# vertical rack and pinion self-stopping

There are many places I see a knob used to set the vertical position of an object (like in a microscope stand), but how does it not just simply slide down due to gravity once the knob is released. It's not like they first have to pull/push out the knob before turning it so as to lock it first. Instead once you turn it, it just remains in that position. I believe it's a rack and pinion mechanism, but I don't understand how it locks itself in place after it's been set.

• Friction is the most likely. If you push on the device will it move down again? – Transistor Mar 5 at 13:54
• @Transistor no. It won't move down if I push on it. But I haven't tried using too much force (don't wanna break anything). – Suryetto Mar 5 at 14:21
• Not all R&P are self-stopping. A typical photog tripod will, when the clamp is released, descend to bottom position rather quickly. – Carl Witthoft Mar 5 at 15:24
• Many microscope supports of this type do actually back-drive (i.e. move and cause the knob to turn fast) if you push harder. So it is simply friction in those cases. If you have a right-angle drive of some kind like a worm gear it may be inherently able to avoid this. – Pete W Mar 5 at 17:59
• Some microscopes (i.e. inverted) move only the objective and that sample support stage doesn't move. I suspect that this is to limit focus "creep." The mass of the objective can be partially supported with a spring so the force the rack & pinion sees is just a small part its mass. – D Duck Mar 5 at 23:26

There are many ways this can be achieved, and it will be depended on the actual implementation.

One very prominent example is the lead (or helix) angle in lead/power/acme screws.

If you do the analysis you will end up with the following

[

where:

• T = torque
• F = load on the screw
• dm = mean diameter
• $$\mu$$ = coefficient of friction
• $$\phi$$ angle of friction
• $$\lambda$$ lead angle
The screw is self-locking when the coefficient of friction $$\tan\phi$$ is greater than the tangent of the lead angle $$\tan\lambda$$. In that case, the torque to lower the load $$(T_{lower})$$ will be either zero (barely moving) or negative ( meaning that either you need to apply torque to keep the screw from moving downwards).