Chapter 7 Usage of Various Functions
7-
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Condition 2
Set up as acceleration slop = 2, max. frequency = 1000, no of pulse out = 5000.
x
If I/D velocity inclination is 2, 2 pulses are output on the 1st step(velocity: 50pps).
Pulse velocity is 50pps. So time consuming is 40ms.
y
4 pulses are output on the 2nd step(velocity: 100pps) and time consumes 20ms
z
By calculation in the same way, the time to reach to 1000pps is 40ms * (20-1) = 760ms,
and the no. of output pulses are 2+4+6...+36+38 = 380 units.
{
Decreasing velocity inclination is 2, thus 380 units of pulses are needed.
|
The no. of pulses in the uniform velocity region are 5000-380-380=4,240 units.
}
Whole spent time is 57,600ms
If the acceleration slop goes bigger, the increasing time and pulse go bigger by direct proportion to inclination.
Then be careful of an occurring of the instruction error when the no. of a/d pulse becomes bigger than the no.
of whole pulse.
Acceleration
Time:760ms
Pulses:380
40ms
50pps
Uniform velocity
Time:4,240ms
Pulses:4,240
Deceleration
Time:760ms
Pulses:380
Example) Acceleration is 2.
Time
velocity
Deceleration
Acceleration step : 19
1st step
2
nd
step
Remark
