[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 Kc1w[EQ  
function sjr=nfdre(~) <~v4BiQ3l^  
(=gqqOOl~  
%系统焦距及各镜间距输入，间距取负正负 YbWz!.WPe  
&+oJPpHi\  
f=input('f:'); Y4,p_6aKJ]  
d1=input('d1:'); .v+J@Y a  
d2=input('d2:'); 4
z~;4  
d3=input('d3:'); &\6(iL  
e(1{W P  
A=f^2/(d3*d2)-f/d1; VTDnh*\5  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); 7OXRR)]V  
C=d3/d2-f/d1; 3]'h(C  
LSXsq}  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 |rhB@k  
a2=d3/(a1*f);%α2 lY,9bSF$  
b2=a1*(1-a2)*f/d2;%β2 Y}yh6r;i  
b1=(1-a1)*f/(d1*b2);%β1 gr.G']9lNq  
:l Z\=2D  
?aTC+\=  
%曲率半径 ^n4aoj  
-\ew,y  
R1=2*f/(b1*b2) ) 54cG  
R2=2*a1*f/(b2*(1+b1)) KqaEHL  
R3=2*a1*a2*f/(1+b2) qf[J-"o  
G2c\"[N1/  
A1=b2^3*(a1-1)*(1+b1)^3; 7VkjnG^!:  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; _wW"Tn]  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; |Df`Aq(eYJ  
Z~|%asjFE  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); TDg<&ND3  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); E J6|y'  
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); L!ms{0rJ  
s6 K~I  
CB=[C1 B1;C2 B2]; Q4N0j' QA  
AB=[A1 B1;A2 B2]; B4m34)EOE  
AC=[A1 C1;A2 C2]; @fVz *  
!|ic{1!_  
%非球面系数 +f_3JL$  
k2=-(det(CB)/det(AB)); r>"l:GZ  
k3=-(det(AC)/det(AB)); 'Q*lp!2>  
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 4Xn-L&0z  
k2=k2 EWVn*xl?  
k3=k3 /B{cL`<  
[
e:ccm  
end AVc|(~V  
W7T2j+]  
%有中间像，焦距输入为正数 \[B#dw#  
i(q a'*  
function sjr=yfdre(~) akgvV~
5  
3%N!omAe  
f=input('f:'); "!Hm.^1  
d1=input('d1:'); WO+>W+|N  
d2=input('d2:'); `
n e9&+  
d3=input('d3:'); %IUTi6P l  
8
..g\ZT  
A=f^2/(d3*d2)-f/d1; D;DI8.4`N  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); #CB`7}jq  
C=d3/d2-f/d1; 09Z\F^*$F  
3.?oG5P#  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); h61BIc@>  
a2=d3/(a1*f); 6a{b%e`  
b2=a1*(1-a2)*f/d2; C7 9~@%T  
b1=(1-a1)*f/(d1*b2); )OQih+#?W  
P[Id[}5Pw  
%曲率半径 :C#(yp  
*{e,< DV  
R1=2*f/(b1*b2) j5 W)9HW:  
R2=2*a1*f/(b2*(1+b1))
$\nAGmp@  
R3=2*a1*a2*f/(1+b2) l9NE
T  
>#xIqxV,  
A1=b2^3*(a1-1)*(1+b1)^3; rPJbbV",+^  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; z
"{Ji{>%=  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; (n@&M!a  
F8*P/<P1cK  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); {
%af  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); R%r<AL5kJk  
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); +~EFRiP]  
a0B,[i  
CB=[C1 B1;C2 B2]; _^]:tL6  
AB=[A1 B1;A2 B2]; hr GfA  
AC=[A1 C1;A2 C2]; xJE26i  
f\vg<lca  
%二次系数 :c&F\Q=  
1Qo2Z;h@  
k2=-(det(CB)/det(AB)); u-X
P`  
k3=-(det(AC)/det(AB)); /y5a~3
  
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 2?9gf,U  
k2=k2 2E=vMAS  
k3=k3 f`,isy[  
zVtNT@1K>u  
end

谢谢分享，学习一下 vip& b}u