[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 `#Yv(a2TY  
function sjr=nfdre(~) :5 zXW;s  
V']{n7a-  
%系统焦距及各镜间距输入，间距取负正负 'v42QJ"{  
o'hwyXy/S  
f=input('f:'); -Y:^<C^^&8  
d1=input('d1:'); l`M7a9*U  
d2=input('d2:'); Gd:fh5u':  
d3=input('d3:');  0+P[
0  
!IUH 5  
A=f^2/(d3*d2)-f/d1; O>~ozW&  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); S@WzvM  
C=d3/d2-f/d1; wFHz<i!jr&  
YYzl"<)c  
a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 NR@n%p  
a2=d3/(a1*f);%α2 ~6HaZlBB  
b2=a1*(1-a2)*f/d2;%β2 _=s{,t &u  
b1=(1-a1)*f/(d1*b2);%β1
5N'Z"C0  
^@a|s Sb  
_R] qoUw;  
%曲率半径 CKau\N7T  
MRn;D|Q  
R1=2*f/(b1*b2) =u,8(:R]s  
R2=2*a1*f/(b2*(1+b1)) N-Jp; D  
R3=2*a1*a2*f/(1+b2) 8E:d!?<^&I  
XFx p^  
A1=b2^3*(a1-1)*(1+b1)^3; pk&;5|cCD  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; $^XCI%DH  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; q8_8rp-@  
I_\#(  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); x6BO%1  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); Tz"Xm/Gy  
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); 6$k"B/k  
=,4iMENm!  
CB=[C1 B1;C2 B2]; (TjY1,f!H  
AB=[A1 B1;A2 B2]; y {PUklq  
AC=[A1 C1;A2 C2]; F=qG+T  
5?{ytNCY  
%非球面系数 TjswB#  
k2=-(det(CB)/det(AB)); ~t` uq  
k3=-(det(AC)/det(AB)); rbEUq.Yk]~  
k1=(k2*a1*b2^3*(1+b1)^3-k3*a1*a2*(1+b2)^3+a1*b2^3*(1+b1)*(1-b1)^2-a1*a2*(1+b2)*(1-b2)^2)/(b1^3*b2^3)-1 K~I%"r|l  
k2=k2 GqT0SP  
k3=k3 w '<8lw  
TcP (?v  
end d77->FX2  
6_.K9;Gd  
%有中间像，焦距输入为正数 km(Mv  
42Vz6 k:  
function sjr=yfdre(~) ofu {g  
;Wws;.~  
f=input('f:'); zv%9?:  
d1=input('d1:'); "")I1iO g  
d2=input('d2:'); 5^}"Tn4I  
d3=input('d3:'); 1$]hyC/f  
n.
a55uy  
A=f^2/(d3*d2)-f/d1; ^-24S#KE  
B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); ]0+5@c  
C=d3/d2-f/d1; j1yW{  
J!5>8I(_wX  
a1=(-B-sqrt(B^2-4*A*C))/(2*A); Eed5sm$H  
a2=d3/(a1*f); QQB\$[M!Z  
b2=a1*(1-a2)*f/d2; `h:$3a:5  
b1=(1-a1)*f/(d1*b2); Jp0*Y-*Y  
!='?+Ysxs  
%曲率半径 RJz$$,RU  
2GHmA_7P  
R1=2*f/(b1*b2) Q|O! cEW/  
R2=2*a1*f/(b2*(1+b1)) xux
j
 
R3=2*a1*a2*f/(1+b2) {5$.:Y  
!@r1B`]j+"  
A1=b2^3*(a1-1)*(1+b1)^3; O!3MXmaO  
B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; aQzu[N  
C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; +~E;x1&'  
sI@y)z  
A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); k4:e0Wd  
B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); 4FWb5b!A=  
C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); g.&n X/  
O e-FI+7  
CB=[C1 B1;C2 B2]; r&u&$"c  
AB=[A1 B1;A2 B2]; XtJ_po  
AC=[A1 C1;A2 C2]; ':}  
y)s+/Teb  
%二次系数 rW.o_z03^  
U]aH4N  
k2=-(det(CB)/det(AB)); V<ZohB?y  
k3=-(det(AC)/det(AB)); <6 LpsM}  
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- 1  
k2=k2 |KV|x^fJ  
k3=k3 m0: IFE($  
w:|BQ,  
end

谢谢分享，学习一下 MmZs|pXk