1. Projection, drawing primitives: [8 marks] An ellipse defined by semi-major and semi-minor axes vectors A, B is to be drawn using orthographic projection with view direction V :

a.      [2 marks] Explain intuitively why the projected curve must also be an ellipse or a straight line.

Possible answers (you will get partial credit for a reasonable attempt at an answer):

The view space transform is a change of basis rotate and translate transform that leaves an equation in x,y,z to have the same or lesser degree. Orthographic projection sets z=0. A quadratic thus projects to a quadratic or linear equation in x,y. If quadratic the ellipse being closed projects to a closed quadratic that is an ellipse. (2 marks).

When (AxB).V=0 or the plane of the ellipse is perpendicular to the view plane the curve projects to a line. (1 mark),

Ellipse eq. x=a*cost, y=bsint z=0. Any view space rotation can be seen as a sequence of 3 rotation around x,y,z axes. Ignore rotation of ellipse around z axes which is the projection plane (ellipse rotated in 2D is an ellipse). After rotation around y by angle m, eq. is x’=x, y’=ycosm, z’=xsinm. After rotation around y by angle n eq. is x’’ =xcosnxsinm*sinn, y’’=ycosm , z’’=xsinn+xsinm*cosn. Throw away z’’ in the equation. We are left with the projected curve x’’= a(cosn-sinm*sinn)cost, y’’= b(cosm)sint which is still the parametric eq. of an ellipse. (2 marks).

b.     [3 marks] Under what conditions can the projected curve be a circle and what is the radius of that circle as a function of A, B, V.

A’=sqrt(||A||2-(A.V) 2). B’=sqrt(||B||2-(B.V) 2). The curve is a circle when A’=B’=radius of the circle.

(1 mark for getting the A’, B’ values, 1.5 marks for saying that when the new semi-maj, semi-minor axes are equal we get a circle, 0.5 marks for pointing out that the radius is A’ or B’.

c.     [3 marks] Write pseudo code for a loop that plots the ellipse defined as (x/a)2 + (y/b)2  =1 as a sequence of n line segments.

ang=2pi(n-1)/n;px=a*cos(ang);py=b*sin(ang);

For (i=0;i<n;i++) {ang=2pi*i/n; x=a*cos(ang);y=b*sin(ang); line(x,y,px,py);x=px;y=py;}

(1 mark for using the parametric eq. of ellipse, 1 mark for loop str., 1 mark for getting the angle incr. right).  Same marking scheme if  you use the explicit equation for 4 quadrants.

d.     EXTRA CREDIT [2 marks] What property of the curve should determine where to sample points on the ellipse so that the n line segments that connect them best approximate the ellipse. What part of your pseudo-code in part c needs to change to reflect this.

The arc length of the ellipse between sample points should be roughly equal for better approximation. Curvature of the ellipse can be used as a measure of this. (1.5 marks). The angle increment is now not fixed but depends on the arc length between samples (0.5 marks).

2.     Illumination: [8 marks] Motion capture systems use a number of cameras, and spherical reflective markers to track the position of the markers in time. The figure below shows how it works. A marker reflection needs to be picked up by 2 or more cameras. Given the position of the camera and reflected spot in the camera image, the viewing ray for the camera is determined. Now the 3D position of the marker is the intersection point of the two viewing rays.

 P

 Y

 Camera1

 X

 Camera2

 Z

a.      [2 marks] Why are the markers spherical in shape and reflective?

We want the markers to show up in the camera images as sharp spots irrespective of the relative angles between the lights, marker and cameras. They are spherical in shape so that they reflect well in all directions and they are reflective so that they show up  as sharp highlights in the camera images.

(1 mark for spherical, 1 mark for reflective).

b.     [4 marks] Two cameras positioned at <0,0,0> and <1,0,7> pick up a marker and their view rays are calculated to be  directed along <1,1,2> and <2,3,-1> respectively. What is the position of the 3D surface point P captured by this setup?

<1,1,2>s = <1,0,7> + <2,3,-1>t.  s=1+2t, s=3t, 2s=7-t =>  t=1, s=3, P=<3,3,6> .

(1 mark for writing the ray equation. 1 mark for equating them. 1 mark for solving s or t. 1 mark for  getting P right).

c.      [2 marks] Typical motion capture systems have many more than 2 cameras spread around in the scene to track the motion of markers. Why?

When objects move, the markers are not always visible to all cameras. If we have more than two cameras the chance that a marker is picked up by at least 2 of the many cameras becomes higher with more cameras.

d. EXTRA CREDIT [2 marks] If view vectors 1..Vn for a marker are calculated for n cameras in the scene how will you calculate the 3D marker position?

Calculate pairwise intersections for all Vi and Vj, Set the point to be the average of these intersection points. …or calculate shortest distance between view vectors and take the midpoint of the shortest distance line as the intersection point if they don’t intersect. Weight average the intersection points based on the shortest distance between the view vectors.

3.     Visibility: [10 marks] Consider a set of polygons, represented in 2D cross-section as line segments. Arrows represent front-facing normals.

a.      [4 marks] Draw the BSP tree obtained by inserting the polygons in alphabetical order. Left subtrees correspond to the front side, and right subtrees to the back. If any polygons are split, label the fragments on the diagram and the tree.

(partial credit for getting parts of the tree right).

b.     [4 marks] Consider a drawing primitive, represented as the hatched region in the figure below, of a cylinder (defined by center C, axis A, radius r and height h with no caps) and  a plane with origin O and normal N such that N.A=0. Give the equations and the test by which one may determine if a point P is on one side of the primitive or the other, similar to the planar equation test for polygons in a BSP.

For the point to be on the front side of hatched region it must be in front of the plane and outside the cylinder. i.e.

plane:  (P-O).N>0

cylinder: let d=(P-C).A. d’= sqrt((P-C).(P-C)-d2).    d’>r

Front/back side test is:  (P-O).N>0 and d’>r.

(2 marks for saying front of plane and outside cyl., 1 mark for plane eq. test., 1 mark for cyl. outside test).

c.     [2 marks] Describe any changes needed to the BSP algorithm to adapt it to be used as visibility ordering algorithm for the partial-cylinder primitive above.

Replace the planar equation test with the test in part b. When primitives intersect use a clipping algorithm that clips to both the plane and the cylinder so that the resulting pieces are all on one side or another.

(1 mark each for pointing out the two changes).

4.     True or False with a reason: [4 marks]

a.      [1 mark] The difference between Gouraud and Phong shading is likely to be more noticeable with diffuse objects than with specular objects.

False. Specular objects tend to be more sensitive to changes in normal vector on the surface than diffuse objects.

b.     [1 mark] The difference between Gouraud and Phong shading is likely to be more noticeable with large polygons than with small polygons (size of the polygon here is its projected size in the 2D image).

True. It is more likely to miss specular highlights on large polygons with Gouraud shading than on small polygons.

c.     [1 mark] Real cameras use a lens to compensate for the fact that apertures larger than a pin-hole make the image blurry.

True. Lenses cause rays through the aperture to focus towards a smaller region on the image plane making the image sharper.

d.     [1 mark] The result of multiple rotations to a point is order dependent but combinations of scaling and translations can be applied in any order with the same result.

False. Take a point at the origin and scale it 2x and translate it to (5,5).

If translated and then scaled the resulting point is (10,10).