Four relative displacements: τi,i+1 = αi+1 ◦ α. −1 i α0 α1 α2 α3 τ01 τ12 τ23 τ30. O.
Bottema, B. Roth. Theoretical Kinematics. Dover Publications 1990 ...
Contributions to Four-Positions Theory with Relative Rotations Hans-Peter Schröcker Unit Geometry and CAD University Innsbruck, Austria
3rd European Conference on Mechanism Science Cluj-Napoca, September 14–18, 2010
Outline
Four-positions theory
Homologous points, planes and lines
Motivation and application
Four-positions theory I
Four rigid-body displacements (positions in space): αi : Σ → Σ 0 ,
I
i ∈ {0, 1, 2, 3}
Four relative displacements: τi,i+1 =
αi+1 ◦ αi−1
O. Bottema, B. Roth Theoretical Kinematics Dover Publications 1990
τ23 α3
α2
τ30
τ12 α0
α1 τ01
Four-positions theory I
Four rigid-body displacements (positions in space): αi : Σ → Σ 0 ,
I
i ∈ {0, 1, 2, 3}
Four relative displacements: τi,i+1 =
αi+1 ◦ αi−1
O. Bottema, B. Roth Theoretical Kinematics Dover Publications 1990
τ23 α3
α2
τ30
τ12 α0
α1 τ01
We require that τi,i+1 is a rotation (“rotation quadrilateral”).
Homologous points, planes and lines Find 1. homologous points X0 , X1 , X2 , X3 on a circle, 2. homologous planes ε 0 , ε 1 , ε 2 , ε 3 tangent to a cone of revolution, and 3. homologous lines `0 , `1 , `2 , `3 which form a skew quadrilateral on a hyperboloid of revolution.
Homologous points on a circle (2D)
Σ2 Σ1 Σ3 Σ0
Homologous points on a circle (3D) I
Theorem (B. Roth) algebraic curve of degree six (for screw quadrilaterals)
Homologous points on a circle (3D) I I
Theorem (B. Roth) algebraic curve of degree six (for screw quadrilaterals) points of the relative revolute axes are solutions
Homologous points on a circle (3D) I I I
Theorem (B. Roth) algebraic curve of degree six (for screw quadrilaterals) points of the relative revolute axes are solutions Theorem (A. M. Schoenflies) Four points of a line ` have homologous images in planes =⇒ all points of ` have homologous images in planes.
Homologous points on a circle (3D) I I I
I
Theorem (B. Roth) algebraic curve of degree six (for screw quadrilaterals) points of the relative revolute axes are solutions Theorem (A. M. Schoenflies) Four points of a line ` have homologous images in planes =⇒ all points of ` have homologous images in planes. Theorem Additionally: Three points of ` have homologous images on circles =⇒ all points of ` have homologous images on circles.
Homologous points on a circle (3D) I I I
I
I
Theorem (B. Roth) algebraic curve of degree six (for screw quadrilaterals) points of the relative revolute axes are solutions Theorem (A. M. Schoenflies) Four points of a line ` have homologous images in planes =⇒ all points of ` have homologous images in planes. Theorem Additionally: Three points of ` have homologous images on circles =⇒ all points of ` have homologous images on circles. points of the two transversals u and v are solutions
Homologous planes tangent to a cone of revolution If ε is a solution plane than so are all planes parallel to ε. =⇒ The set of solution planes consists of pencils of parallel lines.
Homologous planes tangent to a cone of revolution If ε is a solution plane than so are all planes parallel to ε. =⇒ The set of solution planes consists of pencils of parallel lines. Transformation of points and planes uTi = uT · Ai−1 ,
u = [ u0 , u1 , u2 , u3 ] T
Homologous planes tangent to a cone of revolution If ε is a solution plane than so are all planes parallel to ε. =⇒ The set of solution planes consists of pencils of parallel lines. Transformation of points and planes uTi = uT · Ai−1 ,
u = [ u0 , u1 , u2 , u3 ] T
Co-punctality condition (degree four) det(u0 , u1 , u2 , u3 ) = f (u1 , u2 , u3 ) + u0 g(u1 , u2 , u3 ) = 0
Homologous planes tangent to a cone of revolution If ε is a solution plane than so are all planes parallel to ε. =⇒ The set of solution planes consists of pencils of parallel lines. Transformation of points and planes uTi = uT · Ai−1 ,
u = [ u0 , u1 , u2 , u3 ] T
Co-punctality condition (degree four) det(u0 , u1 , u2 , u3 ) = f (u1 , u2 , u3 ) + u0 g(u1 , u2 , u3 ) = 0 Cone condition (degree three) g ( u1 , u2 , u3 ) = 0
Homologous planes tangent to a cone of revolution
f (u1 , u2 , u3 ) = 0,
(degree four)
g(u1 , u2 , u3 ) = 0,
(degree three)
Homologous planes tangent to a cone of revolution
f (u1 , u2 , u3 ) = 0,
(degree four)
g(u1 , u2 , u3 ) = 0,
(degree three)
The system has twelve solutions: 1. planes orthogonal to r01 , r12 , r23 , r30 ,
Homologous planes tangent to a cone of revolution
f (u1 , u2 , u3 ) = 0,
(degree four)
g(u1 , u2 , u3 ) = 0,
(degree three)
The system has twelve solutions: 1. planes orthogonal to r01 , r12 , r23 , r30 , 2. planes orthogonal to the transversals u or v,
Homologous planes tangent to a cone of revolution
f (u1 , u2 , u3 ) = 0,
(degree four)
g(u1 , u2 , u3 ) = 0,
(degree three)
The system has twelve solutions: 1. planes orthogonal to r01 , r12 , r23 , r30 , 2. planes orthogonal to the transversals u or v, 3. six pencils of planes whose homologous images are parallel to a straight line (spurious)
Homologous lines on a hyperboloid of revolution
Solution: ` = u, ` = v
Summary: Interesting solutions
“Future” research v
u r30 r23
r12 r01
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“Future” research v
u r30 r23
r12 r01
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