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APPENDIX
APPENDIX
App. 1 Selection example of direct drive motor
App. 1.1 Selection conditions
(1) Machine configuration
D
Direct drive motor
(DD motor)
(2) Direct drive motor speed
θ
60
N
=
×
o
(t
- t
- t
360
0
p
270
=
×
360
(0.45 - 0.125 - 0.1)
t
: Settling time (Here, this is assumed to be 0.1 s.)
s
(3) Operation pattern
App. 1.2 Selection of direct drive motor
(1) Load moment of inertia
1
2
J
=
× D
× W
L
T
8
1
-3
=
× (300 × 10
8
(2) Acceleration/deceleration torques of load
N
/ 60
o
T
= J
× 2π ×
a
L
t
p
J
× N
L
o
=
60
t
×
p
2π
0.214 × 200
=
9.55 × 0.125
= 35.9 [N•m]
Table mass
T
Rotary table diameter
Rotation angle per cycle
Positioning time
Acceleration/deceleration time
Operation cycle
Load torque
)
s
60
= 200 [r/min]
Speed
[r/min]
200
0.125
t
psa
t
0
2
2
)
× 19 = 0.214 [kg•m
]
0.125
0.1
t
t
psd
S
0.45
1 cycle t
2.0
f
App. - 1
W
= 19 [kg]
D
= 300 [mm]
T
θ
= 270 [degree]
t
= 0.45 [s] or less
0
t
= t
= t
= 0.125 [s]
p
psa
psd
t
= 2.0 [s]
f
T
= 0 [N•m]
L
Time [s]
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