I know about the function sph2cart, but I feel like I must be using it wrong. In a Calculus textbook, I saw that this following spherical equation:
ρ = 1 + 1/5*sin(6θ)*sin(5Φ)
produces something that looks like this:
I wanted to reproduce this in a Matlab graph, so I wrote the following code
[q,t] = meshgrid(linspace(0,pi,100),linspace(0,2*pi,100));
rho = 1+1/5*sin(6*t)*sin(5*q);
[x,y,z] = sph2cart(t,q,rho);
surf(x,y,z)
axis square, axis equal
And I got the following graph:
Why is this happening? Why am I not getting the bumpy sphere my calc textbook shows?
There are two problems:
You need .* rather than * between your sin(6*t) and sin(5*q) since you want per element multiplication not matrix multiplication.
You want q to run from 0-PI and t to run from 0-2PI
This code produces the result you're looking for.
[q,t] = meshgrid(linspace(0,2*pi,100),linspace(0,pi,100));
rho = 1+(1/5*sin(6*t).*sin(5*q));
[x,y,z] = sph2cart(t,q,rho);
surf(x,y,z)
axis square, axis equal
Related
i'm currently trying to plot this function:
Z^2 = (X.^2+Y.^2+2*w.*Z.*Y.*a)./(1-w^2*a^2)
Geogebra gives a lightcone https://en.wikipedia.org/wiki/Light_cone but crushes if i change the parameters a bit. I tried then matlab:
[X,Y] = meshgrid(-10:.5:10);
a=2;
w=1;
Z = (X.^2+Y.^2+2*w.*sqrt(abs(Z)).*Y.*a)./(1-w^2*a^2);
surf(X,Y,Z)
zlim([-5,5])
And it has too few points. I wish i could add some changing meshgrid, like (-5:.1:5), but it gives:
Arrays have incompatible sizes for this operation.
Probably due to sqrt(abs(Z)) in the equation. I don't know how to fix it.
Thanks
It's easier to directly generate the cone data with command cylinder
t = 0:pi/10:2*pi;
r =linspace(2,0,numel(t))
[X,Y,Z] = cylinder(r);
figure
hs1=surf(X,Y,Z)
Note that I have added the handle hs1 which is output of surf.
This way you can change any property of the generated cone surface explained in detail here :
https://uk.mathworks.com/help/matlab/ref/matlab.graphics.chart.primitive.surface-properties.html
I want to plot a surface in MATLAB using surf. I have this equation: x = y^2 +4z^2.
What I am doing is the following:
[x,y] = meshgrid(-4:.1:4, -4:.1:4);
z = sqrt((x - y.^2)./4); % Basically I'm just clearing for z
surf(x,y,z)
But with this I am getting the error: Error using surf X,Y,Z and C cannot be complex. I know there is a complex number because of the values that x and y have, plus the square root. Is there another way to plot a surface in MATLAB? because I really don't know what to do, and my skills are very basics.
Why do you feel that you need to grid x and y, and not use the form of the original equation itself?
This seems to work perfectly fine
[y,z] = meshgrid(-4:.1:4, -4:.1:4);
x = y.^2 + 4*z.^2;
surf(x,y,z)
to produce
MATLAB's surf command allows you to pass it optional X and Y data that specify non-cartesian x-y components. (they essentially change the basis vectors). I desire to pass similar arguments to a function that will draw a line.
How do I plot a line using a non-cartesian coordinate system?
My apologies if my terminology is a little off. This still might technically be a cartesian space but it wouldn't be square in the sense that one unit in the x-direction is orthogonal to one unit in the y-direction. If you can correct my terminology, I would really appreciate it!
EDIT:
Below better demonstrates what I mean:
The commands:
datA=1:10;
datB=1:10;
X=cosd(8*datA)'*datB;
Y=datA'*log10(datB*3);
Z=ones(size(datA'))*cosd(datB);
XX=X./(1+Z);
YY=Y./(1+Z);
surf(XX,YY,eye(10)); view([0 0 1])
produces the following graph:
Here, the X and Y dimensions are not orthogonal nor equi-spaced. One unit in x could correspond to 5 cm in the x direction but the next one unit in x could correspond to 2 cm in the x direction + 1 cm in the y direction. I desire to replicate this functionality but drawing a line instead of a surf For instance, I'm looking for a function where:
straightLine=[(1:10)' (1:10)'];
my_line(XX,YY,straightLine(:,1),straightLine(:,2))
would produce a line that traced the red squares on the surf graph.
I'm still not certain of what your input data are about, and what you want to plot. However, from how you want to plot it, I can help.
When you call
surf(XX,YY,eye(10)); view([0 0 1]);
and want to get only the "red parts", i.e. the maxima of the function, you are essentially selecting a subset of the XX, YY matrices using the diagonal matrix as indicator. So you could select those points manually, and use plot to plot them as a line:
Xplot = diag(XX);
Yplot = diag(YY);
plot(Xplot,Yplot,'r.-');
The call to diag(XX) will take the diagonal elements of the matrix XX, which is exactly where you'll get the red patches when you use surf with the z data according to eye().
Result:
Also, if you're just trying to do what your example states, then there's no need to use matrices just to take out the diagonal eventually. Here's the same result, using elementwise operations on your input vectors:
datA = 1:10;
datB = 1:10;
X2 = cosd(8*datA).*datB;
Y2 = datA.*log10(datB*3);
Z2 = cosd(datB);
XX2 = X2./(1+Z2);
YY2 = Y2./(1+Z2);
plot(Xplot,Yplot,'rs-',XX2,YY2,'bo--','linewidth',2,'markersize',10);
legend('original','vector')
Result:
Matlab has many built-in function to assist you.
In 2D the easiest way to do this is polar that allows you to make a graph using theta and rho vectors:
theta = linspace(0,2*pi,100);
r = sin(2*theta);
figure(1)
polar(theta, r), grid on
So, you would get this.
There also is pol2cart function that would convert your data into x and y format:
[x,y] = pol2cart(theta,r);
figure(2)
plot(x, y), grid on
This would look slightly different
Then, if we extend this to 3D, you are only left with plot3. So, If you have data like:
theta = linspace(0,10*pi,500);
r = ones(size(theta));
z = linspace(-10,10,500);
you need to use pol2cart with 3 arguments to produce this:
[x,y,z] = pol2cart(theta,r,z);
figure(3)
plot3(x,y,z),grid on
Finally, if you have spherical data, you have sph2cart:
theta = linspace(0,2*pi,100);
phi = linspace(-pi/2,pi/2,100);
rho = sin(2*theta - phi);
[x,y,z] = sph2cart(theta, phi, rho);
figure(4)
plot3(x,y,z),grid on
view([-150 70])
That would look this way
It seems to be very basic question, but I wonder when I plot x values against y values, what interpolation technique is used behind the scene to show me the discrete data as continuous? Consider the following example:
x = 0:pi/100:2*pi;
y = sin(x);
plot(x,y)
My guess is it is a Lagrangian interpolation?
No, it's just a linear interpolation. Your example uses a quite long dataset, so you can't tell the difference. Try plotting a short dataset and you'll see it.
MATLAB's plot performs simple linear interpolation. For finer resolution you'd have to supply more sample points or interpolate between the given x values.
For example taking the sinus from the answer of FamousBlueRaincoat, one can just create an x vector with more equidistant values. Note, that the linear interpolated values coincide with the original plot lines, as the original does use linear interpolation as well. Note also, that the x_ip vector does not include (all) of the original points. This is why the do not coincide at point (~0.8, ~0.7).
Code
x = 0:pi/4:2*pi;
y = sin(x);
x_ip = linspace(x(1),x(end),5*numel(x));
y_lin = interp1(x,y,x_ip,'linear');
y_pch = interp1(x,y,x_ip,'pchip');
y_v5c = interp1(x,y,x_ip,'v5cubic');
y_spl = interp1(x,y,x_ip,'spline');
plot(x,y,x_ip,y_lin,x_ip,y_pch,x_ip,y_v5c,x_ip,y_spl,'LineWidth',1.2)
set(gca,'xlim',[pi/5 pi/2],'ylim',[0.5 1],'FontSize',16)
hLeg = legend(...
'No Interpolation','Linear Interpolation',...
'PChip Interpolation','v5cubic Interpolation',...
'Spline Interpolation');
set(hLeg,'Location','south','Fontsize',16);
By the way..this does also apply to mesh and others
[X,Y] = meshgrid(-8:2:8);
R = sqrt(X.^2 + Y.^2) + eps;
Z = sin(R)./R;
figure
mesh(Z)
No, Lagrangian interpolation with 200 equally spaced points would be an incredibly bad idea. (See: Runge's phenomenon).
The plot command simply connects the given (x,y) points by straight lines, in the order given. To see this for yourself, use fewer points:
x = 0:pi/4:2*pi;
y = sin(x);
plot(x,y)
I apologize for the ambiguous title, but I am not entirely sure how to phrase this one. So bear with me.
I have a matrix of data. Each column and row represents a certain vector (column 1 = row 1, column 2 = row 2, etc.), and every cell value is the cosine similarity between the corresponding vectors. So every value in the matrix is a cosine.
There are a couple of things I want to do with this. First, I want to create a figure that shows all of the vectors on it. I know the cosine of the angle between every vector, and I know the magnitude of each vector, but that is the only information I have - is there some algorithm I can implement that will run through all of the various pair-wise angles and display it graphically? That is, I don't know where all the vectors are in relation to each other, and there are too many data points to do this by hand (e.g. if I only had three vectors, and the angles between them all were 45, 12, and 72 degrees it would be trivial). So how do I go about doing this? I don't even have the slightest idea what sort of mathematical function I would need to do this. (I have 83 vectors, so that's thousands of cosine values). So basically this figure (it could be either 2D or multidimensional, and to be honest I would like to do both) would show all of the vectors and how they relate to each other in space (so I could compare both angles and relative magnitudes).
The other thing I would like to do is simpler but I am having a hard time figuring it out. I can convert the cosine values into Cartesian coordinates and display them in a scatter plot. Is there a way to connect each of the points of a scatter plot to (0,0) on the plot?
Finally, in trying to figure out how to do some of the above on my own I have run into some inconsistencies. I calculated the mean angles and Cartesian coordinates for each of the 83 vectors. The math for this is easy, and I have checked and double-checked it. However, when I try to plot it, different plotting methods give me radically different things. So, if I plot the Cartesian coordinates as a scatter plot I get this:
If I plot the mean angles in a compass plot I get this:
And if I use a quiver plot I get something like this (I transformed this a little by shifting the origin up and to the right just so you can see it better):
Am I doing something wrong, or am I misunderstanding the plotting functions I am using? Because these results all seem pretty inconsistent. The mean angles on the compass plot are all <30 degrees or so, but on the quiver plot some seem to exceed 90 degrees, and on the scatter plot they extend above 30 as well. What's going on here?
(Here is my code:)
cosine = load('LSA.txt');
[rows,columns]=size(cosine);
p = cosine.^2;
pp = bsxfun(#minus, 1, p);
sine = sqrt(pp);
tangent = sine./cosine;
Xx = zeros(rows,1);
Yy = zeros(rows,1);
for i = 1:columns
x = cosine(:,i);
y = sine(:,i);
Xx(i,1) = sum(x) * (1/columns);
Yy(i,1) = sum(y) * (1/columns);
end
scatter(Xx,Yy);
Rr = zeros(rows,1);
Uu = zeros(rows,1);
for j = 1:rows
Rr(j,1) = sqrt(Xx(j,1).^2 + Yy(j,1).^2);
Uu(j,1) = atan2(Xx(j,1),Yy(j,2));
end
%COMPASS PLOT
[theta,rho] = pol2cart(Uu,1);
compass(theta,rho);
%QUIVER PLOT
r = 7;
sx = ones(size(cosine))*2; sy = ones(size(cosine))*2;
pu = r * cosine;
pv = r * sine;
h = quiver(sx,sy,pu,pv);
set(gca, 'XLim', [1 10], 'YLim', [1 10]);
You can exactly solve this problem. The dot product calculates the cosine. This means your matrix is actually M=V'*V
This should be solvable through eigenvalues. And you said you also have the length.
Your only problem - as your original matrix the vectors will be 83 dimensional. Not easy to plot in 2 or 3 dimensions. I think you are over simplifying by just using the average angle. There are some techniques called dimensionality reduction - here's a toolbox. I would suggest a sammon projection on 1-cosine (as this would be the distance of points on the unit ball) to calculate the vectors for such a plot.
In the quiver plot, you are plotting all of the data in the cosine and sine matrices. In the other plots, you are only plotting the means. The first two plots appear to match up, so no problem there.
A few other things. I notice that in
Uu(j,1) = atan2(Xx(j,1),Yy(j,2));
Yy(j,2) is not actually defined, so it seems like this code should fail.
Furthermore, you could define Yy and Xx as:
Xx = mean(cosine,2);
Yy = mean(sine,2);
And also get rid of the other for loop:
Rr = sqrt(Xx.^2 + Yy.^2)
Uu = atan2(Xx,Yy)
I still have to think about your first question, but I hope this was helpful.