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I want to plot a two-dimensional function of the polar coordinates r and theta in three-dimensional cartesian coordinates. I have that (sorry about bad maths formatting, LaTeX not compatible, it seems)
f(r,theta) = r/2 * (cos(theta - pi/4) + sqrt(1 + 1/2 * cos(2*theta)))
Converting r and theta to cartesian coordinates
x = r * cos(theta), y = r * sin(theta)
Further, the domain is -1<r<1 and 0<theta<2 * pi, which I define by
r = -1:2/50:1;
and
theta = 0:2*pi/50:2*pi;
giving me two vectors of the same dimensions.
I can define the x and y values used for plotting as row vectors by
x = r. * cos(theta);
and
y = r. * sin(theta);
So now I need to define the z values, which will depend on the values of x and y. I thought I should make a 101x101 where each matrix element contains a data point of the final surface. But how should I do this? I thought about using a double for loop:
for i=1:numel(r)
for j=1:numel(theta)
z(i,j) = r(i)/2 .* cos(theta(j) - pi/4) + r(i).*sqrt(1 + 1/2 * cos(2.*theta(j)));
end
end
Then simply surf(z)
While this definitely gives me a surface, it gives me the incorrect surface! I don't know what is happening here. The incorrect surface is given in Figure 1, while the correct one is given in Figure 2. Can anyone help me out? For reference, the correct surface was plotted with GeoGebra, using
A = Function[<expression 1>, <Expresison 2>, <Expression 3>, <var 1>, <start>, <stop>, <var 2>, <start>, <stop>]
Figure 1. Incorrect surface.
Figure 2. Correct surface.
As others have said, you can use meshgrid to make this work.
Here's your example using gridded r and theta and an anonymous function to replace the double loop:
r = -1:2/50:1;
theta = 0:2*pi/50:2*pi;
% define anonymous function f(r,theta)
f = #(r,theta) r/2 .* (cos(theta - pi/4) + sqrt(1 + 1/2 .* cos(2.*theta)));
% generate grids for r and theta
[r, theta] = meshgrid(r,theta);
% calculate z from gridded r and theta
z = f(r,theta);
% convert r,theta to x,y and plot with surf
x = r.*cos(theta);
y = r.*sin(theta);
surf(x,y,z);
You need to use meshgrid to get matrix coordinates if you want to use surf. Taking your x and y (lower case), call
[X,Y] = meshgrid(x,y);
Then X and Y (upper case) will have the same values as you gave it, but laid out in the two-dimensional array as expected by surf. Loop over the indices here and compute your Z, which should have all(size(Z) == size(X)).
https://www.mathworks.com/help/matlab/ref/meshgrid.html
I have polar coordinates, radius 0.05 <= r <= 1 and 0 ≤ θ ≤ 2π. The radius r is 50 values between 0.05 to 1, and polar angle θ is 24 values between 0 to 2π.
How do I interpolate r = 0.075 and theta = pi/8?
I dunno what you have tried, but interp2 works just as well on polar data as it does on Cartesian. Here is some evidence:
% Coordinates
r = linspace(0.05, 1, 50);
t = linspace(0, 2*pi, 24);
% Some synthetic data
z = sort(rand(50, 24));
% Values of interest
ri = 0.075;
ti = pi/8;
% Manually interpolate
rp = find(ri <= r, 1, 'first');
rm = find(ri >= r, 1, 'last');
tp = find(ti <= t, 1, 'first');
tm = find(ti >= t, 1, 'last');
drdt = (r(rp) - r(rm)) * (t(tp) - t(tm));
dr = [r(rp)-ri ri-r(rm)];
dt = [t(tp)-ti ti-t(tm)];
fZ = [z(rm, tm) z(rm, tp)
z(rp, tm) z(rp, tp)];
ZI_manual = (dr * fZ * dt.') / drdt
% Interpolate with MATLAB
ZI_MATLAB = interp2(r, t, z', ri, ti, 'linear')
Result:
ZI_manual =
2.737907208525297e-002
ZI_MATLAB =
2.737907208525298e-002
Based on comments you have the following information
%the test point
ri=0.53224;
ti = pi/8;
%formula fo generation of Z
g=9.81
z0=#(r)0.01*(g^2)*((2*pi)^-4)*(r.^-5).*exp(-1.25*(r/0.3).^-4);
D=#(t)(2/pi)*cos(t).^2;
z2=#(r,t)z0(r).*D(t) ;
%range of vlaues of r and theta
r=[0.05,0.071175,0.10132,0.14422,0.2053, 0.29225,0.41602,0.5922,0.84299,1.2];
t=[0,0.62832,1.2566,1.885, 2.5133,3.1416,3.7699,4.3982,5.0265,5.6549,6.2832];
and you want interplation of the test point.
When you sample some data to use them for interpolation you should consider how to sample data according to your requirements.
So when you are sampling a regular grid of polar coordinates ,those coordinates when converted to rectangular will form a circular shape that
most of the points are concentrated in the center of the cricle and when we move from the center to outer regions distance between the points increased.
%regular grid generated for r and t
[THETA R] = meshgrid(t ,r);
% Z for polar grid
Z=z2(R,THETA);
%convert coordinate from polar to cartesian(rectangular):
[X, Y] = pol2cart (THETA, R);
%plot points
plot(X, Y, 'k.');
axis equal
So when you use those point for interpolation the accuracy of the interpolation is greater in the center and lower in the outer regions where the distance between points increased.
In the other word with this sampling method you place more importance on the center region related to outer ones.
To increase accuracy density of grid points (r and theta) should be increased so if length of r and theta is 11 you can create r and theta with size 20 to increase accuracy.
In the other hand if you create a regular grid in rectangular coordinates an equal importance is given to each region . So accuracy of the interpolation will be the same in all regions.
For it first you create a regular grid in the polar coordinates then convert the grid to rectangular coordinates so you can calculate the extents (min max) of the sampling points in the rectangular coordinates. Based on this extents you can create a regular grid in the rectangular coordinates
Regular grid of rectangular coordinates then converted to polar coordinated to get z for grid points using z2 formula.
%get the extent of points
extentX = [min(X(:)) max(X(:))];
extentY = [min(Y(:)) max(Y(:))];
%sample 100 points(or more or less) inside a region specified be the extents
X_samples = linspace(extentX(1),extentX(2),100);
Y_samples = linspace(extentY(1),extentY(2),100);
%create regular grid in rectangular coordinates
[XX YY] = meshgrid(X_samples, Y_samples);
[TT RR] = cart2pol(XX,YY);
Z_rect = z2(RR,TT);
For interpolation of a test point say [ri ti] first it converted to rectangular then using XX ,YY value of z is interpolated
[xi yi] = pol2cart (ti, ri);
z=interp2(XX,YY,Z_rect,xi,yi);
If you have no choice to change how you sample the data and only have a grid of polar points as discussed with #RodyOldenhuis you can do the following:
Interpolate polar coordinates with interp2 (interpolation for gridded data)
this approach is straightforward but has the shortcoming that r and theta are not of the same scale and this may affect the accuracy of the interpolation.
z = interp2(THETA, R, Z, ti, ri)
convert polar coordinates to rectangular and then apply an interpolation method that is for scattered data.
this approach requires more computations but result of it is more reliable.
MATLAB has griddata function that given scattered points first generates a triangulation of points and then creates a regular grid on top of the triangles and interpolates values of grid points.
So if you want to interpolate value of point [ri ti] you should then apply a second interpolation to get value of the point from the interpolated grid.
With the help of some information from spatialanalysisonline and Wikipedia linear interpolation based on triangulation calculated this way (tested in Octave. In newer versions of MATLAB use of triangulation and pointLocation recommended instead of delaunay and tsearch ):
ri=0.53224;
ti = pi/8;
[THETA R] = meshgrid(t ,r);
[X, Y] = pol2cart (THETA, R);
[xi yi] = pol2cart (ti, ri);
%generate triangulation
tri = delaunay (X, Y);
%find the triangle that contains the test point
idx = tsearch (X, Y, tri, xi, yi);
pts= tri(idx,:);
%create a matrix that repesents equation of a plane (triangle) given its 3 points
m=[X(pts);Y(pts);Z(pts);ones(1,3)].';
%calculate z based on det(m)=0;
z= (-xi*det(m(:,2:end)) + yi*det([m(:,1) m(:,3:end)]) + det(m(:,1:end-1)))/det([m(:,1:2) m(:,end)]);
More refinement:
Since it is known that the search point is surrounded by 4 points we can use only those point for triangulation. these points form a trapezoid. Each diagonal of trapezoid forms two triangles so using vertices of the trapezoid we can form 4 triangles, also a point inside a trapezoid can lie in at least 2 triangles.
the previous method based on triangulation only uses information from one triangle but here z of the test point can be interpolated two times from data of two triangles and the calculated z values can be averaged to get a better approximation.
%find 4 points surrounding the test point
ft= find(t<=ti,1,'last');
fr= find(cos(abs(diff(t(ft+(0:1))))/2) .* r < ri,1,'last');
[T4 R4] = meshgrid(t(ft+(0:1)), r(fr+(0:1)));
[X4, Y4] = pol2cart (T4, R4);
Z4 = Z(fr+(0:1),ft+(0:1));
%form 4 triangles
tri2= nchoosek(1:4,3);
%empty vector of z values that will be interpolated from 4 triangles
zv = NaN(4,1);
for h = 1:4
pts = tri2(h,:);
% test if the point lies in the triangle
if ~isnan(tsearch(X4(:),Y4(:),pts,xi,yi))
m=[X4(pts) ;Y4(pts) ;Z4(pts); [1 1 1]].';
zv(h)= (-xi*det(m(:,2:end)) + yi*det([m(:,1) m(:,3:end)]) + det(m(:,1:end-1)))/det([m(:,1:2) m(:,end)]);
end
end
z= mean(zv(~isnan(zv)))
Result:
True z:
(0.0069246)
Linear Interpolation of (Gridded) Polar Coordinates :
(0.0085741)
Linear Interpolation with Triangulation of Rectangular Coordinates:
(0.0073774 or 0.0060992) based on triangulation
Linear Interpolation with Triangulation of Rectangular Coordinates(average):
(0.0067383)
Conclusion:
Result of interpolation related to structure of original data and the sampling method. If the sampling method matches pattern of original data result of interpolation is more accurate, so in cases that grid points of polar coordinates follow pattern of data result of interpolation of regular polar coordinate can be more reliable. But if regular polar coordinates do not match the structure of data or structure of data is such as an irregular terrain, method of interpolation based on triangulation can better represent the data.
please check this example, i used two for loops, inside for loop i used condition statement, if u comment this condition statement and run the program, u'll get correct answer, after u uncomment this condition statement and run the program, u'll get wrong answer. please check it.
% Coordinates
r = linspace(0.05, 1, 10);
t = linspace(0, 2*pi, 8);
% Some synthetic data
%z = sort(rand(50, 24));
z=zeros();
for i=1:10
for j=1:8
if r(i)<0.5||r(i)>1
z(i,j)=0;
else
z(i,j) = r(i).^3'*cos(t(j)/2);
end
end
end
% Values of interest
ri = 0.55;
ti = pi/8;
% Manually interpolate
rp = find(ri <= r, 1, 'first');
rm = find(ri >= r, 1, 'last');
tp = find(ti <= t, 1, 'first');
tm = find(ti >= t, 1, 'last');
drdt = (r(rp) - r(rm)) * (t(tp) - t(tm));
dr = [r(rp)-ri ri-r(rm)];
dt = [t(tp)-ti ti-t(tm)];
fZ = [z(rm, tm) z(rm, tp)
z(rp, tm) z(rp, tp)];
ZI_manual = (dr * fZ * dt.') / drdt
% Interpolate with MATLAB
ZI_MATLAB = interp2(r, t, z', ri, ti, 'linear')
Result:
z1 =
0.1632
ZI_manual =
0.1543
ZI_MATLAB =
0.1582
So, I'm really new to MatLab, and I was trying to make a cylinder using [X Y Z] = cylinder;.Then I got these 3 matrices: X Y and Z that generate an actual cylinder if I mesh them. Now, what I need help to do is rotate this [X Y Z] cylinder 90 degrees clockwise in the y axis. I know there is this command called rotate but my teacher wants me to use rotation and translation matrices explicitly. How could I create these matrices and multiply them to the cylinder?Is there a better way to make the cylinder? I'm really not used to matlab, if you could explain in a detailed form, I would be very thankful.
You should use a rotation matrix for the R^3 which serves as a linear map. There are built in fucntions in MATLAB for that but I guess you are not allowed to use them.
Here is a quick and dirty solution:
[X Y Z] = cylinder;
figure;
surf(X,Y,Z);
% set up rotation matrix:
angle_in_degrees = 90;
angle_in_rad = angle_in_degrees* pi/180;
rotationMatrix = [cos(angle_in_rad) 0 sin(angle_in_rad); 0 1 0; -sin(angle_in_rad) 0 cos(angle_in_rad)];
% get points at the two rings and rotate them separately:
positionOld1 = [X(1,:)',Y(1,:)',Z(1,:)'];
positionOld2 = [X(2,:)',Y(2,:)',Z(2,:)'];
positionNew1 = positionOld1*rotationMatrix;
positionNew2 = positionOld2*rotationMatrix;
% reassemble the two sets of points into X Y Z format:
X = [positionNew1(:,1),positionNew2(:,1)];
Y = [positionNew1(:,2),positionNew2(:,2)];
Z = [positionNew1(:,3),positionNew2(:,3)];
figure;
surf(X,Y,Z);
I have this code, which takes a meshgrid, and applies a transformation to every point:
function [newx, newy] = transform(x, y)
newx = 10 * x + y*y;
newy = 5 * y;
end
[x, y] = meshgrid(1:5, 1:5);
[u, v] = arrayfun(#transform, x, y);
I want to plot the new mesh in 2D. The closest I can get is to do so in 3D by adding a Z component of 0:
mesh(u, v, zeros(size(u)))
How can I get matlab/octave to just show this plot on a 2d set of axes?
Maybe I'm missing the point here, but what's wrong with a simple plot(u,v,'b-x',u',v','b-x')?
I have a problem which is twofold:
How do I plot a toroidal surface in MATLAB, given a major radius R and a minor radius a? To avoid confusion, it is the toroidal/poloidal coordinate system, illustrated in the picture below, that I'm talking about.
Now, in any point (phi, theta) on this surface, the minor radius will be distorted by some value that I have stored in a matrix. How do I plot this distorted surface? This might be easy once I have the answer to part 1, but this is my actual goal, so any solution to part 1 that cannot handle this is pretty useless to me.
If anyone can tell me how to make the image appear smaller here, please do =)
Addressing your first question: first you will need to define the coordinates of your torus in toroidal coordinates (seems natural!) and then convert to Cartesian coordinates, which is how MATLAB expects all plots to be constructed (unless you are making a polar plot, of course). So we start of by defining our toroidal coordinates:
aminor = 1.; % Torus minor radius
Rmajor = 3.; % Torus major radius
theta = linspace(-pi, pi, 64) ; % Poloidal angle
phi = linspace(0., 2.*pi, 64) ; % Toroidal angle
We just want a single surface of the torus, so the minor radius is a scalar. We now make a 2D mesh of these coordinates:
[t, p] = meshgrid(phi, theta);
and convert to 3D Cartesian coordinates using the formulas given on the Wikipedia page linked to in the question:
x = (Rmajor + aminor.*cos(p)) .* cos(t);
y = (Rmajor + aminor.*cos(p)) .* sin(t);
z = aminor.*sin(p);
Now we have our torus defined in 3D, we can plot it using surf:
surf(x, y, z)
axis equal
Edit: To address your second question, it depends how you have your distortions stored, but say you have a matrix defined at each toroidal and poloidal point, you will just have to multiply the constant that is the minor radius by the distortion. In the following I create a distortion matrix that is the same dimensions as my toroidal and poloidal angle coordinate matrices and that is normally distributed about a mean of 1 with a FWHM of 0.1:
distortion = 1. + 0.1 * randn(64, 64);
x = (Rmajor + aminor .* distortion .*cos(p)) .* cos(t);
y = (Rmajor + aminor .* distortion .* cos(p)) .* sin(t);
z = aminor.* distortion .* sin(p);
surf(x, y, z)
The result of which is:
You can do this with surf - just create matrices with the x,y,z coordinates. The page you linked to has the trig equations to do this (they are exactly what you'd come up with on your own - I wrote the code below before checking your link).
R = 10;
a = 3;
tx=nan(41,21);
ty=nan(41,21);
tz=nan(41,21);
for j=1:21
for i=1:41
phi = (i-1)*2*pi/40;
theta = (j-1)*2*pi/20;
tx(i,j)= cos(phi) * (R+cos(theta)*a);
ty(i,j)= sin(phi) * (R+cos(theta)*a);
tz(i,j)= sin(theta)*a;
end
end
figure
surf(tx,ty,tz)
axis equal
To distort the surface, replace the constant a with a matrix of the desired minor radius values, and index into that matrix - i.e. tz(i,j) = sin(theta)*distortion(i,j) (but in all dimensions, obviously).