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This question may be beyond the scope of what this StackExchange is for. But I shall ask it.

I am a CNC programmer for a Woodworking company. One of the problems I periodically encounter and have never been able to solve to my own satisfaction is this.

Say you have a desk, that is curved along a specific radius. On its face is an attached a panel, of a known thickness, that is bent around this radius, but also angled, again at a known angle.

The problem comes in, when you take an arc length at the top of the panel, and the bottom of the panel, and take an angled height of the panel, you end up with a Trapezoid shape. But when this trapezoid is bent around the radius, it will always be off. It would seem the trapezoid is the incorrect shape. It would seem to me that the shape would have to be curved along its top and bottom edges. But finding the mathematical system necessary to compute this is eluding me.

enter image description here

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    $\begingroup$ What is the question, exactly? Is it "What is the shape of the outer panel if you flattened it?" In that case, if I understand the rendering correctly, the outer panel is a section of a cone: Its bottom and top are arcs of circles centered on the axis of the cone and its sides are parts of rays on the cone. "Unrolling" this shape gives a sector of an annulus: i.sstatic.net/FEppO.png $\endgroup$ Commented Feb 6, 2017 at 15:54
  • $\begingroup$ The first thing that comes to mind is a spline profile shape. $\endgroup$
    – Jean Marie
    Commented Feb 6, 2017 at 16:33
  • $\begingroup$ I think Travis may have it here. I am rusty on my conics. Would the top and bottom arcs just be the radii at those heights. Unrolling is basically what i am asking for yes. $\endgroup$
    – Aries7
    Commented Feb 6, 2017 at 17:04

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Your shape is a sector of an annulus (see figure below). If $L_2$ and $L_1$ are the longer and shorter length of panel borders (top and bottom), and $h$ the distance between those borders, then you can compute $d$ from a simple proportion: $d=h\cdot L_1/(L_2-L_1)$ and angle $\alpha$ is given by

$$ \alpha={L_1\over d}{180°\over\pi}\ \hbox{degrees}. $$ enter image description here

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  • $\begingroup$ Many Thanks for the help. $\endgroup$
    – Aries7
    Commented Feb 6, 2017 at 20:41

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