Trigonometry/Woodworking filter: Need to get angles of complicated object

June 26, 2009 9:42 AM

I am looking to make Five Intersecting Tetrahedra out of wood. I have made the paper model using the instructions here - when it's done, it looks like this. However, I am at a loss when it comes to figuring out the angles at the apex of each point where three units meet.

When making it out of paper, the three "legs" that make up the points have sides that can compress and move a bit - much easier to build.

When doing it with wood, the outside angles will be concave, much like the paper, but where they meet, the 3 parts will have a slight convex peek that will fit together like this (see the picture on the right of the "top view") - these are the angles that I need to figure out because this is where I would glue the legs together to form the point.

If they're not exactly right, the model will not fit together properly. I can't figure out what they should be from the paper model because they connect differently.

I realize that this might be a difficult thing to answer in this medium but any help would be appreciated.

posted by alrightokay to sports, hobbies, & recreation (10 comments total)

You should ask the guys at http://www.lumberjocks.com someone there will no for sure, but what you are doing is creating a compound miter, there is a calc here for figuring out the angles
posted by zeoslap at 10:11 AM on June 26

(that's know, not no..) and lumberjocks
posted by zeoslap at 10:12 AM on June 26

This guide is quite useful too
posted by zeoslap at 10:16 AM on June 26

Interesting solid geometry problem. Just to be clear: are you planning to make the "beams" have a solid trianglar cross-section?

Also: I think you mean "peak", not "peek".
posted by Johnny Assay at 10:26 AM on June 26

Johnny Assay: Right, if I cut the beam in half, I would have a triangular cross-section. Near the tips of the beams where they meet though, that cross-section is not going to be triangular.
posted by alrightokay at 10:42 AM on June 26

For the sake of definiteness, let's define some vertices on your individual piece. The angle between two faces of a tetrahedron can be shown to be cos^-1(1/3) ≈ 70° 32', so the cross-section of your rods will be a triangle with angles of approximately 70°, 55°, and 55°. Envision one of your pieces as a spear, pointing upward. Let point A be the "highest" point of the spear; this will end up at the vertex of the tetrahedron when assembled, and is the endpoint of the edge of the prism with the 70° angle. The other two edges of the prism will end somewhat lower, due to the mitring; call these points B and C. The face of the prism opposite the 70° angle (which faces "into" the tetrahedron) will have a triangular "peak" at point D, with an edge of the mitre running "up" to point A. (When you assemble the tetrahedron, the three pieces that meet at a vertex will all have their edges AD together; this line runs "into the screen" in your right-hand diagram. The triangular faces of the mitre, ABD and ACD, will each abut a corresponding face of another piece.) Finally, let point E be some point other than A (it doesn't matter where) along the 70° angle of the prism.

Which angles do you need to know? By construction, the angles EAB and EAC are 30°, exactly; the angle between the faces ABD and ACD is 120°, exactly; and the angle EAD can be shown to be 1/2 * cos^-1(1/3), or about 35°. The other angles shouldn't be too hard to find, but I'm not sure exactly what you need.
posted by Johnny Assay at 11:06 AM on June 26

Lee Krasnow makes tetrahedra out of wood.

I saw this Maker's fair interview with him. He makes all his cuts on a table saw.

Good luck!
posted by orme at 11:08 AM on June 26

I'm not sure how your machine is set up, so I'll try to be very explicit.

Each beam starts out with equilateral triangles as cross sections.
Lay the beam down on the table.
Mark the point-- on the edge of the beam which is not touching the table-- that you want to end up being the peak vertex.
Draw a line segment: Starting at the marked point, and ending at a point on the edge of the beam which is touching the table, such that the angle that line makes with the edge not touching the table is precisely thirty degrees.
Repeat for the other side. So now we have two line segments and three points.
Draw another line segment, along the side of the beam touching the table, connecting the two points which were not previously connected.
So now we have drawn a triangle wrapping around the beam.
Next, draw a triangle entirely contained in the bottom face of the beam, as follows.
The first edge of the triangle is the line segment we drew connecting the points on the bottom face.
The other two edges are the same length, intersecting in a point in the center of the bottom face, such that:
The angle between the two edges of the same length is precisely sixty degrees.
(there are two such choices; chose the one "between" the three points we already had- it is obvious from the picture which one to choose.)
So now we have drawn two triangles, and we have four points.
We are ready to make the first cut.
The intersection of the cut and the beam is going to be a triangle.
The vertices of that triangle are: the peak vertex, the point on the center of the bottom face, and one of the points on the edge of the bottom face.
Make the cut.
Make another cut by choosing the other point on the edge of the bottom face.
So now we've made two cuts.
The input data was two angles: 30 degrees and 60 degrees.
Maybe you could enter those into the machine somehow to simplify the procedure.
posted by metastability at 2:47 PM on June 26

Oh and for completeness I should have given the input data in terms of r and q, where r is number of edges per face, and q is the number of faces intersecting in a vertex. For the tetrahedron, r and q are both 3. The number 30 in the above was (1/2)*360/r. The number 60 appearing in the above was 360/q. If you want to make the other Platonic solids, just change r and q.

Tetrahedron: r = 3, q = 3.
Octahedron: r = 3, q = 4.
Icosahedron: r = 3, q = 5.
posted by metastability at 4:05 PM on June 26

after some reflection...I'm not sure the beams are initially equilateral triangles. That goes for the other platonic solids too. The rest of the procedure remains the same.

As for how to do this on a table saw...
Start out with the blade parallel to the beam. Rotate the blade 30 degrees along the axis perpendicular to the table. Then rotate it X degrees along one of the axes perpendicular to the axis which is perpendicular to the plane of the blade. It is obvious which axis to choose from the picture. Your question is: What is X? Well, let's think. If X was zero, then the cut would go through the one of the two points on the edge of the beam touching the table, and the central point on the bottom face, but not the vertex point. So how much do we need to rotate the blade to make it go through the vertex point. remember it's not necessarily an equilateral triangle.

Ah, I leave it there for someone else to pick up...
posted by metastability at 6:46 AM on June 27