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exam-2/exam2.md

@@ -14,6 +14,8 @@ author: |
 \newcommand{\now}[1]{\textcolor{blue}{#1}}
 \newcommand{\todo}[0]{\textcolor{red}{\textbf{TODO}}}
 
+[ 1 2 3 4 6 7 8 9 10 ]
+
 ## Reflection and Refraction
 
 1. \c{Consider a sphere $S$ made of solid glass ($\eta$ = 1.5) that has radius $r =
@@ -33,7 +35,7 @@ author: |
    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}(-\frac{1}{3})}$.
 
    c. \c{(3 points) What is the direction of the reflected ray?}
 
@@ -62,6 +64,27 @@ author: |
 
    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 $v = (0, 6, 6)$.
+
+   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 $\tan(\theta_t) \times (0, \frac{1}{\sqrt{2}},
+   \frac{1}{\sqrt{2}})$.
+
+   The final step is then adding these vectors together and subtracting the
+   interersection point: $(\frac{1}{3}, -\frac{2}{3}, \frac{2}{3}) +
+   \tan(\theta_t) \times (0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}) - (1, 4,
+   8)$.
+
+   \todo all the numbers are wrong
+
 ## Geometric Transformations
 
 2. \c{(8 points) Consider the airplane model below, defined in object
@@ -72,7 +95,26 @@ author: |
    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
+   The order we want is (1) do all the rotations, and then (2) translate to the
+   spot we want. The rotation is done in multiple steps:
+
+   - First we want to make sure the nose of the plane points in the correct
+     direction in the $xz$ plane (rotating around the $y$ axis). The desired
+     resulting direction is $(2, y, -2)$, so that means for a nose currently facing
+     the $+z$ direction which is $(0, y, 1)$, we want to rotate around by around
+     $-\frac{3}{4}\pi$. The transformation matrix is:
+
+     $$
+     M_1 = \begin{bmatrix}
+        1 & 0 & 0 & 0 \\
+        0 & 1 & 0 & 0 \\
+        0 & 0 & 1 & 0 \\
+        0 & 0 & 0 & 1 \\
+     \end{bmatrix}
+     $$
+
+   - Then we want to rotate the plane vertically, so it's pointing in the right
+     direction.
 
    $$
    \begin{bmatrix}
@@ -93,7 +135,56 @@ author: |
    Since the direction of flight was originally $(0, 0, 1)$, we have to
    transform it to $(2, 1, -2)$.
 
-3.
+3. \c{Consider the earth model shown below, which is defined in object
+   coordinates with its center at $(0, 0, 0)$, the vertical axis through the
+   north pole aligned with the direction $(0, 1, 0)$, and a horizontal plane
+   through the equator that is spanned by the axes $(1, 0, 0)$ and $(0, 0, 1)$.}
+
+   a. \c{(3 points) What model transformation matrix could you use to tilt the
+   vertical axis of the globe by $23.5^\circ$ away from $(0, 1, 0)$, to achieve
+   the pose shown in the image on the right?}
+
+   You could use a 3D rotation matrix. Since the axis of rotation is the
+   $z$-axis, the rotation matrix would look like:
+
+   $$
+   M =
+   \begin{bmatrix}
+     \cos(23.5^\circ) & \sin(23.5^\circ) & 0 & 0 \\
+     -\sin(23.5^\circ) & \cos(23.5^\circ) & 0 & 0 \\
+     0 & 0 & 1 & 0 \\
+     0 & 0 & 0 & 1 \\
+   \end{bmatrix}
+   $$
+
+   b. \c{(5 points) What series of rotation matrices could you apply to the
+   globe model to make it spin about its tilted axis of rotation, as suggested
+   in the image on the right?}
+
+   One way would be to rotate it back to its normal orientation, apply the spin
+   by whatever angle $\theta(t)$ it was spinning at at time $t$, and then put it
+   back into its $23.5^\circ$ orientation. This rotation is done around the
+   $y$-axis, so the matrix looks like:
+
+   $$
+   M' =
+   M^{-1}
+   \begin{bmatrix}
+     \cos(\theta(t)) & 0 & \sin(\theta(t)) & 0 \\
+     0 & 1 & 0 & 0 \\
+     -\sin(\theta(t)) & 0 & \cos(\theta(t)) & 0 \\
+     0 & 0 & 0 & 1 \\
+   \end{bmatrix}
+   M
+   $$
+
+   c. \c{[5 points extra credit] What series of rotation matrices could you use
+   to send the tilted, spinning globe model on a circular orbit of radius $r$
+   around the point $(0, 0, 0)$ within the $xz$ plane, as illustrated below?}
+
+   In the image, the globe itself does not rotate, but I'm going to assume it
+   revolves around the sun at a different angle $\phi(t)$. The solution here
+   would be to
 
 ## The Camera/Viewing Transformation
 
@@ -253,7 +344,39 @@ author: |
 
 ## The Projection Transformation
 
-6.
+6. \c{Consider a cube of width $2\sqrt{3}$ centered at the point $(0, 0,
+   -3\sqrt{3})$, whose faces are colored light grey on the top and bottom $(y =
+   \pm\sqrt{3})$, dark grey on the front and back $(z = -2\sqrt{3} \textrm{and}
+   z = -4\sqrt{3})$, red on the right $(x = \sqrt{3})$, and green on the left $(x
+   = -\sqrt{3})$.}
+
+   a. \c{Show how you could project the vertices of this cube to the plane $z =
+   0$ using an orthographic parallel projection:}
+
+      i) \c{(2 points) Where will the six vertex locations be after such a
+      projection, omitting the normalization step?}
+
+      ii) \c{(1 points) Sketch the result, being as accurate as possible and
+      labeling the colors of each of the visible faces.}
+
+      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
+      multiplied.}
+
+   b. Show how you could project the vertices of this cube to the plane $z = 0$
+   using an oblique parallel projection in the direction $d = (1, 0, \sqrt{3})$:
+
+      i) (3 points) Where will the six vertex locations be after such a projection,
+      omitting the normalization step?
+
+      ii) (2 points) Sketch the result, being as accurate as possible and labeling
+      the colors of each of the visible faces.
+
+      iii) (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.
 
 7.
 
@@ -300,6 +423,8 @@ author: |
    \end{bmatrix}}
    $$
 
+   \todo the numbers are wrong lmao
+
    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?}
 
@@ -369,6 +494,36 @@ author: |
     but you should not omit any steps that are essential to the process of
     generating an image of a provided list of triangles.}
 
+    The most essential steps in the scan-conversion process are:
+
+    - First, the input is given as a bunch of geometries. This includes
+      triangles and quads. Spheres aren't really part of the primitives, not
+      sure why historically but I can imagine it makes operations like
+      clipping monstrously more complex.
+
+    - Then, vertices of the geometries are passed to vertex shaders, which are
+      custom user-defined scripts that are run in parallel on graphics hardware
+      that are applied to each individual vertex. This includes things like
+      transforming it through coordinate systems (world coordinates vs. camera
+      coordinates vs. normalized coordinates) in order for the rasterizer
+      to be able to go through and process all of the geometries quickly.
+
+    - Clipping is done as an optimization to reduce the amount of things outside
+      the viewport that needs to be rendered. There are different clipping
+      algorithms available, but they all typically break down triangles into
+      smaller pieces that are all contained within the viewport. Culling is also
+      done to remove faces that aren't visible and thus don't need to be
+      rendered.
+
+    - The rasterizer then goes through and gives pixels colors based on which
+      geometries contain them. This includes determining which geometries are in
+      front of others, usually done using a z-buffer. When a z-buffer is used,
+      the color of the closest object is stored and picked. During this process,
+      fragment shaders also influence the output. Fragment shaders are also
+      custom user-defined programs running on graphics hardware, and can modify
+      things like what color is getting output or the order of the depth buffer
+      based on various input factors.
+
 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

exam-2/exam2.py

@@ -33,7 +33,17 @@ def problem_1():
     sin_theta_i = np.sin(np.arccos(cos_theta_i))
     print("sin theta_i =", sin_theta_i)
 
-    print("approx answer for part d", np.arcsin(1.0 / 1.5 * sin_theta_i))
+    theta_t = np.arcsin(1.0 / 1.5 * sin_theta_i)
+    print("approx answer for part d", theta_t)
+
+    uin = unit(n_norm)
+    print("unit inv normal", uin)
+
+    uperp = unit(v2)
+    print("uperp", uperp)
+
+    answer_1e = uin + math.tan(theta_t) * uperp - p
+    print("1e answer", unit(answer_1e))
 
 def problem_4():
     print("part 4a.")