[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 gy =`cMS@  
function sjr=nfdre(~) bOz\-=au  
<uTsXv  
%系统焦距及各镜间距输入，间距取负正负 peqFa._W  
Ic=V:  
f=input('f:'); 2FVKgyV  
d1=input('d1:'); 8&C(0H]1  
d2=input('d2:'); +~fu-%,k  
d3=input('d3:'); (Z"Xp{u  
ESrWRO f9  
A=f^2/(d3*d2)-f/d1; E7eVg*Cvi  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); &HXSO,@  
C=d3/d2-f/d1; fd,~Yj$R?  
g?$9~/h :;  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 >Ed^dsb&  
a2=d3/(a1*f);%α2 Z
],"<[E  
b2=a1*(1-a2)*f/d2;%β2 FoG<$9  
b1=(1-a1)*f/(d1*b2);%β1 >rFvT>@NU  
\34|9#*z-  
BkZ%0rw%  
%曲率半径 F>b6fUtR  
-KNJCcBJ  
R1=2*f/(b1*b2) E7h}0DX  
R2=2*a1*f/(b2*(1+b1)) R$@.{d&:w  
R3=2*a1*a2*f/(1+b2) |TUpv*pq  
{PVu3W  
A1=b2^3*(a1-1)*(1+b1)^3; :> q?s  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; G2^DukK.  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; #] GM#.  
j>b OnCp~  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); ]s -6GT  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); `P5"5N\h  
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); u9gr@06  
XGoy#h  
CB=[C1 B1;C2 B2]; {;}8Z$  
AB=[A1 B1;A2 B2]; $($SQZK&  
AC=[A1 C1;A2 C2]; ~+np7  
Vx*q'~4y!|  
%非球面系数 ;dFe >`~  
k2=-(det(CB)/det(AB)); $ vjmW! O  
k3=-(det(AC)/det(AB)); \Cs<'
(=  
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 5VTbW   
k2=k2 2bJFlxEU  
k3=k3 *Z:PB%d5  
E nvs[YZe  
end !:t9{z{Ixg  
'oHR4O*  
%有中间像，焦距输入为正数 biG9?  
Xdq, =;  
function sjr=yfdre(~) mCGcM^21-x  
vRq xZN  
f=input('f:'); bevT`D  
d1=input('d1:'); `-H:j:U{  
d2=input('d2:'); C#~MR+;  
d3=input('d3:'); 5q ,  
<C$<(Dw5  
A=f^2/(d3*d2)-f/d1; >&&xJ5  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); -"zu"H~t4  
C=d3/d2-f/d1; }SV3PdE  
4cQ5E9  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); zFtGc  
a2=d3/(a1*f); X4$e2f  
b2=a1*(1-a2)*f/d2; /=@vG Vp6  
b1=(1-a1)*f/(d1*b2); JLu0;XVK  
+I<Sq_-  
%曲率半径 c Vn+~m_%  
hQm4R]a  
R1=2*f/(b1*b2) V8yX7yx  
R2=2*a1*f/(b2*(1+b1)) B0i}Y-Z  
R3=2*a1*a2*f/(1+b2) >y9o&D  
lAk1ncx  
A1=b2^3*(a1-1)*(1+b1)^3; 'u[o`31.  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; fqb$_>3Ol  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; }BJ1#<  
j>*SJtq7  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); nEm7&Gb  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); RC(D=6+[C  
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); ge?-^s4M  
%20-^&zZ  
CB=[C1 B1;C2 B2]; V@:=}*E  
AB=[A1 B1;A2 B2]; XQOprIJ U  
AC=[A1 C1;A2 C2]; i8YgG0[)  


-mJ&N  
%二次系数 }qv-lO  
dCP Tpm  
k2=-(det(CB)/det(AB)); 6B/"M-YME  
k3=-(det(AC)/det(AB)); -^H5z+"^  
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{0-H+  
k2=k2 O{#Cddt:r  
k3=k3 Nox
z kpMF  
23$hwr&G\  
end

谢谢分享，学习一下 R%_H\-wo