[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 5mSXf"R^  
function sjr=nfdre(~) jv2l_  
yWzvE:!)  
%系统焦距及各镜间距输入，间距取负正负 qgZ(o@\  
QQ2xNNF[  
f=input('f:'); @Jm$<E  
d1=input('d1:'); DP;B*s4{U  
d2=input('d2:'); Slv}6at5  
d3=input('d3:'); :\80*[=;Z  
1MB  
A=f^2/(d3*d2)-f/d1; "Wi`S;  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); S<p "k]  
C=d3/d2-f/d1; DVSYH{U4  
?'T>/<(  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 00;=6q]TA  
a2=d3/(a1*f);%α2 $6y1';A  
b2=a1*(1-a2)*f/d2;%β2 D<xPx  
b1=(1-a1)*f/(d1*b2);%β1 .\U+`>4av  
I">z#@CT  
9!9 Gpi  
%曲率半径 {+C>^b  
6Ir ?@O1'!  
R1=2*f/(b1*b2) Q8:u1$
}  
R2=2*a1*f/(b2*(1+b1)) VKN^gz  
R3=2*a1*a2*f/(1+b2) h)y"?Jj  
ORtl~V'  
A1=b2^3*(a1-1)*(1+b1)^3; TP^.]IO-  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; W;KHLHp-  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; y~-?   
Mm6 (Q  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); W{-N,?z  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); ; $y.+5 q  
C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); sgCIY:8  
8:t-I]dzk  
CB=[C1 B1;C2 B2]; -CL
7^  
AB=[A1 B1;A2 B2]; d9Q%GG0]  
AC=[A1 C1;A2 C2]; E#HO0]S  
E0)v;yRcw  
%非球面系数 p^L6uM  
k2=-(det(CB)/det(AB)); t#P)KcWOt  
k3=-(det(AC)/det(AB)); x7NxHTL  
k1=(k2*a1*b2^3*(1+b1)^3-k3*a1*a2*(1+b2)^3+a1*b2^3*(1+b1)*(1-b1)^2-a1*a2*(1+b2)*(1-b2)^2)/(b1^3*b2^3)-1 mDM]RAub)  
k2=k2 @Rx/]wyH  
k3=k3 ?<nz2 piP,  
}>Os@]*'^(  
end KO5
Q;H  
Kjf#uU.7  
%有中间像，焦距输入为正数 RisrU  
J|'T2g  
function sjr=yfdre(~) ;bA
y7  
Y^6=_^  
f=input('f:'); 5X`.2q=d  
d1=input('d1:'); D6ck1pxkx  
d2=input('d2:'); zM@iG]?kc  
d3=input('d3:'); =
} vG|  
4uu*&B  
A=f^2/(d3*d2)-f/d1; B' :ZX-Q)  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); mR0@R;
,p  
C=d3/d2-f/d1; T.(SBP  
Jhj]rsGk  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); Jp= (Q]ab  
a2=d3/(a1*f); o&CvjE  
b2=a1*(1-a2)*f/d2; 94a_ W9  
b1=(1-a1)*f/(d1*b2); ZDVaKDqZ_  
hqBwA1](a  
%曲率半径 1i>)@{P&BN  
S((8DSt*  
R1=2*f/(b1*b2) }Ns_RS$  
R2=2*a1*f/(b2*(1+b1)) ~(&xBtg:}  
R3=2*a1*a2*f/(1+b2) f a\cLC  
/NkZ;<uxJ  
A1=b2^3*(a1-1)*(1+b1)^3; ]3I_H+hU  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; T4f:0r;^f*  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; RV($G8U  
}>OE"#si  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); >)5vsqGZaK  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); ~z'0~3  
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); Tl1?5  
=k8
A7P  
CB=[C1 B1;C2 B2]; P,%|(qB
  
AB=[A1 B1;A2 B2]; PAc~p8S  
AC=[A1 C1;A2 C2]; d@l;dos),  
8ziYav  
%二次系数 %idnm  
!l\pwfXP&%  
k2=-(det(CB)/det(AB)); _ Po9pZ  
k3=-(det(AC)/det(AB)); :heJ5*!,  
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 (%r:PcGMEV  
k2=k2 9z-"JnM  
k3=k3 3n/L;T,X  
w%ip"GT,  
end

谢谢分享，学习一下 lu{}j4