[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) +=XDNSw  
function sjr=nfdre(~) HFJna2B`  
;.=ZwM]C  
%系統(tǒng)焦距及各鏡間距輸入，間距取負正負 =%p%+F@RlW  
-h5yg`+1N\  
f=input('f:'); n=HId:XT  
d1=input('d1:'); q*^Y8s~3I  
d2=input('d2:'); X8F@U ^@  
d3=input('d3:'); C'HW`rh.^  
Ca: jN0  
A=f^2/(d3*d2)-f/d1; -&PiD  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); F9hh- "(Z  
C=d3/d2-f/d1; fl{wF@C6  
c:  /Wk  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 byj}36LN62  
a2=d3/(a1*f);%α2 ]K]$FX<f  
b2=a1*(1-a2)*f/d2;%β2 ~$#"'Tl4J  
b1=(1-a1)*f/(d1*b2);%β1 WaU+ZgDrG  
8PQn=k9  
]9xuLJ)  
%曲率半徑 'A0.(a5  
7j9:s>D  
R1=2*f/(b1*b2) >900I4]I  
R2=2*a1*f/(b2*(1+b1)) P@gVzx)M  
R3=2*a1*a2*f/(1+b2) ^DL}J>
F9G  
w"s;R8  
A1=b2^3*(a1-1)*(1+b1)^3; )7U^&I,  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; OnNWci|7  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2;