[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 e&f9/r
fx  
function sjr=nfdre(~) ~<Z;)e  
<=8REA?  
%系统焦距及各镜间距输入，间距取负正负 ZSq7>}  
7QP%Pny%  
f=input('f:'); Tl=cniy]  
d1=input('d1:'); tS,nO:+x  
d2=input('d2:'); brJ_q0@  
d3=input('d3:'); 5k.NZ  
W HO;;j  
A=f^2/(d3*d2)-f/d1; N+x0"~T}I  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); kf+]bV  
C=d3/d2-f/d1; kH1hsDe|&y  
]Mi ~vG q  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 %3scz)4$  
a2=d3/(a1*f);%α2 vJDK]p<}  
b2=a1*(1-a2)*f/d2;%β2 24"Trg\WK[  
b1=(1-a1)*f/(d1*b2);%β1 {4Y@DQ-  
!7!xJ&/V  
\]}|m<R  
%曲率半径 {.$5:<8aC  
y0>as
l  
R1=2*f/(b1*b2) tWQ_.,ld  
R2=2*a1*f/(b2*(1+b1)) LziEF-_  
R3=2*a1*a2*f/(1+b2) }$3eRu +  
%se4aeOrX  
A1=b2^3*(a1-1)*(1+b1)^3; L<!}!v5ja  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; ~n%~ Z|mMF  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; 8 $0D-z  
F+Rto
q|  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); X=_pQ+j`^  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); j*>+^g\Q6  
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); h`dtcJ0  
e>~g!S}G  
CB=[C1 B1;C2 B2]; TSqfl/UI  
AB=[A1 B1;A2 B2]; OiX
:h#  
AC=[A1 C1;A2 C2]; ,~8:^*0s  
t m?[0@<s  
%非球面系数 ;vvO#3DWM  
k2=-(det(CB)/det(AB)); /PG+ s6  
k3=-(det(AC)/det(AB)); K[0.4+  
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 jZeY^T)f"  
k2=k2 N&7= hni  
k3=k3 XJy~uks,  
"OF4#a17  
end |m7
U^  
~K}iVX  
%有中间像，焦距输入为正数 R>SS\YC'X  
w naP?|/  
function sjr=yfdre(~) ]|62l+  
)=l~XV  
f=input('f:'); &C<K|F!j!  
d1=input('d1:'); z(2pl}  
d2=input('d2:'); dfY(5Wc+f  
d3=input('d3:'); O=UXe]D  
_@9[c9bO  
A=f^2/(d3*d2)-f/d1; WZO8|hY  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); L;zwqdI  
C=d3/d2-f/d1; *,<A[XP  
WV&T   
a1=(-B-sqrt(B^2-4*A*C))/(2*A); I Y%M5
(&Q  
a2=d3/(a1*f); x8k7y:
 
b2=a1*(1-a2)*f/d2; yG\^PD  
b1=(1-a1)*f/(d1*b2); M#X8Rs1`  
j#QJ5(
#  
%曲率半径 LVKvPi  
-
V0_%Smc  
R1=2*f/(b1*b2) 4-;"w;  
R2=2*a1*f/(b2*(1+b1)) Fw5|_@&k  
R3=2*a1*a2*f/(1+b2) |S.G#za  
O 4zD >O  
A1=b2^3*(a1-1)*(1+b1)^3; Q^X  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; ap=m5h27  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; G2 A#&86J{  
4?Pdld  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); FsQeyh>  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); %B?@le+%  
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); u3 k%  
Cbu/7z   
CB=[C1 B1;C2 B2]; `)V1GR2 ES  
AB=[A1 B1;A2 B2]; S:)Aj6>6  
AC=[A1 C1;A2 C2]; :5Vk+s]8  
K~'!JP8@  
%二次系数 & $E[l'  
F.5'5%  
k2=-(det(CB)/det(AB)); e??tp]PLn  
k3=-(det(AC)/det(AB)); X`i'U7%I  
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 mdjPKrF<  
k2=k2 B1<:nl  
k3=k3 0K/Pth"*  
oSVo~F  
end

谢谢分享，学习一下 _|<kKfd?