Math formula(e) needed: tesseract model construction?

Although I CAN do this without the math formula(e) to figure it all out, I would like to have it for a quicker method of figuring out size, etc. so I didn't have to use up so much material in guessing.

I would like to construct the tesseract I have pictured here, using rigid clear plastic sheets or panels.  

Given a said size for the "inner cube" say 2 inches (approx. 5 cm or about 51 mm) what formula(e) must I use to calculate the panels' dimensions to connect to the larger cube (knowing full well this will depend on the size of the larger cube....say approx. 4 inches across (about 10 cm or 102 mm)?

The reason I'd like the formula(e) is that I may need to increase the size of the center cube by an inch or so, and this would allow me to do so without having to go through the "trial and error" thing so many times.

I would, of course, want the center cube to be "centered", if possible or shall I say as close as possible.


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kelseymh6 years ago
For all the math and geometry folks out there, this problem demonstrates very nicely the generalization of Pythagoras' Theorem to N dimensions: The body diagonal H of an N-dimensional rectangle is given by

H2 = SUM si2

where s_i is the length of the ith side. For the case of an N-cube with side s, this reduces to s*sqrt(N).

For your nested cubes, Goodhart, with sides L and M (L > M), each of your red lines is just half the difference of the two body diagonals:

[sqrt(3)L - sqrt(3)M] / 2 = (L-M) sqrt(3)/2

which is BritLuv's solution.
Goodhart (author)  kelseymh6 years ago
Thank you for that. Understanding more about the "mechanics" of how things (especially math) works helps me use it better (and check myself).
The Jamalam6 years ago
Trigonometry and pythagoras' theory would be helpful. If you work out the distance from the corner to the point equally as high as the inner corner, and then between the outer line to the inner line, you can work out the hypotenuse (longest side) because there should be a right angle between the two distances you've worked out. If we name these two distances 'a' and 'b', and the hypotenuse 'c':
a(squared) + b(squared) = c(squared)
All you need to do is work out the square root of c and you have the length of the red lines.
Providing the cube in the middle is perfectly central, all of the red lines would be the same.
Hope this helps.
Goodhart (author)  The Jamalam6 years ago
Thanks, in effect it is the same answer as BrittLiv, only from a different direction. This is what I love about geometry, unlike algebra, you normally have multiple ways to figure out the problem, and several ways to check it.
I haven't done much geometry in my time, but there are many equations and stuff and it's very logical. For me, i really like algebra. My favourite area of maths because it just makes sense without me having to think about it much at all.
Goodhart (author)  The Jamalam6 years ago
For me, algebra becomes confusing because, when I asked people to "show me another way" to get the the same answer (for checking purposes), I never could get any answers. With geometry, I could figure things out with the minimum of information....it is very logical for us lateral thinkers :-)
BrittLiv6 years ago
so you just want to know how long a red line is? (sry, I don't really know what you mean with panels...)
b is the side length of the bigger cube and a of the smaller.
Goodhart (author)  BrittLiv6 years ago
Thank You ! When I have more patches to give, I want to send this one to you:
Ui, let me guess: an Interplanetary Magnetic Field? Very cool, thanks a lot!
Goodhart (author)  BrittLiv6 years ago
Well, it looks like I will have to wait until my Subscription is up for renewal to get mor patches....sorry...
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