numerical solution of equation

2 次查看（过去 30 天）
safi58
safi58 2017-2-20
评论： Walter Roberson 2017-3-15
syms m_c_theta1 m_c0 M theta1 j_L0 j_L_theta1 l
syms m_c_gama k1 gama j_L_gama
j_L_theta1=1.5;
theta1=120;
k1=0.423;
gama=180;
m_c_gama=0.9;
j_L_gama=0.3;
l=0.218;
eqn1=m_c_theta1==(m_c0-1/M+1)*cos(theta1)+j_L0*sin(theta1)+1/M-1;
eqn2=j_L_theta1==(-m_c0+1/M-1)*sin(theta1)+j_L0*cos(theta1);
%eqn3=m_m_theta1==1;
%eqn3=j_Lm_theta1==j_Lm0+l*theta1;
%eqn4=m_c_theta2==j_L_theta1*sin(theta2-theta1)+(m_c_theta1+1)*cos(theta2-theta1)-1;
%eqn5=j_L_theta2==j_L_theta1*cos(theta2-theta1)-(m_c_theta1+1)*sin(theta2-theta1);
%m_m_theta2=1;
%qn6=j_Lm_theta2==j_Lm0+l*theta2;
eqn7=m_c_gama==(1/k1)*j_L_theta1*sin(k1*((gama-theta1)*(pi/180))*(180/pi))+m_c_theta1*cos(k1*((gama-theta1)*(pi/180))*(180/pi));
eqn8=j_L_gama==j_L_theta1*cos(k1*((gama-theta1)*(pi/180))*(180/pi))-k1*m_c_theta1*sin(k1*((gama-theta1)*(pi/180))*(180/pi));
%eqn9=j_Lm_gama==j_L_gama;
%m_m_gama=(-m_c_theta2*cos(k1*(gama-theta2))-(1/k1)*j_L_theta2*sin(k1*(gama-theta2)))/(1+l)
m_c_gama=-m_c0;
j_L_gama=-j_L0;
j_L_theta1=(theta1*(pi/180)*l)/2;
j_L_gama=j_L_theta1-l*(gama-theta1)*(pi/180);
%j_L_theta2=j_Lm_theta2;
%j_Lm0=-j_Lm_gama;
%j_L_gama=j_Lm_gama;
eqns = subs([eqn1, eqn2, eqn7, eqn8]);
sol = vpasolve(eqns, m_c0, j_L0, m_c_theta1,M)
Hi all I want solve these equations numerically and I have given all the values required. but it is showing something like that: % m_c0: [0x1 sym] j_L0: [0x1 sym] m_c_theta1: [0x1 sym] M: [0x1 sym]
I am not sure what is the problem. Can anyone help please?

请先登录，再回答此问题。

采纳的回答

Walter Roberson
Walter Roberson 2017-2-20
You have
theta1=120;
and
eqn1=m_c_theta1==(m_c0-1/M+1)*cos(theta1)+j_L0*sin(theta1)+1/M-1;
is theta1 in degrees or in radians?
Your expressions are really polluted with pi/180 and 180/pi . You should work entirely in radians and only covert to degrees if the user interface demands it.
If you do convert theta1 from degrees to radians you will still get no solution. Your four equations in eqns are not independent. If you solve the first three for m_c0, j_L0, m_c_theta1 and substitute the results into the last one, the remaining equation is independent of the final variable, M. Instead it has constants expressions on both sides, and the values on the two sides are quite different, one being 3/10 and the other being a value that is close to (but not exactly) pi/2
  15 个评论
显示 13更早的评论
safi58
safi58 2017-3-15
I have declared m_c_gama=1.06 and m_c0 should be -1.06 rather I am getting -0.39
Walter Roberson
Walter Roberson 2017-3-15
Your code has
m_c_gama=Q(1.06);
j_L_gama=Q(0.196);
m_c_gama=-m_c0;
Notice you overwrite m_c_gama. Later you have
m_c0=Q(-0.39289591378964274585530032992682);
so that is where the 0.39* is coming from.

请先登录，再进行评论。

更多回答（1 个）

John D'Errico
John D'Errico 2017-2-20
I recall an old phrase: "If wishes were horses, then beggars would ride."
Just wanting a solution to a set of nonlinear equations to exist does not make that so. Not all such systems of equations must have a solution. MATLAB has told you that no solution was found, although that does not ensure none exists.
  2 个评论
safi58
safi58 2017-2-20
I am getting the solution symbolically but when I am trying to do numerically it is returning empty matrix.
John D'Errico
John D'Errico 2017-3-14
The empty matrix indicated that no solution existed.

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Calculus 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by