Homework #8: Lighting Solution

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For this homework, suppose that a point on the surface of an object lit by a single light: psurface normal n = [1; 1;                                                                1]T = 3 [0:58; 0:58; 0:58]T p

lighting coming from the direction l = [2; 3;    1]T =  14    [0:53; 0:80;   0:27]

 

direct light strength (coloring) s = 0:9 [1:0; 1:0; 0:8]T (this is slightly yellowish, like an incandescent bulb)

 

ambient light strength (coloring) samb = 0:1 [1:0; 1:0; 0:8]T (like the direct lighting but weaker)

 

viewing direction v = [0; 0;  1]

 

surface diffuse reflectance coefficients mdiff = [0:1; 0:2; 0:5]T (a slightly greenish blue diffuse coloring)

 

surface specular reflectance coefficients mspec = [0:5; 0:5; 0:5]T (reflects half the incoming light, no change of lighting color)

surface specular glossy exponent mgls = 4

 

surface ambient reflectance equal to the diffuse reflectance mamb = mdiff

 

 

 

  1. What is the RGB triplet for the direct lighting reflected diffusely at that point on the surface? (In the slides this is cdiff.)

 

 

 

  1. What is the RGB triplet for the specular reflection off that point on the surface? (In the slides this is cspec.)

 

 

 

  1. What is the RGB triplet for the ambient lighting reflected off that point on the surface? (In the slides this is camb.)

 

 

 

 

4. What is the RGB triplet for the total light reflected off that point on the surface?


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