Matrix dimensions must agree.

where is the error please ???

clear,clc
cp=1005;
r=287;
d=1.2;
v=300;
tx=284;
px=98;
to=298;

py=100:200;
a1=((d*v*r).^2)./(2*cp.*py);
a2=1;
a3=-to;
p=[a1 a2 a3];
ty=roots(p);
ds=(cp.*log(ty./tx))-(r.*log(py./px));

Answer 1

[Sean de Wolski]

Error using - Matrix dimensions must agree. That is an error with '-' (minus). Going to the only minus:

Compare sizes:

(r.*log(py./px))

and

(cp.*log(ty./tx))

There is your problem.

[mohamed saber]

what should i do ???

Answer 2

[Walter Roberson]

roots() returns a column vector, but everything else is a row vector.

Your a1 vector is the same length as py, but when you put a2 and a3 on the end of that, your p vector becomes 2 elements longer than py. roots() returns a vector one element shorter than its input vector, so roots() is going to return a vector one element longer than py. You then try to subtract between that vector of length of py + 1 and the vector of length of py.

[mohamed saber]

what about that ??

clear,clc

cp=1005;

r=287;

d=1.2;

v=300;

tx=284;

px=98;

to=298;

py=100:200;

a1=((d*v*r).^2)./(2*cp.*py);

a2=1;

a3=-to;

p=[a1 a2 a3];

ty=roots(p);

Ty=ty(1:end-1);

ds=(cp.*log(Ty./tx))-(r.*log(py./px));

[Walter Roberson]

Ty=ty(1:end-1) .';

in order to get the row vector you need.

Answer 3

[mohamed saber]

could you please write the syntax ???

Answer 4

[Andrei Bobrov]

cp=1005; 
r=287; 
d=1.2; 
v=300; 
tx=284; 
px=98; 
to=298;

py=(100:200).'; 
a1=((d*v*r).^2)./(2*cp.*py); 
a2=1; 
a3=-to; 
p=[a1; a2; a3]; 
ty=roots(p); 
Ty=ty(1:end-1); 
ds=(cp.*log(Ty./tx))-(r.*log(py./px));