I'm Davide and I have a problem with the derivation of a function that later should be given to ode15i in Matlab.
Basically I've derived a big symbolic expression that describe the motion of a spececraft with a flexible appendice (like a solar panel). My goal is to obtain a function handle that can be integrated using the built-in implicit solver in Matlab (ode15i).
The problem I've encounter is the slowness of the Symbolic computations, especially in the function "daeFunction" (I've run it and lost any hope for a responce after 3/4 hours had passed).
The system of equations, that is derived using the Lagrange's method is an implicit ode.
The complex nature of the system arise from the flexibility modelling of the solar panel.
I am open to any suggestions that would help me in:
running the code properly.
running it as efficiently as possible.
Thx in advance.
Here after I copy the code. Note: Matlab r2021a was used.
close all
clear
clc
syms t
syms r(t) [3 1]
syms angle(t) [3 1]
syms delta(t)
syms beta(t) [3 1]
mu = 3.986e14;
mc = 1600;
mi = 10;
k = 10;
kt = 10;
Ii = [1 0 0 % for the first link it is different thus I should do a functoin or something that writes everything into an array or a vector
0 5 0
0 0 5];
% Dimension of satellite
a = 1;
b = 1.3;
c = 1;
Ic = 1/12*mc* [b^2+c^2 0 0
0 c^2+a^2 0
0 0 a^2+b^2];
ra_c = [0 1 0]';
a = diff(r,t,t);
ddelta = diff(delta,t);
dbeta = diff(beta,t);
dddelta = diff(delta,t,t);
ddbeta = diff(beta,t,t);
R= [cos(angle1).*cos(angle3)-cos(angle2).*sin(angle1).*sin(angle3) sin(angle1).*cos(angle3)+cos(angle2).*cos(angle1).*sin(angle3) sin(angle2).*sin(angle3)
-cos(angle1).*sin(angle3)-cos(angle2).*sin(angle1).*cos(angle3) -sin(angle1).*sin(angle3)+cos(angle2).*cos(angle1).*cos(angle3) sin(angle2).*cos(angle3)
sin(angle2).*sin(angle3) -sin(angle2).*cos(angle1) cos(angle2)];
d_angle1 = diff(angle1,t);
d_angle2 = diff(angle2,t);
d_angle3 = diff(angle3,t);
dd_angle1 = diff(angle1,t,t);
dd_angle2 = diff(angle2,t,t);
dd_angle3 = diff(angle3,t,t);
d_angle = [d_angle1;d_angle2;d_angle3];
dd_angle = [dd_angle1;dd_angle2;dd_angle3];
omega = [d_angle2.*cos(angle1)+d_angle3.*sin(angle2).*sin(angle1);d_angle2.*sin(angle1)-d_angle3.*sin(angle2).*cos(angle1);d_angle1+d_angle3.*cos(angle2)]; % this should describe correctly omega_oc
d_omega = diff(omega,t);
v1 = diff(r1,t);
v2 = diff(r2,t);
v3 = diff(r3,t);
v = [v1; v2; v3];
[J,r_cgi,R_ci]= Jacobian_Rob(4,delta,beta);
% Perform matrix multiplication
for mm = 1:4
vel(:,mm) = J(:,:,mm)*[ddelta;dbeta];
end
vel = formula(vel);
dr_Ccgi = vel(1:3,:);
omega_ci = vel(4:6,:);
assumeAlso(angle(t),'real');
assumeAlso(d_angle(t),'real');
assumeAlso(dd_angle(t),'real');
assumeAlso(r(t),'real');
assumeAlso(a(t),'real');
assumeAlso(v(t),'real');
assumeAlso(beta(t),'real');
assumeAlso(delta(t),'real');
assumeAlso(dbeta(t),'real');
assumeAlso(ddelta(t),'real');
assumeAlso(ddbeta(t),'real');
assumeAlso(dddelta(t),'real');
omega = formula(omega);
Tc = 1/2*v'*mc*v+1/2*omega'*R*Ic*R'*omega;
% kinetic energy of all appendices
for h = 1:4
Ti(h) = 1/2*v'*mi*v+mi*v'*skew(omega)*R*ra_c+mi*v'*skew(omega)*R*r_cgi(:,h)+mi*v'*R*dr_Ccgi(:,h)+1/2*mi*ra_c'*R'*skew(omega)'*skew(omega)*R*ra_c ...
+ mi*ra_c'*R'*skew(omega)'*skew(omega)*R*r_cgi(:,h)+mi*ra_c'*R'*skew(omega)'*R*dr_Ccgi(:,h)+1/2*omega'*R*R_ci(:,:,h)*Ii*(R*R_ci(:,:,h))'*omega ...
+ omega'*R*R_ci(:,:,h)*Ii*R_ci(:,:,h)'*omega_ci(:,h)+1/2*omega_ci(:,h)'*R_ci(:,:,h)*Ii*R_ci(:,:,h)'*omega_ci(:,h)+1/2*mi*r_cgi(:,h)'*R'*skew(omega)'*skew(omega)*R*r_cgi(:,h)+mi*r_cgi(:,h)'*R'*skew(omega)'*R*dr_Ccgi(:,h)...
+ 1/2*mi*dr_Ccgi(:,h)'*dr_Ccgi(:,h);
Ugi(h) = -mu*mi/norm(r,2)+mu*mi*r'/(norm(r,2)^3)*(R*ra_c+R*R_ci(:,:,h)*r_cgi(:,h));
end
Ugc = -mu*mc/norm(r,2);
Ue = 1/2*kt*(delta)^2+sum(1/2*k*(beta).^2);
U = Ugc+sum(Ugi)+Ue;
L = Tc + sum(Ti) - U;
D = 1/2 *100* (ddelta^2+sum(dbeta.^2));
%% Equation of motion derivation
eq = [diff(jacobian(L,v),t)'-jacobian(L,r)';
diff(jacobian(L,d_angle),t)'-jacobian(L,angle)';
diff(jacobian(L,ddelta),t)'-jacobian(L,delta)'+jacobian(D,ddelta)';
diff(jacobian(L,dbeta),t)'-jacobian(L,beta)'+jacobian(D,dbeta)'];
%% Reduction to first order sys
[sys,newVars,R1]=reduceDifferentialOrder(eq,[r(t); angle(t); delta(t); beta(t)]);
DAEs = sys;
DAEvars = newVars;
%% ode15i implicit solver
pDAEs = symvar(DAEs);
pDAEvars = symvar(DAEvars);
extraParams = setdiff(pDAEs,pDAEvars);
f = daeFunction(DAEs,DAEvars,'File','ProvaSum');
y0est = [6778e3 0 0 0.01 0.1 0.3 0 0.12 0 0 0 7400 0 0 0 0 0 0 0 0]';
yp0est = zeros(20,1);
opt = odeset('RelTol', 10.0^(-7),'AbsTol',10.0^(-7),'Stats', 'on');
[y0,yp0] = decic(f,0,y0est,[],yp0est,[],opt);
% Integration
[tSol,ySol] = ode15i(f,[0 0.5],y0,yp0,opt);
%% Funcitons
function [J,p_cgi,R_ci]=Jacobian_Rob(N,delta,beta)
% Function to compute Jacobian see Robotics by Siciliano
% N total number of links
% delta [1x1] beta [N-1x1] variable that describe position of the solar
% panel elements
beta = formula(beta);
L_link = [1 1 1 1]'; % Length of each link elements in [m], later to be derived from file or as function input
for I = 1 : N
A1 = Homog_Matrix(I,delta,beta);
A1 = formula(A1);
R_ci(:,:,I) = A1(1:3,1:3);
if I ~= 1
p_cgi(:,I) = A1(1:3,4) + A1(1:3,1:3)*[1 0 0]'*L_link(I)/2;
else
p_cgi(:,I) = A1(1:3,4) + A1(1:3,1:3)*[0 0 1]'*L_link(I)/2;
end
for j = 1:I
A_j = formula(Homog_Matrix(j,delta,beta));
z_j = A_j(1:3,3);
Jp(:,j) = skew(z_j)*(p_cgi(:,I)-A_j(1:3,4));
Jo(:,j) = z_j;
end
if N-I > 0
Jp(:,I+1:N) = zeros(3,N-I);
Jo(:,I+1:N) = zeros(3,N-I);
end
J(:,:,I)= [Jp;Jo];
end
J = formula(J);
p_cgi = formula(p_cgi);
R_ci = formula(R_ci);
end
function [A_CJ]=Homog_Matrix(J,delta,beta)
% This function is made sopecifically for the solar panel
% define basic rotation matrices
Rx = #(angle) [1 0 0
0 cos(angle) -sin(angle)
0 sin(angle) cos(angle)];
Ry = #(angle) [ cos(angle) 0 sin(angle)
0 1 0
-sin(angle) 0 cos(angle)];
Rz = #(angle) [cos(angle) -sin(angle) 0
sin(angle) cos(angle) 0
0 0 1];
if isa(beta,"sym")
beta = formula(beta);
end
L_link = [1 1 1 1]'; % Length of each link elements in [m], later to be derived from file or as function input
% Rotation matrix how C sees B
R_CB = Rz(-pi/2)*Ry(-pi/2); % Clarify notation: R_CB represent the rotation matrix that describe the frame B how it is seen by C
% it is the same if it was wrtitten R_B2C
% becouse bring a vector written in B to C
% frame --> p_C = R_CB p_B
% same convention used in siciliano how C sees B frame
A_AB = [R_CB zeros(3,1)
zeros(1,3) 1];
A_B1 = [Rz(delta) zeros(3,1)
zeros(1,3) 1];
A_12 = [Ry(-pi/2)*Rx(-pi/2)*Rz(beta(1)) L_link(1)*[0 0 1]'
zeros(1,3) 1];
if J == 1
A_CJ = A_AB*A_B1;
elseif J == 0
A_CJ = A_AB;
else
A_CJ = A_AB*A_B1*A_12;
end
for j = 3:J
A_Jm1J = [Rz(beta(j-1)) L_link(j-1)*[1 0 0]'
zeros(1,3) 1];
A_CJ = A_CJ*A_Jm1J;
end
end
function [S]=skew(r)
S = [ 0 -r(3) r(2); r(3) 0 -r(1); -r(2) r(1) 0];
end
I found your question beautiful. My suggestion is to manipulate the problem numerically. symbolic manipulation in Matlab is good but is much slower than numerical calculation. you can define easily the ode into a system of first-order odes and solve them using numerical integration functions like ode45. Your code is very lengthy and I couldn't manage to follow its details.
All the Best.
Yasien
I want to speed up my code. I always use vectorization. But in this code I have no idea how to avoid the for-loop. I would really appreciate a hint how to proceed.
thank u so much for your time.
close all
clear
clc
% generating sample data
x = linspace(10,130,33);
y = linspace(20,100,22);
[xx, yy] = ndgrid(x,y);
k = 2*pi/50;
s = [sin(k*xx+k*yy)];
% generating query points
xi = 10:5:130;
yi = 20:5:100;
[xxi, yyi] = ndgrid(xi,yi);
P = [xxi(:), yyi(:)];
% interpolation algorithm
dx = x(2) - x(1);
dy = y(2) - y(1);
x_ = [x(1)-dx x x(end)+dx x(end)+2*dx];
y_ = [y(1)-dy y y(end)+dy y(end)+2*dy];
s_ = [s(1) s(1,:) s(1,end) s(1,end)
s(:,1) s s(:,end) s(:,end)
s(end,1) s(end,:) s(end,end) s(end,end)
s(end,1) s(end,:) s(end,end) s(end,end)];
si = P(:,1)*0;
M = 1/6*[-1 3 -3 1
3 -6 3 0
-3 0 3 0
1 4 1 0];
tic
for nn = 1:numel(P(:,1))
u = mod(P(nn,1)- x_(1), dx)/dx;
jj = floor((P(nn,1) - x_(1))/dx) + 1;
v = mod(P(nn,2)- y_(1), dy)/dy;
ii = floor((P(nn,2) - y_(1))/dy) + 1;
D = [s_(jj-1,ii-1) s_(jj-1,ii) s_(jj-1,ii+1) s_(jj-1,ii+2)
s_(jj,ii-1) s_(jj,ii) s_(jj,ii+1) s_(jj,ii+2)
s_(jj+1,ii-1) s_(jj+1,ii) s_(jj+1,ii+1) s_(jj+1,ii+2)
s_(jj+2,ii-1) s_(jj+2,ii) s_(jj+2,ii+1) s_(jj+2,ii+2)];
U = [u.^3 u.^2 u 1];
V = [v.^3 v.^2 v 1];
si(nn) = U*M*D*M'*V';
end
toc
scatter3(P(:,1), P(:,2), si)
hold on
mesh(xx,yy,s)
This is the full example and is a cubic B-spline surface interpolation algorithm in 2D space.
I saw other topics about this error but I couldn't figure it out. The error "In an assignment A(I) = B, the number of elements in B and I must be the same" occurs at the second for loop. How can I change my code to avoid this error?
h1 = [70 31.859 15 5.774 3.199 2.15 1.626];
h2 = [31.859 15 5.774 3.199 2.15 1.626 1.415];
b = [1253 1253 1253 1253 1253 1253 1253];
R = [455.4 425.6 377.6 374.9 371.3 273.7 268.3];
r = [0.5448714286 0.5291754292 0.6150666667 0.4459646692 0.3279149734 0.2437209302 0.1297662977];
k = [200 200 200 200 200 200 200];
s = sqrt(r/(1-r));
v2 = [20 0 0 0 0 0 0];
v1 = [0 0 0 0 0 0 0];
Ch1 = [0 0 0 0 0 0 0];
Ch2 = [0 0 0 0 0 0 0];
C = [100 100 100 100 100 100 100];
F = b .* k .* sqrt(R-(h1-h2))- R.*sin((acos((R-((h1-h2)./2))./R))) .* (pi/2) .* (1./sqrt(r./(1-r))) .* (atan(sqrt(r./(1-r))))-(pi/4) - (1./(sqrt(r./(1-r)) .* sqrt(h2./R))).* log((h2+R.*((sqrt(h1./R).*tan(1/2 .* atan(sqrt(r./(1-r)).*sqrt(h1./r).*log(1./(1-k))))).^2).*sqrt(1-r))./h2)
M = -R.*R.*(k./2).*(.2*(sqrt(h2./R)*tan(0.5*(atan(s)))-(pi/8).*sqrt(h2./R).*log(1./1-r)))-(acos((R-((h1-h2)./2))./R))
for i=1:6
v1(i) = ((v2(i)*h2)/h1);
v2(i+1) = v1(i);
end
vr = ((v1.*h1)./h2)./(((tan(0.5.*((atan(s)))-(pi/8).*sqrt(h2./R).*log(1./(1-r)))).^(2))+1)
%--------------------------------------------------------------------------
% Calculating E
w = (((2.*R.*h2).^(3/2))./(300.*(b.^2)))
if (w <= (3*10^-4));
E = ((0.0821.*((log(w))^2))+(1.25.*log(w))+4.89)
end
if ((3*10^-4)<= w <= (2.27*10^-3));
E = ((0.0172.*((log(w)).^2))+(0.175.*log(w))+0.438)
end
if (w > (2.27*10^-3))
E = 0.01
end
%--------------------------------------------------------------------------
% Calculating Ch:
y = ((((2.*R).^(0.5)).*((h2).^(1.5)))./(b.^2))
N1 = (0.5-(1/pi).*atan((log(y)+8.1938)./(1.1044)))
N = ((h2./h1).*N1)
for i=1:1;7
Ch2(i) = (h2.*((N.*((Ch1(i)./h2)-(C./h2)))+(C/h2)))
Ch1(i+1) = Ch2(i)
end
DeltaStrain = (E.*((Ch2./h2)-(Ch1./h1)))
if DeltaStrain > 0;
Stepp = ((2/pi).*(sqrt(DeltaStrain))))
Control = 2;
else
Stepp = ((2/pi).*(sqrt(-DeltaStrain))
Control = 0;
end
In the line
Ch2(i) = (h2.*((N.*((Ch1(i)./h2)-(C./h2)))+(C/h2)))
h2 is a vector, and Ch2(i) is a scalar. You cannot assign the value of a vector to a scalar. I suspect you want to replace the entire for loop. Right now you have
for i=1:1;7
Ch2(i) = (h2.*((N.*((Ch1(i)./h2)-(C./h2)))+(C/h2)))
Ch1(i+1) = Ch2(i)
end
(?? what is the meaning of 1:1;7? Is that a typo? I am thinking you want 1:7...
Since you seem to be using the result of one loop to change the value of Ch1 which you are using again in the next loop, it may be tricky to vectorize; but I wonder what you are expecting the output to be, since you really do have a vector as the result of the RHS of the equation. I can't be sure if you want to compute the result for one element at a time, or whether you want to compute vectors (and end up appending results to Ch1 and Ch2). The following line would run without throwing an error - but it may not be the calculation you want. Please clarify what you are hoping to achieve if this is an incorrect guess.
for i = 1:7
Ch2(i) = h2.*(N.*((Ch1(i) - C(i))./h2(i))) + C(i)./h2(i);
Ch1(i+1) = Ch2(i);
end
I'm doing a homework assignment for scientific computing, specifically the iterative methods Gauss-Seidel and SOR in matlab, the problem is that for a matrix gives me unexpected results (the solution does not converge) and for another matrix converges.
Heres the code of sor, where:
A: Matrix of the system A * x = b
Xini: array of initial iteration
b: array independent of the system A * x = b
maxiter: Maximum Iterations
tol: Tolerance;
In particular, the SOR method, will receive a sixth parameter called w which corresponds to the relaxation parameter.
Here´s the code for sor method:
function [x2,iter] = sor(A,xIni, b, maxIter, tol,w)
x1 = xIni;
x2 = x1;
iter = 0;
i = 0;
j = 0;
n = size(A, 1);
for iter = 1:maxIter,
for i = 1:n
a = w / A(i,i);
x = 0;
for j = 1:i-1
x = x + (A(i,j) * x2(j));
end
for j = i+1:n
x = x + (A(i,j) * x1(j));
end
x2(i) = (a * (b(i) - x)) + ((1 - w) * x1(i));
end
x1 = x2;
if (norm(b - A * x2) < tol);
break;
end
end
Here´s the code for Gauss-seidel method:
function [x, iter] = Gauss(A, xIni, b, maxIter, tol)
x = xIni;
xnew = x;
iter = 0;
i = 0;
j = 0;
n = size(A,1);
for iter = 1:maxIter,
for i = 1:n
a = 1 / A(i,i);
x1 = 0;
x2 = 0;
for j = 1:i-1
x1 = x1 + (A(i,j) * xnew(j));
end
for j = i+1:n
x2 = x2 + (A(i,j) * x(j));
end
xnew(i) = a * (b(i) - x1 - x2);
end
x= xnew;
if ((norm(A*xnew-b)) <= tol);
break;
end
end
For this input:
A = [1 2 -2; 1 1 1; 2 2 1];
b = [1; 2; 5];
when call the function Gauss-Seidel or sor :
[x, iter] = gauss(A, [0; 0; 0], b, 1000, eps)
[x, iter] = sor(A, [0; 0; 0], b, 1000, eps, 1.5)
the output for gauss is:
x =
1.0e+304 *
1.6024
-1.6030
0.0011
iter =
1000
and for sor is:
x =
NaN
NaN
NaN
iter =
1000
however for the following system is able to find the solution:
A = [ 4 -1 0 -1 0 0;
-1 4 -1 0 -1 0;
0 -1 4 0 0 -1;
-1 0 0 4 -1 0;
0 -1 0 -1 4 -1;
0 0 -1 0 -1 4 ]
b = [1 0 0 0 0 0]'
Solution:
[x, iter] = sor(A, [0; 0; 0], b, 1000, eps, 1.5)
x =
0.2948
0.0932
0.0282
0.0861
0.0497
0.0195
iter =
52
The behavior of the methods depends on the conditioning of both matrices? because I noticed that the second matrix is better conditioned than the first. Any suggestions?
From the wiki article on Gauss-Seidel:
convergence is only guaranteed if the matrix is either diagonally dominant, or symmetric and positive definite
Since SOR is similar to Gauss-Seidel, I expect the same conditions to hold for SOR, but you might want to look that one up.
Your first matrix is definitely not diagonally dominant or symmetric. Your second matrix however, is symmetric and positive definite (because all(A==A.') and all(eig(A)>0)).
If you use Matlab's default method (A\b) as the "real" solution, and you plot the norm of the difference between each iteration and the "real" solution, then you get the two graphs below. It is obvious the first matrix is not ever going to converge, while the second matrix already produces acceptable results after a few iterations.
Always get to know the limitations of your algorithms before applying them in the wild :)