Storing multiple matrices from a loop into a single variable without cell function

Hello guys I need help in storing the A matrix for the loop at each iteration. I want to know if I can store this A matrix without using the cell function. Here is the code:

clear
clc
clear all

n = 26;
q1(1) = 0.002*n;
q2(1) = 0.001*n;
q3(1) = 0.005*n;
q4(1) = sqrt(1-(q1(1)).^2-(q2(1)).^2-(q3(1)).^2);
wx(1) = 0.0002+0.0001*n;
wy(1) = 0.0003+0.0001*n;
wz(1) = 0.0004+0.0001*n;
jx = 2.1*10^-3;
jy = 2*10^-3;
jz = 1.9*10^-3;
delta_t = 0.1;
Nt = 3.6*10^-10;
t(1) = 0;

     
phi(1) = atan((2*(q2(1)*q3(1)+q1(1)*q4(1)))/(1-2*(q1(1).^2+q2(1).^2)));
theta(1) = asin(2*(q4(1)*q2(1)-q1(1)*q3(1)));
ksi(1) = atan((2*(q4(1)*q3(1)+q1(1)*q2(1)))/(1-2*(q2(1).^2+q3(1).^2)));

    

i = 1;
n_iteration = 54000;

while i<= n_iteration
    
    t(i+1)= t(i)+delta_t;
    
    wx(i+1) = wx(i)+(delta_t/jx)*(jy-jz)*wz(i)*wy(i)+(delta_t/jx)*Nt;
    wy(i+1) = wy(i)+(delta_t/jy)*(jz-jx)*wx(i)*wz(i)+(delta_t/jy)*Nt;
    wz(i+1) = wz(i)+(delta_t/jz)*(jx-jy)*wx(i)*wy(i)+(delta_t/jz)*Nt;
    q1(i+1) = q1(i)-0.5*delta_t*(q2(i)*wx(i)+q3(i)*wy(i)+q4(i)*wz(i));
    q2(i+1) = q2(i)+0.5*delta_t*(q1(i)*wx(i)-q4(i)*wy(i)+q3(i)*wz(i));
    q3(i+1) = q3(i)+0.5*delta_t*(q4(i)*wx(i)+q1(i)*wy(i)-q2(i)*wz(i));
    q4(i+1) = q4(i)-0.5*delta_t*(q3(i)*wx(i)-q2(i)*wy(i)-q1(i)*wz(i));
    
    
    phi(i) = atan((2*(q2(i)*q3(i)+q1(i)*q4(i)))/(1-2*(q1(i).^2+q2(i).^2)));
    theta(i) = asin(2*(q4(i)*q2(i)-q1(i)*q3(i)));
    ksi(i) = atan((2*(q4(i)*q3(i)+q1(i)*q2(i)))/(1-2*(q2(i).^2+q3(i).^2)));

    A1_1(i) = cos(theta(i)).*cos(ksi(i));
    A1_2(i) = cos(theta(i)).*sin(ksi(i));    
    A1_3(i) = -sin(theta(i));
    A2_1(i) = -cos(phi(i)).*sin(ksi(i))+sin(phi(i)).*sin(theta(i)).*cos(ksi(i));
    A2_2(i) = cos(phi(i)).*cos(ksi(i))+sin(phi(i)).*sin(theta(i)).*sin(ksi(i));   
    A2_3(i) = sin(phi(i)).*cos(theta(i));
    A3_1(i) = sin(phi(i)).*sin(ksi(i))+cos(phi(i)).*sin(theta(i)).*cos(ksi(i));
    A3_2(i) = -sin(phi(i)).*cos(ksi(i))+cos(phi(i)).*sin(theta(i)).*sin(ksi(i));
    A3_3(i) = cos(phi(i)).*cos(theta(i));
    
    
    i=i+1
    
end

Answer 1

[Chendi Lin]

Hi Ilker,

If I understand your question correctly, A is a 3x3 matrix. For each i in the iteration, you want to store A without using cell.

Have you tried to store everything in a 3D matrix? For example,

As = zeros(n_iteration,3,3);
while i<= n_iteration
    As[i,:,:] = A;
end

Best,

CD

[Walter Roberson]

As = zeros(n_iteration,3,3);
while i<= n_iteration
  %compute A
  As(i,:,:) = A;
  i=i+1;
end

for loop would be more natural though.

[Ilker Enes Çirkin]

Hi, thank you for your reply. If i understand correctly you are suggesting that i can store the 3x3 A matrices up to down. If I am correct shouldn't I do,

As = zeros((n_iteration)*3,3,3

after that how would I put the A1_1 and A1_2 .... A3_3 values into this As matrix. Thank you.

[Chendi Lin]

As = zeros(n_iteration,3,3) initializes a 3D matrix. You can imagine that each As[i] is a 3x3 matrix, which matches the dimension of A.

I assume what you want is

As = zeros(n_iteration,3,3);
for i = 1:n_iteration
  %compute A
  As(i,1,1) = A1_1;
  As(i,1,2) = A1_2;
  ... etc
  i=i+1;
end

[Ilker Enes Çirkin]

Thank you for replying again. When I do this I get an error that says "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-54000." I don't get this.

[Chendi Lin]

Sorry that I was probably not clear about the modified code. In my code, A1_1 is a scalar value, while in your original code, A1_1 is a vector containing all values. If you try

As = zeros(n_iteration,3,3);
for i = 1:n_iteration
  %compute A
  A1_1 = cos(theta(i)).*cos(ksi(i));
  A1_2 = cos(theta(i)).*sin(ksi(i));  
  As(i,1,1) = A1_1;
  As(i,1,2) = A1_2;
  ... etc
  i=i+1;
end

Would that give you the correct answer?

[Ilker Enes Çirkin]

It did give me an answer but I wasn't satisfied with it. I can share the answer that worked better for me as:

I appriciate your help. Thank you.

A(3*i+1:3*i+3,1:3) = [cos(theta(i)).*cos(ksi(i))  cos(theta(i)).*sin(ksi(i))   -sin(theta(i));
                          -cos(phi(i)).*sin(ksi(i))+sin(phi(i)).*sin(theta(i)).*cos(ksi(i))   cos(phi(i)).*cos(ksi(i))+sin(phi(i)).*sin(theta(i)).*sin(ksi(i))   sin(phi(i)).*cos(theta(i));
                          sin(phi(i)).*sin(ksi(i))+cos(phi(i)).*sin(theta(i)).*cos(ksi(i))   -sin(phi(i)).*cos(ksi(i))+cos(phi(i)).*sin(theta(i)).*sin(ksi(i))   cos(phi(i)).*cos(theta(i));];

[Walter Roberson]

I think you will find that the code works much better if you use 3D indexing such as A(i,:,:) = [that array]

You can reshape() or permute() afterwards .