Netmos 9865 pci parallel port help
28 Jun 2016 15:25 #76745
by BigJohnT
Maximum speed is related to the machine, helps to know what the maximum practical RPM is on the axes motors and the gear reduction. With that info you can calculate theoretical maximum speed. Maximum acceleration in my limited experience falls between 10 and 20 times maximum speed. So for an example if your maximum speed is 5 a good place to start with maximum acceleration is 50. I'm not saying these are sane numbers for your machine just an example.
JT
Replied by BigJohnT on topic Netmos 9865 pci parallel port help
What is a good starting point for max_acceleration and max_velocity? Thanks.
Maximum speed is related to the machine, helps to know what the maximum practical RPM is on the axes motors and the gear reduction. With that info you can calculate theoretical maximum speed. Maximum acceleration in my limited experience falls between 10 and 20 times maximum speed. So for an example if your maximum speed is 5 a good place to start with maximum acceleration is 50. I'm not saying these are sane numbers for your machine just an example.
JT
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- Todd Zuercher
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28 Jun 2016 16:11 #76750
by Todd Zuercher
Replied by Todd Zuercher on topic Netmos 9865 pci parallel port help
That may be true for most metal working machines, but not for most wood working machines where feed rates and max speeds can be very high, but accelerations must be lower because of weaker structures, and lower torque/vel ratios. Often the acceleration to max velocity ratio will be between 10:1 and 1:1. On these sorts of machines max speeds are usually over 500ipm and often over 1000.
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28 Jun 2016 19:05 #76761
by PCW
Replied by PCW on topic Netmos 9865 pci parallel port help
The way I think about this is: Time to full speed = V / A
so if you have A as 10X velocity, that means you get to full speed in 1/10 of a second
As the machine velocities gets higher you typically need to reduce the acceleration as Todd mentions
A slow mill getting to 3 IPS (180 IPM) in 1/10 second is only ~.08 Gs
but a fast router getting to 20 IPS (1200IPM) in 1/10 second is 0.52 Gs
so if you have A as 10X velocity, that means you get to full speed in 1/10 of a second
As the machine velocities gets higher you typically need to reduce the acceleration as Todd mentions
A slow mill getting to 3 IPS (180 IPM) in 1/10 second is only ~.08 Gs
but a fast router getting to 20 IPS (1200IPM) in 1/10 second is 0.52 Gs
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