
#1
Mar2310, 03:50 PM

P: 38

Suppose we've got the setup as shown in the figure (see attachment).
The idea is that the motor transfers its speed and force (rotary) to the actuator force and speed (linear) via some gears and a spindle. Here: R = radius [m] J = inertia [kg m^2] n = rotary to linear transmission [] If Im not mistaken, then the speed of the motor [tex] \phi_{motor} [/tex] is related to the speed of the actuator [tex] \phi_{actuator} [/tex] as follows: [tex] \phi_{actuator} = \phi_{motor} \left( \frac{R_{motor}}{R_{spindle}} n_{actuator} \right) [/tex] The force of the actuator [tex] F_{actuator} [/tex] is related to the torque of the motor [tex] T_{motor} [/tex] as [tex] F_{actuator} = T_{motor} \left( \frac{R_{spindle}}{R_{motor}} \frac{1}{n_{actuator}} \right) [/tex] And my main problem is the following: what is the total inertia [tex] J_{tot} [/tex] seen by motor? Is that [tex] J_{tot} = J_{motor} + \frac{J_{spindle}}{ \left( \frac{R_{spindle}}{R_{motor}} \right)^2 } [/tex] or [tex] J_{tot} = J_{motor} + \frac{J_{spindle}}{ \left( \frac{R_{spindle}}{R_{motor}} \frac{1}{n_{actuator}} \right)^2} [/tex] If someone could confirm/correct my formula, that would be very helpful. Thanks in advance. Bob 



#2
Mar2310, 06:37 PM

PF Gold
P: 946

The first equation should, as far as I can see, give you the equivalent rotational inertia as if all the spindle inertia had been moved to the motor shaft. You would still need to combine all the gear ratios when transforming the motor torque or angular speed to the linear force or linear speed.




#3
Mar2410, 05:47 AM

P: 38

Ok thanks.



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