261 lines
11 KiB
Markdown
261 lines
11 KiB
Markdown
---
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geometry: margin=2cm
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output: pdf_document
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title: Exam 2
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subtitle: CSCI 5607
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date: \today
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author: |
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| Michael Zhang
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| zhan4854@umn.edu $\cdot$ ID: 5289259
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---
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\renewcommand{\c}[1]{\textcolor{gray}{#1}}
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\newcommand{\now}[1]{\textcolor{blue}{#1}}
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## Reflection and Refraction
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1. \c{Consider a sphere $S$ made of solid glass ($\eta$ = 1.5) that has radius $r =
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3$ and is centered at the location $s = (2, 2, 10)$ in a vaccum ($\eta =
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1.0$). If a ray emanating from the point $e = (0, 0, 0)$ intersects $S$ at a
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point $p = (1, 4, 8)$:}
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a. \c{(2 points) What is the angle of incidence $\theta_i$?}
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The incoming ray is in the direction $I = p - e = (1, 4, 8)$, and the normal at
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that point is $N = p - s = (1, 4, 8) - (2, 2, 10) = (1, -2, 2)$. The angle can
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be found by taking the opposite of the incoming ray $-I$ and using the
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formula $\cos \theta_i = \frac{-I \cdot N}{|I| |N|} = \frac{(-1, -4, -8)
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\cdot (1, -2, 2)}{9 \cdot 3} = \frac{-1 + 8 - 16}{27} = -\frac{1}{3}$. So the
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angle $\boxed{\theta_i = \cos^{-1}(-\frac{1}{3})}$.
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b. \c{(1 points) What is the angle of reflection $\theta_r$?}
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The angle of reflection always equals the angle of incidence, $\theta_r =
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\theta_i = \boxed{cos^{-1}(-\frac{1}{3})}$.
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c. \c{(3 points) What is the direction of the reflected ray?}
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The reflected ray can be found by first projecting the incident ray $-I$ onto
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the normalized normal $N$, which is $v = N \times |-I|\cos(\theta_i) =
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(\frac{1}{3}, -\frac{2}{3}, \frac{2}{3}) \times 9 \times \frac{1}{3} = (-1,
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2, -2)$. Then, we know the point on N where this happened is $p' = p + v =
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(1, 4, 8) + (-1, 2, -2) = (0, 6, 6)$.
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Now, we can subtract this point from where the ray originated to know the
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direction to add in the other direction, which is still $(0, 6, 6)$ in this
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case since the ray starts at the origin. Adding this to the point $p'$ gets
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us $(0, 12, 12)$, which means a point from the origin will get reflected to
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$(0, 12, 12)$.
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Finally, subtract the point to get the final answer $(0, 12, 12) - (1, 4, 8)
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= \boxed{(-1, 8, 4)}$.
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d. \c{(3 points) What is the angle of transmission $\theta_t$?}
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Using Snell's law, we know that $\eta_i \sin \theta_i = \eta_t \sin \theta_t
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= 1.0 \times \sin(\cos^{-1}(-\frac{1}{3})) = 1.5 \times \sin(\theta_t)$. To
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find the angle $\theta_t$ we can just solve: $\theta_t =
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\sin^{-1}(\frac{2}{3} \times \sin(\cos^{-1}(-\frac{1}{3}))) \approx
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\boxed{0.6796}$ (in radians).
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e. \c{(4 points) What is the direction of the transmitted ray?}
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## Geometric Transformations
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2. \c{(8 points) Consider the airplane model below, defined in object
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coordinates with its center at $(0, 0, 0)$, its wings aligned with the $\pm
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x$ axis, its tail pointing upwards in the $+y$ direction and its nose facing
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in the $+z$ direction. Derive a sequence of model transformation matrices
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that can be applied to the vertices of the airplane to position it in space
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at the location $p = (4, 4, 7)$, with a direction of flight $w = (2, 1, -2)$
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and the wings aligned with the direction $d = (-2, 2, -1)$.}
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The translation matrix is
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$$
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\begin{bmatrix}
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1 & 0 & 0 & x \\
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0 & 1 & 0 & y \\
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0 & 0 & 1 & z \\
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0 & 0 & 0 & 1 \\
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\end{bmatrix}
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=
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\begin{bmatrix}
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1 & 0 & 0 & 4 \\
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0 & 1 & 0 & 4 \\
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0 & 0 & 1 & 7 \\
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0 & 0 & 0 & 1 \\
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\end{bmatrix}
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$$
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Since the direction of flight was originally $(0, 0, 1)$, we have to
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transform it to $(2, 1, -2)$.
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3.
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## The Camera/Viewing Transformation
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4. \c{Consider the viewing transformation matrix $V$ that enables all of the
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vertices in a scene to be expressed in terms of a coordinate system in which
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the eye is located at $(0, 0, 0)$, the viewing direction ($-n$) is aligned
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with the $-z$ axis $(0, 0, –1)$, and the camera's 'up' direction (which
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controls the roll of the view) is aligned with the $y$ axis (0, 1, 0).}
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a. (4 points) When the eye is located at $e = (2, 3, 5)$, the camera is
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pointing in the direction $(1, -1, -1)$, and the camera's 'up' direction is
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$(0, 1, 0)$, what are the entries in $V$?
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$$\begin{bmatrix}
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0 & 1 & 0 & d_x \\
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\end{bmatrix}$$
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b. (2 points) How will this matrix change if the eye moves forward in the
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direction of view? [which elements in V will stay the same? which elements
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will change and in what way?]
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c. (2 points) How will this matrix change if the viewing direction spins in
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the clockwise direction around the camera's 'up' direction? [which elements
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in V will stay the same? which elements will change and in what way?]
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d. (2 points) How will this matrix change if the viewing direction rotates
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directly upward, within the plane defined by the viewing and 'up' directions?
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[which elements in V will stay the same? which elements will change and in
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what way?]
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5.
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## The Projection Transformation
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6.
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7.
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8. \c{Consider the perspective projection-normalization matrix $P$ which maps
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the contents of the viewing frustum into a cube that extends from -1 to 1 in
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$x, y, z$ (called normalized device coordinates).}
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\c{Suppose you want to define a square, symmetric viewing frustum with a near
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clipping plane located 0.5 units in front of the camera, a far clipping plane
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located 20 units from the front of the camera, a $60^\circ$ vertical field of
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view, and a $60^\circ$ horizontal field of view.}
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a. \c{(2 points) What are the entries in $P$?}
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The left / right values are found by using the tangent of the field-of-view
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triangle: $\tan(60^\circ) = \frac{\textrm{right}}{0.5}$, so $\textrm{right} =
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\tan(60^\circ) \times 0.5 = \boxed{\frac{\sqrt{3}}{2}}$. The same goes for the
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vertical, which also yields $\frac{\sqrt{3}}{2}$.
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$$\begin{bmatrix}
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\frac{2\times near}{right - left} & 0 & \frac{right + left}{right - left} & 0 \\
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0 & \frac{2\times near}{top - bottom} & \frac{top + bottom}{top - bottom} & 0 \\
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0 & 0 & -\frac{far + near}{far - near} & -\frac{2\times far\times near}{far - near} \\
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0 & 0 & -1 & 0
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\end{bmatrix}$$
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$$= \begin{bmatrix}
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\frac{2\times 0.5}{\frac{\sqrt{3}}{2} - (-\frac{\sqrt{3}}{2})} & 0 & \frac{\frac{\sqrt{3}}{2} + (-\frac{\sqrt{3}}{2})}{\frac{\sqrt{3}}{2} - (-\frac{\sqrt{3}}{2})} & 0 \\
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0 & \frac{2\times 0.5}{\frac{\sqrt{3}}{2} - (-\frac{\sqrt{3}}{2})} & \frac{\frac{\sqrt{3}}{2} + (-\frac{\sqrt{3}}{2})}{\frac{\sqrt{3}}{2} - (-\frac{\sqrt{3}}{2})} & 0 \\
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0 & 0 & -\frac{20 + 0.5}{20 - 0.5} & -\frac{2\times 20\times 0.5}{20 - 0.5} \\
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0 & 0 & -1 & 0
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\end{bmatrix}$$
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$$= \boxed{\begin{bmatrix}
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\frac{1}{\sqrt{3}} & 0 & 0 & 0 \\
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0 & \frac{1}{\sqrt{3}} & 0 & 0 \\
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0 & 0 & -\frac{41}{39} & -\frac{40}{39} \\
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0 & 0 & -1 & 0
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\end{bmatrix}}$$
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b. \c{(3 points) How should be matrix $P$ be re-defined if the viewing window
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is re-sized to be twice as tall as it is wide?}
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c. \c{(3 points) What are the new horizontal and vertical fields of view
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after this change has been made?}
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## Clipping
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9. \c{Consider the triangle whose vertex positions, after the viewport
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transformation, lie in the centers of the pixels: $p_0 = (3, 3), p_1 = (9,
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5), p_2 = (11, 11)$.}
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Starting at $p_0$, the three vectors are:
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- $v_0 = p_1 - p_0 = (9 - 3, 5 - 3) = (6, 2)$
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- $v_1 = p_2 - p_1 = (11 - 9, 11 - 5) = (2, 6)$
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- $v_2 = p_0 - p_2 = (3 - 11, 3 - 11) = (-8, -8)$
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The first edge vector $e$ would be $(6, 2)$, and the edge normal would be
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that rotated by $90^\circ$.
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a. \c{(6 points) Define the edge equations and tests that would be applied,
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during the rasterization process, to each pixel $(x, y)$ within the bounding
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rectangle $3 \le x \le 11, 3 \le y \le 11$ to determine if that pixel is
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inside the triangle or not.}
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b. \c{(3 points) Consider the three pixels $p_4 = (6, 4), p_5 = (7, 7)$, and
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$p_6 = (10, 8)$. Which of these would be considered to lie inside the
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triangle, according to the methods taught in class?}
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10. \c{When a model contains many triangles that form a smoothly curving surface
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patch, it can be inefficient to separately represent each triangle in the
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patch independently as a set of three vertices because memory is wasted when
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the same vertex location has to be specified multiple times. A triangle
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strip offers a memory-efficient method for representing connected 'strips'
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of triangles. For example, in the diagram below, the six vertices v0 .. v5
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define four adjacent triangles: (v0, v1, v2), (v2, v1, v3), (v2, v3, v4),
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(v4, v3, v5). [Notice that the vertex order is switched in every other
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triangle to maintain a consistent counter-clockwise orientation.] Ordinarily
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one would need to pass 12 vertex locations to the GPU to represent this
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surface patch (three vertices for each triangle), but when the patch is
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encoded as a triangle strip, only the six vertices need to be sent and the
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geometry they represent will be interpreted using the correspondence pattern
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just described.}
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\c{(5 points) When triangle strips are clipped, however, things
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can get complicated. Consider the short triangle strip shown below in the
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context of a clipping cube.}
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- \c{After the six vertices v0 .. v5 are sent to be clipped, what will the
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vertex list be after clipping process has finished?}
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- \c{How can this new result be expressed as a triangle strip? (Try to be as
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efficient as possible)}
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- \c{How many triangles will be encoded in the clipped triangle strip?}
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## Ray Tracing vs Scan Conversion
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11. \c{(8 points) List the essential steps in the scan-conversion (raster
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graphics) rendering pipeline, starting with vertex processing and ending
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with the assignment of a color to a pixel in a displayed image. For each
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step briefly describe, in your own words, what is accomplished and how. You
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do not need to include steps that we did not discuss in class, such as
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tessellation (subdividing an input triangle into multiple subtriangles),
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instancing (creating new geometric primitives from existing input vertices),
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but you should not omit any steps that are essential to the process of
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generating an image of a provided list of triangles.}
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12. \c{(6 points) Compare and contrast the process of generating an image of a
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scene using ray tracing versus scan conversion. Include a discussion of
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outcomes that can be achieved using a ray tracing approach but not using a
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scan-conversion approach, or vice versa, and explain the reasons why and why
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not.}
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With ray tracing, the process of generating pixels is very hierarchical.
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The basic ray tracer was very simple, but the moment we even added shadows,
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there were recursive rays that needed to be cast, not to mention the
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jittering. None of those could be parallelized with the main one, because in
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order to even figure out where to start, you need to have already performed
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a lot of the calculations. (For my ray tracer implementation, I already
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parallelized as much as I could using the work-stealing library `rayon`)
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But with scan conversion, the majority of the transformations are just done
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with matrix transformations over the geometries, which can be performed
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completely in parallel with minimal branching (only depth testing is not
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exactly) The rasterization process is also massively parallelizable. This
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makes it faster to do on GPUs which are able to do a lot of independent
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operations.
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