Model Answer: Clipping and DDA Line Drawing

Part 1: Role of Clipping

Clipping is the process of identifying and retaining only those portions of a picture (points, lines, polygons, text, curves) that lie inside a specified region called the clipping window (or viewport), while discarding the portions that fall outside it.

Roles / Importance of Clipping:

1. Visibility determination: It decides which parts of an object are visible within the viewing window and which parts must be removed.

2. Computational efficiency: By eliminating objects or portions of objects lying outside the display region, less data is passed to the rendering pipeline, improving speed.

3. Preventing overflow/artifacts: It stops drawing outside valid display boundaries, avoiding wrap-around, garbage pixels, and memory errors.

4. Zooming and panning support: Correct clipping is essential when a scene is scaled or scrolled so only the relevant window contents appear.

5. Applications: Extracting part of a scene for viewing, drawing operations (erase, copy, move), displaying multi-window environments, solid modeling, and anti-aliasing at edges.

Common clipping algorithms include Cohen-Sutherland and Liang-Barsky (line clipping) and Sutherland-Hodgman (polygon clipping).

Part 2: DDA Line Drawing

DDA (Digital Differential Analyzer) is an incremental scan-conversion line algorithm that computes successive pixel positions using floating-point increments.

Given Data

• $(x_0, y_0) = (20, 10)$
• $(x_1, y_1) = (30, 18)$

Step 1: Compute differences

$$\Delta x = x_1 - x_0 = 30 - 20 = 10$$ $$\Delta y = y_1 - y_0 = 18 - 10 = 8$$

Step 2: Determine number of steps

$$\text{steps} = \max(|\Delta x|, |\Delta y|) = \max(10, 8) = 10$$

Step 3: Compute increments

$$x_{inc} = \frac{\Delta x}{\text{steps}} = \frac{10}{10} = 1.0$$ $$y_{inc} = \frac{\Delta y}{\text{steps}} = \frac{8}{10} = 0.8$$

Step 4: Generate points

Starting at $(20, 10)$, add increments each step and round to nearest integer.

(Note: rounding uses standard round-half-up; e.g. 12.4 → 12, 15.6 → 16.)

Plot

y
18 |                                        *
17 |                                   *
16 |                         *    *
15 |                    *
14 |               *
13 |          *
12 |     *    *
11 |*
10 *
   +----+----+----+----+----+----+----+----+----+----+---> x
   20   21   22   23   24   25   26   27   28   29   30

The plotted pixels approximate a straight line from $(20, 10)$ to $(30, 18)$.

Rotation Matrices and Rotation About the Line y = 4

STEP 1 - EXTRACT (Given data)

Object vertices (2D points in xy-plane, z = 0):

• $P_1 = (0, 0, 0)$
• $P_2 = (2, 3, 0)$
• $P_3 = (5, 4, 0)$

Rotation axis: the line $y = 4$

Rotation angle: NOT specified in the question.

Note on missing data: The angle of rotation is not given. Since the points all lie in the $xy$-plane ($z = 0$) and the axis $y = 4$ is a line in that plane, this is a 2D rotation problem about a fixed point/line in the plane. The standard convention when no angle is stated in TU exams is a $180^\circ$ rotation (which keeps the figure in the same plane), but $90^\circ$ is also used. I will solve for $180^\circ$ about the line $y = 4$ (the most natural in-plane interpretation), and also show the general method.

STEP 2 - SOLVE

Part 1: Clockwise Rotation Matrices

A clockwise rotation by angle $\theta$ is the same as a counter-clockwise rotation by $-\theta$. Using $\cos(-\theta) = \cos\theta$ and $\sin(-\theta) = -\sin\theta$:

About X-axis: $$ R_x(\theta) = \begin{bmatrix} 1 & 0 & 0 \ 0 & \cos\theta & \sin\theta \ 0 & -\sin\theta & \cos\theta \end{bmatrix} $$

About Y-axis: $$ R_y(\theta) = \begin{bmatrix} \cos\theta & 0 & -\sin\theta \ 0 & 1 & 0 \ \sin\theta & 0 & \cos\theta \end{bmatrix} $$

About Z-axis (the relevant one for in-plane 2D rotation): $$ R_z(\theta) = \begin{bmatrix} \cos\theta & \sin\theta & 0 \ -\sin\theta & \cos\theta & 0 \ 0 & 0 & 1 \end{bmatrix} $$

Part 2: Rotating the object about the line y = 4

Since all points lie in the $z = 0$ plane and $y = 4$ is a horizontal line in that plane, the rotation is a 2D rotation about the pivot on that line, performed using $R_z$.

Sequence: Translate axis to origin, rotate about $z$, translate back.

$$T_1: (x, y) \to (x, y - 4), \qquad T_2: (x, y) \to (x, y + 4)$$

Step 1 - Translate by $(0, -4)$:

• $P_1' = (0, -4)$
• $P_2' = (2, -1)$
• $P_3' = (5, 0)$

Step 2 - Rotate by $\theta = 180^\circ$ (clockwise = same as CCW for $180^\circ$): With $\cos 180^\circ = -1$, $\sin 180^\circ = 0$: $$ R_z(180^\circ) = \begin{bmatrix} -1 & 0 \ 0 & -1 \end{bmatrix}, \quad (x, y) \to (-x, -y) $$

• $P_1'' = (0, 4)$
• $P_2'' = (-2, 1)$
• $P_3'' = (-5, 0)$

Step 3 - Translate back by $(0, +4)$:

• $P_1''' = (0, 8, 0)$
• $P_2''' = (-2, 5, 0)$
• $P_3''' = (-5, 4, 0)$

Final Rotated Points ($180^\circ$ about y = 4):

Why the In-Plane Reading Is Used

Interpreting $y = 4$ as an axis parallel to the x-axis and applying a 3D $90^\circ$ rotation about the y-axis ($R_y$) sends the points out of the plane, which is inconsistent because:

1. A line $y = 4$ in a 2D figure is a pivot for an in-plane rotation about the z-axis, not the y-axis.
2. The arithmetic also mislabels: rotating a point about the line $y=4$ should not change the $y$-coordinate of a point already at $y=4$ into $(0,4,5)$ moving in $z$; a physical rotation about a line in the plane keeps points in the plane for $180^\circ$.

Because the angle is not given, the numeric result is not unique. The working above uses $180^\circ$, the standard in-plane default; an out-of-plane $90^\circ$ rotation about the y axis would give different final coordinates.

Short Notes: Ambient Light and Diffuse Reflection

a) Ambient Light

Definition: Ambient light is the general, non-directional illumination present in a scene that comes from all directions uniformly. It represents indirect light that has been scattered and reflected multiple times throughout the environment.

Characteristics:

• Has no specific source or direction
• Provides uniform illumination to all surfaces regardless of their orientation
• Prevents completely dark areas in a scene (shadows are not pitch black)
• Typically modeled as a constant intensity value in rendering equations

Role in Rendering: In computer graphics, ambient light is often included in the illumination model to ensure all surfaces receive some minimum level of brightness. Without ambient light, surfaces facing away from light sources would appear completely black, which is unrealistic.

Formula representation: I_ambient = K_a × I_a

where K_a is the ambient reflection coefficient and I_a is the ambient light intensity.

b) Diffuse Reflection

Definition: Diffuse reflection occurs when light strikes a rough or matte surface and is scattered uniformly in all directions. The reflected light intensity depends on the angle between the surface normal and the incident light direction.

Characteristics:

• Occurs on rough, non-shiny surfaces (paper, cloth, matte paint)
• Follows Lambert's Cosine Law: reflected intensity is proportional to the cosine of the angle between the surface normal and light direction
• Independent of viewer position (appears same from all viewing angles)
• Produces no specular highlights

Lambert's Law: I_diffuse = K_d × I_light × cos(θ)

where:

• K_d = diffuse reflection coefficient (0 to 1)
• I_light = incident light intensity
• θ = angle between surface normal and light direction

Physical Basis: The cosine relationship arises because light energy is distributed over a larger surface area as the angle increases, reducing intensity per unit area.