Title: Constraints on curve networks suitable for G^2 interpolation
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Title: Constraints on curve networks suitable for G^2 interpolation
Physical Description: Book
Language: English
Creator: Hermann, T.
Peters, J.
Strotman, T.
Publisher: Department of Computer & Information Science & Engineering, University of Florida
Place of Publication: Gainesville, Fla.
Copyright Date: 2010
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Volume ID: VID00001
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Constraints on Curve Networks suitable for
G2 Interpolation


T. Hermann, J. Peters and T. Strotman

No Institute Given



Abstract. When interpolating a network of curves to create a C1 surface from
spline patches, the network has to satisfy an algebraic condition, called the vertex
enclosure constraint. We show the existence of a similar, additional constraint
that governs the admissibility of curve networks for G2 interpolation by smooth
patches.


1 Introduction

One much-studied paradigm of geometric design is surface interpolation of a given net-
work of 02 curve segments (see Figure 1). While many 02 constructions exist that
join n patches (e.g. [Hah89,GH95,Ye97,Rei98,Pra97,YZ04,LS08,KPN1]), these con-
structions generate the boundary curves that emanate from the common point, i.e. rely
on full control of these curves. In many practical scenarios, however, the curves are
feature curves. That is, they are given and may only be minimally adjusted. It is well-
known, that interpolating a network of curves by smooth patches to create a C1 surface
is not always possible when the number of curves is even, since an additional algebraic
constraint must hold for the normal component of the curve expansion at the com-
mon point. This is thefirst-order vertex enclosure constraint [Pet91]. Here we discuss
whether curve nets have to meet additional second-order vertex enclosure constraints to
allow for their G2 interpolation by smooth surface patches.
In particular, we want to determine constraints, if any, on n boundary curves yJ, j
1,..., n so that consecutive patches surrounding a vertex can join with C2 continuity
after reparameterization by some regular map

S:r 2, I j(u,v)=: []. (1)

We formalize this setup as follows.

Definition 1 (Smooth Network Interpolation). Let

yJ :]R- ]R3, t--yJ(t), j Z, {1,...,n} (2)

be a sequence ofn regular Ck 1 curves in R3 that meet at a common point p in a plane
with oriented normal n and at angles -J less than 7 (cf Figure 1):


yJ(0) = p, t := (yJ)'(0) I n, 0 < j := Z(tj, tj ) < .






















1 X

j/y-1

Fig. 1., I i Network of curve segments. This paper focuses on (right) local network interpolation
(see also Definition 1): curves yJ, j E meeting at a point p are given and pairwise interpo-
lating patches x' are sought. The arrow-labels 1 and 2 indicate the domain parameters associated
with the boundary curves of the patches, e.g. 91xj+l (t, 0) = 2xj (0, t).


A Gk surface network interpolation of{yJ } is a sequence ofpatches

xR :R2 R3, (u, v) x(u, v), j C Z, (4)

that are regular and Ck at p, that interpolate the curve network according to

S(t, 0) = y l(t), xj(0, t) = yj(t), (5)

(with index modulo n) and that connect pairwise so that Gk constraints hold:

at (u, 0) li '+ D =2 92 (xj o ~), for0 < ki < k. (6)

Smooth Network Interpolation restricted to the neighborhood of p is called local net-
work interpolation

Section 2 introduces the constraints for k 2, resulting from expanding (6) at
(0, 0). In Section 3, we review the first-order vertex enclosure constraint in order to
gauge what remaining degrees of freedom can be leveraged to enforce the G2 con-
straints. Section 4 then classifies the G2 constraints at the vertex and analyzes their
solvability for a fixed curve network. Theorem 2 establishes the existence of second-
order vertex enclosure constraints. We conclude with a conjecture on the properties
of a matrix that holds the key to the complete characterization of second-order vertex
enclosure constraints.


2 Notation and constraints

As illustrated in Figure 1, we consider n curves yJ : R R3 that start at a point
p E R3, and we aim at filling-in between the curves using patches xj : R2 R3,
j e Z,. We note that the angle yj corresponds to patch x+I1 and assume for notational
simplicity that the curves are arclength-parameterized at t 0 so that t3 is a unit














vector. Since the patches of interest are at least C2 for the case under investigation,
d012x3 021x3.
Since our focus is on curvature continuity at p = x3(0, 0), we abbreviate the kth
derivative of yJ evaluated at 0 as y( and write

X 2 : ( )(0 ,0 ), 2 d k22 )(0, 0), I 2 : ( 1,' ')(0,0).
(7)
We drop superscripts whenever the context makes them unambiguous, e.g. we write

XkXi2 := Xk Xlkik2 : k Xk2X3 .1 (8)
: yk xo, y : y 1 xo, y :- yJ XO. (9)

That is x k2 i is a vector in R3 and not a vector of vectors [..., x ...].
We also tag the equations arisingfrom (6) for a specific choice of (ki, k2) and j as
(ki, k2)3. Again, to minimize ink, we leave out the superscript when possible. By (5),
,3 has the expansion

+ p +...)v +(.,7_,+ +. +.
+ 3(u, v) [: i +([ (10)
U+ (1+ 1 +...)v +(.,.+ U+...) +...
Substituting the curves according to (5), we obtain from (6) at (0, 0), via the chain rule,
the G1 constraints

Y1 YI oi + Y1 0o (0,1)
x1 = y, ii + xll1i + y201 + yi~11 (1,1)
x2 = 2x11y11 + 2y1 21 + x12o01 + Ys301 + 2Y2011 + 2yi921 (2,1)
x3 3x1i211 + 6x11321 + 6Y1 Y3 + x13o01 + Y4901 + 3y311
+ 'L + 6yi31 (3,1)
and the G2 constraints

y Y2 2 + 2 oix11ol Y + Yo002 + Y221 + Y1Yo2 (0,2)
x2 = 2p11y2-Yol + 211x1i)01 + yi 12 +X21 ,, + 2o01X12o01
+ 11~02 + y3g 1 + Y202 + 2yx11X1101 + 2011Y2o01 + Y1912 (1,2)
X+ 421iy2~01 + 4yY3so01 + 4yllXni, i + 4q21X11i 01 + 4l11x1201o
+ 4021x11o01 + 2y01X13y01 + 4YP21y2 01 + 4ll1x2101 + 2y2012
+ 4)11x12001 + 2x11012 + 2y1 22 + x1202 + y3s02 + 2Y1022
+ 2y2 + x__ .,X + y4 21 + 2y2 (2,2)


3 First-order vertex enclosure

For an even number of patches, the compatibility condition together with the first-order
continuity contraints result in a rank deficient circulant system that is only solvable if















we restrict its right hand side (see e.g. [Wat88,vW86]). Du and Schmitt [DS91] gave a
sufficient condition for this vector-valued constraint to hold and Peters [Pet91] observed
that the components of the system in the tangent plane can always be enforced so that
the essential part of the constraint is its normal component. Theorem 1 from [HPS09]
re-interprets this algebraic scalar-valued constraint geometrically in terms of angles and
normal curvatures.

Theorem 1 (first-order vertex enclosure constraint). Ifn is odd then local network
interpolation is always possible. Ifn is even then local network interpolation is possible
if and only if the vertex enclosure constraint holds:
n
0 ( 1)(cot j l+ cotj)Kj, i :- ny{. (11)
j-1
Two geometric conditions are known to imply the first-order vertex enclosure con-
straint: first, the existence of a symmetric (embedded Weingarten) matrix W e R3x3
such that the boundary curves yJ all satisfy n djdkX = (tj)tWtk [Pet91]; and
second, the 'Euler condition' that the normal curvatures Kj satisfy Kj k= cos2 rT +
k2 sin2 Tf for constants ki, k2 and angles af such that j+l1 f = 7- [HPS09].
To understand the first-order vertex enclosure constraint in more detail, we take the
scalar product of Equation (0,1)J respectively with y yi and y+ (see [HPS09]) and
eliminate. This yields the leading coefficients of the reparameterization:
C, s


1 i .... 1,0 sin (7- + 7 ) sin -
sin 7 sin 7

C ,-s
Here we abbreviated 01o -= and sin(yJ 1 ) sin( +) and used s
sin3J 1 sin7
sin 7y 1, etc. in the illustration. Since 0 < 7- < T7, sin -7 / 0. Reordering the terms in
(1,1), and using (0,1), gives the system

x+ + x Y : (o01Y2 + 11Y1+ 11Y) (13)
sin 7 sin 7 sin 7
which can be solved by



sin 1 -k+ (- 1)+ 13 J, if nodd.

(14)
That is if n is even, x can be chosen freely but, by (13), the alternating sum


(-1) y j (-1)J(cot + cot -)y + (-1) (15)
=1jsin














has to vanish. This imposes the C1 vertex enclosure constraint
n
E(-1)(cot'j +c+ot y)n.y- 0 (11)
j=1

on the boundary curves. And it constrains the 2n degrees of freedom 3i, and qp1 to
satisfy the additional two constraints of (15) on the tangent directions. In the following,
we therefore assume that all xn0 and all but possibly one x,1 are fixed satisfying (0, 1)J
and (1, 1)J. We have also chosen

,, < i,k < 1 when n is even and
S. for < (16)
0 < i + k < 1 when n is odd.


4 Constraints on boundary curves arising from G2 continuity

Our main goal is to solve the local G2 constraints (ki, k2)j of Section 2 for ki +
k2 < 4 in terms of the higher-order derivatives, x31, x,2, x31, x3, xi3 and for the
reparameterization .7 2 not fixed by (16).
First, we consider the equations (ki, k2)3 when ki + k =2 4.


Lemma 1. The equations (ki, k2)j, where ki + k = 4, can always be solved in terms
ofX31, x13, X22.
Proof We have more vector-valued variables, x31, x13, x22, than constraints: (3,1),
(2,2). Equation (3,1) expresses x31 in terms of x13 so that we can focus on solving
(2,2) in terms of x13 and x22. Equation (2,2) can be arranged as

X2 = x__-+ 21oi'1X13 + f(y, x11, x12, x2), (2,2)

where f(y, x11, x12, x12) is the collection of terms on the boundary or appearing in
lower-order equations. Clearly, we can solve n -1 of these equations for x+. In general,
this is all we can hope for since, for equal angles 7-, the analysis in [Peters92] shows
that the constraint matrix for solving (2,2) in terms of just x22 is rank-deficient by 1.
If the tangents do not form an X configuration, i.e. not all consecutive pairs of angles
add to 7, then, by (12), at least one il /7 0. Let 0'U /4 0. Then, for any choice of x,
we can solve (2,2), for x2, ... x,, x1.
If the tangents form an X, possibly with segments of differing length, then the va-
lence must be n 4. We can then choose 011 1/ 0 and solve for x2, ... x4, xl since
Lemma 4 below shows that in this case the constraints for ki + k2 = 3 leave xi1 free
to choose. U

Our analysis should therefore focus on the case ki + k2 = 3, of the number of
derivatives adding to 3. If the corresponding constraints are solvable then no second-
order vertex enclosure constraint exists and a construction is possible. However, the
situation is not that simple as the next lemma shows.














Lemma 2. The equations (ki, k2)j, where ki + k2 3, can be solved in terms ofx12
and x31 if and only if the following n x n system of equations has a solution:

sin 7, k j 1
M2sin(- +7), k = j
Mh = r, Mjk : (17)
sin k = j 1
0, else,


r3 := (2o01o l + 211no01 + 702)X11 (18)
001

+ ((0ol) Y3 + 2o01 11Y2 + 2o01j11Y2 + 02Y2 + 012Y1 + 12Y1)
Yoi
+ol1(2 lxl, + 01 Y3 + 2011y + 2- 2y, + 2' vj 2).

Proof We eliminate x21 by substituting (2,l)J into (1,2)J to obtain
x1+ x., ., 0 2+ 01x1201 + g(y, xll), (19)

where g(y, x11) collects the terms depending on y and x11. We divide both sides by

-70o1 i:= to obtain for h3 :
sm7 sin lJ 1

siny'hj + 2sh( +2in( j)h +sin y hj+l g(y, xl) =:r3. (20)
o01

This is Equation (17). U

Although, generically, we can freely choose all 12 and 2 for kl + k2 > 1,
rank deficiency of the matrix M could lead to an additional constraint on the boundary
curves when we consider a higher-order saddle point. For a higher-order saddle point,
n y3 = 0 for k 1, 2 and this can force n xi3 0 so that

j (0Ol)2
n- r3 --n y3 + 001001n Y3.
o01
If f e R" is a left null-vector of M, i.e. fM = 0, then we obtain the second-order
vertex enclosure constraint


SE ( i +1 j ,+1 j+)n y (21)
j ,,

We therefore focus on the rank of M. The next lemma partly characterizes rank(M)
and hence explains in what cases a second-order vertex enclosure constraint can exists
or where no second-order vertex enclosure constraint exists because M is of full rank.














Lemma 3 (rank of M).
The rank of M is at least n 2. The matrix M is offull rank (rank(M) = n) if either
all angles are equal, and n {3, 4, 6}; or if all angles are less than 7/3.
Proof Since all sin y > 0, we can solve (20) for j = 1,..., n 2, i.e. the rank-
deficiency in the general case is at most 2. Discrete Fourier analysis in [Peters92] shows
M to be of full rank if all angles are equal, and n {3, 4, 6}. If all angles are less than
7/3 then the matrix is strictly diagonally dominant and therefore invertible. U

We will see below that, for n 4 and equal angles, rank(M) 2; and for n = 5,
when three angles are 7/2, then rank(M) 3. Discrete Fourier analysis in [Peters92]
showed rank(M) n 1 ifn E {3, 6} and all angles are equal; and rank(M) n- 2
when all angles are equal and n = 4. In the general case, however, the analysis is more
complex.
As for the first-order vertex enclosure constraint, we can focus exclusively on the
normal component of the constraints since, in the tangent plane, we can always choose
11 and 11n to solve (1,1) for an arbitrary choice of xi1. Then we can use the tangent
component of xn1 and the free choice of 0o2 to solve the tangent component of (20)1
and (20)" (while (20)J is solved in terms of x 2, j = 2, ..., n 1). Focussing on the
normal component, we note that the right hand side simplifies to

n r3 = (2o01il + 2011o01 + 02)n x1n + 201ol11n xn1 (22)
001
1
+ ((0oi)2n y3 + (2o01n11 + 0o2)n y2)
001
S+.',, '., Y3 + 2(11 + l0111)n Y2 .


Lemma 4. Ifn = 3 or n = 4, no second-order vertex enclosure constraint exists for
any choice of J.
Proof For n = 3, sin(Q + 7) = sin 7+ and M simplifies to
-2sin 2 sin73 sin 71
M sin72 -2sin33 sin7 I. (23)
sin72 sin73 -2sin71J
Since 0 < sin yj < 1, multiples of := [1, 1, 1] are the only null-vectors of M; that is,
rank(M) 2. We have a solution iff
r + r2 + r3 0. (24)
If we choose = .' = 0 for k + 1 > 1 then we have a solution since

rl+r +r3= + .'


E j sin(Q- + ) + sin7- Y+ 0.
J- sin
j=l














That is, we can choose x12 freely and enforce all (1, 2)J by choice of x2 and x12. Then
x3l is uniquely determined by (2, 1)- 1 and all constraints for ki + k -2 3 hold.
If n = 4, the determinant of M is
2 sin(74 + 1) sin y4 0 sin y1
S sin72 2 sin(1 +2) sin ly 0
D =sin' (25)
0 sin y3 2sin(y2 +73) sin72
sin 3 0 sin 4 2 sin(3 +4)
S(4 sin(y4 + 7) sin(-y + 72) + siny sin 3 sin 2 sin 74)2 (26)

3 sin 2 sin + 2 (27)

9 sin2 (72 + 3) sin2( + 72). (28)
The last equation holds because > y = 27. That is D = 0 if and only if -l + 72 =
and therefore 3 + 4 i7; or 72 + K3 7 and therefore 4 + 71 t7. That is D = 0
if and only if at least one pair of tangents, t ,t3 or t2,t4, is parallel.
If 71 + 72 = 7 and 2 + 73 = i.e. the tangents form an X then sin = s,
j = 1, 2, 3, 4, for some scalar 0 < s < 1. The matrix

sOsO
M 0 sOsO (29)
OsOs


is of rank 2 and has left null-vectors [1, -c, -1, c] and [-c, -1, c, 1] for any c, for
example c : 2 cos 4. Without loss of generality, we choose C1 := [1, 0, -1, 0] and
f2 := [0, -1, 0, 1]. Since o01 1 and 0oi 0 and, by (1,1)J 1, n x1 -n x1
n r3 := (211 + 002 201)n x1. + 02n y2 + 2(011 + )n Y2. (30)
Choosing, for example, all 0 11= -911, 002 211 211 and 902 = 0, we have
n r3 0 and the constraints can be satisfied.
If 1 + 2 = 7 but 72 + 73 /f 7 then si := sin72 = sin-y and S4 := sin 3 =
sin 74 and hence
2 sin(4 + 71) sin 4 0 sin
sin 1 0 sin y 0
M =
0 sin 4 -2 sin(y + 4) sin / 1
sin 4 0 sin 4 0
For this M, rank(M) =3. Since sin(4 + -1) = cos 74 sin + cos y1 siny 4,
M =0 for := [1, -2cos 4, -1, -2cosy1]. (32)
Since the first-order vertex enclosure constraint holds, one n xi1 can be chosen freely
and we can set
n -r = n r 2cos 4n r2 n r3 2cos71n r4 =0 (33)
by judicious choice of .,














Our main result, however, proves that a second-order vertex enclosure constraint
exists for a higher valence for some choice of 7.

Theorem 2 (second-order vertex enclosure constraint). For n = 5 and some choice
of 3, a second-order vertex enclosure constraint exists.

Proof For n = 5, we compute

det M 18 sin(-J + 7+1). (34)

Abbreviating sj := sin 7 and cj := cos and assuming, without loss of generality
that 71 + 72 7 and therefore 73 + 74 + 75 7, we get

S:= [S3, 2ss(c2s3/s2 + c3) S4, -5, S2, -82]. (35)

If we choose y( so that n y 0 then (14) implies n xl 0. With sj 1, :
sin(Q 1 + 'i), the second-order vertex enclosure constraint (21) has to hold (note
again that sj > 0 for all j):

0 .. ol' {'., + ', Ifj+l)n "yj3
j ,,

= 1s (sj ij' + Sj+l +l)n y. (36)
-Sjsj 1

Specifically, for 7 [3, 3, 2, 2, 2]
[.. ,Sj, ...] [ 1 1 2- ], [ ... Cj,...] [.. . ] ,

[ j_,S j .... , '] [1 0 1 e], [ 3 0 3 1 1].

Then the second-order vertex enclosure constraint is

0 [1,0,1,0, 0][..., n. y,...]t n .y + n .y. (37)

That is, for the two terms corresponding to the curves with opposing tangents, n y
-n y has to hold. U

We note that the case n = 5 yields a doubly rank-deficient matrix M when =
[2, 2, 2, 1, 1].


5 Higher valences

Theorem 2 established the existence of a second-order vertex enclosure constraint. An
explicit proof for valences n > 6 requires exhibiting the null-vector f and hence a full
understanding of the rank of M in its general form. We have not been able to establish
the rank in generality. But we hazard a conjecture.















Conjecture 1. If, for some j both 2 sin(-y +J- 1) < sin -j +sin 1 and 2 sin(Q +
J- 1) < sin -y + sin -j 2 then there is a choice of the remaining angles for which
M is rank-deficient.

The conjecture draws on Lemma 3 which proves full rank when M is diagonally
dominant. Above, we conjecture that when both the row and column of an index are not
diagonally dominant then additional angles can be found so that the determinant of M
is zero.
We conclude with some examples supporting the conjecture. The following choices
of n angles '7, yield a matrix M with zero determinant:

Examples supporting Conjecture 1.
n ..., j,.... -
6 [2, 3,1, 3, h, 3 h], h := 6lUii- 1.636886845
7 [2, 2, 2, 1, 1, 1, 3],
7 |[3,2,1,2,2, h, 2 h], h:= iLi1 0.1139327031
8 |[2,2, 11 1,1,, h, 4-], h:], h al: 0.8337394914
12 7h[4, 1,...,1, h, 11 h], h 2.237657840


6 Conclusion

We established the existence of a second-order vertex enclosure constraint that governs
the admissibility of curve networks for G2 interpolation by smooth patches. We fully
analyzed the practically important cases of valence 3,4 and 5 and characterized the
second-order vertex enclosure constraint for valence 5. In all other cases, lacking an
exact characterization of the null-space of M, Lemma 3 establishes bounds on the angle
distribution that guarantee admissability of any curve network for G2 interpolation.

Acknowledgements.


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