On 07/20/2020 09:00 PM, Tom Smart wrote:
I forgot to ask how would I calculate the loss of abilities of the mill if i
don't drive the servos with their maximum rated power and amps? If I were to
get drives that weren't rated to 180vdc what losses will i have and how do i
calculate the loss?
First, find the rated speed of the motor. If the motor is
rated at 1800 RPM and directly drives a
5 TPI leadscrew, then it will give 360 IPM linear speed at
rated voltage. If you then run the servo
amps at half the motor's rated voltage, you would get half
that speed, or 180 IPM.
Then, if the motor has a peak torque rating at some
amperage, but your servo amp can only deliver
half the rating, you would only get half that torque. if
the motor only shows a continuous torque at
some current, you can guess the peak rating is 2-4 times higher.
I'll use those horrid imperial units, as that's what my
cheat sheet was done in.
If the motor is rated in oz-in, then you can convert that to
lb-ft by dividing by
12*16. So, a 100 Oz-In motor provides 0.52 lb-ft or torque.
A 5 TPI leadscrew advances the axis 0.2 inch per turn. This
is equivalent to a spool with a radius of
0.0318". So, that 100 oz-in motor (0.52 lb-ft) would
produce 0.52/0.0318 = 16.35 lbs of linear force (neglecting
friction).
You can plug in your own numbers to calculate it for your
own motors and servo amps.
So, if your machine has a 200 Lb table, and the leadscrew
were to produce 1000 Lbs linear force,
it would accelerate at 5 G.
Jon
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