Last time we collected together the equations we would need for predicting the performance of the Mabuchi 380 electric motor that we hoped would be close enough to the one in the Grasshopper radio controlled model car to enable us to predict its performance.

Our immediate task is to find two coefficients – K1 and K2 – that we need to do this.

In the previous posting, the equation for the first of these coefficients was given as:

K1 = Ts/Is               where Ts is stall torque and Is is stall current

From the Mabuchi website,  the stall current was 24A, and the stall torque was 75.6mNm

I’m not sure how familiar followers are  with standard abbreviations and their meanings, so I will explain them as we go.

A	Amperes (amps)	fundamental unit of electrical current
mNm	milli Newton-metres	Newton-metres is the fundamental unit of torque (twist), milli in front divides it by 1000 (as in metres and milli-metres)
N	Newton	Fundamental unit of force (yes, I was converted to SI units in the late 1960s)

Now we can substitute the numbers into the equation, and find K1. I made it:

K1 = 3.15mNm/A                         

And we also saw the equation for K2. It was:

K2 = (V – I0*R)/n0                     where R is the resistance of the armature windings
                                                      in Ohms  and n0 is the no-load motor speed in rpm

Greek omega	Ohm	Unit of electrical resistance (voltage/current)
n	rotational speed	two common units are: revolutions per minute, and radians per second (2*pi radians is 1 revolution)

You could use different units for speed, but you would need keep track of them. 

From the Mabuchi website, the no load speed is 18000rpm

We can get resistance from the stall test.

Resistance = voltage/(stall current)

Inserting data  from the Mabuchi website gave:

R = 6/24 Ohms 

or 

R = 0.25 Ohm

Substituting into the equation for K2, we get:

K2 = (6-0.8*0.25)/18000       

or

K2 = 0.000322   V/rpm                       as usual, it’s important to keep track of the units                 

Now we can build our model;

Torque developed by the windings in the armature is:

T = K1*( V – E*n)/R

But this does not all get past the output shaft.

The no load test showed that there was a loss of torque in the motor itself. That was why there was a current at no load. We can correct for this loss torque  by deducting it from the gross torque.

T= K1*(V- K1*n)/R -K1*I0

or

T = 3.15*(V – 0.000322*n)/0.25 – 3.15*0.8  mNm 

Notice that I have retained the milli prefix.

This simplifies to:

T = 12.6*V – 0.00406*n – 2.52  mNm

There is a conundrum here. 

Substituting V = 6 V and n = 18000 rpm gives

T= – 2.52 mNm.

But this was supposed to be the no load condition!

It all comes down to what the loss torque is at stall. If the loss torque were constant over the speed range, then the measured torque at stall would be the torque from the windings less the loss torque. K1 describes the torque from the windings. So this correction would mean that 

K1 = (75.6+2.52)/24 mNm/A

or

K1 = 3.255 mNm/A

I did not make this adjustment because it seemed to me that the loss torque at stall was indeterminate.

This can predict torque for whatever speed and voltage is required.

The graph is for a  constant 6 volt supply over the full speed range.

In the next post I will compare my predictions with the Mabuchi data for maximum efficiency.