1. Bézier surfaces with linear isoparametric lines Imre Juhász, Ágoston Róth Department of Descriptive Geometry, University of Miskolc, Hungary Department of Computer Science, Babes-Bolyai University, Romania

2. We provide control point based necessary and sufficient conditions for (n, m) Bézier surfaces to have linear isoparametric lines.

3. A surface is called ruled surface if it can be generated by a one-parameter family of straight lines.

4. (1,2) Bézier surfaces (2,2) Bézier surfaces

5. Necessary conditions (1) Control points of the u isoparametric line corresponding to v = 0 are ai (0) = bi0 (i = 0, 1,..., n), therefore control points b00,b10,...,bn0 have to be collinear. We will represent its direction by the vector e0 = b10 − b00. (2) Control points of the u isoparametric line corresponding to v = 1 are ai (1) = bim (i = 0, 1,..., n), therefore control points b0m,b1m,...,bnm have to be collinear. We will represent its direction by the vector em = b1m − b0m.

6. Sufficient conditions Lemma 1. If control points b ij (i = 0, 1,..., n) are collinear ∀ j and these lines are parallel, then u isoparametric lines are linear. Proof: Thus control points a i (v) are collinear. ∀ v

7. Lemma 2. If threads of control points b ij (i =0,1,..., n) are collinear ∀ j and ∃ λ i ∈ R (i = l, 2,..., n) for which b ij = b 0j ＋ λ i (b ij 一 b 0j ), ∀ j then u isoparametric lines are linear. Proof: Thus control points a i (v) are collinear. ∀ v This case can be obtained from (1,m) Bézier surfaces by degree elevation.

8. The case n=m=2, α 2, β 2 ∈ R Control points a 0, a l, a 2 are collinear if ( a 2 - a 0 ) × ( a l - a 0 ) = 0. then

9. Since Bernstein polynomials are linearly independent A=0 B+4C=0 D=0

10. (i)α2 =β2 and b01,b11,b21 are collinear,e1=b11-b01 A=0 implies D=0 implies Therefore, either (b21 一 b01 ) = α 2 (b11 一 b01 )and directions e0, el, e2 are arbitrary (cf. Lemma 2), or (cf. Lemma l). (ii) and control points b01,b11,b21 are collinear the direction of which we denote by e1. Due to the conditions A = 0 and D = 0 we have (cf. Lemma l). (iii) Vectors b21 - b01 and b11 - b01 are parallel to the plane spanned by the linearly independent vectors e0, e2,moreover α 2 and β 2 are arbitrary.

11. In this case ∃ μ 1, ν 1, μ 2, ν 2 ∈ R such that b ll 一 b 0l = μ 1 e 0 + ν 1 e 2, b 2l 一 b 0l = μ 2 e 0 + ν 2 e 2. A=0 B+4C=0 D=0

12. Theorem 3. The u isoparametric lines of a (2, 2 ） Bézier surface are linear if and only if one of the following conditions is fulfilled: control points satisfy the conditions of Lemma l (cylindrical surface); control points satisfy the conditions of Lemma 2 (the ruled surface can be obtained from a (l, 2) Bézier surface by degree elevation); vectors e0 and e2 are linearly independent, vectors bil - b0l (i = l, 2) and the rulings are parallel to the plane spanned by e0 and e2 (conoidal ruled surface).

13. The case n> 2, m = 2 Control points a0, al, a2 are collinear if (a2 一 a0 ) × (al 一 a0 ) = 0. then (i)αi =βi and b01,b11,bi1 are collinear,e1=b11- b01,Thus, either (bi1 - b01 ) = αi (b11- b01 )and directions e0, el, e2 are arbitrary (cf. Lemma 2), or (cf. Lemma l). (ii) and control points b01,b11,bi1 are collinear the direction of which we denote by e1. Consequently (cf. Lemma l).

14. (iii) Vectors bi1 - b01 and b11 - b01 are parallel to the plane spanned by the linearly independent vectors e0, e2,moreover αi and βi are arbitrary. ∃ μi, νi ∈ R, from this we get Then (i=1,2,…… ， n) So the surface is conoidal.

15. A (4,2) Bézier surfaces with linear isoparametric lines

16. The case n> 2, m > 2 Control points a0, al, a2 are collinear if (a2 - a0 ) × (al - a0 ) = 0. then

20. Theorem 4. The u isoparametric lines of an (n, m ） Bézier surface are linear if and only if one of the following conditions is fulfilled: control points satisfy the conditions of Lemma l (cylindrical surface); control points satisfy the conditions of Lemma 2 (the ruled surface can be obtained from a (l, m) Bézier surface by degree elevation); det,vector e0 and em are linearly independent, vector bij-b0j (i=2,3,…,n) (j=1,2, …,m-1) are parallel to the plane spanned by e0 and em, moreover coefficients uij,vij of these vectors in the coordinate system with origin b0j and basis vector e0,em satisfy equalitites(23)(24)(25). (conoidal ruled surface) det, the system is solvable and the solution satisfies equality(12).

21. A special singular case (n>=2, m>2) Proposition 5. If,i=2,3,…,n, and the constants appearing in the linear system are given by Then the rank of the matrix is 1 and system is solvable. The control points can be expressed with user-defined vectors e0, em, vj=b1j-b0j, j=1,2,…,m-2, as follows

23. example a (2, 3) ruled Bézier surface is shown which was generated by the following parameters: e0 = (0.30, 0.00, −0.25), e3 = (0.30, 1.00, 0.50), b00 = (−1.00, −1.50, 0.00), b01 = (−1.00, −0.50, 1.00), b02 = (−0.50, 0.50, −1.00)T, b03 = (−1.00, 0.50, 0.00)T, v1 = (1.00, 0.50, 0.00)T, α2 = 3.32, β2 = 1.87, c = 1.00. As parameter c ∈ R\{0} varies control points b21, b12 and b22 move along the curves

25. Thank you!