An error of unable to perform assignment because the size of the left side is 1-by-1-by-3 and the size of the right side is 1-by-3-by-3 when I am using the following:

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Ioannis 2024-3-5

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https://ww2.mathworks.cn/matlabcentral/answers/2090741-an-error-of-unable-to-perform-assignment-because-the-size-of-the-left-side-is-1-by-1-by-3-and-the-si

评论： Ioannis 2024-9-3，11:01

% Vectors and Matrices Initialization:
v1_2=zeros(360,76,3); % each element of the 360x76 matrix is a velocity vector
v2_prime_2=zeros(360,76,3);
r1=zeros(360,3);
r2_prime=zeros(360,76,3);
delta_t=zeros(360,76);
string='pro';
% Main Code:
for i=1:1:360
    for j=1:1:76
        
        % The inputs in the following function are non-zero(
        % r1, r2_prime, delta_t and string). Previous calculations have been
        % made inside the double for loop with these matrices and vectors, but they do not affect the error
        % of this code and thus they are not mentioned.
        
        [v1_2(i,j,:), v2_prime_2(i,j,:)] = lambert_equation(r1(i,:), r2_prime(i,j,:), delta_t(i,j), string);
        
    end
end
Unrecognized function or variable 'lambert_equation'.
%The error is probably in the calculations made for V1 and V2 of the following function:
function [V1, V2] = Lambert(R1, R2, t, str)
%   This function solves Lambert's problem.
%   INPUTS: 
%       R1 = position vector at position 1 (km)
%       R2 = position vector at position 2 (km)
%       t  = time of flight  (s) 
%   OUTPUTS: 
%       V1 = velocity vector at position 1 (km/s)
%       V2 = velocity vector at position 2 (km/s)
%
%   VARIABLES DESCRIPTION:
%       r1, r2     - magnitudes of R1 and R2 
%       c12        - cross-product of R1 and R2 
%       theta      - change in true anomaly between position 1 and 2 
%       z          - alpha*x^2, where alpha is the reciprocal of the
%                    semimajor axis and x is the universal anomaly
%       y(z)       - a function of z
%       F(z,t)     - a function of the variable z and constant t
%       dFdz(z)    - the derivative of F(z,t)
%       ratio      - F/dFdz
%       eps        - tolerence on precision of convergence
%       C(z), S(z) - Stumpff functions
%% Gravitational parameter:    
global mu
%% Magnitudes of R1 and R2:
r1 = norm(R1);
r2 = norm(R2,'fro');
%% Theta:
c12 = cross(R1, squeeze(R2));
theta = acos(dot(R1,squeeze(R2))/r1/r2);
%Assume prograde or retrograde orbit:
if strcmp(str,'pro')
    if c12(3) <= 0
        theta = 2*pi - theta;
    end
elseif strcmp(str,'retro')
    if c12(3) >= 0
        theta = 2*pi - theta;
    end
end
%% Solve equation for z:
A = sin(theta)*sqrt(r1*r2/(1 - cos(theta)));
%Check where F(z,t) changes sign, and use that 
%value of z as a starting point:
z = -100;
while F(z,t) < 0
    z = z + 0.1;
end
%Tolerence and number of iterations:
eps= 1.e-8;
nmax = 5000;
%Iterate until z is found under the imposed tolerance:
ratio = 1;
n = 0;
while (abs(ratio) > eps) && (n <= nmax)
    n = n + 1;
    ratio = F(z,t)/dFdz(z);
    z = z - ratio;
end
%Maximum number of iterations exceeded
if n >= nmax
    fprintf('\n\n **Number of iterations exceeds %g \n\n ',nmax)
end
%% Lagrange coefficents:
f = 1 - y(z)/r1;
g = A*sqrt(y(z)/mu);
g_dot = 1 - y(z)/r2;
%% V1 and V2:
V1 =1/g*(R2 - f*R1);
V2 =1/g*(g_dot*R2 - R1);
return
%% Subfunctions used: 
function dum = y(z)
    dum = r1 + r2 + A*(z*S(z) - 1)/sqrt(C(z));
end
function dum = F(z,t)
    dum = (y(z)/C(z))^1.5*S(z) + A*sqrt(y(z)) - sqrt(mu)*t;
end
function dum = dFdz(z)
    if z == 0
        dum = sqrt(2)/40*y(0)^1.5 + A/8*(sqrt(y(0)) + A*sqrt(1/2/y(0)));
    else
        dum = (y(z)/C(z))^1.5*(1/2/z*(C(z) - 3*S(z)/2/C(z)) ...
        + 3*S(z)^2/4/C(z)) + A/8*(3*S(z)/C(z)*sqrt(y(z)) ...
        + A*sqrt(C(z)/y(z)));
    end
end
%Stumpff functions:
function c = C(z)
    if z > 0
        c = (1 - cos(sqrt(z)))/z;
    elseif z < 0
        c = (cosh(sqrt(-z)) - 1)/(-z);
    else
        c = 1/2;
    end
end
function s = S(z)
    if z > 0
        s = (sqrt(z) - sin(sqrt(z)))/(sqrt(z))^3;
    elseif z < 0
        s = (sinh(sqrt(-z)) - sqrt(-z))/(sqrt(-z))^3;
    else
        s = 1/6;
    end
end
end

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Walter Roberson 2024-3-5

We are missing lambert_equation

It is not the same as your Lambert -- your Lambert function is not designed to handle vectors.

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Answer 1

Drishti 2024-9-3，8:20

在 MATLAB Online 中打开

Hi Loannis,

I reproduced the error on my end by considering ‘Lambert’ as ‘lambert_equation’. The error you're encountering is due to a discrepancy in the dimensions of the variables being assigned from the 'lambert_equation' function.

I managed to fix the error by adjusting the shape of the variables. You can refer to the implemented code:

% Main Code:
for i=1:1:360
    for j=1:1:76
        % The inputs in the following function are non-zero(r1, r2_prime, delta_t and string). 
        % Ensure r2_prime(i,j,:) is correctly reshaped to a 1-by-3 vector
        r2_prime_vector = squeeze(r2_prime(i, j, :))';
        [v1_2(i,j,:), v2_prime_2(i,j,:)] = lambert_equation(r1(i,:), r2_prime_vector, delta_t(i,j), string);
    end
end