finding the right matrix

ni(i,1:2) is supposed to be a 2x2 vector but i keep getting a 1x2, help please
x = [0 0; 200 0; 200 25; 25 25; 25 200; 0 200;0 0];
n = size(x,1); % size will be equal to 7
n = n-1;
% initialize properties
Asum = 0 ; % a scalar
psum = zeros(1,2); % 1 x 2 matrix
R = zeros(2,2); % 2 x 2 matrix
% 3. loop over sides
for i = 1:n % 5 points
h(i) = norm(x(i,:));
% from 3a as well
ni(i,1:2) = x(i,:)'; %------------------------------------
b(i) =norm(x(i+1,:)-(x(i,:)));
mi(i,1:2) = (x(i+1,:)-x(i,:))';
if h>0
ni = ni/h(i);
end
if b(i)>0
mi(i,1:2) = mi(i,1:2)/b(i);
end

5 件のコメント

Walter Roberson
Walter Roberson 2019 年 8 月 4 日
In order for ni(i, 1:2) to be 2x2 then i would need to be a vector of length 2. But you are in a for i loop so i is a scalar.
alexis cordova
alexis cordova 2019 年 8 月 4 日
so how will i go about making that a matrix while storing it as well
Walter Roberson
Walter Roberson 2019 年 8 月 4 日
編集済み: Walter Roberson 2019 年 8 月 4 日
Do you want ni(i,1:2) to end up being two copies of x(i,:)' with both columns the same? If not then which two rows should be used, and should you be using two rows for norm(x(i,:));
alexis cordova
alexis cordova 2019 年 8 月 4 日
so when the loop is ran once im trying to get this below for mi and ni and aslo store it
Capture.PNG
but instead im getting
Capture.PNG
Walter Roberson
Walter Roberson 2019 年 8 月 4 日
mi(i,1:2) = (x(i+1,:)-x(i,:))';
The left hand side is one row and two columns. Perhaps you want
mi(1:2, i) = (x(i+1,:)-x(i,:)).';

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