Hi,
Don't you have to change the value of the motor adjustment according to the unit?
No, you do not change the motor settings with a change in units of a job.
Choose your machines native units. These units will be what Mach uses internally. You adjust the steps per unit value to suit your chosen unit....and then leave it alone
for the life of the machine. You should never need touch them again.
The ancient Greeks used a unit called a cubit, approximatelty the length of your arm from elbow to wrist. Not very scientific but it might help to illustrate how Mach works.
Lets say that Mach had the opportunity to use cubits as native units. Further lets say that the machine takes 3427 steps to travel one cubit. Thats what you would enter in
the motor tuning page.
Lets also assume that mach knows there are 300mm in a cubit or 11.811 inches in a cubit.
When someone gives you a job say in millimeters then put a G21 at the top of the code. Mach now knows to machine a straight line of 78.4mm say, but it also knows there
are 300mm in a cubit, its native unit, so 78.4mm/300=.261333 cubits and so the trajectory planner knows to issue 0.2613333 x 3427 =895.589 steps. As the trajectory planner output is integer
it will be rounded to 896 steps.
Now let guess the same job was given to you but in inches, so 78.4mm becomes 3.0866 inches. Mach calculates 3.0866/11.811=0.261333 cubits or 0.26133 x 3427= 895.589 steps just as before.
Mach converted whatever number its was given to its native units, and then calculates the number of steps required. It does this internally.....so really it does not matter what native units you use.
My machine has metric ballscrews, 5mm pitch, so it makes sense to use metric native units. My servos are set up to require 5000 steps to rotate precisely on turn, and the axis would move precisely 5mm.
My steps per unit is 5000/5 =1000 steps per unit. So easy. If you had inch ballscrews, say 0.200/rev or 0.25/rev then using inch units would make sense. Does not matter to Mach, it does not get
confused like you or I.
Craig