This is my first post in the CNCzone forums so please excuse the stupid questions... (I do realise that this post could belong in any number of forums)
Background: Recently I purchased a Sherline mill and installed a number stepper motors with a non-gecko driver and the system resonated like crazy above 5 revs/sec (which equates to 12 IPM). I finally bit the bullet and purchased a G540 and now achieving 89 IPM on the rapids (max out Mach3 at 60kHz rate, haven't tried a higher kernel speed yet...)
Getting decent speeds is one thing but under actual cutting load conditions without loosing a step is another matter, hence the issue of load testing (preferably without having to mount up a block of aluminium and making chips).
I have given this some thought and believe I might have a method which is:
a) reproduces the maximum machine load while cutting
b) easy to implement
c) cost effective
As some people might be aware the spindle motor of a Sherline machine is rated for 60 Watts continuous load. Assuming the spindle is running at 1000 RPM with a 1/2" bit the force on an axis can be calculated with a bit of maths:
Power (Watts) = torque (N.m) x rotational rate (radians/sec)
60 = torque (N.m) x (2 x Pi x 1000/60)
Torque = 0.573 N.m
Force (N) = torque (N.m) / cutter radius (m)
Force (N) = 0.573 / 0.00635 = 90.2 N (equiv to 9.2 kg)
Note: This is valid for either the X or Y axis
For load testing I would need to apply this force on the axis under test. Given the force is equivalent to the weight of 10kg, a bucket of water connected to the mill axis by a string over a pulley should do the trick. For a more sizable mill you may need to use an anvil for load testing.
It should be noted that the force is only applied in one direction so this test is somewhat artifical however it does have its advantages. When cycling an axis backwards and forwards it should only loose steps in the direction where the axis is loaded. Hence cycling the axis while lifting and lowering the load will accumulate missed steps in one direction only. A simple dial gauge should be able to pick this up.
Can anyone see any faults in the above test arrangement?
Now for the bit where I am struggling with the number crunching. When I attempt to convert the force into a stepper motor torque I end up with answers that appear to be way too small; using the force calculated in the above case the required torque is 1/30th of the motor holding torque of 2 N.m (270 oz-in). Either stepper motor torque drops dramatically when you get any decent speed or my steppers are seriously oversized.
David Campbell
PS: I am trying to figure out if the current 24 Volt supply for the G540 is enough or go for a higher voltage supply like 42 volts. The Gecko FAQ roughly states higher voltage => higher speed however there is very little in the way of guidelines about where you have gone off into the realm of an over-engineered solution.
EDIT: I thought the thumbs down icon was a question mark. Changed to "No Icon" instead.