� � Lecturers: J. Kautz, A. Steed, T. Weyrich � � Demonstrator: James Tompkin � � Lab Time – � TBA in 4.17 Jan Kautz Anthony Steed Tim Weyrich � � Assessment – � Written Examination (2.5 hours, 75%) – � Coursework Section (2 pieces, 25%) � � CW1: Deadline Friday 12/02/2010 by 12 noon � � CW2: Deadline TBA Tim Weyrich � � Radiometry (radiant power, radiance etc) � � Introduction – � Measurement of light energy � � Spectral distributions � � Photometry (luminance etc) – � Measurement including response of visual system � � Simple Model for the Visual System � � Generally C( � ) defines spectral colour distribution � � Simple Model for an Emitter System �� [ � a , � b ] = � � � Generating Perceivable Colours � � In computer graphics C is usually radiance. � � CIE-RGB Colour Matching Functions
Spectral radiant power � � � ( � ) = 0, �� 0 � � � � ( � ) d � = 1 The image cannot be displayed. Your computer may not have enough memory to open the image, or the image may have been corrupted. Restart your computer, and then open the file again. If the red x still appears, you may have to delete the image and then insert it again. � � � � (t) f(x-t)d t = f(x) � � C( � ) = � ( � - � 0 ) is spectral distribution for pure colour with wavelength � 0 � 700 400 Wavelength nm Spectral Energy Distribution Schematic Representation of Colour Spectra 400nm 700nm � � We do not ‘see’ C( � ) directly but as filtered � � Space of all visible colours equivalent to set through visual system. of all functions C : �� R � � Two different people/animals will ‘see’ C( � ) – � C( � ) � 0 all � differently. – � C( � ) > 0 some � . � � Different C( � )s can appear exactly the same to one individual (metamer). � � (Ignoring all ‘higher level’ processing, which basically indicates “we see what we expect to see”).
� � Colour space is infinite dimensional � � Visual system filters the energy distribution through a finite set of channels � � Constructs a finite signal space (retinal level) � � Through optic nerve to higher order processing (visual cortex ++++). Human Eye Schematic � � Rods – 130,000,000 night vision + � � l = � C( � )L( � )d � peripheral (scotopic) � � m = � C( � )M( � )d � � � Cones – 5-7,000,000, daylight vision + � � s = � C( � )S( � )d � acuity (one point only) � � C � (l,m,s) (trichromatic theory) � � Cones � � LMS(C) = (l,m,s) – � L-cones – � M-cones � � LMS(C a ) = LMS(C b ) then C a , C b are – � S-cones metamers . � � Generates chromatic light by mixing streams of energy of light of different spectral distributions � � Finite number (3) and independent of each other
� � C E ( � )= � 1 E 1 ( � )+ � 2 E 2 ( � ) + � 3 E 3 ( � ) � � For a given C( � ) problem is to find the � � E i are the primaries (i.e., the display uses them) intensities � i such that C E ( � ) is metameric to C( � ) � � � i are called the intensities. � � CIE-RGB Primaries are: � � First Method to be shown isn’t used, but – � E R ( � ) = � ( � - � R ), � R = 700nm illustrative of the problem. – � E G ( � ) = � ( � - � G ), � G = 546.1nm – � E B ( � ) = � ( � - � B ), � B = 435.8nm � � CIE = Commission Internationale de L’Eclairage � � For a metamer we require (considering L only): � � Write In principle C, L and E i will be known. � � Expanding: � � And do the same derivation for M and S: � � Or � � Previous method relied on knowing L, M, and S response curves accurately. � � Better method based on colour matching functions. � � This gives 3 equations in 3 unknowns which can � � Define how to get the colour matching be solved for the unknown intensities. functions � i ( � ) relative to a given system of � � But in practice the L, M and S curves cannot be primaries (e.g., RGB). directly observed, so other techniques are used.
� � Let � 0 be a monochromatic colour, and � i ( � 0 ) (i=1,2,3) be the � � This further simplifies to: intensities, then: � � Also, � � and so: � � By substitution: � � Recall that the � i ( � ) were the intensities for � � Now replace � 0 by � , multiply throughout monochromatic colours. by colour C( � ), and integrate: � � The result says that we can find the intensities for a metamer for an arbitrary colour based on these. � � How can we estimate these � i ( � )? � � With further rearrangement, we get the � � This can be done with a perceptual colour- result: matching experiment. Mixing of 3 primaries Target colour overlap Adjust intensities to match the colour
� � Not all monochrome colours can be � � RGB intensities sampled at 5nm intervals represented with positive � i ( � ). between 390nm and 830nm. � � They are ‘2 degree’ color matching functions because the observer only sees a field of view of 2 degrees. � � In that case we add one beam to the target and try to adjust the other two beams to � � 2 degree ones are used in computer graphics match the new colour: because of the relatively narrow field of view when looking at a display. � � Compute the radiance distribution C( � ) � � CIE-RGB Chromaticity Space � � Find out the colour matching functions for the � � CIE-XYZ Chromaticity Space display � i ( � ) � � Converting between XYZ and RGB � � Perform the 3 integrals � � i ( � )C( � )d � to get the intensities for the metamer for that colour on the � � Colour Gamuts and Undisplayable Colours display. � � Summary for Rendering: What to do in � � …. practice � � Except that’s not how it is done … � � ….to be continued…. � � Consider 1 st only monochromatic colours: � � Consider CIE-RGB primaries: – � C( � ) = � ( � - � 0 ) – � For each C( � ) there is a point ( � R , � G , � B ): � � Let the CIE-RGB matching functions be � � C( � ) � � R E R ( � ) + � G E G ( � ) + � B E B ( � ) – � Considering all such possible points – � r( � ), g( � ), b( � ) � � Then, eg, � � ( � R , � G , � B ) – � Results in 3D RGB colour space – � � R ( � 0 ) = � � ( � - � 0 ) r( � )d � = r( � 0 ) – � Hard to visualise in 3D � � Generally – � so we’ll find a 2D representation instead. – � ( � R ( � 0 ), � G ( � 0 ), � B ( � 0 )) = (r( � 0 ), g( � 0 ), b( � 0 ))
� � As � 0 varies over all wavelengths � � It is easy to show that projection of ( � R , � G , � B ) onto � R + � G + � B = 1 is: – � (r( � 0 ), g( � 0 ), b( � 0 )) sweeps out a 3D curve. – � ( � R /D, � G /D, � B /D), � � This curve gives the metamer intensities for � � D = � R + � G + � B all monochromatic colours. � � Show that interior and boundary of the � � To visualise this curve, conventionally curve correspond to visible colours. project onto the plane � � CIE-RGB chromaticity space. – � � R + � G + � B = 1 � � Suppose � 1 and � 2 are two 3D points corresponding to spectral functions C 1 and C 2 . � � Consider the line segment joining them: – � (1-t) � 1 + t � 2 , t � [0,1] � � It is easy to see that for any such t this must correspond to another spectral function. � � When we project the points and line segment to the plane, the line projects to the 2D line joining them. � � All points on the curved boundary and within the curve represent visible colours. � � Define: – � V( � ) = � 1 L( � ) + � 2 M( � ) + � 3 S( � ) � � For specific constants � i this is the – � Spectral Luminous Efficiency curve � � The overall response of visual system to C( � ) is – � L(C) = K � C( � ) V( � )d � � � For K=680 lumens/watt, and C as radiance, L is called the luminance (candelas per square metre)
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