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Michael Zhang 2023-04-30 19:44:00 -05:00
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commit 6efa4a0075
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2 changed files with 209 additions and 22 deletions

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@ -12,6 +12,7 @@ author: |
\renewcommand{\c}[1]{\textcolor{gray}{#1}} \renewcommand{\c}[1]{\textcolor{gray}{#1}}
\newcommand{\now}[1]{\textcolor{blue}{#1}} \newcommand{\now}[1]{\textcolor{blue}{#1}}
\newcommand{\todo}[0]{\textcolor{red}{\textbf{TODO}}}
## Reflection and Refraction ## Reflection and Refraction
@ -99,31 +100,156 @@ author: |
4. \c{Consider the viewing transformation matrix $V$ that enables all of the 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 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 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 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).} controls the roll of the view) is aligned with the $y$ axis (0, 1, 0).}
a. (4 points) When the eye is located at $e = (2, 3, 5)$, the camera is 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 pointing in the direction $(1, -1, -1)$, and the camera's 'up' direction is
$(0, 1, 0)$, what are the entries in $V$? $(0, 1, 0)$, what are the entries in $V$?}
$$\begin{bmatrix} First we can calculate $n$ and $u$:
0 & 1 & 0 & d_x \\
\end{bmatrix}$$
b. (2 points) How will this matrix change if the eye moves forward in the - 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 direction of view? [which elements in V will stay the same? which elements
will change and in what way?] will change and in what way?]}
c. (2 points) How will this matrix change if the viewing direction spins in If the eye moves forward, the eye _position_ and everything that depends on it
the clockwise direction around the camera's 'up' direction? [which elements will change, while everything else doesn't.
in V will stay the same? which elements will change and in what way?]
d. (2 points) How will this matrix change if the viewing direction rotates | $n$ | $u$ | $v$ | $d$ |
| ---- | ---- | ---- | --------- |
| 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? 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 [which elements in V will stay the same? which elements will change and in
what way?] what way?]}
5. 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 ## The Projection Transformation
@ -147,32 +273,36 @@ author: |
\tan(60^\circ) \times 0.5 = \boxed{\frac{\sqrt{3}}{2}}$. The same goes for the \tan(60^\circ) \times 0.5 = \boxed{\frac{\sqrt{3}}{2}}$. The same goes for the
vertical, which also yields $\frac{\sqrt{3}}{2}$. vertical, which also yields $\frac{\sqrt{3}}{2}$.
$$\begin{bmatrix} $$
\begin{bmatrix}
\frac{2\times near}{right - left} & 0 & \frac{right + left}{right - left} & 0 \\ \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 & \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 & -\frac{far + near}{far - near} & -\frac{2\times far\times near}{far - near} \\
0 & 0 & -1 & 0 0 & 0 & -1 & 0
\end{bmatrix}$$ \end{bmatrix}
$$
$$= \begin{bmatrix} $$
= \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 \\ \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 & \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 & -\frac{20 + 0.5}{20 - 0.5} & -\frac{2\times 20\times 0.5}{20 - 0.5} \\
0 & 0 & -1 & 0 0 & 0 & -1 & 0
\end{bmatrix}$$ \end{bmatrix}
$$
$$= \boxed{\begin{bmatrix} $$
= \boxed{\begin{bmatrix}
\frac{1}{\sqrt{3}} & 0 & 0 & 0 \\ \frac{1}{\sqrt{3}} & 0 & 0 & 0 \\
0 & \frac{1}{\sqrt{3}} & 0 & 0 \\ 0 & \frac{1}{\sqrt{3}} & 0 & 0 \\
0 & 0 & -\frac{41}{39} & -\frac{40}{39} \\ 0 & 0 & -\frac{41}{39} & -\frac{40}{39} \\
0 & 0 & -1 & 0 0 & 0 & -1 & 0
\end{bmatrix}}$$ \end{bmatrix}}
$$
b. \c{(3 points) How should be matrix $P$ be re-defined if the viewing window 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?} 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 c. \c{(3 points) What are the new horizontal and vertical fields of view
after this change has been made?} after this change has been made?}

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@ -37,6 +37,62 @@ def problem_1():
def problem_4(): def problem_4():
print("part 4a.") print("part 4a.")
up = np.array([0, 1, 0])
viewing_dir = np.array([1, -1, -1])
n = unit(viewing_dir)
print(f"{n = }")
u = unit(np.cross(up, n))
print(f"{u = }")
v = np.cross(n, u)
print(f"{v = }")
print(math.sqrt(1 / 6.0))
print(math.sqrt(2 / 3.0))
def build_translation_matrix(vec):
return np.array([
[1, 0, 0, vec[0]],
[0, 1, 0, vec[1]],
[0, 0, 1, vec[2]],
[0, 0, 0, 1],
])
def problem_5():
b1 = build_translation_matrix(np.array([0, 0, 5]))
theta = math.radians(-90)
sin_theta = round( math.sin(theta), 5)
cos_theta = round(math.cos(theta), 5)
b2 = np.array([
[cos_theta, 0, sin_theta, 0],
[0, 1, 0, 0],
[-sin_theta, 0, cos_theta, 0],
[0, 0, 0, 1],
])
b3 = build_translation_matrix(np.array([0, 0, -5]))
print("b1", b1)
print("b2", b2)
print("b3", b3)
M = b3 @ b2 @ b1
print("M", M)
ex1 = np.array([1, 1, -4, 1])
print("ex1", ex1, M @ ex1)
up = np.array([0, 1, 0])
n = np.array([1, 0, 0])
u = unit(np.cross(up, n))
v = np.cross(n, u)
print(f"{up = }, {n = }, {u = }, {v = }")
eye = np.array([5, 0, -5])
dx = -(np.dot(eye, u))
dy = -(np.dot(eye, v))
dz = -(np.dot(eye, n))
print(f"{dx = }, {dy = }, {dz = }")
def problem_8(): def problem_8():
def P(left, right, bottom, top, near, far): def P(left, right, bottom, top, near, far):
@ -67,3 +123,4 @@ def problem_8():
print("\nPROBLEM 4 -------------------------"); problem_4() print("\nPROBLEM 4 -------------------------"); problem_4()
print("\nPROBLEM 8 -------------------------"); problem_8() print("\nPROBLEM 8 -------------------------"); problem_8()
print("\nPROBLEM 1 -------------------------"); problem_1() print("\nPROBLEM 1 -------------------------"); problem_1()
print("\nPROBLEM 5 -------------------------"); problem_5()