CSC 418/2504 Fall 2001 Assignment 2: Frequently asked questions

Questions about:

Question 1.
Question 2.
Question 3.
Question 4.
Question 5.
Question 6.

Question 1

Q:I don't know how to get started for part b) and c).
A: If you know the coordinates of a point, in reference frame A, and want the coordinates in reference frame B, you'll have to multiply it by a special transformation matrix. Follow this rule:
- The matrix that converts points from frame A to frame B is the same matrix that moves frame B to frame A.
notice that the order changes. By the way, this was done in question 2 of the
tutorial on geometric transformations #1.
Q:Coordinate system A is not square!
A:True. You'll have to do a shear operation.

Question 2

Q:How do I find vanishing points?
A:Build the WCS->CCS transformation, and apply it to the direction vector in question (a vector, in homogeneous coordinates, has ZERO, not one, for its fourth coefficient). Then divide by the fourth coefficient as usual.

Question 3

Q: For part a), do we just give the coordinates of the points?
A: Yes.
Q: For part b), I wrote some code to calculate the normals. Do I have to submit it?
A: No, but say HOW you obtained the normals.
Q:But how do I compute the normal?
A: In case you missed it in class, use cross products:

N = (+/-)(P₂-P₁) X (P_n-P₁)

Question 4

Q: Rows and columns are swapped in xy2rc!
A: True. Sorry about that. The function should read:

void xy2rc (double x, double y, int *row, int *col) {
  *col = (int) ((x-xMin) / (xMax-xMin) * winWidth);
  *row = (int) (winHeight - (y-yMin) / (yMax-yMin) * winHeight);
}

Q: The diagram is upside down!
A: True. Sorry about that. In generateDiagram, initialize the pixel pointer as follows:
```
        p = ((winHeight-1-row)*winWidth + col) * 3;
      
```
Q: For the bonus question, the blackbody formula you gave is wrong!
A: True. Sorry about that. There's a lambda in the denominator of the exponentiated term. It should read:
- P(λ) = (2πc²h/λ⁵)/(e^(hc)/(λkT)-1)
Q: What are all those constants?
- h = 6.626x10^-34 Joule second (Planck's constant)
- k = 1.381x10^-23 Joule Kelvin (Boltzmann's constant)
- c = 2.998x10⁸ meter/second (speed of light)
- λ = wavelength (in the table, they're given in nanometers (10^-9 meter)).

Q: How do I invert a matrix?
A: On the CDF system, you can use matlab:

      eddie% matlab
      ...
      >> a=[1 2;3 4]
      a =
           1     2
           3     4

      >> inv(a)
      ans =
         -2.0000    1.0000
          1.5000   -0.5000

      >> quit

Question 5

Q: There's not enough information to describe the camera!
A: True. Sorry about that. I've corrected the skeleton code. Please have another look at the question 5 page.
Q: Where's the eye point?
A: Hint: everything's in camera coordinates.
Q: How do I intersect a ray and a cylinder?
A: Get the parametric equation for a ray, and plug it into the equation for a cylinder's sides. The latter is just the implicit equation for a 2D circle (just for x and z), with the third coordinate (y) unspecified (why?). To intersect the ray with the two caps of the cylinder, use the ray-plane intersection method described in class.
NOTE this answer was corrected. The previous answer (a circle on x and y, with z unspecified), was for a cylinder parallel to the Z axis.
Q: All the test rays give HIT the cylinder!
A: A new set of test inputs is available. See the question web page.
Q: I still don't get it. How do I intersect a ray and cylinder?
A: The equations for a ray and an infinite cylinder are, respectively,
1. Q(x,y,z) = P(x,y,z) + t * V(x,y,z)
2. (x - x_c)² + (z-z_c)² = r²
Substitute a) into b), and you'll get a quadratic equation in t. The value of t tells you where you hit the cylinder (if the equation has real roots), when you plug t back into a).
Note that the hit point may be above or below the cap. You have take care of that case.
To intersect the ray with the top and bottom caps of the cylinder, use the ray-plane method given in class and in the notes, and discard points that are:
- on the plane of the top or bottom cap, but
- outside the circle bounding the cap.

Question 6

Q:It's REALLY hard to see the robot clearly in the handout!
A: True. There's now a better picture on the question web page.
Q:Some faces look black.
A: The ambient (overall) light is too low. In initLights, change the first line to:
```
          GLfloat ambient[] = { 1.0, 1.0, 1.0, 1.0 };
      
```
Q:How do I draw a cylinder?
A: Draw an approximate cylinder. Find some points on a circle:
x = r cos(θ)
y = r sin(θ)
θ = 0, Δθ, 2Δθ, 3Δθ, ... 2π
This should give you the top and bottom of the cylinder, for different values of Z. You can also join top and bottom points with rectangles to form the sides of the rectangle.
Note: keep in mind that sin() and cos() expect radians, not degrees.
Q: Why are there two "head turn" controls?
A: Sorry. One of them should be "head tilt" (see the updated ui.cpp code).
Q: How come rightShoulderAngle moves the robot's LEFT arm?
A: Hmm... You noticed. Treat left and right as the camera sees them, at its initial position. It's not the robot's way of seeing things, but it's clearer for the user.

Alejo Hausner, Dept of Computer Science, University of Toronto

Last modified: Tue Oct 23 16:52:50 EDT 2001