Electricity + Control July 2017

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13 Electricity + Control JULY 2017
<<Author>>
Glyn Craig is a
director of Techlyn.
+27 (0) 11 835 1174
Force = mass X acceleration. Note the use of
mass and not weight. Therefore, in a weightless
situation the above equation will apply. The equiv-
alent electrical analogue is impedance. Rotational
systems have rotary inertia.
This is easily calculated if the dimensions of the
object and its density are known. The SI (Systeme
International) unit for inertia is metre.kilogram2
although kg.cm2 is commonly used as results in
more easily visualised numbers.
1 kg.m2 = 10 000 kg.cm2
A 3,4 inch stepper motor rotor is 1,2 kg.cm2.
The inertia of the load needs to bematched to the driv-
ing motor’s rotor inertia. Conventional wisdom states
that a mismatch up to 10:1 is permissible. Note, that
the reflected inertia is reduced by the square of the
gearbox ratio. Thus, a 3:1 gearbox will reduce the re-
flected load inertia by 3 squared, which is then 9:1.
This is the same as a transformer which changes load
impedance by the square of the turns ratio.
Worked example
We conclude with a real life example.
A laboratory carousel carries 24 test tubes and has to
position them one by one beneath a dosing needle.
The following information is available
Carousel dimensions:
• 300 mm diameter
• 10 mm thickness
• Material – aluminium
Friction of carousel: 0,1 Nm
Time available for move: 1 s
Positioning accuracy: 0,10
Move times are moderate, therefore a stepper
solution will be used.
Step 1: Calculate the load inertia:
We use the empirical formula for an aluminium disc
J = D4L/3 800
= 30 4 X 1/3 800
= 213 kg.cm2
Where:
J = inertia in kg.cm2
D = diameter (cm)
L = thickness (cm)
This value far exceeds the inertia of moderately
priced motors.
Step 2: Choose a possible speed reducer:
• The required positioning accuracy is 0,10
(i.e. 360/0,1 = 3 600 increments per revolution)
• If we base the ratio on a motor resolution of
200 steps per rev. (1,80), the required ratio =
3 600/200 = 18:1. This is a popular worm gear-
box ratio
• The motor can then be half stepped (400 steps/
rev) or quarter stepped (800 steps/rev) to give
additional smoothness and correct small errors
• Assuming a maximum motor speed of 10 rev/
sec (600 rpm), the table will rotate at 600/18 =
33,3 rpm which is more than fast enough
Step 3: Select a suitable motor. The load inertia is
reduced by the square of the ratio.
Reflected inertia = 213/182
= 0,66 kg.cm2
This looks like an encouraging number. Referring
to a motor catalogue reveals that a single stack
34 frame motor has a rotor inertia 0f 0,6 kg.cm2.
This is a popular, moderately priced motor.
Available torque is 1,2 Nm, which is more than ad-
equate.
The reflected inertia should not exceed the
motor rotor inertia of a stepper by more than
about 10:1, as this gives rise to resonance prob-
lems.
In this case, the worm gearbox will also have
high internal friction, and this slightly oversize motor
will ensure that there is adequate torque available.
Very often an iterative process is needed
where steps 2 and 3 above are repeated until a
satisfactory compromise is reached.
Conclusion
After all this we see that speed reducer selection
is not a simple process.
At the forefront
of successful
designs lies
the correct
specification of
the mechanical
components
used.
DRIVES, MOTORS + SWITCHGEAR