# Some exceptional point in the solution space

 Our goal is to find the maximal/minimal angles for t1 and t2 and the intersection points with the axes.

We substitute {Cos[t1] -> x1, Cos[t2] -> x2} in our equation for t1 and t2 to get a simpler looking expression:

ex=(Cos[t1] Cos[t2] (3+Cos[2 ts])+2 Cos[ts] (Cos[ts]-2 Sin[t1] Sin[t2])+2 (Cos[t1]+Cos[t2]) Sin[ts]^2==0/.{Sin[t1]->Sqrt[1-Cos[t1]^2],Sin[t2]->Sqrt[1-Cos[t2]^2]})/.{Cos[t1]->x1,Cos[t2]->x2}

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

sx=FullSimplify[Solve[ex,x2]/.{x1 -> Cos[t1], x2 -> Cos[t2]},Assumptions->{0<ts<Pi/2,-Pi<t1<Pi}]

{{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.
t1 is minimal or maximal, when a vertical line only has one intersection with the "ellipse". This means that both solutions for Cos[t2] are the same which only is the case if
2+Cos[2 ts]+2 Cos[t1] Sin[ts]^2=0.

(FullSimplify[Solve[2+Cos[2 ts]+2 Cos[t1] Sin[ts]^2==0,t1]]//Normal)/. C[1]->0

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

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.

Solve[(Cos[t2] /. sx[[2]]) == 1, t1]

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.

References

1. Doris Schattschneider and Wallace Walker, M.C. Escher Kaleidocycles, 1977. ISBN 0-906212-28-6
2. Baker, J.E., 1980. An analysis of Bricard linkages. Mechanism and Machine Theory 15, 267–286.
3. Chen, Y., You, Z., and Tarnai, T., 2005, “Threefold-Symmetric Bricard Linkages for Deployable Structures,”Int. J. Solids Struct.,42(8), pp. 2287–2300.
4. J. Leech, Some properties of the isosceles tetrahedron,Math. Gazette34(1950)269–27
5.Safsten C, Fillmore T, Logan A, Halverson D, Howell L (2016) Analyzing the stability properties of kaleidocycles. J Appl Mech 83:051001.