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Postby zero742 » Tue Feb 01, 2005 11:40 am

Hey, so I'm just starting to grasp the concept of the fourth dimension. I have read a bit about the mathematics of how the 4th dimension works and about some of the different geometries such as hyperbolic and elliptical and have a fairly good idea about how these work, since I'm a engineer/physics major. So its about 5:30a.m. and I'm up instead of asleep or studying for my exam tomorrow and I was just reading about the story about how Bob messed with Fred by rotating the cube(square from Bob's point of view) and then Emily essentially did the same thing to Bob by rotating the tesserat in Bob's world. So basically, what each lower dimension sees(Bob has a higher dimension than Fred) after a rotation is a mirror image of the original image before rotation. The concept I am having trouble with at this point is maybe not so much a concept as maybe just a lack of previous knowledge. I understand that the cube placed in Fred's world has 6 sides, which he can only see one of these, and the tesserat placed in Bob's world (here comes the problem) has "n" sides, of which Bob can see 6. So exactly how many sides are there to a hypercube? Or is there no definitive answer? Thanks for nuturing the curiousity(and perhaps future expert) of a college student.
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Postby jinydu » Tue Feb 01, 2005 3:08 pm

Hi zero742, and welcome to the forum.

What do you mean by side? Do you mean, 1-dimensional line segment?

In any case, quoting from: http://mathworld.wolfram.com/Hypercube.html

A tesseract has 16 polytope vertices, 32 polytope edges, 24 squares, and eight cubes. The dual of the 4-hypercube is the 16-cell.
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Postby houserichichi » Tue Feb 01, 2005 5:33 pm

If you mean "sides" as in the faces of a cube (the part that we see at any given time when looking at a box) then the 4-cube has 24 of those. :wink: Jinydu's so smart - always beat me to the new posts! hah
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Postby wendy » Wed Feb 02, 2005 2:44 am

You take the polynomial (x+2)**n, and the terms of the polynomial, along with the indices, give you the dimension and number of surtopes.

(x+2)**2 = x2 + 4x + 4 = four edges + 4 vertices)

(x+2)**3 = x3 + 6x2 + 12x + 8 = six squares, 12 edges 8 vertices

(x+2)**4 = x4+8x3 + 24x2 + 32x + 16 = 8 cubes, 24 sq, 32 edg 8 vert

(x+2)**5 = x5+10x4 + 40x3 + 80x2 + 80x + 32 = 10 tesseract + 40 cubes + 80 squares + 80 edges + 32 vertices,

Simple, really. The tesseract is the fourth power of a line. What more can you say,
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Postby jinydu » Wed Feb 02, 2005 3:13 am

wendy wrote:You take the polynomial (x+2)**n, and the terms of the polynomial, along with the indices, give you the dimension and number of surtopes.

(x+2)**2 = x2 + 4x + 4 = four edges + 4 vertices)

(x+2)**3 = x3 + 6x2 + 12x + 8 = six squares, 12 edges 8 vertices

(x+2)**4 = x4+8x3 + 24x2 + 32x + 16 = 8 cubes, 24 sq, 32 edg 8 vert

(x+2)**5 = x5+10x4 + 40x3 + 80x2 + 80x + 32 = 10 tesseract + 40 cubes + 80 squares + 80 edges + 32 vertices,

Simple, really. The tesseract is the fourth power of a line. What more can you say,


You mean 16 vertices.
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Postby wendy » Wed Feb 02, 2005 4:44 am

Yeah - 16 is correct. Just doing it all in me head.

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Postby zero742 » Wed Feb 02, 2005 5:40 am

Fantastic
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Postby papernuke » Fri Jul 28, 2006 12:02 am

It would have 48 sides because look here http://tetraspace.alkaline.org/forum/vi ... =tesseract the tesseract is just 8 cubes put together somehow, and each cube has 6 sides so 8x6=48 sides.
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Postby moonlord » Fri Jul 28, 2006 7:10 am

A cube is made of six squares, counting up to 24 segments. Yet the cube only has 12, exactly half of that number. Why? Because every segment is common to exactly two squares. The same for the tessie.
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