Anti-gravity. Well isn't that extra-special?

Discussion in 'Physics & Math' started by babelfish, Sep 11, 2001.

  1. SISGroup,

    Gravity isn't in the formula for electric/magnetic conversion. The strength of the resulting magnetic field is the result of only two factors: The strength of the electric field, and the speed of the field(or the speed at which it changes).

    Tom
     
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  3. James R Just this guy, you know? Staff Member

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    Tom,

    Try this explanation in terms of forces if you have trouble understanding my previous explanation:

    Imagine holding a bicycle wheel by its axle. Each hand is on one side of the wheel, and the wheel is spinning so that the top is moving away from you and the bottom is moving towards you. Now, suppose you want to turn the wheel so that the axle is still horizontal but the left side is closer to your body than the ride side. What force do you need to apply?

    Common sense will tell you that you need to push forwards with your right hand and pull back with your left hand. But your knowledge of precession (and the experiment, if you do it) will tell you that is wrong. What you actually need to do is push upwards with your right hand and down with your left - then the wheel will move around at right angles to the applied force. Here's the explanation:

    Consider a point at the top of the wheel. Instantaneously, it is moving away from you. What you want it to do is to move more to the left. Looking at the point at the bottom of the wheel, you want it to move around to the right a bit (remember it is moving towards you). Extending this type of reasoning, we can tell that all points on the top half of the wheel need to move to the left, and all points on the bottom half need to move to the right. The only way to change the velocity of any part of the wheel in a certain direction is to apply a force in that direction. To apply the right kind of force here you need to push your left hand down and right hand up, then the wheel turns to the left.

    The explanation for the top is very similar to this. The explanation with torques explains the same facts in a different way. Either way, no z-force is needed.
     
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  5. SISGroup Registered Senior Member

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    Joeblow93132
    Yes, you right.
    The formula for magnetic field not consist of mass of electron.

    But, I still cannot understand, how motion of positive mass produce Z-field. (I'm sorry, I 'jump' from my basic question)

    Motion of positive mass will have kinetic energy. And yes, kinetic energy makes a positive cannot be pulled by gravity. This is like moon and earth. Earth cannot pull the moon since velocity of moon is strong enough against gravity of earth. No Z force acting on this phenomenon, i think.

    BTW, If movement of positive mass create Z-field, therefore our earth will be influenced by Z-force of moon. Can you prove that there is a Z-field caused by moon on the earth?
     
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  7. SISGroup,

    There is no Z-force influence on the earth by the moon. This is why:

    1. The Z-force created from a spinning object is at right angles to the spin of the object and at right angles to it's gravitational field. So only the gradient of the gravitational field in the X and Y axis of the Earth is being converted to the Z-force. The resulting Z-force forms only at the Z axis of the Earth. In other words, the Z-force created, is exiting at the north and south poles of the Earth.

    Since the same applies to the moon, you can see that the Earths and the moons Z-forces do not intersect. Therefore, their Z-forces do not influence each other.

    2. The Z-force is a weak force(as weak as gravity). In order to generate a powerful Z-force, you need to either spin a very heavy mass, or a smaller mass very quickly. The Earth does not spin fast enough to create a significant Z-force.

    Tom
     
  8. SeekerOfTruth Unemployed, but Looking Registered Senior Member

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    358
    If your hypothesis is true then it is testable for even if it generated an extremely small force, over time that force would impact the orbitals of circumpolar satellites to such an extent that major course corrections would be needed.

    I do not believe that that is the case for any circumpolar satellite that I am aware of.
     
  9. SeekerOfTruth,

    Interesting. I'll have to think about that.

    Tom
     
  10. SeekerOfTruth,

    When the satellites go over the northern pole, Z-force may pull the satellites toward earth, but when they cross the southern pole, the opposite Z-force pushes them back to their original orbit.
    At either point, the difference between the original orbit and the forced orbit would probably be under one meter.

    Tom
     
  11. James R,

    I read your article about the spinning wheel a couple of times, but I still don't understand your explanation. How do you explain the fact the secondary force is always at a 90 degree angle to the primary force, never more or less? According to your explanation, shouldn't the angle be less than 90 degrees if the wheel spins slower and more if the wheel spins faster?

    Tom
     
  12. James R Just this guy, you know? Staff Member

    Messages:
    30,644
    Tom,

    It's a simple matter of changing the velocity of points on the wheel. If you want to add a velocity component in a particular direction, you need to apply a force in that direction. That's all there is to it.
     
  13. SeekerOfTruth Unemployed, but Looking Registered Senior Member

    Messages:
    358
    But this would imply that there is total symmetry between the polor components of the Z force. Because the distribution of mass in the earth is not symmetrical, and because the earth is rotating, the Z force and its vector/magnitude at the north pole will be different from the Z force at the south pole at some given point in time after the satellite crosses the north pole. Remember, the satellite takes some finite amount of time to traverse from the north pole to the south, therefore the vector/magnitude of the Z force would have changed and there would not be a symmetry of forces to cancel each other out.
     
  14. SeekerOfTruth,

    If the distribution of mass on Earth is not symmetrical, why aren't satellites pulled out of their orbits when they pass over a mountain?

    The fact is that it doesn't matter if the mass of the Earth is symmetrical or not. Only when you take a fragment of a satellites orbit does symmetry matter, when you take the entire orbit of the satellite into consideration it no longer matters because all the forces balance out. The same applies to the Z-force, at some points around the Earth, the push and pull of the Z-force will have different strengths. But once the satellite makes a complete orbit the total pull forces of the Z-force will be equal to all the push forces of the Z-force, thereby putting the satellite back into it's original orbit. This will always be the case as long as the satellite's orbit dissects the center of the Earth.

    Tom
     
  15. SISGroup Registered Senior Member

    Messages:
    49
    Equator area has longer distance with spining axis of earth than pole area. So, equator area has bigger centrifugal force than pole area. Since direction of centrifugal force is opposite of gravitation force, therefore it can decrease the gravitation force. So, both earth poles is not excited by Z force.
    SeekerOfTruth has told us that mass of earth is not symetrical. I thing the north area has bigger mass than south one, because most continents lies in north area. So, the center of earth's mass should be closer with north ole then south one.
    According to formula of gravitation force, therefore north pole will have different number of gravitation force than south one.
     
    Last edited: Jan 10, 2002
  16. tomtomberry Registered Member

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  17. Chagur .Seeker. Registered Senior Member

    Messages:
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    Hi tomtomberry ...

    Welcome to Sciforums.

    Might want to check out this site:

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    Take care

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  18. tomtomberry Registered Member

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  19. TIME02112 Registered Senior Member

    Messages:
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