Brushless experts please read

[Scoob]

I am looking to learn exactly what makes different turn (KV) motors run the way they do. I usually come here to learn the technical stuff because people here seem to know what they are talking about. I think I have a pretty good handle on this stuff but I'm not sure.
Just let me know if something is incorrect and please explain.

1. RPM times torque = mechanical power.
2. Voltage times amp draw = the total wattage pulled by the motor.
3.Total wattage output by the motor will depend on the efficiency as to how much of the power (wattage) pulled, is put out in mechanical power. Example, A motor pulling 400 W @ 80% efficiency will output 320 W.
4. Given two motors that are the same but with diferent turn and KV, like MM4600 vs. MM7700,the low turn high KV motor should always be capable of more power on low voltage such as 2s lipo. The low KV motor even if geared really tall is limited by higher internal ressistance. You can't force it to make up for the RPM difference with torque through tall gearing, due to the resistance (any other reasons ?)
5. More torque requires more current, so the overgeared low KV motor above would only produce as much torque as it's resistance would allow when over geared. The rest of the current forced on the motor by the load of overgearing will turn into heat so the motor has become very inefficient. In other words, it may pull the same current and therefore the same wattage at that voltage as the higher KV motor but will output a lower percentage of that power.
6. Let's say we have two motors that are geared the same and given a voltage that gives them the same RPM such as a 4600 on 11.1V and a 6900 on 7.4v. The 4600 should run more efficiently as a system because there will be less power lost to heat from the resistance of the system. Starting with the IR of the battery, the resistance of the wiring from batt to ESC, the resistance of the ESC itself, and wiring running from ESC to motor, less current flowing through them means less loss in the high V system.

Thanks for any input.

[kufman]

Statement #6

Losses are the current squared times the resistance. So... reducing the current draw by a factor of two lets say, reduces resistive losses by a factor of 4! So what you have said in #6 is true, the higher voltage system will run more efficiently.

Statement #5

Torque is proportional to the amp-turns of the motor. so what you have said in #5 isn't always true. a 10 turn motor running 5A will have similiar torque to a 5 turn motor running 10A.

#4 sounds correct.

#3 I think is correct.

#2 Is correct.

#1 I believe to be correct.

[Serum]

1 yes, basicly it is. but this number gives you the amount of mass (torque) that can be moved in a certain time (rpm)
2 yeah, dead on, Ohms law; P(power)=U (voltage) x I (current) Current and voltage will vary; the voltage will drop under a higher load.
3 Yes, output power of the motor; the entire drivetrain will eat his share of power too. like 400watts on the motor, and 300 watts at the wheels (gears, bearings , time/energy it takes to get/keep parts moving)
4 in large lines; the lower the KV of the motor (from the same series) if they are wind at the same method (delta or wye) the lower KV will have more torque and can be geared taller. the 7700 can be geared lower with about the same poweroutput at the wheels.
5 this is something weird; A lower KV is usually ran at a higher voltage; if a high KV motor is used at 7V and should deliver 700watts, the current that's needed will be 100A, if the low KV motor runs at 10V, the 700 watts take 70A. (P=UxI)
6 Yes, The higher the voltage, the less the internal resistance becomes a factor. Thats why you can use aluminum as a conductor for high voltage long distance energy transport.

[Scoob]

Originally Posted by Serum

4 in large lines; the lower the KV of the motor (from the same series) if they are wind at the same method (delta or wye) the lower KV will have more torque and can be geared taller. the 7700 can be geared lower with about the same poweroutput at the wheels.

So would you say as long as the gear ratio can be achieved for a low KV motor to match the high KV, there should be no need for the high KV motor even on a low voltage? The reason for the constant voltage issue is that I race 1/10th and voltage cannot be upped. Does a higher turn motor have higher resistance or am I off base there?

This confuses me only because I have tried to match MM5700s with my MM4600 on 2s lipo in my T4 by gearing up as I wasn't allowed more voltage. I could get the same top speed but the 4600 got hot quicker than my friends 5700s and seemed more sluggish. We were all running TP8000 2s lipos. This is why I assumed it was having trouble accepting the current the extra gearing was asking for due to resistance. I will say though that it was a one time experiment and I bought a 5700 shortly after for that large track. There may have been something else going on with my setup.

Thanks a lot for the info.:dft009:

[kufman]

To get the same performance out of a lower Kv motor you need to have higher voltage. You are correct when you say that the higher turn motor has more resistance. So.... at a given voltage you can't get as much current to flow through a higher turn motor. V/R=I

There does seem to be a non-scientific "sweet spot" for a given motor geometry. For the MM maybe that spot is the winds used in the 5700. I know in the Lehner basic series, the sweet spot seemed to be with the 4200 version of the motor.

[Scoob]

Alright that makes sense Kufman. I think I may have misread what Serum was saying anyway.

Thanks.

[SpEEdyBL]

To get the same power from 4600 (with the same voltage), it needs to draw at least some amount more current than the 5700. You can do that just by gearing higher, as I see you did, but it will heat up more (due to higher resistance) and it will take longer to get to top speed. Basically, you may never be able to get the amperage required to match the 5700 taking off from a standstill, but you will be able to maintain a higher current throughout the powerband. Although, depending on the motor, this could be the better way to get power, because resistance is not the only contributer to power loss. There is also iron loss which is the voltage times amps the motor pulls at no load. So a motor running at 7 volts that draws 5 amps will always waste at least 35 watts of energy no matter what. Lower kv motors have lower iron losses which makes them more efficient, not because they draw less amps. Even though heat = I^2*r, a 2000kv motor has four times the resistance as a 4000kv motor of the same model, so drawing half the current doesn't really do much in that perspective. Now you can see why the feigaos seem to heat up at all gearing levels and why lower kv motors are needed to achieve more power. Of course, iron loss varies from motor to motor, so a 4000kv neu motor could have less iron loss than a 2000kv feigao motor at equal voltage on each.