[原创]在框架结构确定的情况下，基于matlab的消四种像差的三反系统初始结构的求解 [复制链接]

%无中间像，焦距输入为负数 #y?iUv  
function sjr=nfdre(~) E\X:VQ9  
z-Ndv;:  
%系统焦距及各镜间距输入，间距取负正负 Q;'{~!=  
k3w(KH@  
f=input('f:'); {e1akg.  
d1=input('d1:'); Aw
C"c '  
d2=input('d2:'); Q`ALyp,9b  
d3=input('d3:'); '/ >7pB  
c(R=f+  
A=f^2/(d3*d2)-f/d1; {cW%i:  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); -/7[\S  
C=d3/d2-f/d1; L
PDx3MS  
zj+.MG04  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 15/lX  
a2=d3/(a1*f);%α2 c^?+"7oO0  
b2=a1*(1-a2)*f/d2;%β2 A:?|\r  
b1=(1-a1)*f/(d1*b2);%β1 Q.$|TbVfds  
nKO4o8js{{  
-D4"uoN
.  
%曲率半径 :d!qZFln  
soTmKqj E  
R1=2*f/(b1*b2) lo!.%PP|  
R2=2*a1*f/(b2*(1+b1)) RAh4#8]  
R3=2*a1*a2*f/(1+b2) N1vPY]8  
T08SGB]  
A1=b2^3*(a1-1)*(1+b1)^3; v{T%`WuPRf  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; FthrI  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; &.ilku/  
ZliJc7lss  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); 'Pm.b}p<  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); Ei Yj`P  
C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); 9 :ubPqt  
Q,`:RF3  
CB=[C1 B1;C2 B2]; KH~o0 W  
AB=[A1 B1;A2 B2]; j-R9=vB2  
AC=[A1 C1;A2 C2]; l, [cR?v  
!L=RhMI  
%非球面系数 DMcH, _(  
k2=-(det(CB)/det(AB)); &6#>a"?"  
k3=-(det(AC)/det(AB)); *MG*]\D  
k1=(k2*a1*b2^3*(1+b1)^3-k3*a1*a2*(1+b2)^3+a1*b2^3*(1+b1)*(1-b1)^2-a1*a2*(1+b2)*(1-b2)^2)/(b1^3*b2^3)-1 ma%PVz`I;9  
k2=k2 7SyysH<H  
k3=k3 gpe/dfyJ9  
{wNNp't7  
end "3r7/>xy  
a#k
=! W  
%有中间像，焦距输入为正数 qTA,rr#p0  
B}&9+2M  
function sjr=yfdre(~) ~@Wg3'&  
O2n[`9*  
f=input('f:'); z^etH/]Sy  
d1=input('d1:'); V(^aG=TaW:  
d2=input('d2:'); ; { MK  
d3=input('d3:'); gl$Ks+od  
*\M$pUS{  
A=f^2/(d3*d2)-f/d1; '@WpJ{]A  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); #j?SdQ  
C=d3/d2-f/d1; >B~vE2^tQ~  
u |EECjJn  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); =rB=! ;  
a2=d3/(a1*f); 6M/*]jLq4  
b2=a1*(1-a2)*f/d2; w~n7l97Pw  
b1=(1-a1)*f/(d1*b2); >cV^f6fH  
RY4b<i3  
%曲率半径 /KCJ)0UU  
xyvG+K&  
R1=2*f/(b1*b2) U^@8ebv  
R2=2*a1*f/(b2*(1+b1)) }R\9ybv  
R3=2*a1*a2*f/(1+b2) jO9w7u6  
\GFFPCi4D  
A1=b2^3*(a1-1)*(1+b1)^3; 97]$*&fH  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; ]5_6m;g  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; TA"4yri=7x  
-{=c T?"+  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); $UX^$gG  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); iL](w3EM  
C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); $X;wj5oj  
ifYC&5}SI  
CB=[C1 B1;C2 B2]; =/6rX"\P  
AB=[A1 B1;A2 B2]; n1mqe*Mvs/  
AC=[A1 C1;A2 C2]; +kXj+2  
Q 6)5*o8n  
%二次系数 `PH*tdYrh  
M*xt9'Yd  
k2=-(det(CB)/det(AB)); t]QGyW A]  
k3=-(det(AC)/det(AB)); { yvKUTq`  
k1=(k2*a1*b2^3*(1+b1)^3-k3*a1*a2*(1+b2)^3+a1*b2^3*(1+b1)*(1-b1)^2-a1*a2*(1+b2)*(1-b2)^2)/(b1^3*b2^3)-1 N)&(&2  
k2=k2 <.N337!  
k3=k3 @[vwqPOL  
G=Qslrtg  
end

谢谢分享，学习一下 Y>N`(