". No way. All you are literally testing for, is that premise of your expectation.
The model has some similarities between the real world , but they are coincidental.
What about my spinning orange? I can do all of the though experiments, the KE experiments, the jump experiments, the van experiments, the laminar flow experiments, and get the same result!
The only difference is the local mechanism that keeps the object where it is. There are other "carts" that will do the same, (or better) and look even more convincing than the orange, or the present cart.
I have not changed my view one iota. Nothing at all. I see the treadmill as being quite simply 'not even wrong'. I have not been struggling to defend myself, or to find ever ingenious ways to avoid you, but to try and use your ideas to show you that you are wrong. That's quite a challenge.
The treadmill has no useful connection with windspeed travel.
We shall see.
Think about his words here. He at times seems to think he'll be coming out with a breathtaking proof that none of us will be able to deny. But he has never offered direct answers to even the most basic questions. He can't imagine how we can't see that we're wrong, and is even frustrated that he can think of no way to demonstrate that we are - even though he's promised to describe a test that would show we're wrong. I suspect in his mind he may have even presented such a test.
And the one hundred times I've said then before.
I know, I think I will put a propeller on either side of the axis of rotation of the orange, and then say "from the frame of reference of the propeller, the orange is going at twice windspeed"
spacediver
Thinker
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Humber, please answer.
I honestly don't think humber is capable of giving you a meaningful answer.
No spacediver, You have the entirely the wrong concept, here. I can't unravel it. Let's say I wait until the next time the train brakes. Why not?
I use my GPS? Why not? The only thing you can ever know about anything, is information that you get about it.
spacediver
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then let me sharpen the question for you.
The hobo is in a cubical enclosure. The four walls are each denoted with a large word:
FRONT
RIGHT
BACK
LEFT
The hobo knows that the cube is traveling in one of those four directions at a constant velocity.
Can the hobo determine which of the four directions it is traveling by tossing something up in the air and watching the way it falls? The hobo is not allowed to use any other information. She cannot wait for the cube to change velocity, and cannot use other technologies, or acoustic cues. She can only use the information provided by the trajectory of the objects she throws into the air.
Assume that the cube is large enough that she can toss things high enough to get good data. If you wish, you can also assume that she has precision measuring tools to help her observe the trajectory of the objects she tosses into the air.
According to your previous claims, this should be easy, because the object will experience a slight disippation due to air resistance, and by noting the direction that the disippation occurs, the hobo can then determine whether the cart is moving forward, backwards, left, or right. So if the cube is moving forwards, any objects tossed up in the air will land slightly towards the BACK wall. If the cube is moving to the right, tossed objects will land slightly closer towards the LEFT wall.
The hobo is in a cubical enclosure. The four walls are each denoted with a large word:
FRONT
RIGHT
BACK
LEFT
The hobo knows that the cube is traveling in one of those four directions at a constant velocity.
Can the hobo determine which of the four directions it is traveling by tossing something up in the air and watching the way it falls? The hobo is not allowed to use any other information. She cannot wait for the cube to change velocity, and cannot use other technologies, or acoustic cues. She can only use the information provided by the trajectory of the objects she throws into the air.
Assume that the cube is large enough that she can toss things high enough to get good data. If you wish, you can also assume that she has precision measuring tools to help her observe the trajectory of the objects she tosses into the air.
According to your previous claims, this should be easy, because the object will experience a slight disippation due to air resistance, and by noting the direction that the disippation occurs, the hobo can then determine whether the cart is moving forward, backwards, left, or right. So if the cube is moving forwards, any objects tossed up in the air will land slightly towards the BACK wall. If the cube is moving to the right, tossed objects will land slightly closer towards the LEFT wall.
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H'ethetheth
fishy rocket scientist
Could you please list the coincidental similarities for me? I'm still trying to figure out your train of of thought.
You repeat this a lot, but that doesn't make it an argument.
Subduction Zone
Muse
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Just a sidenote for humber who is fascinated by the action of rolling oranges in his warehouse job. It does not take much of a slope for a ball to keep rolling in one place on a conveyor belt. A dip that is not easily perceivable to the naked eye would do. If the rolling resistance of the ball (or orange) was the same as the component of the force of gravity parallel to the belt then the ball could roll endlessly in one spot. The rolling resistance of a ball is very small and would be a constant at the speed that a conveyor belt is typically run at.
Notice the lame attempt he had in trying to answer the hobo question? Basically his answer is "if I cheat I can figure it out".
humber, I don't think spacediver will mind if I refine the question a bit more, you we will assume are the hobo. You are nude, no GPS, no super handy pocket gyroscope, nothing. You are in the aforementioned boxcar. There are various boxes and balls and other nontechnological items lying about. The evil scientist announces on a loud speaker that you have 10 minutes to find the "front" of the car, or else the car will be filled with poisonous gas. How would you do it? Earlier you said it could be done easily so ten minutes should be plenty of time. I will start to time you at the beginning of your next post.
Notice the lame attempt he had in trying to answer the hobo question? Basically his answer is "if I cheat I can figure it out".
humber, I don't think spacediver will mind if I refine the question a bit more, you we will assume are the hobo. You are nude, no GPS, no super handy pocket gyroscope, nothing. You are in the aforementioned boxcar. There are various boxes and balls and other nontechnological items lying about. The evil scientist announces on a loud speaker that you have 10 minutes to find the "front" of the car, or else the car will be filled with poisonous gas. How would you do it? Earlier you said it could be done easily so ten minutes should be plenty of time. I will start to time you at the beginning of your next post.
H'ethetheth
fishy rocket scientist
What's with the morbid experiments people think up around here? In this discussion we've seen people on skateboards getting run over by cars, people being shot lest they run away at muzzle velocity, and now hobos or humber being gassed to death?
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Michael C
Graduate Poster
Imagine that we have a cart (an ordinary wheeled trolley, no propeller or anything complicated like that!) with mass m near the top of a sloping hill. As we're viewing it the hill slopes down from upper left to lower right where the ground gently levels out at a height of h metres (vertically) lower than the initial position of the cart. The cart is initially at rest on the hill.
We want to look at this in terms of conservation of energy.
Initially (say at time 0) the cart has no kinetic energy (w.r.t. ground) and potential energy of mgh (if we take the level area at the bottom of the hill as "zero" to keep things simple). When the cart is released, it begins to roll down the hill and ends up rolling at constant velocity v along the level area at some fixed time (t) after being released. At that stage it has zero potential energy and kinetic energy of mv^2/2. Our cart is very streamlined and has very low friction bearings, etc., so we'll ignore the tiny losses from that stuff during the roll down the hill.
At time 0, total energy = mgh
At time t, total energy = mv^2/2
Therefore, conservation of energy tells us that:
mgh = mv^2/2
Pretty much a standard high school type problem so far.
Now we'll look at the same thing again but this time use a frame of reference that moves to the right with the same velocity (v) that the cart ends up with in the earlier description. This means that in our new frame the cart's initial velocity (at the top of the hill) becomes -v, which also means it has non-zero kinetic energy to start with. At time t it will have zero velocity (it's at rest in our new frame) and so its kinetic energy is also zero at that point. Our new energy budget looks like this:
At time 0, total energy = mgh + mv^2/2
At time t, total energy = 0
Therefore, if energy was conserved we'd have:
mgh + mv^2/2 = 0
Clearly that can't be right. The cart had non-zero energy at the top of the hill but zero at the end. What has gone wrong?
If anyone else has a nice physics brainteaser, maybe you could post it in a new thread?
I most strongly object. I am NOT close to the answer, not even in the same universe and I will NEVER get to the answer. I will remain fundamentally wrong.
That is, according to humber, ofcourse.
How did you like me showing you, as you asked, exactly where you were wrong in post #675? You are welcome.
Dan O.
Banned
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Humber cannot figure out if he is moving relative to the ground without resorting to high tech gadgetry. Yet somehow, JB's little 3 wheel cart can do it in a breeze. It seems the answer is that humber isn't as smart as a 3 wheel cart.
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This has been bothering me for a while. I'd like to make sure I understand what is being said in this example.
This appears to show that using conservation of momentum to solve the question gives a different answer than using KE. If the KE doesn't tally up exactly and needs a "fudge factor" of over ten percent attributed to an undefined loss, doesn't that say that the use of KE in this example is flawed?
sol invictus
Philosopher
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What do you mean by "the use of KE in this example is flawed"? KE wasn't "used", it was computed.
Total momentum is always conserved, period. However if you fail to keep track of the momentum of some of the stuff involved, the momentum of what remains might not be conserved. Here Dan is telling us that all the mass is accounted for after the collision; therefore, its momentum is conserved (and that's what he used).
Total energy is always conserved, period. However if you fail to keep track of all forms of energy, what remains might not be. For example, KE is NOT conserved in general, because for example it can be converted into thermal energy. Dan hasn't told us anything about the temperatures before or after, but the information he gave us suffices to tell us that the temperatures have increased due to friction, and in fact we know that the thermal energy increased by 42 joules (assuming no radiation or other means of carrying some of it away).
(Note to humber: "conserved" does not mean what you think it means. Neither does "energy", "mass", "momentum", "acceleration", or "physics".)
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