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- From: harrison@cs.ubc.ca (James Harrison)
- Newsgroups: sci.physics
- Subject: Statics question (forces, moments...)
- Date: 21 Jan 1993 22:55:47 -0800
- Organization: Computer Science, University of B.C., Vancouver, B.C., Canada
- Lines: 57
- Distribution: world
- Message-ID: <1jo5pjINN6c9@shiner.cs.ubc.ca>
- NNTP-Posting-Host: shiner.cs.ubc.ca
-
- My apologies if this isn't the correct place to ask this, but I've
- been racking my brain over this one for a while now:
-
- I'm trying to modeling a "clock" spring that provides a Moment of
- force through it's axis that is proportional to its radial "winding":
-
- M= K(theta-rested - theta)
-
- Where is K is in Newton Meter / Radians. So as the spring is wound, it
- provides a restoring force. Still with me?
-
- So attach a long "massless" rod that pivots freely about the axis of the
- spring, and has a force applied to the end of it. The moment of the force
- about the axis of the spring is M = r X F, where `X' is the cross product
- between the two vectors r and F.
-
- Now to the simulation, take M, calculate the radial displacement, rotate the
- rod by a fraction of this amount to keep everything "accurate", iterate
- on the new force F that may now exist. So that all works.
-
- So now I have a wound spring, I can calculate the Moment of the spring,
- and using ||M|| = ||r|| ||F||, I can calculate the magnitude of a transverse
- force that must be applied to the rod at a distance r from the axis to
- keep the spring wound.
-
- But I'd really like to have an inverse to the original M = r X F, to
- calculate the 3-space components of the force F, so that I can sum all of
- the forces applied to the end of the rod, and go from there. For
- example if I had the same point connected to two of these rod-spring
- systems, I would want to iterate the system until it relaxed to its
- static configuration. That is one spring's moment, is another spring's
- force (well kind of..).
-
- Ultimately, I even want the rods to be linear springs. This means
- that the length of the rod will change, and it will be necessary to
- calculate the transverse force created by a radial spring, it won't be
- enough to keep a copy of the old transverse force used to calculate
- the winding of the spring. (If you haven't realized it I'm building a
- model of _thin_ deformable surfaces, but I'm not interested in FEMs.)
-
- I know that I can't just try to solve for the XYZ components of F from
- M = r X F, because the matrix formed by the cross product is linearily
- dependent. Is there a simple way around this? I also know that if F
- and M are orthogonal, then one can be "replaced" with the other.
-
- Have I missed something? Do I have to calculate the XYZ components of
- F by using the fact that M and r are orthogonal, and that r and F are
- in the same plane, so that F is orthogonal to M also?
-
- I really want a quick equation (or set of...) that takes the _vectors_ M
- and r, and gives me the _vector_ F. Any pointers?
-
- Thanks in advance,
- James
- --
- -------------------------------------------------------------------------------
- James Harrison@cs.ubc.ca `RPCing as fast as the network will let me.'
-