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Practice Final CMSC 427 Distributed Tuesday, December 11, 2007 Review Session, Monday, December 17, 5:00pm, 4424 AV Williams Final: 10:30 AM Wednesday, December 19, 2007
General Guidelines: The final will focus on topics that have been discussed in class. I will not ask questions about material from the text book that has not also been discussed in class or in a problem set. However, there will be no questions about OpenGL or programming problems. You will be expected to know some basic equations, but nothing very obscure. I’ll try to indicate below the key equations you should know.
The final will be closed book, with no notes or calculators. This means that problems should not require the use of a calculator; if you feel you need one when you are doing a problem this probably means there is an easier way to do the problem.
The final will be comprehensive. It will focus more (about 3/4 emphasis) on the material covered since the midterm, but you are still responsible for knowing material from the first half of the semester.
Topics and things to know about them:
- Geometry (needed for problems on the next two topics) a. How to take an inner product, and what this does. b. How to take a cross product, and what this does. c. What a unit vector is, and how to find the magnitude of a vector. d. How to write the equations for a 3D line, or for a plane in 3D, or for a sphere.
- Transformations a. How to write rotation about the x, y, or z axis as a matrix. b. How to write translation as a matrix. c. How to combine these. For example, rotate about a point not at the origin. d. How to rotate or translate a line. e. How to relate the rows of a rotation matrix to the axes of a new coordinate system This includes writing the coordinates of points relative to a new coordinate system. Given a position, lookat vector, an upvector, you should be able to write a matrix that transforms the world so that the viewer is at the origin, with the lookat direction in the z direction, and the upvector in the y direction.
- Projection a. Perspective projection i. How to find the image coordinates of a 3D point ii. Some consequences of that, such as properties of vanishing points, knowing when a line in 3D does not project to a line in the image, ….. b. Orthographic projection i. How to find the image coordinates of a 3D point.
ii. Consequences of orthographic projection, such as that it preserves parallelism and ratios of areas.
- Intersections and Relations between lines and planes and points. a. How to use the equation of a line or plane to tell which side of the line/plane a point is on. Example: Given a plane and two points, are the points separated by the plane, or on the same side of it? b. How to find the intersections of spheres, axial rectanguloids, triangles. c. How to intersect a line with a plane, triangle, or sphere
- Discretization a. How to represent a line with pixels. b. How to fill a polygon. i. How to tell if a point is inside a polygon. 1. Implicit function test (testing that it is on the appropriate side of each line). 2. Crossing number test. ii. Flood fill algorithm.
- Sampling and Aliasing a. Creating an image by sampling b. Aliasing – what it means, what causes it. i. Spatial ii. Temporal c. Anti-aliasing i. Smoothing ii. Supersampling and smoothing
- Interpolation a. Linear, bilinear b. Cubic c. Bezier curves -- Given three control points, you should know how to construct points on a Bezier curve, and know some of its basic properties.
- Texture a. Perlin noise algorithm – General idea
- Morphing and interpolation a. Morphing algorithm that you implemented. b. Describing points in a new coordinate system.
- Color a. Superposition b. Color Spaces – CIE, RGB, CMYK, HSV.
- Visibility a. Culling – Basic idea b. Painter’s algorithm i. For example, how to tell if a triangle 1 is in front of, or behind, the plane that contains triangle 2, and how knowing this might help you in the painter’s algorithm. c. Z buffer.
We can describe the line the ray is on with the equations x = z/2, y = 0. The sphere has the equation:
(x-3)^2 + (y-2)^2 + (z-12)^2 = 9 (^2 means squared)
Combining these equations we get:
x^2 – 6x + 9 + 4 + 4x^2 – 48x + 144 = 9 5x^2 – 54x + 148 = 0 The solution to this quadratic equation is: x = (54+- sqrt(-44))/10. That means that both solutions are imaginary, so there is no true intersection point.
- Suppose the camera has a focal point at (0,0,0), and you cast a ray through a pixel whose center is at (1/2, 0, 1). This ray definitely intersects a sphere centered at (3, 2, 12) with a radius of 5. What are the coordinates and surface normal of the first point that the ray intersects?
We get the same equation for a line, and for a sphere we get:
(x-3)^2 + (y-2)^2 + (z-12)^2 = 25 (^2 means squared)
Combining these equations we get:
x^2 – 6x + 9 + 4 + 4x^2 – 48x + 144 = 25 5x^2 – 54x + 132 = 0 The solution to this quadratic equation is: x = (54 +- sqrt( 276))/10 = 7.1, 3.7. These produce the points (7.1, 0,14.2), (3.7, 0, 7.4). The ray starts at the origin, and x and z increase as we go along, so it first reaches the point with the smallest values of x and z, which is (3.7,0,7.4). The normal of this point is a unit vector in the direction from the center of the sphere to this point, which is (3.7,0,7.4) – (3,2,12) = (.7, 2, -4.6)/5. (There’s a bit of round-off error here).
- Consider the BSP tree discussed in class:
a. Suppose we look at this scene from the right side, instead of the left. Use the BSP tree to determine the order in which the triangles should be rendered.
From the right, we’re behind 3, so we want to first render things in front of 3. We’re in front of 2, so we first render things behind it, which is 1. Then we render 2, then anything in front of it, which is 5a. Then 3. Then we are behind 4, so we first render things in front of it. But there is nothing there, we render 4, then finally 5b.
b. Construct a valid BSP tree starting with triangle 1 instead of triangle 3.
Starting with 1 as the root, 3, 4 and 5 are in front, and 2 is behind. We could pick any of 3, 4 or 5 to be the child of 1. Let’s pick 4. Then 3 and 5 are in back of it. Pick its child to be 5, and 3 is the final child, behind 5.
- Suppose we construct a model by taking a sphere centered at (10,0,10) with a radius of 2, and subtracting a cylinder. The cylinder is centered also at (10,0,10). It has a circular cross-section with a radius 1 when we slice it with a plane that has a constant z value (eg., if we intersect the cylinder with the z = 10 plane, we’ll have a circle centered at (10,0,10) with a radius of 1). The cylinder has a length of 5. Suppose we cast a ray from (0,0,0) in the direction (5,0,4). Will it intersect the model? If so, where?
We can describe an infinite cylinder that includes ours with the equation (x-10)^2 + y^2 = 1. We will see whether the ray intersects this infinite cylinder in its actual, finite extent. The line the ray is on has the equation: (x,y,z) = (0,0,0) + t(5,0,4) = (5t,0,4t). Substituting, we get: (5t-10)^2 = 1. So 5t - 10 = +-1 and t = 9/5 or 11/5. So the ray intersects the cylinder at:
from different angles. How might you use these images to determine what parameters you should use to model its reflectance? How many images do you think you would need?
If you take a picture in any direction other than that of the mirror reflection, you’ll just get the diffuse component of reflection. In the mirror direction you get lambertian plus mirror reflection. If you subtract the first image from the second, you get the pure mirror part of the reflection. This gives both Lambertian and mirror separately. Note that a real car hood reflects light in a way that is more complicated than Lambertian plus mirror; this is just a simple model.
9. Create a simple scene with a viewer looking in a specific direction, one or two simple objects, and a point source of light. Then work out how to render or ray trace this scene. Figuring out what kind of scene will create a reasonable set of problems will also be helpful to your studying. This is an Uber- problem. If you can do this, you have studied much of what is needed. a. Transformations i. Create a matrix that will transform the world so that the viewer is at the origin and the viewing direction is in the z direction. ii. Apply this matrix to an object in the scene to determine its coordinates after this transformation. iii. Apply the equations of perspective projection to determine the image location of the object. b. Ray Tracing. i. Find the equation for the image plane. ii. Determine a ray that goes from the viewer to a vertex in the scene, and figure out where it intersects the image plane. iii. Determine a ray that goes through the center of the image, and figure out which object in the scene it first intersects. iv. Suppose a ray strikes a surface that is a bit shiny (that is, it reflects light with some Phong and some Lambertian reflectance), but it is not at all transparent. If we are allowing for multiple bounces and shadows, what rays will we cast from the point where the original rays hits the surface. Give the direction of each ray. c. Lighting i. Find a ray from the viewer that intersects an object in the scene. Suppose the object is Lambertian and white. Determine the intensity it will produce in the image. ii. Now do the same, assuming the object has Phong reflectance. d. Shading i. The surface normal of a vertex isn’t really well-defined. One way to define it is to take the average of the normals of the sides of the object that form the corner. Using this approach, find the normals for the vertices of a polygon in the scene.
ii. As in (c), find the intensity that each of these vertices will produce in the image. Then determine the intensity that you will get in the middle of the polygon using Gouraud or Phong shading.
Consider the following scene.
- Imagine the ground is at y = 0.
- Cube1: There is a cube with corners at (20,0,20), (18,0, 18), (16, 0, 20), (18,0,22), (20,2,20), (18,2,18), (16,2,20), (18,2,22).
- Cube2: There is another cube with corners at (15, 0, 21), (17,0,21), (17,0,23), (15,0,23), (15, 2, 21), (17,2,21), (17,2,23), (15,2,23).
- Sphere: There is a sphere with a radius of 1, centered at (14, 1, 20).
- Viewer: There is a viewer located at (18,0,10), who is looking directly at the left corner of the first cube, that is, at the point (16,0,20). The camera has a focal length of 1.
- Light: There is a point light source located at (4,0,10). 1. Transformations a. Create a matrix that will transform the world so that the viewer is at the origin.
We just need to translate the viewer. We can do this with:
d a b
d^ c
b a c
Viewer
2. Ray Tracing. In this problem, imagine that we are rendering the scene using ray tracing. a. Keep in mind that the focal length of the camera is 1, and that we know what direction we are looking at. Find the equation for the image plane.
As stated before, the viewing direction is (-2,0,10)/sqrt(104). Call this vector n, because it is normal to the image plane. Then, an equation for the image plane is <n,(x,y,z)> = <n, (18,0,10)> + 1 = 64/sqrt(104) + 1. We can write this as: -2x/sqrt(104) + 10z/sqrt(104) = 64/sqrt(104) + 1. That is, the image plane is the set of points that are all the same distance from the origin in the direction n, and this distance is the distance to the viewer, plus one (for the focal length).
b. Suppose we cast a ray towards the lower corner of Cube1, (18,0, 18). Where would this ray intersect the image plane?
This ray is going from the focal point to (18,0,18), so it’s in the direction: (18,0,18)- (18,0,10) = (0,0,8). The equation for this ray is:
(18,0,10) + t(0,0,1) = (x,y,z). If we substitute into the equation for the image plane, we get:
-36/sqrt(104) + 10(10+t)/sqrt(104) = 64/sqrt(104) + 1. -36 + 100 + 10t = 64 + sqrt(104) 10t = sqrt(104) t = sqrt(104)/ (x,y,z) = (18, 0, 10 + sqrt(104)/10)
c. Suppose we cast a ray from the focal point in the direction (-.2, 0, .98). Which surface will this ray intersect? What will be the location of the intersection point? Where will this ray intersect the image plane?
First, let’s get a sense of what’s going on. This ray is almost entirely in the z direction. Let’s look at some points on it. It starts at (18,0,10). The lowest point of the cubes is z = 18, so let’s see what happens when we go +8 in the direction of the ray. We wind up at the point (16.4, 0, 17.84). So by the time the ray reaches the lowest point in Cube 1, it is to the left of side C. Going a bit further, the ray will get to (16, 0, 19.8). So it is also going to miss side D of cube 1. Looks like it’ll hit side D of cube 2. To check this, we must see what the x value is when z = 21. Since we have (x,y,z) = (18,0,10) + t(-.2, 0, .98), when z = 21, 11 = t.98, so t = 11.2245. So when z = 21 on the ray, x = 18 - .211.2245 = 15.75. Side D of cube 2 goes from (15,0,21) to (17,0,21), so there is an intersection.
d. Suppose cube1 is a bit shiny (that is, it reflects light with some Phong and some Lambertian reflectance), but it is not at all transparent. We cast a ray from the camera’s focal point which hits sideD of cube1 at the point (17,1,19). If we are
allowing for multiple bounces and shadows, what rays will we cast from (17,1,19)? Give the direction of each ray.
First, the shadow ray will go from the intersection point towards the light. That is, its direction will be (4,0,10) – (17,1,19) = (-13, -1, -9). We make this a unit vector by dividing by sqrt(251).
We also need to cast a ray in the direction of mirror reflection. First, we need to know the ray in the direction to the viewer. This is (18,0,10) – (17,1,19) = (1,1,-9). The unit vector is (1,1,-9)/sqrt(83). Because the cube is at a 45 degree angle, its normal vector is (-1,0,-1)/sqrt(2). Now we can use the equation (v + r) = 2n<n,v>. This gives us: r + (1,1,-9)/sqrt(83) = 28/sqrt(166)(-1,0,-1)/sqrt(2).
3. Lighting a. Suppose the two cubes are Lambertian and white (they have an albedo of 1). What will be the intensity of the mid-point of each side of each cube? I mean the center points, which have a y value of 1. For example, the midpoint of Cube1- sideD is (17,1,19). Take into account shadows and the fact that light falls of as 1/d^2. To get started, we suppose that the center point of sideD in cube1 has an intensity of 100.
This is a bit complicated. If the point is in shadow, it’s intensity is zero. First of all, one can see that most of the cube sides are facing away from the light, and so they will be attached shadows with intensity of 0. Only sides a and d are facing towards the light. We know from the problem that side D of cube 1 isn’t in shadow. So we have to check on the others. Specifically, is the sphere shadowing any of them? You’ll have to calculate the midpoints, and see whether a line from the midpoint to the light intersects the sphere.
For the ones that aren’t in shadows, side D cube 1 allows us to calculate the intensity of the light. Then we need to compute the normal of each side, the inner product between this and the direction to the light (don’t forget to make it a unit vector) and scale this by the light intensity and 1/d^2.
4. Shading a. The surface normal of a cube at its corner isn’t really well-defined. One way to define it is to take the average of the normals of the three sides of the cube that form the corner. Using this approach, find the normals for the four corners of sideD of cube1. b. Suppose cube1 is a white, Lambertian object, and the light has intensity of 1. Ignoring any cast shadows from the sphere, determine the intensity of light reflected at each of these four corners. c. Suppose we use Gouraud shading. What will be the intensity of the point (16.5, .5, 19.5), which is on sideD of cube1, in the image? d. What would be the answer to problem (c) if we use Phong shading?
