A do-it-yourself experiment:

Question: How many action potentials (spikes) fit in a 'j'?

Step 1. Type the 'j' key (any key will do, really, just as long it produces a character. Shift, for example, does not), with one finger, in the editor of your choice, in intervals of t seconds, in this case t=4s.

Do it as fast as you can. (It is in principle not necessary. But in that case, in the end of the calculation you find the minimum number of spikes per jay. In math, it's cool to find the minimum).

jjjjjjjjjjjjjjjjjjjjjjjj
jjjjjjjjjjjjjjjjjjjjjjjjjjjj
jjjjjjjjjjjjjjjjjjjjjjjjj
jjjjjjjjjjjjjjjjjjjjjjj
jjjjjjjjjjjjjjjjjjjjjj
jjjjjjjjjjjjjjjjjjj

Step 2. Count the numbers of 'j''s (or your preferred character) per line

jjjjjjjjjjjjjjjjjjjjjjjj 24
jjjjjjjjjjjjjjjjjjjjjjjjjjjj 28
jjjjjjjjjjjjjjjjjjjjjjjjj 25
jjjjjjjjjjjjjjjjjjjjjjj 23
jjjjjjjjjjjjjjjjjjjjjj 22
jjjjjjjjjjjjjjjjjjj 19

Step 3. Calculate the geometric mean of j's per trial (see comments why), divide that by trial duration, convert to ms/j (1000/number). Here's the empirical formula:


where:
- nt is number of trials (number of lines)
- t is the duration of the trial
- ji is the number of j's in a trial

My Result:
170,2 ms per jay.

So, if the brain were sequential, 170 action potentials would fit in a jay.

Of course, the brain is parallel, so at any given milisecond, a number of neurons will be active. Say that the mean number of neurons activated per milisecond is between 10 000 and 100 000.