Input shaft speed from open circuit voltage of PMDC machine?

I’m investigating using the open circuit voltage of a small PMDC machine to measure the input speed to the shaft. The actual project relates to measuring the speed of a swinging pendulum.

The PMDC generator is connected to the rotating hinge of the pendulum and the open circuit voltage of the generator is recorded using a data logger. I’m getting the expected output as shown below but I’m not sure how to convert the voltage to the speed.

I know Voltage ∝ Speed, but I’m not sure how to calculate the other parameters using the information I have to complete the equation and make the speed the subject of the equation.

I’m using equations for separately excited DC machines from a book “Electric Machinery Fundamentals” – Stephen J. Chapman.

From the equation Ea = Vt – IaRa where Ea is the internal voltage and Vt is the terminal voltage, I’m assuming that open circuit voltage (which is what I’m measuring) is equal to the terminal voltage as no current will be flowing. If this is true then Ea = Vt.

Ea = K’ϕn where n is the speed. Can I work out K’ and ϕ from the information listed below? Or by using it as a motor can I work out the values?

Equations from book:
K = ZP/ 2(Pi)a
a = number of current paths
P = Number of poles
Z = number of conductors

Information from manufacturer:
Motor type (PM, Series, Shunt), PM
permanent magnet flux (if PMDC),
The exact permanent magnet flux is hard to provide for 5P motor. The center Gs is 1100+/-5%
armature/field resistances, ---
number of poles, 5P
armature winding type (duplex lap-wound, simplex wave wound, etc), simplexlapwinding
number of armature coils/turns per coil, 32T
resistance per turn 0.34 ± 10%Ω

Motor characteristics under no shaft load
6V input -> 72rpm
9V input -> 108 rpm
12V input -> 144 rpm

Is it as simple as using these values in the equation Ea = K’ϕn to work out K’ϕ?

Why I’m measuring the speed using this method instead of other means is a long story but I’m stuck with it and I’m committed to it.
Any help would be much appreciated. Thanks!

Your attached graph, shows rightly, a damped simple harmonic motion. Motion of a pendulum is simple harmonic.

It is losing half of the amplitude in 5 cycles: that is too much to lose. Perhaps the loading is coming from the PMDC generator: theoretically speaking, an oscillating magnet or a dipole is capable of electromagnetic radiation and lose energy.

Zeros in the graph correspond to the pendulum at the extreme ends and the peaks are, you have guessed rightly, corresponding to the centre.

By the way, I forgot what was the question?

c_mitra said:

Your attached graph, shows rightly, a damped simple harmonic motion. Motion of a pendulum is simple harmonic.

It is losing half of the amplitude in 5 cycles: that is too much to lose. Perhaps the loading is coming from the PMDC generator: theoretically speaking, an oscillating magnet or a dipole is capable of electromagnetic radiation and lose energy.

Zeros in the graph correspond to the pendulum at the extreme ends and the peaks are, you have guessed rightly, corresponding to the centre.

By the way, I forgot what was the question?

Click to expand...

Thanks for the reply mitra. It's actually a 50mm square, 2mm thick, 2m long square steel bar pendulum and its hinge is bent out of shape. This probably explains the extra losses.

That graph represents the open circuit output voltage I got after swinging the pendulum. I want to convert that voltage to the speed of rotation of the pendulum.

In my first post I mentioned the equations Ea = Vt – IaRa and Ea = K’ϕn where Vt is the terminal voltage and Ea is the internal voltage. Since I'm measuring the OC voltage of the generator I assumed I could disregard the IaRa component and Ea = Vt simplifying things. In this case Vt = K’ϕn where n is the speed of rotation and K’ϕ are the unknown quantities I need to know to convert the output voltage to input speed.

Using the machine as an unloaded motor I get a linearly increasing speed output with voltage. I believe it makes sense to use these motor values to work out the K’ϕ constants and then rearrange the equation to make speed the subject: n = Vt/K’ϕ but I wanted to confirm there was no problem with this method.

Alternatively, with the other information provided can I work out the values of K’ and ϕ by some other means.

9V input -> 108 rpm
12V input -> 144 rpm

Is it as simple as using these values in the equation Ea = K’ϕn to work out K’ϕ?

Click to expand...

Basically yes. Unloaded Ea is slightly higher than the value calculated from the rpm, need to consider an IaRa term.

Ea = K’ϕn...

Click to expand...

This is clearly wrong; the motor voltage output is not the important parameter. You can see that the voltage is of the form of sin(wt) and the w is the angular frequency of the pendulum.

The actual waveform has two time constant: one is the period and the other is the damping time const. For simplicity these two can be considered independent (they are actually related by the differential equation describing the friction).

How to get the frequency? V=A*sin(wt) and you see dV2/dt2=a*w2*V; hence the second derivative is proportional to w2.

In software, you can look for zero crossing points.

This is clearly wrong; the motor voltage output is not the important parameter. You can see that the voltage is of the form of sin(wt) and the w is the angular frequency of the pendulum.

Click to expand...

The equation can't be wrong, the shaft speed itself is following a sine waveform, so does the motor e.m.f.

I may have got the problem wrong, but as far as I understand, the OP is just asking for the relation of instantaneous shaft speed and generated voltage because he wants to calibrate the pendulum measurement.

FvM said:

The equation can't be wrong, the shaft speed itself is following a sine waveform, so does the motor e.m.f.

I may have got the problem wrong, but as far as I understand, the OP is just asking for the relation of instantaneous shaft speed and generated voltage because he wants to calibrate the pendulum measurement.

Click to expand...

Yes this is correct. I'm looking for the instantaneous shaft speed and want to use the output voltage of the generator to get it.

Yes, you are right if you are looking for the instantaneous shaft speed and the instantaneous voltage produced. For a fixed geometry (motor and the physical coupling) they will be linearly related over a very long range.

I was confused because of "speed of rotation" - I assumed it to mean the period or frequency of the pendulum.