csci5607/exam-2/exam2.md

geometry	output	title	subtitle	date	author
margin=2cm	pdf_document	Exam 2	CSCI 5607	\today	| Michael Zhang | zhan4854@umn.edu $\cdot$ ID: 5289259

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Reflection and Refraction

1. \c{Consider a sphere S made of solid glass (\eta = 1.5) that has radius $r = 3$ and is centered at the location s = (2, 2, 10) in a vaccum ($\eta = 1.0$). If a ray emanating from the point e = (0, 0, 0) intersects S at a point p = (1, 4, 8):}

   a. \c{(2 points) What is the angle of incidence \theta_i?}

   The incoming ray is in the direction I = p - e = (1, 4, 8), and the normal at that point is N = p - s = (1, 4, 8) - (2, 2, 10) = (1, -2, 2). The angle can be found by taking the opposite of the incoming ray -I and using the formula $\cos \theta_i = \frac{-I \cdot N}{|I| |N|} = \frac{(-1, -4, -8) \cdot (1, -2, 2)}{9 \cdot 3} = \frac{-1 + 8 - 16}{27} = -\frac{1}{3}$. So the angle \boxed{\theta_i = \cos^{-1}(-\frac{1}{3})}.

   b. \c{(1 points) What is the angle of reflection \theta_r?}

   The angle of reflection always equals the angle of incidence, $\theta_r = \theta_i = \boxed{cos^{-1}(-\frac{1}{3})}$.

   c. \c{(3 points) What is the direction of the reflected ray?}

   The reflected ray can be found by first projecting the incident ray -I onto the normalized normal N, which is $v = N \times |-I|\cos(\theta_i) = (\frac{1}{3}, -\frac{2}{3}, \frac{2}{3}) \times 9 \times \frac{1}{3} = (-1, 2, -2)$. Then, we know the point on N where this happened is $p' = p + v = (1, 4, 8) + (-1, 2, -2) = (0, 6, 6)$.

   Now, we can subtract this point from where the ray originated to know the direction to add in the other direction, which is still (0, 6, 6) in this case since the ray starts at the origin. Adding this to the point p' gets us (0, 12, 12), which means a point from the origin will get reflected to (0, 12, 12).

   Finally, subtract the point to get the final answer $(0, 12, 12) - (1, 4, 8) = \boxed{(-1, 8, 4)}$.

   d. \c{(3 points) What is the angle of transmission \theta_t?}

   Using Snell's law, we know that $\eta_i \sin \theta_i = \eta_t \sin \theta_t = 1.0 \times \sin(\cos^{-1}(-\frac{1}{3})) = 1.5 \times \sin(\theta_t)$. To find the angle \theta_t we can just solve: $\theta_t = \sin^{-1}(\frac{2}{3} \times \sin(\cos^{-1}(-\frac{1}{3}))) \approx \boxed{0.6796}$ (in radians).

   e. \c{(4 points) What is the direction of the transmitted ray?}

Geometric Transformations

2. \c{(8 points) Consider the airplane model below, defined in object coordinates with its center at (0, 0, 0), its wings aligned with the $\pm x$ axis, its tail pointing upwards in the +y direction and its nose facing in the +z direction. Derive a sequence of model transformation matrices that can be applied to the vertices of the airplane to position it in space at the location p = (4, 4, 7), with a direction of flight w = (2, 1, -2) and the wings aligned with the direction d = (-2, 2, -1).}

   The translation matrix is

   \begin{bmatrix} 1 & 0 & 0 & x \ 0 & 1 & 0 & y \ 0 & 0 & 1 & z \ 0 & 0 & 0 & 1 \ \end{bmatrix}

   \begin{bmatrix} 1 & 0 & 0 & 4 \ 0 & 1 & 0 & 4 \ 0 & 0 & 1 & 7 \ 0 & 0 & 0 & 1 \ \end{bmatrix}

   Since the direction of flight was originally (0, 0, 1), we have to transform it to (2, 1, -2).

The Camera/Viewing Transformation

4. \c{Consider the viewing transformation matrix V that enables all of the vertices in a scene to be expressed in terms of a coordinate system in which the eye is located at (0, 0, 0), the viewing direction (-n) is aligned with the -z axis (0, 0, -1), and the camera's 'up' direction (which controls the roll of the view) is aligned with the y axis (0, 1, 0).}

   a. \c{(4 points) When the eye is located at e = (2, 3, 5), the camera is pointing in the direction (1, -1, -1), and the camera's 'up' direction is (0, 1, 0), what are the entries in V?}

   First we can calculate n and u:

   • Viewing direction is (1, -1, -1).
   • Normalized n = (\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}).
   • u = up \times n = (-\frac{1}{\sqrt{2}}, 0, -\frac{1}{\sqrt{2}}).
   • v = n \times u = (\frac{\sqrt{6}}{6}, \frac{\sqrt{6}}{3}, -\frac{\sqrt{6}}{6})

   \begin{bmatrix} -\frac{1}{\sqrt{2}} & 0 & -\frac{1}{\sqrt{2}} & d_x \ 1 & -1 & -1 & d_y \ \end{bmatrix}

   \todo

   b. \c{(2 points) How will this matrix change if the eye moves forward in the direction of view? [which elements in V will stay the same? which elements will change and in what way?]}

   If the eye moves forward, the eye position and everything that depends on it will change, while everything else doesn't.

   n	u	v	d
   same	same	same	different

   The n is the same because the viewing direction does not change.

   c. \c{(2 points) How will this matrix change if the viewing direction spins in the clockwise direction around the camera's 'up' direction? [which elements in V will stay the same? which elements will change and in what way?]}

   In this case, the eye position stays the same, and everything else changes.

   n	u	v	d
   different	different	different	same

   d. \c{(2 points) How will this matrix change if the viewing direction rotates directly upward, within the plane defined by the viewing and 'up' directions? [which elements in V will stay the same? which elements will change and in what way?]}

   In this case, the eye position stays the same, and everything else changes.

   n	u	v	d
   different	different	different	same

5. \c{Suppose a viewer located at the point (0, 0, 0) is looking in the -z direction, with no roll ['up' = $(0, 1 ,0)$], towards a cube of width 2, centered at the point (0, 0, -5), whose sides are colored: red at the plane x = 1, cyan at the plane x = -1, green at the plane y = 1, magenta at the plane y = -1, blue at the plane z = -4, and yellow at the plane $z = -6$.}

   a. \c{(1 point) What is the color of the cube face that the user sees?}

   \boxed{\textrm{Blue}}

   b. \c{(3 points) Because the eye is at the origin, looking down the -z axis with 'up' = (0,1,0), the viewing transformation matrix V in this case is the identity I. What is the model matrix M that you could use to rotate the cube so that when the image is rendered, it shows the red side of the cube?}

   You would have to do a combination of (1) translate to the origin, (2) rotate around the origin, and then (3) untranslate back. This way, the eye position doesn't change.

   M = \begin{bmatrix} 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 1 & -5 \ 0 & 0 & 0 & 1 \ \end{bmatrix} \cdot \begin{bmatrix} 0 & 0 & -1 & 0 \ 0 & 1 & 0 & 0 \ 1 & 0 & 0 & 0 \ 0 & 0 & 0 & 1 \ \end{bmatrix} \cdot \begin{bmatrix} 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 1 & 5 \ 0 & 0 & 0 & 1 \ \end{bmatrix}

   \boxed{\begin{bmatrix} 0 & 0 & -1 & -5 \ 0 & 1 & 0 & 0 \ 1 & 0 & 0 & -5 \ 0 & 0 & 0 & 1 \ \end{bmatrix}}

   To verify this, testing with an example point (1, 1, -4) yields:

   \begin{bmatrix} 0 & 0 & -1 & -5 \ 0 & 1 & 0 & 0 \ 1 & 0 & 0 & -5 \ 0 & 0 & 0 & 1 \ \end{bmatrix} \cdot \begin{bmatrix} 1 \ 1 \ -4 \ 1 \end{bmatrix}

   \begin{bmatrix} -1 \ 1 \ -4 \ 1 \end{bmatrix}

   c. \c{(4 points) Suppose now that you want to leave the model matrix M as the identity. What is the viewing matrix V that you would need to use to render an image of the scene from a re-defined camera configuration so that when the scene is rendered, it shows the red side of the cube? Where is the eye in this case and in what direction is the camera looking?}

   For this, a different eye position will have to be used. Instead of looking from the origin, you could view it from the red side, and then change the direction so it's still pointing at the cube.

   • eye is located at (5, 0, -5)
   • viewing direction is (-1, 0, 0)
   • n = (1, 0, 0)
   • u = up \times n = (0, 0, -1)
   • v = n \times u = (0, 1, 0)
   • d = (-5, 0, -5)

   The final viewing matrix is $\boxed{\begin{bmatrix} 0 & 0 & -1 & -5 \ 0 & 1 & 0 & 0 \ 1 & 0 & 0 & -5 \ 0 & 0 & 0 & 1 \ \end{bmatrix}}$. Turns out it's the same matrix! Wow!

The Projection Transformation

8. \c{Consider the perspective projection-normalization matrix P which maps the contents of the viewing frustum into a cube that extends from -1 to 1 in x, y, z (called normalized device coordinates).}

   \c{Suppose you want to define a square, symmetric viewing frustum with a near clipping plane located 0.5 units in front of the camera, a far clipping plane located 20 units from the front of the camera, a 60^\circ vertical field of view, and a 60^\circ horizontal field of view.}

   a. \c{(2 points) What are the entries in P?}

   The left / right values are found by using the tangent of the field-of-view triangle: \tan(60^\circ) = \frac{\textrm{right}}{0.5}, so $\textrm{right} = \tan(60^\circ) \times 0.5 = \boxed{\frac{\sqrt{3}}{2}}$. The same goes for the vertical, which also yields \frac{\sqrt{3}}{2}.

   \begin{bmatrix} \frac{2\times near}{right - left} & 0 & \frac{right + left}{right - left} & 0 \ 0 & \frac{2\times near}{top - bottom} & \frac{top + bottom}{top - bottom} & 0 \ 0 & 0 & -\frac{far + near}{far - near} & -\frac{2\times far\times near}{far - near} \ 0 & 0 & -1 & 0 \end{bmatrix}

   = \begin{bmatrix} \frac{2\times 0.5}{\frac{\sqrt{3}}{2} - (-\frac{\sqrt{3}}{2})} & 0 & \frac{\frac{\sqrt{3}}{2} + (-\frac{\sqrt{3}}{2})}{\frac{\sqrt{3}}{2} - (-\frac{\sqrt{3}}{2})} & 0 \ 0 & \frac{2\times 0.5}{\frac{\sqrt{3}}{2} - (-\frac{\sqrt{3}}{2})} & \frac{\frac{\sqrt{3}}{2} + (-\frac{\sqrt{3}}{2})}{\frac{\sqrt{3}}{2} - (-\frac{\sqrt{3}}{2})} & 0 \ 0 & 0 & -\frac{20 + 0.5}{20 - 0.5} & -\frac{2\times 20\times 0.5}{20 - 0.5} \ 0 & 0 & -1 & 0 \end{bmatrix}

   = \boxed{\begin{bmatrix} \frac{1}{\sqrt{3}} & 0 & 0 & 0 \ 0 & \frac{1}{\sqrt{3}} & 0 & 0 \ 0 & 0 & -\frac{41}{39} & -\frac{40}{39} \ 0 & 0 & -1 & 0 \end{bmatrix}}

   b. \c{(3 points) How should be matrix P be re-defined if the viewing window is re-sized to be twice as tall as it is wide?}

   c. \c{(3 points) What are the new horizontal and vertical fields of view after this change has been made?}

Clipping

9. \c{Consider the triangle whose vertex positions, after the viewport transformation, lie in the centers of the pixels: $p_0 = (3, 3), p_1 = (9, 5), p_2 = (11, 11)$.}

   Starting at p_0, the three vectors are:

   • v_0 = p_1 - p_0 = (9 - 3, 5 - 3) = (6, 2)
   • v_1 = p_2 - p_1 = (11 - 9, 11 - 5) = (2, 6)
   • v_2 = p_0 - p_2 = (3 - 11, 3 - 11) = (-8, -8)

   The first edge vector e would be (6, 2), and the edge normal would be that rotated by 90^\circ.

   a. \c{(6 points) Define the edge equations and tests that would be applied, during the rasterization process, to each pixel (x, y) within the bounding rectangle 3 \le x \le 11, 3 \le y \le 11 to determine if that pixel is inside the triangle or not.}

   b. \c{(3 points) Consider the three pixels p_4 = (6, 4), p_5 = (7, 7), and p_6 = (10, 8). Which of these would be considered to lie inside the triangle, according to the methods taught in class?}

10. \c{When a model contains many triangles that form a smoothly curving surface patch, it can be inefficient to separately represent each triangle in the patch independently as a set of three vertices because memory is wasted when the same vertex location has to be specified multiple times. A triangle strip offers a memory-efficient method for representing connected 'strips' of triangles. For example, in the diagram below, the six vertices v0 .. v5 define four adjacent triangles: (v0, v1, v2), (v2, v1, v3), (v2, v3, v4), (v4, v3, v5). [Notice that the vertex order is switched in every other triangle to maintain a consistent counter-clockwise orientation.] Ordinarily one would need to pass 12 vertex locations to the GPU to represent this surface patch (three vertices for each triangle), but when the patch is encoded as a triangle strip, only the six vertices need to be sent and the geometry they represent will be interpreted using the correspondence pattern just described.}

    \c{(5 points) When triangle strips are clipped, however, things can get complicated. Consider the short triangle strip shown below in the context of a clipping cube.}

    • \c{After the six vertices v0 .. v5 are sent to be clipped, what will the vertex list be after clipping process has finished?}

    • \c{How can this new result be expressed as a triangle strip? (Try to be as efficient as possible)}

    • \c{How many triangles will be encoded in the clipped triangle strip?}

Ray Tracing vs Scan Conversion

11. \c{(8 points) List the essential steps in the scan-conversion (raster graphics) rendering pipeline, starting with vertex processing and ending with the assignment of a color to a pixel in a displayed image. For each step briefly describe, in your own words, what is accomplished and how. You do not need to include steps that we did not discuss in class, such as tessellation (subdividing an input triangle into multiple subtriangles), instancing (creating new geometric primitives from existing input vertices), but you should not omit any steps that are essential to the process of generating an image of a provided list of triangles.}

12. \c{(6 points) Compare and contrast the process of generating an image of a scene using ray tracing versus scan conversion. Include a discussion of outcomes that can be achieved using a ray tracing approach but not using a scan-conversion approach, or vice versa, and explain the reasons why and why not.}

    With ray tracing, the process of generating pixels is very hierarchical. The basic ray tracer was very simple, but the moment we even added shadows, there were recursive rays that needed to be cast, not to mention the jittering. None of those could be parallelized with the main one, because in order to even figure out where to start, you need to have already performed a lot of the calculations. (For my ray tracer implementation, I already parallelized as much as I could using the work-stealing library rayon)

    But with scan conversion, the majority of the transformations are just done with matrix transformations over the geometries, which can be performed completely in parallel with minimal branching (only depth testing is not exactly) The rasterization process is also massively parallelizable. This makes it faster to do on GPUs which are able to do a lot of independent operations.