Siemens SINUMERIK 840C Installation Instructions Manual page 670

09.95
curve have been correctly calculated and/or have been entered in the correct
input format (caution: MD 1252* uses a format factor 100 larger than MDs 1244*
and 1248*!)
Example for setting a. To derive the actual acceleration
the characteristic
b. Entering the characteristic break points
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SINUMERIK 840C (IA)
9.5.3 Conventional quadrant error compensation (as from SW 2)
The acceleration when passing through zero speed in a circular path is calcu-
lated as follows:
2
a = v /r
A radius of 10 mm and a circular velocity of 1 m/min = 16.7 mm/s produces an
2
acceleration a = 16.7 /10 [mm/s
The following accelerations were determined as the characteristic break point:
2
a = 1.11 mm/s , a = 27.78 mm/s
1 2
The position control resolution 0.5  10 – 4 mm was selected, resulting in:
1000 units [MS] = 1 mm
The characteristic break points are therefore:
2
a = 11100 units/s , a = 277800 units/s
1 2
The following values must therefore be entered in the machine data in the
given order:
MD 1252* = 695, MD 1248* = 2778, MD 1244* = 111
If unsatisfactory results are obtained for very low speed
values,
a. increase the position control resolution
b. raise the smoothing time constant (MD 1256*), values
 100 ms are recommended.
c. set MD 1824* bit 0 to 1. However, it must be
remembered that compensation if performed on small
traversing movements (e.g. with
with this parameterization.
9 Drive Servo Start-Up Application (as from SW 3)
2 2
] = 27.78 mm/s .
2 2
, a = 695 mm/s
3
2
, a = 6950000 units/s
3
incremental mode)
2
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