Error: Unable to perform assignment because the size of the left side is 1-by-1 and the size of the right side is 1-by-41.

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Ashmika Gupta 2021-7-20

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https://ww2.mathworks.cn/matlabcentral/answers/882483-error-unable-to-perform-assignment-because-the-size-of-the-left-side-is-1-by-1-and-the-size-of-the

回答： DGM 2021-7-20

h=0:0.5:20
m=size(h,2)
E=0.01*10^9;
A=0.001;
n=6;
a=20;
r=20/(2.*sind(180/n));
Lab=a;
Lbc=a*(sind(38)/sind(30));
Lac=a*(sind(38+30)/sind(30));
U=zeros(1,m);
for i=1:m
    phi=acosd(h/(a*(sind(38)/sind(30))));
    lab=2.*r.*sind(180/n);
    lbc=sqrt(h(1,i).^2-2.*r.^2*cosd(phi)+2.*r.^2);
    lac=sqrt(h(1,i).^2-2.*r.^2*cosd(360/n+phi)+2.*r.^2);
    eab=(lab-Lab)/Lab;
    ebc=(lbc-Lbc)/Lbc;
    eac=(lac-Lac)/Lac;
    Ltotal=(Lab+Lbc+Lac);
    U(1,i)=(1/(2.*Ltotal)).*(Lab.*eab.^2+Lbc.*ebc.^2+Lac.*eac.^2)
end

I am not sure why I am getting this error. I want all variables except for U to be reassigned in each for loop. Please advise.

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Answer 1

DGM 2021-7-20

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It's unclear what your intentions are. If I assume you're trying to do this all elementwise WRT h, then you missed an index:

phi = acosd(h(i)/(a*(sind(38)/sind(30))));

... but if that's what you're trying to do, then you don't even need the loop at all anyway.

h=0:0.5:20;
m=size(h,2);
E=0.01*10^9;
A=0.001;
n=6;
a=20;
r=20/(2.*sind(180/n));
Lab=a;
Lbc=a*(sind(38)/sind(30));
Lac=a*(sind(38+30)/sind(30));
phi=acosd(h/(a*(sind(38)/sind(30))));
lab=2.*r.*sind(180/n);
lbc=sqrt(h.^2-2.*r.^2*cosd(phi)+2.*r.^2);
lac=sqrt(h.^2-2.*r.^2*cosd(360/n+phi)+2.*r.^2);
eab=(lab-Lab)/Lab;
ebc=(lbc-Lbc)/Lbc;
eac=(lac-Lac)/Lac;
Ltotal=(Lab+Lbc+Lac);
U=(1/(2.*Ltotal)).*(Lab.*eab.^2+Lbc.*ebc.^2+Lac.*eac.^2)

If instead, you're trying to calculate lac and lbc elementwise WRT h, and phi as a vector each time in the loop, then the result U cannot be a vector, but will need to be a mxm matrix.