[edit: fix typos]
You do not have to set maxNumM=250 just because your sampling rate is 250 Hz rather than 100 Hz. You can leave
maxNumM=100;
It is just a coincidence that maxNumM=Fs in the example.
maxNumM is how many different averaging times there will be in the analysis. The actual numer of different averaging times will be a bit smaller than maxNumM, because the averaging time must be an integer number of samples, and with logartihmically-distributed averaging times, there would be some duplicates, so after you remove the duplicates, there are fewer than maxNumM remaining. In the exaple case, the original gyro signal has 2.16e6 samples. The max averaging time should be the largest power of 2 that is less than or equal to half the number of samples. In the example, that means maxM=2^20=1048576. By choosing maxNumM=100, it means there wil be 100 different durations used to measure variance. The durations (vector m) are equally spaced on a log scale from 1 sample to maxM samples. This means m=[1, 1.15, 1.32, 1.52, ..., 792499.6, 911561.1, 1048576], in the example case. Then each duration is rounded up: m=[1, 2, 2, 2, ..., 792500, 911562, 1048576]. Vector m still has maxNumM=100 elements at this point. The duplicates are removed: m=[1, 2, 3, ..., 792500, 911562, 1048576]. Now vector m has 93 elements, and it stays that way for the rest of the Allan variance analysis.
