matlab,labels,polar-coordinates

The polar function itself does all the hard work of converting the coordinates. So pull the values directly from the plot: h = polar(PSA,PST,'.'); % easiest way to get handle to plot x = get(h,'XData'); y = get(h,'YData'); text(x,y, ' \leftarrow foo'); % puts same text next to every point...

python,numpy,3d,coordinates,polar-coordinates

In the function make_spot_3d_spherical you got the sin and cos mixed up in your definition of x0: x0 = int(rho*np.sin(theta)*np.cos(phi)) should be x0 = int(rho*np.cos(theta)*np.sin(phi)) Now it works. Here's if you vary phi: S_p = np.asarray([make_spot_3d_spherical(1000,2, 30,0,phi) for phi in np.linspace(0, 2*np.pi, 20)]) set_of_Sp0 = np.sum(np.sum(S_p, axis =0), axis =...

python,numpy,plot,polar-coordinates

Do not convert to a cartesian coordinate system if you simply want two 2D arrays that specify coordinates r and theta on a polar grid. To clarify, the error you're seeing is because you can not perform element-wise multiplication between two arrays of unequal shape, which is what you have....

matlab,graph,plot,equation,polar-coordinates

Utilize the output of your polar call to address and change line properties. In this case we're interested in the LineWidth property. w = 0.1; SAU = 0.1; r = 0:100; Phi = 2*pi/w*(1-SAU)*r/500; h.myplot = polar(Phi, r); h.myplot.LineWidth = 8; % Adjust as necessary This assumes you have R2014b...

python,matplotlib,plot,polar-coordinates

I recommend not using the polar plot but instead setting up axis artists. This allows you to set up a 'partial' polar plot. This answer is based on an adjustment of the 3rd example from: axes_grid example code: demo_floating_axes.py import numpy as np import matplotlib.pyplot as plt from matplotlib.transforms import...

r,ggplot2,maps,polar-coordinates

You don't seem to actually use the time in your plot, but the issue is the longitudes wrapping around -180/180. This can be solved using coord_map rather than coord_polar and ensuring that the longitudes don't wrap around. Load packages and generate sample data library("ggplot2") library("dplyr") south_map <- map_data("world") %>% group_by(group)...

python,matplotlib,plot,polar-coordinates

It looks like matplotlib is choking because you are giving it a negative as a radius. I tried your code and got the same results. I changed your line r = map(f, theta) to be r = map(abs(f), theta) And got this plot: The "polar plot" from Wolfram is a...

c++,polar-coordinates,cartesian-coordinates

You are converting to cartesian the points which are in cartesian already. What you want is: std::cout << "Cartesian Coordinates:" << std::endl; std::cout << to_cartesian(to_polar(a)) << std::endl; std::cout << to_cartesian(to_polar(b)) << std::endl; //... Edit: using atan2 solves the NaN problem, (0, 0) is converted to (0, 0) which is fine....

python,numpy,matplotlib,polar-coordinates

The line proprty solid_capstyle (docs). There is also a dash_capstyle which controls the line ends on every dash. import matplotlib.pyplot as plt import numpy as np x = y = np.arange(5) fig, ax = plt. subplots() ln, = ax.plot(x, y, lw=10, solid_capstyle='round') ln2, = ax.plot(x, 4-y, lw=10) ln2.set_solid_capstyle('round') ax.margins(.2) This...

python-2.7,image-processing,mapping,polar-coordinates

There are a few differences / errors: They use the centre of the image as the origin They scale the axis appropriately. In your example, you're plotting your angle (between 0 and in your case, pi), instead of utilising the full height of the image. You're using the wrong atan...

java,animation,graphics,polar-coordinates

x and y will always = 0. Understand that they are 0 to begin with, and then you add a sine or cose to them which is guaranteed to be < 1. x = (int) (x + Math.cos(angleInRadians)); y = (int) (x+Math.sin(angleInRadians)); So 0 + a number < 1 will...

You can enable datalabels, set color and adapt y parameter. marker:{ radius:10 }, dataLabels:{ enabled:true, color: 'white', y:7 }, http://jsfiddle.net/6upMk/3/...

matlab,circle,mesh,polar-coordinates

Explanation: Your call to meshgrid is exactly what you want. You just have to generate the data matrix that corresponds to the entries of th and r. Your data depends on the values of r, and as this is a matrix of increasing values in the rows, r = [rho(1),...

Gnuplot gets confused if you use xrange and yrange setting which contradict the rrange setting. Thats probably why the yrange settings are ignored. Then, you must also use set size ratio -1 in order to get the same scaling in x and yrange. When plotting only the upper two quadrants,...

matlab,colors,plot,gnuplot,polar-coordinates

Here is one possible solution with gnuplot. That uses the circles plotting style to draw the overlapping wedges at the origin with a specified radius. That requires you to have your data sorted by descending maximum radius, and that you have no gaps. Here is a possible script: set xrange...

python,matplotlib,scatter-plot,polar-coordinates

The problem is that you're only converting the edges of the array. By converting only the x and y coordinates of the edges, you're effectively converting the coordinates of a diagonal line across the 2D array. This line has a very small range of theta values, and you're applying that...

image,matlab,image-processing,polar-coordinates

I've managed to do this as follows: I added zeros to top of im matrix to define the radius of curvature (in my case 1018 pixels i.e ROC) and polar origin correctly: im = IML'; % scan_lines (rotated for now) sector=75; % [degrees] im_depth=16; % [cm] (depth + ROC) ROC=3.98;...

I made it transformation from Cartesian coordinate system to Polar coordinate system 'in hand' and then I made a graph like that: rho = linspace(0,0.9,50); phi = linspace(0,2*pi,50); [RHO,PHI] = meshgrid(rho,phi); u = (-1+RHO.^2)./(-RHO.^2+2*RHO.*sin(PHI)-1); figure('units','normalized','outerposition',[0 0 1 1]) set(gcf,'Color',[1,1,1]) box on; surf(RHO,PHI,u) xlabel('\rho'); ylabel('\phi'); zlabel('u(\rho,\phi)'); What do you think? Correct?...

python,matplotlib,polar-coordinates

The plot requires radians. Adding the following line and replotting shows correctly: ra = [x/180.0*3.141593 for x in ra] ...