# Length and diameter of a spiral of nanotube

I was reading from this popular article (in french). Talking about nanotubes of carbon the author says (my translation):

The diameter of the nanotube is of the order of a millionth of a millimiter. This value is difficult to conceive for the human mind. Imagine a nanotube long like the earth-moon distance, roll this nanotube on itself and the nanotube would be the volume of an orange seed.

Is this true?

• if you mean to wind the nanotube onto a spiral disc, then practically the area of the disc will be the same as the total cross-setional area of the tube = 3,84 cm^2, i.e. r = ca. 1 cm Jun 14, 2016 at 16:38
• Sure. But I don't know the radius from the size and length of the nanotube. Is my question unclear? Jun 14, 2016 at 17:12

The length and width of the nanotube are $L=3.844\times10^{8}m$ and $W=10^{-9}m$ for a cross-sectional area of $\pi R^2=3.844\times10^{-1}m^2$ for the area of the circle when the nanotube is rolled up.

Thus $R^2=1.244\times10^{-1}m^2$ giving $R=3.53\times10^{-1}m$ or $35$cm, and that's just the radius. The diameter would be $71$cm.

So assuming "orange grain" means "orange seed" this is off from the article by a couple of orders of magnitude. Have I made an error?

Perhaps they mean when the nanotube is rolled up into a ball?

Then we would have

\begin{equation} \tfrac{4}{3}\pi R^3=\pi r^2L \end{equation}

or

\begin{equation} R^3=\tfrac{3}{4}r^2L=0.75(5\times10^{-10})^2(3.844\times10^8)=7.21\times10^{-11} \end{equation}

giving $R=4.16\times10^{-4}m=0.416\,mm$ for a sphere the diameter of $0.83\,mm$.

• Yes, the original article definitely says the volume of an orange seed, so it is not wound into a spiral but wadded up into a ball. But it is more like the volume of a mustard seed. Jun 14, 2016 at 17:35
• That was a poor translation. Orange "seed" and not "grain" indeed. I removed the reference to the spiral too (sorry, my bad) to not make any interpretation over their claim. Jun 14, 2016 at 17:55