Here is something interesting I saw on the bench today:
I measured this with an external shunt type meter at the battery. Same ESC
95% duty cycle, no load current on a 63100 190kv motor from Torque Boards.
12s 190kv - 3.52A @ 46.8v = 164w
18s 190kv - 4.67A @ 70.2v = 327w
So it takes 163w in waste heat just for the privilege to run 190kv on 18s at full RPMs due to iron losses.
What about copper losses? High Kv motors have lower resistance so have less copper losses at the same current. If we assume that iron losses match RPMs when the motor is the same size, and we choose a motor that reaches the same RPMs at 18s as a 12s 190kv motor, then we end up at 130kv.
Using the resistance values from the Lacroix motors, we can calculate copper losses at 100A
So if two dual motor boards were geared for 40mph top speed, and both drawing 100A at full speed, the 18s 130kv setup would be putting down ((455-343)x2=) 224 additional watts to the wheels compared to an 18s 190kv setup. The 18s 190kv setup would be pumping 224 more watts into heating the motors vs the 18s 130kv.
I donāt have a dyno to test all of this in the real world, but I think itās why companies who do have some sort of dyno choose 130kv for 18s and 190kv for 12s. Longer range, less motor heat and more power to the wheelsā¦ however this is all at full power and full RPMs. I wonder if running 18s 190kv at 50% duty and 20A would be more efficient than 18s 130kv at 50% duty and 20A.
Need someone smarter than me to double check all this lol @Dinnye @Pedrodemio
EDIT: Waitā¦ if you used the āExperimentsā feature in the VESC tool, and checked iron losses at say 5% duty cycle steps for various motors, factored in the winding resistance, then figured out how many mechanical watts are needed to overcome the rolling resistance, wind resistance and hill resistance, factoring in gearing for torque, you would basically have the esk8 equivalent of the Grin motor simulator.
Is it really that easy? Is a mechanical dyno not needed? Where does magnetic saturation come into play?