Hi,
I've done the calclations for the acceleration potential of my machine including rotational inertia. I've had my calculation checked
by peteeng, a professional mechanical engineer. Additionally I have Hiwins calculation formulas and can double check all the calculations
and they all point to the same conclusion.
My previous calculations are widly optimistic.
The full calculation is:
Torque= Ieff. dw/dt
dw/dt is the angular acceleration.
Ieff is the total effective first moment of inertia.
Ieff is the sum of the individual components.
1)The rotational inertia of the armature of the servo. For my 750W 34size Delta servo that is (from the spec sheet)=1.13 .10-4 kg.m2
2)The rotational inertia of the ballscrew. For my 32mm dia screw of 650mm length =5.252 .10-4 kg.m2
Note that I got this figure from the THK spec sheet but both I and peteeng calculated it from first principles and all agree.
3)The effective rotational inertia implied by the axis mass. This forumla I derived from first principles and agrees with the theoretical treatment
published by Hiwin and also agrees with peteengs analysis.
Ilinear= m.(p/2pi)2 where p=pitch in meters and m is the axis mass.
Using numbers for my machine, p=0.005m and m=110kg:
Ilinear=110. (.005/2 x 3.141)2
=0.69 kg.m2
Ieff is the sum of these components:
Ieff= (1.13 + 5.252 + 0.69) .10-4
=7.07 .10-4 kg.m2
dw/dt= 2.4 /7.07 .10-4
=3395 rad/s2
And to convert that to linear acceleration:
Alinear= dw/dt x p/2pi
=2.7 m/s2
This figure is less than 1/10th of what I had calculated previously, and is my error. Note how despite the heavy axis, its contribution
to the first moment of inertia is small, even the first moment of the servo armature exceeds it, but both are dwarfed by the first moment of the ballscrew.
I had not made allowance for that factor previously.
The take away feature is that with my machine at least 'the rotational mass of the ballscrew dominates the acceleration equation'.
Craig