leslie1719 |
2011-12-26 09:48 |
我這邊有一個zpl可以直看出來~ q
o'1Pknz 如下: [bAv|; !RIH: CHIEF RAY HEIGHT ON IMA qYE -z(i (t <Um
Vd !CRA: CHIEF RAY ANGLE IN IMA SPACE KjLj "ey~w=B$M !和取最?視角度 IgVxWh# ?wmr~j MAXFIELD = MAXF() Cu}Rq!9i I)6)~[:' IF (MAXFIELD = = 0.0) THEN maxfield = 1.0 9%4rO\q 2D
"mq~V !獲曲面數 .; :[sv) R \iU)QP n = NSUR() >8ePx,+! J=()
A+ !主光線錐激 hNQ,U{`;^ oYu5]ry b.$Gc!g MVV<&jho{^ RAYTRACE 0, 1, 0, 0, PWAV() Fd2zvi x
ha!.&DO 67d0JQTu mWtwp- MAX_RIH = RAYY(n) MLUq"f~ N t.NG]ejZ BONM:(1 gX);/;9mm+ MAX_CRA = ACOS(RAYN(n-1))*180/3.1416 tvI~?\Ylj @n<WM@|l 4%B${zP(.} Ix"uk6 h PRINT "MAX FIELD: " , maxfield, c" yf>0 ZYg="q0x& PRINT "MAX_CHIEF_RAY_ANGLE : " , MAX_CRA , ^G15]Pyw P\SE_*& PRINT "MAX CHIEF RAY HEIGHT ON IMA : ",MAX_RIH `6UW?1_Z5 /+%1Kq.hP fY\QI
= (ZDRjBth[ PRINT "FIELD ", }nuhLt1 C5F}*]E[y PRINT "CHIEF RAY ANGLE," V+_L9 jh9^5"vQ PRINT "CHIEF RAY HEIGFHT ON IMA." `XQM)A C%l~qf1n 'R= r9_% 6X)8vQH !將最大試廠分為20, 設置試場數據步常 FHY=j/20 , for B2VUH..am jRzR`>5 &`{%0r[UD# jPhOk>m For j, 0, 20, 1 TR|G4l? sy4$!,W: FHY = j/20 om|M=/^ Es1Yx\/: RAYTRACE 0, FHY, 0, 0, PWAV() PoQ@9
A Bm1yBKjO CRA = ACOS(RAYN(n-1))*180/3.1416 dX` _Y rJ K~kKG OLDX = CRA #NwlKZ- U_Id6J]8 OLDY = RAYY(n) p\~ lPXK ^<7)w2ns PRINT FHY*maxfield, " , ",OLDX," ",OLDY OP{ d(~+ H;%a1 xqX~nV#TB %.[t(F $D1Pk 1P@&xcvS\ NEXT =D<46T=(RB Ay/ "2pDZ !繪出該點數據 OiZPL" Q(K {])F%Q_#cD JmtU>2z\ ^3 F[^#" GRAPHICS
&CG3_s<2 esWgYAc3{ FX4](oM l0 rZril RAYTRACE 0, 1, 0, 0, PWAV() M n3cIGL [-=PK\ B MAX_RIH = RAYY(n) lmgMR|v _\1wLcFj MAX_CRA = ACOS(RAYN(n-1))*180/3.1416 dq[j.Nmq z{7&= $ X_WIDE = XMAX() /1.6 ;a*i*{\Rm J+kxb"#d Y_WIDE = YMAX() /1.6 [89#8|+ QB7E:g& | |