On 11 June 2016 at 17:52, <janoscharl...@gmail.com> wrote: > > Yes, exactly, its the linear bearings that can be at different locations and > force therefore the board to different positions, those are the ones that i > am interested in!
Rather than thinking about x, y and theta think about the three pin positions. Give them position vectors r1, r2 and r3. Each is constrained by a linear bearing and so e.g. r1 = a1 + t1*b1 where a1 and b1 are known vectors and t1 is the unknown line parameter. We have then three unknown scalars t1, t2, and t3. Although we don't know r1, r2, or r3 we do know their pairwise distances d12, d13 and d23. This gives three equations e.g. |r1-r2|**2 = d12**2. Substitute for r1, r2 and r3 into those and we get 3 bivariate quadratic equations for t1, t2, and t3. I would expect that sympy can solve that quickly for numeric coefficients (if you can give numeric values for a1, b1, d12, etc.). In your example it's clear from symmetry that if phi is a solution then so is phi+pi so you should expect to get multiple solutions here. This method will also give upside down solutions that you may need to prune. Presumably also you'll need to check the solutions for consistency with the limits on the line bearings. -- Oscar -- You received this message because you are subscribed to the Google Groups "sympy" group. To unsubscribe from this group and stop receiving emails from it, send an email to sympy+unsubscr...@googlegroups.com. To post to this group, send email to sympy@googlegroups.com. Visit this group at https://groups.google.com/group/sympy. To view this discussion on the web visit https://groups.google.com/d/msgid/sympy/CAHVvXxTjyOum-k7Cv-55fTWa8YkV5iokY4uKEnCWHY0Vh0GUtw%40mail.gmail.com. For more options, visit https://groups.google.com/d/optout.
