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TR-Electronic GmbH 2007, All Rights Reserved
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Page 182 of 260
TR - EMO - BA - DGB - 0013 - 00
05/16/2007
Examples:
In the following examples, the default values for speed encoder resolution and
position encoder resolution are used as the basis.
1. Ramp calculation:
•
Gear reduction = 1:1
•
Covered distance s = 700000 degrees
•
Acceleration from standstill = 150 rpm/sec
•
Further acceleration from 150 rpm/sec to 2640 rpm, then constant
•
Deceleration = 150 rpm/sec, until stationary
How long does the ramp processing take, and which sections are covered ?
The speed factor results from Formula 9.
The acceleration phase and the braking phase each last
t
acc
= t
dec
= 2640/150 = 17.60 sec.
VelocityToPositionUnitFactor = [(2
31
/300000)* 2
16
]/[ (2
16
/360)* (2
31
/5000)]
= 5000*360/300000 = 6,
AccelerationToPositionUnitFactor = 6.
The speed of 2640 rpm is as follows in position units:
v
end
= 2640*6 = 15840 degrees/sec.
The acceleration and deceleration of 150 rpm/sec results in
a = b= 150*6 = 900 degrees/sec²,
During the acceleration phase, the distance covered is
s
a
= 0.5* a* t
acc
2
= 0.5*900*17.6² = 139392 degrees.
The deceleration distance is likewise
s
b
= t
dec
* v
end
– 0.5* b* t
dec
2
= 17.6*15840
– 0.5*900*17.6² = 139392 degrees.
Consequently, for travel at constant speed v
end
s – (s
a
+ s
b
) = 700000 – 2*139392 = 421216 degrees still remain.
This phase therefore lasts 421216/15840 = 26.6 sec.
In total, the ramp is (theoretically) traveled in 2*17.6+26.6 = 61.8 sec.
