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DMC-1700/1800 Chapter 7 Application Programming • 171
5 in/sec = 200,000 count/sec
an acceleration rate of 0.1g equals
0.1g = 38.6 in/s2 = 1,544,000 count/s
2
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 Feedrate
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