## The kinematics of the linkage |

To analyze the movement of the kaleidocycle in a first approach we just have a look at the underlying linkage. Due to the 3-fold rotational symmetry only two angles t1 and t2 do occur. Our goal is to find the relation between t1 and t2. Now if we want to map h6 onto h1, we rotate along the z-axis by angle t1, rotate along the x-axis by angle ts and do a translation along the x-axis by 1. If we repeat this procedure 3 times, t1 is again in its original position. This we can exploit to find the relation between t1 and t2. We define the rotation along the z-axis by angle w in homogenous coordinates:
The rotation along the x-axis by w and the following translation in x-direction by 1 is given by
As explained above id must be the Identity matrix.
(-1+Cos[t1] Cos[t2]-Cos[ts] Sin[t1] Sin[t2]) (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
Cos[t1] Sin[t2/2] Sin[ts] (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 From both relations together we can conclude that 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 In [3] we find the relation Cos[ts]^2+ Sin[ts]^2 (Cos[t1]+Cos[t2])+(1+Cos[ts]^2)Cos[t1]Cos[t2]-2Cos[ts]Sin[t1]Sin[t2]=0 in [5] we find (Cos[t1]-Cos[t1]Cos[t2]-1+Cos[t2])Cos[ts]^2+2Sin[t1]Sin[t2]Cos[ts]-(Cos[t1]+Cos[t1]Cos[t2]+Cos[t2])=0 All three relations show the same solution space for a given skew angle ts when visualized. |

We use our relation for the visualization of the relation between t1 and t2.
The smallest possible skew angle ts is π/3 and gives
The hinge angles periodically change from −π to π and back to -π. ts = π/2 can be seen as the largest possible skew angle because ts and π − ts just belong to mirrored versions.With a skew angle ts = π/2 the faces of the diphenoid are isosceles triangles and this arrangement is used in almost all the models in Schattschneider [1]. Here we get:
We will show soon that the hinge angles t1 and t2 are in the range −2 π/3 and +2 π/3 in this case. And finally here is a plot of the general case for values between the two extremes: During a complete cycle of the kaleidocycle, the point P(t1|t2) performs a complete revolution on the "ellipse". The maximum of the hinge angle is somewhere between 2/3π and π. The knowlegde of this maximum angle is important to decide if the kaleidocycle is physically realizible if disphenoids are used instead of the linkage bars. |

References 1. Doris Schattschneider and Wallace Walker, |

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