9/16/2017 4:20 PM  
Joined: 4/2/2016 Last visit: 7/13/2022 Posts: 413 Rating: (6) 
Hi, I have SINAMICS S120, PM340 30kW6SL32101SE260UA0 SYNCHRONOUS MOTOR (+ENCODER) 2500rpm, 30kW 1PH81312DL000BA1 BRAKING RESISTOR 27Ω, 1.2kW/24kW6SE64004BD212DA0 Control Units for singleaxis drivesCU3102PN I need to make one revolution of the motor (360 degrees) to stop and reverse 360 degrees back. for what minimum time can Sinamics do this? we speak without load! can I think so? rated speed motor 1PH8 2500rmp, this means one motor revolution = 60sec/2500=24ms. but this is in theory…the ramp of start and stop is not instantaneous…braking what do you think in practice for what minimum time can Sinamics do this? if I need the maximum speed about what I should think…take a faster motor (at maximum rated speed, eg. servo 1FT..), replacement CU310 > CU320 It makes sense? 
Last edited by: _oleksandr_ at: 9/16/2017 4:26:51 PMLast edited by: _oleksandr_ at: 9/16/2017 4:27:54 PMLast edited by: _oleksandr_ at: 9/17/2017 11:31:45 AMLast edited by: _oleksandr_ at: 9/24/2017 9:56:22 PM 

9/17/2017 4:31 AM  
Posts: 1529 Rating: (200) 
Find the inertia of the motor from the motor data sheet.. At 150% torque, calculate the acceleration time from 0 to full speed. Using this time, calculate the distance traveled (degrees traveled). Double this number to account for deceleration. Calculate the remaining distance traveled at full speed (360  2 * accel time). Calculate the time to travel that distance. Your total time to travel 360 degrees (accel + constant speed + decal) is the shorted time. If you cannot reach top speed (accel distance is > 180 degrees, calculate the time to travel to 180 degrees and double it. CU 310 or 320 have nothing to do with it unless your braking resistor can't handle the duty cycle. You need to calculate that too. If it cannot, then you need a CU320 and either an SLM or an ALM. You should have learned all of the formulas you need to know in high school physics. 
9/17/2017 8:44 AM  
Joined: 4/2/2016 Last visit: 7/13/2022 Posts: 413 Rating: (6) 
Moment of inertia 0.04460 kgm²
Mn=96Nm; 150%=1.5*96=144 Nm
acceleration time = (inertia*speed)/(9.55*150%_torque)=(0.04460 kgm²*2500rpm) / (9.55*144 Nm)=0.081 s (from 0 to 2500rpm)
acceleration = (2500rmp  0 rpm)/0,081 = 30834 rmp/s distance traveled = (acceleration * time*time)/2 = (30834*0.081*0.081)/2=101 Revolutions =101 * 360 = 36485 degrees deceleration = 36485 * 2=72971 degrees
3602*0,081=359.... Help as the next??? I will ask for a simple  this drive can do a half revolutionin on 50ms. 
Last edited by: _oleksandr_ at: 9/17/2017 8:47:23 AMLast edited by: _oleksandr_ at: 9/17/2017 9:21:10 AMLast edited by: _oleksandr_ at: 9/17/2017 9:46:00 AMLast edited by: _oleksandr_ at: 9/17/2017 9:59:21 AMLast edited by: _oleksandr_ at: 9/18/2017 11:36:43 AMLast edited by: _oleksandr_ at: 9/18/2017 4:20:19 PM 

Follow us on