Re: Ball screw reduction ratio

"Padu" <padu@xxxxxxxxxxxx> wrote:

How can I calculate the reduction ration of a ball screw?

For example, let's say I have a DC motor spinning at 3000RPM, with a maximum
of 2000oz-in torque. If I couple the motor to a ballscrew that is 200mm (8")
long and has 6mm (0.236") screw lead, it's intuitive that if 3000RPM is
mantained, then the screw will travel throughout its extension in 0.66
seconds.

Let me just double check my math... 3000RPM = 50 revolutions per second. The
ball screw needs close to 33 revolutions (200mm/6mm per rev.) to travel the
200mm, therefore 33/50 = 0.66s. Therefore 3000RPM is roughly equivalent to
303mm/s.

What about torque?

Let's see if I can remember any mechanics...

Force, rather than torque (as the output is linear rather than
rotational). You need to know the diameter of the screw as well as the
pitch (screw lead). That'll let you work out the angle of the threads
to the axis of the screw, and convert the torque of the motor into a
force at the edge of the screw. You already know the input force is
normal (at right angles) to the axis of the screw. The reaction (as in
Newton's 3rd law) of the screw to the input force is normal to the
threads of the screw. The output force is along the axis of the screw.
Now you can draw a diagram like this:


F _
|<-----/|
| /
| I /
| / R
| /
| /
V

I (down) is your input force. R (up and to the right) is the reaction,
normal to the screw threads. F (left) is the output force, along the
axis of the screw. (I hope the diagram makes sense, it's a bit tricky
with ASCII).

Plug in I (calculated from your motor torque and the radius of the
screw), the angle of R to I (from your screw diameter and pitch), and
do a bit of trigonometry to work out F, your output force. This
neglects friction of course, but I expect you can find a typical
efficiency value for ball screws on the web somewhere, and multiply
your answer for F by that.


Tim
--
Did I really still have that sig?
.

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