1 - CAO PDM PLM - Formations initiale et continue

Modèles mathématiques
Catia V5
http://www.diderot.org/
61, rue David d’Angers
75019 PARIS
Pierre Vinter 1

SOMMAIRE
Contexte du travail présenté
Etude des modèles sur Catia V5
-Courbes
- Surfaces
- Analyses
Prochains Objectifs
Conclusions
Remarques préliminaires :
- Travail non terminé !!
- Document Franco-Anglais
- Valider sur certains points par D. S.
Pierre Vinter 2

Formations, diplômes préparés
Lycée Diderot
‣ Microtechniques
‣ E.D.P.I : Etude et définition de produits industriels
‣ Artisanat et métiers d’art option : Horlogerie
http://www.gita.greta.fr/
Pierre Vinter 3

Licence Professionnelle « Métiers de la production »
option CONCEPTION COLLABORATIVE – MAQUETTE VIRTUELLE
Partenariat avec l’IUT de Saint Denis
En contrat d’apprentissage
Conception de
Produit
Conception
Forme de style
Gestion de la
connaissance
Travail Collaboratif
Maquette Numérique
Conception et
Calculs
SMARTEAM
CATIA V5
Gestion de données
techniques
Conception et
Fabrication
Utiliser le potentiel de Catia V5 peu utilisé en BTS:
- Generative Shape Design et Freestyle
- Knowledge Advisor, Expert ….
- Generative Structural Analysis
Pierre Vinter 4

Licence Professionnelle
CONCEPTEUR NUMERIQUE EN DESIGN ET TECHNIQUE AUTOMOBILE
Demandeurs PSA et Renault
Début
septembre 2006
Définition de surfaces Class A
à partir de courbes de Bézier
Pierre Vinter 5

OBJECTIFS :
Formation de niveau BAC + 3
Comprendre les informations données
par le logiciel
Appliquer les notions mathématiques en
manipulant sur Catia V5
Type of geometry posted
Definition
NupbsCurve
Curve NUPBS no rational
NupbsSurface
Surface NUPBS no rational
NurbsCurve
Curve NURBS rational
NurbsSurface
Surface NURBS rational
PNupbs
Parametric Curve rational
PNurbs
Parametric Curve on surface
PSpline
Parametric Curve
Pierre Vinter 6

Evolution jusqu’aux NURBS
Modeleur paramétrique variationnel
PARAMETRAGE
NURBS
POLYNOMIALS CURVES
CUBIC - QUINTIC SPLINE
BEZIER CURVE
NO UNIFORM B-SPLINE
B-SPLINE
RATIONAL
BEZIER CURVE
Pierre Vinter 7

Representations
Il n’y a pas que la
représentation paramétrique
Explicit representation
Explicit form of a curve 2D : y = f(x)
Explicit form of a surface : z = f(x, y)
Implicit representation
Implicit form of a curve 2D: f(x, y) = 0
Implicit form of a surface 3D: f(x, y, z) = 0
Parametric représentation
• Curve x(u), y(u), z(u)
• Surface x(u, v), y(u, v), z(u, v)
Pierre Vinter 8

CONIQUES
Implicite equation
y 2 -2.p x - (1 -e 2 ) x 2 = 0
Explicit equation
Parametric equation ???
0 < Parameter < 0,5 → Ellipse
Parameter = 0,5 → Parabola
0,5 < Parameter < 1 → Hyperbola
PARAMETRE ≠ EXCENTRICITE
Pierre Vinter 9

Paramétrage : Droite - Cercle
D (u) = ( 1 – u ) OA + u OB
u parameter varying from 0 to 1
B
A (u) = R ( cos U e 1 + sin U e 2 ) + OC
A
Paramétrage en
abscisse curviligne
OM = OA + s
AB
AB
With 0 < s < AB
A (s) = R ( cos (s/R) e 1 + sin (s/R) e 2 ) + OC
Pierre Vinter 10

AVANTAGES
Trièdre de Frenet
R c
: curvature radius
R t
: torsion radius
κ : curvature
τ : torsion
B = T ∧ N
N =1
dOM
2
ds
2
= 1/Rc
Avec un paramétrage en abscisse
curviligne, il est plus facile de
déterminer les rayons de courbure
d’une courbe 3D
Pierre Vinter 11

Curvature – torsion helix
T
x= - R/c sin(s/c)
y = R/c cos(s/c)
z = p / c
- cos(s/c)
N - sin (s/c)
0
-R/c 2 cos(s/c)
dT/ ds -R/c 2 sin (s/c)
0
p/c sin(s/c)
B - p/c cos (s/c)
R/c
dB/ds
p/c 2 cos(s/c)
p/c 2 sin (s/c)
0
T . N = 0
B = T ∧ N
c = √ R 2 + p 2
Curvature radius
Rc = (R 2 +p 2 ) /R
Torsion radius
Rt = - (R 2 +p 2 ) /p
Rt = 252,9 mm
R = 20 mm
p = 10 / 2π
Pierre Vinter 12

C(u)
T = d OM/du × du / ds =
N =
Example 2 : -4 < u < 5
x = u 2 + u + 1
y = u 2 – 2u + 2
-2(u - 1) du/ds
(2u + 1) du/ds
ds/du = √ 8u 2 – 4u + 5
(2u +1) du/ds
2(u-1) du/ds
N =
dT/
ds
dT/
ds
1/R c =
6
(8u 2 –4u + 5) 3/2
-4 < u < 5
Pierre Vinter 13

SPLINE
SPLINE CUBIQUE OU QUINTIQUE ??
Courbe polynomiale
par morceaux
Sub SetSplineType( long iSplineType)
Sets the spline type.
Parameters: iSplineType
The spline type
Legal values: Cubic spline or WilsonFowler ????
Pierre Vinter 14

SPLINE CUBIQUE
OM = (2u 3 –3u 2 +1) OP 0
+ (-2u 3 + 3u 2 ) OP 1
+ (u 3 –2u 2 + u) dOP 0
/du + (u 3 –u 2 ) dOP 1
/du
SPLINE QUINTIQUE
OM = H 1
(u) OP 0
+ H 2
(u) OP 1
+ H 3
(u) dOP 0
/du + H 4
(u) dOP 1
/du + H 5
(u) d 2 OP 0
/du 2 +H 6
(u) d 2 OP 0
/du 2
M(u) = [u][M][P]
Spline
with Catia
[M] =
−6
+ 15
−10
0
0
1
+ 6
−15
+ 10
0
0
0
−3
+ 8
−6
0
1
0
−3
+ 7
−4
0
0
0
−1/2−
1/2
+ 1,5 −1
−1,5
+ 1/2
0,5 0
0 0
0 0
Spline quintic
calculed
Pierre Vinter 15

TENSION
Point 1 (0,0,0); Point (3,2,0); Point 3 (0,3,0)
Line at the point 2 : angle 45°
C 2 (u)
C 1 (u)
x = 3 - u - 6 u 2 + 4 u 3
y = 2 + u + u 2 -u 3
x= u + 8 u 2 -6 u 3
y = 5 u 2 -3 u 3
M(u) = [u][M][P]
2 -2 1 1
0 0 0
-6 –3 0
C 1 (u) [M][P] =
-3 3 -2 -1
0 0 1 0
3 2 0
1 0 0
=
8 5 0
1 0 0
1 0 0 0
-1 1 0
0 0 0
Pierre Vinter 16

2 -2 1 1
-3 3 -2 -1
0 0 1 0
0 0 0
3 2 0
3 0 0
=
-6 -1 0
6 3 0
3 0 0
C 1 (u) become
X = 3 u + 6 u 2 -6 u 3
y = 3 u 2 - u 3
1 0 0 0
-3 3 0
0 0 0
Tension s
0 0 0
3 2 0
Tension
calculée = 1
s 0 0
-s s 0
Tension
calculée = 3
Pb : Tension Catia = 1
Pierre Vinter 17

Boundary Conditions
Une spline passant par trois points et dont les
tangentes V1 et V3 sont connues
Inconnue
- To impose the tangency at the ends of the curve
- To impose a curvature given at the two ends
(Null curvature → natural Cubic Spline)
Deux courbes de degré 3 dans le plan XY
→ 2 × 8 =16 unknown factors
Three points (2×4) + Condition of tangency and
curvature between the 2 curves → 12
Equations
It is necessary to add 4 conditions
Pierre Vinter 18

Least Energy
Minimizes the internal strain energy
so the minimum curvature
δ
2
= ∫κ
ds
minimized
κ = curvature
s = curvilinear abscissa
circle
4 constraints
Spline degree 5
⇒ Least energy
Pierre Vinter 19

y
Curve with
=
R N 0
c
+1
0
x
What solution to draw the curve
without all the conditions ?
The curvature direction is projected
on a plane perpendicular to the
tangency direction y
=
R N 0
c
-1/2
0
x
Pierre Vinter 20

BEZIER CURVE
Limits of splines
The numerical definition of a Bézier curve
Take a cube and draw a curve, intersection of two
cylinders
BÉZIER Pierre 1910-1999
OM =
n
∑
k = 0
B
k ,
n
( u)
OA
k
Pierre Vinter 21

angency and Curvature
T =
T(u=0)
OM(
u)
du
d
OM(
u)
du
d
=
OP1−OP
OP1−OP
0
0
Displacement of the second pole horizontally
(according to the direction of tangency at the point)
κ =
2
( d
d
OM)
∧(
2
dt dt
d
dt
We can write
OM
OM)
3
d
du
2
2
[ OM(0)]
= n⋅n−1[(
⋅ OP0−OP1)
−(
OP1+
OP2)]
Pierre Vinter 22

Tangency
T(u=0)
=
OP1−OP
OP1−OP
0
0
Displacement of the second pole horizontally (according to
the direction of tangency at the point).
You can change the position of the third point, as you want
Curvature( u=0)=
n −1
( OP
0 −OP
1)
∧ ( OP 2−OP
1)
n OP
1
−OP
3
0
Curvature
You can change the position of the third point but :
not in all direction
with a ratio between the 3 first points
The n-th derivative of a bezier curve for one as of the his ends
depends only on the n+1 points of control nearest to this end
Pierre Vinter 23

Continuity of position G 0 :
With Catia
Continuity of class G 1 :
3 points aligned ⇒
curvature = 0
3 points no aligned ⇒ curvature ≠ 0
n−1(
OP
0 −OP1)
∧(
OP2−OP1)
Curvature( u=0)= 3
n OP1−OP
Pierre Vinter 24
0

Rational form of the Bezier ‘s arc
Influence of λ i
:
The parameter λ i
acts like a weight on
the associated top by attracting the
curve towards this top.
Three tops P 0
, P 1
, P 2
OM =
λ0B
0,2
define the characteristic polygon of the curve of degree 2.
OP0+
λ1B1,2OP2+
λ2B
λ0B0,2+
λ1B1,2+
λ2B2,2
2
2
λ0(
1−u
) OP0+
2λ1u(1−u ) OP1+
λ2u OP
= 2
2
λ0(
1−u
)
+ 2λ1u(1−
u)
+
λ2u
2,2
OP
2
2
λ i →∞
Pierre Vinter 25

2
2
λ0(
1−u
) OP0+
2λ1u(1−u ) OP1+
λ2u OP
OM= 2
2
λ0(
1−u
) + 2λ1u(1−u
) + λ2u
If λ 0
= 1-λ ; λ 1
=λ ; λ 2
=1-λ,
the denominateur become = 2(1-2λ)u 2 -2(1-2λ)u+1-λ
the discrimant ∆ for finding the roots of the denominateur is = 2λ -1
2
When λ > ½, the denominator has different
real roots, the curve have asymptotes
When λ < ½, the denominator does not have
any real roots, the curve does not have an
asymptote
0 < Parameter ω < 0,5 → Ellipse
Parameter ω = 0,5 → Parabola
0,5 < Parameter ω < 1 → Hyperbola
When λ = ½, the denominator become constant
If λ=0 one line and λ=1; two lines
Pierre Vinter 26

OI = (1-ω)OH + ωOP 1
93,333 × 0,8 = 74,664 !!!
ω = Linear
interpolation
between H
and P 1
Pierre Vinter 27

Basic-splines
Polynomial parametric shape per pieces → change of the position of a control
point will have a limited impact on the function obtained.
Definition : A B-spline curve of degree d (ou d'ordre k = d + 1) is thus defined by:
-a knots vector T = (t 0
, t 1
,..., t m
), m+1 réels t i
- n+1 control points P i
- n+1 weight functions N i, k
defined recursively on intervals [t i
, t i+1
] :
OM(u)=
n
∑ N
i=0
i, d(
u)
OP
i
To build a B-spline curve of degree d from n+ 1 points P i , it is thus necessary to be given
m + 1 knots or m = n + d + 1, allowing to define the basic functions N i,d (u).
Pierre Vinter 28

Example 1
Displacement of control point
Uniform Basic
Spline
No Uniform
Basic Spline
Pierre Vinter 29

NUPBS CATIA
Assistant de conversion
By defect, the degree of the curves
created by Catia is 5
ordre k = d + 1
Pierre Vinter 30

NUPBS CATIA
Degree 2
6 points
7 points
Pierre Vinter 31

Where are NURBS in Catia V5
workshop DOCUMENTATION
PowerFit creates a NURBS surface
Creation of 3D NURBS curves
NURBS Formats in APT Output
Réponses DS :
« Elles sont supportées dans CATIA mais on n'offre pas de moyen d'en créer en
interactif. Quand elles existent, elles proviennent d'un import »
« L'interactif ne permet pas de travailler avec un degré inférieur à 5 ni de choisir
l'espacement paramétrique entre 2 noeuds (en V4 on pouvait faire tout ça) »
Pierre Vinter 32

SIMPLE SURFACES
The C(v) curves are lines,
circles,polynomials or others which are
used for the generation of surfaces
Revolution of a profile around an axis
S rev (u,v) = C(v) ( cos U e 1 + sin U e 2 ) + v e 3 + OA
Extrusion of a profile in a given direction.
S éti (u,v) = C 1 (u) + v e3 + OPo
Pierre Vinter 33

NUPBS CATIA
Extruded surfaces
OM(u,v)=
In the v-direction, controls points
OP i,0 = OP i and OP i,1 = OP i+ ∆ n
A curve extrude in the direction of the
unit vector n, through a distance ∆
n
1
∑∑
i= 0 j=
0
N i, d(
u)
Nj,1(
v)
OPi,
j
Knots vector
( v 0 , v 1 , v 2 , v 3 ) = ( 0, 0, 1, 1 )
Pierre Vinter 34

Swept surface
Profil C(u) degree d
Guide B(v) degree e
NUPBS
OM(u,v)=
n
p
∑∑
i= 0 j=
0
N i, d(
u)
Nj,
e(
v)
OPi,
j
Controls points
OP i,j = OC i + OB j
P 23
NURBS
C 2 B 3
Pierre Vinter 35

Ruled
Surfaces
Points Ai sur la courbe C 1
C 1
C 2
Points Bi sur la courbe C 2
OM(u,v)=
n
1
∑∑
i= 0 j=
0
N i, d(
u)
Nj,1(
v)
OPi,
j
Controls points
OP i,0 = OA i
OP i,1 = OB i
The specified curves C i have the same degree and the same knot vector
Pierre Vinter 36

Surfaces of revolution
Circle
Axe
Pierre Vinter 37

Analyze the surfacique curvature
kn
Plan P o
=
Ldu
Edu
2
2
N
T
+ 2Mdudv
+ Ndv
+ 2Fdudv
+ Gdv
C(u,v)
Surface
2
2
with
L
E
=
k n normal curvature on 1 point
d
∂C
∂C
•
∂u
∂u
2
∂ C
= • N
2
∂u
2
M
C(
u,
v)
ds
• N =
k n
Pierre Vinter 38
2
N : normal
vector of the
surface
∂C
∂C
F=
•
∂u
∂v
=
2
∂ C
• N
∂u∂v
Evolution of the curvature k n for the curves C(u,v) on the surface according to
(du/dv) when the plan P o containing the normal N carry out a rotation around N.
N
=
G
N
∂C
∂u
∂C
∂u
=
∂C
Λ
∂v
∂C
Λ
∂v
∂C
∂v
2
∂C
•
∂v
∂ C
= • N
2
∂v

Solutions The extremum values of k N are solution of :
( EG − F
2
) k
2
− ( EN + GL − 2FM
) k + LN − M
2
n
n
=
0
2H
K =
= k
k
n
EN + GL − 2FM
min + kn
max =
=
2
EG − F
2
LN − M
n max . kn
min = =
2
EG
−
F
mean _ curvature
Gaussian_
curvature
Pierre Vinter 39

Analysis with reflect lines
A reflection line on the surface X(u,v) is the mirror image of
the light line L on X(u,v) while looking from fixed eye ep.
X(u,v) surface parametrized
N(u,v) normal vector of the surface
It shows the geometry for a point on the surface where an observer sees a
reflection. For such a point, the display program would then assign a bright white
color to the surface. For points where a reflection is not seen, the standard color
of the surface is chosen. This color is darker than the white reflection color. The
simulation now determines the correct color value for a large number of points on
the surface
Pierre Vinter 40

Analysis with isophotes
Definition of isophotes using an eye direction e p : all points with a
constant angle α between e p and the normal N(u,v) lie on an isophote.
Light direction vector = r
r . N(u,v) = Constant
Isophote depends
only on normal, not
position
Line or surface joining the
points of equal brightness or
luminous intensity of a
given source
Pierre Vinter 41

CONCLUSIONS
Niveau mathématique insuffisant des
étudiants Bac +3 pour suivre jusqu’au bout,
Encore certains points à développer :
NUPBS,
Cohérence entre les calculs et la modélisation
Analyses
Fiabilité de la documentation DS !
Plus facile en V4
Freestyle à orienter « Plan de forme »
Pierre Vinter 42

Pierre Vinter 43