[求助]求教CHIEF RAY ANGLE在Zmax中如何可以看出来? [复制链接]

求教CHIEF RAY ANGLE在Zmax中如何可以看出来? 5o 4\Jwt  

我這邊有一個zpl可以直看出來~ DD`Bl1)  
如下: r>4HF"Nm  
!RIH: CHIEF RAY HEIGHT ON IMA qRkY-0vBP  
;i*<HNQ  
!CRA: CHIEF RAY ANGLE IN IMA SPACE RKoM49W  
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!和取最?視角度 0Q?XU.v  
`yYoVu*  
MAXFIELD = MAXF() e
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6-8,qk  
IF (MAXFIELD = = 0.0) THEN maxfield = 1.0 #9-qF9M  
%5#ts/f  
!獲曲面數 'B dZN  
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n = NSUR() N>kY$*  
b&[bfM<  
!主光線錐激 a*?bnw?  
Fk(nf9M%  
Y+vG]?D  
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RAYTRACE  0, 1, 0, 0, PWAV() hKVj\88  
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MAX_RIH = RAYY(n) !C13E lf  
e ]-fb{oVH  
Er /:iO)_  
j#:IG/)GL  
MAX_CRA = ACOS(RAYN(n-1))*180/3.1416   D%[yAr;r  
qiEw[3Za]'  
wq K:=  
r<< ]41  
PRINT   "MAX FIELD: " , maxfield, n h
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PRINT   "MAX_CHIEF_RAY_ANGLE : " , MAX_CRA , z~8`xn,  
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PRINT   "MAX CHIEF RAY HEIGHT ON IMA : ",MAX_RIH d$DNiJ ,  
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p;#@#>h  
X#v6v)c  
PRINT "FIELD ",   p>Z18  
CMu/n]?c  
PRINT "CHIEF RAY ANGLE," `Hlv*" w$  
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PRINT  "CHIEF RAY HEIGFHT ON IMA." ^w|D^F=o  
}EJAC*W,  
ENoGV;WG  
dgbqMu"  
!將最大試廠分為20, 設置試場數據步常  FHY=j/20 , for UdGa#rcNW  
Hs9; &C  
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For j, 0, 20, 1 YCBp]xuE  
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FHY = j/20 {
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RAYTRACE  0, FHY, 0, 0, PWAV() #_(t46  
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CRA = ACOS(RAYN(n-1))*180/3.1416 T:<mme3v  
%imI.6  
OLDX = CRA T)CEcz  
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OLDY = RAYY(n) D MzDV_  
{7F?30: ]  
PRINT FHY*maxfield, " ,  ",OLDX,"   ",OLDY $u"*n\k>  
 b79z<D  
1uwzo9Yg  
`4Db( ~
  
mMt~4(5  
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NEXT =6YffXa_s  
U
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!繪出該點數據 6}/m~m  
;NoD4*  
=w,cdU*  
W)LtnD2 w  
GRAPHICS sUe<21:  
W{!Slf  
zZE@:P&lf  
wJ>.I<F6B  
RAYTRACE 0, 1, 0, 0, PWAV() GZx?vSoHh  
^2f2g>9j_C  
MAX_RIH = RAYY(n) eVvDis  
yt5'2!jc  
MAX_CRA = ACOS(RAYN(n-1))*180/3.1416 Yn>y1~  
@%[ dh@oY  
X_WIDE = XMAX() /1.6 6\5"36&/rQ  
i_'u:P<t  
Y_WIDE = YMAX() /1.6  K0 6 E:  
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SEGMENT = 40 @ak3ZNor  
u^X,ASkQ  
FOR j, 0, SEGMENT-1,1 ,b${3*PPQ  
r1]DkX <6  
o|njgmF;\  
J_@`:l0,z  
FHY = j/SEGMENT fa#5pys  
wK*b2r}0/  
RAYTRACE 0, FHY, 0, 0, PWAV() ;n2b$MB?nM  
L9bIdiB7  
CRA = ACOS(RAYN(n-1)) * 180/3.1416 )BX-Y@fpA  
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RIH = RAYY(n) \0bZ1"  
41s[p56+@  
OLDX = X_WIDE * (CRA/MAX_CRA+2/20) .NX>d@ Kc  
OE8H |?%  
OLDY = Y_WIDE * (1-RIH/MAX_RIH+2/20) N1l&$#Fr!s  
/OsTZ"*.2/  
no UXRQ  
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g\  
|$QL>{81  
;Y@"!\t}  
RAYTRACE  0, (FHY+1/SEGMENT),  0,  0, PWAV() y
^tp^  
(cpaMn@)g  
CRA = ACOS(RAYN(n-1)) * 180/3.1416 !<AY0fpY  
ffibS0aM  
RIH = RAYY(n)       ?]Z EK8c  
eT ZQ[qMp  
NEWX = X_WIDE * (CRA/MAX_CRA+2/20) !vwx0  
'boAv%1_sa  
NEWY = Y_WIDE * (1-RIH/MAX_RIH+2/20) ]>"q>XgnI  
oP`yBX  
YA'_Ba(v)  
wJb"X=i*  
LINE OLDX,OLDY,NEWX,NEWY 1Zi(5S)  
h_?#.z0ih;  
A$ = "0" ?D]qw4J  
lBcRt)_O7  
gtext OLDX,OLDY,0, A$ )WR*8659e  
TkjPa};R  
NEXT [R9!Tz  
1u\kxlZ  
gtext NEWX, NEWY, 0 " THE LAST DATA POINT " vp#r:+=  
^{(i;IVG  
gtitle "THE CHIREF RAY ANGLE VS. REAL IMAGE HEIGHT" m<wEw-1.  
NX5NE2@^qH  
F)Qj<6  
O,A}p:Pgs  
GRAPHICS OFF [ Cu3D
  
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楼上的回答非常棒！谢谢分享这个宏！ TR([u  
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。 <1EmQ)B  

学习了！！！！！！！！！！！！！！！！！！！

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