adamou, lalo, mario universidad de...

BisectorsGoal 2.1.3

Adamou, Lalo, MarioUniversidad de Cantabria

Castro Urdiales - December 2013

Outline of the talk

I The true and the untrimmed bisector

I Equations of the untrimmed bisector surface

I The method to find parametrizations

I Bisector of a circular cylinder and a quadric

I Trimming the bisector of two curves

I Trimming the bisector of two surfaces


True and untrimmed bisectors

Let O1,O2 be two geometric objects (point, curve, surface).

Def. of true bisector: The true (or trimmed) bisector of O1 andO2, B(O1,O2), is the set of points which are equidistant from O1

and O2.

B(O1,O2) = {B | dist(B,O1) = dist(B,O2)}

If the curves or surfaces are regular, the distances are measuredalong the normals. (Beware of singularities)

Def. of untrimmed bisector: The bisector B of two curves(surfaces) S1 and S2 is the set of centers of circles (spheres) whichare tangent to S1 and S2 simultaneously.


Bisector of two surfaces

Equations of the untrimmed bisector

B = (X ,Y ,Z )T ∈ R3 is in the bisector if

I the point B is in the normal lines of S1 and S2:

〈(X ,Y ,Z )− S1(s, t), ∂sS1(s, t)〉 = 0,〈(X ,Y ,Z )− S1(s, t), ∂tS1(s, t)〉 = 0,〈(X ,Y ,Z )− S2(u, v), ∂uS2(u, v)〉 = 0,〈(X ,Y ,Z )− S2(u, v), ∂vS2(u, v)〉 = 0,

(1)

I the point B is at equal distances from S1 and S2:

〈(X ,Y ,Z ), 2(S2(u, v)−S1(s, t))〉+‖S1(s, t)‖2−‖S2(u, v)‖2 = 0.(2)


The equations (1) can be written in matrix form as follows

AB = R, (3)

where

A =

∂sS1x ∂sS1y ∂sS1z

∂tS1x ∂tS1y ∂tS1z

∂uS2x ∂uS2y ∂uS2z

∂vS2x ∂vS2y ∂vS2z

, R =

〈S1, ∂sS1〉〈S1, ∂tS1〉〈S2, ∂uS2〉〈S2, ∂vS2〉

.Goals: 1. To obtain a parametrization(s) of the bisector:

I Rational

I Non-rational

I Rational reparametrization, if possible.I Use it for approximation.


2. Find the implicit equation of the untrimmed bisector.(Presented in Siguensa)

3. Determine which parametrizations (and the correspondingsubset of the domain) correspond to the trimmed bisector:Trimming.

Applied to bisector of:

I plane and quadric, assuming that the quadric is parametrized by thePN parametrization, if possible

I plane and torus

I circular cylinder and quadric

I two cylinders

I cylinder and cone

I two cones


The method to find parametrizations

1. Solve the System AB = R, for B in terms of u, v , s, t,multiplying by the generalized inverse matrix:

B(u, v , s, t) = A+R,

A+ = −a−1k AT

[(AAT)

k−1+ a1(AA

T)k−2

+ · · ·+ ak−1In]

where ai and k are such that:

det(w In − AAT) = a0wn + a1w

n−1 + · · ·+ an−1w + an with a0 = 1

(k 6= 0 is the largest such that ak 6= 0) or ( k = 0 and A+ = 0).

[Decell, H. P., An application of the Cayley-Hamilton theorem to generalized

matrix inversion, SIAM Review. 7 : 526-528(1965).]

2. Eliminate two of the four parameters u, v , s, t from

G (u, v , s, t) = det(|A,R|) = 0,

F (u, v , s, t) =〈B, 2(S2(u, v)− S1(s, t))〉+ ‖S1(s, t)‖2 − ‖S2(u, v)‖2 = 0.


If we eliminate s and t we obtain a parametrization of the bisectorsurface of the form

Bi(u, v) =

xi (u, v)yi (u, v)zi (u, v)

, i = 1, . . . ,m.

where, xi (u, v), yi (u, v), zi (u, v) are in general non-rational.

The case of curves: The method is formally analogous, butsimpler.


Quadric (or Torus) Parametrization Max degree implicit eqn

Parabolic-Cylinder rational 6Circular-Cylinder rational 4Elliptic-Cylinder non rational 8Hyperbolic-Cylinder non rational 8Circular-Cone rational 4Elliptic-Cone non rational 8Sphere rational 4Ellipsoid rational 12Elliptic-Paraboloid rational 10Hyperbolic- Paraboloid rational 10Hyperboloid-One-Sheet rational 12Hyperboloid-Two-Sheets rational 12Torus rational 8

Agreement with M. Peternell [Peternell, M., Geometric properties of bisector

surfaces, Graphical Models 62 (3) (2000), 202-236.]


Circular cylinder and quadric or torus

Cylinder: C(s, t) =

2 rs

1+s2

r(1−s2)1+s2

t

, S(u, v) =

Sx(u, v)Sy (u, v)Sz(u, v)

G (u, v , s, t) is linear in t and quadratic in s.

Two cases

1. Assume S is a surface of revolution with the same axis as C .

S(u, v) =

A(u) 1−v2

1+v2

A(u) 2v1+v2

B(u)

,From G (u, v , s, t): s1 = 1−v

1+v and s2 = −1+v1−v .


Cylinder and surface of revolution with same axis

Theorem: Let σs(u, v) = ‖∂uS × ∂vS‖.

1. The equations F1(u, v , t) = F (u, v , s1, t) = 0 andF2(u, v , t) = F (u, v , s2, t) = 0, are quadratic in t, then thereare four parametrized components of the bisectorBi , i = 1, . . . , 4, which may include a square root of a positiveexpression (coming from σs).

2. If σs is rational, the bisector parametrizations Bi (u, v) arerational. Morover, if σs ∈ Q[δ](u, v), then Bi ∈ Q[δ](u, v).

Corollary: The bisector of a circular cylinder and a quadric ofrevolution, or a torus, with the same axis of revolution, is rational.


Case 2. Suppose S is not a surface of revolution, or it does nothave the same axis as the circular cylinder C.

From G (u, v , s, t), obtain t = t(u, v , s).

Theorem: Let σs(u, v) = ‖∂uS × ∂vS‖.The equation F0(u, v , s) = F (u, v , s, t(u, v , s)) = 0, is quartic in s,and it may be factorized in the following way:

F0(u, v , s) = F1(u, v , σs , s)F2(u, v , σs , s) = 0

where F1(u, v , σs , s) and F2(u, v , σs , s) are quadratic in s. Thus,the parametrization of four components of the bisector Bi areobtained, which may contain square roots.


Example: circular cylinder and ellipsoid

S1(s, t) =

2 s

1+s2 + 4

t1−s2

1+s2

, S2(u, v) =

2u

1+u2+v2

v1+u2+v2

2(1−u2−v2)1+u2+v2


Bi (u, v) =

−2

u(

4 (si(1+u2+v2))2−4+3 si(u2−1+v2)

)(1+u2+v2)(siu2−2 usi 2+2 u+siv2−si )

v((si(16 u2−6 u+16 v2+16))

2−1+15 ui(u2−1+v2))

(1+u2+v2)(siu2−2 usi 2+2 u+siv2−si )((ui(u2−1+v2))

2−1)(2 u2+3 u+2+2 v2)

(1+u2+v2)(siu2−2 usi 2+2 u+siv2−si )

,

i = 1, 2, 3, 4, where

s1 = a1+b1σ+r116c , s2 = −a2+b2σ+r1

16c , s3 = a3+b3σ+r216c , s4 = −a4+b4σ+r2

16c ,

r1 =√α1 + β1 σ and r2 =

√α2 + β2 σ, ai , bi , c , αi , βi ∈ R[u, v , σ]

are very long expressions, and

σ =√

u4 + 14 u2 + 2 u2v2 + 62 v2 + 1 + v4.


TrimmingBisector of two rational curves in the plane(Method of Farouki and Johnstone)

Starting from the parametrization of the untrimmed bisector:

1. Find the points at infinity and cusps, and sort them by theirparameter values.

2. For each segment on the untrimmed bisector determined bytwo consecutive such points, compare the distances of thesegment midpoint to both curves, and discard the segment ifthese distances are different.

3. Find the self-intersection points of the untrimmed bisector.

4. Split each remaining segment of the untrimmed bisector atthese self-intersections. For each of the resulting segments,compare the distances of the midpoint to both curves, anddiscard the segment if they are different.

The remaining segments constitute the true bisector.


Trimming

Bisector of two rational regular oriented disjoint surfaces

S1(u, v), (u, v) ∈ D1 et S2(s, t), (s, t) ∈ D2.

Assume that each Si divides the space in two or more connectedcomponents (e.g. plane, quadric, torus), so there is a componentdetermined by S1 that contains S2, and viceversa.

1. Choose an orientation: Choosethe normal vector N1(u, v) atS1(u, v) pointing into thecomponent that contains S2.Similarly for N2(s, t).


Let B be a point in the bisector,equidistant from P1 = S1(u, v) andP2 = S2(s, t).Normal vectors: N1(u, v) and N2(s, t).

P1,P2 and B form an isosceles triangle.

Thus ⟨−−→P1B,

−−−→P1P2

⟩> 0 and

⟨−−→P2B,

−−−→P2P1

⟩> 0. (4)

Denote ρ(u, v , s, t) =−−−→P1P2 = −

−−−→P2P1

Let t0 = t(u, v , s), si = si (u, v), be the expressions obtained byeliminating t and s from G (u, v , s, t) = 0, F (u, v , s, t) = 0.Therefore, if Bi0(u, v) is in the true bisector, it must verify:{

〈N1(u, v), ρ(u, v , si0 , t0(si0))〉 > 0

〈N2(si0 , t0(si0)) ,−ρ(u, v , si0 , t0(si0))〉 > 0(5)


The example of two disjoint spheres

S1(u, v) =

4 u

1+u2+v2

4v1+u2+v2

2 (1−u2−v2)1+u2+v2 + 1

, S2(s, t) =

6 s

1+s2+t2

6 t1+s2+t2

3 (1−s2−t2)1+s2+t2 − 5

,

N1(u, v) =

32 u

(1+u2+v2)3

32 v

(1+u2+v2)3

−16 (−1+u2+v2)

(1+u2+v2)3

N2(s, t) =

72 s

(1+s2+t2)3

72 t

(1+s2+t2)3

−36 (−1+s2+t2)

(1+s2+t2)3

.Castro Urdiales - December 2013

Elimination: t0 = s vu , s1(u, v) = − 1

11 u, s2(u, v) = 57

uu2+v2

Two parametrizations

B1(u, v) =

11 u

−11+u2+v2

11 v−11+u2+v2

− 9 u2+9 v2+112 (−11+u2+v2)

, B2(u, v) =

35 u

−5+7 u2+7 v2

35 v−5+7 u2+7 v2

− 21 u2+21 v2−252 (−5+7 u2+7 v2)

.

ρ(u, v , s, t) =−−−→P1P2 =

−2(−3 s−3 su2−3 sv2+2 u+2 us2+2 t2u)

(1+s2+t2)(1+u2+v2)

−2(−3 t−3 tu2−3 v2t+2 v+2 vs2+2 t2v)(1+s2+t2)(1+u2+v2)

−(5+u2+v2+11 s2+11 t2+7 s2u2+7 s2v2+7 t2u2+7 t2v2)(1+s2+t2)(1+u2+v2)

.


Conditions

1. For t = t0 =s v

uand s = s1(u, v) = − 1

11u :

β1 = 〈N1(u, v) , ρ(u, v , s1, t0)〉

= 16

(55 + 7 u2 + 7 v2

) (u2 − 11 + v2

)(121 + u2 + v2) (1 + u2 + v2)3

> 0

β2 = 〈N2(s1, t0) , −ρ(u, v , s1, t0)〉

= −527076

(55 + 7 u2 + 7 v2

) (u2 − 11 + v2

)(1 + u2 + v2) (121 + u2 + v2)3

< 0

β1 et β2 have opposite signs. Thus, the component B1 is not partof the true bisector.


2. For t = t0 =s v

uand s = s2(u, v) =

7u

5(u2 + v2):

γ1 = 〈N1(u, v) , ρ(u, v , s2, t0)〉

= 16

(7 u2 − 5 + 7 v2

) (55 + 7 u2 + 7 v2

)(49 u2 + 25 + 49 v2) (1 + u2 + v2)3

γ2 = 〈N2(s2, t0) , −ρ(u, v , s2, t0)〉

=86436

(7 u2 − 5 + 7 v2

) (55 + 7 u2 + 7 v2

) (u2 + v2

)2

(1 + u2 + v2) (49 u2 + 25 + 49 v2)3

γ1 et γ2 have the same sign, the sign of 7 u2 − 5 + 7 v2. Thus, theimage of B2 contains the true bisector.The true bisector is composed by the points B2(u, v) such that

γ1 > 0, γ2 > 0 =⇒ 7 u2 − 5 + 7 v2 > 0


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D. Dutta, C. Hoffman, On the skeleton of simple CSG objects. ASME J. Mech.Des. 115 (1993), 8794.

G. Elber, M-S. Kim, Computing Rational Bisectors. Computer Graphics andApplications, IEEE, Vol. 19, No. 6, (Nov/Dec 1999), 7681.

G. Elber, M-S. Kim, Rational bisectors of CSG primitives. Proc. 5th ACM/IEEESymposium on Solid Modeling and Applications, Ann Arbor, Michigan, (June1999), 246257.

G. Elber, M-S. Kim, A computational model for non-rational bisector surfaces:curve-surface and surface-surface bisectors. Proc. Geometric Modeling andProcessing 2000, Hong Kong, (April 2000), 364372.

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M-S. Kim, G. Elber, J-K. Seong. Geometric Computations in Parametric Space.Spring Conference on Computer Graphics Bundmerice Castle, Slovak Republic,May 12-14, (2005).


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Thanks.


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