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exam-2/exam2.md

@@ -69,7 +69,7 @@ author: |
    in the $+z$ direction. Derive a sequence of model transformation matrices
    that can be applied to the vertices of the airplane to position it in space
    at the location $p = (4, 4, 7)$, with a direction of flight $w = (2, 1, -2)$
-   and the wings aligned with the direction $d = (-2, 2, –1)$.}
+   and the wings aligned with the direction $d = (-2, 2, -1)$.}
 
    The translation matrix is
 
@@ -92,25 +92,89 @@ author: |
    Since the direction of flight was originally $(0, 0, 1)$, we have to
    transform it to $(2, 1, -2)$.
 
-8. Consider the perspective projection-normalization matrix P which maps the
+3.
-   contents of the viewing frustum into a cube that extends from -1 to 1 in $x,
-   y, z$ (called normalized device coordinates).
 
-   Suppose you want to define a square, symmetric viewing frustum with a near
+## The Camera/Viewing Transformation
-   clipping plane located 0.5 units in front of the camera, a far clipping plane
-   located 20 units from the front of the camera, a 60 ̊vertical field of view,
-   and a 60 ̊horizontal field of view.
 
-   a. \c{(2 points) What are the entries in P?}
+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
+   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
+   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
+   pointing in the direction $(1, -1, -1)$, and the camera's 'up' direction is
+   $(0, 1, 0)$, what are the entries in $V$?
 
    $$\begin{bmatrix}
+     0 & 1 & 0 & d_x \\
    \end{bmatrix}$$
 
-   b. (3 points) How should be matrix P be re-defined if the viewing window is
+   b. (2 points) How will this matrix change if the eye moves forward in the
-   re-sized to be twice as tall as it is wide?
+   direction of view? [which elements in V will stay the same? which elements
+   will change and in what way?]
 
-   c. (3 points) What are the new horizontal and vertical fields of view after
+   c. (2 points) How will this matrix change if the viewing direction spins in
-   this change has been made?
+   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?]
+
+   d. (2 points) How will this matrix change if the viewing direction rotates
+   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
+   what way?]
+
+5.
+
+## The Projection Transformation
+
+6.
+
+7.
+
+8. \c{Consider the perspective projection-normalization matrix $P$ which maps
+   the contents of the viewing frustum into a cube that extends from -1 to 1 in
+   $x, y, z$ (called normalized device coordinates).}
+
+   \c{Suppose you want to define a square, symmetric viewing frustum with a near
+   clipping plane located 0.5 units in front of the camera, a far clipping plane
+   located 20 units from the front of the camera, a $60^\circ$ vertical field of
+   view, and a $60^\circ$ horizontal field of view.}
+
+   a. \c{(2 points) What are the entries in $P$?}
+
+   The left / right values are found by using the tangent of the field-of-view
+   triangle: $\tan(60^\circ) = \frac{\textrm{right}}{0.5}$, so $\textrm{right} =
+   \tan(60^\circ) \times 0.5 = \boxed{\frac{\sqrt{3}}{2}}$. The same goes for the
+   vertical, which also yields $\frac{\sqrt{3}}{2}$.
+
+   $$\begin{bmatrix}
+     \frac{2\times near}{right - left} & 0 & \frac{right + left}{right - left} & 0 \\
+     0 & \frac{2\times near}{top - bottom} & \frac{top + bottom}{top - bottom} & 0 \\
+     0 & 0 & -\frac{far + near}{far - near} & -\frac{2\times far\times near}{far - near} \\
+     0 & 0 & -1 & 0
+   \end{bmatrix}$$
+
+   $$= \begin{bmatrix}
+     \frac{2\times 0.5}{\frac{\sqrt{3}}{2} - (-\frac{\sqrt{3}}{2})} & 0 & \frac{\frac{\sqrt{3}}{2} + (-\frac{\sqrt{3}}{2})}{\frac{\sqrt{3}}{2} - (-\frac{\sqrt{3}}{2})} & 0 \\
+     0 & \frac{2\times 0.5}{\frac{\sqrt{3}}{2} - (-\frac{\sqrt{3}}{2})} & \frac{\frac{\sqrt{3}}{2} + (-\frac{\sqrt{3}}{2})}{\frac{\sqrt{3}}{2} - (-\frac{\sqrt{3}}{2})} & 0 \\
+     0 & 0 & -\frac{20 + 0.5}{20 - 0.5} & -\frac{2\times 20\times 0.5}{20 - 0.5} \\
+     0 & 0 & -1 & 0
+   \end{bmatrix}$$
+
+   $$= \boxed{\begin{bmatrix}
+     \frac{1}{\sqrt{3}} & 0 & 0 & 0 \\
+     0 & \frac{1}{\sqrt{3}} & 0 & 0 \\
+     0 & 0 & -\frac{41}{39} & -\frac{40}{39} \\
+     0 & 0 & -1 & 0
+   \end{bmatrix}}$$
+
+   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?}
+
+
+
+   c. \c{(3 points) What are the new horizontal and vertical fields of view
+   after this change has been made?}
 
 ## Clipping
 

exam-2/exam2.py

@@ -1,4 +1,6 @@
 import numpy as np
+from sympy import *
+import math
 
 unit = lambda v: v/np.linalg.norm(v)
 
@@ -33,6 +35,9 @@ def problem_1():
 
     print("approx answer for part d", np.arcsin(1.0 / 1.5 * sin_theta_i))
 
+def problem_4():
+    print("part 4a.")
+
 def problem_8():
     def P(left, right, bottom, top, near, far):
         return np.array([
@@ -42,9 +47,23 @@ def problem_8():
             [0, 0, -1, 0],
         ])
 
-    near, far = left, right = bottom, top = -1, 1
+    near = 0.5
-    print("part 8a", P(left, right, bottom, top, near, far))
+    far = 20
+
+    def compute_view(vfov, hfov):
+        left = -math.tan(hfov) * near
+        right = math.tan(hfov) * near
+        bottom = -math.tan(vfov) * near
+        top = math.tan(vfov) * near
+        return left, right, bottom, top
+
+
+    print("part 8a")
+    vfov = hfov = math.radians(60)
+    left, right, bottom, top = compute_view(vfov, hfov)
+    print(P(left, right, bottom, top, near, far))
     print()
 
-problem_8()
+print("\nPROBLEM 4 -------------------------"); problem_4()
-problem_1()
+print("\nPROBLEM 8 -------------------------"); problem_8()
+print("\nPROBLEM 1 -------------------------"); problem_1()