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Chapter 3: Problem 6

Describe the character drawn by the following three-piece Bezier curve: \((0,1)(0,1)(0,0)(0,0)\) \((0,0)(0,1)(1,1)(1,0)\) \((1,0)(1,1)(2,1)(2,0)\)

Short Answer

Expert verified
Based on the given control points and the analysis of the three Bezier curves, the character formed by connecting them is an "N". The first curve is a straight vertical line segment, while the second and third curves represent smooth arches moving from the bottom to the right.

Step by step solution

01

1. Understanding Cubic Bezier Curves

A cubic Bezier curve is defined by four control points (P0, P1, P2, P3) and the parametric equation for the curve is given by: \(B(t) = (1-t)^3\times P0 + 3\times (1-t)^2\times t \times P1 + 3\times (1-t) \times t^2\times P2 + t^3 \times P3\) Where \(B(t)\) is the point on the curve, \(P0, P1, P2, P3\) are the control points and t is the parameter ranging from 0 to 1.
02

2. Find Equations for the Given Bezier Curves

We have three sets of control points for each of the Bezier curves. Let's find the parametric equations for each curve. 1. First Curve: \((0,1)(0,1)(0,0)(0,0)\) X components: \(B_x(t) = (1-t)^3\times 0 + 3\times (1-t)^2\times t \times 0 + 3\times (1-t) \times t^2\times 0 + t^3 \times 0 = 0\) Y components: \(B_y(t) = (1-t)^3\times 1 + 3\times (1-t)^2\times t \times 1 + 3\times (1-t) \times t^2\times 0 + t^3 \times 0\) So, \(B_1(t) = (0, (1-t)^3+3(1-t)^2t)\) 2. Second Curve: \((0,0)(0,1)(1,1)(1,0)\) X components: \(B_x(t) = (1-t)^3\times 0 + 3\times (1-t)^2\times t \times 0 + 3\times (1-t) \times t^2\times 1 + t^3 \times 1\) Y components: \(B_y(t) = (1-t)^3\times 0 + 3\times (1-t)^2\times t \times 1 + 3\times (1-t) \times t^2\times 1 + t^3 \times 0\) So, \(B_2(t) = ((1-t)3(t^2), (1-t)^2(3t)+3(1-t)t^2)\) 3. Third Curve: \((1,0)(1,1)(2,1)(2,0)\) X components: \(B_x(t) = (1-t)^3\times 1 + 3\times (1-t)^2\times t \times 1 + 3\times (1-t) \times t^2\times 2 + t^3 \times 2\) Y components: \(B_y(t) = (1-t)^3\times 0 + 3\times (1-t)^2\times t \times 1 + 3\times (1-t) \times t^2\times 1 + t^3 \times 0\) So, \(B_3(t) = (3(1-t)t^2+t^3, (1-t)^2(3t)+3(1-t)t^2)\)
03

3. Describe the Character

Combining the parametric equations for each curve, and by analyzing the control points: 1. The first curve starts at (0,1) and ends at (0,0), and it is a degenerate curve, representing a straight vertical line segment. 2. The second curve starts at (0,0) and ends at (1,0), sweeping across the control points (0,1) and (1,1), representing a smooth arch moving from the bottom to the right. 3. The third curve starts at (1,0) and ends at (2,0), and it is a mirror image of the second curve, representing another smooth arch moving from the bottom to the right. From the above description, the character drawn by the three-piece Bezier curve is an "N".

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cubic Bezier Curves
To represent curves in computer graphics and animations, **Cubic Bezier Curves** are frequently used. They are a special type of Bezier curve defined by four control points:
  • **P0**: The starting point
  • **P1**: The first control point, which influences the direction and 'pull' of the curve as it leaves the starting point
  • **P2**: The second control point, influencing the approach towards the endpoint
  • **P3**: The endpoint
The curve is not limited to simply connecting P0 and P3. It is influenced by P1 and P2 to shape the overall trajectory and smoothness. This makes them ideal for generating complex and beautiful curves, like letters or intricate shapes, with minimal data input. The movement along the curve is determined by a parameter, usually denoted by **t**, ranging from 0 to 1. At **t = 0**, the curve is at the starting point; at **t = 1**, it is at the endpoint.
Parametric Equations
**Parametric Equations** are mathematical expressions used to define the coordinates of the points along a curve. In the case of a Bezier curve, the coordinates are determined using the parametric equation:

\[B(t) = (1-t)^3 P0 + 3(1-t)^2t P1 + 3(1-t)t^2 P2 + t^3 P3\]This equation calculates a weighted sum of the control points, where the weights are determined by the binomial coefficients and powers of **t** and **1-t**. These weights smoothly transition from starting point to endpoint as **t** changes from 0 to 1.

  • At **t = 0**, the entire weight is concentrated on **P0**, starting at the beginning.
  • As **t** progresses towards 1, the influence shifts towards **P3**, with **t = 1** making the curve reach the endpoint.
This results in an elegant and flexible way to define and describe the shape of complex curves based on simple mathematical principles.
Control Points
The placement and arrangement of **Control Points** determine the characteristics of a Bezier curve. These crucial points guide the curve and are composed of:
  • **P0** and **P3**: Start and endpoint of the curve. They define the boundary of where the curve exists in the plane.
  • **P1**: Influences the curve's initial direction and shape as it leaves the start position. Its position relative to **P0** will determine how steeply or softly the curve inclines.
  • **P2**: Primarily affects how the curve reaches the endpoint. It curves the path toward **P3** and contributes to the curvature towards the end.
These control points do not sit on the curve except for the start and endpoint. Modifying them will adjust the curve's shape entirely and can create anything from gentle slopes and arches to sharp turns or loops. Understanding how each point controls the curve allows designers and engineers to craft paths accurately and artistically.

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Most popular questions from this chapter

(a) Find \(b_{1}\) and \(c_{3}\) in the cubic spline $$ S(x)= \begin{cases}-1+b_{1} x-\frac{5}{9} x^{2}+\frac{5}{9} x^{3} & \text { on }[0,1] \\ \frac{14}{9}(x-1)+\frac{10}{9}(x-1)^{2}-\frac{2}{3}(x-1)^{3} & \text { on }[1,2] \\ 2+\frac{16}{9}(x-2)+c_{3}(x-2)^{2}-\frac{1}{9}(x-2)^{3} & \text { on }[2,3]\end{cases} $$ (b) Is this spline natural? (c) This spline satisfies "clamped" endpoint conditions. What are the values of the two clamps?

Can a cubic spline be both natural and parabolically terminated? If so, what else can you say about such a spline?

Discuss the existence and uniqueness of a not-a-knot cubic spline when \(n=2\) and \(n=3\).

(a) Find a polynomial \(P(x)\) of degree 3 or less whose graph passes through the four data points \((-2,8),(0,4),(1,2),(3,-2)\). (b) Describe any other polynomials of degree 4 or less which pass through the four points in part (a).

Find the knots and control points for the following B茅zier space curves. (a) \(\left\\{\begin{array}{l}x(t)=1+6 t^{2}+2 t^{3} \\ y(t)=1-t+t^{3} \\\ z(t)=1+t+6 t^{2}\end{array} \quad\right.\) (b) \(\quad\left\\{\begin{array}{l}x(t)=3+4 t-t^{2}+2 t^{3} \\ y(t)=2-t+t^{2}+3 t^{3} \\ z(t)=3+t+t^{2}-t^{3}\end{array}\right.\) (c) \(\left\\{\begin{array}{l}x(t)=2+t^{2}-t^{3} \\ y(t)=1-t+2 t^{3} \\\ z(t)=2 t^{3}\end{array}\right.\)
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