### Bible Wheel Book

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Originally Posted by Richard Amiel McGough
Attachment 1032

Here's what I'm talking about. The proportions were distorted in the icosahedron you drew.

And they are wrong in this picture too because the back is stretched to match the front whereas in a real projection it would appear shorter.

You can see what a real icosahedron looks like in a real cube on this site, and it's nothing like the pictures above. You can rotate it on the site to get different views. Here's a snapshot that is supposed to be like the pic you drew (and the one above):

Attachment 1033

People who get into "sacred geometry" are not usually very good thinkers when it comes to geometry and mathematics.
I cannot open http://www.cut-the-knot.org/Curricul...onInCube.shtml

But from your snanpshot I would say it matches the picture I drew. The icosahedron touches every side of the cube with one point, on 1/3 from the corner (i don't know how to say more exactly) , so fitting to a cube 6 x6 x6 (or maybe also to cube 3 x 3 x 3)

2. Originally Posted by sylvius
I cannot open http://www.cut-the-knot.org/Curricul...onInCube.shtml

But from your snanpshot I would say it matches the picture I drew. The icosahedron touches every side of the cube with one point, on 1/3 from the corner (i don't know how to say more exactly) , so fitting to a cube 6 x6 x6 (or maybe also to cube 3 x 3 x 3)
The snapshot doesn't match at all. The angles and points of contact between the cube and the icosahedron are totally different.

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Originally Posted by Richard Amiel McGough
The snapshot doesn't match at all. The angles and points of contact between the cube and the icosahedron are totally different.

Here you can see how an icosahedron fits within a cube.

I you turn the icosahedron a little you can make it fit in a cube 6 x 6 x 6, touching each face with one vertex instead of two (as pictured above.)

The touching point being here:

Maybe one could find the formula for it.
Last edited by sylvius; 12-28-2013 at 03:25 AM.

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Wikipedia presents this graphic form as inherent to the icosahedron:

http://en.wikipedia.org/wiki/Icosahedron

This construction can be geometrically seen as the 12 vertices of the 6-orthoplex projected to 3 dimensions.
That's very nice, especially sincce it is the birth-image I drew for my first child:

I saw it by then as picture of the Alpha and the Omega and the name of God and the letter Alef (as 10 - 6 - 10).
Last edited by sylvius; 12-28-2013 at 06:39 AM.

5. Originally Posted by sylvius
Here you can see how an icosahedron fits within a cube.

I you turn the icosahedron a little you can make it fit in a cube 6 x 6 x 6, touching each face with one vertex instead of two (as pictured above.)

The touching point being here:

Maybe one could find the formula for it.
That's exactly correct. It's the same as this pic I posted earlier which obviously doesn't match the pics you have been drawing:

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Originally Posted by Richard Amiel McGough
That's exactly correct. It's the same as this pic I posted earlier which obviously doesn't match the pics you have been drawing:

Here you can see how the isocahedron fits within cube 6 x 6 x 6 = Metatron's cube:

It must be proof enough!

I did put it together from this, just that my granddaughter ruined it a bit (and I don't like to do it over again):

http://www.korthalsaltes.com/model.p...en=icosahedron

Last edited by sylvius; 12-29-2013 at 02:28 AM.

7. rdelmonico101@gmail.com Guest

## Star of David

If we use circles we can make a 2D star of David with 37/73.
If we use spheres we can make a 3D version with 220/300.

8. Originally Posted by sylvius
Here you can see how the isocahedron fits within cube 6 x 6 x 6 = Metatron's cube:

It must be proof enough!

I did put it together from this, just that my granddaughter ruined it a bit (and I don't like to do it over again):

http://www.korthalsaltes.com/model.p...en=icosahedron
You didn't have to go through all the effort to make a model. We already knew that the icosahedron fits within the cube. Your model is the same as this pic I posted (except is shows the precise points of contact):

The point is that this does not match the drawings you have done, where you overlap the icosahedron with a 2D projection of a cube.

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Originally Posted by Richard Amiel McGough
You didn't have to go through all the effort to make a model. We already knew that the icosahedron fits within the cube. Your model is the same as this pic I posted (except is shows the precise points of contact):

The point is that this does not match the drawings you have done, where you overlap the icosahedron with a 2D projection of a cube.
On the pic you posted all 12 vertices touch the cube.

In my figure 6 vertices touch the cube, it being like Mr. Gilchrists isocahedron that fits within Metratron's cube.

I wanted to contend that Metratron's cube is the perfect cube 6 x 6 x 6, to which does fit star 73.

10. Originally Posted by sylvius
On the pic you posted all 12 vertices touch the cube.

In my figure 6 vertices touch the cube, it being like Mr. Gilchrists isocahedron that fits within Metratron's cube.

I wanted to contend that Metratron's cube is the perfect cube 6 x 6 x 6, to which does fit star 73.

That is correct - all twelve vertices touch the sides. We can see this in the pic you posted earlier:

None of the pictures you have posted showing the real icosahedron in a cube match the pics you have draw projecting the icosahedron onto the projection of the cube in the star. That's my point. They just don't match up. The 2D drawings are wrong.

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