- #1

10mm

- 4

- 0

I always get my foot-lbs and inch lbs confused and dont know if I am applying the correct unit. If I wanted to know how much linear force I can create from a given amount of torque I apply to a thread, I know that I need to find the mechanical advantage of the thread and the torque applied. I am just wanting to make sure the torque to linear force conversion is correct so I am going to leave out any friction loss etc..

I have a 1.50-12 threaded ring and say the od of this ring is 1.75". First, I find the mechanical advantage of the thread by calculating (1.50 X Pi)/ thread pitch.

1.50 x 3.1416 / .0833=56 mechanical advantage.

Now, here is where I get confused. Since I am working in inches, do I use inch lbs? If I want to apply 100 foot-lbs torque to the ring, do I use inch pounds and convert the 100 foot-lbs to 1200 inch-lbs?

Now, I am applying a torque of 1200 inch-lbs to a threaded ring with a od of 1.75 which is a .875 radius. If I divide 1200 inch-lbs by .875", I get 1371 lbs. Is this correct? Do I then multiply the 1371lbs by the mechanical advantage of 56 which is 76,776 lbs? This seems very high. I still think I am missing something when trying to get a linear force from a know applied torque.

Again, I am using a threaded ring as a example. The ring can be screwing onto a threaded rod that is welded onto a plate and the ring will butt up against the plate and I simply want to know how much linear force I am pulling on this threaded rod.

Thanks for any input.

I have a 1.50-12 threaded ring and say the od of this ring is 1.75". First, I find the mechanical advantage of the thread by calculating (1.50 X Pi)/ thread pitch.

1.50 x 3.1416 / .0833=56 mechanical advantage.

Now, here is where I get confused. Since I am working in inches, do I use inch lbs? If I want to apply 100 foot-lbs torque to the ring, do I use inch pounds and convert the 100 foot-lbs to 1200 inch-lbs?

Now, I am applying a torque of 1200 inch-lbs to a threaded ring with a od of 1.75 which is a .875 radius. If I divide 1200 inch-lbs by .875", I get 1371 lbs. Is this correct? Do I then multiply the 1371lbs by the mechanical advantage of 56 which is 76,776 lbs? This seems very high. I still think I am missing something when trying to get a linear force from a know applied torque.

Again, I am using a threaded ring as a example. The ring can be screwing onto a threaded rod that is welded onto a plate and the ring will butt up against the plate and I simply want to know how much linear force I am pulling on this threaded rod.

Thanks for any input.

Last edited: