Did I miss something? I specified a 1" spool, mind you I screwed up and what I looked at was a 1" fulcrum, ie radius.
That would be 2*2*Π (circumference) / 200 (steps per rev) 10 (gear ratio) = 0.003"/step=159 steps/½"=2.6 steps/second
On further thought, you would need a solid frame to bare the weight since 30Lb would be too much load on the bearings of such a small motor. If it is indeed a simple lift, the rate and force seems like the telescope thing restated, then a counter balance could eliminate the gear ratio since then all the motor needs to do is overcome friction and inertia.[/QUOTE
Do you agree that the original specification was to have the output velocity be 0.5 in. per minute? .... That is how I read it.
Then, for a 'large' output gear radius of 10", v=r*ω, so 0.5 in/min=10" * ω, and the output gear rotational velocity is therefore 0.05 rad/min.
If this is correct, the smaller gear rotational velocity is 10 times the large gear rotational velocity...just the ratio of the gear diameters ..... or 0.5 rad/min...which is the stepper motor speed...
The motor specification is 200 steps/2*pi rad ....or 31.8 steps/rad.
Therefore the motor would turn at 0.5 rad/min * 31.8 steps/rad = 15.9 steps/min, which is equivalent to 0.26 steps/sec...or about 1 full step every 4 sec.
So your stepping rate is 10 times more than mine ..... Apparently we differ on whether to divide or multiply by the gear ratio....I am saying that the rotational rate of the larger output gear will be 10 times slower than that of the stepper motor.... or that the stepper motor speed will be 10 times faster than the output gear.
In any event, this set up would not be a smooth rotational rate, and would depend on the application as to its acceptability. A 20 in dia. output gear might be difficult to find, or justify the cost....As you mentioned, the present considerations do not include friction losses, or safety margins...or transient effects.