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Question

Asked by: Nathaniel Gardner
Subject: Gyroscope Behavior
Question: Okay, I have a complicated question about gyroscopes. Lets so we have a gyroscope system inside a sphere in a Zero-G environment. Now this system contains 2 gyroscopes, both located equidistant from the equator of this sphere, therefore making the northern and southern hemispheres of equal mass. Now, if the bottom gyroscope were to spin rapidly in the clockwise direction at a rate of 50,000 rpm and the top gyroscope were to spin counterclockwise at the same speed, would the forces cancel each other out? Next, if the bottom spun at 50,000 rpms clockwise and the 2nd spun the same speed in the same direction, would the force be doubled? Next, if the bottom spun at 50,000 rpm clockwise and the top spun half that speed counterclockwise what would be the result?

Next scenario. Let's say you have a gyroscope-like device, except there is one difference. In this gyroscope, the disk is cut in half and placed equidistant from the center of the gyroscope so that if you lowered the top half and raised the bottom half to the center, it would make a complete disk. Now lets say you spun the gyroscope, would it still excert the exact same properties?

Last Scenario. Lets say you had a motorized gyrocope in the center of a sphere with a constant rate of spin--lets just say it spins VERY fast--in a Zero-G environment. Now, equidistant from the gyroscope and located above and below the center of the sphere, let's say you had to motors that had a spinning half-disk, both facing the same direction, so that if the gyroscope were turned off, it would cause the system to wobble. What would happen if the gyroscope were spinning, and then the 2 half disks, were to begin spinning, what would be the result? What would the motion of the entire sphere be if the 2 disks spun at a low rate. What would be the motion of the entire sphere if they spun very fast? Please assume that the Northern Hemisphere and Southern Hemisphere of the sphere are equal in mass and are mirror images of each other.

If these questions do not make sense, please e-mail me and I will try to explain it better. Thanks!
Date: 20 June 2006
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Answers (Ordered by Date)


Answer: Jimmy James - 18/07/2006 20:16:06
 For the first question I think it depends on how you attempt to move the system.

If you try to twist the sphere around it's cetner (roll it, let's say) I believe the force will be multiplied because the configuration as rotation symetry. In other words, the top part goes to the left and the bottom part goes to the right and the gyroscopes are also resisting in opposite directions. But if you tried to move the whole system around a point outside the shpere (like tie the sphere at a pole above on of the gyros and swing it in an arc) the gyros forces should cancel each other out to some degree.

For the wobble question, it all depends on the conguration of the gyro. You haven't described whether it's axis is the same as the half-discs or if they are perpendicular.

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Answer: Nathaniel Gardner - 20/07/2006 19:47:40
 Okay, I will just cut down to what I am trying to accomplish. After mulling over the properties of a gyroscope as well as newton's laws of physics, I believe it is possible to create a system that "orbits" around a fixed point in space, without the need for gravity. In the center of the sphere, you would have a complete gyroscope, and then over the central disk (A), you would have a system of weights of equal mass spinning at different speeds(B), and on the bottom, you would have another system of weights spinning at different speeds in the same direction as the top system(C), but both systems B and C would be spinning in the opposite direction and half the speed of (A). Now the weights on top and on bottom would look like slices of pie and every 6.5 hours they would all come together to form a complete circle (but of course on different levels of the sphere). Each piece of the circle would spin at a different speed than another, thereby causing the entire mass of the object to shift toward the direction in which the weights are spinning. System A and B are actually a complete gyroscope, except broken into about 8 pieces. And I know this is a lot of speculation, but if this works, I believe it is possible to create a system for spacecraft to make wide turns without the need for enormous amounts of fuel. This system depends an numerous amounts of variables, but maybe it will work. See what ya'll think. but it is important that if this idea does work that you inform me of it. I do not wish to discover that my idea was taken because someone read my post. But the fact of the matter is, I have not the means to carry out an experiment such as this. And if anyone is interested, I can send the design to you for the experiment as long as credit is given where credit is due. Apart from that, knock yourselves out with this question, I know I sat up for hours thinking about it. Hope you enjoy this one!

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Answer: Jerry Volland - 12/08/2006 13:43:03
 Nathaniel,

This is a very interesting, and potentially valuable, concept. However, if anyone is to notify you of future success, perhaps at a later date, you'll have to post your email address.

Here's mine:
jerryvolland@spaceoffice.us

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Answer: Arthur Dent - 14/08/2006 10:43:01
 Angular momentum can be represented by an axial vector, so you can answer the original questions yourself. If the vectors cancel, then the system will not act as a gyroscope. This fact was once used to disprove the old fallacy that gyroscopic effects are responsible for bicycle stability: an oppositely rotating wheel on the same axis was used to cancel any gyroscopic effect of the bicycle's own wheel. The bicycle's stability was unaffected.
If the vectors are in the same direction, the angular momentum is indeed doubled, and partially cancelling vectors will give a smaller angular momentum.
An interesting case occurs when the oppositely rotating wheels are a very long distance apart. One then has to take account of the delay in 'signal transfer' between the 2 wheels and one has entered the domain of relativistic effects. Strange things (even propulsion) can then be expected to happen. Such a system (oppositely rotating wheel on the same axis, in turn rotated over another mass) has been used to try to detect the Lense-Thirring effect. Unfortunately, such effects will be of the same order as that of gravitational radiation, and nobody has ever succeeded in measuring any; even from supernovae. So what do you think the chances really are of getting any propulsive effect?

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Answer: Nathaniel Gardner - 19/08/2006 05:04:35
 ARGH! I made a mistake in my explaination! Okay, what you would have is a sphere, and inside the sphere is a central disk which spins counterclockwise at X rpm (C). Above (A) and below(B) that you have a system of weights that, when positioned at the start of the process look like a complete disk, but are broken up into 8 pie shaped fragments. Each fragment would have a weight at the end. #1 will spin at 1000 rpm, #2 at 2000 rpm, number 3 at 3,000...etc until you get to number 8 at 8,000 rpm. Fragment numbers 8, 5, 4, and 1 will be located in section A while Fragment numbers 7, 6, 3, and 2 will be located in section B. Both section's A and B will be spinning clockwise. (that is where I made the mistake). Section C is designed to counter the gyroscope effects of sections A and B as well as create gyroscope effects in the opposite direction. C will rotate with a velocity that creates a force that is equal to the force of A + the force of B + 1000 rpm. This will cause the sphere to rotate counterclockwise at 1000rpm. The purpose for the weights is this: Ever so often, all of the weights will for a very brief moment, be on 1 side of the sphere, causing the sphere to move in that direction, but because of the circular motion of th weights, it will cause it to curve. Every 5.25 hours, the disks will all be in different spots, making it appear to be a full disk if looked at from a top view. I believe by this point it will have already begun orbiting around a fixed spot in space. I also believe it will take 5.25 hours for the entire system to make one complete orbit. However, I am unsure of how determine how large the radius of the orbit will be, whether it will make a circle or spiral outward--infinitely approaching a perfect circle, or simply not work at all. My email address is havok15x@hotmail.com. There are many uses I have thought of for a system of this nature. One thing it most definitely could be used for is an alternate means for a spacecraft to turn. It would be unnecessary in normal circumstances for a spacecraft to make a sharp turn, but if this system could be employed to work quickly enough, it would actually turn the spacecraft by the shear force of the system. Also, during long range missions, the spacecraft could simply "arc" it's way to its destination. It could be also used to stabalize a satellite in a low orbit, or even a high orbit where gravity would normally be either too strong or too weak. And, it could also be a means, in the future, for crafts which hover above the ground to turn. There may be other uses, but those are the ones of which I have thought.

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Answer: Jerry Volland - 19/08/2006 14:36:57
 Nathaniel,

I agree with your assesment that the system will begin to orbit when the weights are all lined up, and probably sometime before and after that, to some extent. But what will happen when the weights are lined up on the other side? Won't the orbit be around some point on the other side? Not that this diminishes the effectiveness of your approach. For instance, once 90o of the spiral (due to dimminishing forces from less alligned sectors) has been traversed, the thruster can be turned off, reset, then opereated with counter rotation. The effect of this type of repetitious cycling would be a basically forwards movement in somewhat of a really large zig zag course. But wouldn't the sphere itself rotate opposite to the direction of the fastest disc?

The value of a system which accelerates on curved trajectories (i.ei., the TIE and the GIT) is clear. Congratulations on coming up with one of your own.

Jerry

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Answer: Nathaniel Gardner - 19/08/2006 17:44:39
 Okay, I have figured out something. In order for this system to successfully orbit, the spin of the entire sphere must allow the disks, when all stacked on top of one another (each fragment on 1 side of the sphere), to happen in 4 different directions. If you look at the sphere from directly above and plot a horizontal (x)and vertical axis (y), then at exactly 2.625 hours, the fragments must be lined up along the positive Y axis. Then, after 5.25 hours, it must be lined up along the positive X axis, then after another 5.25 hours, it must be lined up along the negative Y axis, and finally, after 5.25 hours, it will be lined up along the negative X axis, and then repeat the process. (These Numbers only work with with the determined velocity of spin from my previous post). The total amount of time it will take for this sphere to make 1 complete orbit is 21 hours. The position and alignment of the fragment is determined by the velocity of the central disk. I now realize that the radius of the orbit will be determined by how long it takes the fragments to align along the Z axis. If rather than 5.25 hours it takes 10.5 hours for the fragments to align, the orbit should be larger. The only way that could be accomplished is by changing the velocity of each fragment. For a large orbit, each disk must not line up for a while (prime number rpm's will be most effective). However, this is all stuff that I logic together. Please inform me of any mistakes or effects that I have misunderstood or did not take into consideration.

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Answer: Nathaniel Gardner - 19/08/2006 17:44:47
 Okay, I have figured out something. In order for this system to successfully orbit, the spin of the entire sphere must allow the disks, when all stacked on top of one another (each fragment on 1 side of the sphere), to happen in 4 different directions. If you look at the sphere from directly above and plot a horizontal (x)and vertical axis (y), then at exactly 2.625 hours, the fragments must be lined up along the positive Y axis. Then, after 5.25 hours, it must be lined up along the positive X axis, then after another 5.25 hours, it must be lined up along the negative Y axis, and finally, after 5.25 hours, it will be lined up along the negative X axis, and then repeat the process. (These Numbers only work with with the determined velocity of spin from my previous post). The total amount of time it will take for this sphere to make 1 complete orbit is 21 hours. The position and alignment of the fragment is determined by the velocity of the central disk. I now realize that the radius of the orbit will be determined by how long it takes the fragments to align along the Z axis. If rather than 5.25 hours it takes 10.5 hours for the fragments to align, the orbit should be larger. The only way that could be accomplished is by changing the velocity of each fragment. For a large orbit, each disk must not line up for a while (prime number rpm's will be most effective). However, this is all stuff that I logic together. Please inform me of any mistakes or effects that I have misunderstood or did not take into consideration.

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Answer: Jerry Volland - 20/08/2006 04:15:22
 Nathaniel,

If I understand your advanced approach correctly, the sphere will be tipping towards a different axis of rotation while all of the fragments are lined up. One effect you've overlooked is that, without counter acting weights opposite the fragments, normal gyroscopical twisting will be absent and the result will be a unidirectional thrust in the direction opposite to the twisting, due to non planar centrifugal acceleration.

You can see what I'm referring to by looking at my file on split gyroscopes at:

http://www.spaceoffice.us/splitgyro.htm

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