MatLab Equation - Two Unknowns

2 次查看(过去 30 天)
Eddy Ramirez
Eddy Ramirez 2021-3-6
I am running the coding below, and under the length correction I have two unknown variables and when I start the debugging portion and I look up equation 2 only shows
0.89761904761904761904761904761905 == 1 - tan(h)
instead of the entire equation
Nl_Final==(1-((tan(h)*(n*s_AR))/(n*s_AR)));
Any ideas as to why? I am sure MatLab is not ignoring the rest of the values (at least thats what I hope) or am i missing something
clear all;
clc;
E1=39.1*1e9;
Vf=0.60;
Vm=1-Vf;
Df=10;
r=Df/2;
Ef=70*1e9;
Em=3.5*1e9;
vm=.36;
Nl=sym('N_l');
h=sym('h');
s_AR=sym('s_f');
lf=sym('l');
r0=r/sqrt(Vf);
Gm=Em/(2*(1+vm));
n=sqrt((2*(Gm))/(Ef*log(r0/r)));
%%BETA VALUES%%
b_eta=sqrt((2*(Gm))/(Ef*r^2*log(r0/r)));
b_eta1=sqrt(n^2/r^2);
%%EFFECTIVE MODULUS finding length%%
equation1=E1==Ef*Vf*Nl+Vm*Em;
solution1=solve(equation1);
Nl_Final=vpa(solution1);
%%LENGTH CORRECTION FINDING h%%
equation2=Nl_Final==(1-((tan(h)*(n*s_AR))/(n*s_AR)));
equation2_1=Nl_Final==(1-((tan(h)*((b_eta*lf)/2))/((b_eta*lf)/2)));
solution2=solve(equation2);
solution2_1=solve(equation2_1);
h_Final=vpa(solution2*1000);
h_Final_1=vpa(solution2_1*1000);

回答(1 个)

Sai Veeramachaneni
Hi,
Nl_Final==(1-((tan(h)*(n*s_AR))/(n*s_AR)));
In the above equation n*s_AR is getting cancelled out in numerator and denominator which results in
Nl_Final==(1-tan(h))
Hence you are getting equation2 values as
0.89761904761904761904761904761905 == 1 - tan(h)
Hope it helps.
  1 个评论
Eddy Ramirez
Eddy Ramirez 2021-3-10
yeah i saw that after and i also had some typos there thank you for your feedback!

请先登录,再进行评论。

类别

Help CenterFile Exchange 中查找有关 Logical 的更多信息

Community Treasure Hunt

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

Start Hunting!

Translated by