Curves

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Curves
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First of all

• You may ask yourselves What did those papers
  have to do with computer graphics?
• Valid question
• Answer I thought they were cool, unique
  applications of computer graphics
• This week were looking at scan conversion of
  curves

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Interpolation

• Bresenhams line drawing algorithm is nothing
  more than a form of interpolation
• Given two points, find the location of points in
  between them
• The function (in this case its a line) must pass
  through the given points, by definition

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Interpolation Usage

• We used Bresenham line drawing (interpolation) to
  scan convert a symbolic representation of a line
  to a rasterized line
• Other applications exist
• Computing calendar years are leap years (believe
  it or not)
• Line Drawing, Leap Years, and Euclid, Harris and
  Reingold, ACM Computing Surveys, Vol. 36, No. 1,
  March 2004, pp. 66-80
• Path planning for key frame animation

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Key Frame Animation

• Key frames are infrequent scenes that capture the
  essential flow of an animation
• Think of them as the endpoints of a line
• In between frames (tweens) may be generated by
  interpolating between two key frames
• Think of these as the points drawn by the
  Bresenham algorithm

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Approximation

• Another possibility for generating tweens is to
  specify tie-points that guide the generation of
  the intermediate path points but the path does
  not pass through the tie-points
• This technique is called approximation or curve
  fitting

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Curves

• Well look at two techniques for generating
  curves (to be used as either paths or drawn
  objects)
• Interpolation
• Approximation
• Well also see that interpolation is very
  restrictive when considering parametric curves
• Approximation is a much better approach

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Parametric Curves

• Recall the parametric line equations
• The parameter t is used to map out a set of (x,
  y) pairs that represent the line

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Parametric Curves

• In the case of a curve the parametric function is
  of the form
• Q(u) (x(u), y(u), z(u)) in 3 dimensions
• The derivative of Q(u) is of the form
• Q(u) (x(u), y(u), z(u))
• Significance of the derivative?
• It is the tangent vector at a given point (u) on
  the curve

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Derivative
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Parametric Curves

• In the case of object drawing, u is a spatial
  parameter (like t was for the line)
• In the case of key frame animation, u is a
  temporal parameter
• In both cases Q(0) is the start of the curve and
  Q(1) is the end, as was the case for the
  parametric line

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Parametric Curves

• Curvature
• Curvature k 1/ ?
• The higher the curvature, the more the curve
  bends at the given point

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Parametric Curves

• From calculus, a function f is continuous at a
  value x0 if
• In laymans terms this means that we can draw the
  curve without ever lifting our pen from the
  drawing surface
• f(x) is continuous over an interval (a,b) if it
  is continuous for every point in the interval
• We call this C0 continuity

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Parametric Curves
Continuous over (a,b) C0
Continuous over (a,b) C0
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Parametric Curves

• From calculus, a functions derivative f is
  continuous at a value x0 if
• In laymans terms this means that there are no
  sharp changes in direction
• f(x) is continuous over an interval (a,b) if it
  is continuous for every point in the interval
• We call this C1 (tangential) continuity

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Parametric Curves
Continuous derivative over (a,b) C1
Discontinuous derivative over (a,b) not C1
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Parametric Curves

• When we need to join two curves at a single point
  we can guarantee C1 continuity across the joint
• For the case when we cant make one continuous
  curve
• Just make sure that the tangents of the two
  curves at the join are of equal length and
  direction
• If the tangents at the joint are of identical
  direction but differing lengths (change in
  curvature) then we have G1 continuity

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Lagrange Polynomials

• To generate a function that passes through every
  specified point, the type of function depends on
  the number of specified points
• Two points ? linear function
• Three points ? quadratic function
• Four points ? cubic function
• Generating such functions makes use of Lagrange
  polynomials

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Lagrange Polynomials

• The general form is (n is the number of points)
• Lets look at an example

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Lagrange Polynomials

• For two points P0 and P1
• For the starting point (t00) and ending point
  (t11)

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Lagrange Polynomials

• For three points P0 , P1 , and P2
• And it only gets worse for larger numbers of
  points
• Suffice it to say, this isnt the most optimum
  way to draw curves
• Too many operations per point
• Too complex if the artist decides to change the
  curve
• But you could do it

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A Better Way

• The problem with Lagrange polynomials lies in the
  fact that we try to make the curve pass through
  all of the specified points
• A better way is to specify points that control
  how the curve passes from one point to the next
• We do so by specifying a cubic function
  controlled by four points
• The four points are called boundary conditions

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Hermite Boundary Conditions

• Two points
• Two tangent vectors
• Two of the points are interpreted as vectors off
  of the other two points

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Cubic Functions

• Generalized form
• Derivative
• Our goal is to solve these equations in closed
  form so that we can generate a series of points
  on the curve

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Cubic Functions

• There are four unknown values in the equation
• a, b, c, and D (remember, a, b, c, and D are
  vectors in x, y, z so there is actually a set of
  3 equations)
• We need to use these equations to generate values
  of x, y, and z along the curve
• We can generate a closed form solution (solve the
  equations for x, y, and z) since we have four
  known boundary conditions
• u 0 ? Q(0) P0 and Q(0) P0
• u 1 ? Q(1) P1 and Q(1) P1

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Solution of Equations

• Go to the white board

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Implementation

• So, all you have to do to generate a curve is to
  implement this vector (x, y, z) equation
• by stepping 0 u 1
• P0, P1, P0 and P1 are vectors in x, y, z so
  there are really 12 coefficients to be computed
  and youll be implementing 3 equations for Q(u)

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Implementation

• Note that youll have to estimate the step size
  for u or(any ideas?)
• use your Bresenham code to draw short straight
  lines between the points you generate on the
  curve (to fill gaps)
• There is no trick (that Im aware of) comparable
  to the Bresenham approach

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Bezier Curves

• Similar derivation to Hermite
• Different boundary conditions
• Bezier uses 2 endpoints and 2 control points
  (rather than 2 endpoints and 2 slopes)

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Bezier Curves
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Implementation

• So, all you have to do to generate a curve is to
  implement this vector (x, y, z) equation
• by stepping 0 u 1
• P0, P1, P2 and P3 are vectors in x, y, z so there
  are really 12 coefficients to be computed and
  youll be implementing 3 equations for Q(u)

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Bezier example
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Hermite vs. Bezier

• Hermite is easy to control continuity at the
  endpoints when joining multiple curves to create
  a path
• But difficult to control the internal shape of
  the curve
• Bezier is easy to control the internal shape of
  the curve
• But a little more (not much) difficult to control
  continuity at the endpoints when joining multiple
  curves to create a path
• Bottom line is, when creating a path you have to
  be very selective about endpoints and adjacent
  control points (Bezier) or tangent slopes
  (Hermite)

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Result

• Go to the demo program
• My code generates a Hermite curve that is of C1
  continuity
• As I generate new segments along the curve I join
  them by keeping the adjoining tangent vectors
  equal

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Homework

• Implement Bezier and Hermite curve drawing
• Parameters should be control points and brush
  width
• Create a video sequence to show in class
• Demonstrate Bezier and Hermite curves of various
  control points and brush widths
• Be creative
• Due date Next week Turn in
• Video to be shown in class
• All program listings
• Grading will be on completeness (following
  instructions) and timeliness (late projects will
  be docked 10 per week)