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Written Assignment 4

Due Wednesday, November 1, 2006 7:15pm (we will collect them at the beginning of the exam).

Question 1: Arc length parameterization.

Write a parametric function for the curve consisting of three line segments connecting the four points (0,0), (1,0), (1,3), (3,3). Have the paremeterization be unit (e.g. from 0 to 1) and arc-length parameterized.

(remember: arc-length parameterization means that the "speed" of the parameterization is constant, not necessarily unit)

Question 2:

Consider two cubic pieces A and B that are Hermite segments. A's points are A1, A2, A3, A4 and B's points are B1, B2, B3, B4.

You are given the positions of A's points. What restrictions are there on B's points are required to achieve:

As originally stated, this question is ambigious - since the ordering of the Hermite control points isn't given. In fact, the example suggests an ordering that is different than either the book or in class. As long as you state what your assumption is, it would be OK, but the best one to use (since it is consistent with the example) is:

  • P1 (e.g. A1 and B1) is the position of the beginning
  • P2 (e.g. A2 and B2) is the derivative at the beginning
  • P3 (e.g. A3 and B3) is the derivative at the end
  • P4 (e.g. A4 and B4) is the position at the end

Note: P1 and P4 must be these definitions otherwise the example doesn't work. (Actually, 1 could be the end and 4 could be the beginning, but that would be kindof silly).

(example): C(0) continuity?

B1 = A4

2A: C(1) continuity?

B1 = A4
B2 = A3

2B: G(1) continuity?

The simple answer is:

B1 = A4
B2 = k A3 (for some arbitrary scalar k)

However, this doesn't get the case where the derivative vanishes correctly. So, you should add the condition that

A3 != 0

If A3 = 0, its actually hard to know exactly what it takes to make G(1) continuity.

2C: C(2) continuity? (warning - this one is tricky)
actually, this one is too tricky - don't bother trying

Question 3:

In this question, we will be making pictures of a cube. The cube is of unit size (that is, all of its edges have length 1), and is placed such that one of its corners is at the origin, and the edges follow the positive axes. There is a letter painted on each side of the cube. The letter “F” is painted on the front, “B” on the back, “L” on the left, “R” on the right, “T” on the top, and “U” on the underside (bottom). The back of the object is the xy plane (z=0), the front of the object is the xy plane with z=1. Left and right are defined as if you were looking at the object from the front. Several views are shown below.

Sketch the view of the cube as seen by the cameras specified, being sure to label each face with the appropriate letter in roughly the right orientation. VRP is the view reference point (the center of the image plane) and Look At is a point along the view plane normal. Assume some reasonable field of view big enough to fit the entire cube in the view, and that the viewport is square. You need not get the view exactly (in fact, you can’t since we didn’t tell you what the field of view is).

(note: if you think of the VRP as the "lookat" or "eye" point, you will get the same answer for this question)

3A) VRP 6,0,0; Look at 0,0,0; VUP 0,1,0.
3B) VRP -6,1,1; Look at: 1,1,1; VUP 0,0,1.
3C) VRP -6,6,6; Look at 0,0,0; VUP 0,0,1.

Note: the axes (that show where the center of the image plane are) are significant!)

The original answer for 3C was wrong. The correct answer is explained below.

Also, since the Z axis needs to be the up vector, the whole thing needs to be rotated on its side (so the L is lying on its "back"). The whole cube would be on the left side of the screen.

To see this, you need to imagine that the camera (in the cube picture) is where the F diagram is (above and to the right).

Getting 3C right:

  1. the lookat point is the origin, so this should go to the center of the screen
  2. in the picture above (part of the assignment) the camera would be above and to the right of the scene (where the picture of the F face is). So the faces we will be able to see are the L, the F, and the T
  3. The L face will have its lower left corner (which is 0,0,0) at the center of the screen. it will have its bottom edge (the Z axis) pointing up in the image.
  4. The T and F faces can be added since we know how they relate to the L. (e.g. the bottom of the T connects to the top of the L)

Question 4: BSP Tree

Consider the object shown which has 8 faces (all faces are rectangles). We will represent the surface object using a BSP tree where each node in the tree can be a quadralateral (not just a triangle).

4A) Build the BSP tree for the object shown, where faces are used as splitting planes in face number order. If a face is split, label the new faces with sub-letters (if face 1 is split, the pieces would be 1a and 1b). Assume that the positive side of the splitting planes is the side that is “outside” the object.

4B) Build the BSP tree for the object shown, this time choosing face 5 for the root of the tree.

4C) If the camera is at (10,10,10) (where the size of the object is less than 10) looking at the origin (so it sees the whole object), what order would the faces be drawn in if you used the tree of part A?

The viewer is on the + side of 2,3,4,5,6
The viewer is on the - side of 1,7,8

On each side, we render the triangles away from the camera first, then the triangle, then the stuff on the camera side.

  • So starting at node 1, we're on the - side. There is nothing on the + side, so we render 1, then traverse to the - side.
  • At node 2, we're on the + side, so we do the negative side first (recurse to node 3)
  • At node 3, we're on the + side, so we do the negative side first (recurse to node 4)
  • At node 4, we're on the + side, so we do the negative side first
  • At node 7b, we're on the - side, so we do the + side first (nothing there), render the node 7b, then do the negative side
  • Node 8b is a leaf, so we just render it
  • Recursion back to node 4, we've done the away side, so we render the face 4 and there's no positive side
  • Back to node three, we've done the away side, so render the face 3 and then recurse through the plus side
  • At 5, we do the - side first
  • At 6, we do the - side first
  • At 7a, we do the + side first, then the node 7a, then the - side
  • 8a is a leaf
  • back to 6
  • back to 5
  • back to 2

So the order we drew things in (back to front) is: 1,7b,8b,4,3,7a,8a,6,5,2

4D) If the camera is at (10,10,10), what order would the faces be drawn in if you used the tree of part B?

The viewer is on the + side of 2,3,4,5,6
The viewer is on the - side of 1,7,8

On each side, we render the triangles away from the camera first, then the triangle, then the stuff on the camera side.

Of course, the order depends on the tree that you built. Using the one I built...

  • 5-
    • 1B + (nothing), 1B,1B-
    • 2B -
      • 6 (do minus first)
        • 7 (do plus first = nothing), 7, then minus side
        • 8B
        • 6 (plus side has nothing)
    • 2B, 2B+ (nothing)
  • 5
  • 5+
    • 1A + (nothing), 1A, 1A-
    • 2A -
    • 4-
      • 8A+, 8A, 8A-
        • 3
    • 4
    • 4+
    • 2A, 2A+ nothing

So... 1B, 7, 8B, 6, 2B, 5, 1A, 8A, 3, 4, 2A (is back to front)

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Page last modified on October 31, 2006, at 09:03 AM