Electricity + Control July 2017

DRIVES, MOTORS + SWITCHGEAR

Step 2: Choose a possible speed reducer: • The required positioning accuracy is 0,1 0 (i.e. 360/0,1 = 3 600 increments per revolution) • If we base the ratio on a motor resolution of 200 steps per rev. (1,8 0 ), 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 = 213/18 2 = 0,66 kg.cm 2 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.cm 2 . 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.

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.kilogram 2 although kg.cm 2 is commonly used as results in more easily visualised numbers.

1 kg.m 2 = 10 000 kg.cm 2 A 3,4 inch stepper motor rotor is 1,2 kg.cm 2 .

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.

At the forefront of successful

designs lies the correct

Worked example We conclude with a real life example.

specification of the mechanical components used.

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: 0,1 0 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 = D 4 L/3 800 = 30 4 X 1/3 800 = 213 kg.cm 2 Where: J = inertia in kg.cm 2 D = diameter (cm) L = thickness (cm) This value far exceeds the inertia of moderately priced motors.

<>

Glyn Craig is a director of Techlyn.

glyn@techlyn.co.za +27 (0) 11 835 1174

Electricity + Control

JULY 2017

13