How to calculate an area with obtained nodes?

Question

shahin sharafi 2021-8-4

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/892432-how-to-calculate-an-area-with-obtained-nodes

评论： shahin sharafi 2021-8-9

采纳的回答： darova

在 MATLAB Online 中打开

Hi Everyone,

I am going to calculate the area covered by nodes in the plot, but I do not know how! Could someone help me?

Thank you,

The code is copied as bellow:

clear all;clc;close all
%%
Mass=60;Lenght=1;g=10;Ja=60;alpha=(Mass*Lenght*g)/Ja;Zeta3=411.67/Ja;
Time_Delay=0.2;
N=7;
%% 
s1=0;
s2=-Time_Delay;
[Phi_0_s,Phi_TD_s,Diff_Phi_TD_s]=Shape_Function(N,Time_Delay,s1,s2);
%%  
 [M,K]=M_K(N,Time_Delay);
%% 
counter=[];
KP=0:0.1:3;
KD=0:0.1:3; 
 for ii=1:length(KP)
     KP(ii)
     for i=1:length(KD)
       
         KP1=Zeta3*KP(ii);
         KD1=Zeta3*KD(i)
        L=Evaluation(KP1,KD1,alpha,Phi_0_s,Phi_TD_s,M,K);
Max_Real_EignValues=max(real(eig(L)));
if Max_Real_EignValues<0
    plot(KP(ii),KD((i)),'r*','MarkerSize',5)
    hold on
       end
     end
 end 
 hold on
%%  Below Lines are the functions related to top calculations
function [Phi_0_s,Phi_TD_s,Diff_Phi_TD_s]=Shape_Function(N,Time_Delay,s1,s2)
Phi_0_s(1)=1;
Phi_0_s(2)=1+2*s1/Time_Delay;  
for k=3:N
Phi_0_s(k)=((2*k-3)*Phi_0_s(2)*Phi_0_s(k-1)-(k-2)*Phi_0_s(k-2))/(k-1);
end 
Phi_0_s=Phi_0_s';
%% 
Phi_TD_s(1)=1;
Phi_TD_s(2)=1+2*s2/Time_Delay;  
for k=3:N
Phi_TD_s(k)=((2*k-3)*Phi_TD_s(2)*Phi_TD_s(k-1)-(k-2)*Phi_TD_s(k-2))/(k-1);
end 
Phi_TD_s=Phi_TD_s';
%% 
syms s
Phi_TTD_s=sym(zeros(1,N));
Phi_TTD_s(1)=1;
Phi_TTD_s(2)=1+2*s/Time_Delay;  
for k=3:N
Phi_TTD_s(k)=((2*k-3)*Phi_TTD_s(2)*Phi_TTD_s(k-1)-(k-2)*Phi_TTD_s(k-2))/(k-1);
end 
Diff_Phi_TD_s=diff(Phi_TTD_s,s);
Diff_Phi_TD_s=eval(subs(Diff_Phi_TD_s,s,-Time_Delay));
Diff_Phi_TD_s=double(Diff_Phi_TD_s);
Diff_Phi_TD_s=Diff_Phi_TD_s';
end 
%%
function  [A,B]=M_K(N,Time_Delay)
Delta=zeros(N,N);
A=zeros(N,N);
B=zeros(N,N);
for i=1:N
    for j=1:N
       if i==j
        Delta(i,j)=1;
       else 
         Delta(i,j)=0;  
       end
    end
    end 
for i=1:N
    for j=1:N
A(i,j)=(Time_Delay*Delta(i,j))/(2*i-1);  
      if i<j
       if rem(i+j, 2) == 1 
           B(i,j)=2; 
       else 
          B(i,j)=0;     
    end 
   end 
  end 
end
end
%%
function [L]=Evaluation(KP1,KD1,alpha,Phi_0_s,Phi_TD_s,M,K)
kp=KP1;
kd=KD1;
%% G1&G2
G1=alpha*Phi_0_s'-kp*Phi_TD_s';
G2=-kd*Phi_TD_s';
%% C3
C3=Phi_0_s'*inv(M)*Phi_0_s;
%% X Matrix
X1=(Phi_0_s*Phi_0_s')/C3;
X2=-((Phi_0_s*Phi_0_s'/M)*K)/C3;
X3=(Phi_0_s*G1)/C3;
X4=(Phi_0_s*G2-(Phi_0_s*Phi_0_s'/M)*K)/C3;
%% L Matrix
L1=K/M+X2/M;
L2=X1/M;
L3=X3/M;
L4=K/M+X4/M;
L=[L1 L2;L3 L4];
end

3 个评论
显示 1更早的评论隐藏 1更早的评论

darova 2021-8-4

Read about boundary and polyarea

shahin sharafi 2021-8-4

Thank you for your response. Could you please use boundary function for my code for an example? I copied the code in the question part. I really need it. Thank you!

请先登录，再进行评论。

请先登录，再回答此问题。