1. ## Mechanical advantage of a screw

I have a 1 3/8 inch acme screw with a pitch of .25 inches. How do I calculate how much linear force it can generate if 1000 inch pounds is used to turn the screw?

2. Simple Trigonometry. You have already defined all of the things you need. If you wanted to get really accurate, then use the Pitch Diameter of the thread, but it will not make much of a difference.

Dave
Generally, I will not give you the answer to your question, but I **will** guide you into discovering how to solve this yourself.

3. Originally Posted by PinkertonD
Simple Trigonometry. You have already defined all of the things you need. If you wanted to get really accurate, then use the Pitch Diameter of the thread, but it will not make much of a difference.

Dave
Generally, I will not give you the answer to your question, but I **will** guide you into discovering how to solve this yourself.
Umm, ok. I never had trigonometry. Can you explain in a little more detail.

4. Well, you are in for a self-search crash course.

Think about one full turn of thread. It has a distance traveled around the perimeter and a distance advanced longitudinally (along the thread). Lay that out flat and you have a triangle which with a little Trigonometry, will provide the means of generating a ratio of forces.

As my signature says, I am not going to do this for you as my view it is much better for you to learn the process for the next time. A favorite old adage, "give a man a fish and he will eat for a day, teach him how to fish and he will eat for a lifetime."

Dave
Generally, I will not give you the answer to your question, but I **will** guide you into discovering how to solve this yourself.

5. Ok, so I need to get the circumference of the rod, 3.14*1 3/8 = 4.3175, multiply that by threads per inch, 4.3175 * 17.27, divide that by the rise, 17.27/1 = 17.27. Is my mechanical advantage is 17.27? Then to get the force I need to get the radius of the rod, 1.375/2=.6875, then divide that by the torque, 1000/.6875= 1454.5454, then multiply that by the mechanical advantage, 1454.5454 * 17.27=25,119. So the linear force is 25,119 pounds? Is any of this correct or is my bad math skills showing?

Dave
Generally, I will not give you the answer to your question, but I **will** guide you into discovering how to solve this yourself.

7. Originally Posted by Kelly Bramble
Tried that and got 4393 pounds which seems much more realistic. However I don't understand what the coefficient of friction at collar is. Is it the friction at the point of rotation, like where the nut hits the material?

8. Originally Posted by 610bob
Tried that and got 4393 pounds which seems much more realistic. However I don't understand what the coefficient of friction at collar is. Is it the friction at the point of rotation, like where the nut hits the material?
Yes. The threads are rubbing each other as the screw's turned.

9. Originally Posted by RWOLFEJR
Yes. The threads are rubbing each other as the screw's turned.
...and at the support point of the nut and base material that the lifting-force is pushing against. Unless of course you are using a roller thrust bearing there where the friction would be negligible.

Dave
Generally, I will not give you the answer to your question, but I **will** guide you into discovering how to solve this yourself.