[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 lU`]yL  
function sjr=nfdre(~) u_[s+J/  
&.,ZU\`zT  
%系统焦距及各镜间距输入，间距取负正负 &9g4/c-?$  
o$8v8="p  
f=input('f:'); +1zCb=;!{  
d1=input('d1:'); Pwn"!pk  
d2=input('d2:'); ~pWbD~aeg  
d3=input('d3:'); P,^`|\#7  
BWamF{\d1a  
A=f^2/(d3*d2)-f/d1; *,A?lX,9A  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); K4b# y~@  
C=d3/d2-f/d1; uO
%G,b  
z=J%-Hq>  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 [#S}L(  
a2=d3/(a1*f);%α2 @0S3`[/
U  
b2=a1*(1-a2)*f/d2;%β2 wg\*FfQn  
b1=(1-a1)*f/(d1*b2);%β1 CZcnX8P'8  
"
E =\Vz  
Bvj-LT=)  
%曲率半径 }u8o*P|,  
x2(!r3a  
R1=2*f/(b1*b2) sS/#)/B  
R2=2*a1*f/(b2*(1+b1)) JY8wo5H  
R3=2*a1*a2*f/(1+b2) 20h|e+3  
]:m>pI*z.  
A1=b2^3*(a1-1)*(1+b1)^3; AQ:cim`  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; ~GZ!;An  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; yNMnByg3?  
0vbiq  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); 28>PmH]7  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); m`l3@Z  
C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); E~Eh'>Y(B  
WDc2Qt  
CB=[C1 B1;C2 B2]; <8nl}^d5  
AB=[A1 B1;A2 B2]; STmn%&  
AC=[A1 C1;A2 C2]; iTJSW  
'~Uo+<v$w  
%非球面系数 a=}JW]  
k2=-(det(CB)/det(AB)); teh$W<C  
k3=-(det(AC)/det(AB)); `oQ)qa_  
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 q|,cMPS3  
k2=k2 gU1E6V-Jm  
k3=k3 o%_MTCANy  
<+ 0cQq=2  
end HH =sq  
lIPy)25~  
%有中间像，焦距输入为正数 
EooQLZ  
rV.
04m,  
function sjr=yfdre(~) VU
6nu4  
#hvLv  
f=input('f:'); cz<8Kb/XV  
d1=input('d1:'); LnZzY0  
d2=input('d2:'); <\uz",e}  
d3=input('d3:'); ns`|G;1vv  
Ln/6]CMl  
A=f^2/(d3*d2)-f/d1; U%oh?g  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); )QS4Z{)U  
C=d3/d2-f/d1; m:|jv|f  
rF C6"_  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); f@U\2r  
a2=d3/(a1*f); Z@ AHe`A  
b2=a1*(1-a2)*f/d2; "J:~Aa%
_  
b1=(1-a1)*f/(d1*b2); ~xI1@^r  
H{Tt>k  
%曲率半径 3Jt_=!qlo  
ZNb;24  
R1=2*f/(b1*b2) GQ<]Sd}
[  
R2=2*a1*f/(b2*(1+b1)) LDDeZY"xd  
R3=2*a1*a2*f/(1+b2) >'IFr9&3  
ANB@cK_  
A1=b2^3*(a1-1)*(1+b1)^3; LB*qL  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; *xm(K+j  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; ]C.x8(2!f  
~<b/%l>h1  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); ~&-8lD];LM  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); ~P&Brn"=Rs  
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); 05eth  
a(}dF?M=  
CB=[C1 B1;C2 B2]; VxD_:USIF  
AB=[A1 B1;A2 B2]; C'8v\C9Ag  
AC=[A1 C1;A2 C2]; @j +8M  
{z)&=v@  
%二次系数 p<>xq
U  
l|#WQXs*c{  
k2=-(det(CB)/det(AB)); 4.]xK2sW  
k3=-(det(AC)/det(AB)); (eX9O4  
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 1 o<l;:  
k2=k2 ,#=ykg*~/  
k3=k3 ,Qvclu8r  
-dX{ R_*  
end

谢谢分享，学习一下 |ocIp/$