3
$\begingroup$

Most of the calculations I know and can find can solve for the radius if the cord length is known or the angle is known along with the height of the arc segment. However I cannot find, nor figure out how to calculate the radius if only the arc segment and the height of the segment is known.

I have looked up several different methods on this and other site and the following are similar but not exactly what I am looking for.

In CAD I can draw a arc like what is shown at (http://mathworld.wolfram.com/CircularSegment.html). If I fix the mid point at (0,0), the set the cord horizontal, then add a dimension the segment height (i.e. h=2") and then dimension the arc length (i.e. s=10"), the geometry is fully constrained. So I know there is a way to solve this.

If I reversed the inputs and had the arc length (s) known and the radius (R) known I can calculate the segment height but I am having trouble reversing the equation.

$\endgroup$
1
  • $\begingroup$ What is the height of the arcsegment? The altitude of the triangle with vertex at the centre of the circle and basis the chord? $\endgroup$
    – Bernard
    Commented Apr 12, 2016 at 20:34

2 Answers 2

1
$\begingroup$

As Aretino answered, you need some numerical methods to solve the equation $$L\big(1-\cos(\theta)\big)=2H\theta$$ (remember that there is no analytical solution to $x=\cos(x)$).

However, for $-\frac \pi 2 \leq \theta\leq\frac \pi 2$, you could use as a quite good approximation (see here) $$\cos(\theta) \simeq\frac{\pi ^2-4\theta^2}{\pi ^2+\theta^2}$$ which makes the equation $$2 H \theta^3-5 L \theta^2+2 \pi ^2 H \theta=0$$ which reduces to a quadratic $$2 H \theta^2-5 L \theta+2 \pi ^2 H =0$$ the solution of which providing a reasonable starting point for the iterative process.

For illustration purposes, let me consider $L=10$, $H=2$. The retained solution of the quadratic is $$\theta=\frac{1}{4} \left(25-\sqrt{625-16 \pi ^2}\right)\approx 0.846955$$ while the "exact" solution is $\approx 0.849952$.

Edit

Setting $\lambda=\frac{2H}L$, the initial equation write $$1-\cos(\theta)=\lambda\theta$$ Assuming $-\frac \pi 2 \leq \theta\leq\frac \pi 2$, the approximation leads to $$\theta_0=\frac{5-\sqrt{25-4 \pi ^2 \lambda ^2}}{2 \lambda }$$ Now, peform one iteration of Newton method using the exact equation; this will lead to $$\theta_1=\frac{\theta_0 \lambda +\cos (\theta_0 )-1}{\sin (\theta_0 )-\lambda }+\theta_0$$ For illustration, using $\lambda=\frac 25$ as before, this leads to $$\theta_0\approx 0.846955$$ $$\theta_1\approx 0.849961$$ while the exact solution woudld be $\approx 0.849952$. Continuing the iterations for many more figures, the itrates would be $$\left( \begin{array}{cc} n & \theta_n \\ 0 & 0.846954969749869 \\ 1 & 0.849960525069669 \\ 2 & 0.849952018812858 \\ 3 & 0.849952018744878 \end{array} \right)$$ which is the solution for fifteen significant figures.

$\endgroup$
0
$\begingroup$

If $2\theta$ is the angle related to the given arc, having length $l$, radius $r$ and height $h$, then you have $l=2\theta r$ and $h=r(1-\cos\theta)$. By eliminating $r$ you are left with $l(1-\cos\theta)=2h\theta$, which in general can be solved for $\theta$ only numerically.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .