Re: formula ?

Hi Jon,

great explanation. I couldn't get the concept that the lead of the
screw was part of the circumference part of the equation.

OF course that makes sense, a 1/2-5TPI 1/2 screw or a 2"-5tpi will
still move the table one inch for 5 revolutions. so the actual
diameter or circumference of the screw is not important.

I noticed that we rate steppers in oz/in but power in inch/lb.
Can I assume that 16oz/inch is the same as 1in/lb?

That would make a 50oz/in stepper 3.125 inch/lb. Fits close to your
10 inch /lb value. Yours seems to be 150oz/in.

so my 50in/oz at 2:1 is 6.25 in/lb on the leadscrew. The 13tpi
(0.0769"circumference/0.02448dia/0.012243radius) then 6.25 / 0.012243
= output force

6.25 / 0.012243 = 510 pounds of force. That kind of force will bend
the noodle of a leadscrew I am using.

What this means is that I have a enough room to trade power for speed.

Thanks again Jon and all.

Dave

--- In CAD_CAM_EDM_DRO@y..., Jon Elson <elson@p...> wrote:
> docholliday01201 wrote:
>
> > once again, who knows the formula for converting rotational
torque at
> > the motor-to linear force at the table ??
>
> Its really simple. Consider the leadscrew to be a drum pulling a
string,
> where the circumference of the drum is equal to the lead of the
screw.
> If you have a 5 TPI screw, that is the equivalent of a drum with
circumference
> of .200 in, or .0637" diameter or .0318" radius. If the motor
produces 10 In-Lb,
> that is the same as 10 Lbs pull at a radius of one inch. This will
translate
> to 10 / .0318 = 314.16 Lbs linear force on the table! Pretty
amazing, these
> ball screws!
>
> Jon