Rack and pinion spring force?

[adam_m]

How do you determine the correct spring tension for a rack and pinion design similar to CNCRouterparts designs?

Is there a concern and detriment for using too much upward force from the pinion into the rack gear?

Thanks.

Adam,

[wcarrothers1]

I never really could sort out what I considered a good hinge mechanism for the "spring" rack deal. So I just hard mount mine while keeping the alignment as precice as possiable.. Really don't see the point in having spring tension if you kept the tollerance between your linear rail and rack mount precice..

b.

[adam_m]

Originally Posted by wcarrothers1 I never really could sort out what I considered a good hinge mechanism for the "spring" rack deal. So I just hard mount mine while keeping the alignment as precice as possiable.. Really don't see the point in having spring tension if you kept the tollerance between your linear rail and rack mount precice.. b.

So your rack doesn't pivot then, it's a fixed distance from the rack? What about the chance of binding any worries about that?

Adam,

[wcarrothers1]

To me if things are aligned properly in the first place you should not have to worry about binding. My first machine was pretty low on the tollerance on the rack axis and still hard mounted bearing blocks and tension.. ..

First and second machine are hard mounted. Just wanted to say it's not always needed.

b.

[adam_m]

Assuming I were to go with a tensioning system how would one determine the force required?

Adam,

[adam_m]

I can't believe no one else has any ideas on how to determine spring force... thoughts, suggestions, ideas, anyone.???

[ger21]

I'm pretty sure I've seen the formula posted here before. Try the google search.

[ger21]

http://www.cnczone.com/forums/mechan...ar_forces.html

http://www.cnczone.com/forums/72304-post3.html

http://www.cnczone.com/forums/420226-post2.html

[adam_m]

Thanks ger21 exactly what I was looking for except I might need a little help in digesting the information.

So, trying to understand the math and the application here...

From (http://www.cnczone.com/forums/72304-post3.html):

"3) The separating force for 20 degree components is .364 x the drive force applied. Thus if you need 100# drive force, the spring engagement force must be greater than 36.4# in order to keep the pinion engaged."

What determines the driving force that is required? Is that the force of the stepper or the maximum force required before slippage?

How does this apply to a NEMA 380 oz. stepper?

Adam,

[moswald80]

I have been thinking about this lately as I am close to installing the R&P drives on my build. I am using the CNCRouterParts NEMA 23 drives and motors so the way I see it:

380 oz.-in. (motor torque) * 3:1 (reduction [belt from motor to pinion]) = 1140 oz.-in. (neglecting any losses from the belt drive)

1140 oz.-in. (pinion torque) / .5 in. (pinion pitch "radius" [half the pitch diameter]) = 2280 oz. = 142.5 lbs. driving force

142.5 lbs * .364 = 51.87 lbs. max separating force

This seems pretty high to me but it is the theoretical maximum and would likely only happen if the axis was run up against a hard stop with the motor trying to push through it. I would guess that half or less of that force would be required in normal practice. Does anyone have any experience with this?

Thanks,

Mickey

[adam_m]

Mickey,

So does that mean that a spring that has 51.87lbs. force keeps the pinion gear in constant pressure of the rack 100% of the time and would stall the motor before skipping a tooth?

Adam,

[moswald80]

Yes, I believe so.