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Problem Set 5
CMSC 426
Assigned Tuesday, November 1, Due Tuesday, November 8
- Projection a. 5 points: Suppose we have a camera with focal point (0,0,0) and image plane z = 1. We observe a point in the scene at the location (18, 6, 3). Where does this point appear in the image?
We divide the x and y coordinates by the z coordinate, to obtain (6,2,1). Note that we would multiply this by the focal length, but this is 1.
b. 10 points: Suppose we have a camera with focal point (3,2,3) and image plane x + y + z = 11. We are looking at a scene point at the location (13,7,8). Where does this appear in the image plane? That is, give the coordinates of a 3D point that is in the image plane.
The point is projected along the line joining the focal point and the scene point, which is: (3,2,3) + t(10,5,5). We need to find the point where this intersects the image plane. This is where 3+10t + 2 + 5t + 3 + 5t = 11, which implies 20t = 3, t = 3/20. So the point is at (4.5, 2.75, 3.75).
c. 10 points: Suppose we have a camera with a focal point at (0,0,0), and an image plane at z = 1. We are looking at the line described by the equations x = 1, y = z. What is the vanishing point of this line?
A point on this line has coordinates (1, z, z). The image of this point is (1/z, 1). As the point gets very far away, that is, as z goes to infinity, this point converges to (0, 1).
d. 10 points: Suppose we have a camera with a focal point at (0,0,0), and an image plane at z = 1. We are looking at the plane x = 1. What is the horizon of this plane?
The horizon is a line. Any point on this plane will have coordinates (1, y, z). For any point that is far away, the image of this point, (1/z, y/z) will approach (0, y/z). y/z could be anything, depending on what y is. So, all points on the horizon will be on the line, x = 0, (in the image plane, z = 1).
e. Challenge problem, 10 points: Suppose we have a camera with a focal point at (0,0,0), and an image plane at z = 1. We are looking at a circle in the plane x = 1. It has its center at the location (1, 0, 10) , and a radius of
1. What shape does the circle produce in the image? Describe this with an equation.
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- Reconstruction a. 5 points: Suppose we have a camera with a focal point at (0,0,0), and an image plane at z = 1. We see an image point at the location (5,-2,1). Give a line in 3D that this point must lie on.
The line goes from the focal point to the image point, (x,y,z) = (0,0,0) + t(5,-2,1).
b. 10 points: Suppose the point is known to be on a ramp, which is described by the equation 3y-z = -140. What are the 3D coordinates of the scene point we are looking at?
We have to find a point at the intersection of 3y-z = -140 and (x,y,z) = (0,0,0) + t(5, -2, 1). Combining these equations we have 3(-2t)-t = -140. So t = 20, and (x,y,z) = (100, - 40, 20).
- Stereo a. 5 points: Suppose we have two cameras. One has a focal point at (0,0,0), the second has a focal point at (10,0,0). Both have the same image plane, z = 1. Suppose we see a scene point at (2, 5, 1) in the first image, and at (10,5,1) in the second image. Where is the scene point?
We know z = T/d. In this setup, T = 10, d = 2, so z = 5. From the first image, we know that the scene point is on the line (0,0,0) + t(2,5,1). This means t = 5, and the scene point is at (10, 25, 5). We can check that this will project to the image point (10,5,1) in the second image.
b. 10 points: Suppose the second camera has an image plane at z = 2. If we see a point in the first image at (0,1,1), what is the corresponding epipolar line in the second image?
The focal points are (0,0,0) and (10,0,0), and we see a point at: (0,1,1). First, let’s find an equation for a plane that goes through these points. In general, we have Ax + By + Cz = D. From the first point, we know D = 0. From the second point, we know A = 0. So we have B + C = 0 from the third point. We can choose any values that satisfy this equation, so let’s say the plane has the equation: y – z = 0. We can check that this fits all three points. Next, we need to intersect this with the image plane for the second image, which is z = 2. This gives us z = 2, y = 2. This describes a line, in which x can take on any value.
c. 10 points: Suppose the second camera has an image plane at x = 9. What is the epipole in the second image?
We just need to find a line through (0,0,0) and (10,0,0), and intersect it with the plane x = 9, which will occur at the point (9,0,0).
