Re: Elastic belt stretch, lead screw twist

Brad:

See below. :-)

At 02:44 AM 4/9/01 +0000, you wrote:

> > I think there is a 2*pi missing here. It should follow the power
>screw
> > equations which give (for no friction):
> >
> > T = (L*W)/(2*pi) = torque in lead screw.
> >
> >
> > Here L is screw lead (.2"), W is table load (500 lb), pi is 3.14159.
> >
> > With this equation the torque on the lead screw is 15.9 in-lb.
>
>I was making the assumption that the lead screw acted at a
>lever/wedge. In other words, a triangle of 0.2" and 1" sides, with
>the screw the hypotenuse. In this analogy, the screw force was
>multiplied by the ratio of the sides. From your comments, I gather
>that this was a false assumption of how screw force's work.
> However, if it only takes 15.9 ftlbs to drive a 500lb load (plus
>that pesky friction) it seems that we would all be using smaller
>motors on our machines.

No, you are correct. With friction ignored the forces are directly related
to this triangle. However the triangle is developed by unwinding one wrap
around the lead screw. In this case one side is 0.2" and the other pi*d,
the circumfrence at the pitch diameter. The 2 in the 2*pi comes from
converting force at a radius of d/2 to torque.

>You're correct, in keeping things simple, I used a 1" pulley, the
>same as the lead screw pitch diameter, probably smaller than that
>most people use. Your 3 " pulley would have 1/3 the load, a better
>design.

Better? It's all trade-offs. The calcs show that accuracy isn't really
affected by the belt drive so all the other factors get involved. The 1"
belt should counter any problems with a "small" pulley.

>Are you sure that the 2pi is in the force calc, and not a power calc?
>(I don't have any of my books nearby, but it just seems that 15inlb
>won't really generate 500lb of thrust on a 0.2" lead screw.)

I believe this is a torque calculation but it always makes me feel better
when someone can verify my calcs. Please let me know if you can verify
or refute them.

From the other direction, 15 in-lb is 240 oz-in. My understanding is that
steppers of about 650 oz-in or so are appropriate for knee-mill sized
machines. I think the oversizing of 650/240=2.7 times is a safety factor
against lost steps. I believe most use a 2:1 belt drive which would give a
5.4 factor. The 500 lb is about the max steady state table load generated
by a 2 HP machine but dynamic loads could increase this. The 5.4 factor
seems reasonable as a safety factor here since we don't want the
possibility of loosing steps under any conditions. But I have no rule of
thumb for this. If anyone does I would like to hear.

Thanks. :-)

Hugh Currin
Klamath Falls, OR