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I need an equation or a some hints to solve the following problem.

Imagine a roller screw drive. I apply a torque of T to translative move my load mass M. I assume my screw has an efficiency of 90%. Now an additional axial force affects my mass in the opposite moving direction. Is this force completely transformed into torque (of course considering the efficiency) or is it possible, that my whole roller screw is moving, because it is not fixed? I just found papers/books/articles for movable slides/loads, but fixed shafts. But in my case motor and shaft are part of an osciallation system.

I'm not a mechanical engineer, so I'm sorry if the answer may is trivial.

I made a little sketch now enter image description here

The process force Fp is pushing my mass, most of the force is transformed into a load torque Tp which acts against my drive torque TD. Some of the energy is lost by friction. The question is, if there is also a partial force Tp? which is affecting the bearing and therefore exciting my chassis.

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  • $\begingroup$ What do you mean "because it is not fixed?" The roller screw drive should affect the motion between two rigid assemblies, and will be affected by any forces between those two assemblies. If neither of those assemblies is nailed down to the ground, then the motion gets more complex. Which gets back to -- what do you mean by "because it is not fixed?" $\endgroup$
    – TimWescott
    Sep 5, 2013 at 21:43
  • $\begingroup$ exactly that, neither of the two is "nailed" to the ground. The shaft is mounted to a bearing, which itself is part of an oscillating system. And I need to model this complex motion. The question is whether also a translative force is affecting the bearing or if the losses are completely consisting of friction. $\endgroup$
    – Robert Seifert
    Sep 5, 2013 at 22:37
  • $\begingroup$ Oi. Now there's not enough information. But to get back to basics: if the ball screw (roller screw -- whatever) is exerting axial force or has force exerted on it, then that force will show up as torque. Note this isn't the same as a plain old jack screw drive -- in that case, when you try to back drive it, the force will be lost to friction in the nut. $\endgroup$
    – TimWescott
    Sep 6, 2013 at 18:21
  • $\begingroup$ I'm having trouble figuring out exactly what you are asking. A diagram, screenshot, or pictures of the setup would help in understanding your problem. $\endgroup$
    – ddevaz
    Sep 6, 2013 at 19:59
  • $\begingroup$ okay I added a sketch (I was logged out, don't be confused) - I hope my question is clearer now. $\endgroup$
    – Robert Seifert
    Sep 8, 2013 at 16:18

1 Answer 1

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OK. as drawn, ignoring mass and accelerations, the force $F_p$ will appear as a torque on your ball screw.

However, the total force on the ball screw, and hence the torque, depends on the mass of the thing you're moving with the ball screw interacting with gravity (if it's being moved in anything other than a horizontal plane), and on whether or not the whole assembly -- frame and load -- is moving at anything other than a steady velocity.

On a bad day, your mass-spring-damper system will have an overall resonance that interacts with your control system, making oscillations happen where you never expected them.

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  • $\begingroup$ I'm not worried about resonances and I'm aware that there will be no steady speed of my frame. I tried to minimize my problem to a simple case, but actually there are other oscillating parts. And the (oscillating) distance between my mass and the "ground" is quite important, even if it's just some millimeters. At the beginning I thought there is a more trivial answer to my question, thats why I posted it here. I think I'm going to ask some experts at the university, the communication than is maybe easier. Thanks anyway! $\endgroup$
    – Robert Seifert
    Sep 9, 2013 at 22:19
  • $\begingroup$ I understood it now, to honor your effort I edited your answer to be satisfactory and I'll accept it. Finally it was quite trivial, but I haven't seen the point before. $\endgroup$
    – Robert Seifert
    Sep 11, 2013 at 7:59
  • $\begingroup$ Further comments from @thewaywewalk: The point where you get stuck, is that torque is a form of energy while force is a potential. Applying Newton's third law, as suggested, $F_p$ is affecting the bearing with 100% - but $F_p$ is also moving the the mass/screw by $x$ and therefore introducing the energy $F_p \cdot x$, which is transformed into the torque $T_p$ ~ $F_p \cdot x$, which finally brakes the motor. $\endgroup$
    – Ian
    Sep 12, 2013 at 16:57
  • $\begingroup$ @Ian: Huh? Torque happens to have the same units as energy, but it essentially just force in rotation. Just as force must be exerted over some distance to transfer energy, torque must be exerted over some angular displacement to transfer energy. The only difference is that the angular displacement involved is in radians, which is a dimensionless quantity, which is why torque has the same dimensions as energy. But exerting one foot-pound of torque on a stationary shaft for a whole year won't lift a pound of rocks any distance at all. $\endgroup$
    – TimWescott
    Sep 13, 2013 at 17:55
  • $\begingroup$ TimWescott: that text was what @thewaywewalk appended onto your answer. As this was not an appropriate edit, I reverted the changes and moved it to a comment. I take no responsibility for the (in)accuracy of the text itself. $\endgroup$
    – Ian
    Sep 13, 2013 at 18:03

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