# Motor and required torque.

1. Oct 12, 2012

### qwertyuiop23

So I need a motor that is able to provide $$$0.412Nm$$$ of continuous torque (to overcome rolling resistance).

A BLDC I have is rated at $$$285W$$$ and $4700Kv (rpm/V)$.

So I was wondering if my calculations are correct.

Using the relationship between Torque and Power and angular speed
$\begin{split} \omega &\approx 700rads^{-1} \approx 6700rpm\\ T &= 0.412Nm \\ P &= 285W \end{split}$

From this get the voltage from the motor specs
$\begin{split} V &= \frac{6700}{4700}\\ &= 1.42V \end{split}$

So now using the fact that
$\begin{split} K_t &= \frac{1}{K_v}\\ T &= K_t i \end{split}$

Therefore

$\begin{split} i &= K_v \times T\\ &= 4700 \times 0.412\\ &= 1936A \end{split}$

This result makes no sense because that is a tonne of current.

Assuming 100% efficiency (P = IV)
$\begin{split} 285 &= i \times V\\ i &= 196A \end{split}$

Which again seems really really high.

Am I missing something somewhere? Will the motor not draw that amount of current at the given loads/speed? I am confused as I think the equations are correct....?

2. Oct 12, 2012

### Staff: Mentor

Is 6700 rpm what you need? What is the rated voltage, rpm and torque of the motor? I think you just proved that the motor is either too small to meet your needs or needs to be geared down.

And it won't produce 285W at any voltage, only at its rated rated voltage. The limiting factor is actually the Amps.

3. Oct 13, 2012

### qwertyuiop23

6700rpm is the needed speed yes (to turn a wheel of 120mm fast enough to give the required forward speed)

I cannot find a rated voltage/rpm/torque of the motor as it is a brushless DC motor. ie I get given a Kv value and a max power.

Say for arguments sake I have a 'magic' power supply that will give any amount of current/voltage I require.

Or i what you are saying that because of the 1936A I calculated the Kv value is too large (hence Kt too small) and thus the motor will not meet the specifications without a gear reduction?

EDIT:My bad I meant Kv too small giving a Kt that is too large. Reducing current = decrease Kt

EDIT2: Just to make it a bit clearer the absolute necessary parameter is the 0.413N. The speed was calculated using the maximum power and this torque so the 6700 can change.

Last edited: Oct 13, 2012
4. Oct 14, 2012

### CWatters

Power = Torque (Nm) x angular velocity (rads/s)

So you need a motor capable of delivering..

RPM = 6700 = 700 rads/s
Torque = 0.412 Nm

so

Power = 0.412 * 700 = 288W

eg I agree with you so far.

Then I would look at data for the motor efficiency. If the motor is say 80% efficient the power drawn will be..

288 x 100/80 = 360W

Will the controller handle that ok?

If the motor you have in mind has a Kv of 4700 then the voltage does indeed work out at

6700/4700 = 1.42V

The current draw would be roughly

Power in/voltage = 288/1.42 = 202A

Now it's unlikely you will find a motor that is 80% efficient at 1.4V and 200A so it sounds like the motor constant is too high for your application. Consider a reduction gearbox to allow you to operate the motor at higher voltages, higher rpm, lower current yet still achieve the required rpm at the output of the gearbox. A ratio of around 10:1 might do it.

Failing that.. What voltage range is the controller designed to operate from and what voltages do you have available?

Suppose your controller and power supply was ok at 12V, then a KV of around

6700/12 = 560

would be better. Perhaps slightly higher to ensure the required max rpm can be achieved under load. Check the data for the motor carefully.

The current draw would then be around..

360/12 = 30A.

If it helps...Rare earth motors generally have lower Kv than ferrite.

EDIT: Actually a 10:1 gearbox would not work. To achieve an output of 6700rpm the motor would need to turn at 67,000rpm which is a bit too high for most motors/controllers.

Last edited: Oct 14, 2012
5. Oct 15, 2012

### qwertyuiop23

Oh so that's where I was going wrong. I was trying to take the rpm and calculate torque rather than setting my torque and rpm then calculating power.

The method you described makes sense. Just to verify something though,

Say my motor has a 80mOhm resistance (most bldc have around this) then

P= I^2 * R

Will get the power loss due to the windings within the motor so at the first 202A the power loss would be approximately 3.26kW, which when I am only getting out 288W my efficiency would become 8.1% which as you said is not 80%.

Using the second approximation (30A) the power losses become
72W which is much more realistic infact = 80% efficicent.