LynnBlakeGolf Forums - View Single Post - On Forces active in the Golf Swing.... Thread: On Forces active in the Golf Swing.... View Single Post #20 03-19-2009, 02:45 PM drewitgolf Lynn Blake Certified Senior Instructor Join Date: Jan 2005 Location: Massachusetts Posts: 1,334 Force Feed Originally Posted by BBax Just if your interested: Where Vfict (r) is the potential responsible for the centrifugal force: fc = −ω×(ω×r) = (ω2−ω ωT)r . Vfict (r) = −1 2 rT(ω2 − ω ωT)r Now, Rs and RJ are parallel and MsRs+MJRJ = 0. [Note that if a particle is moving then addi- tional Coriolis forces act that are not mentioned in @ @rVeff (r), so we can’t determine stability from Veff . Any questions? Yes Rob, are you sure that your calculation 2 rT(ω2 − ω ωT)r should not be 1.8 rT(ω2 − ω ωT)r ? Just wondering. __________________ Drew Let Your Motion Make the Shot. drewitgolf View Public Profile Send a private message to drewitgolf Visit drewitgolf's homepage! Find all posts by drewitgolf