[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 Oq*a4_R'YV  
function sjr=nfdre(~) Kl(}s{YFn.  
_=UXNr8S  
%系统焦距及各镜间距输入，间距取负正负 OTN"XKa$  
g\rujxHlH  
f=input('f:'); A"vI6ud>  
d1=input('d1:'); `"GD'Oa  
d2=input('d2:'); wLkHU"'  
d3=input('d3:'); x~Se-#$  
#xYkG5`lm  
A=f^2/(d3*d2)-f/d1; dMRwQejY{7  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); QHP^1W`  
C=d3/d2-f/d1; aFIet55o  
lCd^
|E  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 =\CbX  
a2=d3/(a1*f);%α2 wKk  
b2=a1*(1-a2)*f/d2;%β2 h=`rZC  
b1=(1-a1)*f/(d1*b2);%β1 [0/?(i|  
)I1LBv
fQ  
<Y~V!9(~{Q  
%曲率半径 rp=?4^(u  
jG)>{D  
R1=2*f/(b1*b2) G|3OB:  
R2=2*a1*f/(b2*(1+b1)) JC>}(yQA  
R3=2*a1*a2*f/(1+b2) C$ZY=UXz!T  
EnwiE  
A1=b2^3*(a1-1)*(1+b1)^3; (e F5?I  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; -BEPpwb<g  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; gMHH3^\VH)  
QXXcJc~  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); I62Yg p$K  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); uA}asm  
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); C$,S#n@  
2GZUMXK  
CB=[C1 B1;C2 B2]; Aqp3amW!  
AB=[A1 B1;A2 B2]; u6(7#n02  
AC=[A1 C1;A2 C2]; K
VQZ  
BOh&Db*  
%非球面系数 fC~WuG3  
k2=-(det(CB)/det(AB)); w`!Yr:dU  
k3=-(det(AC)/det(AB)); f3v/Y5)  
k1=(k2*a1*b2^3*(1+b1)^3-k3*a1*a2*(1+b2)^3+a1*b2^3*(1+b1)*(1-b1)^2-a1*a2*(1+b2)*(1-b2)^2)/(b1^3*b2^3)-1 >vP^l {SD  
k2=k2 N3x}YHFF  
k3=k3 K.X%
Q,XD  
k{@z87+&  
end SxOM@A  
vP,WV9Q1u  
%有中间像，焦距输入为正数 [oKB1GkA  
=#y&xWxL  
function sjr=yfdre(~) |/p^e  
)'fIrBT  
f=input('f:'); %>JqwMK  
d1=input('d1:'); \u)s Zh  
d2=input('d2:'); f5sk,Z  
d3=input('d3:'); OW #pBeX99  
r@ejU'uz  
A=f^2/(d3*d2)-f/d1; P!]DV$
o  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); D_, 2z  
C=d3/d2-f/d1; _a.Q@A4'  
j&44wuf  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); iqOd]H]v  
a2=d3/(a1*f); Ge97e/CY  
b2=a1*(1-a2)*f/d2; aZBaIl6I  
b1=(1-a1)*f/(d1*b2); y<;#*wB  
}*BY!5  
%曲率半径 nk-?$'i9q  
i5_gz>  
R1=2*f/(b1*b2) xNm<` Y?  
R2=2*a1*f/(b2*(1+b1)) B*?v`6  
R3=2*a1*a2*f/(1+b2) 3J:!8Gmk  
kM9E)uT>(<  
A1=b2^3*(a1-1)*(1+b1)^3; Y$,]~Qzq  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; &}P62&  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; koAc-o  
sS+9ly{9J  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); -M/ny-;`}  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); 43P?f+IYrk  
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); g(<@r2p  
Xw![}L>  
CB=[C1 B1;C2 B2]; *_^AK=i  
AB=[A1 B1;A2 B2]; 0}w>8L7i{  
AC=[A1 C1;A2 C2]; .|o7YTcR
:  
dc:|)bK M  
%二次系数 o3uv"# C  
P/ug
'  
k2=-(det(CB)/det(AB)); ?MN?.O9-  
k3=-(det(AC)/det(AB)); "lUw{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 ? ZN8Ku  
k2=k2 %=_Iq\lC  
k3=k3 *;4r|#LG  
Nf3Kz#!B  
end

谢谢分享，学习一下 JQ0KXS Nr