You can run your own calculations using "the other" capacitor equation. With this, if you know the motors rated voltage and current, and the desired run time, you can calculate the capacitor size needed. Note that this is an approximation, good in this application to approx. +/-5%. The tradeoff is that it is multiplication and division, not exponentials and natural logarithms. The equation is based on the capacitor being discharged by a constant-current load, a load whose current draw does not change even though the capacitor voltage is steadily decreasing. Since a real fan's current does change with the applied voltage, we approximate the average current over the run time.
ec=it
Voltage drop times capacitor value equals average current times the run time
Example:
Fan - 5 V nominal, stops running at 3 V -- e = 2 V
Current at 5 V is 50 mA, current at 3 V is 40 mA -- average current i = .045 A
6 hour run time -- t = 21,600 seconds
Rearranging:
c = (i x t) / e = (.045 x 21,600) / 2 = 486 F
So, you would need about 500 farads (not microfarads) of capacitance charged up to 5 V to run the fan for 6 hours.
ak