[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 $]`rWSYtv`  
function sjr=nfdre(~) ^S%xaA9  
C^^AN~ZD  
%系统焦距及各镜间距输入，间距取负正负 \fD)|  
t|jp]Vp  
f=input('f:'); hzcSKRm  
d1=input('d1:'); =Sp+$:q*  
d2=input('d2:'); [0M`uf/u  
d3=input('d3:'); 4A{6)<e  
y:Ne}S*ncE  
A=f^2/(d3*d2)-f/d1; 7]`l"=/z  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); &`^PO$  
C=d3/d2-f/d1; @yj$  
s,;7m  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 h
%UM<TZ]"  
a2=d3/(a1*f);%α2 L4t(Y7  
b2=a1*(1-a2)*f/d2;%β2 h7a/]~  
b1=(1-a1)*f/(d1*b2);%β1 .:I^O[k  
vBLs88  
6'[gd  
%曲率半径 3n3$?oV  
Z(Z$>P&4  
R1=2*f/(b1*b2) ]]eI80u[  
R2=2*a1*f/(b2*(1+b1)) yE_T#FN  
R3=2*a1*a2*f/(1+b2) F"| ;  
Qu_=K_W  
A1=b2^3*(a1-1)*(1+b1)^3; m>{I>:sq  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; n3" @E<rW  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; K0'= O  
w/,A@fLL  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); 
*ORa@x  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); $g&_7SJ@  
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); ZU68\cL  
<0btw
sv}  
CB=[C1 B1;C2 B2]; vtw97G  
AB=[A1 B1;A2 B2]; }Dx5W9Ri"  
AC=[A1 C1;A2 C2]; aHkt K/  
G{u(pC^  
%非球面系数 +#Ga}eCM  
k2=-(det(CB)/det(AB)); Txa 2`2t
7  
k3=-(det(AC)/det(AB)); fi  
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 Xk?Y  
k2=k2
5h[<!f=  
k3=k3 ^~kfo|  
RHu4cK!5  
end orZwm9#].  
)CoJ9PO7  
%有中间像，焦距输入为正数 y3Ul}mVhA  
?6V U4nK/*  
function sjr=yfdre(~) ,?c=v`e  
X3j|J/  
f=input('f:'); lFV\Go  
d1=input('d1:'); $VJE&b  
d2=input('d2:'); X&gXhr#dL\  
d3=input('d3:'); BmFtRbR  
<Q@{6  
A=f^2/(d3*d2)-f/d1; r"W<1Hu  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); bC,SE*F\  
C=d3/d2-f/d1; us )NgG  
:a*F>S!  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); )(h<vo)-zX  
a2=d3/(a1*f); 49Hgq/uO  
b2=a1*(1-a2)*f/d2; asL!@YE  
b1=(1-a1)*f/(d1*b2); 5 J 7XVe>  
RPZ -  
%曲率半径 98Im/v  
"u3 N9  
R1=2*f/(b1*b2) F-<c.0;6  
R2=2*a1*f/(b2*(1+b1)) BWLeitS/  
R3=2*a1*a2*f/(1+b2) ]D@y""{--s  
ZW ZKyJQ  
A1=b2^3*(a1-1)*(1+b1)^3; -c p)aH)  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; 1i 7p'  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; N6h.zl&04  
RJE<1!{  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); 
DG/<#SCF  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); Q32GI,M%B  
C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); 66'AaA;0^i  
rp!>rM] s  
CB=[C1 B1;C2 B2]; J\Z\q  
AB=[A1 B1;A2 B2]; tRXR/;3O  
AC=[A1 C1;A2 C2]; vJg^uf)  
fH 0&Wc3yC  
%二次系数 0kL tL!3  
V&>mD"~MP  
k2=-(det(CB)/det(AB)); XB+Juk&d  
k3=-(det(AC)/det(AB)); bX`VIFc  
k1=(k2*a1*b2^3*(1+b1)^3-k3*a1*a2*(1+b2)^3+a1*b2^3*(1+b1)*(1-b1)^2-a1*a2*(1+b2)*(1-b2)^2)/(b1^3*b2^3)-1 3M[5_OK  
k2=k2 J72YZrc  
k3=k3 r~sQdf  
YA@OA$`E  
end

谢谢分享，学习一下 h7[PU^m