# Solve Spherical Shell for Radius given Thickness and Volume

I'm looking to calculate the outer radius of a spherical shell of a desired volume and thickness. I don't know if the years have knocked some obvious obstacle out of my perception, but here's what i'm struggling with.
Given: Thickness T, Radii R and r, and Volume V and the following relationships: $$T=R-r$$ $$V=\frac{4}{3}\pi R^3 - \frac{4}{3}\pi r^3$$ How do I solve for R in terms of V and T?

I get as far as:

$$\frac{3V}{4\pi}=R^3-(R-T)^3$$ Which comes down to... $$\frac{3V}{4\pi} = 3R^2-3RT+T^2$$

Which i'm lost on breaking apart. For more clarification, this is supposed to be just a fun exercise for a D&D game. The rules i'm working with are limited by volume, and the integrity of created objects would be limited by thickness. Thus, these are the terms i wanted to control, and then calculate the resulting outer radius of what the character can create.

Thanks for any help that can be provided!

• If your last line is correct, then simply use the quadratic formula here! – Sawarnik Apr 28 '14 at 16:32

This is the right way to do it (except you have an algebraic error in your last equation, which you could deduce from dimensional analysis). Now you have a simple quadratic in $R$: $$R^2 - TR + \left(\frac{T^2}{3} - \frac{V}{4\pi T}\right) = 0.$$
$$R = \frac{T + \sqrt{T^2 - 4\left(\frac{T^2}{3} - \frac{V}{4\pi T}\right)}}{2} = \frac{T + \sqrt{T^2 - \frac{4}{3}T^2 + \frac{V}{\pi T}}}{2} = \frac{T+\sqrt{\frac{V}{\pi T} -\frac{T^2}{3} }}{2}$$
• @user146407 Also, assuming you have $T$ in inches, remember to convert to feet before applying this formula, or the answer you get won't make any sense. – Paul Z Apr 28 '14 at 17:13