Loop not working properly

2 vues (au cours des 30 derniers jours)
Tom
Tom le 11 Août 2019
Modifié(e) : the cyclist le 11 Août 2019
I have a piece of code which creates a 3 x 46 array
theta = 0;
phi = pi/4;
r = 1 : 0.2 : 10;
theta = repmat(theta,size(r));
phi = repmat(phi, size(r)); %obtain position vectors for the points along a radial line where the velocities are evaluated%
[x,y,z] = sph2cart(theta,phi,r);
p = [x; y; z];
I then have a loop where you evaluate a function of the distance from each point represented in this array to a set of other points forming a 3 x 100 array:
for II=1:46
for IJ=1:100
normal=(p(1:3,II)/(norm(p(1:3,II))))';
r=(p(1:3,II)-sing(1:3,IJ))';
A((1+(II-1)*3):(3+(II-1)*3),(1+(IJ-1)*3):(3+(IJ-1)*3))=gradletSlip(r,in)';
end
end
However, something seems to be wrong as when I loop through IJ the number changes as it is a different distance each time, but when II changes from 1 to 2, it is supposed to be at another point in the 3 x 46 array and yet the normal is not changing, but this is just the radial vector, so it seems like it is not moving to the new point in the 3 x 46 array as it does the loop. Any idea how to fix this?

Réponse acceptée

the cyclist
the cyclist le 11 Août 2019
What you are observing seems correct to me.
theta is 0, and phi is constant. This means that even as r increases
  • y is zero (because theta is 0)
  • x and z increase in proportion to each other
(You can see this is true in the xyz array, if you place a breakpoint and look at the values).
Therefore, the normal vector always points in the same direction -- and is constant because it is also normalized.
All seems well.
Side note:
You can take the calculation
normal=(p(1:3,II)/(norm(p(1:3,II))))';
outside of the IJ loop, because its value doesn't depend on IJ. Just calculate it once for ever II loop.
  2 commentaires
Tom
Tom le 11 Août 2019
Hmm, I've tried setting theta to pi/2, for example, and still get that the normal is always the same, I suppose if theta and phi are both constant, the normal always points in the same direction and then after normalisation, you are going to have the same vector as for the unit sphere?
the cyclist
the cyclist le 11 Août 2019
Modifié(e) : the cyclist le 11 Août 2019
Sounds right to me. By construction, xyz is a vector where only the radius increases, so the normal vector would remain constant (but different for different choices of theta and phi).

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