- y is zero (because theta is 0)
- x and z increase in proportion to each other
Loop not working properly
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Tom
el 11 de Ag. de 2019
Editada: the cyclist
el 11 de Ag. de 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?
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the cyclist
el 11 de Ag. de 2019
What you are observing seems correct to me.
theta is 0, and phi is constant. This means that even as r increases
(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 comentarios
the cyclist
el 11 de Ag. de 2019
Editada: the cyclist
el 11 de Ag. de 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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