CSC 308 Graphics Programming

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CSC 308 Graphics Programming

• Curves
• Information modified from Fergusons Computer
  Graphics via Java, and
• Principles of Three-Dimensional Animation,
  Third Ed. by M. ORourke

Dr. Paige H. Meeker Computer Science Presbyterian
College, Clinton, SC
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First, a few announcements

• Program Questions?
• REMINDER Midterm next Wednesday, 10/4
• Whats on it?
• Review class notes and
• Shirley, Ch. 1
• Shirley, Ch. 2
• Shirley, Ch. 6.1
• Shirley, Ch. 15
• Java Coding Graphics Related Questions

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What are we doing today?

• Weve been working in the world of straight lines
  this not being particularly natural, lets take
  a look at what curves can do!

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Types of Lines

• Lines can be straight or curved
• Differences between these?
• Mathematical description
• Behavior as they are used to model
• Type of structures (2d and 3d) that are created
• Visual appearence

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Straight Lines

• No curvature
• Defined by 2 endpoints
• Has slope, does not have change of angularity
• Often called Polygonal Lines

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Curved Lines

• Important in the real world, esp. the natural
  world
• Can be 2d or 3d
• Approaches
• Linear approximation
• Splines

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Linear Approximation

• aka Polyline technique
• Curves are represented as a series of points
  Dot to Dot
• First degree curve

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Linear Approximation

• Advantages?
• Disadvantages?

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Linear Approximation

• Advantages SIMPLE
• Disadvantages
• Awkward to edit
• Number of points required for good approximation
  may be large
• Curve is never truly smooth

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Linear Approximation
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Splines

• The term spline comes from the shipping era,
  when ships were made from wood. To build a plank
  of wood that would conform to the ships hull,
  shipbuilders would force the wooden plank between
  several fixed posts, called ducks. The result
  was a curved plank called a spline. The
  placement of the ducks determined how much
  curvature the finished spline would have.

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Splines

• Computer graphics version
• Curve wood
• Control points ducks
• Hull straight lines that connect the control
  points

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Splines
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Splines

• Do they clear up the disadvantages of linear
  approximation?
• (the answer is yes!)
• How?

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Splines

• Easy to reshape pulling a single control point
  can modify an entire section of the curve and do
  so smoothly

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Splines

• Truly, mathematically curved no matter how
  close you zoom in, the curve doesnt fall apart

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Splines

• Programming representation is compact and
  efficient a handful of control points define
  the entire curve.

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Categories of Splines

• Interpolating
• Approximating

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Interpolating Splines

• Spline passes through each of the control points

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Interpolating Splines

• Advantages?
• Disadvantages?

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Interpolating Splines

• Advantages?
• Disadvantages?
• There is a direct relationship between the
  control points and the final curve.

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Interpolating Splines

• Cardinal Spline type of interpolating spline
  where the curve passes through all the control
  points except the first and last 2nd degree
  curve.

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Approximating Splines

• Spline passes near but not through the control
  points
• Examples
• B-spline (Basis Spline)
• Bezier Spline

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B-Spline

• Curve begins at approximately the 2nd control
  point and ends at approximately the 2nd to last
  control point
• Third degree curve

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Bezier Spline

• Defined by 4 control points
• Contains tangent vectors at its ends to direct
  the curve
• Can increase by attaching more segments
• Third degree curve

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NURBS

• Non-Uniform Rational B-Splines
• Shares features of previous splines
• Goes through first and last control points
  (interpolating splines)
• Does not go through intermediate control points
  (approximating splines)
• Has a set of edit points or knots that lie
  exactly on the curve in addition to control
  points
• Edit control points if need curve to be smooth
• Edit knots if you need the curve in an exact
  position

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Implementation of Curves in Java?

• Lets begin by defining the knots (some points
  along the curve), and then calculate the other
  points inbetween.

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Polynomial parametric eqns

• To represent a straight line - linear parametric
  equation (i.e. one where the highest power of t
  was 1).
• For curves, we need polynomial equations. Why?
  Because the graphs of polynomial equations
  wiggle!

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Cubic parametic eqns

• The general form of a cubic parametric equation
  (in t) is
• xt a0 a1t a2t2 a3t3
• similarly for y
• yt b0 b1t b2t2 b3t3

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Why focus on cubic equations?

• Quadratic equations are not usually wiggly
  enough
• Higher powers are usually too wiggly and
  require much more computation.

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How do we make it wiggle where we want it to?

• Think about those (just consider the x eqn)
• xt a0 a1t a2t2 a3t3
• We want xt
• We control t
• We therefore need values for a0 , a1, a2 a3
• Likewise for y

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Obtaining a0, a1, a2 a3

• How do we make it wiggle where we want?
• Consider two contiguous segments Si and Si1

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Eqn A

• Each segment is represented by a separate cubic
  parametric eqns (only x shown)
• In Si
• In Si1
• Consider what happens at Ni

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3 Important Points

• The segments meet
• The last x and y values in Si are equal to the
  first x and y values in Si1
• the join is continuous
• The gradient at the end of Si is equal to the
  gradient at the start of Si1
• .and smooth
• The gradient of the gradient at the end of Si is
  equal to the gradient of the gradient at the
  start of Si1

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1. The lines meet

• Thus
• At Ni in segment Si t1
• At Ni in segment Si1 t0
• so

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2. The joint is continuous

• Gradient at the end of segment Si gradient at
  the beginning of Si1. What is the gradient? To
  find it
• First differentiate eqn A w.r.t. t (giving eqn
  B)

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2. The joint is continuous

• First differentiate eqn A w.r.t. t (giving eqn
  B)

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2. The joint is continuous

• Again consider what happens at Ni
• At Ni in segment Si t1
• At Ni in segment Si1 t0

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3. The joint is smooth

• Differentiate (eqn B) w.r.t. t

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3. The joint is smooth

• Again consider what happens at Ni
• At Ni in segment Si t1
• At Ni in segment Si1 t0

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Final boundary condition

• Currently have 3 eqns in 4 unknowns
• Need one more boundary condition
• From endpoints of whole line
• Fixed gradient
• give a value for end(s) of curve
• Free end
• 2nd derivative is 0
• Contour
• Ends of line meet (loop) gradient at 1st point
  gradient at last point

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Solve set of simultaneous eqns

• To find the values of a0a3 for each segment
• Gaussian elimination
• x 2(0.5-t)(1-t)xNi-1 4t(1-t)xNi
  2t(t-0.5)xNi1
• Similarly for y

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What have we achieved?

• If we know three points on a curve we can
  calculate any others
• Consider a curve 3 points at a time (red ones)
• Interpolate to generate sufficient intermediate
  points (white)
• Join with short straight lines