[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 K7_UP&`=J  
function sjr=nfdre(~) ohGJ1  
6_GhO@lOG  
%系统焦距及各镜间距输入，间距取负正负 > PRFWO  
/=nJRC3.  
f=input('f:'); 2j[=\K]  
d1=input('d1:'); -:+|zF@f  
d2=input('d2:'); @e.C"@G  
d3=input('d3:'); fgp]x&5Q  
l`lk-nb  
A=f^2/(d3*d2)-f/d1; v!~fs)cdE|  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); U`(ee*}o  
C=d3/d2-f/d1; i &nSh ]KK  
f643#1  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 {L{o]Ii?g  
a2=d3/(a1*f);%α2 mp1@|*Sn  
b2=a1*(1-a2)*f/d2;%β2 XJ;57n-?  
b1=(1-a1)*f/(d1*b2);%β1 g*AWE,%=|  
#jvtUS\  
yLvDMPj  
%曲率半径 jp%S3)  
;WQve_\  
R1=2*f/(b1*b2) 2`K=Hby  
R2=2*a1*f/(b2*(1+b1)) <44G]eb  
R3=2*a1*a2*f/(1+b2) BA:VPTZq  
SwGx?U  
A1=b2^3*(a1-1)*(1+b1)^3; Z"xvh81P  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; Di6?[(8  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; Q~ w|#  
`g=J%p  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); Jq-]7N%k/  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); Naf0)3q>!  
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); q"lSZ; 'E  
,?%Zc$\LW  
CB=[C1 B1;C2 B2]; Ty?cC**  
AB=[A1 B1;A2 B2]; V~3a!-m\  
AC=[A1 C1;A2 C2]; GR_-9}jQP  
 +SU8 +w  
%非球面系数 b{&)6M)zo  
k2=-(det(CB)/det(AB)); p?OoC  
k3=-(det(AC)/det(AB)); By!o3}~g  
k1=(k2*a1*b2^3*(1+b1)^3-k3*a1*a2*(1+b2)^3+a1*b2^3*(1+b1)*(1-b1)^2-a1*a2*(1+b2)*(1-b2)^2)/(b1^3*b2^3)-1 -`h)$&,  
k2=k2 jvL[ JI,b  
k3=k3 Ax7[;|2  
""G'rN_=Bi  
end -uG+BraI  
6<QQ@5_  
%有中间像，焦距输入为正数 ?);v`]  
&U#|uc!+  
function sjr=yfdre(~) sY&IquK^  
Ee! 4xg  
f=input('f:'); .|70;  
d1=input('d1:'); =8. ,43+  
d2=input('d2:'); T.BW H2gRP  
d3=input('d3:'); aB&&YlR=n<  
AQvudx
)@"  
A=f^2/(d3*d2)-f/d1; k="i;! Ge  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); F^;ez/Gl  
C=d3/d2-f/d1; |u<7?)mp  
R&k<AZ  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); m<Dy<((_I  
a2=d3/(a1*f); &j"?\f?  
b2=a1*(1-a2)*f/d2; YjKxb9  
b1=(1-a1)*f/(d1*b2); ",; H`V  
C_JNX9wv  
%曲率半径 '-~~-}= sJ  
k?^z;Tlvw  
R1=2*f/(b1*b2) (hbyEQhF  
R2=2*a1*f/(b2*(1+b1)) }Zn}  
R3=2*a1*a2*f/(1+b2) ]{@-HTt  
$<EM+oJ|ER  
A1=b2^3*(a1-1)*(1+b1)^3; Z@!+v19^  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; pl?`8@dI  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; H<,gU`&R  
<PH#[dH  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2);  x'<X!gw  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); +ye3HGD  
C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); Kn1a>fLaJ_  
W^l-Y%a/o  
CB=[C1 B1;C2 B2]; 1oGw4kD^x  
AB=[A1 B1;A2 B2]; >|UOz&  
AC=[A1 C1;A2 C2]; fuySN!s  
@:#eb1<S  
%二次系数 zQ
d 2  
UZ+<\+q3^  
k2=-(det(CB)/det(AB)); 2P0*NQ  
k3=-(det(AC)/det(AB)); #@Jq~$N|  
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 uKHxe~  
k2=k2 r;N|)  
k3=k3 HG^'I+Yn  
+23xev  
end

谢谢分享，学习一下 4Fr