[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 ;)Rvk&J5  
function sjr=nfdre(~) 'G1~\CT  
yQ\K;  
%系统焦距及各镜间距输入，间距取负正负 O8BxXa@5  
cgKK(-$ny  
f=input('f:'); @ yJ/!9?^  
d1=input('d1:'); ?DP]#9/4  
d2=input('d2:'); #fgRF  
d3=input('d3:'); B"N8NVn  
l:NEK`>i  
A=f^2/(d3*d2)-f/d1; O+Z[bis`  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); SI
_{%~k*B  
C=d3/d2-f/d1; fuJ6 fmT  
S-^y;#=  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 g Z3VT{  
a2=d3/(a1*f);%α2 P;jl!o$  
b2=a1*(1-a2)*f/d2;%β2 b 62 o  
b1=(1-a1)*f/(d1*b2);%β1 p"^^9'`=  
}9T$XF~  
SF*!Z2K  
%曲率半径 12)~PIaF  
,ry2J,IT7  
R1=2*f/(b1*b2) X$,#OR  
R2=2*a1*f/(b2*(1+b1)) #BK\cIr  
R3=2*a1*a2*f/(1+b2) [~$Ji&Dd  
M ,.++W\  
A1=b2^3*(a1-1)*(1+b1)^3; ]/;0  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; hg7`jE&2  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; f:L%th  
42:~oKiQ$"  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); vPuPSE%M  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); FY"!%)TV  
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); b;\qF&T  
>O[# 661  
CB=[C1 B1;C2 B2]; <Q)6N!Tp^  
AB=[A1 B1;A2 B2]; kQlXcR  
AC=[A1 C1;A2 C2]; Q7]:vs)%  
RW 23lRA6  
%非球面系数 cad1eOT'  
k2=-(det(CB)/det(AB)); >*%ySlZbs  
k3=-(det(AC)/det(AB)); MNip;S_j  
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 4&/u1u0  
k2=k2 b\~rL,7(  
k3=k3 mMO:m8W  
rly3f  
end 2&fIF}vk>m  
-[OGZP`8  
%有中间像，焦距输入为正数 N,|:=gD_  
Y}(#kqh>  
function sjr=yfdre(~) '/ 3..3k  
eG
26m_S=  
f=input('f:'); Ty\&ARjb 8  
d1=input('d1:'); !pl<  
d2=input('d2:'); jeMh  
d3=input('d3:'); WQ4:='(  
04NI.Jv  
A=f^2/(d3*d2)-f/d1; sAS\-c'6  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); s5FyP"V  
C=d3/d2-f/d1; cF_`m  
P7d" E  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); dUeM+(s1  
a2=d3/(a1*f); +g ovnx  
b2=a1*(1-a2)*f/d2; LoUi Yf  
b1=(1-a1)*f/(d1*b2); esmQ\QQ^1  
Y~RPspHW  
%曲率半径 H?ssV^k  
MdT'xYomzQ  
R1=2*f/(b1*b2) uc~PKU?tO  
R2=2*a1*f/(b2*(1+b1)) N8:?Z#z  
R3=2*a1*a2*f/(1+b2) mzTF2K  
Ac<V!v71  
A1=b2^3*(a1-1)*(1+b1)^3; f33'2PYl  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; (.7_`T6QG  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; q5:-?|jXJ  
3nf+imAF  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); G\tTwX4  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); vV.'&."g  
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); ftF?T.dx  
a9Lf_/w{&  
CB=[C1 B1;C2 B2]; ^#e:q  
AB=[A1 B1;A2 B2]; K)$.0S9d  
AC=[A1 C1;A2 C2]; MGX %U6  
N5csq(  
%二次系数 y.5mYQA4=[  
K,%H*1YKK  
k2=-(det(CB)/det(AB));
(:].?o  
k3=-(det(AC)/det(AB)); vG#|CO9  
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 wlBdA  
k2=k2 j(6: 
 
k3=k3 &$jg *Kr  
]oP2T:A  
end

谢谢分享，学习一下 8qp!S1Qnv