[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 @1tv/W  
function sjr=nfdre(~) O>Ao#_*hOb  
JB(P-Y#yyA  
%系统焦距及各镜间距输入，间距取负正负 ` EgO&;1D)  
;&U! g&  
f=input('f:'); <:Z-zQp)?  
d1=input('d1:'); YMIX|bj6Y  
d2=input('d2:'); jjU("b=  
d3=input('d3:'); $+Zj)V(  
*pKj6x  
A=f^2/(d3*d2)-f/d1; RMO6kbfP  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); 2:Rxyg@'  
C=d3/d2-f/d1; DgRA\[c  
'qnnZE  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 n;0x\Q|S  
a2=d3/(a1*f);%α2 X)k+BJ  
b2=a1*(1-a2)*f/d2;%β2 iVqa0Gl+}  
b1=(1-a1)*f/(d1*b2);%β1 p'`pO"EO  
Fc.1)yh.  
Sp^jC Xu  
%曲率半径 z\r|5Z  
.qG*$W2f  
R1=2*f/(b1*b2) (H-
Y-Lk+  
R2=2*a1*f/(b2*(1+b1)) =VM4Q+'K  
R3=2*a1*a2*f/(1+b2) 3FpSo+  
$][$ e  
A1=b2^3*(a1-1)*(1+b1)^3; (#%R'9Rv  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; U8s&5~IPn  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; ju%t'u\'  
PXJ`<XM  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); Y
MTB4|{  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); Mg}8 3kS  
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); _M9-n  
1pHt3Vc(G  
CB=[C1 B1;C2 B2]; K_t!P  
AB=[A1 B1;A2 B2]; $jtXNE?  
AC=[A1 C1;A2 C2]; FQGh+.U  
R6/vhze4L2  
%非球面系数 sPUn"7  
k2=-(det(CB)/det(AB)); +/Q?<*[  
k3=-(det(AC)/det(AB)); -+w^"RBV  
k1=(k2*a1*b2^3*(1+b1)^3-k3*a1*a2*(1+b2)^3+a1*b2^3*(1+b1)*(1-b1)^2-a1*a2*(1+b2)*(1-b2)^2)/(b1^3*b2^3)-1 hY-;Vh0J  
k2=k2 )Y:9sd8g7  
k3=k3 PDH00(#;+  
y&8kORz;?  
end yqKERdm  
p.J+~s4G  
%有中间像，焦距输入为正数 -)2sR>`A%  
#}U*gVYe  
function sjr=yfdre(~) t:<dirw,o  
H<bK9k)E  
f=input('f:'); XP'7+/A  
d1=input('d1:'); 9Di@r!Db  
d2=input('d2:'); g4fe(.?c,  
d3=input('d3:'); 5\|u] ~b  
h,WF'X+  
A=f^2/(d3*d2)-f/d1; "G,$Sqi@  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); nax(V  
C=d3/d2-f/d1; ;mYj`/Yj  
D%=FCmL5@=  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); 9X6l`bo'  
a2=d3/(a1*f); Wu9))Ir  
b2=a1*(1-a2)*f/d2; l{b*YUsz>  
b1=(1-a1)*f/(d1*b2); lqe71](sK8  
t)hAD_sf  
%曲率半径 @g~hYc  
IU*w'a  
R1=2*f/(b1*b2) GDMg.w4Yk  
R2=2*a1*f/(b2*(1+b1)) J&bhR9sF  
R3=2*a1*a2*f/(1+b2) :?.RZKXQF  
-g;cg7O#(  
A1=b2^3*(a1-1)*(1+b1)^3; f.cQp&&]r  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; [O52Bn  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; !Hp H  
E6FT*}Q  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); 0~0OQ/>7  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); |#6))Dh  
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); bEz1@"~ p  
aHvsgp]  
CB=[C1 B1;C2 B2]; a~nErB  
AB=[A1 B1;A2 B2]; *tD`X(K  
AC=[A1 C1;A2 C2]; D)*OQLHW  
&+Iv"9  
%二次系数 jQc.@^#+x  
k-uwK-B}v+  
k2=-(det(CB)/det(AB)); ?\D=DIN-r  
k3=-(det(AC)/det(AB)); g)@d(EYY  
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 }#h>*+Q  
k2=k2 dY'mY~Tv  
k3=k3 68k  
'8|y^\  
end

谢谢分享，学习一下 K3WhF