133 lines
5.9 KiB
Markdown
133 lines
5.9 KiB
Markdown
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geometry: margin=2cm
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output: pdf_document
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---
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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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see where it would intersect objects.
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### Creating a virtual screen
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The virtual screen is determined first using the eye's position and where it's
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looking. This gives us a single 3d vector, but it doesn't give us a 2d screen in
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the world. This is where the field of view (FOV) comes in; the FOV determines
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how many degrees the screen should take up.
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{width=180px}
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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. An
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example is shown here:
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{width=180px}\ {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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project rays. As the two black rectangles in the diagram above demonstrates,
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changing the distance would still allow the viewer to see the same amount of the
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scene. (using the word _amount_ very loosely here to mean percentage of the
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landscape, rather than # of pixels, which is determined by the actual image
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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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{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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{width=180px}\ {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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example; simply move the eye position to be incredibly close to the object that
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we are observing, and increase the field of view to cover the entire object.
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Because the rays are going in much different directions and travelling different
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distances, the corners of the image will seem more stretched than if we were
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observing the object from afar and all the rays are in approximately the same
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part of the virtual screen.
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One other point to make is that we're currently using a rectangle for our
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virtual screen, which automatically does a bit of the distortion. If instead we
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were to use a curved lens-like shape, then the rays pointing to any pixel of the
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screen would be travelling the same distance. Moving the eye position closer to
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the object would still generate distortion, but to a lesser extent.
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### Mapping image pixels
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After the rectangle has been determined, we can simply pick one corner to start
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as an anchor, and then find out what pixel values would correspond to it. For
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example, in the image below:
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{width=240px}
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I would pick a starting point like $A$, and then take the vector $B-A$ and
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subdivide it into 4 pieces, letting $\Delta x = \frac{B-A}{4}$. Then, same thing
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for the $y$ direction, I would set $\Delta y = \frac{D-A}{4}$. Taking $A +
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x_i \times \Delta x + y_i \times \Delta y$ yields the precise coordinate
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location for any pixel.
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(Technically really we would want the middle of the pixel, so just add
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$\frac{\Delta x + \Delta y}{2}$ to the point to get that)
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### Parallel Projection Notes
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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. 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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First, we will transform the current point into the vector space of the
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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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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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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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