Contributions to Four-Positions Theory with Relative Rotations

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

“Discrete Gliding Along Principal Curves” 14th International Conference on Geometry and Graphics Kyoto, 08/2010

“Future” research v

u r30 r23

r12 r01

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