[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 rNhS\1-  
function sjr=nfdre(~) H2|'JA#v  
F^{31iU~CX  
%系统焦距及各镜间距输入，间距取负正负 &iu]M=Yb  
'%/=\Q`  
f=input('f:'); ~6t<`&f  
d1=input('d1:'); V?yQm4  
d2=input('d2:'); DXD+,y\=  
d3=input('d3:'); ,k_ b-/  
;Egl8Vhr  
A=f^2/(d3*d2)-f/d1; dpE^BWv3  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); 7R[7
M%H  
C=d3/d2-f/d1; o% Q7 el$f  
5q@s6_"{  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 v(Kj6'  
a2=d3/(a1*f);%α2 M^\`~{*T  
b2=a1*(1-a2)*f/d2;%β2 Q1*_l  
b1=(1-a1)*f/(d1*b2);%β1 ~rI2 RJ  
lFf>z}eLy  
au19Q*r9  
%曲率半径 `0%;Gz%}  
0x5\{f  
R1=2*f/(b1*b2) E3p$^['vx  
R2=2*a1*f/(b2*(1+b1)) 1O,5bi>t7  
R3=2*a1*a2*f/(1+b2) bHm/ZZx  
l#C<bDw  
A1=b2^3*(a1-1)*(1+b1)^3; 0?t;3z$n  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; 0VQBm^$(  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; E4\HI+  
:Nf(:D8  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); 19[oXyFI  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); D]fgBW-  
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); @$kzes\  
,Gbc4x  
CB=[C1 B1;C2 B2]; #]^C(qmb:  
AB=[A1 B1;A2 B2]; r
2tE!gMC  
AC=[A1 C1;A2 C2]; d^54mfgI  
V%;dTCq  
%非球面系数 9G6ZKqum  
k2=-(det(CB)/det(AB)); e/x 9@1s#  
k3=-(det(AC)/det(AB)); J"h
2"$v,  
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 IDpLf*vSG  
k2=k2 !|V_DsP  
k3=k3 Iao?9,NL9O  
*qL'WrB1  
end 1,zc8>M  
Ui"$A/  
%有中间像，焦距输入为正数 yYe>a^r4R  
BZTj>yd  
function sjr=yfdre(~) HD{u
#~8{  
OXZK|C;M}  
f=input('f:'); x]hG2on!  
d1=input('d1:'); jR:
\D_:  
d2=input('d2:'); !s$1C=z5u  
d3=input('d3:'); R)N^j'R~=  
im+g|9@%  
A=f^2/(d3*d2)-f/d1; gkTwGI+w  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); ;H8`^;  
C=d3/d2-f/d1; \I6F;G6  
{C?$osrr  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); t:oq'
t  
a2=d3/(a1*f); `'r~3kP*NT  
b2=a1*(1-a2)*f/d2; #~m^RoE  
b1=(1-a1)*f/(d1*b2); N&G(`]  
/_expSPHl  
%曲率半径 QCDica `+*  
}9B},  
R1=2*f/(b1*b2) N|"q6M!ZL  
R2=2*a1*f/(b2*(1+b1)) =d~pr:.F  
R3=2*a1*a2*f/(1+b2) ;InMgo,  
A? jaS9 &)  
A1=b2^3*(a1-1)*(1+b1)^3; F3q<j$y  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; .*595SuF  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; 1PWi~1q{Q  
{@x-T  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); i*4v!(E  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); XmVst*2=  
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); Y>E` 7n  
6v}q @z  
CB=[C1 B1;C2 B2]; p}I\H ^"8+  
AB=[A1 B1;A2 B2]; Q>\DM'{:4  
AC=[A1 C1;A2 C2]; FW3E UC)P  
6_rgRo&  
%二次系数 c":2<:D&  
Kn?h  
k2=-(det(CB)/det(AB)); }43qpJe8U  
k3=-(det(AC)/det(AB)); )VG>6x  
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 BlT)hG(M>  
k2=k2 KL(sVj^e  
k3=k3 |,TBP@  
|Q%nnN  
end

谢谢分享，学习一下 t
!FC)iY