## Some exceptional point in the solution space |
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We substitute {Cos[t1] -> x1, Cos[t2] -> x2} in our equation for t1 and t2 to get a simpler looking expression:
2 Cos[ts] (-2 Sqrt[1-x1^2] Sqrt[1-x2^2]+Cos[ts])+x1 x2 (3+Cos[2 ts])+2 (x1+x2) Sin[ts]^2==0 Then we solve this eqation for x2 and resubstitute
{{Cos[t2]->-((2 ((Cos[t1] (3+Cos[2 ts])+2 Sin[ts]^2) (Cos[ts]^2+Cos[t1] Sin[ts]^2) − 4 Abs[Sin[t1]] Cos[ts] Sqrt[2+Cos[2 ts]+2 Cos[t1] Sin[ts]^2]))/(3+Cos[2 ts]+2 Cos[t1] Sin[ts]^2)^2)}, {Cos[t2]->-((2 ((Cos[t1] (3+Cos[2 ts])+2 Sin[ts]^2) (Cos[ts]^2+Cos[t1] Sin[ts]^2) + 4 Abs[Sin[t1]] Cos[ts] Sqrt[2+Cos[2 ts]+2 Cos[t1] Sin[ts]^2]))/(3+Cos[2 ts]+2 Cos[t1] Sin[ts]^2)^2)}} We can immediately get an expression for the maximal angle of t1.
Minimum and maximum for t1: In [5] we find the same formula written as . The value for t2 if t1 is minimal/maximal computes to ± ArcCos[7 - 6 Csc[ts]^2]. Due to the symmetry of the graph the roles of t1 and t2 can be reversed obviously. We define |
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For the intersection with the t1-axis we have t2=0 and Cos[t2]=1. With both solutions sx[[1]] and sx[[2] we get the same result.
This is interesting because the result is independent of the skew angle ts. On the other side it is quite logical, because t2=0 means that our linkage is an equilateral triangle.
Again t1 and t2 are interchangeable due to symmetry reasons. |
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References 1. Doris Schattschneider and Wallace Walker, |
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