Do you mean a shaft at one end of the table, a shaft at the other end of the table with a timing belt around a pulley on each of the shafts. Then yes, that is very much like a rack and pinion. I've used that setup to mimic a belt drive in the Cooperhill software. The efficiency of a belt is very high. 95+ %.
Dave On 9/11/2015 6:50 AM, John Thornton wrote: > I mean two jack shafts with the belt fixed to the gantry and the servo > driving a jack shaft. > > JT > > On 9/11/2015 1:40 AM, Milosz K. wrote: >> 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 > > ------------------------------------------------------------------------------ > _______________________________________________ > Emc-users mailing list > [email protected] > https://lists.sourceforge.net/lists/listinfo/emc-users --- This email has been checked for viruses by Avast antivirus software. https://www.avast.com/antivirus ------------------------------------------------------------------------------ _______________________________________________ Emc-users mailing list [email protected] https://lists.sourceforge.net/lists/listinfo/emc-users
