1 in/sec = 40,000 count/sec
5 in/sec = 200,000 count/sec
an acceleration rate of 0.1g equals
0.1g = 38.6 in/s2 = 1,544,000 count/s2
Note that the circular path has a radius of 2" or 80000 counts, and the motion starts at the angle of
270
°
and traverses 360
°
in the CW (negative direction). Such a path is specified with the
instruction
CR
80000,270,-360
Further assume that the Z must move 2" at a linear speed of 2" per second. The required motion is
performed by the following instructions:
Instruction Function
#A Label
VM XY
Circular interpolation for XY
VP 160000,160000
Positions
VE
End Vector Motion
VS 200000
Vector Speed
VA 1544000
Vector Acceleration
BGS Start
Motion
AMS
When motion is complete
PR,,-80000
Move Z down
SP,,80000 Z
speed
BGZ Start
Z
motion
AMZ
Wait for completion of Z motion
CR 80000,270,-360
Circle
VE
VS 40000
Feed rate
BGS Start
circular
move
AMS
Wait for completion
PR,,80000
Move Z up
BGZ Start
Z
move
AMZ
Wait for Z completion
PR -21600
Move X
SP 20000
Speed X
BGX Start
X
AMX
Wait for X completion
PR,,-80000 Lower
Z
BGZ
AMZ
CR 80000,270,-360
Z second circle move
VE
VS 40000
BGS
AMS
PR,,80000 Raise
Z
154
•
Chapter 7 Application Programming
DMC-1600