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Dimensions of arrays being concatenated are not consistent.

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Frane
Frane on 5 Jul 2021
Commented: Stephan on 6 Jul 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.

Accepted Answer

Stephan
Stephan on 5 Jul 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 Comments
Stephan
Stephan on 6 Jul 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

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