1. Consider the canonical viewing space with COP = (0,0,-1), and the usual clipping planes x = -(z+1), x = (z+1), y = -(z+1), y = (z+1), z = 0, z = 1 (assuming that the front and back clipping planes are normalised to 0 and 1). The view plane is the XY plane.
(a) We have a line segment from p0 = (1,1,-4) to p1 = (2,4,4). What is the projection of this line onto the view plane?
The projection in this case is x' = x/(z+1) and y' = y/(z+1). Therefore the projection of the two end-points is: (-1/3, -1/3) and (2/5, 4/5). However, since part of the line segment is behind the COP, the projected line is an external line segment: (-1/3,-Y) to (-1/3,-1/3) and (2/5,4/5) to (2/5, Y) where Y is infinity.
(b) Find the intersection between this line and the z = 0 plane.
The line equation is (1 + t, 1 + 3t, -4 + 8t) t = 0 to 1. Where this meets the z=0 plane, -4 + 8t = 0, so t = 1/2. So the intersection is (3/2, 5/2, 0).
(c) Suppose there is a third point p2 = (3,1,6). Find the polygon formed by clipping the polygon p0,p1,p2 against the z = 0 plane.
p0 to p1 has p0 outside and p1 inside => outputs intersection (3/2,5/2,0) and p1 p1 to p2 has p1 inside and p2 inside => outputs p2 p2 to p0 has p2 inside and p0 outside => outputs ints(p2,p0). The line from p2 to p0 is (3-2t, 1, 6-10t). At the intersection: 6-10t = 0, so t = 3/5. Therefore, the intersection is (9/5,1,0). The clipped polygon is (3/2,5/2,0), p1, p2, (9/5,1,0).
2. In WC there is a line from (0,0,0) to (1,1,1). Find camera setups to achieve each of the following results:
(a) The display shows is a single point in the centre.
There are many possible answers. VRP = (0,0,0) VPN = (1,1,1) VUV = (0,0,1) VPWindow = [-a a -a a] (a > 0) COP = (0,0,-b) (b > 0). FCP = 0, BCP = c > 0.
(b) The display shows a horizontal line all the way across and half way between the bottom and top of the display.
VRP = (1/2, 1/2, 1/2) VPN = (-1,1,0) VUV = (0,-1,1) VPWindow = [-a a -a a] (a > 0) COP = (0,0,-b) (b > 0). FCP = 0, BCP = c > 0.(c) The display shows a vertical line from top to bottom and half way across.
VRP = (1/2, 1/2, 1/2) VPN = (-1,1,0) VUV = (1,1,1) VPWindow = [-a a -a a] (a > 0) COP = (0,0,-b) (b > 0). FCP = 0, BCP = c > 0.(d) The line is not visible at all.
VRP = (0,0,0) VPN = (0,-1,0) VRP = (0,0,1) FCP = a > 0 etc.
3. Under what conditions will clipping in Projection Space lead to the same result as clipping in canonical viewing space?
Provided that when transformation to canonical viewing space has occurred that there is nothing behind the plane through the COP that is parallel to the view plane.
4. (a) The point (1,1) is on the view plane in PS, and belongs to the polygon that has plane equation x - y + z = 0.5. What is the z-depth corresponding to this point on that polygon?
z = 4 - x + y = 0.5(b) Suppose that there is another polygon, with plane equation 2x + y - z = 2.6, which of the two polygons is visible at the point (1,1) on the view plane?For the second one, z = 2x + y - 2.6 = 0.4. Since the z-depth of the second one is smaller, the second would be visible at this point.