[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 rtE,
SN  
function sjr=nfdre(~) c$fM6M }  
/Ql}jSKi  
%系统焦距及各镜间距输入，间距取负正负 L{p-'V  
l:f sZO4  
f=input('f:'); 3vU (4}@  
d1=input('d1:'); j]&{ @Y  
d2=input('d2:'); Q~_x%KN/`  
d3=input('d3:'); oD\+ 5[x  
;xwQzu%M>5  
A=f^2/(d3*d2)-f/d1; V}qmH2h  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); 9To6Rc;  
C=d3/d2-f/d1; @1@WB]mQQ  
m@~x*+Iz
 
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 )zo ;r!eP  
a2=d3/(a1*f);%α2 cC.DBYV+-  
b2=a1*(1-a2)*f/d2;%β2 2_X0Og8s[  
b1=(1-a1)*f/(d1*b2);%β1 "g-
NUl`'  
y>8?RX8  
<@u6*]  
%曲率半径 gBu4`M  
jq{
Ix  
R1=2*f/(b1*b2) >B7OTGw  
R2=2*a1*f/(b2*(1+b1)) 9MxGyGz$  
R3=2*a1*a2*f/(1+b2) YyIt-fPZ  
A4
' aB0^  
A1=b2^3*(a1-1)*(1+b1)^3; {IWb:p#I]  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; )3sb2 #  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; Kq$1lPI  
6! 'Xo:p  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); NY x4& *le  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); =Frr#t!(w0  
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); @H
I5;z  
S}
X:LHr*  
CB=[C1 B1;C2 B2]; 2|BE{91  
AB=[A1 B1;A2 B2]; sa*-B  
AC=[A1 C1;A2 C2]; tO7{g  
rej[G!  
%非球面系数
uE(w$2Wi  
k2=-(det(CB)/det(AB)); '! (`?  
k3=-(det(AC)/det(AB)); 1~
Nz6  
k1=(k2*a1*b2^3*(1+b1)^3-k3*a1*a2*(1+b2)^3+a1*b2^3*(1+b1)*(1-b1)^2-a1*a2*(1+b2)*(1-b2)^2)/(b1^3*b2^3)-1 M .,|cx  
k2=k2 z/b*]"g,  
k3=k3 =xoTH3/,>  
)f Rh^6  
end {y'kwU  
&kvVMnok  
%有中间像，焦距输入为正数 Gj=il-Po  
srL,9)OC  
function sjr=yfdre(~) D#0}/  
zVu}7v()  
f=input('f:'); V 6F,X`7  
d1=input('d1:'); q9Q4F  
d2=input('d2:'); ;q
Z2V  
d3=input('d3:'); @k
w=0  
kel48B  
A=f^2/(d3*d2)-f/d1; #LiC@>  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); F/8y p<_r  
C=d3/d2-f/d1; #7{a~-S  
N*fN&0r  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); @ 55Y2  
a2=d3/(a1*f); M4| L  
b2=a1*(1-a2)*f/d2; Yh=Zn[U  
b1=(1-a1)*f/(d1*b2); I,D=ixK  
_ 0-YsD  
%曲率半径 3?:}lY<,  
":OXs9Yg  
R1=2*f/(b1*b2) ScEM#9T|  
R2=2*a1*f/(b2*(1+b1)) R-ci?7dt3  
R3=2*a1*a2*f/(1+b2) 2sngi@\  
Ch3##-  
A1=b2^3*(a1-1)*(1+b1)^3; F>OYZOC]  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; Liof
v4![  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; pS0T>r  
Ab`Gb  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); YpJzRm{Ra  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); c c:xT0Y  
C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); j2+&B9(  
(U)=t$=o  
CB=[C1 B1;C2 B2]; Nbr{)h  
AB=[A1 B1;A2 B2]; ?G@%haqn6  
AC=[A1 C1;A2 C2]; n}2}4^  
[yz;OoA:;  
%二次系数 R0Qp*&AL  
\C#Vh7z"2&  
k2=-(det(CB)/det(AB)); E)Dik`Ccl  
k3=-(det(AC)/det(AB)); ~34$D],D  
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 T"O!  
k2=k2 @I%m}>4Jm  
k3=k3 DGcd|>q  
[X|P(&\hQd  
end

谢谢分享，学习一下 ,\VNs'j