Hi, generatin random points inside a cuboid problem

4 vues (au cours des 30 derniers jours)
fatima-zahra achbah
fatima-zahra achbah le 9 Oct 2017
Hi, I need to generate a total I of points inside a specific cuboid(a*b*c) and plot it then after havings these points and their coordinations generate a sphere for each point.I already did a script but unfortunately, it gives me a cube results :
clc
close all
a=10; % mm
b=10; % mm
depth=0.3; %mm
i= 1145916; %total number of pores and thus of points inside the parallelepiped
v=i*[];
Vt=a*b*depth; %volume expression mm^3
s = 10;
n = bsxfun(@plus,0,s.*rand(i,3));
r=normrnd(0.005,0.001,[1 i]); %mu=0,005 mm ==>5µm
x=n(:,1);
y=n(:,2);
z=n(:,3);
scatter3(x,y,z,r,'fill')
for k=1:i
v(k)=(4*pi*(r(k)^3))/3;
end
figure
plot(v)
Best regards
  2 commentaires
KSSV
KSSV le 9 Oct 2017
You want random points inside sphere?
fatima-zahra achbah
fatima-zahra achbah le 9 Oct 2017
no, I want random points inside my cuboid and then draw a sphere on each point

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Walter Roberson
Walter Roberson le 9 Oct 2017
n = bsxfun(@plus,0,s.*rand(i,3));
The bsxfun to add 0 is not doing anything useful there. You might as well have
n = s .* rand(i,3);
However, the reason you get a cube is that you are not weighting the coordinates according to the size of their dimension. I would suggest,
n = rand(i, 3) * diag([a, b, depth]); %* rather than .* is deliberate
Or if you are using R2016b or later, you could use
n = rand(i, 3) .* [a, b, depth] ; %needs R2016b or later.
The bsxfun equivalent would be
n = bsxfun(@times, [a, b, depth], rand(i,3) );
In my R2017b system, the time required for bsxfun(@times, data, s) and data .* s are pretty much indistinguishable, and the time for data * diag(s) is about 50% more.
  1 commentaire
fatima-zahra achbah
fatima-zahra achbah le 9 Oct 2017
Oh my god, it works, thank you so much, I truly appreciate your help

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