CPU. Rasterization. Per Vertex Operations & Primitive Assembly. Polynomial Evaluator. Frame Buffer. Per Fragment. Display List.


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1 Elmntary Rndring Elmntary rastr algorithms for fast rndring Gomtric Primitivs Lin procssing Polygon procssing Managing OpnGL Stat OpnGL uffrs OpnGL Gomtric Primitivs ll gomtric primitivs ar spcifid by vrtics Dsign of Lin lgorithms GL_POINTS GL_LINES GL_LINE_STRIP GL_LINE_LOOP GL_POLYGON GL_TRINGLES GL_TRINGLE_STRIP GL_QUDS GL_QUD_STRIP GL_TRINGLE_FN Why Lins? Lin lgorithms in th OpnGL rchitctur Lins: Most common D primitiv  don 0s or 00s of tims ach fram, vn 3D wirframs ar vntually D lins! Lins ar compatibl with vctor displays but nowadays most displays ar rastr displays. ny rndr stag bfor viz might nd discrtization. Optimizd algorithms contain numrous tricks/tchniqus that hlp in dsigning mor advancd algorithms for lin procssing. PU Polynomial Evaluator Display List Pixl Oprations Pr Vrtx Oprations & Primitiv ssmbly Rastrization Txtur Mmory Pr Fragmnt Oprations Fram uffr
2 Lin Rquirmnts asic Math Rviw Must comput intgr coordinats of pixls which li on or nar a lin or circl. Pixl lvl algorithms ar invokd hundrds or thousands of tims whn an imag is cratd or modifid must b fast! Lins must crat visually satisfactory imags. Lins should appar straight Lins should trminat accuratly Lins should hav constant dnsity Lin algorithm should always b dfind. Pointslop Formula For a Lin Givn two points (X 1,Y 1 ), (X, Y ) onsidr a third point on th lin: P = (X,Y) Slop = (Y  Y 1 )/(X  X 1 ) = (Y  Y 1 )/(X  X 1 ) Solving For Y Y = [(Y Y 1 )/(X X 1 )]*(XX 1 )+ Y 1 or, plug in th point (0, b) to gt th Slopintrcpt form: Y = mx + b artsian oordinat Systm 6 P = (X,Y) 5 4 P = (X,Y) 3 1 P1 = (X1,Y1) RISE YY1 SLOPE = = RUN XX1 Othr Hlpful Formulas Lngth of lin sgmnt btwn P and P 1 : L = ( x y x1) + ( y 1) Midpoint of a lin sgmnt btwn P and 1 P 3 : P = ( (X 1 +X 3 )/, (Y 1 +Y 3 )/ ) Two lins ar prpndicular iff 1) M 1 = 1/M ) osin of th angl btwn thm is 0. Using this information, what ar som possibl algorithms for lin drawing? Paramtric Form Givn points P 1 = (X 1, Y 1 ) and P = (X, Y ) X = X 1 + t(x X 1 ) Y = Y 1 + t(y Y 1 ) t is calld th paramtr. Whn t = 0 w gt (X 1,Y 1 ) t = 1 w gt (X,Y ) Nw algorithm idas basd on paramtric form? s 0 < t < 1 w gt all th othr points on th lin sgmnt btwn (X 1,Y 1 ) and (X,Y ).
3 Simpl DD* Lin lgorithm DD Exampl void DD(int X1,Y1,X,Y) { int Lngth, I; float X,Y,Xinc,Yinc; Lngth = S(X  X1); if (S(Y  Y1) > Lngth) Lngth = S(YY1); Y1); Xinc = (X  X1)/Lngth; Yinc = (Y  Y1)/Lngth; *DD: Digital Diffrntial nalyzr X = X1; Y = Y1; whil(x<x){ Plot(Round(X),Round(Y)); X = X + Xinc; Y = Y + Yinc; DD crats good lins but it is too tim consuming du to th round function and long oprations on ral valus. omput which pixls should b turnd on to rprsnt th lin from (6,) to (,1). Lngth =? Xinc =? Yinc =? DD Exampl Fast Lins Midpoint Mthod Lin from (6,) to (,1). Lngth := Max of (S(6), S(1)) = 5 Xinc := 1 Yinc := 0.6 Valus computd ar: (6,), (7,.6), (8,.), (,.8), (,.4), (,1) Simplifying assumptions: ssum w wish to draw a lin btwn points (0,0) and (a,b) with slop m btwn 0 and 1 (i.. lin lis in first octant). Th gnral formula for a lin is y = mx + whr m is th slop of th lin and is th yintrcpt. From our assumptions m = b/a and = 0. y = (b/a)x > f(x,y) = bx  ay = 0 is an quation for th lin. x +y y +x Fast Lins (cont.) Fast Lins (cont.) For lins in th first octant, givn on pixl on th currnt pixl lin, th nxt pixl is to th right (E) or to th right and up (NE). NE = (x i + 1, y i + 1) P = (x i,y i ) E = (x i + 1, y i ) possibl nxt pixls Having turnd on pixl P at (x i, y i ), th nxt pixl is NE at (x i +1, y i +1) or E at (x i +1, y i ). hoos th pixl closr to th lin f(x, y) = bx  ay = 0. Th midpoint btwn pixls E and NE is (x i + 1, y i + ½). Lt b th upward distanc btwn th midpoint and whr th lin actually crosss btwn E and NE. If is positiv th lin crosss abov th midpoint and is closr to NE. If is ngativ, th lin crosss blow th midpoint and is closr to E. To pick th corrct point w only nd to know th sign of. NE = (x i + 1, y i + 1) (x i +1, y i + ½ + ) (x i +1, y i + ½) P = (x i,y i ) E = (x i + 1, y i )
4 Th Dcision Variabl Dcision Variabl calculation f(x i +1, y i + ½ + ) = 0 (point on lin) = b(x i + 1)  a(y i + ½ + ) = b(x i + 1)  a(y i + ½) a f(x i + 1, y i + ½) = a = f(x i + 1, y i + ½)  a Lt d i = f(x i + 1, y i + ½) = a; d i is known as th dcision variabl. Sinc a = 0, d i has th sam sign as. Thrfor, w only nd to know th valu of d i to choos btwn pixls E and NE. If d i = 0 choos NE, ls choos E. ut,, calculating d i dirctly ach tim rquirs at last two adds, a subtract, and two multiplis > too slow! lgorithm: alculat d 0 dirctly, thn for ach i >= 0: if d i = 0 Thn hoos NE = (x i + 1, y i + 1) as nxt point d i+1 = f(x i+1 + 1, y i+1 + ½) = f(x i , y i ½) = b(x i )  a(y i ½) = f(x i + 1, y i + ½) + b  a = d i + b  a ls hoos E = (x i + 1, y i ) as nxt point d i+1 = f(x i+1 + 1, y i+1 + ½) = f(x i , y i + ½) = b(x i )  a(y i + ½) = f(x i + 1, y i + ½) + b = d i + b Knowing d i, w nd only add a constant trm to find d i+1! Fast Lin lgorithm Th initial valu for th dcision variabl, d 0, may b calculatd dirctly from th formula at point (0,0). d 0 = f(0 + 1, 0 + 1/) = b(1)  a(1/) = b  a/ Thrfor, th algorithm for a lin from (0,0) to (a,b) in th first octant is: x = 0; y = 0; d = b  a/; for(i = 0; i < a; i++) { Plot(x,y); if (d = 0) { y = y + 1; d = d + b  a; ls { d = d + b Not that th only nonintgr valu is a/. If w thn multiply by to gt d' = d, w can do all intgr arithmtic. Th algorithm still works sinc w only car about th sign, not th valu of d. rsnham s Lin lgorithm W can also gnraliz th algorithm to work for lins bginning at points othr than (0,0) by giving x and y th propr initial valus. This rsults in rsnham's Lin lgorithm. {rsnham for lins with slop btwn 0 and 1 a = S(xnd  xstart); b = S(ynd  ystart); d = *b  a; Incr1 = *(ba); Incr = *b; if (xstart > xnd) { x = xnd; y = ynd ls { x = xstart; y = ystart for (i = 0; i<a; i++){ Plot(x,y); if (d = 0) { y = y + 1; d = d + incr1; ls d = d + incr; Optimizations Spd can b incrasd vn mor by dtcting cycls in th dcision variabl. Ths cycls corrspond to a rpatd pattrn of pixl choics. Th pattrn is savd and if a cycl is dtctd it is rpatd without rcalculating. 16 Th aliasing problm liasing is causd by finit addrssability of th display. pproximation of lins and circls with discrt points oftn givs a staircas apparanc or "Jaggis" liasd rndring of th lin Dsird lin di=
5 ntialiasing  solutions liasing can b smoothd out by using highr addrssability. liasing / ntialiasing Exampls If addrssability is fixd but intnsity is variabl, us th intnsity to control th addrss of a "virtual pixl". Two adjacnt pixls can b b usd to giv th imprssion of a point part way btwn thm. Th prcivd location of th point is dpndnt upon th ratio of th intnsitis usd at ach. Th imprssion of a pixl locatd halfway btwn two addrssabl points can b givn by having two adjacnt pixls at half intnsity. n antialiasd lin has a sris of virtual pixls ach locatd at th propr addrss. ntialiasd rsnham Lins Lin drawing algorithms such as rsnham's can asily b modifid to implmnt virtual pixls. W us th distanc ( = d i /a) valu to dtrmin pixl intnsitis. Thr possibl cass which occur during th rsnham algorithm: > 0 0 > > 0.5 < 0.5 = = 1  abs(+0.5) = 0 = = 1  abs(+0.5) = 0 = 0 = 1  abs(+0.5) = Lin Rndring Rfrncs rsnham, J.E., "mbiguitis In Incrmntal Lin Rastring," IEEE omputr Graphics nd pplications, Vol. 7, No. 5, May 187. Eckland, Eric, "Improvd Tchniqus For Optimising Itrativ Dcision Variabl lgorithms, Drawing ntiliasd liasd Lins Quickly nd rating Easy To Us olor harts," S 46 Projct Rport, Dpartmnt of omputr Scinc, North arolina Stat Univrsity (Spring 187). Foly, J.D. and. Van Dam, Fundamntals of Intractiv omputr Graphics, ddisonwsly 18. Nwman, W.M and R.F. Sproull, Principls Of Intractiv omputr Graphics, McGraw Hill, 17.
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