Yes, inefficiency (waste heat) is current squared times winding resistance. But resistance goes up exponentially in the windings of the lower kv motor, which effectively cancels out the impact of the change in current. From your own link (Mr Vedder):
Let’s look at an example: Suppose an 8 turn motor has one ohm winding resistance. The winding resistance is proportional to the wire area times wire length. Making the same motor with 4 turns would allow twice as thick wire. Since that wire also is half as long, the resistance is four times lower: 0.25ohm. Further, since current times turns is proportional to torque, we need twice as high current with 4 turns to produce the same torque as with 8 turns. The copper losses are the voltage across the windings times the current: UI. The voltage across them is RI, so the losses are RII. This square relation means that doubling the current will produce four times as much losses, however, since we got four times lower resistance the losses are the same for the 4t motor with double the current as for the 8t motor with half the current. Also note that the 4t motor will spin twice as fast as the 8t motor at the same voltage, so it will have double the kv. Putting this together, the 4t motor is equivalent to the 8t motor, but at half the voltage and double the current.
If the motor itself is not a factor in the power efficiency, since the attributes cancel each other out (motor kV), then there are still other parts that don’t change to counter the heat losses, the ESC mosfets and wiring.
This is also from Vedder’s post.
From the ESC perspective, we should run on as high voltage and low current as possible. Now, one interesting fact about MOSFETs is that the lower voltage they are designed for, the lower resistance they tend to have. So if I make an ESC for a lower voltage, the FETs will also have lower resistance. However, the PCB traces always have the same resistance and MOSFET resistance does not seem to decrease as fast as their voltage decreases, meaning that an ESC designed for higher voltage tends to be more efficient in general.
I wish I had actual practical test data to test this argument… Now that I’ve thought about it more, it seems to be as Vedder said. I will continue pondering and researching this topic …
@SimosMCmuffin, every motor has a torque constant, Kt. Kt= The amount of torque produced per amp. So in a motor torque is proportional to current.
Kt = 1 / Kv these two are linked. This means that a lower Kv motor will produce more torque per amp or that a higher Kv motor needs more amps to produce the same torque.
Ok, I might have been a bit confused earlier with the statement and I admit I might have had the wrong understanding of the mechanics.
But now I’m really wondering how does efficiency come into play in real life situation? Shouldn’t the efficiency and range be the same if you have two same energy capacity batteries?
One with lower voltage, higher capacity (4S 10000 mAh) and the other with higher voltage, lower capacity (8S 5000 mAh).
Two otherwise identical motors, except for the kV, (150 and 300 for example) would be matched to get the same no-load top speed with the proper battery.
I forget who it was but someone here did do a test which was similar in which they simply cut their battery voltage from 12s to 6s and did the same steep route at about 15mph and compared the watt hours spent. It was close but the 12s system was a bit less efficient. I’d understood that the 6s run would be more efficient as the speed of the climb would be closer to the no load speed at 6s max throttle than 12s max throttle. I was thinking of the no-load speed being a fixed number solely based on battery and kv, but recently discovered that the no-load is constantly changing as the throttle setting changes. Ex: half throttle is an effectively half the max voltage of he battery and so the no-load speed is now half as high! This was a surprise to me so I don’t know why either would be more efficient than the other
I now understand the difference between the no-load speed and the speed the motor is running in relation to that is purely determined by the load and size of the motor
But they didn’t change the motor? So he ran the route at half the battery voltage at half the speed basically?
What I’m wondering is what if they change both battery voltage AND motor kV so that the no-load max speed stays the same. That way the max speed should still be same.
Of course running slower is more efficient on just pure aerodynamics alone, also the mosfets and wiring will have more I^2*R heat losses with higher amps so that also decreases efficiency, but he did use the SAME motor.
Same speed up hill (he thinks). Same motor and kv. And the batty variation only.
The mosfets and the current they experience I think is more related to the motor side and would need to apply the same current to get the same torque with the same kv
I switched from 6s to 10s and I noticed that at 6s, with the same settings (with motor max A maxed out because I didn’t care about battery life) changing different lipos (6S vs 10S and then 8S), the 6S climbed more than the 10S. I generally had hotter ESC but colder motor at 6S because I didn’t make the motor stress under load when doing hills (because I had more torque?)… I decided then to go with a 8s (because I’m using 97mm wheels that’s why it can’t go uphill with me 200lb+/~95kg with a single 637x motor with a 10S battery) and I still have a good max speed
Am I right or wrong to think that you should aim to stay near your desired top speed and try to use the highest possible voltage (for the VESC to be efficinent) but that doesn’t shift your top speed so that you have a lot that you wouldn’t use? Otherwise you lose acceleration and efficiency under load? …is this right?
(at 6S has a top speed that is too low for me that’s why I went with 8S)
basically if I went 10S I had ~11km/h - 6mph of speed more that I wouldn’t use, and, even if the ESC would have ran cooler, I would have had less torque, the other way to go at 10S would be to increase my gear ratio like 14T/40T but I’ll need a belt idler for that, do you think it’s worth it or should I keep running at 8S ? I guess the only thing is to try 10S or 12S on another board before rearranging these cells
sorry for hijaking the thread a bit my question is really Does Lower V (battery S, number of cell in series) mean more torque? (if you can still hit your desired max speed) Yes?
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