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Anonymous Poster

Extremely Dangerous Gravitational Levitation Experiment

07/21/2006 3:39 PM

Jaime Soto Figueroa writes:
Extremely Dangerous Gravitational Levitation Experiment:

Imagine a sphere or a disk of radius R[m] = 1[m], mass M[Kg] = 1[Kg] rotating around it's natural rotation axis pointing to the Earth mass center at a super high angular velocity of W[r/s] near the Earth surface, this is at some 6375500 [m] from the Earth Mass Center.

Let "K" be a proportion constant for the Moment of Inertia.

For a sphere K = 2/5
For a disk K = 1 / 2

In the case of the Sphere:

Sphere Moment of Inertia = M * R^2 * K = M * R^2 * 2 / 5
Moment of Inertia of a Disk = M * R^2 * K = M * R^2 * 1 / 2

G = Gravitational Acceleration = 9.8[m/s^2]
H = Distance to the Earth Mass Center = 6375000 [m]

Ep [J] = Potential Energy of the Sphere with respect to the Earth Mass Center
Ep [J] = M*G*H = 62475000 [J]

Ec [J] = Kinetic Energy of rotating Sphere
Ec [J] = M *K *R^2 * W^2

¿How much W[r/s] is needed to get Ep = Ec and get gravitational levitation at equilibrium of a sphere? (Or a disk)

I guess that to reach levitation at equilibrium, Ec must be equal to Ep

If Ec = Ep

Then

M*G*H = M *K *R^2 * W^2

W^2 = G*H/(K * R^2)

W = SQRT(G*H / K) / R

Let's see the case of the Sphere, with K = 2/5

If

Ec[J] = Ep[J] = 62475000 [J]

Then

W[r/s] = 12497 [r/s]

This means 1989 revolutions per second

Then the equatorial speed Se[m/s] of the rotating sphere would be

Se[m/s] = W * R = 12497 [m/s].

Note that the necessary equatorial speed is independent of the mass M and the radius R, so smaller spheres should have higher angular speeds.

To get levitation the mass M doesn't matter, the only thing is about the danger of explosion on experimenting ultra high-speed angular velocity, because the higher the mass M the higher the kinetic energy Ec.

Another shape may be a disk, with K = 1/2, but the important thing is to determine which shape is less fragile and less prone to explode, the sphere or the disk.

A disk with R = 1[m] should rotate at 1779 revolutions per second, thus the necessary equatorial speed would be 11178 [m/s], somewhat lower than that for the sphere.

Note that for any revolution body, sphere, disk, cylinder, ring, etc., the necessary equatorial speed is the same for any radius, so for smaller radius the angular speed should be higher, and note that the necessary angular speed is independent of both, mass M and radius R.

In any case the surface quality of the bodies must be premium mirror like, to avoid air resistance and melting with air friction, or the bodies should float in vacuum chambers, and that the material should be hard enough to cope with the strain and stress of centrifugal forces at that super-ultra high angular speeds.

¿Someone dares to do this experiment? One idea is mounting this gadget on top of a tower in the middle of a desert for if it explodes, then the debris fly away well over our heads.

One previous experiment is to measure if the weight of a rotating body decreases with angular speed, if this is true, then it is only a matter how to reach the necessary angular speed for gravitational levitation.

Please if you have budget to do this, invite me to participate. I have a design for this whole experiment for both cases, the disk and the sphere.

Yes, I admit that I may be crazy.

Thank you.

Jaime Soto Figueroa
http://www.matharts.cl/

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#1

Levitation?

07/22/2006 5:26 AM

Excuse me being thick, but why should Ep=Ec produce levitation?

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Anonymous Poster
#2
In reply to #1

Re:Levitation?

07/22/2006 2:22 PM

Honestly I don't know why Ec should be equal to Ep, it is just a guessing. I am not an expert in physics, just a practicing engineer. How ever, for some reason I believe that this experiment should be done.

The only thing is that if mass is affected by speed, then the mass of a rotating disk or sphere should experiment a variation.

Jaime Soto Figueroa

For example one shape may be two counter rotating thin disks. I we view this body from far away we can consider it as one single body or particle without rotation. But the mass of the whole body should increase with speed, ¿so hat about the weight?

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#3
In reply to #2

Re:Levitation?

07/22/2006 3:31 PM

I'm afraid that adding rotational energy to a disk will increase its mass and its weight (through E=mc^2), so forget about getting levitation out of it!

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#10
In reply to #2

Re:Levitation?

07/24/2006 10:22 AM

Is this a formula you created or did you find it somewhere else? I'm inclined to belivee that it came from somewhere else since there are progressions you don't understand (e.g., Ec=Ep). So in that case, where did it come from? If it's from another website, maybe they can shed some light on what is happening here.

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#4

Extremely Dangerous Gravitational Levitation Exper

07/23/2006 3:00 AM

I can't remember the name of the scientist, but there was one in the late 1800's that was facinated with algular momentum. It is said that this guy would place fast-spinning flywheels into suitcases, leave them laying flat at railroad stations, and then howl at the response of the porter when he picked up the suitcase by the handle and the case would rise in the air without pivoting down. Supposedly a true story.

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#5
In reply to #4

Re:Extremely Dangerous Gravitational Levitation Ex

07/23/2006 3:41 AM

Gyroscope effects of the spinning flywheels could surely have had strange effects. Could not have been true levitation though!

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#6
In reply to #5

Re:Extremely Dangerous Gravitational Levitation Ex

07/23/2006 4:57 AM

I know it has nothing to do with levitation... But it must have been a real hoot in the early 1900's!

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#7
In reply to #6

Re:Extremely Dangerous Gravitational Levitation Ex

07/24/2006 2:52 AM

Yes, pretty cool! The gyro's precession effect probably made the one end of the suitcase pitch up (not rising 'flat')?

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#8
In reply to #7

Re:Extremely Dangerous Gravitational Levitation Ex

07/24/2006 3:35 AM

At the very least, I don't think I'd want to try this stunt in any airport baggage area for awhile!

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Anonymous Poster
#9

re: your post

07/24/2006 10:03 AM

why do you think it would explode? speed of light= 300 million m/s your rotational speed is well below that. what is your force translator for your experiment?

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Anonymous Poster
#11
In reply to #9

Re: your post

07/24/2006 1:17 PM

"Errata:

I wrote

"and note that the necessary angular speed is independent of both, mass M and radius R."

The correct phrase is

"and note that the necessary angular speed is independent of mass M and inversely proportional to radius R." (Because equatorial speed is independent of both, mass M and radius R)

The experiment would be extremely dangerous within a populated area because the centrifuge forces would produce stress and strain well beyond the linear range of any known material, the range where Young modulus is almost constant, and the accumulated energy for the case of a body with a mass of 1 Kg would be about 62475000 [J]. If this energy were liberated, it would translate into the kinetic energy of debris fling toward nearby homes.

Well, I admit that this is only a guessing, but one idea would be to measure the weight of a fast rotating body just to demonstrate the attainable precision of high resolution measurement of weight.

Thank you for your interest and of course I admit that my ideas may be erroneous, just that I would love to do this experiment on top of a tower in the middle of a desert.

Jaime Soto Figueroa
http://www.matharts.cl/

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#12
In reply to #11

Re: your post

09/08/2006 4:49 PM

i don't get da piont ... levitaion is impossible dis way .. dis is clear for sure ..... tough u can atain a consant uniform rotaional acceleration if levitate a magnet nd den introduse a force to make it rotate .... the whole thing (magnet) will store the whole energy without ne idiling loss .... nd b cautious bout the shattering of the material.... dis is hazardous.... plz keep in mind dis few factors dis m8 help u/.. Moment of Inertia = (Rim Density) (Rim Volume) (Rim Radius)2 ENERGY = (1/2) (Moment of inertia) (Spin Speed)^2 Tensile Stress = (Rim Density) (Radius)^2 (Spin Speed)^2 ID = initial diametre OD = diametre reached under tension Maximum stored energy, vs. weight and total volume, is usually given only for a thin hoop or solid disk. For practical aspect ratios, it is related to the rim's ID/OD ratio by slightly complicated relationships. The ID/OD dependence (below) was derived by combining the above equations with the integral of the differential expression: d(Inertia) = (Radius)2 (Rim Density) d(Volume) This leads to the result: Max ENERGY = (ID/OD Factor) (Rim Volume) (Max Tensile Stress) where: (ID/OD Factor) = (1/4) {1 - (ID/OD)^4} / {1 - (ID/OD)^2} From this, we get: (Max ENERGY) / Weight = (ID/OD Factor) (Max Tensile Stress) / (Rim Density) and (where Total Volume includes the space inside the rim): (Max ENERGY) / (Total Volume) = (1/4) {1 - (ID/OD)^4} (Max Tensile Stress)

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#13

Re: Extremely Dangerous Gravitational Levitation Experiment

09/20/2006 10:07 AM

Look at some ion storage rings where velocity for charge particles is near speed of light and there is enough charge particle density going around several km diameter charge storage rings.

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