[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 fgNEq  
function sjr=nfdre(~) n[<Vj1n  
| .PLfc;  
%系统焦距及各镜间距输入，间距取负正负 W6Hiqu+  
+f+\uObi:  
f=input('f:'); )Aj~ xA  
d1=input('d1:'); F](kU#3"S  
d2=input('d2:'); %9IM|\ulp  
d3=input('d3:'); ?wmr~j  
hHdC/mR  
A=f^2/(d3*d2)-f/d1; E|No$QO
)  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); +s`n]1HC  
C=d3/d2-f/d1; $ _ gMJ\{  
ruoiG?:T  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 2D "mq~V  
a2=d3/(a1*f);%α2 .; :[sv)  
b2=a1*(1-a2)*f/d2;%β2
ce@(Ct  
b1=(1-a1)*f/(d1*b2);%β1 6,k}v:  
L./
UgeZ  
rK];2[U  
%曲率半径 $Cte$jg{;  
7[Y<5T]  
R1=2*f/(b1*b2) u ;I5n  
R2=2*a1*f/(b2*(1+b1)) &D/_@\ 0  
R3=2*a1*a2*f/(1+b2) TK'(\[E  
1<lLE1fk  
A1=b2^3*(a1-1)*(1+b1)^3; J|s4c`=  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; KnlVZn[3t  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; ^58'*13ZL  
4%B${zP(.}  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); Ix"uk6 h  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); p3X>  
C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); N~SG=\rP;o  
R:11w#m7w  
CB=[C1 B1;C2 B2]; vQztD_bX%  
AB=[A1 B1;A2 B2]; JI(8{ f  
AC=[A1 C1;A2 C2]; zL1H[}[z+  
 F`f#gpQ  
%非球面系数 &FGz53fd4  
k2=-(det(CB)/det(AB)); yt$V<8a  
k3=-(det(AC)/det(AB)); kpEES{f  
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 Aj-}G^>#  
k2=k2 X=-pNwO   
k3=k3 \3x,)~m  
^oM*f{9  
end 'H:lR1(,  
Z?X ^7<  
%有中间像，焦距输入为正数 pS9CtQqvgy  
B2VUH..am  
function sjr=yfdre(~) jRzR`>5  
]x! vPIyq  
f=input('f:'); amOBUD5Ld`  
d1=input('d1:'); "h\{PoG  
d2=input('d2:'); ^BW V6  
d3=input('d3:'); ]e 81O#t3  
Bx2E9/S3  
A=f^2/(d3*d2)-f/d1; }wz )"  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); u.R:/H<>~  
C=d3/d2-f/d1; J=5G<  
J %URg=r  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); $}N'm  
a2=d3/(a1*f); @:X~^K.  
b2=a1*(1-a2)*f/d2; gQh;4v  
b1=(1-a1)*f/(d1*b2); 3%>"|Y
e}A  
76(&O  
%曲率半径 pJ1G
B  
TAIcp*)ZM  
R1=2*f/(b1*b2) 4>gkXfTF  
R2=2*a1*f/(b2*(1+b1)) <5G*#0gw  
R3=2*a1*a2*f/(1+b2) @zW'!Ol  
=DUsQN!  
A1=b2^3*(a1-1)*(1+b1)^3; R2-OT5Ej  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; s9zdg"c'  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; JuKj  
PKty'}
KF  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); -(@dMY  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); K'7i$bl%  
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); Kmk<  
8}nA8J  
CB=[C1 B1;C2 B2]; #P<v[O/rA  
AB=[A1 B1;A2 B2]; Hi|'  
AC=[A1 C1;A2 C2]; C 3b  
RV.*_FG  
%二次系数 -%NT)o  
xLPyV&j-  
k2=-(det(CB)/det(AB)); ;q59Cr75  
k3=-(det(AC)/det(AB)); Ay22-/C|@  
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 W1i
Kn  
k2=k2 $*{PUj  
k3=k3 /)e&4.6  
~W_m<#K(  
end

谢谢分享，学习一下 VX2KE@