Computer Aided Geometric Design

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Computer Aided Geometric Design. Class Exercise #4 Differential Geometry of Curves. 1. Q.1. Let be an arc-length parameterized curve. Prove: is a straight line. is a plane curve. 2. Solution - Part 1 (. Recall that is the magnitude of the curvature vector: Therefore,

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Computer Aided Geometric Design

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1. Computer Aided Geometric Design Class Exercise #4 Differential Geometry of Curves 1

2. Q.1 Let be an arc-length parameterized curve. Prove: is a straight line. is a plane curve. 2

3. Solution - Part 1 ( Recall that is the magnitude of the curvature vector: Therefore, and by integration: for some real constants 3

4. Solution - Part 1 ( This already means: The direction of the curve is in fact a constant direction. To verify that this is a line, we integrate again and: 4

5. Solution - Part 1 ( and we obtained the parameterization of the straight line passing through with the direction (Conversely – the same with differentiation!) 5

6. Solution - Part 1 ( Question: Where did we use the fact that the curve was arc-length parameterized? Is a general parameterization with not necessarily a straight line? 6

7. Solution - Part 1 ( Answer: This should also be true for non arc-length parameterization… Recall that for non arc-length: 7

8. Solution - Part 1 ( if then: is the zero vector. When does a cross product vanish? 8

9. Solution - Part 1 ( when and are the same (or opposite) directions. We understand this as follows: The tangent never changes direction, only magnitude. 9

10. Solution - Part 2 ( How can we show that a curve is planar? (without the knowledge of this being equivalent to which we are trying to prove….) 10

11. Solution - Part 2 ( We must show that the curve is contained in some plane. A general (implicit) representation of a plane: where is a constant (non-zero) vector, and . 11

12. Solution - Part 2 ( In our case: we show that: where is the bi-normal at some point. 12

13. Solution - Part 2 ( let for some . since: we have: which means is a constant vector. 13

14. Solution - Part 2 ( Now we look at: and therefore: and is planar. 14

15. Solution - Part 2 ( conversely: If a curve is planar then it is in fact contained in the (constant) osculating plane. This means that the bi-normal is constant: 15

16. Solution - Part 2 ( Points to think about: Is this the case for non arc-length parameterizations? What if ? Is the torsion well defined? 16

17. Q.2 Let be an arc-length parameterized curve. Prove that if all normal vectors to the curve pass through a point, then is an arc of a circle. 17

18. Solution It suffices to show that: is planar. has constant curvature. 18

19. Solution The assumption on the normal vectors means there exists a scalar function such that: Where is a constant point in . 19

20. Solution Differentiating gives: Which means: 20

21. Solution or: 21

22. Solution We expressed the zero vector as a linear combination of at every . Therefore: 22

23. Solution In the case , we get , and the entire curve is a point. In the other case: , as required. 23

24. Solution Some points to think about: Any space curve contained in a sphere can have the radius from any point pass through the center of the sphere, and yet it is not necessarily a circle. So is our proof right or not?? 24

25. Solution Shouldn’t something bother us about writing: 25

26. Q.3 Let be a parameterized quadratic curve. Prove that is planar. 26

27. Solution Again, it suffices to show that the bi-normal is constant. This shall imply that the osculating plane is constant, and the curve never leaves it. 27

28. Solution is quadratic. We may assume: 28

29. Solution Differentiating gives: 29

30. Solution We know that has the same direction as the unit tangent . Does has the same direction as the unit normal ? 30

31. Solution Not necessarily. It is true for arc-length parameterizations. (in fact for any constant speed…) But: is contained in the plane spanned by . 31

32. Solution This can be shown analytically. Geometrically, it is explained by the fact that changes the tangent in direction and in magnitude. Therefore: the direction is in the bi-normal direction! 32

33. Solution It suffices to check that is constant. For example, the first component of this cross product is: which does not depend on . Others are shown similarly. 33

34. Solution Thus, the bi-normal direction is constant – and the curve is planar. 34

35. Q.4 Consider the curves: and: 35

36. Q.4 Find rigid motions on such that it will be pieced together with and the resulting curve is: Continuous (). Geometrically . . 36

37. Solution The curves are the following: 37

38. Solution has a tangent: and at 38

39. Solution is the lower right quarter of the unit circle, inflated to a radius of length . For continuity we simply translate by: 39

40. Solution This gives: 40

41. Solution For Geometrically we need first to rotate such that the tangents have the same direction. Since: and: 41

42. Solution The angle between these tangents is given by: 42

43. Solution Which gives: or: It remains to rotate and translate: 43

44. Solution We have the rotation angle, but what is , the translation vector? After the rotation we have: 44

45. Solution The image of under the rotation is: Then: 45

46. Solution Finally: 46

47. Solution In order for the resulting curve to be , we need to change the magnitude of the tangent. Denote by the speed ratio: 47

48. Solution Re-parameterize as follows: Instead of: take: This results in the same trace, but the speed is factored by , as required. 48

49. Solution Re-parameterize as follows: Instead of: take: This results in the same trace, but the speed is factored by , as required. 49

50. Solution Some points to think about: How do we re-parameterize as a single curve? In practice – how do we rotate without explicitly invoking ? 50