[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 _~ v-:w  
function sjr=nfdre(~) CP +4k.)*O  
MM7"a?y)  
%系统焦距及各镜间距输入，间距取负正负 R3$@N  
PDhWFF  
f=input('f:'); \kxh#{$z?  
d1=input('d1:'); NP "ylMr7P  
d2=input('d2:'); L4~ W/6A  
d3=input('d3:'); 7iP5T  
NJ<N%hcjK  
A=f^2/(d3*d2)-f/d1; ):S!Nl  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); $1~c_<DN  
C=d3/d2-f/d1; =w6}\ 'X  
\Y9I~8\gB  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 gd K*"U  
a2=d3/(a1*f);%α2 _1P`]+K\D$  
b2=a1*(1-a2)*f/d2;%β2 YN/u9[=`  
b1=(1-a1)*f/(d1*b2);%β1 q4)Ey  
9T$u+GX'  

f -7S:,  
%曲率半径 I :@|^PYw  
)&Mq,@  
R1=2*f/(b1*b2) J)#S-ZB+'k  
R2=2*a1*f/(b2*(1+b1)) g**5z'7  
R3=2*a1*a2*f/(1+b2) 1#]B^D  
.D!WO  
A1=b2^3*(a1-1)*(1+b1)^3; ;7L;  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; $7Cgo&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; r,0@~;zA  
i+
I0k~wY  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); L@?e:*h  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); V,m3-=q  
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); :mP9^Do2;  
6m@0;Ht  
CB=[C1 B1;C2 B2]; ??j&i6sp  
AB=[A1 B1;A2 B2]; &y\igX1  
AC=[A1 C1;A2 C2]; a
Dh|48}X  
i(Vm!Y82  
%非球面系数 yuv4*  
k2=-(det(CB)/det(AB)); S2Vxe@b)  
k3=-(det(AC)/det(AB)); +<{m45  
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 !vVT]k[N  
k2=k2 },s_nJR:8  
k3=k3 ;Yv14{T!  
9(Jy0]E~  
end 2LC w*eT{)  
+KP_yUq[  
%有中间像，焦距输入为正数 FB_NkXR  
kP?_kMOx  
function sjr=yfdre(~) e5 3,Rqi)@  
kV+ R5R  
f=input('f:'); ^vM_kArA  
d1=input('d1:'); 7@3sUA_Go  
d2=input('d2:'); EZ{\D!_Y  
d3=input('d3:'); A%n?}  
rs_h}+6"s  
A=f^2/(d3*d2)-f/d1; | k}e&Q_/G  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); 0 u2Ny&6w  
C=d3/d2-f/d1; \4QH/e  
b5S7{"<V  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); x`L+7,&n  
a2=d3/(a1*f); MQ9vPgh  
b2=a1*(1-a2)*f/d2; .<.#aY;N  
b1=(1-a1)*f/(d1*b2); }^9paU  
](nH{aY!  
%曲率半径  Fs)  
4m\([EO  
R1=2*f/(b1*b2) Gk 6fO  
R2=2*a1*f/(b2*(1+b1)) ib; yu_  
R3=2*a1*a2*f/(1+b2) 7H#2WFQ7  
Xb0!( (A  
A1=b2^3*(a1-1)*(1+b1)^3; 3Y(9\}E@`  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; qK,V$l(4#  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; KC]tY9 FK  
F9O`HFVK  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); D:)~%wu Lt  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); XR+rT  
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); Wp`C:H  
s-%J5_d f  
CB=[C1 B1;C2 B2]; :P,2K5]y  
AB=[A1 B1;A2 B2]; Uuz?8/w}#  
AC=[A1 C1;A2 C2]; 2+Zti8  
k% NrL@z  
%二次系数 -$9~xX  
iJ4<f->t  
k2=-(det(CB)/det(AB)); }k| g%HJ  
k3=-(det(AC)/det(AB)); | +fwvi&a  
k1=(k2*a1*b2^3*(1+b1)^3-k3*a1*a2*(1+b2)^3+a1*b2^3*(1+b1)*(1-b1)^2-a1*a2*(1+b2)*(1-b2)^2)/(b1^3*b2^3)-1 O{nC^`X  
k2=k2 bO
e<\Y$  
k3=k3 _jo$)x+'x  
_0m}z%rI  
end

谢谢分享，学习一下 ?`"<DH~:0B