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exam-2/exam2.md
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exam-2/exam2.md
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@ -12,6 +12,7 @@ author: |
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\renewcommand{\c}[1]{\textcolor{gray}{#1}}
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\newcommand{\now}[1]{\textcolor{blue}{#1}}
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\newcommand{\todo}[0]{\textcolor{red}{\textbf{TODO}}}
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## Reflection and Refraction
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@ -99,31 +100,156 @@ author: |
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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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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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a. \c{(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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$(0, 1, 0)$, 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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First we can calculate $n$ and $u$:
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b. (2 points) How will this matrix change if the eye moves forward in the
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- Viewing direction is $(1, -1, -1)$.
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- Normalized $n = (\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}})$.
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- $u = up \times n = (-\frac{1}{\sqrt{2}}, 0, -\frac{1}{\sqrt{2}})$.
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- $v = n \times u = (\frac{\sqrt{6}}{6}, \frac{\sqrt{6}}{3}, -\frac{\sqrt{6}}{6})$
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$$
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\begin{bmatrix}
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-\frac{1}{\sqrt{2}} & 0 & -\frac{1}{\sqrt{2}} & d_x \\
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1 & -1 & -1 & d_y \\
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\end{bmatrix}
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$$
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\todo
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b. \c{(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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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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If the eye moves forward, the eye _position_ and everything that depends on it
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will change, while everything else doesn't.
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d. (2 points) How will this matrix change if the viewing direction rotates
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| $n$ | $u$ | $v$ | $d$ |
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| ---- | ---- | ---- | --------- |
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| same | same | same | different |
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The $n$ is the same because the viewing direction does not change.
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c. \c{(2 points) How will this matrix change if the viewing direction spins
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in the clockwise direction around the camera's 'up' direction? [which
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elements in V will stay the same? which elements will change and in what
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way?]}
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In this case, the eye _position_ stays the same, and everything else changes.
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| $n$ | $u$ | $v$ | $d$ |
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| --------- | --------- | --------- | ---- |
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| different | different | different | same |
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d. \c{(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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what way?]}
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5.
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In this case, the eye _position_ stays the same, and everything else changes.
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| $n$ | $u$ | $v$ | $d$ |
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| --------- | --------- | --------- | ---- |
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| different | different | different | same |
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5. \c{Suppose a viewer located at the point $(0, 0, 0)$ is looking in the $-z$
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direction, with no roll ['up' = $(0, 1 ,0)$], towards a cube of width 2,
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centered at the point $(0, 0, -5)$, whose sides are colored: red at the plane
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$x = 1$, cyan at the plane $x = -1$, green at the plane $y = 1$, magenta at
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the plane $y = -1$, blue at the plane $z = -4$, and yellow at the plane $z =
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-6$.}
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a. \c{(1 point) What is the color of the cube face that the user sees?}
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\boxed{\textrm{Blue}}
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b. \c{(3 points) Because the eye is at the origin, looking down the $-z$ axis
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with 'up' = $(0,1,0)$, the viewing transformation matrix $V$ in this case is
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the identity $I$. What is the model matrix $M$ that you could use to rotate
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the cube so that when the image is rendered, it shows the red side of the
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cube?}
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You would have to do a combination of (1) translate to the origin, (2) rotate
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around the origin, and then (3) untranslate back. This way, the eye position
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doesn't change.
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$$
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M =
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\begin{bmatrix}
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1 & 0 & 0 & 0 \\
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0 & 1 & 0 & 0 \\
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0 & 0 & 1 & -5 \\
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0 & 0 & 0 & 1 \\
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\end{bmatrix} \cdot
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\begin{bmatrix}
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0 & 0 & -1 & 0 \\
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0 & 1 & 0 & 0 \\
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1 & 0 & 0 & 0 \\
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0 & 0 & 0 & 1 \\
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\end{bmatrix} \cdot
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\begin{bmatrix}
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1 & 0 & 0 & 0 \\
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0 & 1 & 0 & 0 \\
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0 & 0 & 1 & 5 \\
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0 & 0 & 0 & 1 \\
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\end{bmatrix}
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=
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\boxed{\begin{bmatrix}
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0 & 0 & -1 & -5 \\
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0 & 1 & 0 & 0 \\
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1 & 0 & 0 & -5 \\
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0 & 0 & 0 & 1 \\
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\end{bmatrix}}
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$$
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To verify this, testing with an example point $(1, 1, -4)$ yields:
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$$
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\begin{bmatrix}
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0 & 0 & -1 & -5 \\
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0 & 1 & 0 & 0 \\
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1 & 0 & 0 & -5 \\
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0 & 0 & 0 & 1 \\
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\end{bmatrix}
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\cdot
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\begin{bmatrix}
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1 \\ 1 \\ -4 \\ 1
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\end{bmatrix}
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=
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\begin{bmatrix}
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-1 \\ 1 \\ -4 \\ 1
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\end{bmatrix}
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$$
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c. \c{(4 points) Suppose now that you want to leave the model matrix $M$ as
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the identity. What is the viewing matrix $V$ that you would need to use to
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render an image of the scene from a re-defined camera configuration so that
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when the scene is rendered, it shows the red side of the cube? Where is the
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eye in this case and in what direction is the camera looking?}
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For this, a different eye position will have to be used. Instead of looking
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from the origin, you could view it from the red side, and then change the
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direction so it's still pointing at the cube.
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- eye is located at $(5, 0, -5)$
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- viewing direction is $(-1, 0, 0)$
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- $n = (1, 0, 0)$
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- $u = up \times n = (0, 0, -1)$
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- $v = n \times u = (0, 1, 0)$
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- $d = (-5, 0, -5)$
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The final viewing matrix is $\boxed{\begin{bmatrix}
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0 & 0 & -1 & -5 \\
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0 & 1 & 0 & 0 \\
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1 & 0 & 0 & -5 \\
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0 & 0 & 0 & 1 \\
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\end{bmatrix}}$. Turns out it's the same matrix! Wow!
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## The Projection Transformation
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@ -147,32 +273,36 @@ author: |
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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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$$\begin{bmatrix}
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$$
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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}$$
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\end{bmatrix}
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$$
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$$= \begin{bmatrix}
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$$
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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}$$
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\end{bmatrix}
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$$
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$$= \boxed{\begin{bmatrix}
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$$
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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}}$$
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\end{bmatrix}}
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$$
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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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@ -37,6 +37,62 @@ def problem_1():
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def problem_4():
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print("part 4a.")
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up = np.array([0, 1, 0])
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viewing_dir = np.array([1, -1, -1])
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n = unit(viewing_dir)
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print(f"{n = }")
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u = unit(np.cross(up, n))
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print(f"{u = }")
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v = np.cross(n, u)
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print(f"{v = }")
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print(math.sqrt(1 / 6.0))
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print(math.sqrt(2 / 3.0))
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def build_translation_matrix(vec):
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return np.array([
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[1, 0, 0, vec[0]],
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[0, 1, 0, vec[1]],
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[0, 0, 1, vec[2]],
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[0, 0, 0, 1],
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])
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def problem_5():
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b1 = build_translation_matrix(np.array([0, 0, 5]))
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theta = math.radians(-90)
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sin_theta = round( math.sin(theta), 5)
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cos_theta = round(math.cos(theta), 5)
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b2 = np.array([
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[cos_theta, 0, sin_theta, 0],
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[0, 1, 0, 0],
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[-sin_theta, 0, cos_theta, 0],
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[0, 0, 0, 1],
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])
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b3 = build_translation_matrix(np.array([0, 0, -5]))
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print("b1", b1)
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print("b2", b2)
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print("b3", b3)
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M = b3 @ b2 @ b1
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print("M", M)
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ex1 = np.array([1, 1, -4, 1])
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print("ex1", ex1, M @ ex1)
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up = np.array([0, 1, 0])
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n = np.array([1, 0, 0])
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u = unit(np.cross(up, n))
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v = np.cross(n, u)
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print(f"{up = }, {n = }, {u = }, {v = }")
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eye = np.array([5, 0, -5])
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dx = -(np.dot(eye, u))
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dy = -(np.dot(eye, v))
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dz = -(np.dot(eye, n))
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print(f"{dx = }, {dy = }, {dz = }")
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def problem_8():
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def P(left, right, bottom, top, near, far):
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@ -67,3 +123,4 @@ def problem_8():
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print("\nPROBLEM 4 -------------------------"); problem_4()
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print("\nPROBLEM 8 -------------------------"); problem_8()
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print("\nPROBLEM 1 -------------------------"); problem_1()
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print("\nPROBLEM 5 -------------------------"); problem_5()
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