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¨ UNIVERSITAT DES SAARLANDES Prof.Dr.Philipp Slusallek Dr.Vincent Pegoraro Computer Graphics Group Piotr Danilewski (danilewski@cs.uni-saarland.de)
12. December2011
Introduction to Computer Graphics Assignment 6
Solutions 6.1 Gammacorrection (25 Points) The worst case is obviously if settingγ= 2.highest accuracy is required near zero input, there we5. The get:
2.5 f(0.00000000b) = 0= 0 2.5 f(0.00000001b) = (1/5128) =.39E−6 −6 To represent the step from 0 to 5.39E-6 we need at least 18 bits for the fractional part (log2(5.39∙10 )= −17.5).
6.2 Transformsin color space (50 points)
a)
We can’t change direction of vectors red, green and blue since they depend on the monitor’s hardware characteristics.So the only way remaining to change the position of the white point is by changing the length of the vectors. Simplyspeaking: Wedimm the maximum brightness of one or more primary colors.
b) Overview: letCrgb= (r, g, b) be the color on the first monitor, then we want to know which color 0 Cwe need to use on the second monitor so that both monitors show the same color.With rgb formulars: 0 0 Ccie=M∙CrgbandCcie=M∙C rgb 0 00 0−1 C⇔C M ⇒M∙Crgb=M∙rgb rgb=∙M∙Crgb
Given:xr, yr, xg, yg, xb, yb, xw, ywand due to the normalization of the white point:Yw= 1 (see slide of computer graphics lecture)
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Color Transformations (Cont.) the constants C •ComputingX – Perdefinition the white point is given as (R, G, B) = (1, 1, 1) X • (w, Y , Z )= M*(1,1,1) w w X xC xC xC w rr gg bb1 = Y yC yC yC1 w rr gg bb [ ] [ ][ ] 1 Z1−x−yC1−x−yC1−x−yC w rr rg gg bb b
(X ,Y ,Z )can be computed from (x , y ) –w w w x x • Usingthe normalization constant Y= 1 w
Computer Graphics WS2009/2010 – Color
YwYwYw Then:Xw=xw∙ ≈0.9505 andZw=zw∙= (1−xw−yw)∙ ≈1.0891 ywywyw and for the second monitor we get:Xw2≈1.2121 andZw2= 0.8182 We need to computeCr, Cg, Cb: In the equation above, setzr= (1−xr−yr), zgandzbanalog for convenience issues.
XwxrCrxgCgxbCb1XwxrxgxbCr10 0 Yw=yrCrygCgybCb1⇔Yw=yrygyb0Cg0 1⇔ ZwzrCrzgCgzbCb1Zwzrzgzb0 0Cb1 −1 XwxrxgxbCrxrxgxbXwCr ⇔Yw=yrygybCg⇔yrygybYw=Cg ZwzrzgzbCbzrzgzbZwCb
Solving these equations yields: for the first white point:⇒Cr= 0.6444, Cg= 1.1919, Cb= 1.2033 for the second white point:⇒Cr= 1.2778, Cg= 0.8763, Cb= 0.8763
0 0−1 Now one can directly computeM, MandM: 0.4124 0.3576 0.1805 0.8178 0.2629 0.1314 0 M= 0.2127 0.7151 0.0722M= 0.4217 0.5258 0.0526 0.0193 0.1192 0.9506 0.0383 0.0876 0.6922 1.63435−0.775302−0.251333 0−1 M=−1.31842,2.55168 0.0563736 0.07642−0.280025 1.45144
0−1 Another way to arrive at the final solution (M M) is to observe that since the primary colors are 0 the sameMandMcan be expressed as:
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0 xrxgxbCr0 0xrxgxbC0 0 r 0 0 M=yrygyb0Cg0, M=yrygyb0C0 g 0 zrzgzb0 0Cbzrzgzb0 0C b Thus : Cr −1−1 000 0 0x x xC0 0r Cr0xrxgxrb rg b C Cg 0−10 000 M M= 0C0yrygybyrygyb0Cg0 =C g g 0 Cb 0 0C zrzgzbzrzgzb0 0Cb b0 00 C b Finally : 0.5043 00 0−1 M M= 01.36015 0 0 01.37315
6.3 ColorSpace (25 Points) For the CIE-XYZ color space one simply computes (X, Y, Z) =M∙(1,0,1) = (0.5929,0.2849,0.9699). By normalizing we get the coordinates in the CIE-xy color space:X/(X+Y+Z) = 0.32,Y /(X+Y+Z) = 0.15. Gammacorrection on (1,0,1) yelds (1,0,1) regardless of the gamma, which is why it is not necessary.
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