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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.
The parameters then became:
Leadscrew RPM = 1 800/5 = 360 RPM
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
result in more easy to envisage numbers. (One kg.m² is equal
to 100² = 10 ).
To calculate the inertia of the longest screw we use the empirical
formula for a cylindrical steel shaft:
J = D4L/1 300
where:
J = moment of inertia,
D = diameter, cm
L = length, cm
In our case:
D = 1.6
L = 80
Therefore
J = 1,6⁴ × 80/1300
= 0,4033
To this must be added the equivalent inertia of the payload.
J = WLp²/4000
where
WL = mass of payload, kg
p = leadscrew pitch, mm
For our case, p = 5 and WL = 15
= 0,0938
Therefore total coupled inertia = 0,4033 + 0,0938
= 0,4971
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 ². 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
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)
e = efficiency with a range of 0 to 1
ENERGY + ENVIROFICIENCY: FOCUS ON DRIVES, MOTORS + SWITCHGEAR
43 March ‘16 Electricity+Control