I'm looking into nanotubes and I thought I'd assure myself that the basic geometry equations are indeed correct. No problems for the radius, I quickly found the known formula:
$$R = \sqrt{3(n^2+m^2+nm)}\frac{d_{CC}}{2\pi}$$
if $d_{CC}$ is the carbon bond length.
For the chiral angle, however, the equation should be
$$\tan{\theta} = \frac{m\sqrt{3}}{2n+m}$$
but I got
$$\tan{\theta} = \frac{m\sqrt{3}}{2n-m},$$
a sign difference. I thought I'd look for the mistake and quickly correct it, but I cannot for the life of me find it. I'll show my derivation below. The starting point is of course the nanotube structure:
I'll simplify this picture to make it more clear what I did:
From the above picture, it's easy to see that
$$\sin{\theta} = \frac{h}{na_1}$$
and
$$\sin{\left(120^{\circ} – \theta\right)} = \frac{h}{ma_2}.$$
This can be rewritten using the trigonometric identity
$$\sin{\left(\alpha-\beta\right)} = \sin{\alpha}\cos{\beta} – \cos{\alpha}\sin{\beta}$$
where $\alpha = 120^{\circ}$ and $\beta = \theta$, giving rise to
$$\frac{\sqrt{3}}{2}\cos{\theta} + \frac{1}{2}\sin{\theta} = \frac{h}{ma_2}.$$
Now, using the fact that $a_1 = a_2 (=\sqrt{3}d_{CC})$, we find the following equations:
$$\left\{\begin{array}{rcl}n\sin{\theta} & = & \frac{h}{a_1}\\ m\left[\frac{\sqrt{3}}{2}\cos{\theta} + \frac{1}{2}\sin{\theta}\right] & = & \frac{h}{a_1}\end{array}\right.$$
which obviously allow for combination:
$$\begin{eqnarray}
\frac{n}{m}\sin{\theta} & = & \frac{\sqrt{3}}{2}\cos{\theta} + \frac{1}{2}\sin{\theta} \\
\left(\frac{n}{m}-\frac{1}{2}\right)\sin{\theta} & = & \frac{\sqrt{3}}{2}\cos{\theta} \\
\frac{2n-m}{2m}\sin{\theta} & = & \frac{\sqrt{3}}{2}\cos{\theta} \\
\tan{\theta} & = & \frac{m\sqrt{3}}{2n-m}.
\end{eqnarray}$$
This differs from the known expression in that it has a minus sign in front of $m$ in the denominator, but I fail to see my mistake. It could be really silly and I'm just being blind… Thanks for taking a look at it.
Best Answer
The angle at $C$ is not $120^\circ-\theta$, it's $180^\circ-(120^\circ+\theta)=60^\circ-\theta$. The rest of the algebra is right, and you can quickly see where the sign comes from.