I have a guy who wants me to build a gantry. The problem I have is that the Z axis will be in the 250 lb range. If I use a 10 TPI lead screw driven by a stepper with 10:1 gearing what will my mechanical advantage be? What size motor would I need to drive Z at 50 IPM?

Go to www.nookind.com and look in their acme catalog. For each of their screws, they give you how much torque is required to lift one pound. Divide this by 10 to allow for your gearing.

They also have a ballscrew section, though their load calculations do not include inertia considerations due to acceleration. Depends on how fast do you want to get up to 50"/min?
Al.

Steppers run much better at low speeds. I have 2 800 oz/in steppers on 1/2" ballscrews with a pitch of 5/in. My gantry is made of aluminum, around 125 lbs. I microstep at 2000 p/r, or 10000 p/in. I max out at around 130 i.p.m. but that is not motor inability rather the limits of my computer. The motors stall at the exact same speed when disconnected from the screw, and the speeds were the same with one motor. The second motor was more for maintaining squareness/rigidity than increasing travel speed

I just found this:

http://www3.wcu.edu/~ballaaron/met366/modules/module8/mod8.htm

About 1/3 of the way down the page there's a stepper-leadscrew example using 250 oz/in.

My 272oz/in on a .1 pitch 1/2 ich leadscrew works out to:


272 oz/in / .25 in = 1088 oz
1088 oz / 16oz/lb = 68 lbs
68x2x3.1416x.25 / .1 = 1068 lbs

Advantage = 1068 / 68 = 15.7

ACME is roughtly 30% efficient I think (variable due to loading methods and nut types/tightness etc) so that gets knocked down to about 4.7. So I can get roughly 1278 oz/inch at the axis (6.66 ft/lbs).

At least, thats the way I read it ;)