[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 N8<Wm>GLX~  
function sjr=nfdre(~) L_>j SP  
Z#-:zD7_  
%系统焦距及各镜间距输入，间距取负正负 '?q \mi  
{=(GY@yU/  
f=input('f:'); 1LgzqRq  
d1=input('d1:'); O23dtH  
d2=input('d2:'); \6UK:'5{  
d3=input('d3:'); 1:3I G=  
MX=mGfoa  
A=f^2/(d3*d2)-f/d1; n4albG4  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); ``~7z;E%@  
C=d3/d2-f/d1; E~=`Ac,G2  
:R/szE*Ak  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 +Om(&\c(6  
a2=d3/(a1*f);%α2 yu3T
5@Ww  
b2=a1*(1-a2)*f/d2;%β2 aY^_+&&G  
b1=(1-a1)*f/(d1*b2);%β1 1-@[th  
>x(3p@6p  
HG(J+ocn   
%曲率半径 +="?[:  
\r_-gn'1b  
R1=2*f/(b1*b2) #6])\  
R2=2*a1*f/(b2*(1+b1)) _N4G[jQLJ  
R3=2*a1*a2*f/(1+b2) H /E.R[\+x  
L-7?:  
A1=b2^3*(a1-1)*(1+b1)^3; UjNe0jt%s  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; vT*z3  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; ,a N8`M  
gbP]!d:I  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); ]!yuD/4A  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); K{]!hm,[3  
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); Z J:h]  
?v]EXV3  
CB=[C1 B1;C2 B2]; gO0X-fN8  
AB=[A1 B1;A2 B2]; S>W_p~@  
AC=[A1 C1;A2 C2]; :i
WW2fY  
JXG%Cx!2}  
%非球面系数 +tbG^w%  
k2=-(det(CB)/det(AB)); w1Z9@*C!  
k3=-(det(AC)/det(AB)); C;#-2^h  
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 b?6-lYE>L  
k2=k2 I]HrtI  
k3=k3 !d@qT
.  
c/fU0cA@  
end 3$fzqFo  
X.u&4SH  
%有中间像，焦距输入为正数 V%F^6ds$]0  
o!q3+Pp;}  
function sjr=yfdre(~) Pr |u_^  
3b1;f)t  
f=input('f:'); F iZe4{(p  
d1=input('d1:'); Qh4@Nl#Ncf  
d2=input('d2:'); xgw[)!g^\  
d3=input('d3:'); q;&\77i$  
OhC%5=a7
 
A=f^2/(d3*d2)-f/d1; /U |@sw4  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); gq[|>Rs75  
C=d3/d2-f/d1; H_*;7/&  
Ns?8N":  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); ^Ht!~So  
a2=d3/(a1*f); Gqe?CM  
b2=a1*(1-a2)*f/d2; MeD}S@H  
b1=(1-a1)*f/(d1*b2); ^gP pmb<x  
LxWnPi ^  
%曲率半径 #G'Y2l  
lX^yd5M&f  
R1=2*f/(b1*b2) J$'Q3k  
R2=2*a1*f/(b2*(1+b1)) T$ <l<.Qd  
R3=2*a1*a2*f/(1+b2) jNKu5"HB  
~RlsgtX"  
A1=b2^3*(a1-1)*(1+b1)^3; [1pWg^  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; tO0MYEx"  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; A~MAaw!YE  
s&T"/4  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); gXrPZ|iS  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); X rBe41  
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); *G'R+_tdE  
={HYwP;  
CB=[C1 B1;C2 B2]; uB;\nj5'D  
AB=[A1 B1;A2 B2]; ^[]q/v'3m!  
AC=[A1 C1;A2 C2]; Nqf6CPXE  
mMp(  
%二次系数 k!9LJ%Xh  
P0R8 f  
k2=-(det(CB)/det(AB)); PtfxF]%H  
k3=-(det(AC)/det(AB)); m tPmVze  
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 :Q~Rb<']{x  
k2=k2 J.nJ@?O+  
k3=k3 "x;|li3;  
BU3VXnqT[  
end

谢谢分享，学习一下 UbWeE,T~S