CS 376 Introduction to Computer Graphics 04 / 25 / 2007 Instructor: Michael Eckmann.

CS 376Introduction to Computer Graphics

04 / 25 / 2007

Instructor: Michael Eckmann

Michael Eckmann - Skidmore College - CS 376 - Spring 2007

Today’s Topics• Questions?

• Piecewise Cubic Parametric curves– Hermite– Bezier– Spline


Curves• There are several families of cubic polynomial curve segments that are

determined by different things– Hermite

• determined by the 2 endpoints and the tangent vector at each of the 2 endpoints

– Bezier• determined by the 2 endpoints P

1 & P

4 and 2 other intermediate

points P2 & P

3 not on the curve

– the tangent vectors directions at the end points are P1P

2 and

P3P

4– which are [P

2 – P

1] and [P

4 – P

3] (note: this direction was

correct) Magnitude is to be defined later this lecture...– Spline

• determined by 4 specific points

• Note: there can be higher degree polynomials of these families of curves

(see a few slides ahead for a 4th degree polynomial Bezier curve segment)

Bezier Curve segment examples

more Bezier Curve segment examples


Recall• x(t) = a

xt3 + b

xt2 + c

xt + d

x

• y(t) = ayt3 + b

yt2 + c

yt + d

y

• z(t) = azt3 + b

zt2 + c

zt + d

z where 0 <= t <= 1

• can be written in matrix form as:

• Q(t) = [x(t) y(t) z(t)] = T C • where T = [t3 t2 t 1] and

[ax

ay

az]

• C = [bx

by

bz]

[cx

cy

cz]

[dx d

y d

z]


Basis Matrix and Geometry Matrix• Q(t) = [x(t) y(t) z(t)] = T C

• where T = [t3 t2 t 1] and

[ax

ay

az]

• C = [bx

by

bz]

[cx

cy

cz]

[dx d

y d

z]

• can be rewritten so that C = M G where – M is a 4x4 matrix called the Basis Matrix and – G is a 4x1 column vector called the Geometry Matrix

• so Q(t) = T M G


Basis Matrix and Geometry Matrix• Q(t) = [x(t) y(t) z(t)] = T M G

• where T = [t3 t2 t 1] and

[m11

m12

m13

m14

]• M = [m

21 m

22 m

23 m

24]

[m31

m32

m33

m34

][m

41 m

42 m

43 m

44]

[G1]

• G = [G2]

[G3]

[G4]

Note: the values of Gi are the conditions that define the curve --- such as

endpoints and tangent vectors at those end points


Blending functions• look at x(t) • x(t) = (t3 m

11+ t2 m

21 + t m

31 + m

41)g

1x + (t3 m

12+ t2 m

22 + t m

32 + m

42)g

2x +

(t3 m13

+ t2 m23

+ t m33

+ m43

)g3x

+ (t3 m14

+ t2 m24

+ t m34

+ m44

)g4x

where g1x

is the x coordinate of G1

similar equations for y(t) and z(t)

the curve is a weighted sum of the elements of the Geometry Matrix where the weights are cubic polynomials of t which are called the Blending Functions

the Blending Functions B are given by B = T M, since Q(t) = T M G,Q(t) = B G


Example with lines• Before going through an example to find the blending functions from a

set of geometry information for parametric cubic polynomials, let's first see how the procedure works with determining the blending functions of a parametric line.


Example with lines• Recall the parametric equation of a line:

• x(t) = x0 + t ( x

end – x

0)

• y(t) = y0 + t ( y

end – y

0)

• z(t) = z0 + t ( z

end – z

0)

• what are the endpoints of the line segment described above?

can be rewritten as: • x(t) = a

xt + b

x

• y(t) = ayt + b

y

• y(t) = azt + b

z


• x(t) = axt + b

x

• y(t) = ayt + b

y

• y(t) = azt + b

z To get this in the form of Q(t) = T C = T M G

• T = [ t 1 ] and • C = [ a

x a

y a

z ]

• [ bx b

y b

z]

• C = M G where

• M = [ m11

m12

] and G = [ g1x

g1y

g1z

]• [ m

21 m

22 ] [ g

2x g

2y g

2z ]

Example with lines


• The geometry matrix is simply the two endpoints • G = [ G

1 ] = [ g

1x g

1y g

1z ] = [ x

0 y

0 z

0 ] = [ P

0 ]

• [ G2 ] [ g

2x g

2y g

2z ] [ x

end y

end z

end ] [ P

1 ]

• we need to find out the Basis Matrix M• M = [ m

11 m

12 ]

• [ m21

m22

]

• Q(t) = T M G = [ t 1 ] M G

• Let's do the rest on the board to figure out M

Example with lines


• Q(t) = [ t 1 ] M G• when t=0, Q(0) is the first endpoint G

1

• when t=1, Q(1) is the first endpoint G2

• Q(0) = G1

= [ 0 1 ] M G and• Q(1) = G

2 = [ 1 1 ] M G

• [ G1 ] = [ 0 1 ] M [ G

1 ]

• [ G2 ] [ 1 1 ] [ G

2 ]

• Therefore, • [ 0 1 ] M = I (the identity matrix)• [ 1 1 ] • So, to get M, take the inverse of [ 0 1 ]• [ 1 1 ]

Example with lines


• So, to get M, take the inverse of [ 0 1 ]• [ 1 1 ]

The inverse of a 2x2 invertible matrix A = [ a11

a12

] =

[ a21

a22

]

1 [ a22

-a12

]

---------------- [ -a21

a11

]

a11

a22

- a12

a21

So,M = (1 / (0-1)) [1 -1] = -1 [1 -1] = [ -1 1 ] [-1 0] [-1 0] [ 1 0 ]

Example with lines


• Q(t) = T M G = [ t 1 ] M G = [ t 1 ] [ -1 1 ] G • [ 1 0 ]

• Recall that T M are the blending functions. • What then, are the blending functions of lines?

• Let's plot them on the board.

Example with lines


• Q(t) = T C = T M G = [ t3 t2 t 1 ] M G• M is the 4x4 Basis matrix and G is the Geometry matrix• For Hermite curve segments, recall that they are defined with 2 endpoints

and the tangents at those endpoints. So, the Geometry matrix is made up of the 2 endpoints and the 2 tangent vectors.

• The tangent vectors are determined by the derivative of the curve with respect to t at each of the end points (when t=0 and when t=1).

• x(t) = axt3 + b

xt2 + c

xt + d

x

• y(t) = ayt3 + b

yt2 + c

yt + d

y

• z(t) = azt3 + b

zt2 + c

zt + d

z where 0 <= t <= 1

• x'(t) = 3axt2 + 2b

xt + c

x

• y'(t) = 3ayt2 + 2b

yt + c

y

• z'(t) = 3azt2 + 2b

zt + c

z

Hermite Basis Matrix and Blending Functions


• x'(t) = 3axt2 + 2b

xt + c

x

• y'(t) = 3ayt2 + 2b

yt + c

y

• z'(t) = 3azt2 + 2b

zt + c

z

So, Q'(t) = [ 3t2 2t 1 0 ] C• agreed?• Let's call the 2 endpoints in the Geometry Matrix P

1 and P

4 and the 2

tangent vectors at those 2 endpoints R1 and R

4.

• [ P1 ]

• G = [ P4 ]

• [ R1 ]

• [ R4 ]



• Let's just work with the x coordinates (the y and z coordinates will be handled similarly)

• So, x(t) = axt3 + b

xt2 + c

xt + d

x = T C

x = [ t3 t2 t 1 ] M G

x

• and x'(t) = 3axt2 + 2b

xt + c

x = T C

x = [ 3t2 2t 1 0 ] M G

x

• when t = 0, x(t) gets us the first endpoint and when t = 1, x(t) gets us the other endpoint.

• x(0) = P1x

= [ 0 0 0 1 ] M Gx

• x(1) = P4x

= [ 1 1 1 1 ] M Gx

• when t = 0, x'(t) gets us the tangent at the first endpoint and when t = 1, x'(t) gets us the tangent at the other endpoint.

• x'(0) = R1x

= [ 0 0 1 0 ] M Gx

• x'(1) = R4x

= [ 3 2 1 0 ] M Gx



• [P1 ] [ 0 0 0 1 ]

• [P4 ] = G

x = [ 1 1 1 1 ] M G

x

• [R1 ] [ 0 0 1 0 ]

• [R4 ]

x [ 3 2 1 0 ]

• just like when we solved for M (the basis matrix) with the line equations, we do the same thing here because clearly

• [ 0 0 0 1 ] • [ 1 1 1 1 ] M = Identity• [ 0 0 1 0 ] • [ 3 2 1 0 ]• so, that matrix on the left is equal to M-1



• There are techniques to get the inverse of a matrix which we won't go into here --- if you've had linear algebra, you should know how.

• The inverse of• [ 0 0 0 1 ] • [ 1 1 1 1 ]• [ 0 0 1 0 ] • [ 3 2 1 0 ]• is• [ 2 -2 1 1 ] • [ -3 3 -2 -1 ]• [ 0 0 1 0 ] • [ 1 0 0 0 ]• This is the Basis Matrix for Hermite curve segments. • Let's verify that it is indeed the inverse of the matrix above. How?



• [ 2 -2 1 1 ] • T M = [ t3 t2 t 1 ] [ -3 3 -2 -1 ]• [ 0 0 1 0 ] • [ 1 0 0 0 ]• The blending functions are:

2 t3 - 3t2 + 1 -2 t3 + 3t2

t3 - 2t2 + t t3 - t2

• These are respectively multiplied by P1, P

4, R

1, and R

4

and then added together to get Q(t) which is the curve



• [ P1 ]

• G1 = [ P

4 ]

• [ R1 ]

• [ R4 ]

• [ P4 ]

• G2 = [ P

7 ]

• [ kR4 ]

• [ R7 ]

• If two Hermite curve segments have the above geometry matrices then we can see that they join up at P

4 and that the tangent vectors at P

4 are

proportional, hence giving G1 continuity. If k=1 then C

1 continuity.

Using the Geometry matrices to join curves together


• As stated earlier, the pieces of information defining a Bezier curve are 4 points. Therefore, the Geometry Matrix are these 4 points.

• P1 and P

4 are the endpoints and P

2 and P

3 are the intermediate points that

do not necessarily (usually don't) live on the curve.• R

1 (the tangent at the first endpoint) = 3 [P

2 – P

1]

• R4 (the tangent at the other endpoint) = 3 [P

4 – P

3]

• [ P1 ]

• G = [ P2 ]

• [ P3 ]

• [ P4 ]

• Q(t) = [ t3 t2 t 1 ] M G

Bezier Basis Matrix and Blending Functions


• The Basis Matrix M for Bezier curves is:

• [ -1 3 -3 1 ] • [ 3 6 3 0 ] • [ -3 3 0 0 ] • [ 1 0 0 0 ]

• Q(t) = [ t3 t2 t 1 ] M G• So, what are the blending functions?



• The Blending functions are:-t3 + 3t2 -3t + 13t3 + 6t2 + 3t-3t3 + 3t2 t3

• These functions are the Bernstein polynomials which are of the form:

• C(n, k) tk(1 – t )n-k

• where C(n,k) is the choose function. C(n,k) = n! / (k! ((n-k)!))

• In the case where n=3 (cubic), we have k among 0,1,2,3, which gives one

function per geometric element (point.)



• A nice feature of Bezier curves is:– Because the blending functions are symmetric to the lines t and 1-t,

the sequence of points used to define a Bezier curve can be reversed without changing the shape of the curve.

• Both Bezier and Hermite curves are easy to make have G1 or C

1

continuity at all the join points.

• It is not easy though to get C2 continuity at the join points for Hermite

and Bezier curves.

• Hermite and Bezier curves interpolate the points (that is, the curves go

through the points).

• Splines are C2 continuous.

Bezier/Hermite Curves vs. Splines


• Splines are C2 continuous.

• Natural cubic splines interpolate the control points and the coefficients of a natural cubic spline are dependent on all n control points.

– expensive to invert an n+1 by n+1 matrix– moving one control point affects the entire curve

• B-splines are defined by m+1 control points, where m>=3. The control

points are named P0 through P

m.

– There are m-2 cubic polynomial, C2 continuous curve segments

joined together. These segments are named Q3 through Q

m.

– The join points (as well as the endpoints) of the B-spline are called knots. There are m-1 knots.

– moving one control point has only a local effect (that is a good thing)– clearly the B-splines approximate (not interpolate) the control points– Let's look at the handout

Splines


• Uniform = spacing between knots are equal (that is the difference in t between knots is uniform)

– the blending functions for each segment have the same shape but are shifted

• Nonuniform = spacing between knots are unequal – more flexibility to control the curve shape than uniform– 2 or more consecutive knots that are the same reduce the continuity

there (e.g., if we have 0 difference between 2 or more knots then continuity is reduced by 1 (i.e., C

2 -> C

1 continuity))

• Rational = each spline curve segment is defined as a ratio of polynomials– are invariant under rotation, scaling, translation AND perspective

transformations of the control points. So, we apply the perspective transform to the control points and then generate the perspectively transformed curve from the transformed control points to obtain the correct view of the curve.

– in addition to the myriad of curves that can be produced, they can precisely define the conic sections (e.g. circle, ellipse, parabola, hyperbola)

• NURBS = NonUniform Rational B-Splines

B-Splines


• NURBS = NonUniform Rational B-Splines– used frequently in graphics packages due to the properties just

described• Nonrational = each spline curve segment is defined as a polynomial (not

a ratio of polynomials)– has the disadvantage (compared to rational B-splines) that they are

not invariant to perspective transformations as well as cannot precisely describe the conics

B-Splines


• Recall that we defined parametric cubic curves as Q(t) = T M G.• For surfaces we define a parametric cubic surface with two parameters,

hence Q(s,t)• for some particular value of s, say s

1, Q(s

1,t) is a parametric curve. Also,

for some particular value of t, say t1, Q(s,t

1) is a parametric curve.

• [ G1(t) ]

• Q(s,t) = S M G(t) = S M [ G2(t) ]

• [ G3(t) ]

• [ G4(t) ]

Curve Surfaces

CS 376 Introduction to Computer Graphics 04 / 25 / 2007 Instructor: Michael Eckmann.

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Transcript of CS 376 Introduction to Computer Graphics 04 / 25 / 2007 Instructor: Michael Eckmann.