solution of 3d nonlinear equation
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ali hassan
el 29 de Sept. de 2020
Editada: Walter Roberson
el 30 de Sept. de 2020
x_p, y_p, z_p=(4, 5, 2)
x_1, y_1, z_1=(8, 9, 5)
x_2, y_2, z_2=(2, 5, 1)
x_3, y_3, z_3=(6, 1, 3)
t_1=5.692820*10^-9
t_2=-2.924173*10^-9
t_3=-12.010097*10^-9
c=3.0*10^8
and my three equations are
eqn1 = sqrt((x(s)-x_p)^2+(y(s)-y_p)^2+(z(s)-z_p)^2)-sqrt((x(s)-x_1)^2+(y(s)-y_1)^2+(z(s)-z_2)^2)-(c*t_1)
eqn2 = sqrt((x(s)-x_p)^2+(y(s)-y_p)^2+(z(s)-z_p)^2)-sqrt((x(s)-x_2)^2+(y(s)-y_2)^2+(z(s)-z_2)^2)-(c*t_1)
eqn3 = sqrt((x(s)-x_p)^2+(y(s)-y_p)^2+(z(s)-z_p)^2)-sqrt((x(s)-x_3)^2+(y(s)-y_3)^2+(z(s)-z_3)^2)-(c*t_1)
where only x(s), y(s), z(s) are unknown are rest all are known
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Walter Roberson
el 29 de Sept. de 2020
Editada: Walter Roberson
el 29 de Sept. de 2020
Perhaps the first equation should involve t1 instead of t_1, and second equation should involve t2 instead of t_1, and the third should involve t3 instead of t_1 ? You do not use t1, t2, or t3 after you define them.
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Walter Roberson
el 29 de Sept. de 2020
x_p = 4; y_p = 5; z_p = 2;
x_1 = 8; y_1 = 9; z_1 = 5;
x_2 = 2; y_2 = 5; z_2 = 1;
x_3 = 6; y_3 = 1; z_3 = 3;
t1 = 5.692820*10^-9;
t2 = -2.924173*10^-9;
t3 = -12.010097*10^-9;
syms xs ys zs %our unknowns
syms c %constant not given in question
eqn1 = sqrt((xs-x_p)^2+(ys-y_p)^2+(zs-z_p)^2)-sqrt((xs-x_1)^2+(ys-y_1)^2+(zs-z_2)^2)-(c*t1);
eqn2 = sqrt((xs-x_p)^2+(ys-y_p)^2+(zs-z_p)^2)-sqrt((xs-x_2)^2+(ys-y_2)^2+(zs-z_2)^2)-(c*t2);
eqn3 = sqrt((xs-x_p)^2+(ys-y_p)^2+(zs-z_p)^2)-sqrt((xs-x_3)^2+(ys-y_3)^2+(zs-z_3)^2)-(c*t3);
sol = solve([eqn1, eqn2, eqn3], [xs, ys, zs]);
disp(sol.xs)
disp(sol.ys)
disp(sol.zs)
the values will be parameterized in c, which you indicate is a known value, but which you did not provide a value for.
There are two solutions for each variable.
You should probably re-substitute the solutions and verify that the values work, as MATLAB warns that some of the solutions produced might no be true solutions.
12 comentarios
Walter Roberson
el 30 de Sept. de 2020
Editada: Walter Roberson
el 30 de Sept. de 2020
That is not 6 possible solutions, that is two solutions with three components each.
possibleSol(:, all(possibleSol>0 & imag(possibleSol)==0, 1))
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