Woodworking geometry question

[Elrond Hubbard]

"MikeWhy" wrote in news:VYuSk.6821$ [e-mail]:

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.

[Hoosierpopi]

"Projecting onto faces of a square stick" is your problem.

The "sticks" would form the edges of the tetrahedron and, by definition, the faces of those sticks would be the same as the faces of the tetrahedron relative, one to the other. It would seem that you would start off with "sticks" that were triangular, not square. The image at:

one built out of "sticks" joined at little balls (as if some sort of kids construction toy) and ignores the edges as such.

In your case, the end of your (triangular) stick, would be cut at the same angle as the faces of the tetrahedron and each stick "end" would form a third of the tetrahedron in miniature. Sticks of equal length, glued together at end points so cut, should create the figure/shape you asked about.

But square sticks won't cut it as the edges of the finished piece need to be co-planer with the face(s) of the figure.

[Hoosierpopi]

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

[MikeWhy]

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. ;)

[Doug Miller]

Ummm....no.

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

[MikeWhy]

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.

[Elrond Hubbard]

[e-mail] (Doug Miller) wrote in news:BkJSk.7598$ [e-mail]:

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.

[Elrond Hubbard]

"MikeWhy" wrote in news:%kISk.5501$ [e-mail]:

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

[Elrond Hubbard]

Hoosierpopi wrote in news: [e-mail]:

Maybe your finished piece needs to have the faces co-planar with the edges, but mine needs to be made from square sticks. The faces are not, nor do they need to be, in the same plane as the triangle formed by the edges. When I'm done, the faces of the sticks will be proud of the plane of the edges. Not a problem - that's what I'm after.

[Doug Miller]

Isosceles? Nope. Try again.

[Doug Miller]

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

[MikeWhy]

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? Polyhedrons are more easily defined in that system than Euclidean.

[RM MS]

I agree with the cardboard model, and makig it wider to gain accuracy, ut instead of cardboard, on our shop we use a lot of clear acetate sheets of .020 "and .040" thickness. It breaks and peels off cleanly on a knifed llne. Then after dry-fitting those pieces, you can trace the angles onto your wood pieces. Thin styrene works well, too.

[MikeWhy]

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.

[Elrond Hubbard]

"MikeWhy" wrote in news:FYPVk.7752$ [e-mail]:

The central angle of a tetrahedron - the angle formed by connecting the ends of any edge to the center of the solid - is 109.472 degrees. If you take your blade tilt angle of 35.264, multiply it by two, and add it to the central angle, you get 180 degrees, which makes sense.

But that information is insufficient for cutting the sticks - it does not give you the 120 degree dihedral that lets three ends come together to a point.

The layout I gave in my previous post:

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.