Harmonic Drive HMA Series, Руководство пользователя

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2-3 Examining the operating status
2-5
S
e
le
c
tion gui
de
lin
e
s
2
Appe
Acceleration time and deceleration time
Calculate the acceleration and deceleration times for the motor tentatively selected using the following
formula.
Acceleration time:
Deceleration time:
t
a
: Acceleration time (s)
t
d
: Deceleration time (s)
k: Acceleration reduction coefficient 1 to 1.5
The total positioning time may become
shorter if the acceleration is lowered for
the purpose of reducing the settling time
after positioning.
J
M
: Inertia moment of motor (kg
･
m
2
)
J
L
: Inertia moment of load (kg
･
m
2
)
N: Rotational speed of motor (r/min)
T
M
: Maximum momentary torque (Nm)
T
F
: Motor friction torque (Nm)
T
F
= K
T
× I
R
-T
R
K
T
: Torque constant (Nm/A)
T
R
: Rated torque (Nm)
IR: Rated current (A)
T
L
: Load torque (Nm): The polarity is positive (+) when the torque is applied in the rotation direction, or
negative (-) when it is applied in the opposite direction.

Calculation example 1
Select a motor that best suits the following operating conditions:
･
Rotational speed: 4800 r/min
･
Inertia moment of load: 2.0×10
-4
kg
･
m
2
･
Since the load mechanism is mainly inertia, the load torque is negligibly small.
(1) According to the conditions above, tentatively select an HMAC08 from the table in section 2-1.
(2) From the rating table, the following values are obtained: J
M
= 0.734 × 10 -4 kg
･
m 2 , T
M
= 1.8 Nm,
T
R
= 0.52 Nm, K
T
=0.35 Nm/A, I
R
= 2.1 A
(3) Based on the above formula, the motor's friction torque T
F
is calculated as 0.35 × 2.1 - 0.52 =
0.215 Nm.
(4) If k = 1.3, the acceleration time and deceleration time can be obtained as follows from the
above formulas:
t
a
= 1.3 × (0.734 + 2.0) × 10
-4
× 2 ×
π
/ 60 × 4,800 / 1.8
≒
0.099 s
t
d
= 1.3 × (0.734 + 2.0) × 10
-4
× 2 ×
π
/ 60 × 4,800 / (1.8 + 2 × 0.215)
≒
0.080 s
(5) If the calculated acceleration/deceleration times are too long, correct the situation by:
･
Reducing the inertia moment of load
･
Selecting a motor with a larger frame size
ta td
N
Time
Rotational speed
( )
L M
L M a
T T
N
60
π 2
J J k t
−
×
×
× + × =
( )
L F M
L M d
T T 2 T
N
60
π 2
J J k t
+ × +
×
×
× + × =