deriving antiprism formulas

Question

According to Wikipedia:

Let $a$ be the edge-length of a uniform $n$-gonal antiprism; then the volume is:

$$V = \frac{n \sqrt{4\cos^2\frac{\pi}{2n}-1}\sin \frac{3\pi}{2n} }{12\sin^2\frac{\pi}{n}}~a^3$$

and the surface area is:

$$A = \frac{n}{2} \left( \cot\frac{\pi}{n} + \sqrt{3} \right) a^2$$

Furthermore, the volume of a right $n$-gonal antiprism with side length of its bases $l$ and height $h$ is given by:

$$V = \frac{nhl^2}{12} \left( \csc\frac{\pi}{n} + 2\cot\frac{\pi}{n}\right).$$

Does anyone know how these formulas are derived? I always prefer to understand where a formula comes from rather than taking Wikipedia's word for it. Pictures are always appreciated.

Bonus: I would also like to know the surface area and volume formulae for crossed and twisted antiprisms and their derivations, if anyone knows them.

Answer 1

The first thing to think about is a coordinatisation of its vertices, such as $(\frac{\cos(k π/n)}{2\sin(π/n)}, \frac{\sin(k π/n)}{2\sin(π/n)}, (-1)^k \frac{h}2)$ for all $k$ integral. (Here the denominators of the first two coordinates just ensure unit side lengths of the base polygons.) Then observe the body center in here is the origin. Accordingly the volume easily gets calculated by the sum of volumes of face-based pyramids pointing (with their tips) each to that body center.

In order to do so, because the volume of a pyramid is $\frac13 hG$, where h is its height and G is the base area, i.e. the area of the according polygon, you surely will have to evaluate those beforehand. But these are either geometrically straight forward, or can in turn be calculated by the same approach one dimension lower: take the center point of each face and calculate the "volume" (= area) of all its "face"-based (= side-based) "pyramids" (= triangles).

Those individual face areas clearly have to be summed up only, when asking for the total surface content.

Finally, note that three consecutive vertices of the coordinates of the first paragraph (i.e. with consecutive $k$) will enclose a lateral triangle of that antiprism. If that one would be asked to become a regular triangle, this imposes a restriction onto $h$, which can be solved for and thus inserted into your latter volume formula (i.e. Volume($n,l,h$)) becoming the volume formula of the uniform antiprism as of your first volume formula (i.e. Volume($n, a$), where $a\equiv l$ for sure).

--- rk

Answer 2

Antiprisms belong to a class of polyhedron known as prismatoid. To compute the volume of a prismatoid, one only need $4$ numbers: its altitude and three areas.

A prismatoid is a polyhedron whose vertices lies on two parallel planes and the side faces (faces not on the two planes) are either triangles or trapezoids.

Given a prismatoid, the distance between the two parallel planes is known as its altitude (or height). Let's denote it by $h$.

Choose a coordinate system where the two planes are $z = 0$ and $z = h$. Let $q$ be the number of side edges (edges not on the two planes). If one intersect the prismatoid with plane $z = t$ for a $t \in (0,h)$, the intersection will be a $q$-gon.

Let $\Delta(t)$ be the area of the $q$-gon and $(x_i(t),y_i(t),t)$ be the coordinates of the intersection between the $i^{th}$ side edge and plane $z = t$. Assume the side edges are ordered correctly, $\Delta(t)$ can be computed using shoelace formula: $$\Delta(t) = \frac12\sum_{i=1}^q (x_i(t) y_{i+1}(t) - y_i(t)x_{i+1}(t))$$

Since the side edges are straight lines, $x_i(t), y_i(t)$ are polynomial in $t$ with degree at most $1$. So $\Delta(t)$ is a polynomial in $t$ with degree at most $2$.

The volume of the prismatoid can be expressed as an integral.

$$\verb/Volume/ = \int_0^h \Delta(t) dt$$ Being a polynomial with degree at most $2$, above integral of $\Delta(t)$ can be evaluated exactly using Simpson's 1/3 rule. The end result is the

prismoidal formula $$\verb/Volume/ = \frac{h}{6}\left(\Delta_{t} + 4\Delta_{m} + \Delta_{b}\right) $$ where $\Delta_t = \Delta(h)$, $\Delta_m = \Delta(\frac{h}{2})$, $\Delta_b = \Delta(0)$ are the areas of the "top" face, the "middle" cross-section and "bottom" face of the prismatoid respectively.

Let's apply this to an antiprism $X_0$ whose top/bottom faces are regular $n$-gon with side $\ell$ and altitude $h$.

Let $\alpha_n$ be the area of a regular $n$-gon with side $1$. Since we can decompose such a $n$-gon into $n$ triangles with base $1$ and height $\frac12\cot\frac{\pi}{n}$, we have $$\alpha_n = \frac{n}{4}\cot\frac{\pi}{n}$$ So $\Delta_t = \Delta_b = \alpha_n \ell^2$. It is not hard to see the middle cross-section of the antiprism is a regular $2n$-gon with side $\frac{\ell}{2}$. This means $\Delta_m = \frac14 \alpha_{2n} \ell^2$. Plug these into prismoidal formula, one obtain the formula for volume of antiprism you want to derive

$$\verb/Volume/(X_0) = \frac{h\ell^2}{6}\left(2\alpha_n + \alpha_{2n}\right) = \frac{nh\ell^2}{12}\left(\cot\frac{\pi}{n} + \cot\frac{\pi}{2n}\right) = \frac{nh\ell^2}{12}\left(2\cot\frac{\pi}{n} + \csc\frac{\pi}{n}\right) $$ For the uniform $n$-gonal antiprism with side $a$, $\ell = a$ and one only need to figure out what $h$ is. If one project its vertices to $xy$-plane, the vertices will form a regular $2n$-gon with circumradius $\frac{a}{2}\csc\frac{\pi}{n}$. The side edges of antiprism get projected to line segments of length $\displaystyle\;s = \frac{2a\sin\frac{\pi}{2n}}{2\sin\frac{\pi}{n}} = \frac{a}{2\cos\frac{\pi}{2n}}\;$. This leads to $$h = \sqrt{a^2 - s^2} = \frac{a}{2\cos\frac{\pi}{2n}}\sqrt{4\cos^2\frac{\pi}{2n} - 1}$$ As a result, the volume of regular $n$-gonal antiprism is

$$V = \frac{na^3}{24}\sqrt{4\cos^2\frac{\pi}{2n} - 1}\underbrace{\frac{\cot\frac{\pi}{n} + \cot\frac{\pi}{2n}}{\cos\frac{\pi}{2n}}}_{\color{blue}{[1]}} = \frac{na^3\sqrt{4\cos^2\frac{\pi}{2n} - 1}\sin\frac{3\pi}{2n}}{12\sin^2\frac{\pi}{n}} $$ Aside from a typo in your original formula, this is what you already know.

About its area, it is simply

$$(2\alpha_n + 2n\alpha_3)a^2 = a^2 \left(2\frac{n}{4}\cot\frac{\pi}{n} + 2n\frac{\sqrt{3}}{4}\right) = \frac{na^2}{2}\left(\cot\frac{\pi}{n} + \sqrt{3}\right)$$

Update - about volume of twisted prisms.

Start with antiprism $X_0$ we have studied before. Let $R = \frac{\ell}{2\sin\frac{\pi}{n}}$. WOLOG, we will assume in cylindrical polar coordinates${}^{\color{blue}{[2]}}$, the vertices of $X_0$ are located at

$$ (\rho,\theta,z) = (R, \frac{2k\pi}{n}, h), (R,\frac{(2k+1)\pi}{n},0) \quad\text{ for }\quad k = 0, \ldots, n-1$$

If one twist the top face by an angle $\phi$, ie. replace vertices $(R,\frac{2k\pi}{n},h)$ by $(R,\phi + \frac{2k\pi}{n},h)$ while keep vertices at bottom face fixed, we obtain a twisted prism. Let's call it $X_\phi$. When $|\phi| < \frac{\pi(n-1)}{n}$, this twisted prism is an ordinary polyhedron and prismoidal formula applies.

Let $\phi_{\pm} = \frac12(\phi \pm \frac{\pi}{n})$. The mid-section is a star-shaped $2n$-gon with vertices located at

$$(\rho,\theta,z) = (R\cos\phi_{-}, \phi_{-} + \frac{2k\pi}{n}, \frac{h}{2}), (R\cos\phi_{+}, \phi_{+} + \frac{2k\pi}{n}, \frac{h}{2}), \quad\text{ for }\quad k = 0, \ldots, n - 1$$

It has area $$\Delta_m = \frac{2n}{2} (R\cos\phi_{+})(R\cos\phi_{-})\sin\frac{\pi}{n} = \frac{n\ell^2}{8\sin\frac{\pi}{n}}(\cos\frac{\pi}{n} + \cos\phi)$$

Plug these into prismoidal formula, the volume of twisted prism $X_\phi$ is given by the formula:
$$\bbox[border:1px solid blue;padding: 1em;]{ \verb/Volume/(X_\phi) = \frac{nh\ell^2}{12}\left(2\cot\frac{\pi}{n} + \csc\frac{\pi}{n}\cos\phi\right)}$$

In particular, when $\phi = \pm \frac{\pi}{n}$, $X_\phi$ reduces to a $n$-gonal prism. Above expression reproduces the volume $\alpha_n h\ell^2$.

When $|\phi| \ge \frac{\pi(n-1)}{n}$, I haven't figure out what the twisted prism becomes. It is probably too complicated for me and I'll stop here.

Notes

• $\color{blue}{[1]}$ - let $c = \cos\frac{\pi}{2n}$ and $s = \sin\frac{\pi}{2n}$, the mess can be simplified to $$\frac{\frac{c^2-s^2}{2sc} + \frac{c}{s}}{c} = 2 \frac{(4c^2-1)s}{(2cs)^2} = 2\frac{\sin\frac{3\pi}{2n}}{\sin^2\frac{\pi}{n}}$$
• $\color{blue}{[2]}$ - In terms of $(\rho,\theta,z)$, Cartesian coordinates $(x,y,z) = ( \rho\cos\theta, \rho\sin\theta, z)$.