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Program Example
Section 10-1
10-1 Program Example
This section shows several motion programs for different typical application.
10-1-1 Positioning with PTP Control
Explanation of the
operation
• Positioning to the 3 points below is executed with PTP control, and the
axes returns to the origin ([J01]0, [J02]0).
• When positioning is completed, 1 will be output to CPU unit using the
general I/O area.
1,2,3...
1.
[J01]10000, [J02]50000
1 is output to the general output (0BA0).
↓
2.
[J01]50000, [J02]75000
↓
3.
[J01]100000, [J02]100000
1 is output to the general output (0BA1).
↓
4.
[J01]0, [J02]0
Program
The numbers 01) to 15) are used only for the sake of explanation. They are
not needed in programming.
01) PROG P001 Q00000003;
02) MOVE [J01]10000 [J02]50000;
03) NOPS;
04) #OW0BA0 = 1;
05) WAIT #IW0B00 = = 1;
06) #OW0BA0 = 0;
07) MOVE [J01]50000 [J02]75000;
08) DWELL T300;
09) MOVE [J01]100000 [J02]100000;
10) NOPS;
11) #OW0BA1 = 1;
12) MOVE [J01]0 [J02]0;
13) NOPS;
14) #OW0BA1 = 0;
15) END;
Explanation of the
program
01) A program No. and axes to be used are specified.
02) Positioning to [J01]10000, [J02]50000 with PTP control is performed.
03) NOPS command will hold the program advance till positioning 02) is com-
pleted (within in-position range).
04) Once positioning is completed, 1 will be output to the general output
(OW0BA0).
05) The program advance will be held till the general input (IW0B00)
becomes 1.
06) The program will advance to the next block after the general output from
04) is cleared.
07) Positioning to [J01]50000, [J02]75000 with PTP control is performed.
08) The program waits for 3 seconds.
09) Positioning to [J01]100000, [J02]100000 with PTP control is performed.
10) to 14) 1 will be output to the general output (OW0BA1) after positioning is
completed and the axes will return to the origin. Once they reach
the origin, the general output (OW0BA1) will be reset.
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