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!
Best Answer
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 $R$: $$R^2 - TR + \left(\frac{T^2}{3} - \frac{V}{4\pi T}\right) = 0.$$
Use the quadratic formula to solve it, and pick the positive root.
$$R = \frac{T + \sqrt{T^2 - 4\left(\frac{T^2}{3} - \frac{V}{4\pi T}\right)}}{2} = \frac{T + \sqrt{T^2 - \frac{4}{3}T^2 + \frac{V}{\pi T}}}{2} = \frac{T+\sqrt{\frac{V}{\pi T} -\frac{T^2}{3} }}{2}$$