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I am a woodworker and want to build boxes that are not at right angles, a pyramid for example. I am looking for the formulas that will allow me to relate the two miter cuts necessary (one that will create a trapezoid shape, for example, and the other that will allow the edges of the trapezoids to be glued [for a normal box it is 45, for a frame it is either 0 or 90] to the exterior angle of the box.

Daniel Robert-Nicoud
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steve barr
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    Hi Steve. An answer will depend on the steepness of the pyramid. Do you want full generality, or do you have a specific box in mind? If you hand-sketched one particular design, you could insert a picture of it into your question and that would help. It would also help to label the angles that you know - for instance if you were making something like the Pyramids in Egypt, the faces are triangles. If you knew what angles you want on those triangles, it helps. Or if you knew what angle you want those faces to make with the floor as they recline, that helps too. – 2'5 9'2 Oct 12 '13 at 20:07
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    You might find this useful: http://www.pdxtex.com/canoe/compound.htm – vadim123 Oct 12 '13 at 20:24
  • Determining whether you want a triangular or square base (other shapes...) will change the outcome as well – abiessu Oct 12 '13 at 20:55
  • http://www.amazon.com/Solid-Mensuration-Proofs-Willis-Kern/dp/B000KDO2ZQ is a great book. Out of print, but used ones are sometimes available. – Fred Daniel Kline Oct 17 '13 at 20:52

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My linear algebra is a bit rusty, but I think it provides a means to solve this if you have Cartesian coordinates for all the verticies.

For a face $A$, with points in Cartesian coordinates $P_{A,1}$, $P_{A,2}$, $P_{A,3}$ going clockwise around it, $\phi_{A,2}$, the internal angle of the face at $P_{A,2}$ is the arccosine of the cross product of the norms of $P_{A,1}$ and $P_{A,3}$ seen as vectors from $P_{A,2}$

$$ \phi_{A,2}=\cos^{-1}\left(\left\|P_{A,1}-P_{A,2}\right\|\times\left\|P_{A,3}-P_{A,2}\right\|\right) $$

For a face $A$, with points in Cartesian coordinates $P_{A,1}$, $P_{A,2}$, $P_{A,3}$ going clockwise around it (any three points on the face if it has more than 3, as long as they are going clockwise), the normal vector $\vec{n}_A$ (the direction the face is facing) is the norm of the dot product of $P_{A,1}$ and $P_{A,3}$ seen as vectors from $P_{A,2}$.

$$ \vec{n}_A=\left\|\left(P_{A,1}-P_{A,2}\right)\cdot\left(P_{A,3}-P_{A,2}\right)\right\| $$

The angle $\theta_{AB}$ between faces $A$ and $B$ with normals $\vec{n}_A$ and $\vec{n}_B$ is the supplement of the angle between the normals, which is the the arccosine of their cross product.

$$ \theta_{AB}=\pi-\cos^{-1}\left(\vec{n}_A\times\vec{n}_B\right) $$

The mitre angle for their shared edge (assuming equal thicknesses of material) is half that $$\frac{\theta_{AB}}{2}$$

smithkm
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