If you mean by an open ended timing belt that is fixed at both ends and the motor is traversing along the belt then, yes. One glaring omission in my initial reply was not accounting for a speed reduction. I'm guess there will be one between your servo and timing belt pulley. In that case you have to account for speed reduction:
w/N alpha/N Jreflected = ((gantry m + motor m) * R^2) /N^2 Jt=Jreflected + Jmotor + Jreducer Trl=(Ffriction * R)/(n*N) 2015-09-10 9:55 GMT-04:00 Dave Cole <[email protected]>: > Yes. > > On 9/10/2015 7:36 AM, John Thornton wrote: > > It's a timing belt drive, would that be close to a rack and pinion? > > > > JT > > > > On 9/9/2015 11:58 PM, Milosz K. wrote: > >> John, > >> > >> Assuming this is a rack & pinion type configuration: > >> > >> angular v (w) = linear vel * pulley radius (R) > >> angular accel (alpha) = linear accel * R > >> Jreflected = (gantry mass + motor mass) * R^2 > >> inertia total (Jt) = Jmotor + Jreflected > >> Torque reflected (Trl) = (Ffriction * R)/efficiency (n) > >> Torque peak(Tp) = Trl + ((Jt*alpha)/n) > >> P = Tp*w > >> > >> > >> > >> > >> > >> > >> > >> 2015-09-09 14:12 GMT-04:00 John Thornton <[email protected]>: > >> > >>> Thanks for all the info, now to chew on this a bit. > >>> > >>> JT > >>> > >>> On 9/9/2015 5:53 AM, John Thornton wrote: > >>>> I need to calculate the power needed to move a gantry. I assume that > >>>> mass and velocity and acceleration are the key factors. Knowing those > >>>> how do you figure out the watts needed. This is not a milling machine > so > >>>> no cutting forces need to be taken into effect. I'll be moving the > >>>> gantry with timing belts. > >>>> > >>>> Thanks > >>>> JT > ------------------------------------------------------------------------------ _______________________________________________ Emc-users mailing list [email protected] https://lists.sourceforge.net/lists/listinfo/emc-users
