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exam-2/exam2.md
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@ -11,10 +11,11 @@ author: |
---
\renewcommand{\c}[1]{\textcolor{gray}{#1}}
\renewcommand{\r}[1]{\textcolor{red}{#1}}
\newcommand{\now}[1]{\textcolor{blue}{#1}}
\newcommand{\todo}[0]{\textcolor{red}{\textbf{TODO}}}
[ 1 2 3 6 8 9 ]
[ 2 3 8 9 ]
## Reflection and Refraction
@ -23,31 +24,44 @@ author: |
1.0$). If a ray emanating from the point $e = (0, 0, 0)$ intersects $S$ at a
point $p = (1, 4, 8)$:}
Legend for the rest of the problem:
![](1.jpg){width=50%}
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 $\boxed{\theta_i = \cos^{-1}(-\frac{1}{3})}$.
The incoming ray is in the direction $I = v_0 = p - e = (1, 4, 8)$, 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:
- $\cos \theta_i = \frac{-I \cdot N}{\|I\|_2 \|N\|_2}
= \frac{(-1, -4, -8) \cdot (-1, 2, -2)}{9 \cdot 3} = \frac{1 - 8 + 16}{27} = \frac{1}{3}$.
So the angle $\boxed{\theta_i = \cos^{-1}\left(\frac{1}{3}\right)}$.
b. \c{(1 points) What is the angle of reflection $\theta_r$?}
The angle of reflection always equals the angle of incidence, $\theta_r =
\theta_i = \boxed{\cos^{-1}(-\frac{1}{3})}$.
\theta_i = \boxed{\cos^{-1}\left(\frac{1}{3}\right)}$.
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 $p' = p + v =
(1, 4, 8) + (-1, 2, -2) = (0, 6, 6)$.
The reflected ray can be found by first projecting the incident ray $v_0$ onto
the normalized normal $n$, which is:
- $proj = n \times |v_0|\cos(\theta_i) = \left(\frac{1}{3}, -\frac{2}{3}, \frac{2}{3}\right) \times 9 \times \frac{1}{3} = (-1, 2, -2)$
Then, we know the point on N where this happened is:
- $nx = p + proj = (1, 4, 8) + (-1, 2, -2) = (0, 6, 6)$.
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
case since the ray starts at the origin. Adding this to the point $nx$ gets
us $(0, 12, 12)$, which means a point from the origin will get reflected to
$(0, 12, 12)$.
@ -56,34 +70,45 @@ author: |
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).
Using Snell's law, we know that
\begin{align}
\eta_i \sin \theta_i &= \eta_t \sin \theta_t \\
\r{1.0} \times \sin \theta_i &= \r{1.5} \times \sin \theta_t \\
1.0 \times \sin(\r{ \cos^{-1}(-\frac{1}{3}) }) &= 1.5 \times \sin \theta_t \\
1.0 \times \r{ \frac{2\sqrt{2}}{3} } &= 1.5 \times \sin \theta_t \\
\r{ \frac{1.0}{1.5} } \times \frac{2\sqrt{2}}{3} &= \sin \theta_t \\
\r{ \frac{4}{9} } \sqrt{2} &= \sin \theta_t \\
\theta_t &= \sin^{-1} \left(\frac{4}{9}\sqrt{2}\right) \approx
\boxed{0.67967 \ldots}
\end{align}
e. \c{(4 points) What is the direction of the transmitted ray?}
We can just add the angle obtained above to the inverse of the normal $n' =
(-1, 2, -2)$ (since that will be transmitted through the material). For the
tangent direction, we can use the vector from the ray's origin to the
incident ray projected onto the normal, which we'll call $v = (0, 6, 6)$.
We can just add together $s - p$ and $v_5$ in the diagram above, these two
make up the orthogonal basis of the transmitted ray $v_4$. The lengths of
these two are not important, but related through the angle $\theta_t$.
Now we know that the relationship between the coefficients of $n'$ and $v$ is
$\tan(\theta_t)$. We can add these to obtain a point along the transmitted
ray (it doesn't actually matter where this occurs since we're going to
normalize this anyway, the proportion stays consistent regardless). So using
the unit length for the inverse normal which is $\frac{n'}{||n'||} =
(\frac{1}{3}, -\frac{2}{3}, \frac{2}{3})$ we know that the perpendicular
vector will be $\tan(\theta_t) \times (0, \frac{1}{\sqrt{2}},
\frac{1}{\sqrt{2}})$.
Also note that in the diagram, the sphere is drawn as a 2D circle. But in
reality, the number of vectors perpendicular to $s - p$ is infinite. The
$v_5$ we're looking for is parallel with $v_1$, so we can just use
$\frac{v_1}{\|v_1\|_2}$ as our unit vector for $v_5$.
The final step is then adding these vectors together and subtracting the
interersection point: $(\frac{1}{3}, -\frac{2}{3}, \frac{2}{3}) +
\tan(\theta_t) \times (0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}) - (1, 4,
8)$.
\begin{align}
v_4 &= (s - p) + \tan \theta_t \times \|s - p\|_2 \times \frac{v_1}{\|v_1\|_2} \\
v_4 &= (\r{ (2, 2, 10) - (1, 4, 8) }) + \tan \theta_t \times \|s - p\|_2 \times \frac{v_1}{\|v_1\|_2} \\
v_4 &= \r{ (1, -2, 2) } + \tan \theta_t \times \|\r{(1, -2, 2)}\|_2 \times \frac{v_1}{\|v_1\|_2} \\
v_4 &= (1, -2, 2) + \tan \theta_t \times \r{3} \times \frac{v_1}{\|v_1\|_2} \\
v_4 &= (1, -2, 2) + \tan \theta_t \times 3 \times \frac{\r{(0, 6, 6)}}{\|\r{(0, 6, 6)}\|_2} \\
v_4 &= (1, -2, 2) + \tan \theta_t \times 3 \times \r{\left(0, \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)} \\
v_4 &= (1, -2, 2) + \tan \theta_t \times \left(0, \r{\frac{3}{2}\sqrt{2}}, \r{\frac{3}{2}\sqrt{2}} \right) \\
v_4 &= (1, -2, 2) + \r{\frac{4}{7}\sqrt{2}} \times \left(0, \frac{3}{2}\sqrt{2}, \frac{3}{2}\sqrt{2} \right) \\
v_4 &= (1, -2, 2) + \r{\left(0, \frac{12}{7}, \frac{12}{7} \right)} \\
v_4 &= \r{\left(1, \frac{-2}{7}, \frac{26}{7} \right)} \\
\end{align}
\todo all the numbers are wrong
This can be normalized into $\boxed{\left(\frac{7}{27}, -\frac{2}{27},
\frac{26}{27}\right)}$.
## Geometric Transformations
@ -380,14 +405,14 @@ author: |
i) \c{(2 points) Where will the six vertex locations be after such a
projection, omitting the normalization step?}
- $[\begin{matrix}-1.732 & -1.732 & -6.928\end{matrix}]$ $\rightarrow$ $[\begin{matrix}-1.0 & -1.0 & 7.0\end{matrix}]$
- $[\begin{matrix}-1.732 & -1.732 & -3.464\end{matrix}]$ $\rightarrow$ $[\begin{matrix}-1.0 & -1.0 & 5.0\end{matrix}]$
- $[\begin{matrix}-1.732 & 1.732 & -6.928\end{matrix}]$ $\rightarrow$ $[\begin{matrix}-1.0 & 1.0 & 7.0\end{matrix}]$
- $[\begin{matrix}-1.732 & 1.732 & -3.464\end{matrix}]$ $\rightarrow$ $[\begin{matrix}-1.0 & 1.0 & 5.0\end{matrix}]$
- $[\begin{matrix}1.732 & -1.732 & -6.928\end{matrix}]$ $\rightarrow$ $[\begin{matrix}1.0 & -1.0 & 7.0\end{matrix}]$
- $[\begin{matrix}1.732 & -1.732 & -3.464\end{matrix}]$ $\rightarrow$ $[\begin{matrix}1.0 & -1.0 & 5.0\end{matrix}]$
- $[\begin{matrix}1.732 & 1.732 & -6.928\end{matrix}]$ $\rightarrow$ $[\begin{matrix}1.0 & 1.0 & 7.0\end{matrix}]$
- $[\begin{matrix}1.732 & 1.732 & -3.464\end{matrix}]$ $\rightarrow$ $[\begin{matrix}1.0 & 1.0 & 5.0\end{matrix}]$
- $[\begin{matrix}- \sqrt{3} & - \sqrt{3} & - 4 \sqrt{3}\end{matrix}]$ $\rightarrow$ $[\begin{matrix}-1 & -1 & 7\end{matrix}]$
- $[\begin{matrix}- \sqrt{3} & \sqrt{3} & - 4 \sqrt{3}\end{matrix}]$ $\rightarrow$ $[\begin{matrix}-1 & 1 & 7\end{matrix}]$
- $[\begin{matrix}\sqrt{3} & - \sqrt{3} & - 4 \sqrt{3}\end{matrix}]$ $\rightarrow$ $[\begin{matrix}1 & -1 & 7\end{matrix}]$
- $[\begin{matrix}\sqrt{3} & \sqrt{3} & - 4 \sqrt{3}\end{matrix}]$ $\rightarrow$ $[\begin{matrix}1 & 1 & 7\end{matrix}]$
- $[\begin{matrix}- \sqrt{3} & - \sqrt{3} & - 2 \sqrt{3}\end{matrix}]$ $\rightarrow$ $[\begin{matrix}-1 & -1 & 5\end{matrix}]$
- $[\begin{matrix}- \sqrt{3} & \sqrt{3} & - 2 \sqrt{3}\end{matrix}]$ $\rightarrow$ $[\begin{matrix}-1 & 1 & 5\end{matrix}]$
- $[\begin{matrix}\sqrt{3} & - \sqrt{3} & - 2 \sqrt{3}\end{matrix}]$ $\rightarrow$ $[\begin{matrix}1 & -1 & 5\end{matrix}]$
- $[\begin{matrix}\sqrt{3} & \sqrt{3} & - 2 \sqrt{3}\end{matrix}]$ $\rightarrow$ $[\begin{matrix}1 & 1 & 5\end{matrix}]$
ii) \c{(1 points) Sketch the result, being as accurate as possible and
labeling the colors of each of the visible faces.}
@ -396,6 +421,8 @@ author: |
sides are not visible because the cube is parallel to the axis, and when you
do an orthographic projection, those faces are lost.
![](6aii.jpg){width=40%}
iii) \c{(2 points) Show how you could achieve this transformation using one or
more matrix multiplication operations. Specify the matrix entries you would
use, and, if using multiple matrices, the order in which they would be
@ -405,14 +432,7 @@ author: |
in this Python script:
```py
def ortho(points):
left = min(map(lambda p: p[0], points))
right = max(map(lambda p: p[0], points))
bottom = min(map(lambda p: p[1], points))
top = max(map(lambda p: p[1], points))
near = min(map(lambda p: p[2], points))
far = max(map(lambda p: p[2], points))
def ortho_transform(left, right, bottom, top, near, far):
step_1 = np.array([
[1, 0, 0, 0],
[0, 1, 0, 0],
@ -421,20 +441,20 @@ author: |
])
step_2 = np.array([
[1, 0, 0, -(left + right) / 2.0],
[0, 1, 0, -(bottom + top) / 2.0],
[0, 0, 1, -(near + far) / 2.0],
[1, 0, 0, -((left + right) / 2)],
[0, 1, 0, -((bottom + top) / 2)],
[0, 0, 1, -((near + far) / 2)],
[0, 0, 0, 1],
])
step_3 = np.array([
[2.0 / (right - left), 0, 0, 0],
[0, 2.0 / (top - bottom), 0, 0],
[0, 0, 2.0 / (far - near), 0],
[(2 / (right - left)), 0, 0, 0],
[0, (2 / (top - bottom)), 0, 0],
[0, 0, (2 / (far - near)), 0],
[0, 0, 0, 1],
])
M_ortho = step_3 @ step_2 @ step_1
return step_3 @ step_2 @ step_1
```
b. \c{Show how you could project the vertices of this cube to the plane $z =
@ -444,14 +464,45 @@ author: |
i) \c{(3 points) Where will the six vertex locations be after such a
projection, omitting the normalization step?}
- $[\begin{matrix}- \sqrt{3} & - \sqrt{3} & - 4 \sqrt{3}\end{matrix}]$ $\rightarrow$ $[\begin{matrix}- \frac{4 \sqrt{3}}{3} - 1 & -1 & 7\end{matrix}]$
- $[\begin{matrix}- \sqrt{3} & \sqrt{3} & - 4 \sqrt{3}\end{matrix}]$ $\rightarrow$ $[\begin{matrix}- \frac{4 \sqrt{3}}{3} - 1 & 1 & 7\end{matrix}]$
- $[\begin{matrix}\sqrt{3} & - \sqrt{3} & - 4 \sqrt{3}\end{matrix}]$ $\rightarrow$ $[\begin{matrix}1 - \frac{4 \sqrt{3}}{3} & -1 & 7\end{matrix}]$
- $[\begin{matrix}\sqrt{3} & \sqrt{3} & - 4 \sqrt{3}\end{matrix}]$ $\rightarrow$ $[\begin{matrix}1 - \frac{4 \sqrt{3}}{3} & 1 & 7\end{matrix}]$
- $[\begin{matrix}- \sqrt{3} & - \sqrt{3} & - 2 \sqrt{3}\end{matrix}]$ $\rightarrow$ $[\begin{matrix}- \frac{2 \sqrt{3}}{3} - 1 & -1 & 5\end{matrix}]$
- $[\begin{matrix}- \sqrt{3} & \sqrt{3} & - 2 \sqrt{3}\end{matrix}]$ $\rightarrow$ $[\begin{matrix}- \frac{2 \sqrt{3}}{3} - 1 & 1 & 5\end{matrix}]$
- $[\begin{matrix}\sqrt{3} & - \sqrt{3} & - 2 \sqrt{3}\end{matrix}]$ $\rightarrow$ $[\begin{matrix}1 - \frac{2 \sqrt{3}}{3} & -1 & 5\end{matrix}]$
- $[\begin{matrix}\sqrt{3} & \sqrt{3} & - 2 \sqrt{3}\end{matrix}]$ $\rightarrow$ $[\begin{matrix}1 - \frac{2 \sqrt{3}}{3} & 1 & 5\end{matrix}]$
ii) \c{(2 points) Sketch the result, being as accurate as possible and
labeling the colors of each of the visible faces.}
![](6bii.jpg){width=40%}
Excuse the poor sketch but the point is the transformation exposes the green
face to the left by doing a shear first.
iii) \c{(4 points) Show how you could achieve this transformation using one
or more matrix multiplication operations. Specify the matrix entries you
would use, and, if using multiple matrices, the order in which they would be
multiplied.}
It's the same as the ortho, except it uses an extra matrix that is applied
before any of the other transformations:
```py
def oblique_transform(left, right, bottom, top, near, far):
step_0 = np.array([
[1, 0, (1 / sqrt3), 0],
[0, 1, 0, 0],
[0, 0, 1, 0],
[0, 0, 0, 1],
])
M_ortho = ortho_transform(left, right, bottom, top, near, far)
return M_ortho @ step_0
```
7. \c{Consider the simple scene shown in the image below, where two cubes, one
of height 1 and one of height 2, are both resting on a horizontal groundplane
($y = -\frac{1}{2}$), with the smaller cube’s front face aligned with $z =

161
exam-2/exam2.py
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@ -1,7 +1,12 @@
import itertools
import numpy as np
from sympy import *
import math
from sympy import N, Number, Rational, init_printing, latex, simplify, Expr
from sympy.vector import CoordSys3D, Vector
init_printing()
import sympy
unit = lambda v: v/np.linalg.norm(v)
@ -57,45 +62,65 @@ def print_bmatrix(arr):
print("\\\\")
def problem_1():
p = np.array([1, 4, 8])
e = np.array([0, 0, 0])
s = np.array([2, 2, 10])
C = CoordSys3D('C')
p = 1 * C.i + 4 * C.j + 8 * C.k
e = 0 * C.i + 0 * C.j + 0 * C.k
s = 2 * C.i + 2 * C.j + 10 * C.k
radius = 3
i = p - e
print("incoming", i)
print("|I| =", np.linalg.norm(i))
n = s - p
print("normal", n)
v0 = p - e
print("v0 (incoming ray) =", v0)
print(" - |v0| =", v0.magnitude())
print()
n_norm = unit(n)
print("normal_norm", n_norm)
cos_theta_i = np.dot(-i, n) / (np.linalg.norm(i) * np.linalg.norm(n))
print("part a = cos^{-1} of ", cos_theta_i)
print(np.arccos(cos_theta_i))
n = p - s
print("n (normal) =", n)
n_norm = n.normalize()
print(" - n_norm (normalized) =", n_norm)
print()
proj = n_norm * np.linalg.norm(i) * cos_theta_i
print("proj", proj)
cos_theta_i = (-v0).dot(n) / (v0.magnitude() * n.magnitude())
theta_i = sympy.acos(cos_theta_i)
print("[1a] theta_i (angle of incidence) =", theta_i)
print()
p_ = p + proj
print("proj point", p_)
proj_len = cos_theta_i * v0.magnitude()
proj = n_norm * proj_len
print("projection of v0 onto n =", proj)
v2 = p_ - e
print("v2", v2)
nx = p + proj
print("nx (proj point) =", nx)
print()
sin_theta_i = np.sin(np.arccos(cos_theta_i))
print("sin theta_i =", sin_theta_i)
v1 = nx - e
print("v1 (from e to nx) =", v1)
theta_t = np.arcsin(1.0 / 1.5 * sin_theta_i)
print("approx answer for part d", theta_t)
z = e + 2 * v1
print("z =", z)
uin = unit(n_norm)
print("unit inv normal", uin)
v3 = z - p
print("v3 =", v3)
print()
uperp = unit(v2)
print("uperp", uperp)
sin_theta_i = sympy.sin(theta_i)
print("sin(theta_i) =", sin_theta_i)
answer_1e = uin + math.tan(theta_t) * uperp - p
print("1e answer", unit(answer_1e))
theta_t = sympy.asin(Rational(2, 3) * sin_theta_i)
print("[1d] theta_t (angle of transmission) =", theta_t)
print(" - approx =", theta_t.evalf())
print()
v5_len = sympy.tan(theta_t) * (s - p).magnitude()
print("v5 len =", v5_len)
v4 = (s - p) + v1.normalize() * v5_len
print("v4 =", v4)
print(" - [1e] normalized =", v4.normalize())
print()
print("s - p =", s - p)
print("|s - p| =", (s - p).magnitude())
print("|v1| =", v1.normalize())
print("tan(theta_t) =", sympy.tan(theta_t))
def problem_4():
camera_pos = np.array([2, 3, 5])
@ -229,14 +254,11 @@ def problem_7():
solve(camera_pos, angle)
def problem_6():
def calculate(points):
left = min(map(lambda p: p[0], points))
right = max(map(lambda p: p[0], points))
bottom = min(map(lambda p: p[1], points))
top = max(map(lambda p: p[1], points))
near = min(map(lambda p: p[2], points))
far = max(map(lambda p: p[2], points))
# sqrt3 = math.sqrt(3)
sqrt3 = sympy.sqrt(3)
width = 2 * sqrt3
def ortho_transform(left, right, bottom, top, near, far):
step_1 = np.array([
[1, 0, 0, 0],
[0, 1, 0, 0],
@ -245,46 +267,75 @@ def problem_6():
])
step_2 = np.array([
[1, 0, 0, -(left + right) / 2.0],
[0, 1, 0, -(bottom + top) / 2.0],
[0, 0, 1, -(near + far) / 2.0],
[1, 0, 0, -((left + right) / 2)],
[0, 1, 0, -((bottom + top) / 2)],
[0, 0, 1, -((near + far) / 2)],
[0, 0, 0, 1],
])
step_3 = np.array([
[2.0 / (right - left), 0, 0, 0],
[0, 2.0 / (top - bottom), 0, 0],
[0, 0, 2.0 / (far - near), 0],
[(2 / (right - left)), 0, 0, 0],
[0, (2 / (top - bottom)), 0, 0],
[0, 0, (2 / (far - near)), 0],
[0, 0, 0, 1],
])
M_ortho = step_3 @ step_2 @ step_1
return step_3 @ step_2 @ step_1
def oblique_transform(left, right, bottom, top, near, far):
step_0 = np.array([
[1, 0, (1 / sqrt3), 0],
[0, 1, 0, 0],
[0, 0, 1, 0],
[0, 0, 0, 1],
])
M_ortho = ortho_transform(left, right, bottom, top, near, far)
return M_ortho @ step_0
def calculate(M, points):
left = min(map(lambda p: p[0], points))
right = max(map(lambda p: p[0], points))
bottom = min(map(lambda p: p[1], points))
top = max(map(lambda p: p[1], points))
near = min(map(lambda p: p[2], points))
far = max(map(lambda p: p[2], points))
for point in points:
point_ = np.r_[point, [1]]
trans_ = M_ortho @ point_
trans_ = M(left, right, bottom, top, near, far) @ point_
trans = trans_[:3]
print("-", print_trans(np.around(point, 3), np.around(trans, 3)))
v = np.vectorize(lambda x: float(x.evalf()))
l = np.vectorize(lambda x: latex(simplify(x)))
point = l(point)
trans = l(trans)
print("-", print_trans(point, trans))
sqrt3 = math.sqrt(3)
width = 2.0 * sqrt3
cube_center = np.array([0, 0, -3.0 * sqrt3])
print("HELLOSU")
cube_center = np.array([0, 0, -3 * sqrt3])
points = []
for (dx, dy, dz) in itertools.product([-1, 1], [-1, 1], [-1, 1]):
point = cube_center + np.array([dx * width / 2, dy * width / 2, dz * width / 2])
for (dz, dx, dy) in itertools.product([-1, 1], [-1, 1], [-1, 1]):
point = cube_center + np.array([
dx * width / 2,
dy * width / 2,
dz * width / 2,
])
points.append(point)
calculate(points)
print("Helosu Ortho")
calculate(ortho_transform, points)
print("Helosu Oblique")
calculate(oblique_transform, points)
def problem_9():
pass
print("\nPROBLEM 8 -------------------------"); problem_8()
print("\nPROBLEM 1 -------------------------"); problem_1()
print("\nPROBLEM 5 -------------------------"); problem_5()
print("\nPROBLEM 9 -------------------------"); problem_9()
print("\nPROBLEM 7 -------------------------"); problem_7()
print("\nPROBLEM 6 -------------------------"); problem_6()
print("\nPROBLEM 4 -------------------------"); problem_4()
print("\nPROBLEM 6 -------------------------"); problem_6()
print("\nPROBLEM 1 -------------------------"); problem_1()