[原創(chuàng)]在框架結(jié)構(gòu)確定的情況下，基于matlab的消四種像差的三反系統(tǒng)初始結(jié)構(gòu)的求解 [復制鏈接]

%無中間像，焦距輸入為負數(shù) nG$+9}\UlP  
function sjr=nfdre(~) Rm}G4Pq  
"5v^6R9e  
%系統(tǒng)焦距及各鏡間距輸入，間距取負正負 V`rxjv}!  
f?TS#jG4}  
f=input('f:'); ==S^IBG  
d1=input('d1:'); !DD4Bqez  
d2=input('d2:'); w,j;XPp  
d3=input('d3:'); ts;^,|h  
bc;?O`I<  
A=f^2/(d3*d2)-f/d1; wEw;],ur  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); \}AJ)v*<  
C=d3/d2-f/d1; CSH*^nk':O  
ANfy+@  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 >3}N
;  
a2=d3/(a1*f);%α2 )x35  
b2=a1*(1-a2)*f/d2;%β2 c<sq0('`  
b1=(1-a1)*f/(d1*b2);%β1 $-.*8*9  
1k`gr&S  
1@9M[_<n5  
%曲率半徑 6n;? :./  
5=b6B=\*~  
R1=2*f/(b1*b2) 
Qn.3B  
R2=2*a1*f/(b2*(1+b1)) f ~bgZ  
R3=2*a1*a2*f/(1+b2) h gu\~}kD  
^])s\a$