How can I simplify the symbolic expressions?
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Ali Almakhmari
le 11 Sep 2023
Modifié(e) : Ali Almakhmari
le 11 Sep 2023
I hate the fact that solve keeps outputting symbolic expressions with extremely large numbers, when clearly it can simplify further and get rid of those numbers. How can I force it to do that? Its okay with me if I can get an approximation up 6 decimal places.
clear
clc
syms q_bar delta_o theta1 theta2 q S CL_alpha CL_delta b e KT
eqn1 = (2*theta1 - theta2 - q_bar.*theta1 + q_bar.*theta2) == (-0.4805020381.*q_bar.*delta_o);
eqn2 = (-theta1 + theta2 - 2.*q_bar.*theta2) == (-0.0051005137.*q_bar.*delta_o);
sol = solve(eqn1,eqn2,theta1,theta2);
L1 = q.*S.*CL_alpha.*sol.theta1;
L2 = q.*S.*CL_alpha.*sol.theta2 + q.*S.*CL_delta.*delta_o;
eqn1 = (0.25.*L1.*b + 0.75.*L2.*b) == 0;
a=solve(eqn1,q_bar)
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Torsten
le 11 Sep 2023
If you need 6 decimal places, why do you supply model parameters with 10 decimal places ?
-0.4805020381*q_bar
-0.0051005137*q_bar
Réponse acceptée
John D'Errico
le 11 Sep 2023
Modifié(e) : John D'Errico
le 11 Sep 2023
clear
clc
syms q_bar delta_o theta1 theta2 q S CL_alpha CL_delta b e KT
eqn1 = (2*theta1 - theta2 - q_bar.*theta1 + q_bar.*theta2) == (-0.4805020381.*q_bar.*delta_o);
eqn2 = (-theta1 + theta2 - 2.*q_bar.*theta2) == (-0.0051005137.*q_bar.*delta_o);
sol = solve(eqn1,eqn2,theta1,theta2);
L1 = q.*S.*CL_alpha.*sol.theta1;
L2 = q.*S.*CL_alpha.*sol.theta2 + q.*S.*CL_delta.*delta_o;
eqn1 = (0.25.*L1.*b + 0.75.*L2.*b) == 0;
a=solve(eqn1,q_bar)
They are simple already. At least, as simple as they can be. You have unknowns in there, inside and outside of the square roots. I suppose you could play around, and make it different, but simpify will not find anything glaring to reduce there.
Can you make the numbers smaller? Well, yes, you can force VPA to round them.
vpa(a,10)
Is that simpler? I guess.
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