I am taking my lonely lunch time at school to write this so take it seriously.
I’ve labeled the givens and conclusions by number so we can accurately dispute the problem. If you have complaints about how I skate, put them here.
Given:
(1) P=F*V=T(torque)/t(seconds)
(2) The faster you go, the more more friction/frictional torque
(3) The power into the motor will always increase the faster you go.
Conclusion:
(4) A motor set to a high gear ratio (20/36) will max out its power input somewhere if it doesn’t reach max speed.
Thanks for reading, have a great day.
Here’s my after comment
My friend told me that not only is there a limit to the power, but the torque is also diminishing as rpm increases, therefore, bottle-necking power output.
Hope this helps anyone in understanding power and torque limitation of DC motors. And of course…
Thanks for reading and have a great day.
It was in the edit. Can you not see it?
I was confused as to why you can’t max out power with a high gear ratio, then someone told me that the motor will lose torque as RPM increase bottlenecking the power output at high speeds. Was that stupid?
I thought so since 45mph requires a lot more torque than starting up. Is there like a graph or datasheet demonstrating the torque/power over RPM/volts, I just want to be able to use as much power as possible. Is 16/36 the way to go?
The language is a bit cryptic and Torque / s is not a physically meaningful quantity. But if I understand correctly what you wanted to say, then you are right: The faster you go, the more power and torque you need (duh). You can use the usual calculators to estimate if what motor specs you need for your case:
http://vesc-project.com/calculators
But I don’t really understand how you connect it to gear reduction. The way it actually connects: if you are not limited by specs of your components, the gear ratio doesn’t matter for your example (speaking from physics point of view) because the motor will get you the same acceleration at a higher torque. The output power is the same, though, because the motors rpm is lower:
P = F * v = T_wheel * w_wheel (T is toque, w is angular velocity)
T_wheel = 1/R * T_motor (R is gear reduction)
w_wheel = R * w_motor
so:
P = F * v
= T_wheel * w_wheel
= T_motor / R * w_motor * R
= T_motor * w_motor
As you see the power is constant and the gear reduction cancels out. But in reality that’s not what matters, because what you didn’t include in your calculation is that the motor efficiency is lower at low rpm and low toque load. That’s why a high gear reduction is more efficient, and means that the INPUT power goes up. But it has nothing to do with friction. Just motor efficiency. So get the highest possible gear reduction / lowest KV that can give you your desired max speed. Then you are most efficient.
No, the opposite. Hub motors are usually inefficient because their KV is too high. They are usually “geared”, to run at 60+ km/h, because you can only get so much copper into a motor of a certain size, which provides a lower limit for the KV. This is the reason why you want to have a gear reduction for a BLDC motor in the first place.
If you run the hub motors at sufficiently low torque load (large motors, many motors, not too many hills or aggressive acceleration, …) the overall efficiency of a hub setup can even be higher, because they make up for lower motor efficiency with their lower rolling friction (no belts).
What! I get 8w/mi efficiency when I go 5mph and only 20w/mi when 25+mph. I don’t know about the motor your using, but I too thought the same, lower the speed, the less efficient. My experience proves that the slower you go, the more efficient. Did you ever test this yourself?
And for most hub motors, I think they are geared at only 48kmph. 60~70kv. kv * 60 * voltage * pi * wheel dia./1000 ← isn’t that how you calculate kmph?
In this case you are comparing different motors. If the amount of cooper on a motor is the same, the ratings are the same, be it torque, power, loses, efficiency ,etc, but they happen at different voltage/current for a given Kv
The case for why hub motors generally are less efficient is because the Kv is not optimal for a given voltage, if you get a hub motor of 300 Kv vs the same motor wound at 300 Kv but with a 3:1 gearing, the hub motor will need more current to achieve the same torque since Kt = 60/(2piKv)
That leads us to the losses, power is P = U I^2, but since we need 3 times the current we have 9 times more cooper loses
But if we match a hub motor Kv to the voltage and desired top speed, we have no reason to say it is less efficient, in fact it can be more since the only mechanical losses is a pair of bearings
This is not what I or @ZackoryCramer were talking about. This if about the power/torque needed at a certain, fixed speed. Meaning doing the exact same riding with a different setup. About the motor efficency.
The overall energy/distance that you are talking about should be independent of speed for low speeds and quadratic with speed at high speeds when the wind resistance kicks in.
