[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 w
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function sjr=nfdre(~) pJ{sBp_$  
~q<UE\H  
%系统焦距及各镜间距输入，间距取负正负 q*2ljcb55  
KNV$9&Z  
f=input('f:'); sx6` g;  
d1=input('d1:'); K]RkKMT,  
d2=input('d2:'); I` /'\cU9  
d3=input('d3:'); |XeuqZa  
f F?=W  
A=f^2/(d3*d2)-f/d1; IKpNc+;p  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); C6GYhG]  
C=d3/d2-f/d1; /q8n_NR  
0%vXPlfnY  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 Nj?,'?'O}  
a2=d3/(a1*f);%α2 &0M^UvO  
b2=a1*(1-a2)*f/d2;%β2 U|,VH-#  
b1=(1-a1)*f/(d1*b2);%β1 )5B90[M|t  
@}#$<6|  
FX\ -Y$K  
%曲率半径 Jyvc(
~x  
d"$ \fL  
R1=2*f/(b1*b2) r3#H]c  
R2=2*a1*f/(b2*(1+b1)) 6zyozJA  
R3=2*a1*a2*f/(1+b2) X)`?P*[  
 u"tv6Qp  
A1=b2^3*(a1-1)*(1+b1)^3; a Z ^SK|E  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; "{|9Yis=  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; BH\qm (X  
aM~M@wS  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); pkhZW8O  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); 4u A;--j  
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); 6MF%$K3  
eo"6 \3z  
CB=[C1 B1;C2 B2]; 87y$=eZ  
AB=[A1 B1;A2 B2]; 9J*m!-hOY  
AC=[A1 C1;A2 C2]; % `\8z  
u[y>DPPx  
%非球面系数 Bx2E9/S3  
k2=-(det(CB)/det(AB)); P5kkaLzG  
k3=-(det(AC)/det(AB)); q f-1}  
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 dX` _Y  
k2=k2 u JGYXlLE  
k3=k3 H"6:!;9,  
Gb\}e}TB[  
end buq3t+0  
pJ1G
B  
%有中间像，焦距输入为正数 $U3s:VQ'  
ybcQ,e  
function sjr=yfdre(~) 5Tkh6s  
4, 8gf2  
f=input('f:'); qK#\k@E  
d1=input('d1:'); c(i-~_  
d2=input('d2:'); ZI-)'  
d3=input('d3:'); P8piXG  
>DpnIWn  
A=f^2/(d3*d2)-f/d1; e=QnGT*b5  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); Jx(%t<2  
C=d3/d2-f/d1; P%(pbG-X.  
/EA4-#uw  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); rU%\ 8T0f  
a2=d3/(a1*f); rer=o S  
b2=a1*(1-a2)*f/d2; +kmPQdO;*/  
b1=(1-a1)*f/(d1*b2); 32:q'
 
52,pCyU  
%曲率半径 >n{(2bcFs  
/m(vIl  
R1=2*f/(b1*b2) iO(9#rV  
R2=2*a1*f/(b2*(1+b1)) X2/
`EN\  
R3=2*a1*a2*f/(1+b2) KzG8K 6wZ  
/)e&4.6  
A1=b2^3*(a1-1)*(1+b1)^3; ~W_m<#K(  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; @I_A\ U{  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; QB7E:g&7  
A!x_R {,yH  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); %DbL|;z1  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); R4%!W~K  
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); TY],H=  
,0[bzk  
CB=[C1 B1;C2 B2]; 3#j%F  
AB=[A1 B1;A2 B2]; ,g;~:  
AC=[A1 C1;A2 C2]; p9y "0A|  
IF(W[J  
%二次系数 Yy@;
U]R  
rc<^6HqD  
k2=-(det(CB)/det(AB)); :w_Zr5H]  
k3=-(det(AC)/det(AB)); { %vX/Ek  
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 b`u
sRoD{+  
k2=k2 D: NBb!   
k3=k3 1>[3(o3t  
cAogz/<S  
end

谢谢分享，学习一下 hc4<`W{