I am trying to plot two different ellipses in Matlab with the same foci. Essentially I am plotting the elliptical orbit before and after an impulse maneuver.
The first (before maneuver) ellipse will always be in-line with the x-axis, whereas the second ellipse (after maneuver) will be at an angle above/below the x-axis (The second one will have a different major/minor axis and eccentricity as well).
In order to save space, the code for plotting one ellipse will be the same as the second ellipse, so I'll only show my code here for plotting one ellipse.
The problem is that it plots from the center of the grid, and not the foci.
Re = 6378.136; %km
e = .65;
x1 = -4.4*Re; %-a
x2 = 4.4*Re; % a
y1 = 0;
y2 = 0;
a = 1/2*sqrt((x2-x1)^2+(y2-y1)^2);
b = a*sqrt(1-e^2);
t = linspace(0,2*pi);
X = a*cos(t);
Y = b*sin(t);
w = atan2(y2-y1,x2-x1);
x = (x1+x2)/2 + X*cos(w) - Y*sin(w);
y = (y1+y2)/2 + X*sin(w) + Y*cos(w);
plot(x,y,'b')
axis equal
grid on
hold on
I'm sure it's something simple but I can't seem to figure it out.
Any help is much appreciated. Thanks.
Edit: David helped solve this issue, thank you.
Related
Perhaps this is a basic question but I haven't been able to find anything specifically like this and I'm wondering how to do it in the best way.
I have two sets of points (x1,y1,z1) and (x2,y2,z2) and I have converted them into polar coordinates. I would like to create a counter-clockwise helix of decreasing radius to reach the second point.
I would also like to specify how many revolutions it takes.
All of the examples I have seen are two points on the x axis and going clockwise.
Any suggestions would be very much appreciated!
Thanks.
This example code generates a counter clockwise spiral from p1 to p2 that isn't on x-axis and you can specify the number of revolutions. However it is in 2D and initial points are in cartesian coordinates. I'm not sure how to do it in 3D but I hope this can help you with the offsetting and the counter-clockwising.
%random 2d points
p1 = [3,5];
p2 = [1,4];
%radius of first point to second
r = norm(p1-p2);
%angle between two point wrt the y-axis
theta_offset = tan((p1(2)- p2(2))/(p1(1)-p2(1)));
rez = 150; % number of points
rev = 5; % number of revolutions
t = linspace(0,r,rez); %radius as spiral decreases
theta = linspace(0,2*pi*rev,rez) + theta_offset; %angle as spiral decreases
x = cos(theta).*t+p2(1);
y = sin(theta).*t+p2(2);
plot(x,y)
I want to plot smooth contour plot from X Y Z matrix.
sf = fit([X Y] Z, 'poly23');
plot(sf);
I have not enought smooth curve..
What I need?
You can use the functions such as griddata and csaps. Together they will lead you to the result as smooth as you wish. The first function adds additional points to your data matrix set. The second one makes the result smoother. The example of the code is below. In the example smoothing is done first in X direction and then in Y direction. Try to play around with the resolution and smoothing_parameter (the current set of these parameters should be OK though).
x = min_x:step_x:max_x;
y = min_y:step_y:max_y;
resolution = 10;
xg = min_x:(step_x/resolution):max_x;
yg = min_y:(step_y/resolution):max_y;
[X,Y] = meshgrid(x,y);
[XG,YG] = meshgrid(xg,yg);
smoothing_parameter = 0.02;
fitted = griddata(X,Y,Z,XG,YG,'cubic');
fitted_smoothed_x = csaps(xg,fitted,smoothing_parameter,xg);
fitted_smoothed_xy = csaps(yg,fitted_smoothed_x',smoothing_parameter,yg);
surf(XG,YG,fitted_smoothed_xy');
EDIT: If you want to get just a contour plot, you can do, for example, as presented below. As I don't have the real data, I will use build-in function peaks to generate some.
[X,Y,Z] = peaks(30);
figure
surfc(X,Y,Z)
view([0 90])
zlim([-10 -8])
Here you just look at your contour plot from above being below the surface.
I would like to create plots using matlab that represent a numerical assessment of quality in a radial fashion.
The best method I've found seems to not work properly. One runs the following code:
theta = (0 : (360/11) : 360)*pi/180;
r = 0 : 2 : 20 ;
[TH,R] = meshgrid(theta,r);
[X,Y] = pol2cart(TH,R);
Z = meshgrid(Data);
surf(X,Y,Z);
Data is a vector of data containing 11 numbers, an example dataset being the following:
Data = 0.884, 0.882, 0.879, 0.880, 0.8776, 0.871, 0.8587, 0.829, 0.811, 0.803, 0.780
the output of surf here is this:
I would like to produce a more refined version of this type of image:
which I have generated with the following code:
for theta = 0 : pi/100 : pi;
v = [InterpolatedImageHeight;LengthVector];
x_center = InterpolatedImageHeight((HorizontalRes+1)/2);
y_center = 0; %InterpolatedImageHeight((HorizontalRes+1)/2);
center = repmat([x_center; y_center], 1, length(InterpolatedImageHeight));
R = [cos(theta) -sin(theta); sin(theta) cos(theta)];
vo = R*(v - center) + center;
x_rotated = vo(1,:);
y_rotated = vo(2,:);
scatter(x_rotated,y_rotated,DotSize,InterpolatedData,'filled'); %x,y,area,color,properties
end
The issue with this is that it is a scatter plot where I am essentially using plot(r,Data), plotting many many copies, and increasing the dot size. The graphic itself has many seams, this takes an enormous amount of memory, and is time intensive where surf or mesh will run extremely fast and take minimal memory.
How does one produce concentric rings with a variable input for color?
There are two completely different plots in your question. The first one represents the data as rays from the origin towards the outside of the circle. The data-points are placed anti-clockwise. A refined version of this can be achieved like this:
Data = [0.884, 0.882, 0.879, 0.880, 0.8776, 0.871,...
0.8587, 0.829, 0.811, 0.803, 0.780];
theta = linspace(0,2*pi,length(Data));
r = linspace(0,20,length(Data));
[TH,R] = meshgrid(theta,r);
Z = meshgrid(Data);
[X,Y,Z] = pol2cart(TH,R,Z);
surf(X,Y,Z);
view(2);
shading interp
Note that I used linspace to generate theta and r to always match the length of Data. Z is also passed trough pol2cart. Then you can use shading interp to remove the lines between the patches and interpolate the color. With view(2) you can set the perspective as you would have a 2d-plot.
This is the result:
It's relatively easy to get a result like in your second example. There the data-points represent concentric circles around the origin and are placed from the origin towards the outside. Therefore, just transpose the meshgrid of Z by using the following line:
Z = meshgrid(Data)';
This is the result then:
based on the code by Darren Rowland in this thread I have come up with the following solution:
x = interp1(1:length(data),datax,(datax(1):datax(end)/f:datax(end)),'linear');
y = interp1(1:length(datay),datay,datay(1):datay(end)/f:datay(end),'spline');
theta = linspace(0,2*pi,n);
xr = x.'*cos(theta);
zr = x.'*sin(theta);
yr = repmat(y.',1,n);
figure;
surf(xy,yr,zr,zr*numcolors);
which is elegant, runs quickly, and produces beautiful figures. This is a sample of the output with some extra chart elements:
I have a 3-dimensional data to be plotted in matlab. The data set are built by stacking 10 exponential curves with different parameters along y directions such as
x = 0:0.01:15;
x0 = 0.5;
y = [beta1, beta2, beta3, beta4, beta5, beta6, beta7, beta8, beta9, beta10];
Z(1, :) = A*exp(-(x-x0).^2/beta1);
Z(2, :) = A*exp(-(x-x0).^2/beta2);
Z(3, :) = A*exp(-(x-x0).^2/beta3);
Z(4, :) = A*exp(-(x-x0).^2/beta4);
...
Z(10, :) = A*exp(-(x-x0).^2/beta10);
% here A could be change based on beta too (no code shown here)
I am trying to plot Z with waterfall except for I don't want the height (i.e. the vertical line) appears on the edge. I don't know if there is any other way to plot the data as waterfall-like curves but without those vertical lines. Thanks
"it is plotted with lines instead of patch with surface".
In other words, you want the boundary lines to be invisible. Well that's no trivial feat as the boundary lines are separate from any color scheme you can directly include. What you need to do is get the data after it drawn then modify it accordingly:
e.g.
[X,Y,Z] = peaks(30);
h = waterfall (X,Y,Z);
CD = get (h, 'CData');
CD(1,:) = nan;
CD(end-2:end,:) = nan;
set (h, 'CData', CD)
note that CD(1,:) is for the "rising" boundary, while CD(end-2:end-1,:) is for the falling boundary, and CD(end,:) is for the bottom.
i know this is an old post, but the below will make the region under the curve transparent:
figure;
[X,Y,Z] = peaks(10);
handle_figure = waterfall( X, Y, Z );
set( handle_figure, 'FaceColor', 'none' );
I have a 3-dimensional data to be plotted in matlab. The data set are built by stacking 10 exponential curves with different parameters along y directions such as
x = 0:0.01:15;
x0 = 0.5;
y = [beta1, beta2, beta3, beta4, beta5, beta6, beta7, beta8, beta9, beta10];
Z(1, :) = A*exp(-(x-x0).^2/beta1);
Z(2, :) = A*exp(-(x-x0).^2/beta2);
Z(3, :) = A*exp(-(x-x0).^2/beta3);
Z(4, :) = A*exp(-(x-x0).^2/beta4);
...
Z(10, :) = A*exp(-(x-x0).^2/beta10);
% here A could be change based on beta too (no code shown here)
I am trying to plot Z with waterfall except for I don't want the height (i.e. the vertical line) appears on the edge. I don't know if there is any other way to plot the data as waterfall-like curves but without those vertical lines. Thanks
"it is plotted with lines instead of patch with surface".
In other words, you want the boundary lines to be invisible. Well that's no trivial feat as the boundary lines are separate from any color scheme you can directly include. What you need to do is get the data after it drawn then modify it accordingly:
e.g.
[X,Y,Z] = peaks(30);
h = waterfall (X,Y,Z);
CD = get (h, 'CData');
CD(1,:) = nan;
CD(end-2:end,:) = nan;
set (h, 'CData', CD)
note that CD(1,:) is for the "rising" boundary, while CD(end-2:end-1,:) is for the falling boundary, and CD(end,:) is for the bottom.
i know this is an old post, but the below will make the region under the curve transparent:
figure;
[X,Y,Z] = peaks(10);
handle_figure = waterfall( X, Y, Z );
set( handle_figure, 'FaceColor', 'none' );