COMP 558 lecture 6 Sept. 27, 2010

Transcription

1 Radiometry We have discussed how light travels i straight lies through space. We would like to be able to talk about how bright differet light rays are. Imagie a thi cylidrical tube ad cosider the amout of light that passes through the tube without hittig the sides. We measure the power 1 of light through the tube. The power of light that gets through the tube, depeds o the legth of the tube ad o the cross-sectioal area the tube (πr 2, where r is the tube radius). If you hold the area costat ad icrease the legth of the tube, you decrease the amout of light that passes through the tube because you will have a tighter restrictio o the directio of rays that ca pass through the tube. Similarly, if you hold the legth of the tube costat, ad you decrease the cross-sectioal area of the tube, you agai decrease the umber of rays that pass through the tube. Thus, the power of light that passes through the tube depeds o the legth ad cross sectioal area. Agle vs. solid agle For a thi tube, cosider the agle subteded by the diameter at the far ed of the tube, whe see from the axis at the opeig of the ear ed of the tube. This agle is approximately the ratio diameter. This agle is i uits of radias, i.e. which is the arclegth o the uit circle. Recall 1 tubelegth radia is degrees, π radias is 90 degrees, etc. 2π 2 To measure light, oe eeds to cosider a agular measuremet called solid agle. Solid agle is the area o a uit sphere that is subteded by a object, whe that object is viewed from a poit at the ceter of the sphere. The uits of solid agle are steradias. The maximum value of solid agle is 4π, which is the etire area of the uit sphere. For example, if you are iside a room, the the solid agle subteded by the walls ad ceilig ad floor of the room is 4π steradias. We said earlier that the power of light that passes through the tube depeds o the legth ad width of the tube. Alteratively, we ca express the power by the area of the ear ed of the tube ad by the solid agle subteded by the far ed of the tube whe see from ear ed. This leads us to the followig defiitio. Radiace Place oe ed of a small hollow tube at some poit x = (X,Y,Z) ad poit it i some directio l. Measure the power of light passig through the tube. Defie the radiace L(x,l) of the ray to be 1 Namely, light eergy per secod, measured i Watts. I am igorig color properties for ow. 1

2 the light power 2 per uit cross sectio area at x per uit steradia of the coe of rays arrivig at x, radiace, L(x, l) power cross sectio area solid agle I tryig to uderstad this defiitio, you should thik about a thi tube, so that the cross sectioal area ad solid agle are both very small. [ASIDE: If there is o scatterig of light i space (e.g. fog), the radiace is costat alog a light ray. Ituitively, if you place your thi tube at a poit X i space ad poit it i some directio l, ad the move the tube i this directio, the power that you get through the tube will ot chage (i.e. the brightess of the light you will see through the tube will ot chage). Thigs that we look at do ot get brighter whe we walk toward them. They get bigger (i terms of their solid agle) but they do t get brighter. Note: small light sources do seem to get dimmer as you move away from them. But the reaso is that the solid agle they subted gets smaller ot that the radiace decreases.] Irradiace Aother importat cocept is the total power of light that arrives at a surface per uit area. This is importat for uderstadig how much illumiatio is received by poits i the world, ad also how much light arrives at a pixel o the sesor plae. Today we ll examie poits i the world ad ext lecture we ll cosider the sesor plae. If you take a poit x o a surface, the light ca arrive at x from a hemisphere of directios. Usually oe talks about the amout of light arrivig i a small area cotaiig that poit, rather tha about the set of rays arrivig at precisely that poit. The reaso for talkig about a small arrival area is that it allows oe to capture the orietatio (ormal) of the surface at that poit. The irradiace E(x) is defied as the light power arrivig per uit area o the surface i a small eighborhood of x. Light ca arrive from a hemisphere of directios, so we divide the hemisphere 2 eergy per uit time 2

3 ito solid agles dω i ad the add up the cotributio of each oe: E(x) = L(x,l i ) (x) l i dω i where (x) is the uit ormal vector of the surface at x, ad dω i is a small solid agle cetered i the directio l i. The l i compoet is eeded because the radiace is defied per uit area o the surface. Cosider a small tube cetered o a ray i directio l i which has oe ed at the surface poit x. If the icomig ray is oblique to the surface, the the surface will itersect the tube at a oblique agle. The light through the tube will be be spread out over a area proportioal to 1/ l i 1. Thus, to get the power per uit area o the surface, rather tha the cross-sectioal area of the tube, we eed to multiply by l i 1. Special case 1: sulight Cosider the special case of a parallel source such as sulight. The su spas a small solid agle Ω su ad the radiace is approximately costat L su over that agle, so we ca write E(x) = L(x,l i )(x) l i dω i su L su (x) l i dω su L su Ω su (x) l su where L su is the radiace of the source, Ω su is the solid agle subteded by the source, ad l su is the directio to the ceter of the su. If you are strugglig with the l effect, thik of the (approximately) parallel rays from the su ad to otice that umber of rays that reach a uit area patch of surface is proportioal to (x) l, amely to the cosie of the agle betwee the surface ormal ad the light source. Cosider the sketch below. The parallel vectors idicate the parallel rays arrivig at the surface. The three boldface lies (bottom left) idicate three surfaces that havig differet surface ormals. The two dotted lies illustrate that as the surface ormal tilts away from the light source, the umber of rays reachig a uit area of the surface decreases. (Oly a slice is show, but you should imagie this picture is the same i each depth slice.) The umber of rays depeds o the legth of the dotted lie, which by simple trigoometry is (x) l. l 3

4 Special case 1: disk skylight i a room This case was discussed i class (see lecture slides). Details will be give i the lecture otes for lecture 7. Surface reflectace The above discussio was cocered with how much light arrives at a surface poit x per uit surface area. To model the itesity of light reflected from a surface poit x, we must specify ot just the irradiace (icomig light), but also say somethig about the surface material. There are may differet kids of surface materials ad they differ i visual appearace i may ways. For example, some surfaces are shiy ad others are ot. Surfaces that are shiy ca have very differet appearaces from each other: compare plastic, metal, velvet, leather, or ski, etc. Here I preset oly basic model for how people describe surface reflectace. This wo t give you much uderstadig of the topic, but it at least exposes you to the existece of the topic! Bidirectioal reflectace distributio fuctio (BRDF) Suppose that we were to illumiate the surface ear a poit x from oly a small set of directios cetered at l i which is a uit vector. This determies the compoet of the irradiace at x that is due to a small set of directios of icidet light. We ca the measure the radiace of the surface i some outgoig directio, say l out. The quatity of iterest is how this outgoig radiace i directio l out depeds o the icomig radiace l i. If we kow this depedece for all pairs l i ad l out, the we have characterized the reflectace properties of the surface. l out l i Oe captures this depedece by defiig a bidirectioal reflectace distributio fuctio(brdf): ρ(x,l i,l out ) which is defied such that: L(x,l out ) = ρ(x,l i,l out ) L(x,l i )(x) l i dω i 4

5 A BRDF ca be measured i a laboratory by sequetially illumiatig a surface at x with lights from each directio l i, ad the measurig the resultig outgoig radiace i each directio l out. (Details omitted.) Measurig ad modellig BRDF s is a very importat problem i computer visio. The applicatios are maily i computer graphics, where oe would like to be able to reder surfaces usig realistic BRDFs models that are obtaied from real surfaces, istead of the oversimplified models (e.g. Phog) that were used i classical computer graphics. Diffuse (matte, Lambertia) reflectace For may surfaces, the BRDF ρ(x,l i,l out ) is idepedet of l out ad l i. I particular, for such a surfaces, the radiace leavig a surface poit x is modelled as L(x,l out ) = ρ(x) L(x,l i )(x) l i dω i The costat ρ(x) specifies whether the surface is light or dark colored. For a black surface, ρ(x) would be ear zero. NoticethattheoutgoigradiaceforLambertiasurfacesdoesotdepedol out. Suchsurfaces, for which the reflected radiace is the same i all directios, are called Lambertia 3, or matte. May computer visio methods assume the surfaces are approximately Lambertia. The mai reaso is that this coditio helps oe to fid matchig poits i images take from differet viewpoits. Why? If a pixel has a particular itesity value from camera positio ad we wish to fid a correspodig pixel whe the same camera is moved to aother positio, the we oly eed to cosider poits that have the same image itesity. 3 after a 18th cetury scietist Joha Lambert who wrote a treatise o light 5