Interpolation method for quaternionic-Bézier curves

We study rational quaternionic–Bézier curves in three dimensional space. We construct the quadratic quaternionic–Bézier curve which interpolates five points, or three points and two tangent vectors.


Introduction
We analyse rational Bézier curves with quaternion control points and weights, called quaternionic-Bézier (QB) curves.This class of curves has two remarkable properties: • a QB-curve of degree n can be converted to the classical rational Bézier curve of degree 2n; • QB-curves are invariant with respect to Möbius transformations.
The QB-curves of degree one are circles.They were discussed in [6].Quadratic QB-curves were described in [7,3].However, this description is not convenient, because the middle control may be in 4-dimensional space and is not clear how to use this point in practise for modelling purposes.In this paper we give two interpolation constructions for quadratic QB-curves: -the interpolation curve through five points in R 3 , -the interpolation curve through three points with prescribed tangent vectors at the endpoints.
We use the idea of Anton Gfrerrer [1] to construct a curve for the interpolation method on the hyperquadric.In order to use this we notice the one-to-one correspondence between quaternionic curves and the curves on Study quadric in R 8 (see also [4]).On the another hand, points on the Study quadric can be represented as the displacements in R 3 , while curves on the Study quadrics mean the motion in R 3 .Therefore, QB-curves could be important in kinematics applications (see [2,4]).

Notations and definitions
Let us denote by R, C, H the set of real numbers, complex numbers and quaternion numbers respectively.The quaternions H can be identified with R 4 where r = Re(q), p = Im(q) denote real and imaginary parts of a quaternion q = [r, p].The multiplication in the algebra H is defined by the formula where p 1 • p 2 , p 1 × p 2 are scalar and vector products in R 3 .Let q = [r, −p] means the conjugate quaternion, and |q| = r 2 + p • p = √ q q denotes the length of the quaternion.The multiplicative inverse of q is q −1 = q/|q| 2 = [r/|q| 2 , −p/|q| 2 ], i.e. qq −1 = q −1 q = 1.We identify the set of pure imaginary quaternions with R 3 :
Geometrically, (p; q) ∈ SQ if and only if the vector p is orthogonal to the vector q in the Euclidean sense.The point on the Study quadric c = (p; q) ∈ SQ defines a point in R 3 : because Re(q p −1 ) = Re(q p/pp) = 4 1 q i p i /pp = 0.An important application of the Study quadric is modelling of rigid body displacements.A dual quaternion c = (p; q) ∈ SQ acts on the quaternion x = [0, x 1 , x 2 , x 3 ] by the formula The first part of the above formula pxp −1 means rotation while the second means translation, therefore this action is an element of SE(3), the group of rigid body displacement (see for details [2]).
The rational quaternionic curve c(t) can be defined as: For applications it is important to describe curves in three dimensional space.We observe that a rational QB-curve in 3D is the same as a curve on the Study quadric.

Interpolation construction for QB-curves
Rational quadratic quaternionic-Bézier curves were considered in [7].Our interpolation construction of the QB-curve was inspired by the idea A. Gfrerrer (see [1]) of interpolation points on the Study quadric.The arbitrary point c = (p; q) ∈ SQ can be presented in the homogeneous form c = (p; q) = pq −1 q; q = (a 0 w 0 ; w 0 ), where a 0 = pq −1 ∈ Im(H) = R 3 , w 0 = q, which we interpret as the point a 0 in R 3 with a quaternion weight w 0 .
Let n 1 be an integer, and let T = [t 0 , . . ., t n ] be n + 1 pairwise different real parameter values.Then we define polynomials of degree n with the following properties Let us fix n + 1 homogeneous points (a 2i w 2i ; w 2i ), i = 0, . . ., n and define quaternionic polynomials We are going to find conditions when the curve C(t) = q(t)p(t) −1 is in Im(H).First of all we note, that for arbitrary weights w i , i = 0, 1, . . ., n (because f j (t i ) = 0, j = i).Let S = [s 1 , . . ., s n ] be different (from the set T ) real parameters (i.e. S ∩ T = ∅) and a 2j−1 ∈ Im(H), j = 1, . . ., n be some points in R 3 .We consider the linear system C(s j ) = a 2j−1 , j = 1, . . ., n.
The quaternion linear system (2) with n equations and n unknowns w 2 , w 4 , . . ., w 2n has a unique solution if the corresponding matrix A = (α ij ) is not singular.
Proof.The non-singular quaternion matrix A has an LU-decomposition [5, Theorem 3.1].There exists a decomposition P A = LU and the matrix A is invertible A −1 = U −1 L −1 P .Therefore, the system (2) has a unique solution for the weights w 2 , w 4 , . . ., w 2n .We can take w 0 = 1 and construct the curve C(t).This curve is in R 3 for the set T ∪ S of cardinality 2n + 1.Hence, according to Lemma 1 below, the curve The rational curve C(t) = q(t)p(t) −1 is an imaginary quaternion for all parameters t if and only if Re(C(β i )) = 0 for some real different parameters β i , i = 0, 1, . . ., 2n.
Proof.We present the rational curve C(t) as follows: The numerator q(t)p(t) is the quaternion polynomial of degree 2n.If Re(C(β i )) = 0 for some real different parameters β i , i = 0, 1, . . ., 2n, then the real polynomial Re(q(t)p(t)) of degree 2n is zero on 2n + 1 different parameter values, i.e.Re(q(t)p(t) and discuss the case n = 2 in greater detail.We choose five different points a 0 , a 1 , . . ., a 4 ∈ R 3 .Then the linear system (2) multiplied by 16 is The above formulas for weights are correct if The expression d −1 43 d 32 d −1 21 d 14 is the cross-ratio of four points a 4 , a 3 , a 2 , a 1 , which is a real number if and only if these four points are on a circle (see for example [8]).Geometrically, the inequality (8) means that four points are not circular or they are circular but the cross-ratio is not equal to −3.
For applications it is also important to know tangent vectors.Let C ′ (t 0 ) be the tangent vector at the starting point a 0 and C ′ (t n ) -the tangent vector at the end point a 4 of the curve.If we replace two conditions C(s i ) = a 2i−1 , i = 1, 2, with the conditions C ′ (t 0 ) = v 0 ∈ Im(H), C ′ (t n ) = v 1 ∈ Im(H) we obtain the linear system Eliminating w 2 , w 4 we get the following solution The above formulas for weights are well defined if According to Remark 4.5 in [8] d 42 d −1 20 d 04 is the tangent vector to the circle c 420 through a 4 , a 2 , a 0 at the point a 4 .Therefore, the condition (11) means that the vector v 1 is not tangent vector to the circle through the point a 4 .
Corollary 2. In the case n = 2 the interpolation curve C(t) defined by (1) with three points C(0) = a 0 , C(1/2) = a 2 , C(1) = a 4 in R 3 and the weights w 2 , w 4 defined in formulas (9), ( 10) is in Im(H) = R 3 .Moreover, the curve C(t) has the tangent vectors C ′ (0) = v 0 and C ′ (1) = v 1 in R 3 at the points a 0 and a 4 , respectively.The formulas for weights are well defined if the inequality (11) holds.In particular, if the vector v 1 is not a tangent vector to the circle through a 4 , a 2 , a 0 at the point a 4 then the weights are well defined by (9) and (10).
Then the curve C(t) can be presented in the Bézier from with new homogeneous points (p i u i , u i ) .
We note that the middle control point p 1 usually is not in 3D (see [7]).