For Dr. Tom Van Flandern

Abhi
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21 years 5 months ago #5828 by Abhi

Replied by Abhi on topic Reply from Abhijit Patil

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>
Yes, the band broke first. I'm estimating about 1/10 sec for the sound to reach us. <hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

Please follow this link.

230nsc1.phy-astr.gsu.edu/hbase/tables/soundv.html#c1

I find that speed of sound in rubber is 1600 m/s and that of in air is 343 m/s. That means speed of sound in rubber is almost five times greater than speed of sound in air.

In such situation, the information regarding breaking of band should reach to you first through rubber to your hand and the information through air will reach to you later through air.

In such situation, you must react to first information reaching to you through rubber band. If the information regarding release of opposite force by pole i.e. breaking of rubber band reaches to you with speed faster than that in air, then I don't understand why should you keep pulling the band until you hear sound wave through air if the information is already reached to you much much before through band.

-Abhi.

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Larry Burford
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21 years 5 months ago #5831 by Larry Burford

Replied by Larry Burford on topic Reply from Larry Burford

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>
[Abhi]
I find that speed of sound in rubber is 1600 m/s and that of in air is 343 m/s. That means speed of sound in rubber is almost five times greater than speed of sound in air.
<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>
Uh huh. And the speed of sound in the steel of the slinky is even faster than that.
<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>
In such situation, the information regarding breaking of band should reach to you first through rubber to your hand and the information through air will reach to you later through air.

In such situation, you must react to first information reaching to you through rubber band. If the information regarding release of opposite force by pole i.e. breaking of rubber band reaches to you with speed faster than that in air, then I don't understand why should you keep pulling the band until you hear sound wave through air if the information is already reached to you much much before through band.
<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>
When you release the free end, the band begins to relax. In general, news of this event travels from each point in the band to every other point in the band at *some speed*. But the relaxation of the band starts at and is most complete at the released end and the energy stored in the rest of the band is now accelerating that free end toward the weighted end.

News of this process (the free end moving) travels down the band at *some speed* to the weighted end and manifests itself there as a drop in the tension felt by the test mass.

But most of the band remains stretched and under tension. True, that stretch (and the resultant tension) are now dropping rapidly but they will not drop to zero for hundreds and hundreds of milliseconds. Enough time for even a human being to notice and react. And the last place it will drop to zero is at the connection to the test mass. Until the band is totally relaxed, there is still tension on the test mass.

If you release the test mass at any time after the free end is released but before the tension in the band drops to zero, the test mass is going to accelerate. The longer you wait the lower the tension is and the lower the aceleration will be.

Or at least, so says theory as I learned it. And, the experiments I've described seem to support that theory.

You have a different theory, which experiment (amusing demostrations, really) seems to contradict. But, what if we are misreading them? You can continue to *argue* that we are wrong, (and we will continue to not believe you) or you can prove it.

Do the experiment, show us the data, and make us eat crow (that's an expression that means force us to realize that we are wrong). If we are, we WILL admit it.

I wouldn't mind if it turns out that you are right (because then we get to explore those things you mentinoed that hang in the air after support is removed). I just don't think you are.

In situatinos like this, the burden of proof is on the challenger (that would be you).

Regards,
LB

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Abhi
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21 years 5 months ago #5832 by Abhi

Replied by Abhi on topic Reply from Abhijit Patil

You said that
<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>On our last run, the band broke next to the knot at the pole. And it broke just before we all released the stretced band. We knew this from the buzzing sound, which startled us and triggered a "reflex release" of the band.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

(1) First: The band broke at the pole.

(2) Second: Then you released the band.

(3) Cause: Sound waves with speed reaching to you with velocity some 343 m/s. It startled you and triggered "reflex release" of band. That buzzing sound could have originated from another source. I do not see any connection between hearing the sound and reflex release of band.

(4) Fact: Velocity of sound waves is far less than velocity of sound in band. The information about breaking of band will reach you first through rubber band and NOT through air. Information regarding breaking of band means information regarding non-existence of opposite force by pole. This will cause you to fall on ground in the direction of pulling the band. And it will cause reflex release of band from your hand because you need your hands free to support your body when your body falls on ground. And as you were in process of being thrown and releasing the band, you heard buzzing sound. The time interval was very very small to distingiush what happened when.

(5) My analysis is that the band broke at t = 0, this information regarding non-existence of opposite force by pole reached to you at t = 0. This caused you to be thrown in direction of pulling of band which caused "reflex release" of band from your hand. But as I said, during this time interval, the "whole" band can remain stretched. So even when you released band, there was tension in band which caused test pilot to accelerate towards pole with acceleration much slower than that of broken free end of band i.e. the same way in your slinky example.

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>Or at least, so says theory as I learned it. And, the experiments I've described seem to support that theory.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

The experiment you described does not even support established theory. The information must reach to your hand first through rubber band and then through air to ear later. You are talking about information reaching to you through air first and you are not talking about information through band. Forget my theory and post your experiment to USENET, it will not stand even applying established theory.

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote> I wouldn't mind if it turns out that you are right (because then we get to explore those things you mentinoed that hang in the air after support is removed). I just don't think you are.
In situatinos like this, the burden of proof is on the challenger (that would be you).
Regards,
LB <hr height=1 noshade id=quote></BLOCKQUOTE id=quote>
I am saying that if established knowledge of physics is correct, then things can hang in air without support. But as I think it is not correct. Hence things must fall.

|A

D|
|B G

> downward
|C

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Larry Burford
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21 years 5 months ago #5833 by Larry Burford

Replied by Larry Burford on topic Reply from Larry Burford

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>
[Abhi]
I am saying that if established knowledge of physics is correct, then things can hang in air without support. But as I think it is not correct. Hence things must fall.
<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

Sorry, it appears that I misunderstood your position on this. I thought you were arguing that things WOULD remain in the air, rather than that they would not.

If I now understand you correctly, you are saying that "established knowledge" (IOW Newtonian Physics or, at high speeds and/or high gravitational potentials, Relativistic Physics) predicts that things don't fall. At least, under certain conditions.

And your point is that they must fall, under all conditions.

Please verify that I am stating your position correctly. But, assuming that I am ...

=========

You present and analyze the following thought experiment:

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>
|A

D|
|B G

> downward
|C

F|

ABC is rod attached to roof. ADEFC is coninuous rod. GE is rod placed at point E. There is very small distance between point B and G.

If I release rod ABC so that whole system falls freely, how this information regarding release of rod ABC will travel to point G?

Established knowledge states that it will travel beginning from point A-->D-->E-->G and C-->F-->E-->G. Obviously it will take some time and during this time as point G has no information at all, it will remain stationary in space. As rod ABC is falling and point G is not moving, then this will obstruct falling of rod ABC and hence system in space.

And as established knowledge claims information travelling through decompression waves, then consider that distance between point B and G is so small that rod ABC does not move in space at all. Then in this situation the molecules in rod will not get enough space to relax. And no information will travel through system. Result is whole system hanging in space without support.
<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>
Your analysis is faulty. It is NOT the case that Established Knowledge makes the prediction you claim it does.

Every atom in the device is subject to the force gravity. As soon as the device is released every atom in it is free to respond to this force by accelerating. And each of them do in fact begin to fall.

But AD and CF have been stretched by the weight of DEF and GE, and this stretch will relax now that the device is in free fall. B may or may not contact G as a result of this relaxation, but it makes no difference. The gap at B/G can be any size, including zero, and Established Knowledge predicts that the device and any gaps will fall.

Depending on the springyness of the device, various parts of it might briefly fall at differing rates, but the Center of Mass of the device will fall at exactly the expected rate.

Take my slinky experiment and turn it vertically, so that the unweighted end is supported by one hand and the weighted end is hanging down, supported by the stretched slinky.

Then let it go. The center of mass will fall at 9.8 m/sec^2 but while it is doing this the slinky will be relaxing to its unstretch length. The "top" and the "bottom" will both move toward the center, so that initially the top will fall faster than the middle and the bottom will fall slower.

Suppose that part of the weight was a rod almost as long as the stretched slinky, so that as soon as the slinky is released the rod prevents the slinky from relaxing. Established Knowledge predicts that it will ... fall.

An object that rotates while falling can do something similar - some parts of it can actually be rising while the center of mass is falling. But if you track any specific point on the object through a full rotation, you see that the average rate of fall for that point is the same as for the Center of Mass.

Regards,
LB

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Abhi
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21 years 5 months ago #5945 by Abhi

Replied by Abhi on topic Reply from Abhijit Patil

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>Sorry, it appears that I misunderstood your position on this. I thought you were arguing that things WOULD remain in the air, rather than that they would not.

If I now understand you correctly, you are saying that "established knowledge" (IOW Newtonian Physics or, at high speeds and/or high gravitational potentials, Relativistic Physics) predicts that things don't fall. At least, under certain conditions.

And your point is that they must fall, under all conditions.

Please verify that I am stating your position correctly. But, assuming that I am ...<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

Yes. You understood what I am saying. Established knowldge of Newtonian mechanics predicts that when we remove support of system so that it fall freely, every point of system do NOT receive information regarding removal of support instantaneously neither with speed of light. It receives information ONLY with speed equal to "speed of sound" in material of system. This results in strange situations like

(1) system hanging in space.

(2) G force mg acting in cycle on body where m is mass of "same" body. That means you can feel your own weight mg on your head. And in every cycle if information propagation, this weight gets added because your head has no way to know whether it is it's own weight or weight of some other body. mg+mg+mg...... = (n)mg where n = infinity.
And amazingly gravity also don't know about it. So it keep acting on same body twice, thrice.....

(3) When we leave support, things can fall with acceleration greater than g or less than g.

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>
Every atom in the device is subject to the force gravity. As soon as the device is released every atom in it is free to respond to this force by accelerating. And each of them do in fact begin to fall.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>

This is root of all the problems. You are saying exactly what I am saying on internet in last one and half years. But no one agrees with me. If you go out there with same statement, they will call you crank, crackpot........as they call me..

Ask Dr. Tom, he will say that "you" are wrong. Every atom will NOT fall as soon as we leave support. It will be "hanging in space" until information reaches to every point with speed of sound.

Dr. Tom will say that when we leave support of rod ABC, this information will not reach to point G instantaneously or with speed of light. So if gap between B and G is very small fraction of millimeter, it will strike to point G and system will be temporarily halted in space. Now the point G will feel force mg where m is mass of whole system(amazingly, "m" also includes mass of molecules at point G. But point G has no way to know whether "m" is mass of system or other body from outside system).

Due to this force mg, point G of rod will be compressed at t = 0 (it will not fall because it has NO information that support is released and it has no information that "m" includes its own mass or mass of other body). And now this information regarding addition of force mg will travel through compression wave from G-->E-->D-->A and G-->E-->F-->C. When it reaches rod ABC, it has no way to make difference between point G and support. So force mg which was supposed to be balanced by support point, now will act on point G at t = L/c where L is length of rods and c is speed of wave.

Now point G has no way again to distinguish between force mg acting at t = 0 and and at t = t. So it will think that another additional force mg is placed on it. It will further gets compressed and this information will run towards support point which is G itself. And this cycle will continue.....until rod GE breaks down in space.

Actually I am constructing a situation in which due to propagation delay, my own weight can act thousand, million times at "same point" which is sufficient enough to cause earthquake.

But even in that situation, physicists will say that,
yes, there will be earthquake......

-Abhi.

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Larry Burford
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21 years 5 months ago #5834 by Larry Burford

Replied by Larry Burford on topic Reply from Larry Burford

Abhi,

I'm almost positive this is (partly) a communications problem. When you tell people "standard Physics predicts that things don't fall under certain conditions" most people, knowing that standard Physics does NOT predict this, will mistakenly believe that what you said was "I (Abhi) belive that under some conditions things don't fall". Or something to that effect.

This is what *I* thought you were saying. Because I know standard Physics says doesn't say this.

I'll bet this (or something like it) is what Dr Van Flandern thought you were saying, and I'll bet the people on USENET thought so, too. Because they know standard Physics doesn't say this.

<BLOCKQUOTE id=quote>quote:<hr height=1 noshade id=quote>[Abhi]
Standard Physics makes the following erroneous predictions:

(1) system hanging in space.

(2) G force mg acting in cycle on body where m is mass of "same" body. That means you can feel your own weight mg on your head. And in every cycle if information propagation, this weight gets added because your head has no way to know whether it is it's own weight or weight of some other body. mg+mg+mg...... = (n)mg where n = infinity.
And amazingly gravity also don't know about it. So it keep acting on same body twice, thrice.....

(3) When we leave support, things can fall with acceleration greater than g or less than g.
<hr height=1 noshade id=quote></BLOCKQUOTE id=quote>
Try thinking about the difference between the center of mass of a "stretchy" device and the various parts that make up the device. As long as the device is hanging by one of its parts, the rest of the device will be distorted by the force of gravity acting on its various parts.

Energy will be stored in the various parts of the device by this distortion.

When the device is dropped, the center of mass will fall smoothly at 9.8 m/Sec^2, but that stored energy will cause some parts of it to oscillate with respect to the center of mass and this oscillatory movement can, under some conditions, cause some of the parts of the device to fall at a rate other than 9.8 m/Sec^2.

But, like my previous example with a rotating object, if you average the motion of any point on the device *over a full oscillation* you will get 9.8 m/Sec^2.

When point B touches point G on your device, standard physics does NOT predict that the device stops falling. Point B pushes on point G and point G starts moving, point B slows down a little, rod GE begins to compress, rod ABC begins to bend a little at point B, which changes the changing stretch that CD and AF are experiencing, etc. And all the while gravity is pulling downward on all of the atoms in the device. The motion of any particular point of the device is the result of all the various forces acting on that particular point, and some of those forces are changing rapidly in magnitude and/or direction.

This is very a complex series of events. It is possible in theory to do a detailed analysis of the motion of every atom in this device, but we probably don't have the technology to actually do it yet. And actually measuring the movement of each atom is definitely not yet possible.

But since the device is now in free fall, the internal oscillations of the various parts of the device are *irrelevant to the fall of the device as a whole*.

======

Standard Physics DOES NOT make the prediction you claim.

======

Regards,
LB

PS - Sanity check request - if anyone else sees anything that I have missed or gotten wrong, I'd appreciate a heads up. Thanks.

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