Dimensions of arrays being concatenated are not consistent.

3 vues (au cours des 30 derniers jours)
Frane
Frane le 5 Juil 2021
Commenté : Stephan le 6 Juil 2021
Hello there,
When I run my programm it says the error message written in the tittle. It says that for the 1st row of matrix A (line 79). Any suggestions?
Thanks.
P.S. I've uploaded the file in the message.

Réponse acceptée

Stephan
Stephan le 5 Juil 2021
clc;
clear all;
close all;
%% Parametri
%% Inercija; {kg.m2}
Juc = 2.6e-2;
Jlc = 1e-3;
%% Krutost; {N.m/rad}
kuc = 114.59;
klc = 687.55;
kr = 400000;
%% Prigusenje; {N.m.s/rad}
duc = 6e-2;
dlc = 6;
dr = 12000;
%% Radijus; {m}
rpin = 7.78e-3;
%% Sila; {N}
Fr = 1;
Flim = 75;
Flim_column = 0.2;
Flim_torsion = 0.01;
%% Masa; {kg}
mr = 10;
%% Moment; {N.m}
MsAmp = 4.5;
%% Kutna brzina; {rad/s}
freq = 2;
%%
Kt = {0.3 , 'N.m/a'};
J = 0.01;
b = 0.1;
L = 0.5;
R = 1;
K = 0.01;
kESF = 8e6;
kESF_column = 7000;
kESF_torsion = 7000;
%% Ulazni podatci
time = 0.01: 0.01: 0.2;
Ms = MsAmp*sin(2*pi*freq*time);
dm = 0;
Jm = 0;
%% Algoritam prema kojem se racuna I iz Ms
if Ms >=3.5
I = 21.29*Ms-69.4;
elseif Ms < 3.5 & Ms >= 2
I = 2.73 * Ms - 4.47;
elseif Ms < 2 & Ms >= 0
I = 0.5 * Ms;
elseif Ms < 0 & Ms >= (-2)
I = 0.5 * Ms;
elseif Ms < (-2) & Ms >= (-3.5)
I = 2.73 * Ms + 4.47;
elseif Ms < (-3.5)
I = 21.29 * Ms + 69.4;
end
%% Matrice A, B, C i D
A = [0 1 0 0 0 0 0 0 0;
-kuc/Juc -duc/Juc kuc/Juc duc/Juc 0 0 0 0 0;
0 0 0 1 0 0 0 0 0;
kuc/Jlc duc/Jlc -kuc/Jlc -duc/Jlc -klc/Jlc -dlc/Jlc klc/Jlc*rpin dlc/Jlc*rpin 0;
0 0 0 0 0 1 0 0 0;
0 0 0 0 0 0 0 1 0;
0 0 0 0 -klc/mr*rpin -dlc/mr*rpin klc/mr*rpin dlc/mr*rpin 0;
0 0 0 0 0 0 0 0 1;
0 0 0 0 0 0 0 0 -dm/Jm];
B = [0 0 0;
1 -K 0;
0 0 0;
0 K/Jlc 0;
0 0 0;
0 0 0;
0 0 1;
0 0 0;
0 K/Jm 0];
C = B.';
D = 0;
  2 commentaires
Frane
Frane le 5 Juil 2021
Modifié(e) : Frane le 5 Juil 2021
It works. Can you tell me what did you change so I know in the future?
Stephan
Stephan le 6 Juil 2021
A = [0 1 0 0 0 0 0 0 0;
-kuc/Juc -duc/Juc kuc/Juc duc/Juc 0 0 0 0 0;
0 0 0 1 0 0 0 0 0;
kuc/Jlc duc/Jlc -kuc/Jlc -duc/Jlc -klc/Jlc - dlc/Jlc klc/Jlc*rpin dlc/Jlc*rpin 0;
% ^
% |
% ---- change this
A = [0 1 0 0 0 0 0 0 0;
-kuc/Juc -duc/Juc kuc/Juc duc/Juc 0 0 0 0 0;
0 0 0 1 0 0 0 0 0;
kuc/Jlc duc/Jlc -kuc/Jlc -duc/Jlc -klc/Jlc -dlc/Jlc klc/Jlc*rpin dlc/Jlc*rpin 0;
% ^
% |
% ---- to this. Otherwise MATLB
% interprets:
% -klc/Jlc - dlc/Jlc
% as one expression, which leads
% to having one column
% missing. Better: use
% comma to seperate
% entries in matrices
% and vectors

Connectez-vous pour commenter.

Plus de réponses (0)

Catégories

En savoir plus sur Programming dans Help Center et File Exchange

Tags

Community Treasure Hunt

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

Start Hunting!

Translated by