Nuno Vasconcelos UCSD

Transcription

1 Radiomety Nuno Vasconcelos UCSD

2 Image fomation two components: geomety and adiomety geomety: pinhole camea point x,y,z in 3D scene pojected into image pixel of coodinates x, y y accoding to the pespective pojection equation: x' y' = f X, Y, Z What is fx,y,z? 2

3 Image fomation two components: geomety and adiomety geomety: pinhole camea point x,y,z in 3D scene pojected into image pixel of coodinates x, y y accoding to the pespective pojection equation: x ' y ' x = f y z z 3

4 Radiomety simplifying assumptions objects: Lambetian sufaces: eflect light equally in all diections eflections detemined by albedo: atio of eflected/incident light light souces: point infinity... by the time they hit the object all ays ae paallel single diection s of light fo the whole scene 4

5 Radiomety adiomety equation: the powe at pixel i is a function of object popeties which? light souce which inteaction between the two? i =? E θ θ Light diection o V i image bightness 5

6 Radiomety adiomety equation: i = E ρ 0 cos θ the powe at pixel i is the poduct of souce powe E, albedo of the object at eflection point, and angle between souce diection and object nomal E θ θ Light diection o V i image bightness 6

7 Radiomety a ule fo any numbe of bounces V o θ 1 o V 2 V 1 θ 2 1 n = E ρ i = 0 i = 1 n i n + 1 cos θ i 7

8 Lambetian sufaces note that on n n +1 = E ρ i = 0 i = 1 n i cos θ i unless all cosines ae close to 1 thei poduct goes to zeo quickly e.g. see decay of cos n θ with n this means that only light that aives fontally to all the bounces gets popagated vey fa such an alignment is vey unlikely we don t eally have to woy about many bounces the pocess becomes tactable 8

9 In vecto fom note that if n is the suface nomal and s the light diection the two vectos have unit nom then cos θ = n.sand i = Eρ n. s 0 0 θ E Light diection θ s n V o i θ 2 image bightness 9

10 Lambetian sufaces note that light diection s is constant but the suface nomal n and the albedo ρ ae functions on the object suface i = Eρ n. s 0 0 θ E Light diection θ s n suface nomal map o 10

11 Nomals the amount of eflected light at depends on the suface nomal n how do I compute this? fist we note that a suface is a function of thee vaiables fx,y,z = 0 suface nomal map e.g., the sphee of adius and cente x 0,y 0,z 0 is descibed by x-x y-y 0 2 +z-z = 0 11

12 Nomals the key is to note that the nomal is othogonal to the suface the suface is the set of points whee the function fx,y,z is constant to walk along the suface I walk along the set of points whee fx,y,z stays constant tg plane infinitesimally, this is the plane fx,y,z=0 tangent to the suface at to walk along the nomal, I have to walk in the diection along which fx,y,z gows most quickly the key is then to find this diection of lagest gowth 12

13 Nomals ecall that a scala function gows the most when its deivative is lagest the genealization to a multivaiate function is the gadient f f f =,, x y f z T f no incease f inceases the most in the diection of its gadient hence, the nomal to suface fx,y,z = 0 at point is the gadient of fx,y,z evaluated at 13

14 Nomals this is an impotant esult that we will use many times in the couse e.g. fo the sphee of cente 0,0,0 f 0,0,0 tg plane x, y z T f = 2, and the nomal at point is simply 2 e.g. n100=200 n1,0,0 2,0,0 fx,y,z=0 2 14

15 Nomals note that this explains the shading of a sphee we have seen a light souce of diection s geneates an image s I 0 = E ρ n 0, s n, s fo the sphee this is just I 0 0, s the dot poduct of s with 0 itself 0 E 15

16 Nomals what about moe complex objects? we appoximate by a tiangle mesh sample a numbe of points on the suface thee ae compute gaphics algoithms fo doing this appoximate suface by the tiangles that connect those points the moe tiangles you use, the bette the appoximation 16

17 Nomals how do I compute the nomal to a tiangle? use the fact that a tiangle is a patch of a plane how do I compute the nomal to a plane? know what is the plane associated with my tiangle? Q1: a plane is a function of the fom ax + by + cz + d = 0 this is the plane of paametes a,b,c,d in this case f x, y, z = ax + by + cz + d 17

18 Nomals and f x, y, z = a, b, c note that this tells us vaious things: all points in the plane have the same nomal this is called the nomal to the plane n = a, b, c T T n=a,b,c ax+by+cz+d=0 the paamete d only has to do with the distance to the oigin note that if 0,0,0 0 0 is on the plane, then d = 0, othewise d = 0 the nomal and d fully specify the plane plane { n = d } =, 18

19 Nomals Q2: how do I know the plane associated with my tiangle note that tiangle = 3 points, 0, 1, 2 thee points define a plane we just have to solve the system of equations 0 1 n, n, 0 1 n, 2 = = d d = d 2 note that multiplying n and d by the same numbe does not change anything. need to enfoce exta constaint that n = 1 19

20 example fom hw thee points n, 1,1,2 0,1,1 1 = d n, 0,0,1 n, 1,0,2 = d = d 102 1,0,2 b = d c b = 0 c = d c = d a = d 2c a = d -1,1,2 0,1,1 k 0,0,1 i j leads to the plane equation dx dz = d z = x

21 lanes fo a plane ou adiomety equation i = E ρ n. s 0 0 simplifies to i = E n, s ρ 0 ρ 0 this means that constant shading is unifom the vaiations that we see ae vaiations in albedo 21

22 Meshes fo a mesh this holds fo each tiangle i = E n, s ρ 0 but <n,s> changes fom tiangle to tiangle diffeent nomals hence, we can still have complex shading effects as the numbe of tiangles inceases 22

23 Moe complexity in summay, we can do a lot with ou simplistic model Lambetian sufaces point infinity i = E ρ 0 cos θ θ E o Light diection θ V i image bightness tuns out that we can wok with much moe complex light souces 23

24 Multiple light souces the key insight that the equation is linea on s E if we have n infinity, we can just assume that s n E i, 0 0 ρ = if we have n infinity, we can just assume that s = s s n = = = k k k k s n E s n E, ρ ρ esulting image is sum of the images due to each souce i k k s k n E, 0 0 ρ 24 esulting image is sum of the images due to each souce

25 Demo we stat fom this image 25

26 Demo adding this + 26

27 Demo we get the image lit by two light souces 27

28 Demo adding this + 28

29 Demo we get the image lit by the thee light souces 29

30 Demo adding this + 30

31 Demo we get the image lit by the fou light souces 31

32 Demos othe possible pattens 32

33 Demos Fom these I can ceate a movie 33

34 Demo and this happens in the eal wold too note combo of geomety and adio- mety 34

35 Moe complexity and what about fluoescent lights, etc.? is this still a point souce? no, but an infinitesimal patch is da the patch da is a point souce we wok with the powe density, instead of powe e.g. da, centeed at x, emits density ExdA in its nomal diection sx 35

36 Moe complexity the contibution of patch centeed at x to the powe that hits object point 0 is x = E x ρ n, s x 0 0 da x 2 sx 2 to compute the oveall powe, due to all patches, we simply integate = E x ρ n, s x A 0 0 da x 1 sx1 and by lineaity of the dot poduct = ρ n, E x s x da 0 0 A 36

37 Moe complexity we thus have x 2 sx 2 = ρ n, E x s x da 0 0 A x 1 sx1 note that the integal does not depend on 0 this is the same as computing E = E x, y, z s x, y, x dxdydz A which is a 3D vecto and assuming a point souce of magnitude E = E and unit diection s = E/ E 37

38 In summay geomety x x ' = f y z y ' z adiomety i = E ρ 0 cos θ θ E o θ Light diection V i image bightness 38

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