So, say I have a sector of a ring. I know the thickness of the ring (in the diagram lines BC, DF, and GE are all this thickness). I also know the width of the sector which is set by the length of the chord which defines the outer arc (DE). And I also know the height between the base of the sector (the chord of the inner arc) and the peak of the ring (the line BH). From this, is it possible to calculate either the radii of the inner and outer arcs, and/or the angle the sector encompasses (the angle DAE)?
Diagram of what I'm trying to explain and the points I refer to.
I hope that makes some sense as a question? Any help would be great, I've tried working it through with what little I know about the geometry of a circular segment, but it's been a while since I dusted off my algebra and unless I'm missing something obvious, I don't know if I'm even looking for the right stuff…
Thanks for any help 🙂
Best Answer
Yes, it is possible. Here is how.
Let us give the names $a,b,c$ to the measurements:
$$\begin{cases}a&:=&DE=CB=GE\\ b&:=&HB \\ c&:=&FG& \text{assumed measurable}\end{cases}\tag{1}$$
(I have prefered to take $FG$ instead of $DE$ because the explanations are simpler like this).
Let us denote by $R,r$ the outer and inner radii resp., and $\theta = \angle BAD \ \iff \ 2 \theta = \angle EAD$.
(1) yields:
$$\begin{cases}a&=&R-r\\ b&=&HB=HC+CB=AC-AH+CB=r-r \cos \theta +a \\ c&=&FH+HG=2 r \sin \theta\end{cases}$$
These 3 equations with three unknowns $R,r,\theta$ can be written in the following way :
$$\begin{cases}R-r&=&a&(i)\\ r(1-\cos \theta)&=&b-a&(ii) \\ 2 r \sin \theta&=&c&(iii)\end{cases}$$
In order to be able to extract in particular $\theta$, let us use classical trigonometry formulas on equations (ii) and (iii):
$$\begin{cases}2 r(\sin \tfrac{\theta}{2})^2&=&b-a \\ 4r \sin \tfrac{\theta}{2}\cos \tfrac{\theta}{2}&=&c\end{cases}$$
Taking their ratio, one gets:
$$\tfrac{1}{2}\tan \tfrac{\theta}{2}=\tfrac{(b-a)}{c} \ \iff \ \theta=2 \tan^{-1}\tfrac{2(b-a)}{c}$$
which is perfectly defined quantity.
Having $\theta$, one can compute the value of $r$, using equation (ii) above, and finally the value of $R$ using equation (i).
Remark: having preferred $DE$ to $FG$, it is important to recall that the connection between these two lengths is very simple:
$$\frac{DE}{FG}=\frac{R}{r}$$