Autotransformer question

Well, maybe. I have very underpowered 1/8 Hp continuous motors on my Series I Bridgeport. Yes, it rounds off corners a bit when trying to make a sharp corner at higher speed. I really don't mind that, it rarely can even be detected. I calculated that neglecting losses, I can deliver about 750 Lbs of linear force to the table. It is able to snap off a 3/8" end mill without the servo amps faulting or getting a following error. (That's why I now do setup at lower speeds - the machine is faster than my eyes can detect interference.)

Iggy mentioned 500 - 600 Lbs force, that might be iffy on a Series II.

Jon

Reply to
Jon Elson
Loading thread data ...

I may be mistaken in my calculation. Let me redo. Stall torque of the motor is 3.2 N-m = 2.5 ft-lbs. 2:1 reduction gives me 5 ft-lbs on the ball screw.

T=F*l/(2pi*nu)

T is torque, L is lead (ratio) of the screw, I think 0.1, nu is efficiency. Say nu is 0.8.

F = T*2*pi*nu/l

If ballscrew is 1 inches in diameter, we get

F = 5*2*3.14*0.8/0.1 = 251 lbs.

Now, I contend that 300 lbs is a lot of force when it comes to milling. It is far more than I am able to exert on my manual Bridgeport.

I may have an error in my calculation. I just did an experiment. I ran my X table at 30 volts. It moved at appx. 1 inch per second at 1.45 amps.

I pushed against the table fairly hard (being somewhat space constrained). Maybe I exerted 70 lbs force. The table did not slow down at all (of course) and the amps rose from 1.45 to something like

1.67. i
Reply to
Ignoramus28517

At the maximum handle force I would apply to a 1/2" end mill, a spring scale between the spindle and table reads 60 lbs.

jsw

Reply to
Jim Wilkins

Makes perfect sense to me.

i
Reply to
Ignoramus8895

This is the spring scale, much capacity for little money:

formatting link
You can multiply the range with rope and pulleys at the cost of less accuracy from pulley friction.

jsw

Reply to
Jim Wilkins

These calculations based on slope of the screw thread make my head hurt. A leadscrew can be made equvalent to a drum with a rope around it. For a .200" lead (5 TPI) the equivalent drum is 0.0318" radius. (This assumes the rope is infinitely thin.) Then you can convert torque to linear force very easily. So, 2.5 Ft-Lbs from the motor is 2.5 * 16 =

40 In-Lbs. With a 2:1 belt reduction, that is 80 In_Lbs at the leadscrew. 80/0.0318 = 2513 linear pounds force on the table. I **think** I have done this calculation correctly, I have been using this scheme for a decade.

Jon

Reply to
Jon Elson

Jon, your calculation makes some intuitive sense. Yesterday at night, I realized that I forgot 2*pi in my calculation, which would then yield a number similar to yours.

i
Reply to
Ignoramus8895

I also made a goof, I multiplied by 16 when it should have been 12 to convert Ft-Lb to In-Lb. So, it should give 1885 Lbs linear force.

Jon

Reply to
Jon Elson

PolyTech Forum website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.