Filter löschen
Filter löschen

Info

Diese Frage ist geschlossen. Öffnen Sie sie erneut, um sie zu bearbeiten oder zu beantworten.

error in build rows of matrix

1 Ansicht (letzte 30 Tage)
mohamad khazaeezade
mohamad khazaeezade am 16 Feb. 2018
Geschlossen: MATLAB Answer Bot am 20 Aug. 2021
i have this code for Computing 'dim' as number of rows of 'M' : But this code gives an error:
Error using zeros Leading inputs must be numeric.
Error in Untitled3 (line 108) M=zeros(dim,9);
the code:
clc
clear all
close all
S0 =0.02;
b = 3;
Q = 29
n = 0.017
g = 9.81;
z = 0
ystart = 2
dy=0.1;
A1 = (b+z*ystart)*ystart;
P1 = b + (2*ystart*sqrt(1+z^2));
R1 = A1/P1;
T1= b+ 2*z*ystart;
D1=A1/T1;
F1 = (T1 * Q^2)/(g*(A1)^3);
Sf1 = (Q^2)*(n^2)/((R1^(4/3))*((A1)^2));
Gy1 = (1-F1)/(S0 - Sf1);
syms yc
A=(b+z*yc)*yc;
T=b+2*z*yc;
P=b+2*yc*sqrt(1+z^2);
R=A/P;
D=A/T;
eqn=(Q ^2)/g==(A^3)/(T);
y=vpasolve (eqn);
if z>0
for ii=1:6
if real(y(ii,1))>=0 , imag(y(ii,1))==0;
yc=y(ii,1);
end
end
else
for ii=1:3
if real(y(ii,1))>=0 , imag(y(ii,1))==0;
yc=y(ii,1);
end
end
end
yc;
fr = Q / (A1*sqrt(g*D1));
if fr > 1
disp('supcr')
elseif fr < 1
disp ('subcr')
elseif fr == 1
disp ('cr')
end
syms yn
An=(b+z*yn)*yn;
Tn=b+2*z*yn;
Pn=b+2*yn*sqrt(1+z^2);
Rn=An/Pn;
Dn=An/Tn;
Eq=Q==(An/n)*S0^(0.5)*(Rn)^(2/3);
yn=vpasolve(Eq,yn,1)
if yn > yc
if ystart > yn
disp ('M1')
y2=yn;
elseif ystart > yc
disp ('M2')
y2=yn;
elseif ystart < yc
disp ('M3')
y2=yc;
end
elseif yc > yn
if ystart > yc
disp ('S1')
y2=yc;
elseif ystart > yn
disp ('S2')
y2=yn;
elseif ystart < yn
disp ('S3')
y2=yn;
end
end
dim=floor((ystart-y2)/dy+1)
M=zeros(dim,9);
  1 Kommentar
Rik
Rik am 17 Feb. 2018
Bearbeitet: Rik am 17 Feb. 2018
As was said in the question that was deleted, you need to say what that line should do before we can give a meaningful answer. I could suggest M=zeros(max([1 dim]),9);, which would remove the error, but I can't say if that makes any sense at all for your code.
Have a read here and here. It will greatly improve your chances of getting an answer.

Antworten (0)

Diese Frage ist geschlossen.

Tags

Community Treasure Hunt

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

Start Hunting!

Translated by