TECHNICAL ASIDE
when the pump runs at idle, it pulls .17A. When the pump output is blocked (or under load) it runs at .21A. So, If I were to do the math, it is a 3V air pump that uses .21A max with 14.28Ohms resistance. Correct?
No. You have a pump which draws typically 0.17A to 0.21A at 3V when new. A motor is not a resistor and simple Ohm's Law doesn't apply.
But if you take account of Ve, the "back e.m.f", the voltage generated by the motor when running, (because motors are also generators), then Ohm's Law makes more sense. The motor winding and brush resistance combination is constant, (ignoring changes due to temperature).
Ve has a linear relationship to speed, (sometimes expressed with a constant Kv in volts per RPM).
Motor torque has a linear relationship to current I, (sometimes expressed with a constant Kt in Nm per Amp)..
With open pump ports, you power the motor. Initially (briefly) the motor will take its stall current (Istall = V / Rm) and develop a high torque to turn the shaft. As the motor accelerates it will generate a back e.m.f, which reduces the voltage across the resistive portion (Rm) of the load. As this voltage difference decreases, the current decreases, (Irun = (V-Ve)/Rm), and the torque decreases. As the torque decreases the acceleration decreases until the motor runs at the point where torque produced matches the mechanical load torque from the pump mechanics.
If you restrict the pump flow, the differential pressure increases. This is mechanically translated to an increased load torque on the motor and the motor slows. As it slows, it develops a lower Ve, so the current rises. The motor reaches a new equilibrium at a slower speed, a lower back e.m.f. voltage, a higher current and a higher torque balancing the increased pressure.
To find the absolute maximum current, Istall, then you'll need to lock the motor shaft (turning it into a resistor), and measure current quickly. Only power the motor briefly, a second or two, to do this. Then you can use Ohm's Law to find motor resistance Rm = V / Istall.
Do I do these calculations with the maximum current? or an average in between?
It depends on what you are doing the calculations for. If it's for component sizing, then a sensible approach would be to take the higher current and add a 50% margin to allow for changes as the pump ages and your tubes fur up!