If you haven't already, sketch it out on a cube as I described earlier. The geometry at each vertex is surprisingly simple, and not entirely obvious. Looking at one corner of the cube, the tetrahedron edges form the prime, 45 degree diagnonals on each of the three faces meeting at that corner.

If you've already done that, I won't be able to help further. Compound miters might as well be rocket science for me. I'm guessing, for a radial arm saw, swing the arm 22.5 degrees to split the plan view 45 deg; tilt the blade 67.5 degrees (probably wrong) to split the 45 degree rise. Repeat for the other side. Maybe someone else can take it from there...

I'll lay it out in solidworks later, see if any more insight forms. It sounds interesting enough to lay out on a small stick and whittle it.

I think I've got it figured out. I did the mental exercise of making the original sticks then cutting off the corners to make the diagonally oriented one. I determined where the original cuts crossed the mid-line of the square stick. The midline of the square stick is the edge of the rotated stick - I can now calculate the angles, knowing the run and rise of those coordinates.

I'll post the results tonight after I make a test piece when I come home from work.

Sounds as if you are trying to make a compound tetrahedron - two
together.

I cut up some square sticks of cherry scrap and pocketed a marking knife and 3/4" butt chisel into the office. A few minutes of (completely, therapeutically relaxing) whittling later, the first piece looks like it's right on. Aside from the woodchips in the keyboard, I can't think of a better way to break up a work day.

For those who might want to take their own whittle break, the dimensions for 1" square stock are 2.000" back along the top spine, and 1.692" back on the two sides. Scale the numbers for your stock size, and mark and connect the dots back to a sharp tip at the front end of the bottom edge. Picture in your mind the two flat facets opening mostly upward and more gently to the sides and forward, meeting in a straight spine connecting the top corner to the pointy tip lying on the desk. Three sticks cut this way form a mitered tetrahedron vertex.

(I'll have to find a better way to hold the workpiece. The corners are all bruised from pressing it hard against the edge of the desk. I can now see a need for a small tool chest in the desk drawer. ;)

Nice!

I solved the math today, and here's the solution:

For those who want to lay the lines out with a protractor, the angles on two adjacent faces, rising from a common vertex, are 30.36 degrees. Continue on the other two faces at 73.67 degrees, and the lines will meet at a point on the opposite edge from where you started. The distance from your starting point, measured horizontally, is twice the thickness of your stick.

Which is a convoluted way of saying exactly what Mike says above.

A plane cut through each pair of lines (one long, one short) will make a 120 degree dihedral angle, three of which make one vertex of the tetrahedron.

If I get a chance to make a model, I'll post pictures of both this version and the flat-edge version on ABPW this weekend.

Thanks to everyone for the suggestions and help!

Scott

I whittled a few more sticks, and gosh darn it, they don't quite make a tetrahedron. That was earlier in the week. I figured at first it was just a sloppy cut here or there; it really wasn't all that far off. But, while figuring my way around Sketchup, I thought I would lay it out again, and see what it has to say. This time, the distances came out a bit more rational, although the geometry of it still escapes me. The distance I wrote earlier as 1.692" is actually 1.707". That lays out to an exact 45 deg. miter, blade tilt 35.264 degrees off horizontal. If you're using sine bars or a TS-Aligner to set tilt, that works out to 0.707" per inch. (It was so much easier to mark out and just whittle the 6 sticks than to set up a sledded jig for the tablesaw. Depends on how many you're making, of course.)

There's really something magical about a tetrahedron's relationship to cubes and 45 degree angles, as evidenced by the root 2 multiples everywhere you look. It's easy to see how Fuller got trapped into this for so many years.

was put together straight off the bandsaw, using that layout and
bandsaw setup.

And yes, it is tres cool that the 45 degree business keeps showing up in a pyramid made of equilateral triangles!

Scott

And yes, it is tres cool that the 45 degree business keeps showing up in a pyramid made of equilateral triangles!

Scott

Ummm....no.

There are no 45-degree angles anywhere on a tetrahedron.

wrote:

You obviously haven't sketched it out, and clearly read too fast to catch the import and meaning of "not entirely obvious" and "surprisingly".

Bucky Fuller spent the latter decades of his life noodling over just that conundrum. For penance, scary sharpen a chisel and cut 45 degree chamfers on two end grain and one long grain edge, and marvel at the isoceles triangle that emerges. Repeat 3 more times just to be sure that there are no 45 degree angles in that triangle.

You obviously haven't sketched it out, and clearly read too fast to catch the import and meaning of "not entirely obvious" and "surprisingly".

Bucky Fuller spent the latter decades of his life noodling over just that conundrum. For penance, scary sharpen a chisel and cut 45 degree chamfers on two end grain and one long grain edge, and marvel at the isoceles triangle that emerges. Repeat 3 more times just to be sure that there are no 45 degree angles in that triangle.

Isosceles? Nope. Try again.

wrote:

Fuller defined a geometry where perpendicular and orthogonal are 60 degrees, not the 90 degrees of Euclidean geometry. I pay homage to his work by nicking off the corner with one more cut, leaving an equilateral triangle where the 3 cuts meet. The relevance? <shrug> Polyhedrons are more easily defined in that system than Euclidean.

Fuller defined a geometry where perpendicular and orthogonal are 60 degrees, not the 90 degrees of Euclidean geometry. I pay homage to his work by nicking off the corner with one more cut, leaving an equilateral triangle where the 3 cuts meet. The relevance? <shrug> Polyhedrons are more easily defined in that system than Euclidean.

snipped-for-privacy@milmac.com (Doug Miller) wrote in

Read it again, but carefully, Doug. He's describing how a tetrahedron fits into a cube. Each edge of a tetrahedron lies on one diagonal of each face of a cube. And, yes indeedy, those diagonals are at 45 degrees to the edges of the cube.

Read it again, but carefully, Doug. He's describing how a tetrahedron fits into a cube. Each edge of a tetrahedron lies on one diagonal of each face of a cube. And, yes indeedy, those diagonals are at 45 degrees to the edges of the cube.

Gotcha. I misunderstood his post the first time around, but, yes, that's correct.

Elrond Hubbard wrote:

Hwy do any more math at all?

Make a jig that holds the sticks rotated 45 degrees. (I.e. cut a V groove in a board.) Then use the jig to hold the sticks and cut them just like you did for your first tetrahedron.

Hwy do any more math at all?

Make a jig that holds the sticks rotated 45 degrees. (I.e. cut a V groove in a board.) Then use the jig to hold the sticks and cut them just like you did for your first tetrahedron.

That's a good idea, and would certainly work. I've also considered taking one of my original sticks and cutting the corners off then measuring the resulting angles - same result. But part of me would really like to get there by calculation rather than by analog means.

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