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Lenze · Drive Solution Designer · Manual · DMS 4.2 EN · 12/2013 · TD23 87
7Applications
7.4 Omega belt drive
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
7.4.3 Calculations
The following applies to a linear axis with an omega belt drive that is moved, as is shown in the
drawing:
The belt pulley has the following effective diameter, where p
Cog
specifies the belt pitch:
[7-26] Equation 1: Diameter of belt pulley
Conversion of translatory variables into rotary variables
[7-27] Equation 2: Angle
[7-28] Equation 3: Angular velocity
[7-29] Equation 4: Angular acceleration
Forces of the linear motion
First the mass that is to be moved linearly is to be calculated. The payload m
L
can adopt different
values during the travel cycle. The mass of carriage m
aux
and the mass of the drive motor and the
gearbox m
D
are taken into consideration separately.
[7-30] Equation 5: Total mass
The friction force F
μ
can for instance occur on the guide rails of the slide. The force acts opposite to
the direction of movement and is taken into consideration by the fraction v/|v| in the following
equation, where at v = 0 the force F
μis
0. The static friction at standstill is not considered.
[7-31] Equation 6: Friction force
If the friction force F
μ
is related to the mass in motion, a specific travelling resistance results, which
contains all parts depending on the mass:
[7-32] Equation 7: Specific travelling resistance of the application
d
Cog
[mm]
p
Cog
[mm] N
Cog
π
-------------------------------------------
=
ϕ
2000 s
d
Cog
--------------------
2000 π s⋅⋅
p
Cog
N
Cog
----------------------------
==
ω
2000 v
d
Cog
--------------------
2000 π v⋅⋅
p
Cog
N
Cog
----------------------------
==
α
2000 a
d
Cog
--------------------
2000 π a⋅⋅
p
Cog
N
Cog
----------------------------
==
m
sum
m
L
m
aux
m
D
++=
F
μ
m
sum
g μ
Gdn
βcos
v
v
-----
⋅⋅ =
F’ g μ
Gdn
βcos⋅⋅=
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