Add more to the writeup
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7 changed files with 128 additions and 25 deletions
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@ -36,7 +36,7 @@ examples/%.ppm: examples/%.txt
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examples/%.png: examples/%.ppm
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convert $< $@
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writeup.pdf: writeup.md
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writeup.pdf: writeup.md $(EXAMPLES_PNG)
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$(PANDOC) -o $@ $<
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clean:
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21
assignment-1/examples/fov-demo-1.txt
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21
assignment-1/examples/fov-demo-1.txt
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@ -0,0 +1,21 @@
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imsize 640 480
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eye 0 0 15
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viewdir 0 0 -1
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hfov 60
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updir 0 1 0
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bkgcolor 0.1 0.1 0.1
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mtlcolor 0 0.5 0.5
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sphere -1 -2 -5 2
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sphere 3 -5 -1 0.5
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mtlcolor 0.5 0.5 1
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sphere 1 2 -3 3
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sphere -6 3 -4 1
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mtlcolor 0.5 0 0.5
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sphere 5 5 -1 1
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sphere -6 -4 -8 7
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mtlcolor 0.5 1 0.5
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cylinder 5 1 -2 1 -2 1 1 2
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21
assignment-1/examples/fov-demo-2.txt
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assignment-1/examples/fov-demo-2.txt
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@ -0,0 +1,21 @@
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imsize 640 480
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eye 0 0 15
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viewdir 0 0 -1
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hfov 30
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updir 0 1 0
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bkgcolor 0.1 0.1 0.1
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mtlcolor 0 0.5 0.5
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sphere -1 -2 -5 2
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sphere 3 -5 -1 0.5
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mtlcolor 0.5 0.5 1
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sphere 1 2 -3 3
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sphere -6 3 -4 1
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mtlcolor 0.5 0 0.5
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sphere 5 5 -1 1
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sphere -6 -4 -8 7
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mtlcolor 0.5 1 0.5
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cylinder 5 1 -2 1 -2 1 1 2
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9
assignment-1/examples/up-dir-demo-1.txt
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assignment-1/examples/up-dir-demo-1.txt
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@ -0,0 +1,9 @@
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imsize 640 480
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eye 0 0 8
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viewdir 0 0 -1
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hfov 60
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updir 0 1 0
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bkgcolor 1 1 1
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mtlcolor 0.5 1 0.5
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cylinder 0 0 -4 1 0 0 1 8
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9
assignment-1/examples/up-dir-demo-2.txt
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assignment-1/examples/up-dir-demo-2.txt
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@ -0,0 +1,9 @@
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imsize 640 480
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eye 0 0 8
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viewdir 0 0 -1
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hfov 60
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updir 1 1 0
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bkgcolor 1 1 1
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mtlcolor 0.5 1 0.5
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cylinder 0 0 -4 1 0 0 1 8
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@ -57,7 +57,7 @@ impl ObjectKind for Cylinder {
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let discriminant = b * b - 4.0 * a * c;
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let mut solutions = match discriminant {
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let possible_side_solutions = match discriminant {
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// Discriminant < 0, means the equation has no solutions.
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d if d < 0.0 => vec![],
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@ -76,6 +76,16 @@ impl ObjectKind for Cylinder {
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_ => unreachable!("Invalid determinant value: {discriminant}"),
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};
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// Filter out solutions that don't have a valid Z position.
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let side_solutions = possible_side_solutions.into_iter().filter(|t| {
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let ray_point = ray.eval(*t);
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let rotated_ray_point = rotation_matrix * ray_point;
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let z = rotated_ray_point.z - rotated_cylinder_center.z;
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// Check to see if z is between -len/2 and len/2
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z.abs() < self.length / 2.0
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});
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// We also need to add solutions for the two ends of the cylinder, which
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// uses a similar method except backwards: check intersection points
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// with the correct z-plane and then see if the points are within the
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@ -96,26 +106,19 @@ impl ObjectKind for Cylinder {
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(o.z + c.z - self.length / 2.0) / r.z,
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]
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};
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// Filter out all the solutions where the z does not lie in the circle
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solutions.extend(possible_z_intersections.into_iter().filter(|t| {
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let end_solutions = possible_z_intersections.into_iter().filter(|t| {
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let ray_point = ray.eval(*t);
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ray_point.x.powi(2) + ray_point.y.powi(2) <= self.radius.powi(2)
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}));
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});
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// Filter out solutions that don't have a valid Z position.
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let solutions = solutions
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.into_iter()
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.filter(|t| {
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let ray_point = ray.eval(*t);
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let rotated_ray_point = rotation_matrix * ray_point;
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let z = rotated_ray_point.z - rotated_cylinder_center.z;
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// Check to see if z is between -len/2 and len/2
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z.abs() < self.length / 2.0
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})
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.filter_map(|t| NotNan::new(t).ok());
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let solutions = side_solutions.into_iter().chain(end_solutions.into_iter());
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// Return the minimum solution
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solutions.min().map(|t| t.into_inner())
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solutions
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.filter_map(|t| NotNan::new(t).ok())
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.min()
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.map(|t| t.into_inner())
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}
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}
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@ -5,6 +5,10 @@ output: pdf_document
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# Raycaster
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#### Michael Zhang \<zhan4854@umn.edu\>
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---
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Determining the viewing window for the raycaster for this assignment involved
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creating a "virtual" screen in world coordinates, mapping image pixels into that
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virtual screen, and then casting a ray through each pixel's world coordinate to
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@ -22,7 +26,15 @@ how many degrees the screen should take up.
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Changing the angle of the field of view would result in a wider or narrower
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screen, which when paired with the aspect ratio (width / height), would produce
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a bigger or smaller viewing screen, like the orange box in the above diagram
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shows. Simply put, FOV affects how _much_ of the frame you're able to see.
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shows. Simply put, FOV affects how _much_ of the frame you're able to see. An
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example is shown here:
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![](examples/fov-demo-1.png){width=180px}\ ![](examples/fov-demo-2.png){width=180px}
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The left image uses an FOV of 60, while the right image uses an FOV of 30. As
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you can see, the left side has a wider range of vision, which allows it to see
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more of the world. (both images can be found in the `examples` directory of the
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handin zip)
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Curiously, distance from the eye actually doesn't really affect the viewing
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screen very much. The reason is the screen is only used to determine how to
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@ -34,10 +46,16 @@ dimensions)
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The up-direction vector controls the rotation of the scene. Without the
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up-direction, it would not be possible to tell which rotation the screen should
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be in:
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be in.
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![Rotation determined by up direction](doc/rot.jpg){width=240px}
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To see what this looks like, consider the following images, where the left side
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uses an up direction of $(0, 1, 0)$, while the right side uses $(1, 1, 0)$ (both
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images can be found in the `examples` directory of the handin zip)
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![](examples/up-dir-demo-1.png){width=180px}\ ![](examples/up-dir-demo-2.png){width=180px}
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Together, all of these parameters can uniquely determine a virtual screen
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location, that we can use to cast rays through and fill pixels. We can change
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any of these to produce an image with a more exaggerated view of the scene for
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@ -76,7 +94,8 @@ $\frac{\Delta x + \Delta y}{2}$ to the point to get that)
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Because of the way I implemented parallel projection, it's recommended to
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either put the eye much farther back, or use `--distance` to force a much bigger
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distance from the eye for the raycaster. This is due to the size of the image.
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distance from the eye for the raycaster. See the `--help` to see how this option
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is used.
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### Cylinder Intersection Notes
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cylinder, so that the cylinder location is $(0, 0, 0)$ and the direction vector
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is normalized into $(0, 0, 1)$.
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Then it's a matter of determining if the $x$ and $y$ coordinates fall into the
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space constrained by the equation $(ox + t\times rx - cx)^2 + (oy + t\times ty -
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cy)^2 = r^2$ and if $z \le L$.
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This can be done by using a rotation matrix (since we are sure this
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transformation is just a rotation). This rotation is actually a 2D rotation,
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around the normal between the cylinder direction and $(0, 0, 1)$. We then
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rotate everything we are working with (the cylinder and the ray) into this
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coordinate system to make calculations easier.
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See the comments in the code for a more detailed explanation of the equation
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used.
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Then it's a matter of determining if the $x$ and $y$ coordinates fall into the
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space constrained by the equation $(o_x + t\times r_x - c_x)^2 + (o_y + t\times
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r_y - c_y)^2 = r^2$ and if $z \le L$. I can solve this using the quadratic
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formula the same way as the sphere case.
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We want a quadratic equation of the form $At^2 + Bt + C = 0$. The values for
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$A$, $B$, and $C$ are:
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- $A = r_x^2 + r_y^2$
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- $B = 2(r_x(o_x - c_x) + r_y(o_y - c_y))$
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- $C = (c_x - o_x)^2 + (c_y - o_y)^2 - r^2$
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Solving this for $t$ yields 0-2 solutions depending on if the equation was
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satisfied or not. Then, we can plug any solutions we get back into the ray
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equation and determine if the $z$-coordinate is in the range of the cylinder
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that we want.
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We will also have to do this for the ends of the cylinder, but just backwards.
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So we would start with the $z$-coordinate, solve for $t$s where the ray hits
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that $z$-plane, and then check the $x$ and $y$ values to see if they satisfy the
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ray equation as well.
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