Why is my code a matrix and not a double

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Zach Harrison
Zach Harrison on 16 Apr 2021
Commented: David Fletcher on 16 Apr 2021
Why is E_n a 4x4 matrix? I want it to be a single value.
Initial Conditions:
M_0 = [0, 30, 90, 135, 180, 200, 350;
0, 30, 90, 135, 180, 200, 350;
0, 30, 90, 135, 180, 200, 350;
0, 30, 90, 135, 180, 200, 350];
M = M_0 .* (pi / 180);
e = [0.01;
0.1;
0.5;
0.9];
E = zeros(4,7);
error = 10;
i = 2;
j = 2;
while j <= 4
while i <= 7
if M(i,j) > -pi && M(i,j) < 0 || M(i,j) > pi
E_n = M(i,j) - e;
elseif M(i,j) < pi && M(i,j) > 0
E_n = M(i,j) + e;
elseif M(i,j) == 0 || M(i,j) == 180
E_n = M(i,j);
end
while error >= 10^-6
E_n_plus_1 = E_n + ((M(i,j) - E_n + e(j) * sin(E_n)) / (1 - e(j) * cos(E_n)));
E_n = E_n_plus_1;
error = E_n / E_n_plus_1;
end
E(i,j) = E_n_plus_1;
i = i + 1;
end
j = j + 1;
end

Answers (2)

David Fletcher
David Fletcher on 16 Apr 2021
Edited: David Fletcher on 16 Apr 2021
The probable source of your error is the first two conditions in the if block - they result in E_n being a 4x1 vector since e is a 4x1 vector. Only the last condition in the if block will result in E_n being a scaler.
...
while j <= 4
while i <= 7
if M(i,j) > -pi && M(i,j) < 0 || M(i,j) > pi
E_n = M(i,j) - e; % <-------------------------------------------- 4x1
elseif M(i,j) < pi && M(i,j) > 0
E_n = M(i,j) + e; % <-------------------------------------------- 4x1
elseif M(i,j) == 0 || M(i,j) == 180
E_n = M(i,j); % Scaler
end
while error >= 10^-6
E_n_plus_1 = E_n + ((M(i,j) - E_n + e(j) * sin(E_n)) / (1 - e(j) * cos(E_n)));
E_n = E_n_plus_1;
error = E_n / E_n_plus_1;
end
E(i,j) = E_n_plus_1;
i = i + 1;
end
j = j + 1;
end

Robert Brown
Robert Brown on 16 Apr 2021
1 e = [0.01;
2 0.1;
3 0.5;
4 0.9];
5while j <= 4
6 while i <= 7
7 if M(i,j) > -pi && M(i,j) < 0 || M(i,j) > pi
8 E_n = M(i,j) - e;
9 elseif M(i,j) < pi && M(i,j) > 0
10 E_n = M(i,j) + e;
In lines 1-4, you define e as a 4x1 matrix
if you execute line 8 or line 10, E_n becomes a 4x1 matrix.
Your first loop executes line 10, and E_n = M(2,2) + e
>> M(2,2)
ans =
0.5236
e =
0.01
0.1
0.5
0.9
E_n =
0.5336
0.6236
1.0236
1.4236
So e, a 4x1 vector, automatically expands your answer (E_n) into a 4x1 vector
Your first loop executes line 10, and E_n = M(2,2) + e
Perhaps you wanted to use a single value from e, to keep E_n as a scalar value?
Did you want to add e(j) instead of just e?
  4 Comments
David Fletcher
David Fletcher on 16 Apr 2021
You are not resetting your inner loop back to the initial i value - so the inner loop will only ever run the first time

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