Screw Drive Torque
Screw Drive Torque
(OP)
Hi all,
What is a formula to figure out how many oz-in is required to drive a certain screw. How many oz-in is required to push a 5TPI screw?
Thanks. :)
What is a formula to figure out how many oz-in is required to drive a certain screw. How many oz-in is required to push a 5TPI screw?
Thanks. :)





RE: Screw Drive Torque
To answer your question we need a lot more information for
instance size of screw, load on screw, type of thread, how fast you want the screw to turn, screw material and mating material.
regards desertfox
RE: Screw Drive Torque
say a 5TPI ball screw pushing 100lbs, at 1000RPM on precision ground ball screw and ball nut.
Thanks. :)
RE: Screw Drive Torque
What are the materials involved and what diameter is the 5tpi thread.
regards desertfox
RE: Screw Drive Torque
For the steady-state condition, the exercise boils down to some simple triganometry and a few assumptions regarding friction.
For the start-up condition you need to work in F=ma.
Desertfox, diameter is irrelevant.
RE: Screw Drive Torque
I cannot find a formula for working out the torque to either
raise or lower a load on a thread without the diameter or for that matter do the trignometry that you mention in your thread.
Perphaps you can show us how you do the trignometry without
knowing the thread diameter.
regards desertfox
RE: Screw Drive Torque
T = F [(p/2?)+(µRt)]
Where
F = desired force
p = thread pitch
µ = coefficient of friction
Rt = mean thread radius
For threads plus nut:
T = F [(p/2?)+µ(Rt+Rn)]
Where
Rn = mean nut bearing radius
RE: Screw Drive Torque
RE: Screw Drive Torque
I stand corrected. Diameter is helpful.
Shigley and Mitchell's Mechanical Engineering Design (4th ed.)provides an example of torque required to drive a power screw in chapter 8.
RE: Screw Drive Torque
Thought I was missing something simple for a minute
RE: Screw Drive Torque
Once the problem is reduced to an inclined plane, there is no need for diameter. However the diameter is necessary to convert from thread pitch to incline angle.
RE: Screw Drive Torque
For your stated problem, the traveling nut on the screw is moving at a constant linear velocity because the screw is turning at a constant angular velocity, ie. 1000 RPM.
The traveling nut is pushing at a constant 100 lb thrust.
Now assume the system is lossless, so work, energy is conserved. For 5 turns of the screw the advance is 1 inch = 5TPI. For rotation Work = T x Theta where T = torque, Theta = radians. For linear motion, Work = Force x Distance.
In this case when the work out = 100 lb x 1 in = 100 in.lbs. at the same time work in = T x 5 x 2 PI, so:
Torque = 100/10PI = 3.1831 in.lbs.
For precision ball screws, one can expect around 90% efficiency (10% loss) so the input torque would probably be closer to:
3.1831 x 1.1 = 3.5 in.lbs.
In other words, the torque(T) thrust(W) relation for a lossless screw system is T = func(TPI) x W