Solving 5 very complicated equations with 5 unknowns
8 views (last 30 days)
Show older comments
Hi, I'm trying to derive a zoom lens and this equation is what i came across. Currently, I'm trying to find 5 variables with 5 equations(feq1, feq2, feq3, feq4, feq5).The equations were generated as shown below.
syms f1 f2 f3 f4 fe p2 p3;
fl = @(f1, f2, d) 1/(1/f1 + 1/f2 - d/(f1 * f2)); %focal length formula
bfl = @(f1, f2, d) (f2 * (d - f1)) / (d - (f1 + f2)); %back focal length formula
f_34 = fl(f4, f3, p3);
bfl_34 = bfl(f4, f3, p3);
f_234 = fl(f_34, f2, bfl_34+p2-p3);
bfl_234 = bfl(f_34, f2, bfl_34 + (p2-p3));
f_1234 = fl(f_234, f1, bfl_234 + (80-p2));
bfl_1234 = bfl(f_1234, fe, bfl_234 + (80 - p2));
f_1234e = fl(f_1234, fe, bfl_1234 + (125 - 80)) % The equation of the lens system
%% substituting p2 and p3 in my equation
feq1 = subs(subs(f_1234e, p2, 63.8654),p3, 44.6468) == 179.0893;
feq2 = subs(subs(f_1234e, p2, 57.8781),p3, 43.4521) == 166.8919;
feq3 = subs(subs(f_1234e, p2, 52.4784),p3, 41.9933) == 151.9181;
feq4 = subs(subs(f_1234e, p2, 49.7862),p3, 40.6357) == 143.6190;
feq5 = subs(subs(f_1234e, p2, 44.6669),p3, 36.5343) == 122.4775;
Which is very complicated.
I've aleady tried solve() and vpasolve(), but their outputs were empty.
Attempt #1:
S = vpasolve([feq1, feq2, feq3, feq4, feq5],[f1 f2 f3 f4 fe]);
Attempt #2:
S = solve([feq1, feq2, feq3, feq4, feq5]);
Please tell me if i've got something wrong or other ways to solve this hard question.
Thanks!
0 Comments
Accepted Answer
Torsten
on 22 Dec 2022
syms f1 f2 f3 f4 fe p2 p3
fl = @(f1, f2, d) 1/(1/f1 + 1/f2 - d/(f1 * f2)); %focal length formula
bfl = @(f1, f2, d) (f2 * (d - f1)) / (d - (f1 + f2)); %back focal length formula
f_34 = fl(f4, f3, p3);
bfl_34 = bfl(f4, f3, p3);
f_234 = fl(f_34, f2, bfl_34+p2-p3);
bfl_234 = bfl(f_34, f2, bfl_34 + (p2-p3));
f_1234 = fl(f_234, f1, bfl_234 + (80-p2));
bfl_1234 = bfl(f_1234, fe, bfl_234 + (80 - p2));
f_1234e = fl(f_1234, fe, bfl_1234 + (125 - 80)); % The equation of the lens system
%% substituting p2 and p3 in my equation
feq1 = subs(subs(f_1234e, p2, 63.8654),p3, 44.6468) == 179.0893;
feq2 = subs(subs(f_1234e, p2, 57.8781),p3, 43.4521) == 166.8919;
feq3 = subs(subs(f_1234e, p2, 52.4784),p3, 41.9933) == 151.9181;
feq4 = subs(subs(f_1234e, p2, 49.7862),p3, 40.6357) == 143.6190;
feq5 = subs(subs(f_1234e, p2, 44.6669),p3, 36.5343) == 122.4775;
f = [feq1;feq2;feq3;feq4;feq5];
f = lhs(f)-rhs(f);
f = matlabFunction(f);
F = @(x)f(x(1),x(2),x(3),x(4),x(5));
f0 = [68;35;-14;96;186];
sol = fsolve(F,f0)
More Answers (1)
Bradley Chen
on 21 Dec 2022
2 Comments
Torsten
on 21 Dec 2022
I suspect there will be many more "answers" depending on the initial guesses for the unknowns. At least "fsolve" showed this behaviour when using it for your problem.
See Also
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
