ok, so I had to go look at this for my own interest again
so a float (non audio) is represented as 11 bits (including sign, and usually we only show the bottom 6 bits, = 64, so allowing some headroom for calculations)
so 2^11 = 2048, so that should be the cycle length, which has to be completed at 3000hz
BUT , and this I dont understand and would like @johannes to clarify
for some reason, the inlet is divided by 4 , (inlet_freq >> 2)
we actually need 4 times the above in one cycle , or 2^13 = 8192
8192 / 3000 ( hz. k-rate) = 2.730666666666667 , hence the number above.
so my question for @johannes, is why divide by 4, is this just to bring into a more 'reasonable range' ?
(it confused me )