July 2013 6.8 The control loop 813
Minimum distance To attain the maximum velocity, a minimum distance s
min
 must be traversed. 
If the traversed distance is greater than s
min
, a movement with constant 
speed is inserted at the time 2T
r
. The minimum distance is:
Example Rapid traverse v
max
 = 30 000 mm/min (= 0.5 m/s); MP1010.x = 30000
Max. jerk with velocity v > 20 000 mm/min (= 0.33 m/s) r
max1
=70m/s
3
; 
MP1090.1 = 70, MP1092 = 20000
Max. jerk r
max2
 = 35 m/s
3
 during machining; MP1090.0 = 35
Maximum attainable acceleration a
max1
 during rapid traverse:
Maximum attainable acceleration a
max2
 during machining (v up to 20 000 mm/
min):
Distance s
min
 required to attain rapid-traverse velocity:
Distance s
Acceleration a
t
T
r
2T
r
3T
r
4T
r
Jerk r
Velocity v
s
min
v
max
s
min
2 v
max
v
max
r
max
------------⋅⋅=
a
max1
v
max
r
max1
⋅ 0,5 
m
s
------ 7 0  
m
s
3
------⋅ 5,92 
m
s
2
------== =
a
max2
v
max
r
max2
⋅ 0,33 
m
s
------ 3 5  
m
s
3
------⋅ 3,40 
m
s
2
------== =
s
min
2 v
max
v
max
r
max
------------⋅⋅ 20,5 
m
s
------
0,5 
m
s
------
70 
m
s
3
-------
---------------⋅⋅ 0,085 m 85 mm== ==
The rectangular jerk curve is rounded through the use of a nominal position 
value filter (MP1096.x ≠ 0). As a result, acceleration is reduced and the 
minimum distance required for attaining the maximum velocity is 
increased.