prmindartmouth wroteVisualize an idealized imaginary setscrew tightened vertically downwards in the arbor.
A setscrew works by pressing its' tip in compression inward against the shaft. This compressive force (call it "C") is resisted by being shared in upward compression along the spiral bottom surface of the tread in the arbor.
The more tread we have from choosing a longer setscrew, the lower the compressive (normal) force per unit area.
The setscrew does not back out immediately due to friction and the resisting friction force is equal to the coefficient of friction for the steel screw on the steel arbor multiplied by the normal force. Since the normal force is less, the friction force holding the set screw is less, but we have more area, so the total friction force remains the same (equal to "C"). So a longer set screw is not more likely to back out.
In the real world, vibration tries to convert static friction to kinetic. :confused: :confused: Maybe jpg could convert this rambling stream of thought to something coherent. I have an appointment and just ran out of time! :D :D
I think that I agree with that analysis. And you mentioned something that spurred another thought.
For a given "compressive force", the holding torque of the setscrew -- i.e. it's ability to resist being back-driven -- is indeed proportional to the coefficient of friction between the threads of the setscrew and of the arbor. Which can vary wildly, depending on whether the screw interface is lubricated, and with what. As in around 10:1.
Dusty, you might want to try cleaning both the setscrew and the mating arbor threads with acetone or denatured alcohol. That will get you a bare steel-on-steel interface, and maximize the coefficient of friction. It may also require you to torque the setscrew somewhat harder in order to achieve the same compressive force.
I'll be very curious to hear if that helps.
