Fuji Electric Frenic-Mini - 2 ] Calculation of the Acceleration Time; 3 ] Calculation of the Deceleration Time

7-10
(3) For a load running horizontally
Assume a carrier table driven by a motor as shown in Figure 7.7. If the table speed is υ (m/s) when the
motor speed is N
M
(r/min), then an equivalent distance from the rotation axis is equal to 60·υ / (2π·N
M
)
m. The moment of inertia of the table and load to the rotation axis is calculated as follows:
)mkg()WW()
N
2
60
(J
2
0
2
M
••
•
•
+
π
υ
=
(7.9)
[ 2 ] Calculation of the acceleration time
Figure 7.9 shows a general load model. Assume that a motor drives a load via a reduction-gear with
efficiency η
G
. The time required to accelerate this load to a speed of N
M
(r/min) is calculated with the
following equation:
)s(
60
)0
N
(2
JJ
t
M
G
LM
G
21
ACC
−π
τ
−
τ
+
=
•
•
η
η
(7.10)
where,
J
1
: Motor shaft moment of inertia (kg·m
2
)
J
2
: Load shaft moment of inertia converted to motor shaft (kg·m
2
)
τ
M
: Minimum motor output torque in driving motor (N·m)
τ
L
: Maximum load torque converted to motor shaft (N·m)
η
G
: Reduction-gear efficiency.
As clarified in the above equation, the equivalent moment of inertia becomes (J
1
+J
2
/η
G
) by
considering the reduction-gear efficiency.
Figure 7.9 Load Model Including Reduction-gear
[ 3 ] Calculation of the deceleration time
In a load system shown in Figure 7.9, the time needed to stop the motor rotating at a speed of N
M
(r/min) is calculated with the following equation:
)s(
60
)
N0
(2
JJ
t
M
G
LM
G
21
DEC
−
π
−
+
=
•
•
•
•
τ
η
τ
η
(7.11)
where,
J
1
: Motor shaft moment of inertia (kg·m
2
)
J
2
: Load shaft moment of inertia converted to motor shaft (kg·m
2
)
τ
M
: Minimum motor output torque in braking (or decelerating) motor (N·m)
τ
L
: Maximum load torque converted to motor shaft (N·m)
η
G
: Reduction-gear efficiency
In the above equation, generally output torque τ
M
is negative and load torque τ
L
is positive. So,
deceleration time becomes shorter.