Bezier Curve-

 

Bezier Curve may be defined as-

• Bezier Curve is parametric curve defined by a set of control points.
• Two points are ends of the curve.
• Other points determine the shape of the curve.

 

The concept of bezier curves was given by Pierre Bezier.

 

Bezier Curve Example-

 

The following curve is an example of a bezier curve-

 

 

Here,

• This bezier curve is defined by a set of control points b₀, b₁, b₂ and b₃.
• Points b₀ and b₃ are ends of the curve.
• Points b₁ and b₂ determine the shape of the curve.

 

Bezier Curve Properties-

 

Few important properties of a bezier curve are-

 

Property-01:

 

Bezier curve is always contained within a polygon called as convex hull of its control points.

 

 

Property-02:

 

• Bezier curve generally follows the shape of its defining polygon.
• The first and last points of the curve are coincident with the first and last points of the defining polygon.

 

Property-03:

 

The degree of the polynomial defining the curve segment is one less than the total number of control points.

 

Degree = Number of Control Points – 1

 

Property-04:

 

The order of the polynomial defining the curve segment is equal to the total number of control points.

 

Order = Number of Control Points

 

Property-05:

 

• Bezier curve exhibits the variation diminishing property.
• It means the curve do not oscillate about any straight line more often than the defining polygon.

 

Bezier Curve Equation-

 

A bezier curve is parametrically represented by-

 

 

Here,

• t is any parameter where 0 <= t <= 1
• P(t) = Any point lying on the bezier curve
• B_i = i^th control point of the bezier curve
• n = degree of the curve
• J_n,i(t) = Blending function = C(n,i)tⁱ(1-t)^n-i where C(n,i) = n! / i!(n-i)!

 

Cubic Bezier Curve-

 

• Cubic bezier curve is a bezier curve with degree 3.
• The total number of control points in a cubic bezier curve is 4.

 

Example-

 

The following curve is an example of a cubic bezier curve-

 

 

Here,

• This curve is defined by 4 control points b₀, b₁, b₂ and b₃.
• The degree of this curve is 3.
• So, it is a cubic bezier curve.

 

Cubic Bezier Curve Equation-

 

The parametric equation of a bezier curve is-

 

 

Substituting n = 3 for a cubic bezier curve, we get-

 

 

Expanding the above equation, we get-

P (t) = B₀J_3,0(t) + B₁J_3,1(t) + B₂J_3,2(t) + B₃J_3,3(t)     ………..(1)

 

Now,

 

 

Using (2), (3), (4) and (5) in (1), we get-

 

P(t) = B₀(1-t)³ + B₁3t(1-t)² + B₂3t²(1-t) + B₃t³

 

This is the required parametric equation for a cubic bezier curve.

 

Applications of Bezier Curves-

 

Bezier curves have their applications in the following fields-

 

1. Computer Graphics-

 

• Bezier curves are widely used in computer graphics to model smooth curves.
• The curve is completely contained in the convex hull of its control points.
• So, the points can be graphically displayed & used to manipulate the curve intuitively.

 

2. Animation-

 

• Bezier curves are used to outline movement in animation applications such as Adobe Flash and synfig.
• Users outline the wanted path in bezier curves.
• The application creates the needed frames for the object to move along the path.
• For 3D animation, bezier curves are often used to define 3D paths as well as 2D curves.

 

3. Fonts-

 

• True type fonts use composite bezier curves composed of quadratic bezier curves.
• Modern imaging systems like postscript, asymptote etc use composite bezier curves composed of cubic bezier curves for drawing curved shapes.

 

PRACTICE PROBLEMS BASED ON BEZIER CURVE IN COMPUTER GRAPHICS-

 

Problem-01:

 

Given a bezier curve with 4 control points-

B₀[1 0] , B₁[3 3] , B₂[6 3] , B₃[8 1]

Determine any 5 points lying on the curve. Also, draw a rough sketch of the curve.

 

Solution-

 

We have-

• The given curve is defined by 4 control points.
• So, the given curve is a cubic bezier curve.

 

The parametric equation for a cubic bezier curve is-

 

P(t) = B₀(1-t)³ + B₁3t(1-t)² + B₂3t²(1-t) + B₃t³

 

Substituting the control points B₀, B₁, B₂ and B₃, we get-

P(t) = [1 0](1-t)³ + [3 3]3t(1-t)² + [6 3]3t²(1-t) + [8 1]t³      ……..(1)

 

Now,

To get 5 points lying on the curve, assume any 5 values of t lying in the range 0 <= t <= 1.

Let 5 values of t are 0, 0.2, 0.5, 0.7, 1

 

For t = 0:

 

Substituting t=0 in (1), we get-

P(0) = [1 0](1-0)³ + [3 3]3(0)(1-t)² + [6 3]3(0)²(1-0) + [8 1](0)³

P(0) = [1 0] + 0 + 0 + 0

P(0) = [1 0]

 

For t = 0.2:

 

Substituting t=0.2 in (1), we get-

P(0.2) = [1 0](1-0.2)³ + [3 3]3(0.2)(1-0.2)² + [6 3]3(0.2)²(1-0.2) + [8 1](0.2)³

P(0.2) = [1 0](0.8)³ + [3 3]3(0.2)(0.8)² + [6 3]3(0.2)²(0.8) + [8 1](0.2)³

P(0.2) = [1 0] x 0.512 + [3 3] x 3 x 0.2 x 0.64 + [6 3] x 3 x 0.04 x 0.8 + [8 1] x 0.008

P(0.2) = [1 0] x 0.512 + [3 3] x 0.384 + [6 3] x 0.096 + [8 1] x 0.008

P(0.2) = [0.512 0] + [1.152 1.152] + [0.576 0.288] + [0.064 0.008]

P(0.2) = [2.304 1.448]

 

For t = 0.5:

 

Substituting t=0.5 in (1), we get-

P(0.5) = [1 0](1-0.5)³ + [3 3]3(0.5)(1-0.5)² + [6 3]3(0.5)²(1-0.5) + [8 1](0.5)³

P(0.5) = [1 0](0.5)³ + [3 3]3(0.5)(0.5)² + [6 3]3(0.5)²(0.5) + [8 1](0.5)³

P(0.5) = [1 0] x 0.125 + [3 3] x 3 x 0.5 x 0.25 + [6 3] x 3 x 0.25 x 0.5 + [8 1] x 0.125

P(0.5) = [1 0] x 0.125 + [3 3] x 0.375 + [6 3] x 0.375 + [8 1] x 0.125

P(0.5) = [0.125 0] + [1.125 1.125] + [2.25 1.125] + [1 0.125]

P(0.5) = [4.5 2.375]

 

For t = 0.7:

 

Substituting t=0.7 in (1), we get-

P(t) = [1 0](1-t)³ + [3 3]3t(1-t)² + [6 3]3t²(1-t) + [8 1]t³

P(0.7) = [1 0](1-0.7)³ + [3 3]3(0.7)(1-0.7)² + [6 3]3(0.7)²(1-0.7) + [8 1](0.7)³

P(0.7) = [1 0](0.3)³ + [3 3]3(0.7)(0.3)² + [6 3]3(0.7)²(0.3) + [8 1](0.7)³

P(0.7) = [1 0] x 0.027 + [3 3] x 3 x 0.7 x 0.09 + [6 3] x 3 x 0.49 x 0.3 + [8 1] x 0.343

P(0.7) = [1 0] x 0.027 + [3 3] x 0.189 + [6 3] x 0.441 + [8 1] x 0.343

P(0.7) = [0.027 0] + [0.567 0.567] + [2.646 1.323] + [2.744 0.343]

P(0.7) = [5.984 2.233]

 

For t = 1:

 

Substituting t=1 in (1), we get-

P(1) = [1 0](1-1)³ + [3 3]3(1)(1-1)² + [6 3]3(1)²(1-1) + [8 1](1)³

P(1) = [1 0] x 0 + [3 3] x 3 x 1 x 0 + [6 3] x 3 x 1 x 0 + [8 1] x 1

P(1) = 0 + 0 + 0 + [8 1]

P(1) = [8 1]

 

Following is the required rough sketch of the curve-

 

 

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