TAILIEUCHUNG - Generalized Curvatures Part 8

Tham khảo tài liệu 'generalized curvatures part 8', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | 148 15 Approximation of the Area of Surfaces 2. In particular if I sin 01 I E and I sin 021 E then t ĨÕE sin ap . sin Y Proof of Lemma 4 It is a simple computation which we do not reproduce here. The complete proof can be found in 66 . Comparing the Length of a Geodesic and Its Chord Proposition 10. Let S be a smooth compact surface of E3 Us be a neighborhood of S where the map pr US S is well defined and p and q be two points on S such that p q c US and pr p q c S dS. Then the distance lpq between p and qonS satisfies -q pq --SpY-q- Proof of Proposition 10 The left inequality is trivial. On the other hand since pr p q is a curve on S its length is larger than the length lpq of the minimal geodesic on S whose ends are p and q. Therefore using the mean value theorem one has lpq l pr p q sup ID pr m I pq. me p q Since Proposition 3 implies that ID pr m I 1 1 1 -Ipr m - mIIhIpr m 1 - S pq Proposition 10 is proved. Comparing the Normals at a Vertex Proposition 11. Let S be a smooth surface t be a triangle closely inscribed in S and p be a vertex oft. Then the angle ap e 0 2 between the normals of S and t at p satisfies ĨÕ t sin ữp . 2 sin p 1 - rnS t where Yp is the angle oft at p. This proposition is a consequence of the following lemma. Lemma 5. Let S be a smooth surface and let p q e S such that p q c S. Then the angle 0 e 0 2 between -q and the orthogonal projection of -q onto TpS satisfies A Bound on the Deviation Angle 149 sin0 i bn 2 where hS denotes the supremum over S of the norm of the second fundamental form ofS and l is the distance on s between p and q. The proof of Lemma 5 is similar to that of Proposition 8. Proof of Proposition 11 Denote by l1 the distance on S between p and p1 and by l2 the distance on s between p and p2. Since T is closely inscribed in s thanks to Corollary 10 we obtain PP1 t t l1 T- l2 - . 1 - S T 1 - 0 s 1 1 - S t Therefore Lemma 5 implies that a_ h pr 1 n S t nS t sin01 z. and sin02 . . 2 -2 1 - MS

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