[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 eS jXaZh  
function sjr=nfdre(~) c.PPVqx  
{ L
JRdV  
%系统焦距及各镜间距输入，间距取负正负 "+ji`{  
J,E'F!{  
f=input('f:'); bewi.$E{  
d1=input('d1:'); %o+VZEH3  
d2=input('d2:'); *Gm
%Dn  
d3=input('d3:'); PU^Z7T);  
;o#R(m@Lx  
A=f^2/(d3*d2)-f/d1; ET`;TfqM  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); |T9p#) ec2  
C=d3/d2-f/d1; 08S|$_  
~J6c1jG  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 dOe|uQXyD  
a2=d3/(a1*f);%α2 ?K/z`E!xhN  
b2=a1*(1-a2)*f/d2;%β2 rK2*DuE  
b1=(1-a1)*f/(d1*b2);%β1 &Fl^&&1C  
?'h<yxu]u0  
c]W]m`:  
%曲率半径 INrUvD/*  
9frS!AQ  
R1=2*f/(b1*b2) c)M_&?J!5  
R2=2*a1*f/(b2*(1+b1)) 8_ns^6XK5p  
R3=2*a1*a2*f/(1+b2) ('xu2 ;<  
v>$GVCY  
A1=b2^3*(a1-1)*(1+b1)^3; Au"[2cG  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; g0^%X9s  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; 2`l$uEI3oJ  
^m{kn8  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); L;'+O u  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); XWbe|K!e  
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); #:3E.=  
.Lp0_R@  
CB=[C1 B1;C2 B2]; 2ZB'WzH.X  
AB=[A1 B1;A2 B2]; y^:g"|q  
AC=[A1 C1;A2 C2]; hsljJvs
 
A[ZJS  
%非球面系数 ^;$f-e  
k2=-(det(CB)/det(AB)); 2HMlh.R(C  
k3=-(det(AC)/det(AB)); t}`|\*a  
k1=(k2*a1*b2^3*(1+b1)^3-k3*a1*a2*(1+b2)^3+a1*b2^3*(1+b1)*(1-b1)^2-a1*a2*(1+b2)*(1-b2)^2)/(b1^3*b2^3)-1 |9!3{3  
k2=k2 OlFls 8#>  
k3=k3 rsvZi1N4w$  
PS@` =Z  
end ,l`q  
s%J|r{F6  
%有中间像，焦距输入为正数 %"DEgIP  
ZV_Z)<  
function sjr=yfdre(~) 6A/Nlk.  
ID5?x8o#k  
f=input('f:'); S0g5Ym ia  
d1=input('d1:'); ![:S~x1  
d2=input('d2:'); +8Zt<snG  
d3=input('d3:'); 'e\m6~u\hm  
~|pVz/s|G  
A=f^2/(d3*d2)-f/d1;
2:/'  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); M0x5s@  
C=d3/d2-f/d1; 7^Jszd:c08  
RWXj)H)w  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); FcsEv {#U  
a2=d3/(a1*f); ^,b*.6t  
b2=a1*(1-a2)*f/d2; aM3%Mx?w  
b1=(1-a1)*f/(d1*b2); E[6JHBE*r  
)-[ 2vhXz  
%曲率半径 yK0Q,   
.F?yt5{5No  
R1=2*f/(b1*b2) )"jG)c^1*  
R2=2*a1*f/(b2*(1+b1)) ||}|=Sz  
R3=2*a1*a2*f/(1+b2) J~DP*}~XK  
_$
wWKJy9  
A1=b2^3*(a1-1)*(1+b1)^3; m^O:k"+!  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; KcfW+>W3  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; 0d:t$2~C  
z>&Py(  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); o]}b#U8S  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); R',|Jf=`  
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); W2yNEiH  
Q]w;o
&eo  
CB=[C1 B1;C2 B2]; Xe*@`&nv@  
AB=[A1 B1;A2 B2]; ?qw&H /R  
AC=[A1 C1;A2 C2]; eP'e_E  
omECes)  
%二次系数 2#*Bw=  
BYVp~!u  
k2=-(det(CB)/det(AB)); ]BP/KCjAI<  
k3=-(det(AC)/det(AB)); :{LVS nG  
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 sG0cN;I]t  
k2=k2 L`x:Y>C(  
k3=k3 QXq~e  
lV`y6{o#T  
end

谢谢分享，学习一下 nrwb6wj