MM Research Preprints, 200–206
MMRC, AMSS, Academia, Sinica, Beijing
No. 21, December 2002
Blending of Triangular Algebraic Surfaces
He Shi and Yongli Sun
Institute of Systems Science
Academy of Mathematics and System Sciences, CAS
Beijing 100080, P.R.China
Email: (hshi,ysun)@mmrc.iss.ac.cn
Abstract. The triangular patches are the most frequently used way to construct the
0
interpolation patches of scattered data. This article applies W u s method to construct
triangular algebraic patches on scattered data, and to construct smooth blending surfaces
of the two adjacent triangular algebraic patches.
Keywords. Irreducible ascending sets, Triangular algebraic surfaces, Smooth blending.
1. Introduction
The scattered data interpolation is a classical research field, mainly dealing with theories
and applications of constructing of the smooth surfaces according to the given scattered
data. The technique has been generally applied to all kinds of scientific research and engineering, such as weather report, geological prospecting, medical science and so on. In 1968,
Shepard discussed the two dimention interpolation function for irregularly spaced data, and
gave the algorithm of the least square distance and weighted interpolation [1] . In 1973,
Bomhill analyzed the interpolation of distributed data from the point of surface shaping
(Barnhill − Birkhof f − Gordon)[2] and gave the BBG form of the triangular. Henceforth,
a lot of professors have given different methods for the construction of interpolation surface
for the scattered data from different point of views. In China, the research for the interpo0
lation of scatterd data started from 80 s, many beautiful results has been reached in both
theoretical and applicational fields,including the research of the theories and applications by
([3] − [18]).
The triangular surface is the fundmental concept for the scattered data interpolation.
For three points P1 , P2 , P3 in R3 , we can construct a triangle whose vertexes are the above
three points, at the same time we define three normals N1 , N2 , N3 at these three points.
We now have a triangular field with definite normals. The triangular patches are the most
frequently used way to construct the interpolation patches of scattered data. This paper applies the method of generic point, founded by Wu([19]), to construct interpolation triangular
algebraic patches on scattered data with definite normals and smooth blending surfaces of
the two adjacent triangular algebraic patches.
Section 2 gives the concept of irreducible algebraic curves and surfaces, and algebraic representation of the generic points; Section 3 presents the main results, i.e. Theorems 11, 14,
Blending of Triangular Algebraic Surfaces
201
and 15. Section 4 gives some discussion and remarks.
2. Fundmental principles and methods
All the following problems are considered in R3 . Suppose that S(F = 0) is the irreducible algebraic surface generated by polynomial F (x, y, z) = 0, the planar irreducible
algebraic curves, generated by the plane h and algebraic surface F , will be represented by
S(h = 0, F = 0). The fundmental method of constructing the scattered data interpolation
surface is the triangulation of scattered data, so the triangular patches are the foundation of
the interpolation surface on scattered data . Suppose that h is a polynomial of x, y, z with
degree 1; i.e. h = 0 represents a plane.
Definition 1 Let S be the surface represented by the polynomial F . If F is irreducible,
then we call S an irreducible algebraic surface.
Definition 2 Let C be a planar algebraic curve generated by plane h = 0 and surface
F = 0. If {Remdr(F, h, z), h} is irreducible ascending set, then C is called an irreducible
planar algebraic curve.
Definition 3 ξ = (x0 , y0 , z0 ) is called the generic point of the planar algebraic curve C,
if it satisfies:
1. ξ on C.
2. For ∀f (x, y, z) ∈ R[x, y, z], if f (x0 , y0 , z0 ) = 0, then f (x, y, z) = 0, ∀(x, y, z) ∈ C.
Let R be the field of real numbers, the irreducible algebraic surface in R3 can be represented by irreducible polynomial P = 0, (P ∈ R[x, y, z]). Suppose that C is the spacial
planar irreducible algebraic curve, defined by the irreducible ascending set IRR = {P1 , P2 },
0
0
and S, S are the irreducible algebraic surfaces defined by the irreducible polynomials P, P .
We have Lemma 4 and Lemma 5 ([19]).
Lemma 4 The surface S contains wholly of C if and only if Remdr(P/IRR) = 0.
0
Lemma 5 Suppose that C is contained wholly in S, S , but is not contained wholly in
the singularity part of them. Let
0
0
D1 = Px Py − Py Px .
0
0
D2 = Px Pz − Pz Px .
0
0
D3 = Py Pz − Pz Py .
0
0
0
0
Where (Px , Py , Pz ), (Px , Py , Pz ) are the correspondent normals. Then S and S will touch
smoothly along C if and only if Remdr(Di /IRR) = 0.
Lemma 6 Let p1 , p2 , p3 be three noncollinear points. Then we have
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He Shi and Yongli Sun
(1) There is an irreducible algebraic surface with degree 2 and two free variables.
(2) There is an irreducible algebraic surface with degree 3 and eight free variables.
Proof. Let
F = a1 x2 + a2 y 2 + a3 z 2 + a4 xy + a5 xz + a6 yz + a7 x + a8 y + a9 z + a10 = 0
G = a1 x3 + a2 y 3 + a3 z 3 + a4 x2 y + a5 x2 z + a6 y 2 x + a7 y 2 z + a8 z 2 x + a9 z 2 y
+a10 xyz + a11 x2 + a12 y 2 + a13 z 2 + a14 xy + a15 xz + a16 yz + a17 x + a18 y + a19 z + a20 = 0
(1) The construction of 2-degree irreducible algebraic surface.
Because a 2-degree reducible algebraic can be obtained as the multiplication of two planes,
so we can add 4 points in the space, through the computation in Maple, we get the 2-degree
surface with two free variables.
(2) The construction of 3-degree irreducible algebraic surface.
Because a 3-degree reducible surface can be obtained as the multiplication of three planes
or a plane and an irreducible surface with degree 2, from (1), we can add 8 points to the
space, then by the computation in Maple, we can construct a surface with eight free variables.
~1 , N
~2 , N
~3 .
Lemma 7 Suppose that p1 , p2 , p3 are three noncollinear points with normals N
~
We need to construct the irreducible surface, passing point pi with normals Ni , i = 1, 2, 3,
we have the following results:
(1) We can construct an irreducible surface with degree 2.
(2) Also we can construct an irreducible surface with degree 3.
Proof. Let 2-degree and 3-degree equations be:
F = a1 x2 + a2 y 2 + a3 z 2 + a4 xy + a5 xz + a6 yz + a7 x + a8 y + a9 z + a10 = 0
G = a1 x3 + a2 y 3 + a3 z 3 + a4 x2 y + a5 x2 z + a6 y 2 x + a7 y 2 z + a8 z 2 x + a9 z 2 y
+a10 xyz + a11 x2 + a12 y 2 + a13 z 2 + a14 xy + a15 xz + a16 yz + a17 x + a18 y + a19 z + a20 = 0
~1 , N
~2 , N
~3 are nonparallel, then we can get nine equations, so we can con(1) If normals N
struct an irreducibel surface with degree 2; when two of the three normals are nonparallel,
we can add two points p, q to the space and they satisfy the conditions: they are not on
~1 , N
~2 , N
~3 , so we get nine equations,
the planes, passing points p1 , p2 , p3 with the normals N
therefore we can get an irreducible algebraic surface with degree 2; when three normals are
nonparallel, by adding 4 points, we still get nine equations, thus we get the required surface.
~1 , N
~2 , N
~3 are nonparallel, we choose three points p4 , p5 , p6 , with normals N
~4 , N
~5 , N
~6 ,
(2) If N
~
~
satisfying that Ni is nonparallel to Nj , i 6= j, i, j = 1, 2, 3, 4, 5, 6. so we get 18 equations,
thus we construct an irreducible surface with degree 3, passeing through point pi with the
~ i and one parameter; if two of the three normals are parallel, we can add 4 points
normal N
~4 , N
~5 , N
~6 , N
~7 , satisfying that Ni is nonparallel to N
~j , i 6= j,
p4 , p5 , p6 , p7 , with normals N
i, j = 1, 2, 3, 4, 5, 6, 7. Then we can construct the correspondent 3-degree irreducible surface;
if the three normals are parallel, in the same way, we get 17equations, with 2 free parameters,
so we can get the required surface.
Blending of Triangular Algebraic Surfaces
203
3. Main results
Definition 8 We call surface F = 0 meets surface G = 0 with C 1 continuity along their
intersection curve X, if F and G have the same partial derivatives at each point along X.
Theorem 9 Let F = 0, G = 0 be two irreducible algebraic surfaces, h = 0 a plane, and
C = S(h = 0, F = 0) an irreducible curve, the following conditions are satisfied:
(1) Remdr(G, {h, F }) = 0.
(2) Remdr(di , {h, F }) = 0, i = 1, 2, 3.
where
d1 = Fx − Gx .
d2 = F y − G y .
d3 = Fz − Gz .
Then the algebraic surface G = 0 meets F = 0 with C 1 continuity along the curve
C = S(h = 0, F = 0).
Proof. For (1) holds, from Lemma 4 we know: algebraic surface G = 0 contains the
curve C = S(h = 0, F = 0); by the definition of C 1 , algebraic surface G = 0 meets F = 0 with
the same first derivative along the curve C = S(h = 0, F = 0), so Di , i = 1, 2, 3 in Lemma 2
turns to the correspondent di , i = 1, 2, 3. From Lemma 5: while (2) holds, algebraic surface
G = 0 meets F = 0 with C 1 continuity along the algebraic curve C = S(h = 0, F = 0).
Lemma 10 Suppose that P1 , P2 , P3 are three points and the definite normals are
~
~2 , N
~3 respectively. Then there exists a 3-degree irreducible algebraic surface F = 0
N1 , N
satisfying:
(1) F (Pi ) = 0, i = 1, 2, 3.
~ i , i = 1, 2, 3.
(2) (Fx , Fy , Fz ) |Pi = N
Proof. The equation of 3-degree surface is
F = a1 x3 + a2 y 3 + a3 z 3 + a4 x2 y + a5 x2 z + a6 y 2 x + a7 y 2 z + a8 z 2 x + a9 z 2 y + a10 xyz + a11 x2
+a12 y 2 + a13 z 2 + a14 xy + a15 xz + a16 yz + a17 x + a18 y + a19 z + a20 = 0
For F passes through the point Pi , we get 3 equations F (Pi ) = 0, i = 1, 2, 3. And from
~ i , i = 1, 2, 3, we get another 9 equations, so we have
(Fx , Fy , Fz ) |Pi = N
(
F (Pi ) = 0.
~ i.
(Fx , Fy , Fz ) |Pi = N
the coeffients of the polynomial F are considered as the unknowns, the number of the equations is 12, the number of the unknowns is 20, so we get the solutions.
Lemma 11 Suppose that F = 0 is an irreducible algebraic patches with degree 3; h = 0
is a plane, and {h, F } is an irreducible ascending set; C = S(h = 0, F = 0) is an irreducible
algebraic curve. Then there exists an algebraic surface G = 0 with degree 3, and G = 0
meets F = 0 with C 1 continuity along the curve C with 4 free variables.
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He Shi and Yongli Sun
Proof. Let
F = a1 x3 + a2 y 3 + a3 z 3 + a4 x2 y + a5 x2 z + a6 y 2 x + a7 y 2 z + a8 z 2 x + a9 z 2 y + a10 xyz + a11 x2
+a12 y 2 + a13 z 2 + a14 xy + a15 xz + a16 yz + a17 x + a18 y + a19 z + a20 = 0
G = b1 x3 + b2 y 3 + b3 z 3 + b4 x2 y + b5 x2 z + b6 y 2 x + b7 y 2 z + b8 z 2 x + b9 z 2 y + b10 xyz + b11 x2
+b12 y 2 + b13 z 2 + b14 xy + b15 xz + b16 yz + b17 x + b18 y + b19 z + b20 = 0
The transversal plane h = 0 is h = ax + by + cz + d = 0. Because G = 0 meets F = 0
with C 1 continuity along the curve C, therefore, from Lemma 1: Remdr(G, {h, F }) = 0.
Thus we get a group of equations with respect to the coefficients of G = 0, i.e. L1 . But from
Theorem 4, we have
d1 = Fx − Gx .
d2 = F y − G y .
d3 = Fz − Gz .
satisfying: Remdr(di , {h, F }) = 0, i = 1, 2, 3, so we get another group of equations L2 with
0
respect to the coefficients of G = 0, i.e. b s. Let L1 , L2 form another group of equations L
with 27 equations, 20 variables. The rank of the coefficient matrix is 16, i.e. there are 4 free
variables, so the algebraic surface G = 0 has 4 adjustable coefficients, we arbitrarily choose
b2 , b7 , b14 , b20 , then we get a correspondent algebraic surface.
Theorem 12 Let F1 = 0, F2 = 0 be two 3-degree irreducible algebraic surfaces, and
they do not intersect. Then we need at most four 3-degree irreducible algebraic surfaces to
blend F1 and F2 with C 1 continuity.
Proof. Suppose that h1 = 0 is a plane, such that F1 = 0 intersects the plane h1 = 0
transversally, and the set {F1 , h1 } is irreducible, then from Lemma 11, there must exist
another irreducible algebraic surface G1 = 0 with degree 3, which meets F1 = 0 with C 1
continuity along the irreducible algebraic curve C1 = S(h1 = 0, F1 = 0), generating 4 free
variables, by using the same method, we get planes h2 = 0, h3 = 0, h4 = 0, h5 = 0, such that
{hi , Gi−1 }, i = 2, 3, 4, 5 are irreducible ascending sets satisfying Gi−1 = 0 intersects hi = 0
transversally, Gi = 0 meets Gi−1 with C 1 continuity along the irreducible algebraic curve
S(hi = 0, Gi−1 = 0), i = 2, 3, 4. And G4 = 0 intersects h5 = 0 transversally, satisfying that
F2 = 0 meets G4 = 0 with C 1 continuity along S(h5 = 0, G4 = 0).
Theorem 13 Suppose that P1 , P2 , P3 , P4 are four noncollinear points. The definite
~ i . Then we need at most six 3-degree irreducible algebraic
normal at Pi (i = 1, 2, 3, 4) is N
surfaces to interpolate the four points, satisfying:
~ i.
(1) The normals of F1 = 0 and F2 = 0 at the above four points are the same with N
1
(2) Fi meets Fi+1 with C continuity, i = 1, 2, 3, 4, 5.
Proof. From Lemma 3, there must exist 3-degree irreducible surfaces F1 = 0, F6 = 0,
such that F1 (Pi ) = 0, i = 1, 2, 3; F6 (Pi = 0), i = 2, 3, 4, at the same time, (F1x , F1y , F1z ) |Pi =
~ i , i = 1, 2, 3; (F6x , F6y , F6z ) |P , i = 2, 3, 4. From Theorem 7, we need at most four irreN
i
ducible algebraic surfaces Fi = 0, i = 2, 3, 4, 5, with degree 3, to blend F1 , F6 smoothly, and
Fi meets Fi+1 with C 1 continuity, i = 1, 2, 3, 4, 5.
Blending of Triangular Algebraic Surfaces
205
4. Remark
This paper solves the scattered data interpolation problem with definite normals. There
remains a lot of such problems for us to solve. For example, the intersection of the interpolation surfaces around a point.
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