Uniformly scale a hollow cylinder, but keep the hole radius constant I'd like to linearly scale a hollow cylinder so that I get some specific volume while keeping the original proportions (i.e., outer radius & height). The problem is that I'd like the size of the hole to stay the same, so I can't just use the cube root of the old-to-new-volume ratio as the scaling factor for all dimensions.
How can this be solved?
Background: I'm making a set of molds for concrete weight plates. The heaviest plate has a standardized diameter and the height (thickness) will be determined by the amount of concrete poured. Lighter weights are arbitrarily sized, and using the same diameter would make them too thin (fragile), while using the same thickness and smaller diameter would make them too fat to stack them all on the end of a barbell. So proportionally scaling down the largest mold seems like the best option.
 A: Let's say our disks have inner radius $r$ (which is constant under scaling), outer radius $R$ (so $r < R$), and thickness $h$. The volume is
$$
V = \pi R^{2}h - \pi r^{2}h = \pi(R^{2} - r^{2})h.
$$
Our goal is to pick a sequence of $R$s and $h$s to get specified volumes/weights.
If we scale the (outer) radius and thickness together, with scale factors $s_{0} = 1$, $s_{1}$, $s_{2}$, $s_{3}$, ..., we get the sequence $R$, $s_{1}R$, $s_{2}R$, $s_{3}R$, ..., and similarly for $h$. The resulting volumes are
\begin{align*}
  V_{0} &= \pi(R^{2} - r^{2})h, \\
  V_{1} &= \pi(s_{1}^{2}R^{2} - r^{2})s_{1}h = \pi [s_{1}^{3}R^{2} - s_{1}r^{2}]h, \\
  V_{2} &= \pi(s_{2}^{2}R^{2} - r^{2})s_{2}h = \pi [s_{2}^{3}R^{2} - s_{2}r^{2}]h, \\
  V_{3} &= \pi(s_{3}^{2}R^{2} - r^{2})s_{3}h = \pi [s_{3}^{3}R^{2} - s_{3}r^{2}]h,
\end{align*}
and so forth.
Note that it might be desirable (aesthetically and/or mechanically) instead to scale the dimensions so that the area and thickness scale together. For example, making a disk half as thick wouldn't make the radius half as large, but about $70$ percent as large, because $\sqrt{1/2} \approx 0.707\dots$. In this case, the sequence of radii is $R$, $\sqrt{s_{1}}R$, $\sqrt{s_{2}}R$, $\sqrt{s_{3}}R$, ..., and the resulting volumes are
\begin{align*}
  V_{0} &= \pi(R^{2} - r^{2})h, \\
  V_{1} &= \pi(s_{1}R^{2} - r^{2})s_{1}h = \pi [s_{1}^{2}R^{2} - s_{1}r^{2}]h, \\
  V_{2} &= \pi(s_{2}R^{2} - r^{2})s_{2}h = \pi [s_{2}^{2}R^{2} - s_{2}r^{2}]h, \\
  V_{3} &= \pi(s_{3}R^{2} - r^{2})s_{3}h = \pi [s_{3}^{2}R^{2} - s_{3}r^{2}]h,
\end{align*}
and so forth. The respective results look like this:

As you note, volume does not scale proportionally to the dimensions because the inner radius is constant. But no matter which strategy we use, one of the above or some other, the next step is to take the desired volumes (weights) and the standarized largest radius $R$, calculate $h$, and then successively calculate $s_{1}$, $s_{2}$, $s_{3}$, ... either numerically or by algebra.
A fringe benefit of the second scheme is that we only need to solve a quadratic equation to get the scale factors. For the first method we need to solve a cubic. This can be done algebraically, but is messier.

The next time you hear someone make a crack about the uselessness of algebra, I hope you'll set them straight. ;)
