2-10
Engineering Guide CDA3000
2 Drive definition
2.3.1 Example 1:
Traction drive
1. Determine power requirement to move the application
2. Select motor
The selected motor must have a power rating higher than P
Drive
. Select
the motor from the list.
3. Calculate gross output
P
Gross
= P
a
+ P
F
+ P
aR
= 264W + 9W + 65W = 338W
For more details on “Selection of inverter modules” refer to sections 3.3 to
3.6.
Example: Z-axis of a manipulator
m = 51.5 kg
a = 3 m/s
2
v = 1.5 m/s
η
= 0.88
ta = 0.5
µ = 0.01
Selected motor:
Type 71L/4, 370W, J
M
= 0.00073 kgm
The motor is to be run at max. 2000 rpm
(70 Hz characteristic).
P
a
m a v
⋅ ⋅
η
------------------
51 5kg
,
3m s
2
⁄
1 5m s
⁄
,
⋅
⋅
0 88
,
---------------------------------------------------------------
264W
=
=
=
P
F
m g
µ
v
⋅ ⋅ ⋅
η
--------------------------
51 5kg
,
9 8m s
s
⁄
,
0 01
,
1 5m s
⁄
,
⋅
⋅
⋅
0 88
,
------------------------------------------------------------------------------------
9W
=
=
=
P
Fahr
P
a
P
F
273W
=
+
=
P
aR
J
M
n
2
M
⋅
91 2
,
t
a
⋅
--------------------
0 00073kgm
2
,
2000
2
min
1
–
⋅
91 2
,
0 5
,
⋅
------------------------------------------------------------------
65W
=
=
=
Содержание CDA3000
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Страница 336: ...Engineering Guide CDA3000 A 45 DE EN FR 1 2 3 4 5 6 A Appendix B Practical working aids for the project engineer ...