acceleration, was:Re: Another router video..

Marliss,
Just for fun I ran these numbers on your motor:
top speed 50 inches/second
travel 24 inches
time 0.5 second
Assuming constant acceleration up to 50 inches/second, then constant speed
after that, two equations can be written:
1/2 a t^2 + 50(1/2-t) = 24 and a t = 50.

Solving these simultaneously gives a = 1250 inches/second^2 and t = .04
second.
1250 inches/second^2 is about 3 g. So the motor accelerates for 40 ms at 3
g and then coasts at constant speed for the remaining 460 ms.

If the size of the screw and the mass of the stage it was moving were known,
it would be easy to calculate how much power was being applied by the motor.
This would say something about the efficiency of the drive if the input
power were known.

If you kept the acceleration constant at 1250 inches/second^2, the stage
would travel the 24" in 200 ms, reaching a velocity of 245 inches/second or
14700 inches/minute.

regards-
Elliot

Date: Thu Jun 1, 2006 9:41 pm (PDT)
From: "Mariss Freimanis" mariss92705@...
Subject: Re: Another router video..


I agree with what you say. Someone's demand to explain how you do
something is not your command and obligation just because you're
asked. You are the one giving as you chose; the one asking is just
that, asking. What's he giving in return?

Everyone gets excited by high-speed anything; it's not a big deal.
Today I was running experimental PID closed-loop steppers from the
G100. Call them 50-pole AC servomotors now with the loop closed.:-)

The motor is on a 24" linear stage (1 TPI screw). The speed was 50"
per second (3,000 IPM at 3,000 RPM) while operating closed-loop. The
motor is NEMA-23, 3A/phase, 24VDC supply and a G201 drive. Not
pushing the motor hard at all.

It is kind of fun to move 24" out and back in less than a second. If
I have the time and people want to see it, I'll take and post a
video. 3,000 IPM steppers.

Mariss
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