Your calculations are correct.
PeakVelocity=\frac { 3\cdot \Delta x }{ 2\cdot \Delta t }
AverageAcceleration=\frac { 9\cdot \Delta x }{ 2\cdot \Delta { t }^{ 2 } }
but when using s-curves
PeakAcceleration=1.5\cdot AverageAcceleration
So your system will need to accelerate at a high rate.
So why use s-curve instead of linear ramps?
Because linear ramps require the acceleration to change instantly which is impossible because to change the acceleration the pressure must be increased first which requires flow which requires the spool to move.
Norm, do you know the natural frequency of the actuator and load?
Now about the AccelerationTime=35/\omega
This rule of thumb is bogus yet I have seen professor use it in their documents. There is no justification for it.
Bosch Rexroth has a similar rule where they replaced the 35 by 18. I have an old Rexroth book that says how they came up with this rule of thumb but it is still bogus.
- Newton did not include natural frequency in his laws of motion.
- The rules doesn’t take into account it will take more time to accelerate to higher velocities.
- The rule does not take into account that the surface area on the piston is not the same on both sides. Obviously a hydraulic actuator can accelerate faster in the extend direction because the surface area on the cap side of the piston is greater than on the rod side of the piston.
Force accelerate the actuator and load. The key is to have the force required to accelerate faster than the peak acceleration required by the motion planner. In Norm’s case this is 0.6g. The peak acceleration occurs when the velocity is half way between 0 and peak velocity. The worst case occurs when accelerating in the retract direction so…
Acceleration=\frac { RodPressure\cdot RodArea-CapPressure\cdot CapArea-Friction }{ mass }
The pressures in the cap and rod side of the cylinder can be calculated using the equations from the VCCM equation thread.
PeakAcceleration=\frac { { P }_{ s }\cdot { A }_{ rod } }{ mass } -\frac { \left( { A }_{ cap }^{ 3 }+{ A }_{ rod }^{ 3 } \right) }{ mass } \cdot \left( \frac { { v }_{ mid } }{ { K }_{ v } } \right) ^{ 2 }-g\cdot sin({ 45 }^{ ° })
Here I am assuming the cylinder is pointed down at a 45 degree angle so the retracting motion must lift the actuator and load.
In your example { v }_{ mid }=50\quad mm/sec
I can’t stress too much that thorough knowledge of the VCCM equation is necessary.