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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:

enter image description here

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.

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    Subtract $\frac{3V}{\pi}$ on both sides and you have a cubic function $p(h)$ of which you want to calculate the root. The formulas for the roots of cubic functions are quite complex. In practice, no one uses them. You would use numerical methods like Newton's method to find the root. – 2017-01-24
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    There might be a chance to find the result. Given $\theta$ and $r_1$, it should be possible to calculate the volume $V_c$ and the height $h_c$ of the cone that completes the frustum to form a larger cone. Let $H = h+h_c$ be the height of the larger cone, then the volume of the larger cone can be expressed as $V_c+V = uH^3$ with some constant $u$ that depends on $\theta$. From this, you can find $H$, and then you have $h = H-h_c$. – 2017-01-24
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    @ReinhardMeier Not sure... since $r_1$ is the "middle" radius and I don't know the distance $h$ to $r_2$ (or $r_2$), I'm not sure how I could find the total height of the final cone. That supposes I'm following you, in which case $V_c$ would be the volume of a cone *starting* at $r_2$, though $r_1$ and all the way *down* to a tip below my drawing above? – 2017-01-24

2 Answers 2

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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 $0$.

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    Gar! So elegant... I definitely used the wrong substitution. I walked through all the cancellations of the expansion just to be sure. Just perfect. – 2017-01-24
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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.

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    A for effort for sure, and thanks for explaining the cone method in the comments. I get it now. Have to hand it to the other answer given how simple it ended up being. Thanks for putting this together! – 2017-01-24