Harmonic Drive HMA Series, Manual

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2-3 Examining the operating status
2-6
S
e
le
c
tion gui
de
lin
e
s
2
Appe
Evaluating effective torque and average rotational speed
One way to check if the heat generated from the motor during operation would present a problem is to
determine if the point of operation, determined by the effective torque and average rotational speed, is
inside the continuous motion range explained in [1-12Operable range].
Using the following formula, calculate the effective torque Tm and average rotational speed N
av
when
the motor is operated repeatedly in the drive pattern shown to the right.
T
a
: Acceleration time from speed 0 to N (s)
t
d
: Deceleration time from speed N to 0 (s)
t
r
: Operation time at constant speed N (s)
t: Cycle time (s)
T m : Effective torque (Nm)
T
a
: Torque during acceleration (Nm)
T
r
: Torque at constant speed (Nm)
T
d
: Torque during deceleration (Nm)
N
av
: Average rotational speed (r/min)
N: Rotational speed at constant speed (r/min)

Calculation example 2
The calculation method is explained below using HMAC08 as an example.
Operating conditions: Accelerate an inertia load and then let it move at a constant speed,
followed by deceleration, based on conditions similar to those used in calculation example 1. The
travel angle per cycle is 3,600° and the cycle time is 0.8 seconds.
(1) The travel angle is calculated from the area of the rotational speed vs. time diagram shown
above. In other words, the travel angle is calculated as follows:
θ
= (N / 60) x {t
r
+ (t
a
+ t
d
) / 2} x 360
Accordingly, tr
= θ/ (6 x N)
– (ta + td) / 2
When θ = 3,600° and ta = 0.099 (s), td = 0.080 (s), N = 4,800 (r/min) in calculation example 1,
are applied to this formula, tr is calculated as 0.035 (s).
(2) Next, calculate the torque during acceleration and torque during deceleration. Based on the
acceleration/deceleration time formulas in the preceding section, the relational expressions for
torque during acceleration and torque during deceleration if k = 1 are as follows:
T
a
= (J
M
+ J
L
) x 2 x
π
/ 60 x N / t
a
+ T
L
T
d
= (J
M
+ J
L
) x 2 x
π
/ 60 x N / t
d
-2 x T
F
- T
L
When the values in calculation example 1 are applied to this formula, T
a
= 1.38 (Nm) and T
d
=
1.29 (Nm) are obtained.
(3) Calculate the effective torque. Apply the values in (1) and (2), T
r
= 0 (Nm), and t = 0.8 (s) to
the above formulas.
(4) Calculate the average rotational speed. Apply the values in (1), N = 4,800 (r/min), and t = 0.8
(s) to the above formulas.
t
t T t T t T
T
d
2
d r
2
r a
2
a
m
× + × + ×
=
t
t 2 N t N t 2 N
N
d r a
av
× + × + ×
=
Nm 0. 64
0. 8
0. 080 1. 29 0. 035 0 0. 099 1. 38
T
2 2 2
m
=
× + × + ×
=
0. 8
0. 080 2 4, 800 0. 035 4, 800 0. 099 2 4, 800
N
av
× + × + ×
=
Time
Tr
ts: Stoppedtime
Ta, Tr, Td: Output torques
ta td
N
tr
Time
Rotational speed
t: Cycle time
ts
Ta
T
or
que
Td
= 750 r/min