I bought a farm jack a few years ago but solved the problem in another way before it arrived so I had a dusty new jack in the corner of the garage. The column is a beefy little castellated I-beam. I looked and they have a different column design now so I don't know if it would still slide in there. I was worried that lifting it with a pair of T-nuts might put too much focal pressure on the slots since the table weighs over 400 lbs. I know I wouldn't wanted to be lifted that way.
adh2000 in Waukesha, I grew up in Oak Creek and was fixing circuit boards at the GE Medical Systems plant at the Hwy 16 exit around the time this mill was produced.
The fish scale I used definitely isn't an optimal solution, just the only one I could come up with on the island. It updates relatively slowly and if the tension isn't increasing fast enough it locks the current measurement. I just have regular torque wrenches, none with digital output, always looking for a reason to buy new tools though.
TNB thank you for taking that measurement. That really helps. What weight increment were you using? Wondering what the last weight was that would not overcome static friction.
I took a series of measurements, not just the one I reported. Initially I made the measurements with blue painters tape wrapped around the rim of the wheel. It wasn't strong enough to turn the wheel once the rotary table was on though. I measured the force (in lbs) at the rim necessary to raise and lower the Z axis when there was no table on the machine, then with the rotary table on, then with the table and a Kurt vise.
Fup Fdn u * Cotangent (A)
No table 9.6 5.5 3.7
table 18.5 8.6 2.7
table + vise 21.0 9.0 2.5
u is supposed to be a "mu" for static coefficient of friction
A is the lead angle of the screw with Cot(A) being an unknown constant.
The screw is treated as an inclined plane initially. That gives a value for the term
u * Cot(A) = (Fup + Fdn) / (Fup - Fdn)
Measuring the force required to go up and down at each weight allows elimination of the weight of the table from the equation, which is unknown for the vertical table that is part of the machine. So the term proportional to the static coefficient of friction appears to decrease with increasing weight. That shows the model is too simple, but it is still useful for understanding the system. The apparent coefficient would decrease with increasing weight if there was another term that became less significant with increasing table weight. I think that would be the case for force necessary to overcome static friction when force on a sliding surface is independent of the weight on the table. I have never seen a jib but I imagine that "tightening the jibs" that increases the force necessary to turn the wheel may fit into the model with the addition of this term. This may all be a bit of nerdy silliness but having a reasonably accurate model of the knee would allow comparison between machines with different tables.
I was struggling with the algebra to calculate that term when the rotary phase converter unexpectedly arrived yesterday. I wasn't prepared with all the odds and ends to wire it in so I ran around town and found parts to get it going. The good news is that the mill and DRO readout for 3 of the 4 axes work. The bad news is that the readout for the rotary table doesn't. The external cable from the table goes through a hole in the base to connect to another cable. That was disconnected and the seller said he didn't disconnect it. The wire at the internal connector looks damaged so it may have been pulled apart by the cable. Simply mating the connectors did not work. Hopefully the connector just has to be repaired.