Motor power question

[I0sens]

I am trying get to my motor physics right. This might be correct or totally wrong:

Lets say I have a motor with a specification of 140 Kv, 1000W.
That means that a 24V battery will drive it theoretically to 24 x 140 = 3660 rpm. (no load)
And let's say that the max speed with that is 30mph.

Then, when going uphill the rpm drops to 800 rpm. The back EMV is then 800 / 140 = 5.7 V
The amps will go drastically up up course but the ESC limits them to 50A, lets say.

Here is the question: Does this mean the motor is now only producing 50 x 5.7 = 285 W?
Or is it the 24V Battery - 5.7V * 50 A = 910W pulling on the motor?

If this is a nominal 24V, 1000 W motor (= 41 Amps) , it will be overloaded already.

Is this how it works? Never let the motor go down in rpm (through load) by more than.. 20%, 30% , 50%?? (permanently)


So if I use a gear for that uphill example, that allows the motor to run at 2500RPM I could go uphill with 16.3mph
if 2500/140 = 18V * 41A = 738W are enough power for that?

Thank you for your consideration!


Oh, bonus question: Is there a difference in this behavior between brushed and brush-less?

 

[xfrankie]

If I remember corrrectly, the difference is

1) "effective output" power, which pulls you up a hill, calculated as bEMF * PhaseAmps
2) total input power, which comes out of your battery, calculated as BatteryVolts * BatteryAmps


For ease of demonstration, lets say the PhaseAmps are twice the BatteryAmps.

so when you go uphill, and ESC limits the current to 50 BatteryAmps, the power input from your battery into the ESC+motor is
24V (BattVolt) * 50A (BattAmps) = 1200 Watts

But the effective output is 5.7V (bEMF) * 100A (PhaseAmps) = 570 Watts of actual work

The rest of the total power (1200-570=630W) is only generating heat inside the motor.
In such case, the motor would be only 52.5% efficient, which is a very rough estimation, since I don't know the actual PhaseAmps.


That said, there must be something wrong with your data, because if 3660rpm = 30mph , then at the same gearing 800rpm = 6.5mph
Getting only 6.5mph with so much power... are you pulling 500 pounds up a vertical wall or what? :lol:

 

[I0sens]

Thank you xfrankie,

That gives me something to research (your "phase Amps").

I did not claim that my data was right, in fact I thought I only would get 285W which might explain the 8mpH.

I am sure there is also a formula that will give you the incline of the hill given the motor amps and weight of the system - after all it's all physics.

 

[I0sens]

Me again.

I have never heard of "PhaseAmps" and Google does not yield any good results either.
So - where are you getting these 100 amps from?
Can you point me into a direction where this is further explained?

So - my questions still stands unanswered.

 

[xfrankie]

TLDR: 20V 50A (DC) from your battery gets converted to 5V 200A (AC) to your motor -- or however much the ESC construction allows -- the AC current to the motor is called "PhaseAmps"

------------------------------------------

The ESC is limited by the mosfets and copper leads inside, they cannot output infinite amperes.
The motor is just a twisted up copper wire, with very low resistance (usually fractions of an Ohm).
If it were fed the full battery voltage at start, the Amps would go through the roof (quite literally for the mosfets).

The controller functions as a step-down inverter (DC to AC converter), with more-or-less constant power between input and output.
It takes the relatively high voltage from your battery and converts it into the relatively low voltage fed to the motor (just a tiny bit higher than the motor bEMF voltage). Which can still develop massive amperes, because of the low resistance.

From another point of view: If the power that comes out of your battery and the power that goes into the motor stay the same (ignore tiny losses in controller), AND the voltage to the motor is much lower than the battery, then the Amps to the motor must be much higher.
These are transferred via the three phase cables to the motor, so they are called Phase Amps.
These are what creates torque in the motor.

The more PhaseAmps, the more torque you get (up to the saturation point). Torque times speed = power out.
And the speed is directly proportional to bEMF. Thus bEMF(speed) times Phase amps(torque) = power out.
The applied voltage is a bit higher than the bEMF voltage -- it has to be, it has to create the amps through the motor -- and this is where the (in)efficiency comes from.

The motor never sees the full battery voltage, unless it is near the max speed. Then it generates enough bEMF to fully counter the incoming voltage from the ESC, resulting in less voltage difference (applied V minus bEMF V).
This leads to less and less amps, thus less nad less torque to overcome friction/losses, until it reaches an equilibrium (max speed). The ESC is a step-down, it can not raise the voltage any higher than that of the battery.

wait... what was the original question again? :lol:

 

[I0sens]

Thanks for the exhausting explanation

The original question, though was what the the optimal gear settings might be.

As you said, if the motor runs at max speed, it has no more torque left and if the gear is so that it runs at a fraction of it's
rpm, there will be low voltage and high amps.

BTW in my opinion you confirmed my conjecture that if the ESC keeps the voltage slightly over the bEMF then
the power of the motor will only be that low voltage time the Amps.

I am not sure how you jumped from 50 Battery Amps to 100 Phase Amps or is there a law that Phase Amps is 2x Battery amps?

There got to be a motor power calculator somewhere...

 

[I0sens]

After much research I can now answer my questions myself.
Useful documents: https://www.maxongroup.com/medias/sys_master/root/8815460712478/DC-EC-Key-Information-14-EN-42-50.pdf?attachment=true
and other Maxon sites and https://www.pcb-3d.com/tutorials/dc-motors-voltage-current-speed-power-losses-and-torque-relationships/

As mentioned above: Torque is proportional to current and speed to Voltage.

Power on the other hand is proportional to rpm x Torque.

That means the (output) power is zero when the motor is blocked while the torque and current is the highest, and Zero again when the speed is at full speed, no load, "no" current.
Power peeks at about 50% of nominal rpm and 50% maximal Torque.
Efficiency is another curve that peaks at about 1/7 of max rpm.

That's a lot of stuff, but can be deducted from the above documents with some math skills.

The bottom line for a bike though is that you need a motor with a lot of amps (50A at the minimum) and a voltage as high as possible 48V. 50 Amps will determine the max output at the motor axle and a high ratio gear that sets the motor RPM down and again helps to get the Torque up for some minimum climbing power.

That is all relatively easy to buy, but now(output) you need a battery that can deliver 50 - 100 Amps.... EXPENSIVE!

 

[E-HP]

That is all relatively easy to buy, but now(output) you need a battery that can deliver 50 - 100 Amps.... EXPENSIVE!

It doesn't have to be expensive, if you're willing to manage the risk. Put three of these in series for 18S, 20Ah and 240 amps, 480A peak:

https://hobbyking.com/en_us/turnigy-high-capacity-battery-20000mah-6s-12c-drone-lipo-pack-xt90.html?queryID=bf4213c8680220922f1c121ebbc105c7&objectID=78412&indexName=hbk_live_magento_en_us_products_hbk_price_stock_2_group_2_asc