User`s manual

Programmed Motion
GFK-1742A Chapter 7 Programmed Motion 7-39
7
t
v
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7
2
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000
7280000 800
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PMOVE
Starts
Third
CMOVE
Starts
Second
CMOVE
Starts
Example 2 below shows how the result desired in Example 1 could be obtained by replacing
Example 1’s single move with four moves. Four moves are required since both the acceleration
and deceleration portions of the profile must each be divided into two moves. To divide the total
acceleration (or deceleration) time in half, we calculate the distance at the midpoint of either
slope, when velocity is 12000, to be 720,000 user units.
The distance traveled during acceleration or deceleration is calculated using the formula:
Change in velocity x Required time
Distance traveled =
2
12,000 x 120
720,000 =
2
(Since 240 seconds is needed to reach a velocity of 24,000, a velocity of 12,000 can be reached in
120 seconds.) The initial CMOVE and the final PMOVE both use this distance. A second
CMOVE “takes over” at the midpoint of the acceleration slope from the first CMOVE and
accelerates to the target velocity of 24,000. A third CMOVE is required for dividing up the
deceleration portion of the profile. The final move, a PMOVE, “takes over” from the third
CMOVE at the deceleration midpoint distance (720,000 user units from the final position). The
third CMOVE, as it approaches its final position, will automatically decelerate to the PMOVE’s
velocity of 12,000. The dashed lines in the Example 2 drawing separate the four moves. To
calculate the distances of the second and third CMOVEs, we subtracted the distances we
calculated for the first CMOVE and final PMOVE (720,000 each for a total of 1,440,000) from
the final distance of 8,000,000. This gave us a remaining distance of 6,560,000, which we
divided equally between the second and third CMOVES (3,280,000 each).
ACCEL 100
VELOC 12000
CMOVE 720000, INCR, LINEAR
VELOC 12000
CMOVE 3280000, INCR, LINEAR
VELOC 24000
CMOVE 3280000, INCR, LINEAR
VELOC 12000
PMOVE 720000, INCR, LINEAR
Figure 7-19. Maximum Acceleration Time Example 2