I would like to generate a 3D binary mask which represents an ellipsoid with centers xc,yc,zc and radiuces xr,yr,zr.
I noticed that the function ellipsoid generates a mesh of points given these parameters. However, I want the data to be represented by a binary matrix (in my case, of size [100,100,100]), and not a mesh.
My Parameters are:
mask = zeros(100,100,100);
xc = 50; yc = 50; zc = 50;
xr = 15; yr = 15; zr = 15;
Thanks in advance!
To generate a binary mask of shapes which can use an equation you can follow the steps:
Generate a mesh (with ndgrid). Make sure the domain limits includes the volume/surface mask, and choose the mesh resolution according to your needs.
Use the volume/surface equation to generate a binary mask, by doing a simple logical comparison of the coordinates with the equation.
Simple 2D example:
Let's define a simple ellipse (in 2D).
%% // Simple 2D example
xc = 5 ; yc = 6 ; %// ellipse center = (5,6)
xr = 3 ; yr = 2 ; %// ellipse radiuses
xbase = linspace(0,10,11) ; %// temporary variable used to send to "ndgrid"
[xm,ym] = ndgrid( xbase , xbase ) ; %// generate base mesh
mask = ( ((xm-xc).^2/(xr.^2)) + ((ym-yc).^2/(yr.^2)) <= 1 ) %// get binary mask
Gives you the binary mask:
mask =
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 1 0 0 0 0
0 0 0 0 0 1 1 1 0 0 0
0 0 0 0 0 1 1 1 0 0 0
0 0 0 0 1 1 1 1 1 0 0
0 0 0 0 0 1 1 1 0 0 0
0 0 0 0 0 1 1 1 0 0 0
0 0 0 0 0 0 1 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
Granted you can hardly recognise an ellipse in the mask but I kept the resolution low to be able to display it as text. You can easily increase the resolution by defining a finer mesh.
3D Ellipsoid:
Well, it's exactly the same method, except we are going to add one dimension to the mesh, and use the 3D equation of the ellipsoid. So for your precise case:
%% // 3D ellipsoid
xc = 50; yc = 50; zc = 50;
xr = 15; yr = 15; zr = 15;
xbase = linspace(1,100,100) ;
[xm,ym,zm] = ndgrid( xbase , xbase , xbase) ;
mask = ( ((xm-xc).^2/(xr.^2)) + ((ym-yc).^2/(yr.^2)) + ((zm-zc).^2/(zr.^2)) <= 1 ) ;
I cannot show you a text output with these kind of 3D arrays, but your mask is now a 3D logical array containing true inside the ellipsoid and false elsewhere.
If I understood your question correctly, this should work:
mask = zeros(100, 100, 100);
%//your ellipsoid properties
xc = 50; yc = 50; zc = 50;
xr = 15; yr = 15; zr = 15;
for x=1:100
for y=1:100
for z=1:100
if ( ((x-xc)/xr)*((x-xc)/xr) + ((y-yc)/yr)*((y-yc)/yr) + ((z-zc)/zr)*((z-zc)/zr) < 1 )
mask(x,y,z) = 1; %//set elements within ellipsoid to 1
end
end
end
end
You can do this with Heaviside functions, this probably needs a bit more thought to get exactly what you want but as a start,
close all
clear all
%Setup Domain
maxdomain = 100;
mindomain = 0.;
step = 1.0;
X = mindomain:step:maxdomain;
Y = mindomain:step:maxdomain;
Z = mindomain:step:maxdomain;
[x,y,z] = meshgrid(X,Y,Z);
xc = 50; yc = 50; zc = 50;
xr = 15; yr = 15; zr = 15;
r2 = xr/2;
r = sqrt((x-xc).^2/xr + (y-yc).^2/yr + (z-zc).^2/zr);
u = heaviside(r-r2);
%Plot Surface of sphere
p = patch(isosurface(x,y,z,u));
isonormals(x,y,z,u,p)
set(p,'FaceColor','red','EdgeColor','none');
camlight ; alpha(0.6);
xlabel('x'); ylabel('y'); zlabel('z');
daspect([1,1,1]); view(3);
axis tight; camlight; camlight(-80,-10);
lighting gouraud;
which for your values above looks like,
and forxr = 15; yr = 45; zr = 15;,
The Heaviside function can be defined using,
function [out]=heaviside(x)
out=0.5.*(sign(x)+1.0);
end
if the Symbolic Math Toolbox is not available.
Related
I tried to make a 3dof(3 degrees of freedom mechanism) in matlab but i get this error and i don't know why.
this is for a school project and i need to simulate a human finger.
the code is running normal but after i enter the values for the angles it says that A,B,C are
undefined and i don't know why
a1 = input('valuarea lui q1(grade):');
a2 = input('valuarea lui q2(grade):');
a3 = input('valuarea lui q3(grade):');
L1=35;
L2=45;
L3=30;
z = [-10 10];
plot(z,10);
grid ON;
O=[0;0;0;1];
m= linspace(pi/2,pi/2+a1*pi/180,100);
n = linspace(-pi/2,a2*pi/180,100);
k=linspace(-pi/2,a3*pi/180,100);
for a=1:100
[A1,B1,C1] = Transform(m(a),n(a),k(a),L1,L2,L3);
x = [O(1) A(1) B(1) C(1)];
y = [O(2) A(2) B(2) C(2)];
Cx(i)= C1(1);
Cy(i) = C1(2);
i=i+1;
Plot = plot(x,y,'r',...1
'LineWidth',1);
title('Sumularea unui deget');
plot(Cx,Cy,'--g',...
'LineWidth',1);
pause(0.075);
delete(Plot);
end
plot(x,y,'r',...
'LineWidth',3);
function [A,B,C ] = Transform( m,n,p,l1,l2,l3 )
P = [0;0;0;1];
T1 = [cos(m) -sin(m) 0 0;sin(m) cos(m) 0 0;0 0 1 0; 0 0 0 1];
T2 = [cos(n) -sin(n) 0 11;sin(n) cos(n) 0 0;0 0 1 0; 0 0 0 1];
T3 = [cos(p) -sin(p) 0 12;sin(p) cos(p) 0 0;0 0 1 0; 0 0 0 1];
T4 = [1 0 0 13;0 1 0 0; 0 0 1 0; 0 0 0 1];
A = T1*T2*P;
B = T1*T2*T3*P;
C = T1*T2*T3*T3*P;
end
In your main function, this is the first use of the variables:
x = [O(1) A(1) B(1) C(1)];
They are never written previously. Instead A1 is written, which is a different variable. I guess you mixed the two up.
I am writing a function in Matlab to model the length of stay in hospital of stroke patients. I am having difficulty in storing my output values.
Here is my function:
function [] = losdf(age, strokeType, dest)
% function to mdetermine length of stay in hospitaal of stroke patients
% t = time since admission (days);
% age = age of patient;
% strokeType = 1. Haemorhagic, 2. Cerebral Infarction, 3. TIA;
% dest = 5.Death 6.Nursing Home 7. Usual Residence;
alpha1 = 6.63570;
beta1 = -0.03652;
alpha2 = -3.06931;
beta2 = 0.07153;
theta0 = -8.66118;
theta1 = 0.08801;
mu1 = 22.10156;
mu2 = 2.48820;
mu3 = 1.56162;
mu4 = 0;
nu1 = 0;
nu2 = 0;
nu3 = 1.27849;
nu4 = 0;
rho1 = 0;
rho2 = 11.76860;
rho3 = 3.41989;
rho4 = 63.92514;
for t = 1:1:365
p = (exp(-exp(theta0 + (theta1.*age))));
if strokeType == 1
initialstatevec = [1 0 0 0 0 0 0];
elseif strokeType == 2
initialstatevec = [0 1 0 0 0 0 0];
else
initialstatevec = [0 0 (1-p) p 0 0 0];
end
lambda1 = exp(alpha1 + (beta1.*age));
lambda2 = exp(alpha2 + (beta2.*age));
Q = [ -(lambda1+mu1+nu1+rho1) lambda1 0 0 mu1 nu1 rho1;
0 -(lambda2+mu2+nu2+rho2) lambda2 0 mu2 nu2 rho2;
0 0 -(mu3+nu3+rho3) 0 mu3 nu3 rho3;
0 0 0 -(mu4+nu4+rho4) mu4 nu4 rho4;
0 0 0 0 0 0 0;
0 0 0 0 0 0 0;
0 0 0 0 0 0 0];
Pt = expm(t./365.*Q);
Pt = Pt(strokeType, dest);
Ft = sum(initialstatevec.*Pt);
Ft
end
end
Then to run my function I use:
losdf(75,3,7)
I want to plot my values of Ft in a graph from from 0 to 365 days. What is the best way to do this?
Do I need to store the values in an array first and if so what is the best way to do this?
Many ways to do this, one straightforward way is to save each data point to a vector while in the loop and plot that vector after you exit your loop.
...
Ft = zeros(365,1); % Preallocate Ft as a vector of 365 zeros
for t = 1:365
...
Ft(t) = sum(initialstatevec.*Pt); % At index "t", store your output
...
end
plot(1:365,Ft);
As the title says, I want to covert a line(or any equation) to 2D matrix.
For example: If I have a line equation y = x, then I want it to be like:
0 0 0 0 0 1
0 0 0 0 1 0
0 0 0 1 0 0
0 0 1 0 0 0
0 1 0 0 0 0
1 0 0 0 0 0
and the length of rows and columns can be variable.
Is there a function or method to implement that?
use meshgrid to get x-y grids:
% set resolution parameters
xmin = 1;
xmax = 100;
dx = 0.1;
ymin = 1;
ymax = 100;
dy = 0.1;
% set x-y grid
[xg, yg] = meshgrid(xmin:dx:xmax, ymin:dy:ymax);
% define your function
f = #(x) (x - 30).^2;
% apply it on the x grid and get close y values
D = abs(f(xg) - yg);
bw = D <= 1;
figure;
imshow(bw);
% flip y axis
set(gca,'YDir','normal')
you get:
and then you can further dilate/erode/skeletonize the output
Given x and y coordinates, how to fill in a matrix with those coordinates? As it is said, just do it in a for loop, or use the "sub2ind" function.
% x,y coordinates
x=0:.01:30;
y=10*sin(2*pi*.1*x);
% add offset so that (x,y)-coordinates are always positive
x=x+abs(min(x))+1;
y=y+abs(min(y))+1;
figure,plot(x,y,'.');axis tight
x=ceil(x); y=ceil(y);
im=zeros(max(y),max(x));
ind=sub2ind(size(im),y,x);
im(ind)=1;
figure,imagesc(im),axis image, axis xy;colormap gray;axis tight
xlabel('x'); ylabel('y')
Why not doing it on your own? You loop through all x-coordinates of the matrix that you (probably scaled) use as the x for your function and get an y out, that you (probably scaled) can round and then use as a y coordinate for your matrix to set the 1.
I currently have a camera alignment model set in camera as origin coordinates:
http://i.imgur.com/UnSCAvG.png
and, since the camera was stationary I'm trying to convert the system to a pattern-fixed model
http://i.imgur.com/OKunxwA.png
Matlab is nice enough to show it to me but alas there I've found no way to extract the data unless it's on a fixed-camera model. It should be a simple matter of applying the inverse of the translation and the rotation to change between systems but unfortunately this is not working and I can't see why.
oldpoint = [0 0 0 1]';
translation = ([1 0 0 18.1043121043; 0 1 0 31.092351038; 0 0 1 -80.0610295707; 0 0 0 1]);
rotation = [eul2rotm([0.0957937074 -0.0234017529 -0.037084526]) zeros(3,1); 0 0 0 1];
newpoint = translation * point;
newpoint = rotation * newpoint;
I've tried several different alternatives to this but thus far none aproximate the coordinates I'm trying to get.
Two things were going wrong, matlab was not using euler angles and Z needed to be inverted.
clc;
clear;
%%
oldpoint = [0 0 0 1]';
newpoints = zeros(13,4);
i = 1;
while( i<= length(translations) )
trans = translations(i,:);
rota = rotations(i,:);
display(trans);
display(rota);
translation = ([1 0 0 -trans(1); 0 1 0 -trans(2); 0 0 1 trans(3); 0 0 0 1]);
rotation = [rotationVectorToMatrix([-rota(1) -rota(2) -rota(3)]) zeros(3,1); 0 0 0 1];
newpoint = translation * oldpoint;
newpoint = rotation * newpoint;
newpoints(i,:) = newpoint;
i = i + 1;
end
I am using Matlab and Euler Angles in order to reorient a 3axes coordinate system. Specifically,
Rz = [cos(ψ) sin(ψ) 0;-sin(ψ) cos(ψ) 0;0 0 1];
Ry = [cos(φ) 0 -sin(φ);0 1 0;sin(φ) 0 cos(φ)];
Rx = [1 0 0;0 cos(θ) -sin(θ);0 sin(θ) cos(θ)];
Rtotal = Rz*Ry*Rz
Then I loop through my old system coordinates (x,y,z) and make a vector coord_old. Then I get the reoriented system with (xn,yn,zn)
for i=1:size(num,1)
coord_old = [x(i,1);y(i,1);z(i,1)];
coord_new = Rtotal*coord_old;
xn(i,1) = coord_new(1,1);
yn(i,1) = coord_new(2,1);
zn(i,1) = coord_new(3,1);
end
My issue is that when θ,φ,ψ≃0 then x->-y and y->x and when θ,φ≃0 and ψ=90 then x and y will not rotate! That means that when x,y should rotate they don't and when they shouldn't rotate they stay as they were!
--EDIT--
For example, when ψ=20.0871, φ=0.0580 and θ=0.0088 I get these results
See that x->-y and y->x while z doesn't change at all!
Any thoughts?
Ok, I see two main problems here:
Rtotal = Rz*Ry*Rz is probably not what you want since Rz is multiplied twice. I think you mean Rtotal = Rz*Ry*Rx.
Your rotation matrix seems to be incorrect. Check this Wikipedia artice to get the correct signs.
Here a corrected rotation matrix:
Rz = [cos(psi) -sin(psi) 0; sin(psi) cos(psi) 0; 0 0 1];
Ry = [cos(phi) 0 sin(phi); 0 1 0; -sin(phi) 0 cos(phi)];
Rx = [1 0 0; 0 cos(theta) -sin(theta); 0 sin(theta) cos(theta)];
Rtotal = Rz*Ry*Rx;
With this matrix I get the correct results:
x=1; y=2; z=3;
psi=0; phi=0; theta=0;
[xn,yn,zn] >> 1 2 3
x=1; y=2; z=3;
psi=90/180*pi; phi=0; theta=0;
[xn,yn,zn] >> -2 1 3
And here a full graphical example of a cube in 3d-space:
% Create cube (not in origin)
DVert = [0 0 0; 0 1 0; 1 1 0; 1 0 0 ; ...
0 0 1; 0 1 1; 1 1 1; 1 0 1];
DSide = [1 2 3 4; 2 6 7 3; 4 3 7 8; ...
1 5 8 4; 1 2 6 5; 5 6 7 8];
DCol = [0 0 1; 0 0.33 1; 0 0.66 1; ...
0 1 0.33; 0 1 0.66; 0 1 1];
% Rotation angles
psi = 20 /180*pi; % Z
phi = 45 /180*pi; % Y
theta = 0 /180*pi; % X
% Rotation matrix
Rz = [cos(psi) -sin(psi) 0; sin(psi) cos(psi) 0; 0 0 1];
Ry = [cos(phi) 0 sin(phi); 0 1 0; -sin(phi) 0 cos(phi)];
Rx = [1 0 0; 0 cos(theta) -sin(theta); 0 sin(theta) cos(theta)];
Rtotal = Rz*Ry*Rz;
% Apply rotation
DVertNew = Rtotal * DVert';
% Plot cubes
figure;
patch('Faces',DSide,'Vertices',DVert,'FaceColor','flat','FaceVertexCData',DCol);
patch('Faces',DSide,'Vertices',DVertNew','FaceColor','flat','FaceVertexCData',DCol);
% Customize view
grid on;
axis equal;
view(30,30);
When I use your code and insert 0 for all angles, I get Rtotal:
Rtotal =
1 0 0
0 1 0
0 0 1
This is the identity matrix and will not change your values.
You have an error in your matrix multiplication. I think you should multiply: Rtotal*coord_old. I think you are missing the _old. depending on what is in you coordvariable, this may be the bug.
When I run:
for i=1:size(1,1)
coord_old = [1;2;3];
coord_new = Rtotal*coord_old;
xn(i,1) = coord_new(1,1);
yn(i,1) = coord_new(2,1);
zn(i,1) = coord_new(3,1);
end
I get the correct result:
coord_new =
1
2
3
Thank you both #Steffen and #Matt. Unfortunately, my reputation is not high enough to vote Up your answers!
The problem was not with Rtotal as #Matt correctly stated. It should be as it was Rz*Ry*Rx. However, both your ideas helped me test my code with simple examples (5 sets of coordinates and right hand rule), and realize where my (amateur) mistake was.
I had forgotten I had erased parts of codes where I was expressing my angles to degrees... I should be using sind & cosd instead of sin and cos.