Computer Aided Geometric Design

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, and by integration: for some real constants 3

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Solution Differentiating gives: Which means: 20

Solution or: 21

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

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

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

Solution Shouldn’t something bother us about writing: 25

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

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

Solution is quadratic. We may assume: 28

Solution Differentiating gives: 29

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

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

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

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

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

Q.4 Consider the curves: and: 35

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

Solution The curves are the following: 37

Solution has a tangent: and at 38

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

Solution This gives: 40

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

Solution The angle between these tangents is given by: 42

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

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

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

Solution Finally: 46

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

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

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

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

Computer Aided Geometric Design