1 个投票

dimensions of X and Y are as given below
X = 1X89000 Y = 1X 89000
[uxmesh, uymesh] = meshgrid(X,Y);
would you please tell me what is wrong in the above statements? Whenever I execute the above statements, MATLAB keeps crashing..

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Jasmina
Jasmina 2024-6-18
1)
Q = 5e-9; %ładunek [C]
a = 0.4; %bok kwadratu[m]
k = 9e9; % stała Coulomba
EPS0 = 8.85e-12; %[F/m]
positions = [
-a/2, -a/2; %Q1
-a/2, a/2; %Q2
a/2, -a/2; %Q3
a/2, a/2; %Q4
];
charges = [-Q, Q, Q, -Q]; %wartości ładunków
x = linspace(-0.3, 0.3, 100);
y = linspace(-0.3, 0.3, 100);
[X, Y] = meshgrid(x, y); %siatka punktów w przestrzeni
V = zeros(size(X));
Ex = zeros(size(X));
Ey = zeros(size(Y));
%obliczanie potencjału i natężenia pola elektrycnzego
for k = 1:lenght(charges)
r = sqrt((X - positions(k, 1)).^2 + (Y - positions(k, 2)).^2);
V = V + charges(k) ./ (4*pi*EPS0*r);
Ex = Ex + charges(k) * (X - positions(k, 1)) ./ (4*pi*EPS0*r.^3);
Ey = Ey + charges(k) * (Y - positions(k, 2)) ./ (4*pi*EPS0*r.^3);
end
figure;
contourf(X, Y, V, 50, 'LineColor', 'none');
colorbar;
title('mapa potencjału elektrycznego');
xlabel('x[m]');
ylabel('y[m]');
hold on;
quiver(X, Y, Ex, Ey);
hold off;
2)
Q = 5e-9; %ładunek [C]
a = 0.4; %bok kwadratu[m]
k = 9e9; % stała Coulomba
EPS0 = 8.85e-12; %[F/m]
%pozycje ładunków
positions = [
-a/2, -a/2; %Q1
-a/2, a/2; %Q2
a/2, -a/2; %Q3
a/2, a/2; %Q4
];
charges = [-Q, Q, Q, -Q]; %wartości ładunków
x = linspace(-0.3, 0.3, 100);
y = linspace(-0.3, 0.3, 100);
[X, Y] = meshgrid(x, y); %siatka punktów w przestrzeni
V = zeros(size(X));
Ex = zeros(size(X));
Ey = zeros(size(Y));
%obliczanie potencjału i natężenia pola elektrycnzego
for k = 1
(charges)
r = sqrt((X - positions(k, 1)).^2 + (Y - positions(k, 2)).^2);
V = V + charges(k) ./ (4piEPS0*r);
s
Ex = Ex + charges(k) * (X - positions(k, 1)) ./ (4*pi*EPS0*r.^3);
Ey = Ey + charges(k) * (Y - positions(k, 2)) ./ (4*pi*EPS0*r.^3);
end
figure;
contourf(X, Y, V, 50, 'LineColor', 'none');
colorbar;
title('mapa potencjału elektrycznego');
xlabel('x[m]');
ylabel('y[m]');
hold on;
quiver(X, Y, Ex, Ey);
hold off;
3)
Q1 = 10e-9; %ładunek 1 [C]
Q2 = -1e-9; %ładunek 2 [C]
a = 0.1; %bok kwadratu [m]
k = 8.99e9; %stała Coulomba
%pozycje ładunków
M1 = [-a/2, a/2];
M2 = [a/2, -a/2];
N = 100;
t = linspace(0, 1, N);
x_path = M1(1) + t * (M2(1) - M1(1));
y_path = M1(2) + t * (M2(2) - M1(2));
W = 0;
for i = 1:N-1
r = sqrt((x_path(i) - 0).^2 + (y_path(i) - 0).^2);
F = k * Q1 * Q2 / r^2; %siła Coulomba
dx = x_path(i+1) - x_path(i);
dy = y_path(i+1) - y_path(i);
dl = sqrt(dx^2 + dy^2);
W = W + F * dl;
end
disp(['praca wykonana przez sile elektryczną: ', num2str(W), 'J']);
3)
Q = 8e-9; %ładunek [C]
a = 0.4; %długość boku trójkąta [m]
k = 8.99e9; %stała Coulomba
EPS0 = 8.85e-12; %przenikalność elek. próżni
h = (sqrt(3)/2)*a; %wysokość trójkąta
positions = [
0, -h/3;
a/2, h*2/3; %pozycje ładunków
-a/2, h*2/3;
];
charges = [Q, -Q, Q];
x = linspace(-0.3, 0.3, 100); %siatka punktów w przestrzeni
y = linspace(-0.3, 0.3, 100);
[X, Y] = meshgrid(x, y);
V = zeros(size(X));
Ex = zeros(size(X));
Ey = zeros(size(Y));
%obliczanie potencjału i natężenia pola elektrycnzgeo
for k = 1:length(charges)
r = sqrt((X - positions(k, 1)).^2 + (Y - positions(k, 2)).^2);
V = V + charges(k) ./ (4*pi*EPS0*r);
Ex = Ex + charges(k) * (X - positions(k, 1)) ./ (4*pi*EPS0*r.^3);
Ey = Ey + charges(k) * (X - positions(k, 2)) ./ (4*pi*EPS0*r.^3);
end
figure;
contourf(X, Y, V, 50, 'LineColor', 'none');
colorbar;
title('mapa potencjału elektrycznego');
xlabel('x[m[');
ylabel('y[m]');
hold on;
quiver(X, Y, Ex, Ey);
hold off;
4)
Q = 8e-9; %ładunek [C]
a = 0.4; %długość boku trójkąta [m]
k = 8.99e9; %stała Coulomba
EPS0 = 8.85e-12; %przenikalność elek. próżni
h = (sqrt(3)/2)*a; %wysokość trójkąta
positions = [
0, -h/3;
a/2, h*2/3; %pozycje ładunków
-a/2, h*2/3;
];
charges = [Q, -Q, Q];
x = linspace(-0.3, 0.3, 100); %siatka punktów w przestrzeni
y = linspace(-0.3, 0.3, 100);
[X, Y] = meshgrid(x, y);
V = zeros(size(X));
Ex = zeros(size(X));
Ey = zeros(size(Y));
%obliczanie potencjału i natężenia pola elektrycnzgeo
for k = 1:length(charges)
r = sqrt((X - positions(k, 1)).^2 + (Y - positions(k, 2)).^2);
V = V + charges(k) ./ (4*pi*EPS0*r);
Ex = Ex + charges(k) * (X - positions(k, 1)) ./ (4*pi*EPS0*r.^3);
Ey = Ey + charges(k) * (X - positions(k, 2)) ./ (4*pi*EPS0*r.^3);
end
figure;
contourf(X, Y, V, 50, 'LineColor', 'none');
colorbar;
title('mapa potencjału elektrycznego');
xlabel('x[m[');
ylabel('y[m]');
hold on;
quiver(X, Y, Ex, Ey);
hold off;

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回答（2 个）

Matt Fig
Matt Fig 2011-6-8

0 个投票

Are you running out of memory? Why not tell us what prompted you to ask the question???
There is nothing syntactically wrong with the statement. However, you better have
8*2*89000^2 bytes (read: LOTS!!) of free memory, assuming X and Y are doubles...

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priya
priya 2011-6-8
It won't mention anything like out of memory. It gets into not responding and the system hangs up.
sometimes, the system becomes slow...
I need to compare true velocities with calculated velocities. The calculated velocities are in a matrix whose dimensions are 246X389 and true velocities are in a matrix whose dimensions are 1X89000. To compare the true velocities and calculated velocities, I did Interpolation using interp2. meshgrid is required for interp2. Hence i did meshgrid for true velocities.
Is there anything wrong in my approach?
Matt Fig
Matt Fig 2011-6-8
What do you mean compare? Are you trying to find the calculated velocity that is closest to a true velocity for each element of the calculated velocity (or vis versa)? Are the velocities sorted?
priya
priya 2011-6-8
I need to find out the difference between true and calculated velocities and find out the error percentage of the velocities.
Matt Fig
Matt Fig 2011-6-8
In that case I would think it would be acceptable to use a FOR loop and loop over the calculated velocities one at a time to find where in the list of true velocities the closest number lies, like:
[err(ii), idx(ii)] = min(abs(cv(ii)-tv));
Walter Roberson
Walter Roberson 2011-6-8
Current versions of interp2() do not require meshgrid(). The documentation says (at least as of 2008b):
XI and YI can be matrices, in which case interp2 returns the values of Z corresponding to the points (XI(i,j),YI(i,j)). Alternatively, you can pass in the row and column vectors xi and yi, respectively. In this case, interp2 interprets these vectors as if you issued the command meshgrid(xi,yi).

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Walter Roberson
Walter Roberson 2011-6-8

0 个投票

Are X and Y vectors? If so, then the above should work, but only if you have enough memory: 89000 x 89000 double precision numbers requires between 59 and 60 gigabytes.

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priya
priya
2011-6-8

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Jasmina
2024-6-18

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