17 events
when toggle format what by license comment
Sep 9 '13 at 21:13 answer TimWescott timeline score: 3
Sep 9 '13 at 16:43 comment added Robert Seifert I had a long talk with a colleague today, he says, that the full force Fp is affecting the bearing. But we weren't sure, whether the screw is self-inhibiting or not, but that would make a huge difference right? Actually a document which shows me the complete force and torque balance of screw would already help me a lot. And if this document would also explain the differences between self-inhibited to uninhibited - perfect! Already thank you to everybody participating!
Sep 9 '13 at 8:13 comment added Robert Seifert The force is actually not ground related, but for my posted simplified example it is. The roller screw's height is fixed related to rotor and chassis by the bearing, but the whole system can oscillate up and down. And I need to know if the whole system is oscillating (obviously it is) caused by the force Fp and how.
Sep 9 '13 at 2:40 comment added Guy Sirton Also Newton's third law applies. So the answer to your question is basically that in a properly designed screw system there will be forces acting on your chassis at the bearing. The servo motor shaft shouldn't see any forces along your Fp direction it's all taken by the bearing.
Sep 9 '13 at 2:31 comment added Guy Sirton When you use the right kind of bearings and mount them properly (e.g. see here: hiwin.com/pdf/bs/ballscrews.pdf ) your screw isn't going to move much when you apply a force. The load is taken by the bearing (and thus obviously also transmitted through your chassis)... (+ what Tim said about backdriving/friction and if it's a servo it will also fight any backdriving)...
Sep 8 '13 at 20:12 comment added James Waldby - jwpat7 Is Fₚ (the process force that is shown as lifting the mass) also ground-relative? Is the roller screw's height fixed, relative to the frame and motor, but can vary relative to ground as the frame moves up or down?
Sep 8 '13 at 16:18 comment added Robert Seifert okay I added a sketch (I was logged out, don't be confused) - I hope my question is clearer now.
S Sep 8 '13 at 16:16 history suggested CommunityBot CC BY-SA 3.0
adding sketch
Sep 8 '13 at 16:15 review Suggested edits
S Sep 8 '13 at 16:16
Sep 6 '13 at 19:59 comment added ddevaz 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.
Sep 6 '13 at 18:21 comment added TimWescott 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.
Sep 5 '13 at 22:37 comment added Robert Seifert 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.
Sep 5 '13 at 21:43 comment added TimWescott 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?"
Sep 5 '13 at 21:17 history tweeted twitter.com/#!/StackRobotics/status/375729444201443328
Sep 5 '13 at 19:08 history edited Ian CC BY-SA 3.0
added 45 characters in body
Sep 5 '13 at 12:59 review First posts
Sep 5 '13 at 18:50
Sep 5 '13 at 12:40 history asked Robert Seifert CC BY-SA 3.0