Knowledge of science

Discussion in 'Pseudoscience' started by adhams, Dec 20, 2014.

  1. zgmc Registered Senior Member

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    Have you taken any math classes?
     
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  3. adhams Registered Senior Member

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    What iam trying to say is that the exploded balls mass will concentrate in the center not for any reason but that the further away u go from center(by one adjusted unit) the more the volume of that new part of radius is more similar to the next one
    Radius of one give volume 4.1 r2 33.5 r3 113 now 33.5/4.1 is far more than 113/33.5 meaning that as u progress in The radius the less the difference is between the volumes
     
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  5. river

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    Well established is not the problem

    The problem is going beyond the " established " thinking
     
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  7. zgmc Registered Senior Member

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    I see what you were doing now. You really need to type out equations properly
     
  8. rpenner Fully Wired Valued Senior Member

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    \(A(r) = 4 \pi r^2 , \; V(r) = \frac{4}{3} \pi r^3 \\ r = 1 \rightarrow V(r) = V(1) = \frac{4 \, \pi}{3} \approx 4.18879 \\ r = 2 \rightarrow V(r) = V(2) = \frac{32 \, \pi}{3} \approx 33.51032 \\ r = 3 \rightarrow V(r) = V(3) = 36 \, \pi \approx 113.09734 \\ \frac{V(2)}{V(1)} = \frac{\frac{32 \, \pi}{3}}{\frac{4 \, \pi}{3}} = 8 = \left( \frac{2}{1} \right)^3 > \left( \frac{3}{2} \right)^3 = \frac{27}{8} = \frac{36\, \pi}{\frac{4 \, \pi}{3}} = \frac{V(2)}{V(1)} \)

    Wrong, because you used addition to define the differences between the radii but multiplication to define the differences between volumes, which is misleading.

    It's clearly misleading because \(\frac{2}{1} > \frac{3}{2}\).

    If we use addition to define both definitions of differences and r,d > 0, we have \( 0 < r < r + d < r + 2 d\) AND
    \( \left( V(r + d) - V(r) \right) = \frac{4 \pi}{3} \left( (r + d)^3 - r^3 \right) = \frac{ 4 d^3+12 d^2 r+12 d r^2 }{3} \pi < \frac{28 d^3+36 d^2 r+12 d r^2}{3} \pi = \frac{4 \pi}{3} \left( (r + 2d)^3 - (r + d)^3 \right) = \left( V(r + 2d) - V(r+d) \right) \).

    If we use multiplication to defined both definitions of differences, which normal people call ratios, and r > 0 and k > 1, we have \( 0 < r < k r < k^2 r \) AND
    \( \frac{V(k r)}{V(r)} = \frac{ (kr)^3}{ r^3} = k^3 = \frac{ (k^2r)^3}{ (kr)^3} = \frac{V(k^2 r)}{V(k r)}\).
     
  9. danshawen Valued Senior Member

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    You can have all the time and energy you want, but matter and space are self-limiting.
     
  10. adhams Registered Senior Member

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    Yah iam misleading iam a fool
     
  11. danshawen Valued Senior Member

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  12. adhams Registered Senior Member

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    I believe that my theory also includes the use of uranium 238 if it's moved against the movement of the earth it would be still in space and therefor would be like uranium 235 in its space absorption then shooting a neutron of the same speed shot at u235 would lead to similar results the u238 already works with fast neutrons pls try it
     
  13. origin Heading towards oblivion Valued Senior Member

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    Holy crap perhaps you should rename this thread Lack of knowledge of science.
     
  14. Aqueous Id flat Earth skeptic Valued Senior Member

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    So: just because you don't understand relativity, the people who do are stupid, and the science is crap? What a moron. This isn't even pseudoscience.

    Cesspool.
     
    Kristoffer likes this.

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