Originally Posted by
cokemaster
I start looking into this in more detail, but it raises more questions than answers. Just for clarification; the difference in this discussion is what your system can deliver without permanent damage (i.e. temperature or current getting to high so you get irreversible damage) and what is required to move a car from A to B in a certain time frame. For the same condition (weight, time) it will take the same energy regardless of what system you have (2S or 3S) to move the car from A to B. This energy is equivalent to what the “car” load your system with in wattage. So for the same load you require the same wattage (power) from your system. So derived from P = U x I (power = voltage x current). This is a simplification, but more about that later. Consequently, when voltage increases the current decreases and vice versa for constant power demand.
It is not a constant power situation with a given motor, the 13.5 originally asked about, as you acknowledge a couple times later in your post. But for discussion, achieving the equal power, and equal A-B times, with significantly different voltages likely requires a different motor to accomplish the current change.
The Ohms law is only true for ohmic system like resistor and wires which is not the case for a brushless motor. If you look up a motor equation you will see that you have to take into account reactive forces like EMF, etc. But if you could apply Ohms law your statement would be true. However, I have stated from the beginning that you cannot increase voltage above your rating without risking permanent damage. If your system is rated for 3S you can run 3S and lower, A 3S system does not run hotter than a 2S if you e.g. use a Traxxas VXL system. Yeah, it will if you do not change your gearing, but that it’s like running any wrong pinion in your 2S car.
The original question is about sizing the ESC, which needs to handle the peak amps involved. Peak amps occur at stall where the more complex aspects of a motor are not terribly significant, and it is mostly resistance. But even at higher RPM where back EMF comes into play, effectively creating complex impedance and reducing current, Ohm's Law still works, just requires complex math to work out and it varies with RPM. Trends remain the same.
Obviously there are other considerations and limits to the system as mentioned.
Obviously, by lowering your voltage you will have less available power,
YES!
and for the old system using NiCd or NiMh batteries with a typical cell voltage of 1.2V that would discharge to around 1.0V would give you a total loss of 1.2V for a six cell pack. That is equal voltage drop for a lipo battery (8.4V -7.2V), so why do we not see this “power loss” from start to end for a brushless system, like we did for the “old” system?
Much of it is due to the greater capacities. With lipo at least in racing cells are frequently not drained very far with a 13.5 in typical race heats. Wasn't always the case with lower capacity nickel cells combined with less efficient brushed motors which used more. A situation where say only 30% of the pack is used compared to 80 or 100% makes a big difference all by itself, regardless of cell chemistry.
To be fair fresh really good nimh cells can come close in discharge flatness, but in practice are very hard to keep that way.
It got me thinking. An ESC is a PMW device (Pulse Width Modulation) and it’s basically using its duty cycle to control the average voltage the motor sees. The longer the cycle, the higher the average voltage which again increase the RPM of your motor. What if the ESC modulates so you never get above 7.2V regardless you battery voltage (8,4V -7,2V). Thus the car will be 100% consistent during your race. Also, that make sense that Novak rate their motors with 7.2V and not 8.4V. But this is just my theory.
Pure hypothetical, that's not the way ESC's work. Although some would like that capability. The 7.2V used for the motor data is likely a remnant of nimh days.
I never stated that voltage did not impact power (well it was not my intention anyway
). I don’t know much about 4S versus 6S, but lower KV gives you higher rating (watts) so would it not be a normal thing to increase your voltage so you can decrease your current which give you less voltage drop and lower temperature (less current give you less temperature rise in a motor).
Point remains it takes a motor/load change to accomplish. Perhaps not in the way you seem to be stating, as all else equal a lower KV motor has a lower watt capability at a given voltage. Reference the Novak chart linked earlier.