- Joined
- Feb 1, 2015
- Messages
- 10,092
Today I was making torque and force measurements on my G0602 lathe. To measure torque, I made a 3" pulley to fit on the lead screw input shaft. I wrapped several turns of some light cord, one end secured through a small hole in the pulley and the other end with a loop to attach a 20 lb. digital scale. The scale reading times the 1.5" radius will be the applied torque in lb.-in. and will measure torque from 0 to 30 lb.-in. or 480 oz.-in.. On the other end, I anchored one side a 50 lb. digital scale to the tailstock and the other side to the carriage.
I had previously decided to use the 2:1 gear reduction in the gear box. My first measurement was of the torque required to turn the lead screw. There was some stiction on startup and some variability in the readings but the dynamic torque ranged between 5 and 10 oz.-in. Next was with the half nuts engaged but no load. Here was a bit of a surprise. The required torque was 91 oz.-in. Considering that the stepper that I am using has a torque of around 300 oz.-in. at 350 rpm, that is a lot of overhead.
After some investigation I discovered that the half nuts had a death grip on the lead screw with the additional friction causing the large torque requirement. Clearly, what I needed to do was to limit the travel of the half nut cam. I pulled the apron, prepared to put some sort of a shim in and discovered that there was an M6 threaded hole directly below the thread dial which exited at the face of the half nut cam when the half nuts were closed. Two M6 sets screws, one for the adjustment and the second as a lock, and I have my adjustable stop. With set screws in place, the amount of backlash increased by about 3 thou. The measured torque with the half nuts engaged and no load was 38 oz.-in.. I
t was interesting to find the threaded hole as I can think of no other reason for it other than my use. The set screw(s) are not on the BOM so it isn't a case of parts missed in manufacturing.
Next was to measure torque with a load. I put the 50 lb. scale in place and engaged the lead screw. At 50 lbs. of force working against the carriage travel, I measured 91 oz.-in. of torque. Given that 38 oz.-on. is due to the frictional losses, that nets out at 53 oz.-in. or approximately an oz.-in. of torque for each lb. of resisting force. Assuming that I really do have 300 pz.-in of available torque, I should net out at around 250 lbs. of available force at the carriage.
I had previously decided to use the 2:1 gear reduction in the gear box. My first measurement was of the torque required to turn the lead screw. There was some stiction on startup and some variability in the readings but the dynamic torque ranged between 5 and 10 oz.-in. Next was with the half nuts engaged but no load. Here was a bit of a surprise. The required torque was 91 oz.-in. Considering that the stepper that I am using has a torque of around 300 oz.-in. at 350 rpm, that is a lot of overhead.
After some investigation I discovered that the half nuts had a death grip on the lead screw with the additional friction causing the large torque requirement. Clearly, what I needed to do was to limit the travel of the half nut cam. I pulled the apron, prepared to put some sort of a shim in and discovered that there was an M6 threaded hole directly below the thread dial which exited at the face of the half nut cam when the half nuts were closed. Two M6 sets screws, one for the adjustment and the second as a lock, and I have my adjustable stop. With set screws in place, the amount of backlash increased by about 3 thou. The measured torque with the half nuts engaged and no load was 38 oz.-in.. I
t was interesting to find the threaded hole as I can think of no other reason for it other than my use. The set screw(s) are not on the BOM so it isn't a case of parts missed in manufacturing.
Next was to measure torque with a load. I put the 50 lb. scale in place and engaged the lead screw. At 50 lbs. of force working against the carriage travel, I measured 91 oz.-in. of torque. Given that 38 oz.-on. is due to the frictional losses, that nets out at 53 oz.-in. or approximately an oz.-in. of torque for each lb. of resisting force. Assuming that I really do have 300 pz.-in of available torque, I should net out at around 250 lbs. of available force at the carriage.
