05-04-2017, 07:16 AM
(This post was last modified: 05-04-2017, 07:21 AM by Ancient Vizier.)

I'm still looking for some data sheets I made where the latitudes of the vertices of additional Platonic Solids were worked out including some formulas I created from scratch since I couldn't locate any other examples even with all the math stuff on the Internets. I'm not even sure I can find most of the formulas at present even though I've got a set of printed sheets with their proportions and ratios. (It needs double-checking and it would be 1 Royal Pain^2 to re-type so still hoping for an electronic version).

I just found this, though - I think it's probably a fairly even mix between some tetrahedral statistics and my attempts to project "generic" values for their proportions using the Radian value 57.29577951 as the Radius value in the equations, then trying to find Pi-related values that would serve to represent them. IIRC, I think the finished products are what you get if you use the Meg Yard value 2.720174976 as e'

excerpt from:

The Enigmatic Matrix Double Tetrahedron & Other Curios

Wed Sep 11 16:34:34 2002

Grid Point msg board

...

Previously, we used this approach

we wish our Circumradius (aka Radius, aka Radian) to be the generic Radian value of the Pyramid Matrix, 57.29577951 (the Radian in arc-degrees). With the Edge Length/Circumradius ratio of 1.632993161855, if the Circumradius is 1, the Edge Length will be 1.632993161855. So... if the Circumradius is 57.29577951, the Edge Length will be 1.632993161855 x 57.29577951, which equals 93.56361609

Then, we used the formulas EL2 x SA. Ratio= Surf. Area and EL3 x Vol. Ratio= Volume

Which we later corroborated using the known surface area and volume of the Sphere and the known ratios between them and the surface area of the tetrahedron (e').

Since our radius (Circumradius), like our diameter, for the "protruding" tetrahedron, is now 1/2 of the figure used for a single circumscribed tetrahedron, we repeat this replacing R (57.29577951) with (R / 2)... 57.29577951 / 2 = 28.65788976; 28.65788976 x 1.632993161855 = 46.78180805, which is the Edge Length of the smaller tetrahedral "protrusion" or "cap," and precisely 1/2 of the Edge Length of the larger single circumscribed tetrahedron 93.56361609 / 46.78180805 = 2.

Previously, for the single (larger) tetrahedron, we did essentially this:

R (57.29577951) x 1.632993161855 = EL 93.56361609

EL^2 (8754.15027) x el^2 / sa ratio sqrt 3 (1.723050808) = SA 15162.63302

For the surface area, and

EL^3 (819069.9538) x el^3 / vol ratio (0.117851130) = VOL 96528.31619

For the Volume.

For the smaller tetrahedral "protrusion", we now do the same

R / 2 (57.2957791 / 2 = 28.64788976) x 1.632993161855 = EL 46.78180805

EL^2 (2188.537564) x el^2 / sa ratio sqrt 3 (1.723050808) = 3790.658255 SA

For the surface area, and

EL^3 (102383.7442) x el^3 / vol ratio (0.117851130) = VOL 12066.03995

For the surface area, as we add a smaller tetrahedral protrusion to the larger tetrahedron, we are actually covering up the bottom side and an equal area of the face of the larger tetrahedron, so what we actually gain in surface area is equal to 1/2 the total surface area of the smaller tetrahedron.

3790.658255 SA / 2 = 1895.3291131 gain in surface area for each small tetrahedron x 4 small tetrahedra added = 7581.316145 total gain in surface area, and adding this to the surface area of the large tetrahedron... 15162.63302 + 7581.316145 = 22743.94947

22743.94947, Surface Area of the Double Tetrahedron

Referring to the Double Tetrahedron inscribed by a sphere of Radius = 57.29577951, our "generic" Pyramid Matrix format)...

For the volume, we simply add four times the volume of the "cap"- one "cap" for each side of the larger tetrahedron, to arrive at the total volume of the double tetrahedron.

VOL 12066.03995 x 4 "caps" = 48264.1598

VOL Tetrahedron 96528.31619 + total Vol of 4 smaller caps 48264.1598 = 144792.4760

14479.4760, the Volume of the Double Tetrahedron

-----------------

It's not likely that these exact figures per se belong to the Pyramid Matrix, and in this case, when we seek close approximations there are a number of possibilities that present themselves. It would probably be most prudent, in spite of clear possibility of multiple intended Matrix-oriented to try to follow our previous approximation style...

"Alternate Tetrahedral Surface Area" ? 15165.55436 for 15162.63301

"Alternate Tetrahedral Volume" ? 96546.91716 for 96528.31961

Where

96528.31961 / 15162.63301 = 6.366197714, the reciprocal of (1/2 Pi / 10)

41252.69124 / 2.720174960 = 15165.55436

& etc.

22743.94947, Surface Area of the Double Tetrahedron

14479.4760, the Volume of the Double Tetrahedron

Probably the simplest is to sample the ratio between the values for the single tetrahedron and the double tetrahedron:

22743.94947 Surface Area of the Double Tetrahedron / Surface Area of the Tetrahedron

15162.63301 = 1.5

144792.4760, the Volume of the Double Tetrahedron / Volume of the Tetrahedron 96528.31961 = 1.5

Preserving this wonderful ratio, and applying this same ratio to the previously proposed Alternate Tetrahedral Surface Area 15165.55436 x 1.5 = 22748.33154, a possible "Alternate Double Tetrahedral Surface Area"

96546.91716 x 1.5 = 144820.3757

The ratio between the Volume of the circumscribing Sphere and the double tetrahedron is now

787873.5293 / 144820.3757 = 5.44034995 = 2.72017496 x 2, or twice Alternate e'

The ratio between the Surface Area of the circumscribing

41252.69124 / 22748.33155 = 1.81449973 = 2.72017496 x 1.5

I just found this, though - I think it's probably a fairly even mix between some tetrahedral statistics and my attempts to project "generic" values for their proportions using the Radian value 57.29577951 as the Radius value in the equations, then trying to find Pi-related values that would serve to represent them. IIRC, I think the finished products are what you get if you use the Meg Yard value 2.720174976 as e'

excerpt from:

The Enigmatic Matrix Double Tetrahedron & Other Curios

Wed Sep 11 16:34:34 2002

Grid Point msg board

...

Previously, we used this approach

we wish our Circumradius (aka Radius, aka Radian) to be the generic Radian value of the Pyramid Matrix, 57.29577951 (the Radian in arc-degrees). With the Edge Length/Circumradius ratio of 1.632993161855, if the Circumradius is 1, the Edge Length will be 1.632993161855. So... if the Circumradius is 57.29577951, the Edge Length will be 1.632993161855 x 57.29577951, which equals 93.56361609

Then, we used the formulas EL2 x SA. Ratio= Surf. Area and EL3 x Vol. Ratio= Volume

Which we later corroborated using the known surface area and volume of the Sphere and the known ratios between them and the surface area of the tetrahedron (e').

Since our radius (Circumradius), like our diameter, for the "protruding" tetrahedron, is now 1/2 of the figure used for a single circumscribed tetrahedron, we repeat this replacing R (57.29577951) with (R / 2)... 57.29577951 / 2 = 28.65788976; 28.65788976 x 1.632993161855 = 46.78180805, which is the Edge Length of the smaller tetrahedral "protrusion" or "cap," and precisely 1/2 of the Edge Length of the larger single circumscribed tetrahedron 93.56361609 / 46.78180805 = 2.

Previously, for the single (larger) tetrahedron, we did essentially this:

R (57.29577951) x 1.632993161855 = EL 93.56361609

EL^2 (8754.15027) x el^2 / sa ratio sqrt 3 (1.723050808) = SA 15162.63302

For the surface area, and

EL^3 (819069.9538) x el^3 / vol ratio (0.117851130) = VOL 96528.31619

For the Volume.

For the smaller tetrahedral "protrusion", we now do the same

R / 2 (57.2957791 / 2 = 28.64788976) x 1.632993161855 = EL 46.78180805

EL^2 (2188.537564) x el^2 / sa ratio sqrt 3 (1.723050808) = 3790.658255 SA

For the surface area, and

EL^3 (102383.7442) x el^3 / vol ratio (0.117851130) = VOL 12066.03995

For the surface area, as we add a smaller tetrahedral protrusion to the larger tetrahedron, we are actually covering up the bottom side and an equal area of the face of the larger tetrahedron, so what we actually gain in surface area is equal to 1/2 the total surface area of the smaller tetrahedron.

3790.658255 SA / 2 = 1895.3291131 gain in surface area for each small tetrahedron x 4 small tetrahedra added = 7581.316145 total gain in surface area, and adding this to the surface area of the large tetrahedron... 15162.63302 + 7581.316145 = 22743.94947

22743.94947, Surface Area of the Double Tetrahedron

Referring to the Double Tetrahedron inscribed by a sphere of Radius = 57.29577951, our "generic" Pyramid Matrix format)...

For the volume, we simply add four times the volume of the "cap"- one "cap" for each side of the larger tetrahedron, to arrive at the total volume of the double tetrahedron.

VOL 12066.03995 x 4 "caps" = 48264.1598

VOL Tetrahedron 96528.31619 + total Vol of 4 smaller caps 48264.1598 = 144792.4760

14479.4760, the Volume of the Double Tetrahedron

-----------------

It's not likely that these exact figures per se belong to the Pyramid Matrix, and in this case, when we seek close approximations there are a number of possibilities that present themselves. It would probably be most prudent, in spite of clear possibility of multiple intended Matrix-oriented to try to follow our previous approximation style...

"Alternate Tetrahedral Surface Area" ? 15165.55436 for 15162.63301

"Alternate Tetrahedral Volume" ? 96546.91716 for 96528.31961

Where

96528.31961 / 15162.63301 = 6.366197714, the reciprocal of (1/2 Pi / 10)

41252.69124 / 2.720174960 = 15165.55436

& etc.

22743.94947, Surface Area of the Double Tetrahedron

14479.4760, the Volume of the Double Tetrahedron

Probably the simplest is to sample the ratio between the values for the single tetrahedron and the double tetrahedron:

22743.94947 Surface Area of the Double Tetrahedron / Surface Area of the Tetrahedron

15162.63301 = 1.5

144792.4760, the Volume of the Double Tetrahedron / Volume of the Tetrahedron 96528.31961 = 1.5

Preserving this wonderful ratio, and applying this same ratio to the previously proposed Alternate Tetrahedral Surface Area 15165.55436 x 1.5 = 22748.33154, a possible "Alternate Double Tetrahedral Surface Area"

96546.91716 x 1.5 = 144820.3757

The ratio between the Volume of the circumscribing Sphere and the double tetrahedron is now

787873.5293 / 144820.3757 = 5.44034995 = 2.72017496 x 2, or twice Alternate e'

The ratio between the Surface Area of the circumscribing

41252.69124 / 22748.33155 = 1.81449973 = 2.72017496 x 1.5

"Work and pray, live on hay, you'll get Pie In The Sky when you die." - Joe Hill, "The Preacher and the Slave" 1911