I'm looking for an equation that satisfies these conditions:

  • Input $90^\circ$, result is $90^\circ$
  • Input $45^\circ$, result is $60^\circ$
  • Input $0^\circ$, result is $45^\circ$

For an input value of P, the closest I have gotten was this:

F(P) = 90 - 45 * cos(P) 

That formula satisfies the first and last input, and comes close to satisfying the middle requirement by producing the value $58.18....$

I'm looking for this formula because I believe it would help me to solve a larger problem - which I will describe. Suppose you have a truncated octagonal pyramid.

truncated octagonal pyramid

Now mentally modify this pyramid so that it no longer has perfect symmetry. Instead of being formed from eight trapezoids, it is instead formed by four rectangles and four trapezoids in an alternating pattern. Furthermore, the base length of the four rectangles can be arbitrarily longer than the base lengths of the four trapezoids.

Depending on the angle of pitch of the pyramid, the base corners of the four trapezoids will have an angle that ranges from $45^\circ$ to $90^\circ$.

  • The angle is $45^\circ$ if the pyramid has been utterly flattened.
  • The angle is $90^\circ$ if the pyramid is perpendicular to the ground, and thus ceases to be a pyramid. That is, in this configuration, the trapezoids have again become rectangles.

Between those two extremes, when the four rectangular faces of the pyramid are at a $45^\circ$ angle to the ground, the base corners of the trapezoids presumably have a $60^\circ$ angle.

Thus for any position between those extremes, I would like to accurately compute the base angles of the trapezoids. Based on the angle the rectangles form with the ground, this information could be used to determine mitre-cuts of the trapezoids so that they fit properly into the structure. I only mention that because I am in fact trying to build something. :)

  • $\begingroup$ The parabola that passes through your 3 points of interest is $y=45+x/6+x^2/270$. $\endgroup$ – Semiclassical Jul 29 '14 at 4:10

I don't think the equation you seek is so simple, although I am likely not seeing some $\cos^{-1}(\;)$ simplification. I computed the trapezoid base corner angle $\tau$ as a function of the angle $\theta$ of the "pitch of the pyramid," as you call it. I get this:

Notice this agrees with your three points: $$\theta=0^\circ,\; \tau=90^\circ$$ $$\theta=45^\circ,\; \tau=60^\circ$$ $$\theta=90^\circ,\; \tau=45^\circ$$ The calculation is based on the 3D situation depicted below, where $a,b,c$ are three consecutive points of the top octagon, and $a_0,b_0,c_0$ their counterparts on the base. $\tau$ is the angle $\angle a_0 b_0 b$, and $\cos \tau$ can be found from $(b-b_0) \cdot (a_0-b_0)$.

  • 1
    $\begingroup$ not sure what he wants, really; note that an extreme case of this degenerates to four squares and four equilateral triangles alternating, and a square on top. I'll see if I can find a name, the full solid is the result of both lopping the corners and chamfering the edges of a cube en.wikipedia.org/wiki/Rhombicuboctahedron $\endgroup$ – Will Jagy Jul 14 '15 at 0:28

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