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
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
(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. ;)
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
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!
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.
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.
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.
firstname.lastname@example.org (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.
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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