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I'm struggling to find the recurrence relation to evaluate the volume of a solid formed by a series of truncated cones one on top of the other. The image below illustrates the problem for 2 truncated cones, with the solid being formed by the complete revolution of the area around the $z$ axis. Each $i$ truncated cone is characterized by its height $H_i$ and the aperture angle $\alpha_i$, while the smallest radius is $R$ and $0\le\alpha<\frac{\pi}{2}$.

enter image description here

Since the volume of a truncated cone is given by: $$V = \frac{\pi H}{3}\left(r_1^2+r_1r_2+r_2^2\right)$$

For 1 truncated cone: $$V_1 = \frac{\pi H}{3}\left[R^2+R(R+H_1\tan\alpha_1)+(R+H_1\tan\alpha_1)^2\right]$$

For 2 truncated cones: $$V_2 = \frac{\pi}{3}\left\{H_1\left[R^2+R(R+H_1\tan\alpha_1) + (R+H_1\tan\alpha_1)^2\right] + H_2\left[(R+H_1\tan\alpha_1)^2 + (R+H_1\tan\alpha_1)(R+H_1\tan\alpha_1+H_2\tan\alpha_2)+(R+H_1\tan\alpha_1+H_2\tan\alpha_2)^2 \right]\right\}$$

And so on. Just by analyzing the expressions, my first thought was to propose the following formula for a series of $n$ truncated cones: $$V_n = \frac{\pi}{3}\sum_{i=1}^{n}{H_i\left[\left(R+H_i\tan\alpha_i+H_{i-1}\tan\alpha_{i-1}+H_{i-2}\tan\alpha_{i-2}+...\right)^2 +\left(R+H_i\tan\alpha_i+H_{i-1}\tan\alpha_{i-1}+H_{i-2}\tan\alpha_{i-2}+...\right)\left(R+H_{i-1}\tan\alpha_{i-1}+H_{i-2}\tan\alpha_{i-2}+...\right) + \left(R+H_{i-1}\tan\alpha_{i-1}+H_{i-2}\tan\alpha_{i-2}+...\right)^2 \right]} $$

But i'm not sure if this is a rigorous solution or the most elegant way to represent this volume. For example, if $i<1$, than both $H_i$ and $\alpha_i$ must be zero. Thanks in advance!

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    $\begingroup$ Why not write a recurrence relation $V_{i+1} = V_i + v_{i+1}$ where $V_i$ represents total volume from $1$ till $i$, $v_i$ represents volume of the $i$th truncated cone. Now express $v_i$ in terms of $R_i, H_i$ and $\alpha_i$. Also $R_{i+1} = R_i + H_i \tan \alpha_i$ $\endgroup$
    – Math Lover
    Mar 2, 2021 at 13:17
  • $\begingroup$ I'd rather express it in a more explicit way since at the end of the problem there's a minimization regarding the $\alpha_i$ angles, but I'll look forward to your suggestion. I edited a mistake regarding $i$ and $n$ as well in the last equation. Thank you! $\endgroup$
    – Mateus Forcelini
    Mar 2, 2021 at 13:23
  • $\begingroup$ ok then what you have would work, my point with building the recurrence was that you could finally turn that into explicit sum expressing with two summations - the inner one goes from $1 \leq j \leq i$ and outer one goes from $1 \leq i \leq n$ $\endgroup$
    – Math Lover
    Mar 2, 2021 at 13:33
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    $\begingroup$ I'm not sure if I got how to express it in 2 sums, could you detail it, please? Following your advice, I expressed it as $V_n=\frac{\pi}{3}\sum_{i=1}^{n}{H_i\left[R_i^2+R_iR_{i+1}+R_{i+1}^2\right]}$ with $R_1 = R$ and $R_{i+1}=R_i+H_i\tan\alpha_i$, which I think would be clearer for my problem. Thank you very much! $\endgroup$
    – Mateus Forcelini
    Mar 2, 2021 at 14:20
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    $\begingroup$ Yes this looks good. Using recurrence that is a clean way to represent but if you want to state it in a single expression as you mentioned earlier then you can use nested summation. See my answer. $\endgroup$
    – Math Lover
    Mar 2, 2021 at 15:15

1 Answer 1

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Here is one way to express it using summations as I mentioned in comments.

$v_1 = \frac{\pi}{3} (r_1^2 + r_1 r_2 + r_2^2) h_1$

Now, $r_1 = r_2 - x$, where $x = h_1 \tan \alpha$

Then we have,

$r_1^2 + r_1 r_2 + r_2^2 = (r_2-x)^2 + r_2(r_2-x)+r_2^2 = 3r_2^2 + x^2 - 3r_2x$

So we can express the volume $V$ as -

$\displaystyle \frac{\pi}{3} \bigg[\sum_{i=1}^n 3 h_i \big(R + \sum_{j=1}^i h_j \tan \alpha_j\big)^2 + h^3_i \tan^2 \alpha_i - 3 h^2_i \tan \alpha_i\big(R + \sum_{j=1}^i h_j \tan \alpha_j \big)\bigg]$

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