| leslie1719 |
2011-12-26 09:48 |
我這邊有一個zpl可以直看出來~ >#"jfjDuR 如下: |k['wqn" !RIH: CHIEF RAY HEIGHT ON IMA W<cW;mO
X:xC>4]gG' !CRA: CHIEF RAY ANGLE IN IMA SPACE iOX4Kl jm#F*F vL !和取最?視角度 ^a:vJ)WB7 wB1-|=K1 MAXFIELD = MAXF() $v?! 6: WwCK K IF (MAXFIELD = = 0.0) THEN maxfield = 1.0 u~naVX\3b &kXGWp !獲曲面數 Oo3qiw
8;+Hou n = NSUR() &<fRej]v {"gyXDE1 !主光線錐激 x3Dg%=R QYf/tQg$ pjmGzK rH2tC=% RAYTRACE 0, 1, 0, 0, PWAV() Up!ZCZ$RC }jyS\drJ uV/HNzC Yt O@n@1 MAX_RIH = RAYY(n) +,{Wcb ()3x%3 CL<KBmW7 E
|GK3 / MAX_CRA = ACOS(RAYN(n-1))*180/3.1416 b*6c.o 6b+ WlIb qqz,~EhC nM (=bEX PRINT "MAX FIELD: " , maxfield, @dT: 1s H[_uVv;}6 PRINT "MAX_CHIEF_RAY_ANGLE : " , MAX_CRA , s:m<(8WRw NbdMec PRINT "MAX CHIEF RAY HEIGHT ON IMA : ",MAX_RIH ^]sMy7X0IK kb}]sj Fgc:6<MGM # 1qVFU PRINT "FIELD ", ^b `>/> S'%cf7Z PRINT "CHIEF RAY ANGLE," eB/hyC1 (&Tb,H)= PRINT "CHIEF RAY HEIGFHT ON IMA." d'x<F[`O 8NF;k5 .^N#|hp^ (-Ct!aW| !將最大試廠分為20, 設置試場數據步常 FHY=j/20 , for EkB6- nz i6g[E4nk EfrkB" )zL"r8si For j, 0, 20, 1 :9rhv{6Wp /Y\E68_Fh FHY = j/20 [nPzhXs ,d [b"]Zy RAYTRACE 0, FHY, 0, 0, PWAV() +O!M> g%q?2Nv CRA = ACOS(RAYN(n-1))*180/3.1416 :'=~/GR )r6SGlE[Y OLDX = CRA UN
.[,%<s D -+)M8bt OLDY = RAYY(n) D'sboOY M@2Qn-I PRINT FHY*maxfield, " , ",OLDX," ",OLDY k.%W8C<Pa ]x(2}h^S m9<[bEO<$ LG@c)H74 !wR{Y[Yu fF-\TW NEXT l^eNZ3:H eze(>0\f !繪出該點數據 D@b<}J>0' uI-76 ^qN1~v=hS 8$jT#\_ GRAPHICS uA/.4 b Sp$x%p0 1PTu3o&3 ~ew**@N RAYTRACE 0, 1, 0, 0, PWAV() r dG2| Tp d
@kLLDP MAX_RIH = RAYY(n) UG?C=Tf `=l{kBZT| MAX_CRA = ACOS(RAYN(n-1))*180/3.1416 NUNn[c J)yy}[Fx X_WIDE = XMAX() /1.6 :iNAXy Jx
;"a\KD Y_WIDE = YMAX() /1.6 Md?bAMnG+} 'St= izhd SEGMENT = 40 jnKM6%z qfH~h g FOR j, 0, SEGMENT-1,1 ${?ex nb$ @G$<6CG\
0S5C7df "%f5ltut3 FHY = j/SEGMENT K,,@', ^Er`{|o6u RAYTRACE 0, FHY, 0, 0, PWAV() A%w]~ chC9 WT!\X["FI$ CRA = ACOS(RAYN(n-1)) * 180/3.1416 <VxpMF FR6I+@ oX~ RIH = RAYY(n) K#sb"x` F#bo4'&>@ OLDX = X_WIDE * (CRA/MAX_CRA+2/20) DMxS-hl
%iS]+Sa.K OLDY = Y_WIDE * (1-RIH/MAX_RIH+2/20) XQY&4tK < | |