# Solving the Pulley Problem for the radius of one of the pulleys

I've been working on a project involving some pulleys, and while trying to work out some of the math I was happy to find it's already a mathematical problem!

with the solution given as (Wikipedia)

The result of this equation gives the length of the required belt, but since my project doesn't involve custom belts (but rather what I can pick up from the hardware store), I'd much prefer to have this solved for $$r_2$$.

Solving for $$r_2$$ is trivial, except for its presence in the definition of $$\theta$$. That's where I got stuck. I also tried plugging it into wolfram alpha, and playing around with treating the trig functions in their complex exponential form, and that's precisely when I decided to reach out for help haha.

Note: In the end, I'm also hoping to treat $$P$$ as $$P=r_1 + r_2 + d$$, where $$d$$ is the clearance between the pulleys. This is a more useful metric for me, as I can assign a constraint that, say, $$d>1in.$$, but that further complicates solving for $$r_2$$ so I refrained from including that in the initial question.

• You need to find $r_2$. What is known? Aug 8, 2022 at 7:23
• @IvanKaznacheyeu I know $r_1$, the length of the belt (not shown, but call it $L$), and if we allow $P=r_1 + r_2 + D$, I know $D$ as well. I believe I know every variable but $r_2$, but I've been told there's no analytic solution to solve for $r_2$ because of its presence in the definition of theta Aug 8, 2022 at 16:39
• It's because equation for $r_2$ is transcendental. But I believe you don't need analytical solution, only method for computing $r_2$ with necessary accuracy for given $D$, $r_1$ and $L$. Is it correct? Aug 8, 2022 at 17:05
• Equation for $r_2$ looks like $$2\sqrt{4r_1r_2+2D(r_1+r_2)+D^2}+2\pi r_1-2(r_1-r_2)\arccos\frac{r_1-r_2}{r_1+r_2+D}-L=0$$ This equation can be solved by numerical methods or by plotting left-hand side for $x=r_2$ and finding intersection with x-axis. Aug 8, 2022 at 17:11
• Looks like that. Plot for $r_1=3$, $D=2$, $L=24$ looks only slightly non-linear for $r_2\in[0;r_1]$. Aug 9, 2022 at 10:52

Rewrite the equation as $$\cos\frac\theta2= \frac{r_1-r_2}{r_1+r_2+d}$$ and then solve for $$r_2$$ $$r_2=\frac{r_1(1-\cos\frac\theta2)+d\cos\frac\theta2}{1+\cos\frac\theta2}$$ which can be further simplified to $$r_2 =\frac d2+(r_1-\frac d2\tan^2\frac\theta4)$$
• @JShoe - I do not understand. I assume that you intent to solve for $r_2$ in terms of $\theta$ Aug 7, 2022 at 19:44
• @JShoe - simply because you have both $\sin \theta$ and $\theta$ Aug 7, 2022 at 19:51