p =(x,y,d) y (0,0) d z Projection plane, z=d

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1 Projections ffl Mapping from d dimensional space to d 1 dimensional subspace ffl Range of an projection P : R! R called a projection plane ffl P maps lines to points ffl The image of an point p under P is the intersection of a projection line through p with the projection plane. The Universit of Teas at Austin 1

2 Parallel Projections ffl All projection lines are parallel. ffl An orthographic projection has projection lines orthogonal to projection plane. ffl Otherwise a parallel projection is an oblique projection ffl Particularl interesting oblique projections are the cabinet projection and the cavalier projection. The Universit of Teas at Austin

3 Perspective Projection ffl All projection lines pass through the center of projection (eepoint). ffl Therefore also called central projection ffl This is not affine, but rather a projective transformation. Projective Transformation ffl Does not preserve angles, distances, ratios of distances or affine combinations. ffl Cross ratios are preserved. ffl Incidence relationships are generall preserved. ffl Straight lines are mapped to straight lines. The Universit of Teas at Austin

4 Perspective Transform in Ee Coordinates ffl Given a point p, find its projection P(p) ffl Convenient to do this in ee coordinates, with center of projection at origin and z = d projection plane ffl Note that ee coordinates are left-handed p =(,,d) (0,0) d z z Projection plane, z=d p=(,,z) ffl Due to similar triangles P(p) = (d=z; d=z; d) ffl For an other point q = (k; k; kz);k = 0 on same projection line P(q) = (d=z; d=z; d) ffl If we have surfaces, we need to know which ones occlude others from the ee position ffl This projection loses all z information, so we cannot do occlusion testing after projection The Universit of Teas at Austin

5 Homogeneous Coordinates ffl Homogeneous coordinates represent n-space as a subspace of n + 1 space ffl For instance, homogeneous -space embeds ordinar -space as the w = 1 hperplane ffl Thus, we can obtain the -d image of an homogeneous point (w; w; wz; w); w = 0 as (; ; z; 1) = (w=w; w=w; wz=w; w=w), that is, b dividing all coordinates b w. ffl Lines in homogeneous space which intersect the w = 1 hperplane project to -space points. ffl Notice that this is just a perspective projection from -d homogeneous space to -space, instead of dividing b z, we are dividing b w. The Universit of Teas at Austin 5

6 The OpenGL Perspective Matri ffl The visible volume in world space is known as the viewing pramid or frustum. ffl Specif with the call glfrustum(l; r;b;t;n;f) ffl In OpenGL, the window is in the near plane ffl l and r are u-coordinates of left and right window boundaries in the near plane ffl b and t are v-coordinates of bottom and top window boundaries in the near plane ffl n and f are positive distances from the ee along the viewing ra to the near and far planes ffl Maps the left and right clipping planes to = 1 and = 1 ffl Maps the bottom and top clipping planes to = 1 and = 1 ffl Maps the near and far clipping planes to z = 1 and z = 1 (1, 1, 1) n (-1, -1, -1) f The Universit of Teas at Austin

7 Shearing the Window to the z ais ffl After appling the modelview matri, we are looking down the z ais. ffl We need to move the ra from the origin through the window center onto the z ais. ffl Rotation won't do since the window wouldn't be orthogonal to the z ais. ffl Translation won't do since we need to keep the ee at the origin. ffl We need differential translation as a function of z, i.e. shear. ffl When z = n, ffi should be r+l n so we get 0 = + r + l n z 0 = + t + b n z z 0 = z and ffi should be t+b n, The Universit of Teas at Austin

8 viewing frustum t l r b (t+b)/ f n (r+l)/ z [-(r-l)/,(t-b)/,-n] f n viewing frustum [(r-l)/,-(t-b)/,-n] 0 0 z = r+l 1 0 n 0 t+b 0 1 n z 1 5 The Universit of Teas at Austin 8

9 Adjusting the Clipping Boundaries ffl For ease of clipping, we want the oblique clipping planes to have equations = ±z and = ±z. ffl This will make the window square, with boundaries l = b = n and r = t = n. ffl This requires a scale to make the window this size. [-(r-l)/,(t-b)/,-n] [-n,n,-n] z z [(r-l)/,-(t-b)/,-n] [n,-n,-n] Thus the mapping is The Universit of Teas at Austin 9

10 0 = n r l 0 = n r l z 0 = z or in matri form: 0 0 z = n r l n 0 t b z 1 5 The Universit of Teas at Austin 10

11 Field of View Frustum Scaling ffl After the frustum is centered on the z ais: window near plane (t-b)/ θ z n viewing frustum n ffl Note that t b = cot ffl This gives the mapping 00 = 0 cot ffl Since the window need not be square, we can define the mapping using the aspect ratio aspect = = (r l) (t b) ffl Then maps as 00 = 0 cot aspect The Universit of Teas at Austin 11

12 ffl This gives us the alternative scaling formulation: 0 0 z = cot aspect cot ffl This is used b gluperspective( ; aspect; n; f) 5 z 1 5 The Universit of Teas at Austin 1

13 Perspective Mapping ffl Recall that we want to map the frustum to a cube centered at the origin. (1, 1, 1) n (-1, -1, -1) ffl OpenGL looks down z rather than z. f ffl Note that when ou specif n and f, the are given as positive distances down z = 1. ffl First we map the bounding planes = ±z and = ±z to the planes = ±1 and = ±1. ffl This can be done b mapping to z and to z. ffl If we set z 0 = 1, this is equivalent to projecting onto the z = 1 plane. ffl However, we want to derive a map for z that preserves lines and depth information. ffl To map to z and to z - ffl First use a matri to map to homogeneous coordinates, then project back to space b dividing (normalizing). The Universit of Teas at Austin 1

14 a c 0 0 b d 5 z 1 5 = az + c bz + d bz+d bz+d az+c bz+d ffl Now we solve for a; b; c and d such that z [n; f] maps to z 0 [ 1; 1]. ffl To map to z, ffl Thus bz + d = z az + c bz + d ) d = 0 and b = 1 becomes az + c z The Universit of Teas at Austin 1

15 ffl Since the near plane is at z = n and the far plane at z = f, our constraints on the near and far clipping planes (e.g., that the map to -1 and 1) give us 1 = 1 = an + c n af n + an f ) c = n + an ) (f + n) = a(n f) ) a = f + n n f ) a = (f + n) f n ) c = n + = (f + n)n f n n(f n) n(f + n) = fn f n f n This gives us (f+n) f n fn f n z 1 5 = z(f +n) f n f n z 5 The Universit of Teas at Austin 15

16 ffl After normalizing we get» z ; z ; z(f + n) fn z(f n) ; 1 T ffl If we multipl this matri in with the geometric transforms, the onl additional work is the divide. ffl After normalization we are in left-handed -dimensional Normalized Device Coordinates The Universit of Teas at Austin 1

17 Complete OpenGL Perspective Matri ffl Combining the three steps given above, the complete OpenGL perspective matri is n r l 0 n 0 t b 0 0 r+l r l 0 t+b t b 0 (f+n) f n fn f n 5 = n r l n 0 (f+n) fn t b f n f n r+l 1 0 n 0 t+b 0 1 n ffl Using gluperspective the matri becomes cot( =) aspect cot( =) (f+n) f n fn f n The Universit of Teas at Austin 1

18 Wh Map Z ffl D! D projections map all z to same value. ffl Need z to determine occlusion, so a D to D projective transformation doesn't work. ffl Further, we want D lines to map to D lines (this is useful in hidden surface removal). ffl The mapping (; ; z; 1)! (n=z; n=z; n; 1) maps lines to lines, but loses all depth information. ffl We could use (; ; z; 1)! (n=z; n=z; z; 1) Thus, if we map the endpoints of a line segment, these end points will have the same relative depths after this mapping. BUT: It fails to map lines to lines ffl The map (; ; z; 1)! n z ; n z ; zf + zn fn z(f n) does map lines to lines, and it preserves depth information. ; 1 The Universit of Teas at Austin 18

19 Mapping Z ffl It's clear how and map. How about z? z! zf + zn fn z(f n) = P (z) ffl We know P (f) = 1 and (P (n) = 1. What maps to 0? ) P (z) = 0 zf + zn fn z(f n) = 0 ) z = fn f + n Note that f + f > fn=(f + n) > fn + n so f > fn f + n > n The Universit of Teas at Austin 19

20 ffl What happens as map z to 0 or to infinit? lim P (z) = fn z!0 + z(f n) = 1 lim z!0 P (z) = fn z(f n) = +1 z(f + n) lim P (z) = z!+1 z(f n) = f + n f n z(f + n) lim P (z) = z! 1 z(f n) = f + n f n The Universit of Teas at Austin 0

21 _ 8 8 _ 0 n fn f+n f f+n f-n The Universit of Teas at Austin 1

22 ffl What happens if we var f and n? z(f + n) fn lim P (z) = f!n z(f n) = (zn n ) z 0 which is not surprising, since we're tring to map a single point to a line segment. zf fn lim P (z) = f!1 zf = z n z ffl But note that this means we are mapping an infinite region to [0,1] and we will effectivel get a far plane due to floating point precision, zf lim P (z) = n!0 zf = 1 i.e., the entire map becomes constant (again, we are mapping a point to an interval). The Universit of Teas at Austin

23 ffl Consider what happens as f and n move awa from each other. We are interested in the size of the regions [n; fn=(f +n)] and [fn=(f + n);f]. When f is large compared to n, we have fn f + n : = n So and f fn f + n fn f + n n : = n : = f n But both intervals are mapped to a regions of size 1. Thus, as we move the clipping planes awa from one another, the far interval is compressed more than the near one. With floating point arithmetic, this means we'll lose precision. In the etreme case, think about what happens as we move f to infinit: we compress an infinite region to an finite one. Therefore, we tr to place our clipping planes as close to one another as we can. The Universit of Teas at Austin

24 Clipping in Homogeneous Space Projection: linear transformations then normalize ffl Linear transformation nr ns f+n f n fn f n z 1 5 = μ μ μz μw 5 ffl Normalization μ μ μz μw 5 = μ= μw μ= μw μz= μw 1 5 = X Y Z 1 5 The Universit of Teas at Austin

25 Region Mapping 1 10 _ n f The Universit of Teas at Austin 5

26 Clipping not good after normalization: ffl Ambiguit after normalization 1» μ; μ; μz μw» +1 Numerator can be positive or negative Denominator can be positive or negative ffl Normalization epended on points that are subsequentl clipped. Clipping in homogeneous coordinates: ffl Compare unnormalized coordinate against μw j μwj» μ; μ; μz» +j μwj The Universit of Teas at Austin