2x 2005 C7 Ballscrew 1450mm Length with Ballnuts + FK/FF mounts and motor coupler

[Jonathan]

I have a gantry that weighs approximately 80kg with spindle. With a decent rate of acceleration let's assume its set to 1000 and 160 steps per in mach with dual RM2010 and 4Nm motors. How much kinetic energy is transferred to the motors when doing a direction change at say 1000rpm or 10m/min? You got to decelerate that 80kg in a very short space of time and I'd hate to think how big this number is going to be or if the motors can even handle that.

Right, so you have said let:
acceleration =1000mm/s^2=1m/s^2
velocity = 10m/min = 10/60= 0.167 m/s (not actually required for the initial calculation)
mass = 80kg

It's just newtons 2nd law, so F=ma=80*1=80 Newtons (equivalent to lifting about 8.2kg), so not far off my earlier estimate.

From the motor's point of view it's not 80N, since we have to consider the ballscrew in between. The ballscrew gives the motors a huge mechanical advantage. The relevant formula, which can be derived from considering the ballscrew as a ramp, is:

T=F*p/(2*pi*e)
Where T is the torque, F is the force applied to the nut, p is the ballscrew pitch (so 10mm=0.01m) and e is the efficiency (about 0.9). So chuck the numers in:
T=80*0.01/(2*pi*0.9)=0.14 Nm

So changing direction at 10m/min and 1m/s, as you describe, only translates to an extra 0.14Nm torque on the ballscrew, compared to going at a constant speed and 80N force on the ballnut.