Essay Abstract

The tessellation of space is considered for the 3-sphere. It is found that there is a matter density $Omega_M = 0.284$ associated with the curvature of the 3-sphere.

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Artist and computer software developer.

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You computed the value as N = 10,000 K = 0.28821, N = 100,000 K= 0.28413, N = 1,000,000 L = 0.28452. However, from N = 100,000 to 1,000,000, the value of K is increasing. Therefore, I could not follow your logic on the finite size scaling $lim_{N to infty} K (N) = 0.284$. Would you tell us the details of your logics?

    Dear Yutaka,

    Thank you for sharing your thoughts.

    To be honest, qhull does not support N larger than 1,000,000. I would not say that it's simply increasing, but more like oscillating.

    Even if my number is off, there is still some curvature K, and there is a matter density associated with it.

    I hope I've answered your question.

    - Shawn

    P.S. Do you know of a software that does what qhull does, but without the problems of qhull?

    Thanks for your guidance.

    P.P.S. Or would you know how to compile qhull, and use long double suppport?

    P.P.P.S. Where N = 1,100,000, K = 0.284381 +/- 0.136794. Oscillating.

    It occurred to me that this curvature leads to just dark matter. Regular matter is above and beyond dark matter. The rest is dark energy.

    - Shawn

      Dear Shawn Halayka,

      looks nice. It is not coincidene. I tink it must lead to something ocillating.

      As my ToE 12*pi*c^3=1^2 (mapping 2D to 3D-time) is about 0,298233... it is a representation of a ocillation that "looks" like random but is not (if we know how to calculate PI)

      Best regards

      Manfred U.E. Pohl

        Dear Manfred,

        Thanks for sharing your observations, and for your confidence that this is not simply a coincidence.

        - Shawn

        4 days later

        I've attached a screenshot of a 2-sphere tessellation made out of triangles.

        I do effectively the same thing with the 3-sphere tessellation made out of tetrahedra. Once I have the tessellation, I calculate curvature based on neighbouring tetrahedra.

        - ShawbAttachment #1: 1_2-sphere.png

          Dear Shawn,

          i am not sure about the final notation on how to calculate cosmos in Pythagoras style. But effectivly there should be two tessellation-3d-volume spheres in equilibrium of force (surface*surface/distance)

          a try attached..

          Best, ManfredAttachment #1: gravity.png

            Dear Manfred,

            Can you please explain in greater detail how this applies to my model?

            - Shawn

            i try,

            - we talk about cosmic matter denisty

            - i don't work with 1,0,1,0,1.. but with 1,-1,1,-1,1..

            (coulomb : to each charge -1 must be a charge +1 in universe)

            flow of information (m/s (1/-1))

            =>

            12 pi c^3=1 => c[meter/second] = 0,298233409... (1st constant)

            fine structure - constant = 1/137,63 [meter / second] (2nd constant)

            (0,298233409..-0,2845...) * 10^-4 = would be approx. fine - structure-constant (mine is 1/137,63)

            (my correction factor is 10^4 to project stream of information from one dimension into 4 dimension)

            but in detail i can't tell regarding your lim K->infinity N , as i don't have the concept of infinity in my mathematic and i didn't ready your code in detail.

            best, manfred

              It occurred to me that there is some kind of collapse:

              The pseudorandomly-placed vertices gravitate, and so their tetrahedra become as regular as possible.

              - Shawn

                P.S. The edge length histogram collapses from a wave shape into a single spike shape.