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uiuo 2009-04-11 21:31

求教CHIEF RAY ANGLE在Zmax中如何可以看出来?

求教CHIEF RAY ANGLE在Zmax中如何可以看出来? (Qa/EkE^*w  
leslie1719 2011-12-26 09:48
我這邊有一個zpl可以直看出來~ CPNL 94x  
如下: B}PIRk@a1  
!RIH: CHIEF RAY HEIGHT ON IMA LpR3BP@At  
gy&[?m6M=  
!CRA: CHIEF RAY ANGLE IN IMA SPACE %Gt .m  
`jwa<N4e@  
!和取最?視角度 v\p;SwI   
^f# F I&  
MAXFIELD = MAXF() &|/| ''A)  
L!=QR8?@E  
IF (MAXFIELD = = 0.0) THEN maxfield = 1.0 V}\~ugN)y  
8 Y5  
!獲曲面數 ! u:Weoz  
,"B+r6}EF  
n = NSUR() ]Kr `9r),  
) /z@vY  
!主光線錐激 G "73=8d  
J~#;<e{\"  
&^W|iXi#  
AhZ8 0!  
RAYTRACE  0, 1, 0, 0, PWAV() HD(.BW7  
m=YU2!Mb  
1Tu *79A  
qh`t-  
MAX_RIH = RAYY(n) F&&$Qn_+  
#5V9o KM  
9uGrk^<t  
=jN *P?  
MAX_CRA = ACOS(RAYN(n-1))*180/3.1416   'nF2aD%A  
St e=&^  
9/nn)soC3  
rwGKfoKI  
PRINT   "MAX FIELD: " , maxfield, )oyIe)  
TA{\PKA)  
PRINT   "MAX_CHIEF_RAY_ANGLE : " , MAX_CRA , '0FhL)x?"T  
U~pV)J  
PRINT   "MAX CHIEF RAY HEIGHT ON IMA : ",MAX_RIH 1Z9qjV%^  
#5f-`~^C{  
Z l*X?5u  
ELj\[&U  
PRINT "FIELD ",   hNzB4 p  
78/Zk}I]  
PRINT "CHIEF RAY ANGLE," 0zi~p>*nJC  
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PRINT  "CHIEF RAY HEIGFHT ON IMA." L31HG H2l  
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NgY =&W,  
7(5 4/  
!將最大試廠分為20, 設置試場數據步常  FHY=j/20 , for g[3)P+  
(vTtDKp@  
l g-X:Z.  
L|,!?cSAT  
For j, 0, 20, 1 9M:O0)s  
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FHY = j/20 ;A*`e$  
ia MUsa{  
RAYTRACE  0, FHY, 0, 0, PWAV() |>#{[wko  
~/P&Tub^  
CRA = ACOS(RAYN(n-1))*180/3.1416 |cR;{Z8?_  
GSRVe/ [  
OLDX = CRA 4y+< dw  
uH(f$A  
OLDY = RAYY(n) 07[_.i.l  
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PRINT FHY*maxfield, " ,  ",OLDX,"   ",OLDY sIxTG y.  
C<P%CG&;  
q,+yqrt  
3J5!oF{H  
fP. 6HF_p_  
HbxL:~:}J  
NEXT 1A\OC  
|fHV2Y`:g  
!繪出該點數據 s mnS DS  
/@,j232  
!s&NT @ S  
LS917ci-  
GRAPHICS _iir<}  
P9vA7[  
]VD|xm:kj  
ayfFVTy1d  
RAYTRACE 0, 1, 0, 0, PWAV() yp({>{u7  
/|D*w^ >  
MAX_RIH = RAYY(n) <x<"n t  
~W{2Jd  
MAX_CRA = ACOS(RAYN(n-1))*180/3.1416 |563D#?cR  
;oy-#p>N%  
X_WIDE = XMAX() /1.6 L{8xlx`  
Mw;sLsu  
Y_WIDE = YMAX() /1.6 BBtzs^C|  
'7i Sp=  
SEGMENT = 40 G{6;>8h  
<psZQdH  
FOR j, 0, SEGMENT-1,1 6x%h6<#xh*  
k@z,Iq8  
Bq$rf < W  
%@6}GmK^  
FHY = j/SEGMENT 1q,{0s_kp  
/N&CaH\;^$  
RAYTRACE 0, FHY, 0, 0, PWAV() 3U7 *>H  
_Nqt21sL  
CRA = ACOS(RAYN(n-1)) * 180/3.1416 U 15H2-`  
~NpA".PB  
RIH = RAYY(n) Vz=PiMO  
Yi$vg  
OLDX = X_WIDE * (CRA/MAX_CRA+2/20) ]%IT|/;9Y  
;{KV /<3  
OLDY = Y_WIDE * (1-RIH/MAX_RIH+2/20) k\NMy#]Zt  
Qm Ce>+  
Ht&:-F+dm  
4X+ifZO  
O vk_\On  
}x07^4$j  
RAYTRACE  0, (FHY+1/SEGMENT),  0,  0, PWAV() V0P>YQq9s  
4\4FolsK  
CRA = ACOS(RAYN(n-1)) * 180/3.1416 Cd>WUw  
p(/dBt[3k  
RIH = RAYY(n)       #0y)U;dA+w  
}M H0L#Tu  
NEWX = X_WIDE * (CRA/MAX_CRA+2/20) f7X6fr<  
z <##g  
NEWY = Y_WIDE * (1-RIH/MAX_RIH+2/20) 8[E!E)4M  
FXi{87F2  
2kIa*#VOJ  
Stu4t==U  
LINE OLDX,OLDY,NEWX,NEWY 8j=}u/T@F  
/XzH?n/{R  
A$ = "0" Ljx(\Cm  
)~HUo9K9  
gtext OLDX,OLDY,0, A$ ucG@?@JENm  
b"vv>Q~U  
NEXT !U'QqnT  
X;"Sx#U  
gtext NEWX, NEWY, 0 " THE LAST DATA POINT " woD>!r>)  
-\ EP.Vtz  
gtitle "THE CHIREF RAY ANGLE VS. REAL IMAGE HEIGHT" {6_|/KE9_  
NqDHCI  
(4dhuT  
3G`aHTWk  
GRAPHICS OFF j]aoR  
2&^]k`Aj6D  
rcasiodu 2011-12-27 09:57
楼上的回答非常棒!谢谢分享这个宏! "hzB9*"t  
还有一个简单的方法,利用Ray Trace观察主光线的ray angle以及imaging height,参见http://www.opticsky.cn/read-htm-tid-51156.html,此贴介绍的很详细。 /oR<A  
paradism 2012-07-01 12:51
学习了!!!!!!!!!!!!!!!!!!!
wanglu079 2012-07-09 13:04
用宏指令看
qq41894351 2012-07-09 22:26
什么事宏指令
oxfocean 2019-05-14 17:10
这个宏厉害了
1014597 2019-05-15 06:04
这个宏厉害了
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