Question |
| Asked by: |
dave brown |
| Subject: |
Help, need a math wizz :) |
| Question: |
If you have read about my formula in my
I shall not polute... thread
I think I may understand what the 4th dimension is telling me.
I may be able to prove that it is indeed spacetime, but for spin, and represented by:
space / time = distance/time = velocity (but circular)
- that is not what I guessed it to be in my thread.
I need someone to calculate, or give me formulae and tips for, calculating different properties of spin.
as it takes 256 of circle with dia. of 1 to do the same as circle with dia. of 4
It may be the ratio of inertia of x mass at radius of 2 vs. x mass at radius of 0.5
appreciated :) |
| Date: |
13 September 2005
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Answers (Ordered by Date)
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| Answer: |
dave brown - 13/09/2005 00:42:34
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| | actually,
4D may give inertia for a ring of x mass at the radius
--- 1D * 3D = @4D
5D may give inertia for a circle of y mass bounded by the radius
--- 2D * 3D = @5D
6D may give inertia for a sphere of z mass
--- 3D * 3d = @6D
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| Answer: |
dave brown - 13/09/2005 00:44:33
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| | omit 2nd post, was premature.
sorry.
1st post is valid
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| Answer: |
dave brown - 13/09/2005 03:54:03
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| | I will have to cease this thought.
I do not know enough about the possibilities to continue.
I do know that it relates to dimensioning. Finding the boundaries.
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| Answer: |
dave brown - 13/09/2005 03:59:47
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| | I realize I have taken Pi*r^2 and 4/3*Pi*r^3 and found them to be dimension based.
1D: (2^1*Pi*r)/(2*1)
2D: (2^2*Pi*r^2)/(2*2)
3D: (2^3*Pi*r^3)/(2*3)
and started working on
4D: (2^4*Pi*r^4)/(2*4)
- glad most of the work is done for me.
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| Answer: |
dave brown - 13/09/2005 04:21:03
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| | I may end up showing that living on the surface of a sphere is in the 6th dimension,
folks in 3 dimensional spacetime on a 3 dimensional sphere.
or
3 dimensional folk in the 1 dimension of spacetime.
we can move in 1, 2, or 3 dimensional directions,
and work with 3 dimensions,
but we live in the 4th.
but who knows.
here i go
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