BCA 4th Sem Notes -Cubic Curves

• UNIT-I 

~The Advantages of Interactive Graphics
~Representative Uses of Computer Graphics 
~Classification of Application Development of Hardware and software for computer Graphics
~Overview, Scan:
~Converting Lines
~Scan Converting Circles
~Scan Converting Ellipses

UNIT-II 

~Hardcopy Technologies
~Display Technologies
~Raster-Scan Display System
~Video Controller
~Random-Scan Display processor
~Input Devices for Operator Interaction
~Image Scanners
~Working exposure on graphics tools like Dream Weaver, 3D Effects etc
~Clipping
~Southland- Cohen Algorithm
~Cyrus-Beck Algorithm
~Midpoint Subdivision Algorithm

UNIT-III 

~Geometrical Transformation
~2D Transformation
~Homogeneous Coordinates and Matrix Representation of 2DTransformations 
~composition of 2D Transformations
~The Window-to-Viewport
Transformations

~Introduction to 3D Transformations Matrix

UNIT-IV 

Representing Curves & Surfaces----

Polygon meshes parametric

Cubic Curves

Quadric Surface

Solid Modeling   Representing Solids

Regularized Boolean Set Operation primitive Instancing Sweep   Representation
Boundary Representations
Spatial Partitioning Representations
Constructive Solid Geometry Comparison of Representations

UNIT-V                           

~Multimedia Definition
~CD-ROM and the multimedia highway
~Computer Animation
(Design, types of animation, using different functions)

UNIT-VI  

~Uses of Multimedia
~Introduction to making multimedia –
~The stage of Project
~hardware & software requirements to make good multimedia skills
~Training opportunities in Multimedia Motivation for Multimedia usage

Cubic Curves

a cubic curve can be uniquely defined by giving 4 points. It would seem in our best interest to somehow specify those 4 points and have the curve pass through them. We can do this, because a cubic equation has 4 unknowns (our $c_i$-s). If we specify 4 constraints, we can find those currently unknown coefficients and thus make $g\left(t\right)$ actually usable.

If we want a cubic curve to pass through all 4 points (called control points), we can specify the constraints:

$g\left(0\right)=c_0+c_1\cdot 0+c_2\cdot 0+c_3\cdot 0=p_0$

$g\left(\frac{1}{3}\right)=c_0+c_1\cdot \frac{1}{3}+c_2\cdot {\left(\frac{1}{3}\right)}^2+c_3\cdot {\left(\frac{1}{3}\right)}^3=p_1$

$g\left(\frac{2}{3}\right)=c_0+c_1\cdot \frac{2}{3}+c_2\cdot {\left(\frac{2}{3}\right)}^2+c_3\cdot {\left(\frac{2}{3}\right)}^3=p_2$

$g\left(1\right)=c_0+c_1\cdot 1+c_2\cdot 1+c_3\cdot 1=p_3$

The control points $p_i$ are called parameters, so a vector $p$ would be a parameter vector.

We can write this system of equations as a matrix-vector operation.

$C\cdot c=p$

We call the matrix $C$ a constraint matrix.

$C=\left(\begin{array}{cccc}1& 0& 0& 0\\ 1& \frac{1}{3}& {\left(\frac{1}{3}\right)}^2& {\left(\frac{1}{3}\right)}^3\\ 1& \frac{2}{3}& {\left(\frac{2}{3}\right)}^2& {\left(\frac{2}{3}\right)}^3\\ 1& 1& 1& 1\end{array}\right)$

There is also the coefficient vector $c$. Those are the unknowns.

$c=\left(\begin{array}{c}{c}_0\\ c_1\\ c_2\\ c_3\end{array}\right)$

And finally a parameter vector $p$.

$p=\left(\begin{array}{c}{p}_0\\ p_1\\ p_2\\ p_3\end{array}\right)$

Given this system we can find the solution to $c$ by taking the inverse of $C$ and multiplying from the left.

$C\cdot c=p$

$C^{-1}\cdot C\cdot c=C^{-1}\cdot p$

$c=C^{-1}\cdot p$

The matrix $C^{-1}$ is called a basis matrix and is usually denoted $B$. This would give us the coefficients that satisfy the equations. We can multiply this equation once more with another vector to get the actual curve equation that we can use. The elements of our new vector $t_{vec}$ depend on our argument value $t$.

$t_{vec}=\left(\begin{array}{cccc}1& t& t^2& t^3\end{array}\right)$

$t_{vec}\cdot c=t_{vec}\cdot B\cdot p$

Notice that $t_{vec}\cdot c$ is exactly our initial curve function. 

$g\left(t\right)=t_{vec}\cdot B\cdot p$

The product $t_{vec}\cdot B$ gives us another vector $b$ that has scalar functions as elements. When we multiply it with the parameter vector $p$ we will get a sum of the products of corresponding elements of $b$ and $p$. So each element of $p$ is taken into account, or blended, by a scalar coefficient from $b$. That is why the elements of $b$ are usually called blending functions. They are functions, because they depend on our argument $t$. Sometimes our curve function is thus written out as:

$$g\left(t\right)=\sum _{i=0}^n{b}_i\left(t\right)\cdot p_i$$,

where $b_i\left(t\right)$ is the $i$-th element of $b$, a function of $t$.

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