Computer Aided Geometric Design

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Computer Aided Geometric Design. Class Exercise #3 Conic Sections – Part 2 5-Point Construction Rational Parameterizations. 1. Q.4. For and , what is the formula for the family of implicit conics passing through and with tangent directions and ?

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Computer Aided Geometric Design

Class Exercise #3

Conic Sections – Part 2

5-Point Construction

Rational Parameterizations

1

Q.4

For and , what is the formula for the family of implicit conics passing through and with tangent directions and ?

Calculate the value of the last degree of freedom which could force the conic to interpolate through an arbitrary point .

2

Solution

Recall the following 5-point construction:

3

Solution

When seeking a conic that interpolates , we write the one parameter family of implicit conics:

4

Solution

where are the implicit line representations:

5

Solution

(Check for yourself:

must interpolate !)

6

Solution

On the other hand:

must be a conic section, since it is a quadratic form.

7

Solution

when and coincide, we have:

8

Solution

Which means we require tangency to and and interpolation of three points.

9

Solution

In our case:

10

Solution

Therefore is:

11

Solution

The degree of freedom is determined after selecting the point :

hence:

12

Solution

…. What if ?

13

Q.5

Given three points and , and a blending constant , determine the discriminant of the resulting conic.

For which values of is it an ellipse? A parabola? A hyperbola?

14

Solution

Recall the coordinates given by:

15

Solution

In this coordinate system:

is an implicit form for :

16

Solution

Therefore, by the same construction as in the previous question, the implicit curve is given by:

or, for we write:

17

Solution

We now find the discriminant with respect to the implicit conic in the coordinates:

18

Solution

Rearranging:

Therefore the discriminant is:

19

Solution

we see that this is a quadratic expression in . What are the possible signs?

which is zero for or .

20

Solution

Going back to the construction we see that gives:

This is a singular case that does not satisfy our geometric demands of tangency at and .

21

Solution

Check yourself:

How can this be?? We constructed with the demand for tangency!

22

Solution

Hint: What happens to the normal in the degenerate implicit form of ?

Does the analyticdemand of orthogonality with the gradient still mean what we hope it means, geometrically?

23

Solution

We conclude that in the coordinates the following holds:

Parabola

Ellipse

Hyperbola.

24

Solution

How, if at all, does the answer change in the coordinate system?

How is the above related to the coordinate change mapping ?

Why were the values of restricted? How is this related to the values of ?

25

Q.6

Let:

Assume , where and are the ratios determined by the tangent to a point on the curve (as in the next slide).

26

Q.6

Recall from lectures that is constant for the entire curve (in fact characterizing it)

27

Q.6

Give a rational parameterization of the curve, with

28

Solution (The Quick Way)

The quick solution:

remember the formula for the special case where :

29

and for our points this gives:

namely:

and we are done…

Solution (The Quick Way)

30

Solution (The Long Way)

31

Consider the points:

As in the figure:

Solution (The Long Way)

32

Parameter values of shall correspond to the varying intersection points of the tangent to the curve with the axes:

Choose:

Solution (The Long Way)

33

and since (special case!):

which gives:

hence:

Solution (The Long Way)

34

and in lectures the following relation between the tangency point and was shown (by differentiation):

Solution (The Long Way)

36

Combining gives:

Solution (The Long Way)

37

By definition of , and from our choice of , we have:

Finally – we can substitute for .

Solution (The Long Way)

38

Calculate the left term:

Solution (The Long Way)

39

A similar calculation gives for the right term, which means we got the same result as in the short solution with the formula.

Solution (The Long Way)

40

which is a combination of three given points on a polygon with special blending coefficients.

This is a special case of much more general representations you shall see later in lectures.

A Final Remark

42

Q.7

Given:

Provide a rational parameterization for the conic section through

Which curve is it?

43

Solution

We shall parameterize with . The following was shown in lectures:

44

Solution

We know all but – the product of the ratios.

To find we recall:

so we need to find …

45

Solution

Remember that are all equivalent ways of specifying the unique conic.

They are determined, in our case, by the fourth given point:

46

Solution

Substituting the forth point in the implicit

equation:

is not yet possible, since it is given in coordinates.

47

Solution

From the usual mapping:

This is an easy linear system in as unknowns, solution of which is:

48

Solution

This can be plugged in:

giving:

49

Solution

Therefore:

This was the last unknown in our parameterization:

50

Solution

This is a hyperbola, since , which is smaller than .

51

Solution