On Sat, Aug 15, 2015 at 10:27:33PM +0100, Will Godfrey wrote: > As a matter of curiosity I wonder if we can work out what the actual difference > is. Would the Xrun occur when there was a one frame difference, one period, or > 1 buffer? It's running at 256 frames per period and 2 periods per buffer. The > xruns occur every 11 minutes (and a few seconds). If you assume that an xrun occurs when the difference in samples is one period, then the difference in sample frequency is P / T, with P the period and T the time in seconds between xruns. With the values you give: dF = 256 / (11 * 60) =~ 0.39 Hz. You can measure this easily using jack_delay (or jack_iodelay). The change in measured latency, in samples, per second is the difference in sample rate. Ciao, -- FA A world of exhaustive, reliable metadata would be an utopia. It's also a pipe-dream, founded on self-delusion, nerd hubris and hysterically inflated market opportunities. (Cory Doctorow) _______________________________________________ Linux-audio-user mailing list Linux-audio-user@xxxxxxxxxxxxxxxxxxxx http://lists.linuxaudio.org/listinfo/linux-audio-user
