Okay, this is probably a theoretical question that has little use, but it has my curiosity about what I "think" I know.

My naive understanding is that when a motor manufacturer lists the specs for a motor and provides a "kV" value, that tells me what the motor RPM performance should theoretically be for a "no load" configuration.  So, for a motor with a kV of 1000, if I were to hook up a 3S LiPo and let it run at full voltage, then it would be spinning at 11,100 +/- RPM.  Since we put a prop on the motor and load it down, as the prop gets bigger and has more pitch increasing the load, keeping the voltage at a constant level, the current draw goes up and max RPM is going to go down up until the motor burns out.

Question 1: Is this understanding close to correct?
Question 2: Are there any "rules of thumb" or experimental understanding about the amount of decrease in measured RPM as the load increases? (Okay, I recognize that this is probably highly dependent on specific motors.)

I've come to these questions from my perspective that we are dealing with an optimization (partial) problem for several different parameters.  Some of these are independent, and some are functions of combinations of other parameters.  Having some insight into practical, available RPM from a motor for a given cell configuration might be useful.

TIA,
Jim Howell
Huntsville, AL

Yes, it is very easy ... motor has internal resistance so for example if the value is 100mohm, loaded motor needs draw 10A, the effective voltage is less of that voltage on the resistance so means 10A * 100mohm = 1V ... so if the Kv was 1000 and the voltage was 10V and no load rpm was 10 000 the loaded motor will turn 9000 rpm. 

Means that the higher is internal resistance the "softer" motor is. Motor is defined not only by Kv, but also by Internal resistance and no load current. If you know those 3 parameters, you can predict what the motor does, how quick it spins, what are loses and also efficiency.

Andy, it is more about size, not quality .. if you load small motor with Ri=200mohm to 10A then rpm drop will be large (before it burns  ), while large motor with only 10mohm will be much stronger :-))

the formula is here:
actual rpm = Kv*(U-Ri*I)

while quality of the motor is defined as maximal possible efficiency:
max eff = (1-SQRT(I0*Ri/U))^2
you can see that smaller no load curren (I0) and smaller resistance gives better result

Yes, it is very easy ... motor has internal resistance so for example if the value is 100mohm, loaded motor needs draw 10A, the effective voltage is less of that voltage on the resistance so means 10A * 100mohm = 1V ... so if the Kv was 1000 and the voltage was 10V and no load rpm was 10 000 the loaded motor will turn 9000 rpm. 

Means that the higher is internal resistance the "softer" motor is. Motor is defined not only by Kv, but also by Internal resistance and no load current. If you know those 3 parameters, you can predict what the motor does, how quick it spins, what are loses and also efficiency.

Igor,

Thank you!  I have observed the other parameters, but had no clue as to how to use the information.  I'm thinking that what you've given me is a good first start for improving my "first choice" motor selection.

Thanks again, I always feel like I learn from your posts, this one especially.

Jim Howell

Igor,

Once again, I am indebted to you.  Your explanations about internal resistance and no load current loss and the relationships for each are quite clear.  I will save this away for future reference.  Thank you.

Jim Howell