G5) cern — 1H sxowance Equation v © being absorbed: a(p, a). Hence the total absorbed density at this point ang )(p, @). If this is now integrated over the volume and directions, the total ab the volume Is ap, 0) (p, dpa the boundary of v a= | fou. oydpdo caa68 Finally the emission density function is specified as e(p, o), the flux dens ‘emitted (photon enersy pet unit time) at point p in direction @. Hence total ain the volume is SY e={ fet ovine gas) ow (292 ny be wn (£038) O48, = 0,404, got to do with computer graphi knew &(p, a) then we would have a compl lighting in computer graph irection in which we might be interested (F) at every tual Why do we care ‘The answer is su solution to the prot energy flowing in every di point throughout a volume of space (V) For example, suppose we had a bye oF camera in the scene, and wished to use that to make an Imag fantom world of projections". Then we would find al the ays of ght Pnter the Tens (Le, a set of o © 0) and reach the retina of fraphie film. This lens would correspond to a set of points p Bach ‘therefore catries energy {image would be formed, on the wed by a person generat lal scene. (Of course, how is book!) response of the eye or camera. AN for “film” plane, which when associated with seeing the the subject of the zest of Now how can we find @(, w)? We could try to soive the int » ‘ould try to solve the integral equation (£Q38) for %(p, a). This is extremely hard and in practice cannot be achieved. of different approximations and between these two extremes. Simplifying assumptions EQ 3:8) a number of most the whole realm of computer graphics. customary to assume that there is n0 inter a sample of. estimate avelengths, and ther fength distribution at a point along a ray. This rules out fluorescence. by which material absorbs light at one wavelength to reflect ferent wavelength within a short period of time. riance, tis assumed that any solution of the equation for the ‘changes in the of energy remains valid overtime unles scene itself (eg, an object moves position), This which occurs when energy absorbed at one moment is emitted ge time delay. Light transports In a vacuum. A very important pace through which light tra thi Cquation become simpler: absorption and outscateing only occur at the Taeindariey of surfaces. Thre is no emission except from objects, There ls no orption, again except is unobstructe led “fee spac selves generate 2 scenes representing b a less reasonable assumption for Out ny kind of lighting effects caused by ‘door scenes - itTHE RADIANCE EQUATION erounce @) Objects are isotropic. When photons partially absor ‘material does not have this property. @ ee See cerca aioe cecy ao 2 Ses 4 photometry. Note also that by rearrang- ), we obtain: shown in Figure 3.1. Then €Q3.12) LL the corresponding flux is: #0 = LdAcostda ‘As we let da and do become vanishingly smé ‘94d, dB) as the @(4A, dB) = LdAcos®,doy ‘However, using (EQ 2.23) on page 56, radiant power tran energy represented photon and the energy # caries but the ray and its associated radiance, “there are three other characterzations of light energy that are important in computer graphics. The fist is called radantintensty I isthe radiant power {or dun) per unit solid angle (steadians). Hence, Tis the radiant intensity, then the associated radiant power is given in (EQ 3.13): 0 = Ido Hence, comparing with (EQ 3.10): @Q313) eQ3.14) T= LdAcos®The second ts called radiosity, a surface, and is usually denoted by with the energy leaving area da, the flux can be recovered as: a = Baa (EQ 3.15) Irradiance is the flux pet unit area that arrives at a surface. by E, and if the irradiance arriving at dA is F then the fl do = Baa 93.16) Suppose L(p, 0) is radiance artiving at point p along direction o, then It follows from (EQ 3.9) that: 2p 0) » $2 = 107, eeor0de ea3.n Reflectance ‘model for lighting. Suppose a ray of {ncldent direction @, The volume over which that energy may be reflected is should the hemisphere whose base is the tangent plant ‘be clear the hemisphere contains the entire which are not obscured by the surface of whi ‘The next question to ask is how much energy ult in a reflected direction @? We introduce the term :DF) which cent direction @, Then: Lip, ©) = fp, & @ ‘we RADIANCE EQUATION (BT) where the constant of proportionality isthe relectivty ofthe ma reflected out across al di ‘also see Feynman et al, 1977, pp. 26-3 t0 26-4.) ‘The radiance equation he lighting problem points and directmp Suoye aouerpes tn Sundnoo my S400} afo axp yt atdurexe - (9 e110" ar ] 2 eceoo . ' (oa) [sor = 0 auojonain Sujofino 205 uonenbs souerper amy Peceon tarosonce dita w |= ponidaynu: souepean ‘© uopoomp Bunmosuy Sue 103 tbo mau # axtzap 0} owen au Buysn pur “Q400q ovsafiins usaq aney eH} suop ]@ nouvow swans wu 40 ssourorauopusdapar Math auspusdap 821A, povapisuoa 51 Supen Ae tH Suerpes aun yo wows jo sodKL TE 1aML Hbomigos 21m 25TE™ oy Auo punoy ae Su aba aouerpes aun jo wonnjos Jo ad&s 218 0% ST eh foquinis © sagt op 105 (03 Burpuodsass09 ay QUE ‘uoRage1 Te refit 070d uN So ‘dr vem 08 A (ee DD wopenbs 20H _aguag ay “vapuadapat dap ota) Yon nfs 2 0 28 (® owns pxenaw Hu. 40 sNoumos rowynba soNwiavy 34x ~ ONANsrr mde wr ree uy passnostp st Auaiss 30 waygord yeoue8 au, EL J=ydeUD UI passmosD st sdope sojudes# aurn-tear eta uoringor ax “zuoa9e OvUI Uayer 9q 03 aary 29 ysis ~ s291g0 30 (Cooueypes aynqunuo> ymns Aer yd ov Aagyqisis sur UTE ms v yBnowp waka os pur ‘yuaq aze sioalgo auaredsuen 134 tlt s90lgo uoe> 281n0> Jo “vodsuen au yumuosjaua uP yBnomn 0d @® unas Prrrtrery 5 “uon mos reqor8 & Aseap st “aha ogpads #105 hex 30 95 ue “uon>ap oF 1adser se 10} paxdope s uopenbs souerper ayy “sorydeS aun TeRL zz dey uti ‘2 Hod uonpasiaut patesouas aie shes Nouynoa s>yvova 3H al