IAI TTA Series - Page 68

Chapter 1 Specications Check1.2 Mechanical Specications
58
Constant velocity range movement distance S =
Stroke St – Acceleration/deceleration movement distance (Sa + Sb) [mm]
= 500 – (123.55 + 123.55) = 252.9
Cycle time tcr = Constant velocity range movement distance S [mm] / Set Velocity V [mm/s] +
Acceleration timeta [s] + Deceleration time tb [s] + Positioning convergence
time [s]
= 252.9/600 + 0.41+0.41+0.15 = 1.39 [s]
Figure out the acceleration / deceleration time ratio tod in d).
Acceleration/Deceleration Time Ratio tod = (cceleration time at operation ta + Deceleration
time at operation tb) /Cycle time tcr (tct) = (0.41+ 0.41) /1.39 = 0.59 (59 %)
In e), the reference of the duty will be 78% when the load factor LF = 100% and acceleration /
deceleration time ratio tod = 66% according to the graph.
Use the actuator with consideration to select the velocity, acceleration / deceleration and payload
so the actual operation point should be at this duty or less.
Example 2) Y-axis
Calculate the duty of when;
Set Payload Mr [kg] : 11 [kg]
Set Acceleration / Deceleration αr [G] : 0.2 [G]
Transported Mass of ZR-axes M [kg] : 7.7 [kg]
Acceleration / Deceleration at Operation α [G] : 0.2 [G]
Stroke St [mm] = Movement Distance : 100 [mm]
Set Velocity V [mm/s] : 600 [mm/s]
Positioning convergence Time [s] : 0 [s]
Check the operation pattern in a).
Reached Velocity Vmax = √[Stroke St ×Set Acceleration α ]
= √[ St [mm] x 9,800 [mm/s
2
] x Set Acceleration α [G] ]
= √[ 100 x 9800 x 0.2] = 443 [mm/s]
Set Velocity V < Reached Velocity Vmax …1) Trapezoid Pattern
Figure out the load factor LF in b).
Load Factor LF = (M x α) / (Mr x αr) [%]
= (7.7 x 0.2 x 9800) / (11 x 0.2 x 9800) = 0.7 (70 [%])
Figure out the acceleration / deceleration time ta (tb) and the cycle time tct in c).
Acceleration / Deceleration Time ta (tb) = √(Movement Distance St [mm] / Acceleration /
Deceleration at Operationα [G] x
9800 [mm/s
2
] )
= √(100 / (0.2 x 9800)) = 0.23 [s]
Cycle Time tct =
(Acceleration / Deceleration Time ta + Deceleration Time tb) [s] + Positioning termination time [s]
= 0.23 + 0.23 + 0.15 = 0.61 [s]
Figure out the acceleration / deceleration time ratio tod in d).