06
Nov
A guy welding together a large cube using angle iron for the edges notices that the cube is not very rigid. So he decides to weld on some diagonal braces, going from any one corner to another. How many braces does he need to make the cube rigid?
2 across the middle should do it I think. Oh, I was assuming the faces themselves were rigid.
If all the corners are completely limp, hmmm
You have 8 corner points. That’s 24 degrees of freedom in 3 dimensions.
You have 12 edge pieces that fix the distance between a pair of points. That sucks up 12 degrees of freedom.
You don’t care about overall position (3 degrees) or orientation (3 more degrees). The cube can be rigid and have those be whatever.
So 24 – 12 – 3 – 3 leaves 6.
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You can check this reasoning on some simpler structures like a triangle or a square.
A triangle has 3 corner points–9 degrees of freedom.
You don’t care about 6 (overall position and orientation).
The three sides take care of 3 degrees.
9 – 6 – 3 = 0
A triangle is inherently rigid. Check.
A square has 4 corner points–12 degrees of freedom.
You don’t care about 6.
The four sides take care of 4
12 – 6 – 4 = 2
You need two braces for a rigid square. Check.
1 acoss the internal space will do it
if u add the diagonal braces the cube will become an iindeterminate 3d truss.u can tell if the cube is rigid or not be finding the displacemnet
i think adding 2 diagonals at each face will make it veery rigid
Two per face is a must for large structure; however one each face may be enough for smaller ones.
5 … Last and final edit.
Looking at Alexander’s picture of a cube, it is clear that you can clip out the two braces on the right hand side and replace them with an interior diagonal from the upper left to the lower right. This makes 5 braces.
[Note: I've also looked at cubes with three and four diagonals and they appear stable -- except they would be very vulnerable to torque so I discarded those.]
(Shameless plagiarism follows)
You have 8 corner points. That’s 24 degrees of freedom in 3 dimensions.
You have 12 edge pieces that fix the distance between a pair of points. That sucks up 12 degrees of freedom.
You don’t care about overall position (3 degrees) or orientation (3 more degrees). The cube can be rigid and have those be whatever.
So 24 – 12 – 3 – 3 leaves 6.
Answer: 6 braceshttp://www.robertinventor.com/cubeetc/cu…