Pyrolytic Carbon And Magnetic levitation

Any links?

Yes it does add to the weight....we don't have any technology that produces weightlessness....

https://www.youtube.com/watch?v=jOaBnJpIRzM

In your photo,I would readily agree,and the conclusion is logical.

However,there is a toy that uses 2 magnets,one above and one below,with 2 pucks of pyrolytic carbon,one below,and one above to suspend another magnet in between them.

This is a little harder to evaluate using common sense,but I would presume the result would be the same if the entire assembly is weighed on one scale,but if the top was weighed separately from the bottom of the assembly,the top assembly should weigh less and the bottom assembly should weigh more when levitating the center magnet.

Or would it?

The top is pushing down the same amount as the bottom is pushing up.

Would that make the apparent weight on the bottom scale equal to twice the weight

of the suspended magnet?

As we all know,common sense and logic need not apply at the quantum level.

See link below:

https://www.youtube.com/watch?v=jOaBnJpIRzM

It makes no difference if the parts are supported by magnetic fields, springs, or whatever. The forces have to add up. The only difference is magnetic fields are invisible, so it looks like magic.

Except for the spinning superconducting disc, discovered by a Russian scientist,which

was verified by an independent researcher at first,then was denied later.

Even the airline companies expressed interest.

Further interviews by researchers were denied or answered with "No comment".

Conspiracy theories abound.

I have one just like this. I believe it works better with spacers between the magnets and a piece of iron on the bottom. The iron removes the air gap at the bottom and the spacers put more magnetic "resistance" between the magnets so there is more field at the top.

Yes, because the magnets take on the weight of the suspended object when they push it up, just as if it was not repelled but lying on them.

Please view the link and question below,and comment.

thanks

I have viewed the link. Not sure what "the question below" is.

But, as you said, "common sense and logic need not apply".

That must mean that God did it.

Let me clarify my question a bit:

In the link, the magnet is suspended between two magnets and carbon plates.

The forces between the upper and lower are balanced so that the center magnet

levitates.

The top magnet is pressing down on the center magnet with a force equal to the

force of the lower magnet pressing up, so does that result in the bottom magnet

reflecting twice as much downward force toward the lower magnet?

I am imagining the center magnet as repelling the bottom carbon,and the top

carbon,and the harder the top carbon presses by closing the gap,the more pressure

is placed on the bottom carbon.

If the top and bottom halves of the device are weighed independently,what will the

top half weight be,as compared to the bottom half when the distance between the

carbon plates decreases?

Will the top part weigh less as it presses harder,and the bottom weigh more by the

same amount?

It would seem that we would not be measuring merely the weight of the center

magnet, but rather the combined repulsive force of the two carbon plates?

Analogous to placing a magnet on a metal scale plate,and then holding another

magnet of the repelling polarity above it and watching the scale increase the

apparent"weight",when actually you are measuring the repulsive force plus the

weight of the magnet.

The repulsive force would be felt by your hand,and applied to the scale,same as pressing with your hand.

Yes. And if you throw the magnet away and push down with your finger the result is the same.

You're not measuring mass or weight. You are now measuring force.

Yes it adds to the total weight, if you want to move heavy loads with little energy, you will have to "tune" a specific anti-frequency (or several, for omnidirectional travel) of the 3D "magnetic" mesh that supports and shapes our universe.

So....have you figured out how to do that already?

Yes, I have one exactly like the one pictured in #2 by Solar Eagle. Any force supporting the levitating component (the pyrolytic carbon, in this case) would add to the total weight. There is nothing magic here, it makes no difference if it is supported by a magnetic field, springs, or whatever.

Consider both scenarios performed inside a black box: it will weigh the same in either case, yes? In the case of levitation, the supporting magnet is bearing the weight of both itself and the supportee. That there is no physical, 'hardware' connection between the two objects is immaterial; the force of the supportee's weight is borne by the supporter's magnetic field which, in turn, pushes against the supporter (for lack of better terms). That, in turn, pushes against the tabletop, scales, whatever is supporting it all.

Alternatively, think of the supportee being suspended by a piece of thread: what's really supporting it, at the atomic level? Electric fields binding the thread's atoms together? Would the assembly weigh less as a result, simply because they're electric fields? Of course not. The fields commute the force just as your magnets' field does, but on the atomic scale where you can't see what's going on (at least not without squinting real hard :-)

Pyrolytic carbon is strongly diamagnetic. Whereas in ferromagnetic materials the the spin of the material's electrons align parallel to the applied field, in diamagnetic materials a portion align in the opposite direction, opposing the field. Actually, they oppose changes in the field; that is, where there's a gradient -where the field intensity changes depending on where you are.

Put two magnets in opposition and they repel. If you hold both in your hands, one above the other, do you weigh less? No.

Yes, but is the cat dead, or alive?

I don't know about Schrodinger's cat being both dead and alive in a box, but Schrodinger's in a box.

If you want to have some real fun, buy yourself a Levitron. Kind of pricey (about eighty bucks, last time I checked) but, for hard-core geeks, well worth it.

If you weigh a running hourglass, does it weigh less? Part of the sand is in freefall.

The same most of the time, but, slightly less at the beginning before the first grains land, and, slightly more at the end after the last grains have left the upper chamber but are still landing. I'm assuming that the scales don't have the accuracy and time resolution to register individual grains landing.

If the hourglass has air in it, the weight will not change,the sand particle simply displaces an equal weight of air as it falls.

If however the sand is falling in a vacuum, the grain is no longer applying pressure to the vessel being weighed,so it should weigh less until the grain lands at the bottom.

A skydiver does not add or subtract to/from the weight of the Earth when in free

fall,but an object falling in a vacuum will only add weight when it enters the

atmosphere,such as a meteor.

In reference to the hourglass:

This is presuming a very accurate and sensitive scale,accurate to at least 1/2 grain of sand.

The sand displaces neither weight nor volume. If you dropped a bowling ball, does it displace an equal weight of air? No. Does it displace an equal volume of air? No, because it is *already* occupying that volume of air; it is moving through the air, but not displacing it in the sense that a submerged submarine displaces an equal volume of water.

If the hourglass were evacuated - all the air removed, leaving only the sand, would it still work? Does sand fall in a vacuum? So, does the air in an hourglass have anything to do with it apart from contributing its own weight? No.

Imagine two people on some large scales (like might be used to weigh a vehicle); one person is standing directly on the scales whilst the other person is on a stepladder, also on the scales.

The person on ladder drops a bowling ball to the person below, who catches it. Whilst the ball is in free-fall, what happens to the scales' reading? Does it change? Yes. It becomes less by an amount equal to the weight of the bowling ball. Same as with a piece of sand falling through an hourglass. Whilst the grain of sand is in free-fall, the scale does not feel its weight.

If you drop a weight in an enclosure like a grain of sand in an hour glass, what happens is that the enclosure (with contents) is lighter for the time that the object is falling and "makes up for it" when it hits bottom. For example, if a weight of 1 lb. falls for 1 second and stops in 1/100 of a second, the total weight will be less by 1 lb for 1 second and 100 lb heavier for 1/100 of a second. This is because of conservation of momentum. Momentum = force x time.

This assumes that air resistance is negligible. If the air is impeding its fall, the object will be slowed by the air before it hits bottom, and that force will be transferred to the enclosure. Nevertheless, weight x time summed up between the time the object is released and when it is brought to rest at the bottom will always be the same.

There is an demonstration of this experiment at the link below.

Weight of an Hourglass

I have been unable to find verification of the statement I am about to make, but it seems that in defining a free body diagram everything within the diagram must be in force equilibrium. In other words, there can be no unopposed/unbalanced forces generated within the free body. To simplify the hourglass situation, imagine a cross arm (aka gallows) with a cannon ball suspended from it. The whole cross arm + cannon ball system is set on a huge scale and the cannon ball is suspended high enough that the scale has time to adjust while the cannon ball is released and is freely falling. I think we can expect the scale to initially show the weight of the cross arm and cannon ball combined. While the cannon ball is free falling, we would expect the scale to show only the weight of the cross arm. And when the cannon ball comes to rest on the scale, the scale would again show the combined weight. If we have drawn the free body diagram to include the cross arm and cannon ball together then there will have been no change to the forces acting externally on the free body and thus there should be no change in the scale reading. But, while the ball is falling, it is in a condition of unbalanced forces i.e. it is falling/accelerating/moving independently of the remainder of the free body components. Therefore, in this short period of time, the cannon ball and the cross arm must be modeled as separate free bodies. So, the levitation toy can be modeled in its entirety as a single free body but the hourglass and sand cannot be modeled as a single free body.

Exactly.

Once the cannonball is released it is no longer part of the system but a free agent. Instead of falling onto the scale where its weight will be felt again, you could drop it off the side where its weight will not be felt again, but it doesn't matter; whilst it is in free-fall it is out of the picture completely. Same with the falling sand in an hourglass. In both cases no forces are being exerted by or on the objects in free fall. It is why they are in free-fall.

Not so for magnetic levitation. A force is being exerted by the magnetic field on both the levitator and levitatee. That the levitated object is not physically touching anything else is irrelevant; at the atomic level they're not touching anyway, even when they're 'in contact.' Nothing is. At that level it's all forces of one kind or another and where 'in contact' loses its meaning completely. Fundamentally little difference here apart from scale.

I gave you and Europium a GA for your answers.

Can't deny the logic of your answers.

Makes sense to me.

An object, the cannon ball, at rest has the potential energy of gravity's pull and therefore adds to the weight indicated on the scale. When in free fall it is no longer applying force to the scale, but building kinetic energy as it falls until it meets resistance by putting a dent in the bed of the scale. The longer the fall the bigger the dent.

I'm sure your already know this, but weight is not produced or changed by potential energy. Raising the ball higher increases the potential energy but does not affect the weight. Increasing potential energy increases the velocity of the drop and thus the force (which creates the dent) at impact. "Weight" is always and only the product of the mass of the object and the gravitational attraction acting on that mass (W=mg).

You are correct. I should have used mass instead of weight. I often interchange the terms for some so as to not have to explain the difference.

HiTek,

Thanks for a great puzzle. I think this is the case:

If you could put imaginary scales, of infinite resolution and accuracy, under the whole mechanism, and under each part, and you had perfectly uniform and constant gravity, you would see that,

If the whole mechanism, including the magnet, weighs x when the adjuster is at it's farthest distance, then the whole mechanism would weigh x + δ when the mechanism is adjusted closely to give suspension. δ is some very tiny number and represents the mass equivalence of the internal strain of the higher magnetic field. δ is pretty theoretical - we think it ought to be there, but who could measure it without a perfect scale? I don't know where we would find δ hiding; we would have to measure the change in volumes to get an estimate of strains to calculate the energies, so let's just forget δ since we can't weigh it anyway.

With the magnet suspended, the bottom part would weigh the original plus the weight of the magnet plus an extra Δ, which is likely a small but definite fraction of the weight of the magnet, and represents the reaction force of the repulsive force exerted on the magnet.

The top would weight Δ less, since the repulsive force exerted on the magnet would tend to lift the top.

The magnet would weigh nothing.

For any curious person - If the magnet were blue, and were made in New South Wales, what would be the maximum size it could be and still stay stable?

When you consider things at atomic scales, every atom in your body is levitated above the ones below it, from your heels to your head. The only difference here is that you can see the space between the magnet and pyrolytic carbon, but the effect is the same. It's all mostly empty space, permeated by fields.

Were you to somehow enlarge a hydrogen atom so that its nucleus - a single proton - is the size of a marble, and then place it in the center of the Rose Bowl, that atom's sole electron would be found somewhere up in the Nosebleed Section. Mostly empty space, by far.

At atomic scales, furthermore, nothing is touching. Absolutely nothing. At that level 'touching' doesn't even mean anything; it's all fields and forces, and even the location of any matter there loses its conventional meaning; there, at that scale where location, momentum, energy and other properties are all represented by probability waves.

Fluff and fields. Unless one is in free-fall, one result is that every atom in one's body is levitated by something below and captive by something nearby, all by means of fields. Levitated through and through.

Downward force on a mass m in our gravitational field is its Weight.

ref...

http://en.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation

F1=F2=G*M1*M2/(r*r)

We simplify this for most "practical" use by setting

g = G*M1/(r*r) = 6.674e-11 * 5.97219e24 / (6374195*6374195) = ~9.81 m/s2

and using

Force = Weight = m * g

For the magnetic levitation toy, all masses are static and weight should be the gravitational force on the sum of the masses. Yes, the toy weighs more with the added suspended magnet.

The hourglass system is a little different in theory.

Define time t<0 sand at top of hourglass and NOT moving.

Define time t>T sand at bottom of hourglass and NOT moving.

For all practical purposes, the weight of an hourglass does not change from the static sand-up (t<0) to the static sand-down state (t>T).

However, theoretically, the center of mass has slightly changed and the gravitational force (weight) should have also changed since r is now a tiny bit shorter.

This difference can be calculated but is in the "noise" out at 8-10 significant places.

Physics theory, posts, and links here have demonstrated that there is a momentary weight fluctuation while the sand is in motion. This is not disputed.

I'm only trying to remind everyone how the F=mg "formula" was derived. Sometimes we use these formulas so often that we forget they are approximations. Going back to basic physics helps us remember.

In the case of the hourglass, only the time when sand is in free-fall was considered, although this was not stated explicitly (likewise for a few other conditions as well) but, yes, when the system is in equilibrium, the weight will be the same even if the CG has moved.

I think perhaps part of the confusion between mass (a fundamental property of matter) and weight (a force, ie, the product of mass *and* acceleration) is that both use the *same units.* This can be confusing to some who assume the same units describe the same things when in fact they are two very different measures altogether. Hence the need for a distinct unit for force - the newton - which is still not in common use outside the science and engineering disciplines. A person's mass would be in kilograms, for example, whilst their weight would be in newtons.