# Math assistance: Building a tesseract/hypercube model

October 12, 2017 11:41 AM Subscribe

Howdy Hive Mind! I'm planning on making a nested cube version of the hypercube/tesseract model in glass. The inner and outer cube measurements are easy, I know the angles of the diagonal pieces are 45 degrees, but for the life of me I can't figure out their length.
Let's say the outer cube is 6" on a side and the inner cube is 3", what are the lengths of the connectors?

"the angles of the diagonal pieces are 45 degrees"

Which angle exactly?

posted by floppyroofing at 12:07 PM on October 12

Which angle exactly?

posted by floppyroofing at 12:07 PM on October 12

You can also imagine it geometrically. It's the hypotenuse of a right-angle triangle, so you just need to figure out the dimensions of the triangle.

Imagine a square projected onto the horizontal plane of the outer cube (for example, in the bottom-left corner of your linked image). It's 1.5" by 1.5" (or 3/2") because of the difference between your inner and outer cubes. The length of the diagonal of this square is

This diagonal forms the lower side of your right-angle triangle.

Now imagine a vertical line between the inner end of this line and the lower-left corner of the inner cube, forming the other side of your triangle. The length of the vertical line is just 3/2", so the length of your connector, forming the hypotenuse of the triangle between these two lines is

posted by figurant at 12:08 PM on October 12 [1 favorite]

Imagine a square projected onto the horizontal plane of the outer cube (for example, in the bottom-left corner of your linked image). It's 1.5" by 1.5" (or 3/2") because of the difference between your inner and outer cubes. The length of the diagonal of this square is

sqrt( (3/2)^2 + (3/2)^2 ) = sqrt(18/4) or (3/2)*sqrt(2)

This diagonal forms the lower side of your right-angle triangle.

Now imagine a vertical line between the inner end of this line and the lower-left corner of the inner cube, forming the other side of your triangle. The length of the vertical line is just 3/2", so the length of your connector, forming the hypotenuse of the triangle between these two lines is

sqrt( ( (3/2)*sqrt(2) )^2 + 3/2^2 )giving you sqrt(27/4) or approximately 2.598". Which fortunately agrees with letourneau's answer =).

posted by figurant at 12:08 PM on October 12 [1 favorite]

Thank you all! I knew the MeFi Hive Mind would save me. Now onward to a test build.

posted by Lighthammer at 12:11 PM on October 12

posted by Lighthammer at 12:11 PM on October 12

*"the angles of the diagonal pieces are 45 degrees"*

Which angle exactly?

Which angle exactly?

If you do an orthonormal projection of your object into 2 dimensions, sure, then they'll look like 45 degrees, but in three dimensions, it's impossible for any line to be 45 degrees from three mutually perpendicular lines. For it to be 45 degrees from two of them, it'd have to be coplanar with them, making it 90 degrees from the remaining one. In fact, angle the cube's diagonal makes with an edge winds up as something messy around 50.7 degrees ( atan(sqrt(3/2) ) ).

Of course that might not matter for assembly - if you make everything the right lengths, and make sure the things supposed to be parallel are parallel, the fiddly angles will be determined for you.

posted by aubilenon at 12:55 PM on October 12

*In fact, angle the cube's diagonal makes with an edge winds up as something messy around 50.7 degrees ( atan(sqrt(3/2) ) ).*

Oops, that's just wrong. It's actually acos( sqrt (2/3 ) ), 35.26 degrees. The cube's diagonal, one face's diagonal, and an edge make a 1, sqrt(2), sqrt(3) right triangle.

posted by aubilenon at 1:07 PM on October 12 [1 favorite]

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posted by letourneau at 11:50 AM on October 12 [7 favorites]