I've got a graph: the velocity as a function of the rpm of a car. The actual graph consists of four subgraphs (one for each of the 4 gears of the car). I conjoined those sub-graphs with a for loop and some if statements. This resulting graph is shown below.
I need to add a graph which is tangent to all the tops of the graphs, like the red line. The final result would look like this:
(It's now still without the red line, obviously...)
These are the coordinates where the tangent line would touch the graph: (17, 5130.36), (28, 3117.98), (39, 2239.37), (51, 1714.72).
Since you already know the tangent points,
x = [17 28 39 51];
y = [5130.36 3117.98 2239.37 1714.72];
plot(x, y, 'b.')
the solution is a simple interpolation
xs = linspace(min(x), max(x), 100);
ys = spline(x, y, xs);
hold on
plot(xs, ys, 'r')
with the result:
I here used Matlab's standard cubic spline interpolation; depending on the properties of your data, other interpolation functions might give better results. In particular, if you knew not only the positions of the tangent points but also the tangent slope, you could use that information to impose constraints on a piecewise polynomial interpolation.
Related
I have solved the heat equation in octave via finite difference and produced the following 3-D plot whose point colors correspond to the temperatures in each element of my three dimensional hamburger.
My computational resources limit the resolution at which I may solve my burger. Thus the only way to get the plot I want is to make my scatter3 points huge blobs of color and it looks kind of bad.
[x,y,z] = meshgrid(1:nx,1:ny,1:nz) % Defines a grid to plot on
scatter3(x(:), y(:), z(:), 40, burgermatrix(:), 's', 'filled')% Point color=value
What I want is a nice gorgeous smooth rectangular prism like this:
So I figure I need to somehow interpolate between the 3D points that I have. Can anyone help me figure out how to do this?
I might be missing something obvious, but here's the example from octave's help slice:
[x, y, z] = meshgrid (linspace (-8, 8, 32));
v = sin (sqrt (x.^2 + y.^2 + z.^2)) ./ (sqrt (x.^2 + y.^2 + z.^2));
slice (x, y, z, v, [], 0, []);
[xi, yi] = meshgrid (linspace (-7, 7));
zi = xi + yi;
slice (x, y, z, v, xi, yi, zi);
shading interp; %addition by me
Isn't this exactly what you need? You have your grid (x,y,z), your solutions (T), so you just need to plot it slicing along [0 0 1] etc. Something like
[xi yi]=meshgrid(unique(x),unique(y));
slice (x, y, z, T, xi, yi, max(z(:))*ones(size(xi)));
and the same for cuts along the two other axes. (Obviously the unique calls should be substituted with the vectors you already have from which you constructed the 3d mesh in the first place.)
NOTE: By the way, you should really consider changing the default (jet) colormap. I was yesterday enlightened by a colleague about the viridis colormap made by the SciPy people, see for instance this post and video link therein. Their reasoning is overwhelming, and their colormap is beautiful. This should define it: viridis, although I haven't tried it myself yet.
(If it wasn't for jet, I'd tell you that your temperature profile seems strongly 1d. Do you happen to have periodic boundary conditions along the vertical walls and homogeneous (i.e. constant) boundary conditions along the horizontal ones?)
I'm currently frustrated by the following problem:
I've got trajectory data (i.e.: Longitude and Latitude data) which I interpolate to find a linear fitting (using polyfit and polyval in matlab).
What I'd like to do is to rotate the axes in a way that the x-axis (the Longitude one) ends up lying on the best-fit line, and therefore my data should now lie on this (rotated) axis.
What I've tried is to evaluate the rotation matrix from the slope of the line-of-fit (m in the formula for a first grade polynomial y=mx+q) as
[cos(m) -sin(m);sin(m) cos(m)]
and then multiply my original data by this matrix...to no avail!
I keep obtaining a plot where my data lay in the middle and not on the x-axis where I expect them to be.
What am I missing?
Thank you for any help!
Best Regards,
Wintermute
A couple of things:
If you have a linear function y=mx+b, the angle of that line is atan(m), not m. These are approximately the same for small m', but very different for largem`.
The linear component of a 2+ order polyfit is different than the linear component of a 1st order polyfit. You'll need to fit the data twice, once at your working level, and once with a first order fit.
Given a slope m, there are better ways of computing the rotation matrix than using trig functions (e.g. cos(atan(m))). I always try to avoid trig functions when performing geometry and replace them with linear algebra operations. This is usually faster, and leads to fewer problems with singularities. See code below.
This method is going to lead to problems for some trajectories. For example, consider a north/south trajectory. But that is a longer discussion.
Using the method described, plus the notes above, here is some sample code which implements this:
%Setup some sample data
long = linspace(1.12020, 1.2023, 1000);
lat = sin ( (long-min(long)) / (max(long)-min(long))*2*pi )*0.0001 + linspace(.2, .31, 1000);
%Perform polynomial fit
p = polyfit(long, lat, 4);
%Perform linear fit to identify rotation
pLinear = polyfit(long, lat, 1);
m = pLinear(1); %Assign a common variable for slope
angle = atan(m);
%Setup and apply rotation
% Compute rotation metrix using trig functions
rotationMatrix = [cos(angle) sin(angle); -sin(angle) cos(angle)];
% Compute same rotation metrix without trig
a = sqrt(m^2/(1+m^2)); %a, b are the solution to the system:
b = sqrt(1/(1+m^2)); % {a^2+b^2 = 1}, {m=a/b}
% %That is, the point (b,a) is on the unit
% circle, on a line with slope m
rotationMatrix = [b a; -a b]; %This matrix rotates the point (b,a) to (1,0)
% Generally you rotate data after removing the mean value
longLatRotated = rotationMatrix * [long(:)-mean(long) lat(:)-mean(lat)]';
%Plot to confirm
figure(2937623)
clf
subplot(211)
hold on
plot(long, lat, '.')
plot(long, polyval(p, long), 'k-')
axis tight
title('Initial data')
xlabel('Longitude')
ylabel('Latitude')
subplot(212)
hold on;
plot(longLatRotated(1,:), longLatRotated(2,:),'.b-');
axis tight
title('Rotated data')
xlabel('Rotated x axis')
ylabel('Rotated y axis')
The angle you are looking for in the rotation matrix is the angle of the line makes to the horizontal. This can be found as the arc-tangent of the slope since:
tan(\theta) = Opposite/Adjacent = Rise/Run = slope
so t = atan(m) and noting that you want to rotate the line back to horizontal, define the rotation matrix as:
R = [cos(-t) sin(-t)
sin(-t) cos(-t)]
Now you can rotate your points with R
[r,t] = meshgrid(linspace(0,2*pi,361),linspace(0,pi,361));
[x,y]=pol2cart(sin(t)*cos(r),sin(t)*sin(r));
%[x,y]=pol2cart(r,t);
surf(x,y);
I played with this addon but trying to find an default function to for this. How can I do the 3D-polar-plot?
I am trying to help this guy to vizualise different integrals here.
There are several problems in your code:
You are already converting spherical coordinates to cartesian coordinates with the sin(theta)*cos(phi) and sin(theta)*sin(phi) bit. Why are you calling pol2cart on this (moreover, we're not working in polar coordinates!)?
As natan points out, there is no third dimension (i.e. z) in your plot. For unity radius, r can be omitted in the spherical domain, where it is completely defined by theta and phi, but in the cartesian domain, you have all three x, y and z. The formula for z is z = cos(theta) (for unit radius).
You didn't read the documentation for surf, which says:
surf(Z,C) plots the height of Z, a single-valued function defined over a geometrically rectangular grid, and uses matrix C, assumed to be the same size as Z, to color the surface.
In other words, your surf(x,y) line merely plots the matrix x and colors it using y as a colormap.
Here's the above code with the mistakes fixed and plotted correctly:
[f,t] = meshgrid(linspace(0,2*pi,361),linspace(0,pi,361));
x = sin(t)*cos(f);
y = sin(t)*sin(f);
z = cos(t);
surf(x,y,z)
i just started with my master thesis and i already am in trouble with my capability/understanding of matlab.
The thing is, i have a trajectory on a surface of a planet/moon whatever (a .mat with the time, and the coordinates. Then i have some .mat with time and the measurement at that time.
I am able to plot this as a color coded trajectory (using the measurement and the coordinates) in scatter(). This works awesomely nice.
However my problem is that i need something more sophisticated.
I now need to take the trajectory and instead of color-coding it, i am supposed to add the graph (value) of the measurement (which is given for each point) to the trajectory (which is not always a straight line). I will added a little sketch to explain what i want. The red arrow shows what i want to add to my plot and the green shows what i have.
You can always transform your data yourself: (using the same notation as #Shai)
x = 0:0.1:10;
y = x;
m = 10*sin(x);
So what you need is the vector normal to the curve at each datapoint:
dx = diff(x); % backward finite differences for 2:end points
dx = [dx(1) dx]; % forward finite difference for 1th point
dy = diff(y);
dy = [dy(1) dy];
curve_tang = [dx ; dy];
% rotate tangential vectors 90° counterclockwise
curve_norm = [-dy; dx];
% normalize the vectors:
nrm_cn = sqrt(sum(abs(curve_norm).^2,1));
curve_norm = curve_norm ./ repmat(sqrt(sum(abs(curve_norm).^2,1)),2,1);
Multiply that vector with the measurement (m), offset it with the datapoint coordinates and you're done:
mx = x + curve_norm(1,:).*m;
my = y + curve_norm(2,:).*m;
plot it with:
figure; hold on
axis equal;
scatter(x,y,[],m);
plot(mx,my)
which is imo exactly what you want. This example has just a straight line as coordinates, but this code can handle any curve just fine:
x=0:0.1:10;y=x.^2;m=sin(x);
t=0:pi/50:2*pi;x=5*cos(t);y=5*sin(t);m=sin(5*t);
If I understand your question correctly, what you need is to rotate your actual data around an origin point at a certain angle. This is pretty simple, as you only need to multiply the coordinates by a rotation matrix. You can then use hold on and plot to overlay your plot with the rotated points, as suggested in the comments.
Example
First, let's generate some data that resembles yours and create a scatter plot:
% # Generate some data
t = -20:0.1:20;
idx = (t ~= 0);
y = ones(size(t));
y(idx) = abs(sin(t(idx)) ./ t(idx)) .^ 0.25;
% # Create a scatter plot
x = 1:numel(y);
figure
scatter(x, x, 10, y, 'filled')
Now let's rotate the points (specified by the values of x and y) around (0, 0) at a 45° angle:
P = [x(:) * sqrt(2), y(:) * 100] * [1, 1; -1, 1] / sqrt(2);
and then plot them on top of the scatter plot:
hold on
axis square
plot(P(:, 1), P(:, 2))
Note the additional things have been done here for visualization purposes:
The final x-coordinates have been stretched (by sqrt(2)) to the appropriate length.
The final y-coordinates have been magnified (by 100) so that the rotated plot stands out.
The axes have been squared to avoid distortion.
This is what you should get:
It seems like you are interested in 3D plotting.
If I understand your question correctly, you have a 2D curve represented as [x(t), y(t)].
Additionally, you have some value m(t) for each point.
Thus we are looking at the plot of a 3D curve [x(t) y(t) m(t)].
you can easily achieve this using
plot3( x, y, m ); % assuming x,y, and m are sorted w.r.t t
alternatively, you can use the 3D version of scatter
scatter3( x, y, m );
pick your choice.
Nice plot BTW.
Good luck with your thesis.
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.