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Math formula(e) needed: tesseract model construction? Answered


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.

Thanks....




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kelseymh (author)2010-12-03

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.

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Goodhart (author)kelseymh2010-12-04

Thank you for that. Understanding more about the "mechanics" of how things (especially math) works helps me use it better (and check myself).

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The Jamalam (author)2010-12-02

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.

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Goodhart (author)The Jamalam2010-12-02

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.

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The Jamalam (author)Goodhart2010-12-03

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.

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Goodhart (author)The Jamalam2010-12-03

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 :-)

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BrittLiv (author)2010-11-28

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.

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Goodhart (author)BrittLiv2010-11-28

Thank You ! When I have more patches to give, I want to send this one to you:

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BrittLiv (author)Goodhart2010-11-28

Ui, let me guess: an Interplanetary Magnetic Field? Very cool, thanks a lot!

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Goodhart (author)BrittLiv2010-12-01

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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BrittLiv (author)Goodhart2010-12-01

Don't worry, I'm not going anywhere (I love this side ;-))

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Goodhart (author)BrittLiv2010-12-01

Schooch gave me some patches to give out :-)

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BrittLiv (author)Goodhart2010-12-01

Thanks a lot to both of you, I love the patch!

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Goodhart (author)BrittLiv2010-12-01

YW :-) it was well deserved

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Jayefuu (author)Goodhart2010-12-01

Or you could pray/hope/wish... :)

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Goodhart (author)BrittLiv2010-11-29

I borrowed it from This site here.      
It will be sent next month when I am granted the ability to send more.

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scoochmaroo (author)Goodhart2010-12-01

You should be all patched up and ready to go!

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Jayefuu (author)2010-11-28

I'll CAD it for you if you like. Then I can make you a load of templates for the surfaces to make up the tesseract.

James

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Jayefuu (author)Jayefuu2010-11-28

My model agrees with BrittLiv.

For an outer cube of 4" and an inner of 2" I get a red line length of 1.732 inches.

Let me know if you want a revised cube size answer or any double checking.

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Jayefuu (author)Jayefuu2010-11-28
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Goodhart (author)Jayefuu2010-11-28

Thank you so much. Now I just need to find some appropriate materials.... :-)

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Jayefuu (author)Goodhart2010-11-28

Acrylic laser cuts very very nicely. If you want the surfaces getting ready for laser cutting I don't mind doing it for you. It probably wouldn't be economical for me to laser cut them then send them to you, though I can if you really like.

If I were doing it I'd cut the surfaces from 3mm acrylic and hold them together with strips of OHP transparency and double sided tape. I think....

What's it for? Art?

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Goodhart (author)Jayefuu2010-11-28

I am thinking about using it for one of a few ideas I have. I will want the inside edges of the small cube visible but not necessarily opaque. Batting around the most practical way to do this without it being too flimsy

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NachoMahma (author)2010-11-28

. As you know, I'm no Mathematician, so just consider this a let's-see-how-close-I-can-get guess.
. It seems to me that if you extend a red line to the center of the inside cube, it becomes the hypotenuse of two right triangles - one part of the inside cube and the other (?colinear? with the first) is part of the outside cube. You know the dimensions of each edge, use Trig to calc the rest.
. Waited with bated breath to see if I'm even in the ballpark.

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yokozuna (author)NachoMahma2010-11-28

This should work... basically, usually the a^2 + b^2 = c^2 method:

outside square [length/2^2 + length/2^2 = length of red line to center of cubes^2]
MINUS
inside square [length/2^2 + length/2^2 = length of red line to center of cubes^2]

so... if your outside cube is 4x4x4 and the inside is 2x2x2, that should make the length of the red lines sqrt8 - sqrt2. Roughly 1.42 inches.

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BrittLiv (author)yokozuna2010-11-28

Sorry, I don't think this is right, have a look at the picture, the green line is sqrt2 times length and not just length...

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yokozuna (author)BrittLiv2010-11-28

So i drew it out and measured the red line... it comes out to 1 and 7/16... seemingly confirming my answer. The green line is the wrong hypotenuse, I'm figuring it to the middle of the cube, not off an outside square. Granted, it's been a long time and I could be wrong, my formula is probably not as simple as it could be, but it _should_ be correct. Interestingly, if you use 2 instead of 3 as the square root in your formula, it comes out to be the same as my answer.

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BrittLiv (author)yokozuna2010-11-28

ok, let me make a better drawing, the orange line is "legth", the green line sqrt2 times length and the blue line sqrt3 times length.

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yokozuna (author)BrittLiv2010-11-28

Okay, it seems you are right. Jayefuu drew it up on the CAD and confirmed your answer. I realized I was using the wrong figure for one of my side lengths, giving me the wrong number to make the square root of. :)

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NachoMahma (author)NachoMahma2010-11-28

. Oops! You would have to extend two diametrically opposed lines to the center of the tesseract to form the hypotenuse for the triangle in the outside cube.

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Goodhart (author)NachoMahma2010-11-28

All my "trig" skills have long since faded into the murkiness of oblivion. I appreciate the effort though :-)

I can work a "known" formula still, but remembering one is becoming harder each year...

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NachoMahma (author)Goodhart2010-11-28

. I know the feeling. It's been ~30 years since I had a Trig class. I bet we could figure it out after a (long) while, but I vote we wait for the engineers to wake up. ;)

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Goodhart (author)NachoMahma2010-11-28

yeah, it's still only 8 am in CA LOL

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