Why do I keep getting Error using + Arrays have incompatible sizes for this operation. Error in solver_DE_1 (line 37) x_c=lsin(theta(ang,1))+rcos(ang_ci);

Question

Drake Campo 2022-9-27

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1813330-why-do-i-keep-getting-error-using-arrays-have-incompatible-sizes-for-this-operation-error-in

评论： Walter Roberson 2022-9-27

在 MATLAB Online 中打开

Error using +

Arrays have incompatible sizes for this operation.

Error in solver_DE_1 (line 37)

x_c=l*sin(theta(ang,1))+r*cos(ang_ci);

Didnt get this error when i was running R2022a but now when i run R2022b it doesn't work. Is there any add ons i need?

clc
clear all
time=0:.01:15;
global l
[T,theta]=ode45(@vibdif,time,[pi/4;0;pi/4;0;pi/4;0;pi/4;0]);
theta(:,1)
figure(1)
plot(time,theta(:,1)*180/pi,'b',time,theta(:,3)*180/pi,'r','linewidth',2)
legend('LDE','NDE')
xlabel('Time(s)','fontsize',15)
ylabel('\theta(deg)','fontsize',15)
grid
% figure(2)
% plot(time,theta(:,5)*180/pi,'b','linewidth',2)
% xlabel('Time(s)','fontsize',15)
% ylabel('\theta(deg)','fontsize',15)
% grid
figure(3)
plot(time,theta(:,1)*180/pi,'b',time,theta(:,7)*180/pi,'r','linewidth',2)
legend('Sta','Mov')
xlabel('Time(s)','fontsize',15)
ylabel('\theta(deg)','fontsize',15)
grid
si=size(theta(:,1));
X_r=[-0.2 0.2 0.2 -0.2];
Y_r=[0 0 0.2 0.2];
ang_ci=0:0.1:2*pi;
r=0.06;
for ang=1:1:si(1,1)
    
    x_c=l*sin(theta(ang,1))+r*cos(ang_ci);
    y_c=-l*cos(theta(ang,1))+r*sin(ang_ci);
    
    x_cn=l*sin(theta(ang,3))+r*cos(ang_ci);
    y_cn=-l*cos(theta(ang,3))+r*sin(ang_ci);
    
    figure(4)
    plot([0;l*sin(theta(ang,1))],[0;-l*cos(theta(ang,1))],'b','linewidth',2)
    hold on
    plot([0;l*sin(theta(ang,3))],[0;-l*cos(theta(ang,3))],'r','linewidth',2)
    hold on
    fill(X_r,Y_r,'g')
    hold on
    text(0,-0.05,'O')
    hold on
    text(-1.5,-1.5,'Simulation for both linear and nonlinear','fontsize',14)
    hold on
    fill(x_c,y_c,'b')
    hold on
    fill(x_cn,y_cn,'r')
    hold off
    
    axis([-2 2 -2 0.3])
    grid
end

3 个评论
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Drake Campo 2022-9-27

在 MATLAB Online 中打开

function dm=vibdif(t,m);
global l
dm=zeros(8,1);
g=9.81;% this is gravity in SI system
l=1;% this is the lebbght of pendulum
dm(1)=m(2);
dm(2)=-(g/l)*m(1);%linearized DE around equli point
dm(3)=m(4);
dm(4)=-(g/l)*sin(m(3));%original nonlinear DE
dm(5)=m(6);
dm(6)=(g/l)*m(5);%inverted pendulum
dm(7)=m(8);
dm(8)=-(g/l)*m(8)+25*sin(5*t);%moving pendulum

Walter Roberson 2022-9-27

在 MATLAB Online 中打开

This code works for me in R2022b.

time=0:.01:15;
global l
[T,theta]=ode45(@vibdif,time,[pi/4;0;pi/4;0;pi/4;0;pi/4;0]);
theta(:,1)
figure(1)
plot(time,theta(:,1)*180/pi,'b',time,theta(:,3)*180/pi,'r','linewidth',2)
legend('LDE','NDE')
xlabel('Time(s)','fontsize',15)
ylabel('\theta(deg)','fontsize',15)
grid
% figure(2)
% plot(time,theta(:,5)*180/pi,'b','linewidth',2)
% xlabel('Time(s)','fontsize',15)
% ylabel('\theta(deg)','fontsize',15)
% grid
figure(3)
plot(time,theta(:,1)*180/pi,'b',time,theta(:,7)*180/pi,'r','linewidth',2)
legend('Sta','Mov')
xlabel('Time(s)','fontsize',15)
ylabel('\theta(deg)','fontsize',15)
grid
si=size(theta(:,1));
X_r=[-0.2 0.2 0.2 -0.2];
Y_r=[0 0 0.2 0.2];
ang_ci=0:0.1:2*pi;
r=0.06;
for ang=1:1:si(1,1)
    
    x_c=l*sin(theta(ang,1))+r*cos(ang_ci);
    y_c=-l*cos(theta(ang,1))+r*sin(ang_ci);
    
    x_cn=l*sin(theta(ang,3))+r*cos(ang_ci);
    y_cn=-l*cos(theta(ang,3))+r*sin(ang_ci);
    
    figure(4)
    plot([0;l*sin(theta(ang,1))],[0;-l*cos(theta(ang,1))],'b','linewidth',2)
    hold on
    plot([0;l*sin(theta(ang,3))],[0;-l*cos(theta(ang,3))],'r','linewidth',2)
    hold on
    fill(X_r,Y_r,'g')
    hold on
    text(0,-0.05,'O')
    hold on
    text(-1.5,-1.5,'Simulation for both linear and nonlinear','fontsize',14)
    hold on
    fill(x_c,y_c,'b')
    hold on
    fill(x_cn,y_cn,'r')
    hold off
    
    axis([-2 2 -2 0.3])
    grid
end
function dm=vibdif(t,m);
    global l
    dm=zeros(8,1);
    g=9.81;% this is gravity in SI system
    l=1;% this is the lebbght of pendulum
    dm(1)=m(2);
    dm(2)=-(g/l)*m(1);%linearized DE around equli point
    dm(3)=m(4);
    dm(4)=-(g/l)*sin(m(3));%original nonlinear DE
    dm(5)=m(6);
    dm(6)=(g/l)*m(5);%inverted pendulum
    dm(7)=m(8);
    dm(8)=-(g/l)*m(8)+25*sin(5*t);%moving pendulum
end