I was rather hoping no-one would ask the 'how many turns' question . . . . There's a simple formula I am sure, but not known to me. Being curious, I'm looking - and if I find an answer I'll post it. It has to relate to the fact that after 2 turns of the crank and one of the camshaft, the idler teeth are 5 teeth out. But as 5 doesn't divide into any of the 22 44 or 49, it's probably a question of factorising or something I forgot before I 'learnt' it ha ha. The engineers and scientists on here will know . . .
However, the philosophy of the 'hunting tooth' - whereby an idler is nearly always a weird number of teeth - is standard practice. What we don't want is the same too/eeth on the idler always taking the load every time a valve opens. So, by giving that gear a prime number's worth of teeth (or at least, like 49, a number not divisible by any of the numbers that divide into the number of teeth of the driver and driven gear (in this case 2 and 11 and 22)), we get to a situation where the load is shared equally between all the teeth of the idler and it, theoretically, lasts a very long time. With consequent benefit to the other pinions in the train, which are not having to mesh with the same ever more worn teeth. Amazing the stuff that goes into making even the apparently simple bits of a machine, n'est-ce-pas?