# Clustering for mathematicians

(Or for people who like a little math, anyway. Impractical but, to some minds, fun math.)

You have twelve 13 mm motor mount tubes. You have a BT-70 airframe tube. Can you get all twelve motor tubes in?

Over on YORF this formula turned up:

$n = \lfloor\frac{\pi}{\arcsin (m/(a-m))}\rfloor$

($\lfloor x\rfloor$, “floor x“, means the largest integer not greater than x; for example, $\lfloor \pi\rfloor = 3$.) That’s a formula for the largest number of motor mount tubes of outside diameter m you can fit inside an airframe of inside diameter a, assuming you arrange them in a single ring around the inside wall. If you do that, and then draw line segments through the center of the airframe tangent to the motor tubes, and line segments connecting the center of the airframe to the centers of the motor tubes, and line segments from the centers of the motor tubes to the tangent points, you can probably persuade yourself the angle subtended by a motor mount tube from the center of the airframe is $2\arcsin (m/(a-m))$. Divide that into $2\pi$ (I’m using radians for the angular measures) and discard the remainder to get the largest number of motor tubes that will fit around the circle. There’s a caveat: if $m > a-m$ then the argument of the arcsin function is $>1$, which isn’t allowed (in real numbers). Then again, in that case m is larger than a/2 and obviously in that case you’ll get only one motor tube into the airframe… or none, if $m > a$!

Now, if you do this with 2, 3, 4, or 5 motor tubes, that’s that, but if you put 6 in:there’ll be room for a seventh too:And as the airframe tube gets larger and larger, not only can you put more and more motor tubes around the outside, but the space in the center gets larger, to the point where you can start to put multiple tubes there. For instance, here are 9 tubes around the outside with room for 2 in the middle:

How much room is there in the middle? Well, it’s obvious that there’s enough room that you could put a (fictitious) tube of diameter $a' = a-2m$ there; that would be a circle to which all the motor tubes are tangent on the outside:

But then into that fictitious tube you could pack up to $n' = \lfloor\frac{\pi}{\arcsin (m/(a'-m))}\rfloor = \lfloor\frac{\pi}{\arcsin (m/(a-3m))}\rfloor$ motor tubes, for a total of $n_{tot} = \lfloor\frac{\pi}{\arcsin (m/(a-m))}\rfloor + \lfloor\frac{\pi}{\arcsin (m/(a-3m))}\rfloor$.

And if that tube gets large enough, then in its center you could put another fictitious tube, concentric, with diameter $a-4m$, and pack tubes into that, and so on. Of course if the center tube turns out to be smaller than 2m, but larger than m, you can put one tube there.

Each of the rings of motor tubes has a width of 2m, so the number of rings you can fit in is $k = \lfloor a/2m\rfloor$. And if $a-2km > m$ you have room for one more tube in the center; putting it another way, you can add $\lfloor a/m-2k\rfloor$ tubes (which is either 0 or 1). Put all this together, and the total number of motor tubes you can fit is:

$\displaystyle\sum_{i=1}^k \lfloor\frac{\pi}{\arcsin (m/(a-(2i-1)m))}\rfloor + \lfloor a/m-2k\rfloor$

Well, that’s the number you can fit assuming you follow the above strategy, of fitting successively smaller rings of tubes inside fictitious tubes that fit inside the larger ring of tubes. Of course the fictitious tubes do not take up all available space, and you might be able to do better with an arrangement that interleaves tubes from a smaller ring between tubes from the larger ring. You might even do better with an asymmetric, irregular arrangement. Better from a mathematical point of view, I mean. Not so good from the point of view of keeping your thrust centered! But mathematicians don’t worry about that.

In fact, optimal packing of circles inside circles is in general an unsolved problem. Suppose you have 5 motor tubes; how small an airframe tube can you get away with and still be able to cluster them? It seems patently obvious to me that the optimum packing must be the symmetric pentagonal one, where $a/m$ works out to 2.701301… But what seems patently obvious is not always easy to prove. It appears this was not proved to be the optimum packing of 5 circles inside a circle until 1968 (by R. L. Graham). The case of 13 circles wasn’t proved until 2003 (by F. Fodor), and the case of 12 circles as far as I know hasn’t been proved yet! Here are pictures of the cases up through 20 circles with the best known packings, proved or otherwise.1

For 12 circles the smallest known value for $a/m$ is 4.029+ (found by Kravitz in 1967; this is a little better than the value of 4.154… obtained using the above formula). The outer diameter of BT-5 is 13.74 mm, so the smallest airframe inner diameter known to work for a 12-motor cluster would be 4.029… $\times$ 13.74 mm = 55.35 mm and change. BT-70 is 55.25 mm inner diameter. So, no — unless you squeeze or distort your tubes, or use some special thin wall motor tubing — it just barely won’t work. Or unless Kravitz overlooked a cleverer solution! Unlikely, but until someone proves it, we can’t say for sure a 12 x 13 mm BT-70 cluster is impossible.

1 Or, if you want to cluster a NewWay rocket, check out this page.