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I wish to calculate the volumes of a truncated cone whith asymetry over all axes and ellipses as base and top - how do I do that?

I have height and radii of the corresponding ellipses. (I assume correctly I need the geometric means of the 2 radii describing the ellipsis.)

Kind Regards

Tony Stark
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Toby El Tejedor
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2 Answers2

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The volume of a truncated right circular cone is given by $$V = \frac{\pi h}{3}\left( \; r_1^2 + r_1 r_2 + r_2^2\; \right)$$ where $h$ is the height, and $r_1$ and $r_2$ the radii of the bases. Let us take $r_1 = 1$ for the "bottom" radius and interpret $r := r_2$ as the ratio of the "top" radius to the bottom. Then $$V = \frac{\pi h}{3} \left( \; 1 + r + r^2 \; \right)$$ To turn the circular bases into elliptical ones, we simply stretch the figure in two orthogonal directions (perpendicular to the axis of the cone). If the radii of the "bottom" base ellipse are to be $a$ and $b$, then these are our scale factors (since we declared the bottom circle to have radius $1$); upon stretching, the "top" ellipse will have radii $ar$ and $br$, and the volume will be scaled by $ab$: $$V = \frac{\pi a b h}{3}\left(\; 1 + r + r^2 \; \right)$$

Blue
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  • Cheers, but does this still hold true when the ellipses are rotated against each other? (Still orthogonal to h) – Toby El Tejedor Jul 30 '13 at 06:00
  • @TobyElTejedor: If your "cone" is well-behaved between the bases, then Cavalieri's Principle guarantees that the volume formula above still holds. By "well-behaved" here, I mean that, at every level, the elliptical cross-section of your "cone" matches (in shape ---and, therefore, area--- but not necessarily orientation) the elliptical cross-section of the cone I describe. – Blue Aug 04 '13 at 19:17
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You can model this shape in $\mathbb{R}^3$ with the equations:

$$\frac{H^2x^2}{a^2}+\frac{H^2y^2}{b^2}=(z-H)^2\\[.2in]z=0\\[.2in]z=h<H$$

where the base ellipse has axes of length $2a$ and $2b$, the top ellipse has axes of length $\frac{(2a)(H-h)}{H}$ and $\frac{(2b)(H-h)}{H}$, and the frustrum has height $h$. Here $H$ represents the height of the elliptic cone from which the frustrum was cut.

Now, notice for any given height $z$, the area of the cross section of our frustrum is:

$$A(z)=\frac{\pi ab(H-z)^2}{H^2}$$

We integrate this from $z=0$ to $z=h$ to obtain

$$V(a,b,h,H)=\frac{\pi abh(h^2-3hH+3H^2)}{3H^2}$$

If you wish to express this in terms of the dimensions of the top ellipse, notice by similar triangles we have the following formula:

$$\frac{a}{H}=\frac{c}{H-h}$$

where $2c$ is the length of the horizontal axis of the top ellipse. It follows that $H=\frac{ah}{a-c}$, so we get:

$$V(a,b,c,h)=\frac{\pi bh(c^2+ac+a^2)}{3a}$$

A similar formula exists if we replace $H$ with a function of $2d$, the length of the vertical axis of the top ellipse. If you want to express this in terms of the geometric means of the radii of the top and bottom, $n$ and $m$, you can check that you will obtain the formula:

$$\boxed{\displaystyle V(n,m,h)=\frac{1}{3}\pi h(n^2+nm+m^2)}$$

Jared
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  • Good, thank you. If I now twist the ellipses against each other, does this still hold true? (Baring in mind the ellipses remain orthogonal to h) – Toby El Tejedor Jul 30 '13 at 06:07