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Computer Aided Geometric Design

Class Exercise #4

Differential Geometry of Curves

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Q.1

Let be an arc-length parameterized curve. Prove:

1. is a straight line.2. is a plane curve.

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Solution - Part 1 (Recall that is the magnitude of the curvature vector:

Therefore,

and by integration:

for some real constants

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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:

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Solution - Part 1 (and we obtained the parameterization of the straight line passing through with the direction

(Conversely – the same with differentiation!)

(𝑏1 ,𝑏2)

(𝑎1 ,𝑎2)

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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?

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Solution - Part 1 (Answer:

This should also be true for non arc-length parameterization…

Recall that for non arc-length:

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Solution - Part 1 (if then:

is the zero vector.

When does a cross product vanish?

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Solution - Part 1 (when and are the same (or opposite) directions.

We understand this as follows: The tangent never changes direction, only magnitude.

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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….)

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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 .

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Solution - Part 2 (In our case: we show that:

where is the bi-normal at some point.

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Solution - Part 2 (let for some . since:

we have:

which means is a constant vector.

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Solution - Part 2 (Now we look at:

and therefore:

and is planar.

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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:

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Solution - Part 2 (Points to think about: Is this the case for non arc-length

parameterizations?What if ? Is the torsion well defined?

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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.

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SolutionIt suffices to show that:

is planar. has constant curvature.

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SolutionThe assumption on the normal vectors means there exists a scalar function such that:

Where is a constant point in .

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SolutionDifferentiating gives:

Which means:

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Solutionor:

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SolutionWe expressed the zero vector as a linear combination of at every .

Therefore:

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SolutionIn the case , we get , and the entire curve is a point.

In the other case: , as required.

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SolutionSome 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??

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SolutionShouldn’t something bother us about

writing:

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Q.3

Let be a parameterized quadratic curve. Prove that is planar.

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SolutionAgain, 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.

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Solution is quadratic. We may assume:

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SolutionDifferentiating gives:

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SolutionWe know that has the same direction as the unit tangent .

Does has the same direction as the unit normal ?

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SolutionNot necessarily. It is true for arc-length parameterizations.

(in fact for any constant speed…)

But: is contained in the plane spanned by .

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SolutionThis 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!

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SolutionIt 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.

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SolutionThus, the bi-normal direction is constant – and the curve is planar.

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Q.4

Consider the curves:

and:

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Q.4

Find rigid motions on such that it will be pieced together with and the resulting curve is:

1. Continuous ().2. Geometrically .3. .

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SolutionThe curves are the following:

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0

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Solution has a tangent:

and at

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Solution is the lower right quarter of the unit circle, inflated to a radius of length .

For continuity we simply translate by:

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SolutionThis gives:

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-15

-10

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0

5

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SolutionFor Geometrically we need first to rotate such that the tangents have the same direction.

Since:

and:

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SolutionThe angle between these tangents is given by:

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SolutionWhich gives:

or:

It remains to rotate and translate:

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SolutionWe have the rotation angle, but what is , the translation vector?

After the rotation we have:

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-15

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0

5

10

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SolutionThe image of under the rotation is:

Then:

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SolutionFinally:

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-15

-10

-5

0

5

10

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20

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SolutionIn order for the resulting curve to be , we need to change the magnitude of the tangent.

Denote by the speed ratio:

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SolutionRe-parameterize as follows:Instead of:

take:

This results in the same trace, but the speed is factored by , as required.

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SolutionRe-parameterize as follows:Instead of:

take:

This results in the same trace, but the speed is factored by , as required.

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SolutionSome points to think about:

How do we re-parameterize as a single curve?

In practice – how do we rotate without explicitly invoking ?