Volume of a frustrum given height, one radius, and slope of wall

Question

I'm trying to figure out the volume of a frustrum, but don't have what reference equations typically provide. Being 10 years out of college, I'm not sure if I should know this or if it's genuinely difficult! I want to calculate the height required to hold a volume, $V$, in a slope-walled vessel knowing the bottom radius, $r_1$ and the angle that dictates the wall slope. Here's an illustration:

For the starting points, we have the formula for volume:

$\displaystyle V = \frac{1}{3} \pi(r_1^2 + r_1r_2 + r_2^2)h$

And since I don't know what $r_2$ will be at the height that provides my volume, I used the fact that the difference between $r_2$ and $r_1$ at height $h$ will yield this substitution:

$r_2 = h \tan \theta + r_1$

Plugging things in, I worked through it like this:

$\displaystyle V = \frac{1}{3} \pi\left(r_1^2 + r_1(h \tan \theta + r_1) + (h \tan \theta + r_1)^2\right)h$

$\displaystyle \frac{3V}{\pi} = (r_1^2 + r_1 h \tan \theta + r_1^2 + h^2 \tan^2 \theta + 2 r_1 h \tan \theta + r_1^2)h$

$\displaystyle \frac{3V}{\pi} = (3r_1^2 + 3 r_1 h \tan \theta + h^2 \tan^2 \theta)h$

$\displaystyle \frac{3V}{\pi} = h^3 \tan^2 \theta + 3 h^2 r_1 \tan \theta + 3 h r_1^2$

I have this gnawing feeling that this should be a standard form I recognize. $\tan \theta$ is just a constant, so in essence I have something in the form of:

$x^3 + x^2y + xy^2$

I'm just not sure how I should go about solving it. Thanks for any corrections/insights.

Answer 1

In $V=\frac{1}{3}\pi(r_1^2+r_1r_2+r_2^2)h$, replace $h$ with $\frac{r_2-r_1}{\tan(\theta)}$ and multiply both sides with $\frac{3\tan(\theta)}{\pi}$: $$ \frac{3V\tan(\theta)}{\pi} = (r_1^2+r_1r_2+r_2^2)(r_2-r_1) = r_2^3-r_1^3 $$ So we have $$ r_2=\sqrt[3]{\frac{3V\tan(\theta)}{\pi} + r_1^3} $$ from which you can calculate $h=\frac{r_2-r_1}{\tan(\theta)}$.

If numerical robustness is an issue, better calculate $h$ as follows: $$ h=\frac{3V}{\pi(r_1^2+r_1r_2+r_2^2)} $$ which can be used even if $\theta$ approaches $0$.

Answer 2

We both seem to get some third order polynomial in $h$ where we can try to look for a fixed point or root numericaly.

For a full circular cone of height $H$ and radius $R$ we have (link) $$ V = \frac{1}{3}\pi R^2 H $$ In this case we extend the frustrum downwards into a full cone and have $$ V_f = \frac{1}{3} \pi r_2^2 H $$ For the angle we have $$ \tan(\theta) = \frac{r_2}{H} $$ For the bottom cone we have $$ V_b = \frac{1}{3} \pi r_1^2 (H - h) $$ so we get $$ V = V_f - V_b = \frac{1}{3} \pi r_2^2 H - \frac{1}{3} \pi r_1^2 (H - h) = \frac{1}{3} \pi \left( (r_2^2 - r_1^2) H + r_1^2 h \right) = \frac{1}{3} \pi \left( (r_2^2 - r_1^2) r_2 \cot(\theta) + r_1^2 h \right) $$ Solving for $h$ gives $$ h = \frac{3 V - \pi (r_2^2 - r_1^2) r_2 \cot(\theta)}{\pi r_1^2} $$ Using $$ r_2 = r_1 + h \tan(\theta) $$ This is \begin{align} h &= \frac{3 V - \pi (2 r_1 h \tan(\theta) + h^2 \tan(\theta)^2) r_2 \cot(\theta)}{\pi r_1^2} \\ &= \frac{3 V - \pi (2 r_1 + h \tan(\theta)) h (r_1 + h \tan(\theta))}{\pi r_1^2} \\ &= F(h) \end{align} where $F$ is a third order polynomial in $h$. This form is a candidate for fixed point iteration.