stepper motor inertia

Thanks to Dan Mauch and Alan KM6W I have rotor inertia for some
steppers, I'l use the MCG IH23014 252 oz-in one in this example.
moment of inertia = 340 g cm^2

the moment of inertia of a cylinder rotated around its axis is
I= m r^2
m= pi r^2 l density
density steel is about 7.8 g cm^-2
so I=12.2 r^4 l

Conside a 1/2" diameter leadscrew 1' long,
I=60 gm cm^2 ; this is much less than the rotor inertia

there are some coupling bits of larger diameter, about 1" diameter
and 2" long altogether,
these have I=160 gm cm^2 by the same formulas,
so total is 220 g cm^2 ; this is still less than the motor inertia.

The stage mass has some inertia too, this is coupled to the motor via
the screw thread. Using a bit of algebra (does anyone want to see
this?) I get

I = p^2 m/2pi , where p is the screw pitch and m is the mass moved by
the stage.

The Taig in my example has a 20 pitch screw and I guess a 5 lb mass
(too lazy to weight it).

therefore I = 36 g cm^2

added to the screw and coupling intertia we get I=256 g cm^2, still
less than the motor inertia.

It wouldn't take much for the screw inertia to increase very rapidly,
it goes by the fourth power of the radius.
A coarser lead screw would increase the inertia load of the stage
mass radidly, for example a 5 pitch screw would quadruple it compared
to the 20 pitch.

Screw friction adds a constant load which should just be subtracted
from the motor torque.

Dan, can you suggest a URL for a motor torque curve? If this can be
put into a simple form, it should be possible to write an equation
that will describe the maximum attainable acceleration for any
condition of screw size, pitch, stage mass and screw friction.

This may be of some small use in setting up acceleration curves in
software. Acceleration could be reduced as speed increased, allowing
the greatest possible acceleration at low speeds.

regards-
Elliot