T. Soient and S. , T(s,t) deux surfàces difTérentiables et (ug,vg) (sO,tO) les coordonnées paramétriques d'un point d'intersection non tangentielle et n'appartenant au bord d'aucune des deux surfaces. Alors la courbe d'intersection passant par ce point adntet une représentation paramétrique dans le voisinage de ce point

. Su, Tr(sg,t6) et T1(sg,t6) une base dans l'espace, sinon notre point est un point d'intersection tangentielle

. Maintenant, sltpposons que les 3 prerliers vecteurs forntent cette base. La fonction différentielle de q par rapport à u, v et s en (1r0, u0 ,tg) est inversible ôu, p.5

. Sans-perte-de-généralité, (a+b)], et on prend t:a ( notons Que e"(t)=1'(t) )On obtient ensuite: -e(t0) e(to ) = e(

. Ceci-entraîne, (to)ll.llfl (a -x)r"r.lo.ll ..:T!ollr"r,{lli r. -'.ro'.ll = I ,rp

S. Le-point, ug,vg) se projetant orthogonalement en T(sg,tg) sur la surface T, est situé à une certaine distance w6 de T(sg,tg)

. Maintenant, Pour cela, on écrit le système définissant I'irnage de (u,v) par la fbnction h

L. 'du-'du-'du, du f. S(u,v) -T(s,t),I(s, t) >= 0

K. Où-la-matrice-carré and . Est-donnée-par, Ku=(I'It-<S-T.Tt i' j = s' t

. De-la-même-manière, ltor,rr déterminer les clérivées de s et t par rapport à v. On dérive par rapport à v le système défirrissant lr(u,v), et on obtient ensttite: --T lds dtl l

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