Torque to linear force conversion and units

[10mm]

I always get my foot-lbs and inch lbs confused and don't 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.

 

[Simon Bridge]

Good grief - no wonder you get confused.

http://en.wikipedia.org/wiki/Screw_(simple_machine)#Torque_form

 

[10mm]

PLease explain. The mechanical advantage of he thread is correct. If I have a 1.50"-12 stub acme thread.
thd diameter X Pi / lead(distance between thds)
1.50 ∏ / .0833 = 56
which is the same as 2∏r/lead that is shown in the link you provided

My actual question was if I have a ring(or nut) with a 1.50-12 thread and the od of the ring is 1.75", and I want to see what kind of linear force I generate from applying 100ft-lbs to the od of the ring, do I use inch pounds (100 x 12=1200 inch lbs)
.875" is the distance from center or radius
1200 in-lbs /.875 =1371lbs

Then I multiply this by the mechanical advantage of 56, this comes to 76,776 lbs

Im not an engineer. I am not even that good in math. I work in QC and ask too many questions. One of our mechanical engineers at work quickly explained to me why he made a change on his design. He quickly explained it to me. I just wanted to know if I understood it. My question was more about the torque to linear force than anything. The MA created by the thread was just something else that I had to show because it was also a variable. I just like to problem solve to my abilities. I posted this to find out if I totally missed the boat, or if I am close.
Again, all feedback is welcome.

 

[Simon Bridge]

The screw thread acts like a ramp - so a small force can lift a larger force.
Although the applied force gets amplified a great deal, it is at the expense of speed ...

The calculation is correct for the given details.
The link I gave you spells it out.

 

[10mm]

thank you Simon. I should have described the threaded rod sliding inside of another tube. The threaded rod is keyed to the tube so it cannot turn. So when torque is applied to the nut, it pulls the threaded rod thru the outer tube. The force that the calculations were giving me seemed so high (76,776 lbs), I didnt think it could be correct. My confusion was based mostly around if I was using inch pounds and converting the torque to a linear force. I find this stuff to be fun.

 

[Simon Bridge]

It would definitely crack a nut.
Explains why a vice is so strong.

it is pretty shocking - 12 turns an inch is not all that fine ... but I would imagine that applying that torque risks stripping the thread.