I am trying to create a "plane", so to speak, of points in MATLAB from a set of initial points. I have so far been able to create only one row of points, with the algorithm shown below:
% Generate molecular orientation and position
a = 4.309; % lattice constant in angstroms
l = 10; % number of lattices desired
placeHolder = [0 0 0 ; a/2 a/2 0; a/2 0 a/2; 0 a/2 a/2]; % centers of molecules
numMol = 4; %input('how many molecules are in the unit cell? \n # molecules = ');
numAtoms = 2; %input('how many atoms per molecule? \n # atoms per molecule = ');
atomPerUC = numMol*numAtoms; % number of atoms per unit cell
dir = zeros(numMol,3); % array for orientations
atomPosition = zeros(numAtoms*l^3,3,numMol); % array for positions of atoms
theta = zeros(numMol,1); % array for theta values
phi = zeros(numMol,1); % array for phi values
b = 1.54; % bond length in angstroms
for kk = 1:numMol % generate unit cell
% disp(['What is the molecular orientation for molecule ',num2str(kk),' ?']);
% dir(kk,1) = input('u = '); % ask for user input for molecular orientations
% dir(kk,2) = input('v = ');
% dir(kk,3) = input('w = ');
dir = [1,1,1;-1,1,1;-1,-1,1;1,-1,1];
u = dir(kk,1); % set variables for theta, phi computation
v = dir(kk,2);
w = dir(kk,3);
theta(kk) = w/sqrt(u^2+v^2+w^2); % theta value for molecule k
if v<0 % phi value for molecule k
phi(kk) = 2*pi - acos(abs(v)/sqrt(u^2+v^2+w^2));
else if v>0
phi(kk) = acos(u/sqrt(u^2+v^2+w^2));
end
end
theta = theta(kk); phi = phi(kk); % set variables for theta, phi for x,y,z computation
xp = placeHolder(kk,1); % cooridnates of center of molecule k
yp = placeHolder(kk,2);
zp = placeHolder(kk,3);
x1 = (b/2)*sin(theta)*cos(phi) + xp; % cooridnates for atoms in molecule
x2 = -(b/2)*sin(theta)*cos(phi) + xp;
y1 = (b/2)*sin(theta)*sin(phi) + yp;
y2 = -(b/2)*sin(theta)*sin(phi) + yp;
z1 = (b/2)*cos(theta) + zp;
z2 = -(b/2)*cos(theta) + zp;
atomPosition(1,:,kk) = [x1 y1 z1];
atomPosition(2,:,kk) = [x2 y2 z2];
end
for k = 1:numMol
x01 = atomPosition(1,1,k); y01 = atomPosition(1,2,k); z01 = atomPosition(1,3,k);
x02 = atomPosition(2,1,k); y02 = atomPosition(2,2,k); z02 = atomPosition(2,3,k);
for ii = 1:l-1
atomPosition(2*ii+1,:,k) = [(atomPosition(2*ii-1,1,k) + a) y01 z01];
atomPosition(2*ii+2,:,k) = [(atomPosition(2*ii,1,k) + a) y02 z02];
end
end
My problem is that I do not know how to, from here, turn this "row" of points into a "plane" of points. This can be thought of as taking the points that are known on the x-axis, and creating similar rows moving up in the y-direction to create an x-y plane of points.
Any help/suggestions would be appreciated!
Though I do not understand exactly what you are trying to do. In a simple case you can move from a row of points to a plane by adding the extra dimension.
ie.
x=[1,2,3,4,5]
y=x^2
changes to
x=[1,2,3,4,5]
y=[1,2,3,4,5]
[x,y] = meshgrid(x,y)
z=x^2+y^2
Related
I am running simulations of a protein/membrane system, and want to quantify the degree of membrane deformation. I have averaged the membrane surface on a grid during the simulation, which results in a text file with three columns, containing the x, y, and z points of the membrane. I then convert this information to a mesh surface in matlab, which I then use to calculate the gaussian and/or mean curvature. The problem is, I'm getting similar values at the very beginning of the simulation, when the surface (membrane) is very flat, and at the end, when it is completely deformed. Unless I'm misunderstanding what curvature is, this is not correct and I believe I am doing something wrong in the matlab portion of this process. Here's the script where I loop over many frames (each of which is a different file containing x,y,z coordinates of the averaged membrane) and convert it to a mesh:
https://pastebin.com/reqWAz01
for i = 0:37
file = strcat('cg-topmem_pos-', num2str(i), '.out');
A = load(file);
x = A(:,1);
y = A(:,2);
z = A(:,3);
xv = linspace(min(x), max(x), 20);
yv = linspace(min(y), max(y), 20);
[X,Y] = meshgrid(xv, yv);
Z = griddata(x,y,z,X,Y);
[K,H,Pmax,Pmin] = surfature(X,Y,Z);
M = max(max(H))
if i == 0
fileID = fopen('Average2-NoTMHS-5nmns.txt', 'a');
fprintf(fileID, '%f %f\n',M);
fclose(fileID);
else
fileID = fopen('Average2-NoTMHS-5nmns.txt', 'a');
fprintf(fileID, '\n%f %f\n',M);
fclose(fileID);
end
end
Then using the following calculates curvature:
https://pastebin.com/5D21PdBQ
function [K,H,Pmax,Pmin] = surfature(X,Y,Z),
% SURFATURE - COMPUTE GAUSSIAN AND MEAN CURVATURES OF A SURFACE
% [K,H] = SURFATURE(X,Y,Z), WHERE X,Y,Z ARE 2D ARRAYS OF POINTS ON THE
% SURFACE. K AND H ARE THE GAUSSIAN AND MEAN CURVATURES, RESPECTIVELY.
% SURFATURE RETURNS 2 ADDITIONAL ARGUMENTS,
% [K,H,Pmax,Pmin] = SURFATURE(...), WHERE Pmax AND Pmin ARE THE MINIMUM
% AND MAXIMUM CURVATURES AT EACH POINT, RESPECTIVELY.
% First Derivatives
[Xu,Xv] = gradient(X);
[Yu,Yv] = gradient(Y);
[Zu,Zv] = gradient(Z);
% Second Derivatives
[Xuu,Xuv] = gradient(Xu);
[Yuu,Yuv] = gradient(Yu);
[Zuu,Zuv] = gradient(Zu);
[Xuv,Xvv] = gradient(Xv);
[Yuv,Yvv] = gradient(Yv);
[Zuv,Zvv] = gradient(Zv);
% Reshape 2D Arrays into Vectors
Xu = Xu(:); Yu = Yu(:); Zu = Zu(:);
Xv = Xv(:); Yv = Yv(:); Zv = Zv(:);
Xuu = Xuu(:); Yuu = Yuu(:); Zuu = Zuu(:);
Xuv = Xuv(:); Yuv = Yuv(:); Zuv = Zuv(:);
Xvv = Xvv(:); Yvv = Yvv(:); Zvv = Zvv(:);
Xu = [Xu Yu Zu];
Xv = [Xv Yv Zv];
Xuu = [Xuu Yuu Zuu];
Xuv = [Xuv Yuv Zuv];
Xvv = [Xvv Yvv Zvv];
% First fundamental Coeffecients of the surface (E,F,G)
E = dot(Xu,Xu,2);
F = dot(Xu,Xv,2);
G = dot(Xv,Xv,2);
m = cross(Xu,Xv,2);
p = sqrt(dot(m,m,2));
n = m./[p p p];
% Second fundamental Coeffecients of the surface (L,M,N)
L = dot(Xuu,n,2);
M = dot(Xuv,n,2);
N = dot(Xvv,n,2);
[s,t] = size(Z);
% Gaussian Curvature
K = (L.*N - M.^2)./(E.*G - F.^2);
K = reshape(K,s,t);
% Mean Curvature
H = (E.*N + G.*L - 2.*F.*M)./(2*(E.*G - F.^2));
H = reshape(H,s,t);
% Principal Curvatures
Pmax = H + sqrt(H.^2 - K);
Pmin = H - sqrt(H.^2 - K);
Any help would be greatly appreciated. I'm afraid that there is some issue between how the mesh is created and how curvature is calculated, but I am not matlab literate and could use some help. Thanks very much.
I am trying to average my CFD data (which is in the form of a scalar N x M x P array; N corresponds to Y, M to x, and P to z) over a subset of time steps. I've tried to simplify the description of my desired averaging process below.
Rotate the grid at each time step by a specified angle (this is because the flow has a coherent structure that rotates and changes shape/size at each time step and I want to overlap them and find a time averaged form of the structure that takes into account the change of shape/size over time)
Drawing a sphere centered on the original unrotated grid
Identifying the grid points from all the rotated grids that lie within the sphere
Identify the indices of the grid points in each rotated grid
Use the indices to find the scalar data at the rotated grid points within the sphere
Take an average of the values within the sphere
Put that new averaged value at the location on the unrotated grid
I have a code that seems to do what I want correctly, but it takes far too long to finish the calculations. I would like to make it run faster, and I am open to changing the code if necessary. Below is version of my code that works with a smaller version of the data.
x = -5:5; % x - position data
y = -2:.5:5; % y - position data
z = -5:5; % z - position data
% my grid is much bigger actually
[X,Y,Z] = meshgrid(x,y,z); % mesh for plotting data
dX = diff(x)'; dX(end+1) = dX(end); % x grid intervals
dY = diff(y)'; dY(end+1) = dY(end); % y grid intervals
dZ = diff(z)'; dZ(end+1) = dZ(end); % z grid intervals
TestPoints = combvec(x,y,z)'; % you need the Matlab Neural Network Toolbox to run this
dXYZ = combvec(dX',dY',dZ')';
% TestPoints is the unrotated grid
M = length(x); % size of grid x - direction
N = length(y); % size of grid y - direction
P = length(z); % size of grid z - direction
D = randi([-10,10],N,M,P,3); % placeholder for data for 3 time steps (I have more than 3, this is a subset)
D2{3,M*N*P} = [];
PosAll{3,2} = [];
[xSph,ySph,zSph] = sphere(50);
c = 0.01; % 1 cm
nu = 8e-6; % 8 cSt
s = 3*c; % span for Aspect Ratio 3
r_g = s/sqrt(3);
U_g = 110*nu/c; % velocity for Reynolds number 110
Omega = U_g/r_g; % angular velocity
T = (2*pi)/Omega; % period
dt = 40*T/1920; % time interval
DeltaRotAngle = ((2*pi)/T)*dt; % angle interval
timesteps = 121:123; % time steps 121, 122, and 123
for ti=timesteps
tj = find(ti==timesteps);
Theta = ti*DeltaRotAngle;
Rotate = [cos(Theta),0,sin(Theta);...
0,1,0;...
-sin(Theta),0,cos(Theta)];
PosAll{tj,1} = (Rotate*TestPoints')';
end
for i=1:M*N*P
aa = TestPoints(i,1);
bb = TestPoints(i,2);
cc = TestPoints(i,3);
rs = 0.8*sqrt(dXYZ(i,1)^2 + dXYZ(i,2)^2 + dXYZ(i,3)^2);
handles.H = figure;
hs = surf(xSph*rs+aa,ySph*rs+bb,zSph*rs+cc);
[Fs,Vs,~] = surf2patch(hs,'triangle');
close(handles.H)
for ti=timesteps
tj = find(timesteps==ti);
f = inpolyhedron(Fs,Vs,PosAll{tj,1},'FlipNormals',false);
TestPointsR_ti = PosAll{tj,1};
PointsInSphere = TestPointsR_ti(f,:);
p1 = [aa,bb,cc];
p2 = [PointsInSphere(:,1),...
PointsInSphere(:,2),...
PointsInSphere(:,3)];
w = 1./sqrt(sum(...
(p2-repmat(p1,size(PointsInSphere,1),1))...
.^2,2));
D_ti = reshape(D(:,:,:,tj),M*N*P,1);
D2{tj,i} = [D_ti(f),w];
end
end
D3{M*N*P,1} = [];
for i=1:M*N*P
D3{i} = vertcat(D2{:,i});
end
D4 = zeros(M*N*P,1);
for i=1:M*N*P
D4(i) = sum(D3{i}(:,1).*D3{i}(:,2))/...
sum(D3{i}(:,2));
end
D_ta = reshape(D4,N,M,P);
I expect to get an N x M x P array where each index is the weighted average of all the points covering all of the time steps at that specific position in the unrotated grid. As you can see this is exactly what I get. The major problem however is the length of time it takes to do so when I use the larger set of my 'real' data. The code above takes only a couple minutes to run, but when M = 120, N = 24, and P = 120, and the number of time steps is 24 this can take much longer. Based on my estimates it would take approximately 25+ days to finish the entire calculation.
Nevermind, I can help you with the math. What you are trying to do here is find things inside a sphere. You have a well-defined sphere so this makes things easy. Just find the distance of all points from the center point. No need to plot or use inpolyhedron. Note line 66 where I modify the points by the center point of the sphere, compute the distance of these points, and compare to the radius of the sphere.
% x = -5:2:5; % x - position data
x = linspace(-5,5,120);
% y = -2:5; % y - position data
y = linspace(-2,5,24);
% z = -5:2:5; % z - position data
z = linspace(-5,5,120);
% my grid is much bigger actually
[X,Y,Z] = meshgrid(x,y,z); % mesh for plotting data
dX = diff(x)'; dX(end+1) = dX(end); % x grid intervals
dY = diff(y)'; dY(end+1) = dY(end); % y grid intervals
dZ = diff(z)'; dZ(end+1) = dZ(end); % z grid intervals
TestPoints = combvec(x,y,z)'; % you need the Matlab Neural Network Toolbox to run this
dXYZ = combvec(dX',dY',dZ')';
% TestPoints is the unrotated grid
M = length(x); % size of grid x - direction
N = length(y); % size of grid y - direction
P = length(z); % size of grid z - direction
D = randi([-10,10],N,M,P,3); % placeholder for data for 3 time steps (I have more than 3, this is a subset)
D2{3,M*N*P} = [];
PosAll{3,2} = [];
[xSph,ySph,zSph] = sphere(50);
c = 0.01; % 1 cm
nu = 8e-6; % 8 cSt
s = 3*c; % span for Aspect Ratio 3
r_g = s/sqrt(3);
U_g = 110*nu/c; % velocity for Reynolds number 110
Omega = U_g/r_g; % angular velocity
T = (2*pi)/Omega; % period
dt = 40*T/1920; % time interval
DeltaRotAngle = ((2*pi)/T)*dt; % angle interval
timesteps = 121:123; % time steps 121, 122, and 123
for ti=timesteps
tj = find(ti==timesteps);
Theta = ti*DeltaRotAngle;
Rotate = [cos(Theta),0,sin(Theta);...
0,1,0;...
-sin(Theta),0,cos(Theta)];
PosAll{tj,1} = (Rotate*TestPoints')';
end
tic
for i=1:M*N*P
aa = TestPoints(i,1);
bb = TestPoints(i,2);
cc = TestPoints(i,3);
rs = 0.8*sqrt(dXYZ(i,1)^2 + dXYZ(i,2)^2 + dXYZ(i,3)^2);
% handles.H = figure;
% hs = surf(xSph*rs+aa,ySph*rs+bb,zSph*rs+cc);
% [Fs,Vs,~] = surf2patch(hs,'triangle');
% close(handles.H)
for ti=timesteps
tj = find(timesteps==ti);
% f = inpolyhedron(Fs,Vs,PosAll{tj,1},'FlipNormals',false);
f = sqrt(sum((PosAll{tj,1}-[aa,bb,cc]).^2,2))<rs;
TestPointsR_ti = PosAll{tj,1};
PointsInSphere = TestPointsR_ti(f,:);
p1 = [aa,bb,cc];
p2 = [PointsInSphere(:,1),...
PointsInSphere(:,2),...
PointsInSphere(:,3)];
w = 1./sqrt(sum(...
(p2-repmat(p1,size(PointsInSphere,1),1))...
.^2,2));
D_ti = reshape(D(:,:,:,tj),M*N*P,1);
D2{tj,i} = [D_ti(f),w];
end
if ~mod(i,10)
toc
end
end
D3{M*N*P,1} = [];
for i=1:M*N*P
D3{i} = vertcat(D2{:,i});
end
D4 = zeros(M*N*P,1);
for i=1:M*N*P
D4(i) = sum(D3{i}(:,1).*D3{i}(:,2))/...
sum(D3{i}(:,2));
end
D_ta = reshape(D4,N,M,P);
In terms of runtime, on my computer, the old code takes 57 hours to run. The new code takes 2 hours. At this point, the main calculation is the distance so I doubt you'll get much better.
I want to use streamline to show a vector field. The vector field is singular in a point. I want to remove regions near the singularity (fo example regions which their distance to singularity is less than 1). I wrote below code but it doesn't show anything. Could anyone help me?
clear all;
close all;
r1 = 1; r2 = 5; % Radii of your circles
x_0 = 0; y_0 = 0; % Centre of circles
[x,y] = meshgrid(x_0-r2:0.2:x_0+r2,y_0-r2:0.2:y_0+r2); % meshgrid of points
idx = ((x-x_0).^2 + (y-y_0).^2 > r1^2 & (x-x_0).^2 + (y-y_0).^2 < r2^2);
x = sort(x(idx));
[x, index] = unique(x);
y = sort(y(idx));
[y, index] = unique(y);
U=cos(x)/sqrt(x.^2+y.^2);
V=sin(x)/sqrt(x.^2+y.^2);
streamslice(x,y,U,V);
The problem with your code is that U and V are all zeros, so you get white space. The reason for that is that you don't use elementwise division with ./. So as a first step you should write:
U = cos(x)./sqrt(x.^2+y.^2);
V = sin(x)./sqrt(x.^2+y.^2);
Now U and V are not zeros but are also not matrices anymore, so they are not a valid input for streamslice. The reason for that is that x and y are converted to vectors when calling:
x = sort(x(idx));
y = sort(y(idx));
My guess is that you can remove all this indexing, and simply write:
r1 = 1; r2 = 5; % Radii of your circles
x_0 = 0; y_0 = 0; % Centre of circles
[x,y] = meshgrid(x_0-r2:0.2:x_0+r2,y_0-r2:0.2:y_0+r2); % meshgrid of points
U = cos(x)./sqrt(x.^2+y.^2);
V = sin(x)./sqrt(x.^2+y.^2);
streamslice(x,y,U,V);
so you get:
I think you misunderstood the concept of streamslice. Is this you expecting?
close all;
r1 = 1; r2 = 5; % Radii of your circles
x_0 = 0; y_0 = 0; % Centre of circles
[xx,yy] = meshgrid(x_0-r2:0.2:x_0+r2,y_0-r2:0.2:y_0+r2); % meshgrid of points
% idx = ((xx-x_0).^2 + (yy-y_0).^2 > r1^2 & (xx-x_0).^2 + (yy-y_0).^2 < r2^2);
% x = sort(xx(idx));
% [x, index] = unique(x);
% y = sort(yy(idx));
% [y, index] = unique(y);
U=cos(xx)./sqrt(xx.^2+yy.^2);
V=sin(xx)./sqrt(xx.^2+yy.^2);
streamslice(xx,yy,U,V);
I am trying to use Matlab's ifftn in 3-dimensions to get solution in physical space. In particular I am trying to use ifftn on 1/k^2. The analytical solution to that in physical space is 1/(4*pi*r). However I am not recovering that. Please note: $r = sqrt(x^2 + y^2 + z^2)$ and $k = sqrt(kx^2 + ky^2 + kz^2)$.
clc; clear;
n = 128; % no. of points for ifft
L = 2*pi; % size of the periodic domain
x = linspace(-L/2,L/2,n); y = x; z = x; % creating vectors for x-y-z direction
[X,Y,Z] = meshgrid(x,y,z); % creating meshgrid for physical space
R = sqrt(X.^2 + Y.^2 + Z.^2); % use for 1/(4*pi*r)
k = (2*pi/L)*[0:n/2 -n/2+1:-1]; % wave vector;
[k1,k2,k3] = meshgrid(k,k,k);
denom = (k1.^2 + k2.^2 + k3.^2); % This is k^2
F = 1./denom; F(1,1,1) = 0; % The first value is set to zero as it is infinite
phi = 1./(4*pi*R); % physical domain solution
phys = fftshift(ifftn(F)); % Using ifftn
ph_abs = abs(phys);
mid = ph_abs(n/2,:,:); % looking at the midplane of the output
mid = permute(mid,[3 2 1]); % permuting for contourplot.
PHI = phi(n/2,:,:); %looking at the midplane of the physical space.
PHI = permute(PHI,[3 2 1]);
figure(1)
surf(x,z,log(mid))
shading flat
colorbar();
figure(2)
surf(x,z,log10(abs(PHI)))
shading flat
colorbar();
Is there a way to display surf in its actual size and aspect ratio? I've been looking in the web but couldn't find anything which worked. I'm also quite new to MATLAB.
Here is my code:
rpos1 = 0; % image row vector pos counter
rpos2 = 0; % existing image vector pos counter
rpos3 = 0; % avg vector pos counter
prompt = {'Image location:','Lowest image #:','Highest image #:','Row of interest:','Background noise reduction:'};
dlg_title = 'Input';
num_lines = 1;
def = {'C:\Users\Moz\Desktop\Hyperspecdata\images','300','390','700','100'};
answer = inputdlg(prompt,dlg_title,num_lines,def);
directory = (answer{1});
x1 = str2num(answer{2});
x2 = x1;
y1 = str2num(answer{3});
y2 = y1-1;
z = str2num(answer{4});
v = str2num(answer{5});
finalslice = zeros(1,1312); % create matrix (imagecount x 1312)
%INSERT SLICES WITH 1 GAP%
for k = x1 : y1
baseFileName = sprintf('image0000000%03d.pgm',k);
fullFileName = fullfile(directory, baseFileName); %fullfile(folder, baseFileName);
A = imread(fullFileName);
A = floor(A./16); % transform back to 12 bit
B = A-v; % remove background noise
rpos1 = rpos1+1; % jump to next row
thisline = B(z,:); % desired row in images
finalslice(rpos1,:) = thisline; % add row vector
rpos1 = rpos1+1; % jump to next row
emptyline = zeros(1,1312); % create empty row vector
finalslice(rpos1,:) = emptyline; % insert empty row vector
end
%INSERT AVERAGES INTO GAPS%
for k = x2 : y2
rpos2 = rpos2+1; % find first existing vector
line1 = finalslice(rpos2,:);
rpos2 = rpos2+2; % find second existing vector
line2 = finalslice(rpos2,:);
avgline1 = (line1 + line2)/2; % average both
rpos3 = rpos3+2;
finalslice(rpos3,:) = avgline1; % insert average vector
rpos2 = rpos2-1; % jump back to second existing vector
end
figure(1)
h = surf(finalslice,'EdgeColor','none','LineStyle','none','FaceLighting','phong');
colormap('jet');
view(2)
I'm loading a bunch of images and taking slices at a specific points and stitching them together. The output doesn't display the actual size and aspect ratio.
SOLVED:
Added daspect()
o = max(max(finalslice));
[m n] = size(finalslice);
figure(1) h = surf(finalslice,'EdgeColor','none','LineStyle','none','FaceLighting','phong'); colormap('jet');
view(2)
daspect([m n o]);