Electricity and Control March 2016

ENERGY + ENVIROFICIENCY: FOCUS ON DRIVES, MOTORS + SWITCHGEAR

Initial calculations Initial assumptions: Ballscrews: 16 mm diameter by 5 mm pitch Z axis screw length: 800 mm Required traverse rate: 1 800 mm/minute Resolution: Around 1,0 micron (Note: Resolution is not the same as accuracy) Mass of the Z payload including the X axis: 15 kg.

To calculate the inertia of the longest screw we use the empirical formula for a cylindrical steel shaft:

J = D 4 L/1 300

where: J

= moment of inertia, kg.cm²

D

= diameter, cm

L

= length, cm

The parameters then became:

In our case: D

= 1.6 = 80

Leadscrew RPM = 1 800/5 = 360 RPM

L

An unexpected result of high leadscrew rotational velocity is the ten- dency to develop destructive vibration at critical speeds. Reference to design charts revealed that the lowest critical speed for a screw of this length was in excess of 2 000 RPM. No problem here. 360 RPM was judged to be on the low side given motors capable of 10 times this speed. It was decided to fit low backlash planetary gearboxes. This would provide the following advantages: • Motor torque increased by 3 • Motor resolution increased by 3 • Reflected inertia on the motor shaft decreased by 3² = 9 At this point we could select a suitable motor. Be aware that this portion of the selection process could require a number of iterations. Available planetary single stage gearboxes were quoted as having amaximumbacklash of 10 arc-minutes. From this we could determine that the lost angular motion would be 10/60 = 1/6 of a degree, or 1/6 × 1/360 = 1/2 160 of a revolution. This then gives lost motion of 5 000/ 2 160 = 2,3 micron. This is small compared to temperature effects which would lengthen or shorten the leadscrew by 11 micron per 1° Celsius change per metre length. At this point a motor could be selected and the following checked: • Reflected inertia connected to the motor. • Linear force generated. Inertia Matching of load inertia to the motor rotor inertia is essential for a stable system. This is a large subject and will be dealt with fully in a future article. For now, it is sufficient to say that the ratio of coupled inertia to rotor inertia should not exceed 3:1. The SI (System Internationale) unit for inertia is kg.m² (kilogram metres squared). However, for small systems such as this the units kg.cm² result in more easy to envisage numbers. (One kg.m² is equal to 100² = 10 0000kg.cm²).

Therefore J

= 1,6⁴ × 80/1300 = 0,4033 kg.cm²

To this must be added the equivalent inertia of the payload.

J = WLp²/4000

where WL

= mass of payload, kg = leadscrew pitch, mm

p

For our case, p = 5 and WL = 15 = 0,0938 kg.cm²

Therefore total coupled inertia = 0,4033 + 0,0938 = 0,4971 kg.cm²

A range of 23 frame (2,3 inch) brushless motors was available from the supplier, and we selected the largest, with a rotor inertia of 0,2302 kg.cm². The inertiamismatchwas then : 0.4971/0.2302 = 2.1592 : 1 This matching would be acceptable without a gearbox, but we chose to stay with the 3 : 1 gearbox in the interests of rigidity. Reduc- tion gearboxes reduce the coupled inertia by the square of the ratio, in the same way as a transformer reduces coupled impedance by the square of the turns ratio. The motor therefore sees a reflected inertia of 4\,4971/3² = 0,0552 kg.cm² At this point we were able to estimate the available linear force which could be produced. We use another empirical formula:

F = T.644.e.g/P N

Where T

= torque in Newton metres (Nm) = efficiency with a range of 0 to 1

e

March ‘16 Electricity+Control

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