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Matrix dimensions must agree.

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mohamed saber
mohamed saber 2012-3-28
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));

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回答（4 个）

Sean de Wolski
Sean de Wolski 2012-3-28
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.
Walter Roberson
Walter Roberson 2012-3-28
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.
  2 个评论
mohamed saber
mohamed saber 2012-3-28
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
Walter Roberson 2012-3-28
Ty=ty(1:end-1) .';
in order to get the row vector you need.

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mohamed saber
mohamed saber 2012-3-29
could you please write the syntax ???
Andrei Bobrov
Andrei Bobrov 2012-3-29
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));

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