solving 5 non linear simultaneous equation
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Bibek Tripathi
am 30 Apr. 2016
Beantwortet: Alex Sha
am 1 Aug. 2019
I have a set of 5 non-linear equations. first 4 equations are h,k,a,b --> in terms of variable 'bsfc'. the fifth equation bsfc^2 in terms of h,k,a,b. So i have 5 equations and 5 variables. This programming is basically finding the radius of the rotated ellipse. I have my x and y values too. However I am not getting the right solution.
The right solution is bsfc=246
I have my code here:
syms bsfc h k a b
y=86;
x=1717;
phi=0;
eqn1= h == 9.514*bsfc;
eqn2= k == 0.0001*(bsfc+1)^2-0.07*bsfc+19;
eqn3= a == 10* atan(bsfc/30 - 5.8) -3.5;
eqn4= b == 0.025 * atan(bsfc/20 - 10.3) -0.022;
eqn5= bsfc^2 == (((x-h)*cos(phi)+(y-k)*sin(phi))^2/(a^2)+ ((x-h)*sin(phi)-(y-k)*cos(phi))^2/(b^2))
sol=solve([eqn1,eqn2,eqn3,eqn4,eqn5],bsfc,h,k,a,b);
hSol = sol.h
kSol = sol.k
aSol = sol.a
bSol = sol.b
bsfcSol = sol.bsfc
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Akzeptierte Antwort
Walter Roberson
am 1 Mai 2016
Given those equations, bsfc approximately 1154.643046927194 or approximately 1325.553297436922
Your proposed value 246 does not come at all close to satisfying eqn5
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Walter Roberson
am 2 Mai 2016
syms bsfc h k a b x y phi;
Q = @(v) sym(v);
eqn1 = h == Q(9.514)*bsfc;
eqn2 = k == Q(0.1e-3)*(bsfc+1)^2-Q(0.7e-1)*bsfc+19;
eqn3 = a == 10*atan((1/30)*bsfc-Q(5.8))-Q(3.5);
eqn4 = b == Q(0.25e-1)*atan((1/20)*bsfc-Q(10.3))-Q(0.22e-1);
eqn5 = bsfc^2 == ((x-h)*cos(phi)+(y-k)*sin(phi))^2/a^2+((x-h)*sin(phi)-(y-k)*cos(phi))^2/b^2;
sol_hkab = solve([eqn2,eqn3,eqn4,eqn5],[h, k, a, b]);
eqn_bsfc1 = subs(eqn1, [h k a b], [sol_hkab.h(1), sol_hkab.k(1), sol_hkab.a(1), sol_hkab.b(1)]);
eqn_bsfc2 = subs(eqn1, [h k a b], [sol_hkab.h(2), sol_hkab.k(2), sol_hkab.a(2), sol_hkab.b(2)]);
xyt = [Q(1256.97), Q(7.076), Q(0.00044)];
bs = linspace(0,750,2000);
s_bsfc1 = subs(eqn_bsfc1, [x y phi], xyt);
eq1f = matlabFunction(feval(symengine,'lhs',s_bsfc1)-feval(symengine,'rhs',s_bsfc1));
eq1fr = @(x) real(eq1f(x));
eq1fi = @(x) imag(eq1f(x));
s_bsfc2 = subs(eqn_bsfc2, [x y phi], xyt);
eq2f = matlabFunction(feval(symengine,'lhs',s_bsfc2)-feval(symengine,'rhs',s_bsfc2));
eq2fr = @(x) real(eq2f(x));
eq2fi = @(x) imag(eq2f(x));
plot(bs, eq1fi(bs), 'b--', bs, eq2fi(bs), 'g--', bs, eq1fr(bs), 'b-', bs, eq2fr(bs), 'g-');
legend({'imag sol1', 'imag sol2', 'real sol1', 'real sol2'});
and then start zooming in on places where the imaginary is 0, looking for real solutions.
Weitere Antworten (2)
Alex Sha
am 31 Jul. 2019
Look this result:
h: -4113.26496133792
bsfc: -432.338129213572
k: 67.8689272162873
a: -18.7135929756326
b: -0.0604868803921316
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Alex Sha
am 1 Aug. 2019
one more solution:
h: 10985.2739483926
bsfc: 1154.64304691955
k: 71.7260719049847
a: 11.9021369375982
b: 0.0167429176020812
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