I want to create a spiral of circle markers which never overlap with each other. This is what I got so far, but it overlaps the first markers, and the last ones are too far apart from each other.
t = pi : pi/20 : 20*pi;
t = asind(1./t);
r = t;
x = r .* cos(t);
y = r .* sin(t);
plot(x,y,'o-');
axis equal; hold on
Plotting without redefining t as asinf(1/t) as follows, is shown in the second plot.
t = pi : pi/20 : 20*pi;
r = t;
x = r .* cos(t);
y = r .* sin(t);
plot(x,y,'o-');
Any ideas on how does the spacing of the angles t must be to accomplish that the markers don't overlap?
You can approximate the arc length, greatly simplifying Gilles-Phillipe's solution. This is a simplification, which means that the distance between the markers is not identical everywhere. However the distances are fairly consistent, especially further out.
The approximation here is to assume that the spiral is, locally, a circle. The arc length then is r*dt at a position in the spiral a distance r from the origin, for a change in angle of dt radian.
We now no longer need to solve symbolic equations. I wrote the code in a loop. I'm sure it's possible to vectorize it, making the whole thing two lines of code, but I'll leave that as an exercise to the reader.
This is the code:
d = 1; % step size
q = 1/(2*pi); % spiral constant -- radius grows by q every 1 radian turn
N = 300; % number of points
t = 0; % initial angle
r = d; % initial radius
p = zeros(100,2);
p(1,:) = [r*cos(t),r*sin(t)]; % first point
for ii=2:N
dt = d/r;
t = t+dt;
r = r+dt*q;
p(ii,:) = [r*cos(t),r*sin(t)];
end
clf
plot(p(:,1),p(:,2),'o-')
axis equal
Try this:
syms s;
scale = 10;
l = scale/2 : scale/2 : 40*scale;
t = double(arrayfun(#(y) vpasolve((0.5*(s*sqrt(1+s^2)+asinh(s)))==y,s), l));
x = t .* cos(t);
y = t .* sin(t);
plot(x,y,'o-');
pbaspect([1 1 1]);
axis(scale*[-5 5 -5 5])
The idea is to parameterize using the arclength of the curve. The arclength of this spiral is l=1/2*(t*sqrt(1+t*t)+asinh(t)) (can be found using Matlab symbolic integration). To place points uniformly, we do a uniform sampling of the arclength, and find the corresponding t by solving the equation. Since it cannot be solved easily symbolically, we use a numerical solver.
Note that the scale and the aspect ratio of the plot is really important for it to look uniform and non-overlapping. This is why I added axis/ratio definition. Since each point is solved numerically, it can take quite some time to evaluate. There may be a faster way to do it, but at least you have a result.
I obtain the following result:
Related
My code is not plotting the correct contour plot for a plane perpendicular to a ring of charge running through its center properly. My problem is that the contour plot is not filling 2D space.
I've made two versions of code, one uses a for loop to calculate a
Riemann sum and the other simply uses the sum command. Both rely on the
'subs' command for substituting values from a meshgrid into my expression for V (electric potential).
Version 1 (using for loop):
%% Computing a symbolic expression for V for anywhere in space
syms x y z % phiprime is angle that an elemental dq of the circular
charge is located at, x,y and z are arbitrary points in space outside the
charge distribution
N = 200; % number of increments to sum
R = 2; % radius of circle is 2 meters
dphi = 2*pi/N; % discretizing the circular line of charge which spans 2pi
integrand = 0;
for phiprime = 0:dphi:2*pi
% phiprime ranges from 0 to 2pi in increments of dphi
integrand = integrand + dphi./(sqrt(((x - R.*cos(phiprime) )).^2 + ((y -
R.*sin(phiprime) ).^2) + z.^2));
end
intgrl = sum(integrand);
% uncessary but harmless step that I leave to show that I am using the
sum of the above expression for each dphi
eps0 = 8.854e-12;
kC = 1/(4*pi*eps0);
rhol = 1*10^-9; % linear charge density
Vtot = kC*rhol*R.*intgrl; % symbolic expression for Vtot
%% Graphing V & E in plane perpedicular to ring & passing through center
[Y1, Z1] = meshgrid(-4:.5:4, -4:.5:4);
Vcont1 = subs(Vtot, [x,y,z], {0,Y1,Z1}); % Vcont1 stands for V contour; 1
is because I do the plane of the ring next
contour(Y1,Z1,Vcont1)
xlabel('y - axis [m]')
ylabel('z - axis [m]')
title('V in a plane perpendicular to a ring of charge (N = 200)')
str = {'Red line is side view', 'of ring of charge'};
text(-1,2,str)
hold on
% visually displaying line of charge on plot
circle = rectangle('Position',[-2 0 4 .1],'Curvature',[1,1]);
set(circle,'FaceColor',[1, 0, 0],'EdgeColor',[1, 0, 0]);
% taking negative gradient of V and finding symbolic equations for Ex, Ey
and Ez
g = gradient(-1.*(kC*rhol*R.*intgrl),[x,y,z]);
%% now substituting all the values of the 2D coordinate system for the
symbolic x and y variables to get numeric values for Ex and Ey
Ey1 = subs(g(2), [x y z], {0,Y1,Z1});
Ez1 = subs(g(3), [x y z], {0,Y1,Z1});
E1 = sqrt(Ey1.^2 + Ez1.^2); % full numeric magnitude of E in y-z plane
Eynorm1 = Ey1./E1; % This normalizes the electric field lines
Eznorm1 = Ez1./E1;
quiver(Y1,Z1,Eynorm1,Eznorm1);
hold off
Version 2 (using sum command):
syms x y z
R = 2; % radius of circle is 2 meters
N=100;
dphi = 2*pi/N; % discretizing the circular line of charge which spans 2pi
phiprime = 0:dphi:2*pi; %phiprime ranges from 0 to 2pi in increments of
dphi
integrand = dphi./(sqrt(((x - R.*cos(phiprime) )).^2 + ((y -
R.*sin(phiprime) ).^2) + z.^2));
phiprime = 0:dphi:2*pi;
intgrl = sum(integrand); % Reimann sum performed here
eps0 = 8.854e-12;
kC = 1/(4*pi*eps0);
rhol = 1*10^-9; % linear charge density
Vtot = kC*rhol*R.*intgrl; % symbolic expression for Vtot
Everything else after that point for version 2 is the same as version 1 (substituting for the symbols x,y,z etc)
I would post images of what the code produces but apparently you need 10 reputation for that. Thanks stackoverflow. This will be much more confusing to understand without the images.
The vector field produced by my code is correct while the contour plot seems to use only a few points around the ends of the ring and connect them with straight lines in a strange diamond shape. I can't get it to fill space.
I receive no error messages. The contour lines accumulate around the ends of the ring (where the potential would approach infinity) in a strange diamond shape but aren't graphed anywhere else. I need the contour plot to fill the 2D grid
I received a solution to this question from MATLAB's community and posted about it here:
https://scicomp.stackexchange.com/questions/32834/graphing-electric-potential-of-a-ring-of-charge-using-matlab-help/32842#32842
I would post here but this "you can't post images because you don't have enough reputation" thing would make my explanation too abstract and difficult to understand so go take a look if you are having MATLAB contour plot issues and want to see my problem and solution
I have to plot the speed vector of an object orbiting around a central body. This is a Keplerian context. The trajectory of object is deduced from the classical formula ( r = p/(1+e*cos(theta)) with e=eccentricity.
I manage into plotting the elliptical orbit but now, I would like to plot for each point of this orbit the velocity speed of object.
To compute the velocity vector, I start from classical formulas (into polar coordinates), below the 2 components :
v_r = dr/dt and v_theta = r d(theta)/dt
To take a time step dt, I extract the mean anomaly which is proportional to time.
And Finally, I compute the normalization of this speed vector.
clear % clear variables
e = 0.8; % eccentricity
a = 5; % semi-major axis
b = a*sqrt(1-e^2); % semi-minor axis
P = 10 % Orbital period
N = 200; % number of points defining orbit
nTerms = 10; % number of terms to keep in infinite series defining
% eccentric anomaly
M = linspace(0,2*pi,N); % mean anomaly parameterizes time
% M varies from 0 to 2*pi over one orbit
alpha = zeros(1,N); % preallocate space for eccentric anomaly array
%%%%%%%%%%
%%%%%%%%%% Calculations & Plotting
%%%%%%%%%%
% Calculate eccentric anomaly at each point in orbit
for j = 1:N
% initialize eccentric anomaly to mean anomaly
alpha(j) = M(j);
% include first nTerms in infinite series
for n = 1:nTerms
alpha(j) = alpha(j) + 2 / n * besselj(n,n*e) .* sin(n*M(j));
end
end
% calcualte polar coordiantes (theta, r) from eccentric anomaly
theta = 2 * atan(sqrt((1+e)/(1-e)) * tan(alpha/2));
r = a * (1-e^2) ./ (1 + e*cos(theta));
% Compute cartesian coordinates with x shifted since focus
x = a*e + r.*cos(theta);
y = r.*sin(theta);
figure(1);
plot(x,y,'b-','LineWidth',2)
xlim([-1.2*a,1.2*a]);
ylim([-1.2*a,1.2*a]);
hold on;
% Plot 2 focus = foci
plot(a*e,0,'ro','MarkerSize',10,'MarkerFaceColor','r');
hold on;
plot(-a*e,0,'ro','MarkerSize',10,'MarkerFaceColor','r');
% compute velocity vectors
for i = 1:N-1
vr(i) = (r(i+1)-r(i))/(P*(M(i+1)-M(i))/(2*pi));
vtheta(i) = r(i)*(theta(i+1)-theta(i))/(P*(M(i+1)-M(i))/(2*pi));
vrNorm(i) = vr(i)/norm([vr(i),vtheta(i)],1);
vthetaNorm(i) = vtheta(i)/norm([vr(i),vtheta(i)],1);
end
% Plot velocity vector
quiver(x(30),y(30),vrNorm(30),vthetaNorm(30),'LineWidth',2,'MaxHeadSize',1);
% Label plot with eccentricity
title(['Elliptical Orbit with e = ' sprintf('%.2f',e)]);
Unfortunately, once plot performed, it seems that I get a bad vector for speed. Here for example the 30th element of vrNorm and vthetaNorm arrays :
As you can see, the vector has the wrong direction (If I assume to take 0 for theta from the right axis and positive variation like into trigonometrics).
If someone could see where is my error, this would nice.
UPDATE 1: Has this vector representing the speed on elliptical orbit to be tangent permanently to the elliptical curve ?
I would like to represent it by taking the right focus as origin.
UPDATE 2:
With the solution of #MadPhysicist, I have modified :
% compute velocity vectors
vr(1:N-1) = (2*pi).*diff(r)./(P.*diff(M));
vtheta(1:N-1) = (2*pi).*r(1:N-1).*diff(theta)./(P.*diff(M));
% Plot velocity vector
for l = 1:9 quiver(x(20*l),y(20*l),vr(20*l)*cos(vtheta(20*l)),vr(20*l)*sin(vtheta(20*l)),'LineWidth',2,'MaxHeadSize',1);
end
% Label plot with eccentricity
title(['Elliptical Orbit with e = ' sprintf('%.2f',e)]);
I get the following result :
On some parts of the orbit, I get wrong directions and I don't understand why ...
There are two issues with your code:
The normalization is done incorrectly. norm computes the generalized p-norm for a vector, which defaults to the Euclidean norm. It expects Cartesian inputs. Setting p to 1 means that it will just return the largest element of your vector. In your case, the normalization is meaningless. Just set vrNorm as
vrNorm = vr ./ max(vr)
It appears that you are passing in the polar coordinates vrNorm and vthetaNorm to quiver, which expects Cartesian coordinates. It's easy to make the conversion in a vectorized manner:
vxNorm = vrNorm * cos(vtheta);
vyNorm = vrNorm * sin(vtheta);
This assumes that I understand where your angle is coming from correctly and that vtheta is in radians.
Note
The entire loop
for i = 1:N-1
vr(i) = (r(i+1)-r(i))/(P*(M(i+1)-M(i))/(2*pi));
vtheta(i) = r(i)*(theta(i+1)-theta(i))/(P*(M(i+1)-M(i))/(2*pi));
vrNorm(i) = vr(i)/norm([vr(i),vtheta(i)],1);
vthetaNorm(i) = vtheta(i)/norm([vr(i),vtheta(i)],1);
end
can be rewritten in a fully vectorized manner:
vr = (2 * pi) .* diff(r) ./ (P .* diff(M))
vtheta = (2 * pi) .* r .* diff(theta) ./ (P .* diff(M))
vrNorm = vr ./ max(vr)
vxNorm = vrNorm * cos(vtheta);
vyNorm = vrNorm * sin(vtheta);
Note 2
You can call quiver in a vectorized manner, on the entire dataset, or on a subset:
quiver(x(20:199:20), y(20:199:20), vxNorm(20:199:20), vyNorm(20:199:20), ...)
How to distribute the points to be like Fig.A
This matlab code for Fig. B :
N = 30; % number of points
r = 0.5; % r = radius
d = 50; % dimension
C_point = 0; % center point
figure, clf
C = ones(1, d) * C_point;
C_rep = repmat( C,N,1);
X = randn(N,d);
s2 = sum(X.^2,2) ;
radius = r * (rand(N,1).^(1/d));
X = X.*repmat(radius./sqrt(s2),1,d) + C_rep;
%% Plot 2D
t = linspace(0, 2*pi, 100);
x = r*cos(t) + C(1);
y = r*sin(t) + C(2);
plot(x,y,'b')
hold on
plot(C(1),C(2),'b.', 'MarkerSize', 10) % center point
hold on
plot(X(:,1), X(:,2),'r.','markersize',10);
axis equal;rotate3d off; rotate3d on;drawnow;shg;
hold on
ax = axis;
Source of the code
What I should change to be like fig. A
The OP's code computes points uniformly distributed within a d-dimensional box, projects those onto a d-dimensional sphere, then samples the radius to move them inside the d-dimensional ball. This is perfect except that the points inside the box, when projected onto the sphere, do not form a uniform distribution on that sphere. If instead you find random points distributed in a Gaussian distribution, you are guaranteed uniform angle distribution.
First compute points with a Gaussian distribution in d dimensions (I do all here with minimal changes to the OP's code):
N = 1000; % number of points
r = 0.5; % r = radius
d = 3; % dimension
C_point = 0; % center point
C = ones(1,d) * C_point;
C_rep = repmat(C,N,1);
X = randn(N,d);
Note that I use randn, not rand. randn creates a Gaussian distribution.
Next we normalize the vectors so the points move to the sphere:
nX = sqrt(sum(X.^2,2));
X = X./repmat(nX,1,d);
These points are uniformly distributed, which you can verify by scatter3(X(:,1),X(:,2),X(:,3)); axis equal and turning the display around (a 2D rendering doesn't do it justice). This is the reason I set d=3 above, and N=1000. I wanted to be able to plot the points and see lots of them.
Next we compute, as you already did, a random distance to the origin, and correct it for the dimensionality:
radius = r * (rand(N,1).^(1/d));
X = X.*repmat(radius,1,d) + C_rep;
X now is distributed uniformly in the ball. Again, scatter3(X(:,1),X(:,2),X(:,3)); axis equal shows this.
However, if you set d=50 and then plot only two dimensions of your data, you will not see the data filling the circle. And you will not see a uniform distribution either. This is because you are projecting a 50-D ball onto 2 dimensions, this simply does not work. You either have to trust the math, or you have to slice the data:
figure, hold on
t = linspace(0, 2*pi, 100);
x = r*cos(t) + C(1);
y = r*sin(t) + C(2);
plot(x,y,'b')
plot(C(1),C(2),'b.', 'MarkerSize', 10) % center point
axis equal
I = all(abs(X(:,3:d))<0.1,2);
plot(X(I,1), X(I,2),'r.','markersize',10);
The I there indexes points that are close to the origin in dimensions perpendicular to the first two shown. Again, with d=50 you will have very few points there, so you will need to set N very large! To see the same density of points as in the case above, for every dimension you add, you need to multiply N by 10. So for d=5 you'd have N=1000*10*10=1e5, and for d=50 you'd need N=1e50. That is totally impossible to compute, of course.
I have an ellipse in 2 dimensions, defined by a positive definite matrix X as follows: a point x is in the ellipse if x'*X*x <= 1. How can I plot this ellipse in matlab? I've done a bit of searching while finding surprisingly little.
Figured out the answer actually: I'd post this as an answer, but it won't let me (new user):
Figured it out after a bit of tinkering. Basically, we express the points on the ellipse border (x'*X*x = 1) as a weighted combination of the eigenvectors of X, which makes some of the math to find the points easier. We can just write (au+bv)'X(au+bv)=1 and work out the relationship between a,b. Matlab code follows (sorry it's messy, just used the same notation that I was using with pen/paper):
function plot_ellipse(X, varargin)
% Plots an ellipse of the form x'*X*x <= 1
% plot vectors of the form a*u + b*v where u,v are eigenvectors of X
[V,D] = eig(X);
u = V(:,1);
v = V(:,2);
l1 = D(1,1);
l2 = D(2,2);
pts = [];
delta = .1;
for alpha = -1/sqrt(l1)-delta:delta:1/sqrt(l1)+delta
beta = sqrt((1 - alpha^2 * l1)/l2);
pts(:,end+1) = alpha*u + beta*v;
end
for alpha = 1/sqrt(l1)+delta:-delta:-1/sqrt(l1)-delta
beta = -sqrt((1 - alpha^2 * l1)/l2);
pts(:,end+1) = alpha*u + beta*v;
end
plot(pts(1,:), pts(2,:), varargin{:})
I stumbled across this post while searching for this topic, and even though it's settled, I thought I might provide another simpler solution, if the matrix is symmetric.
Another way of doing this is to use the Cholesky decomposition of the semi-definite positive matrix E implemented in Matlab as the chol function. It computes an upper triangular matrix R such that X = R' * R. Using this, x'*X*x = (R*x)'*(R*x) = z'*z, if we define z as R*x.
The curve to plot thus becomes such that z'*z=1, and that's a circle. A simple solution is thus z = (cos(t), sin(t)), for 0<=t<=2 pi. You then multiply by the inverse of R to get the ellipse.
This is pretty straightforward to translate into the following code:
function plot_ellipse(E)
% plots an ellipse of the form xEx = 1
R = chol(E);
t = linspace(0, 2*pi, 100); % or any high number to make curve smooth
z = [cos(t); sin(t)];
ellipse = inv(R) * z;
plot(ellipse(1,:), ellipse(2,:))
end
Hope this might help!
I'm trying to create a matrix of 0 values, with 1 values filling a ellipse shape. My ellipse was generated using minVolEllipse.m (Link 1) which returns a matrix of the ellipse equation in the 'center form' and the center of the ellipse. I then use a snippet of code from Ellipse_plot.m (from the aforementioned link) to parameterize the vector into major/minor axes, generate a parametric equation, and generate a matrix of transformed coordinates. You can see their code to see how this is done. The result is a matrix that has index locations for points along the ellipse. It does not encompass every value along the outline of the ellipse unless I set the number of grid points, N, to a ridiculously high value.
When I use the MATLAB plot or patch commands I see exactly the result I'm looking for. However, I want this represented as a matrix of 0 values with 1s where patch 'fills in' the blanks. It is apparent that MATLAB has this functionality, but I have yet to find the code to execute it. What I am looking for is similar to how bwfill of the image processing toolbox works (Link 2). bwfill does not work for me because my ellipse is not contiguous, so the function returns a matrix filled completely with 1 values.
Hopefully I have outlined the problem well enough, if not please comment and I can edit the post to clarify.
EDIT:
I have devised a strategy using the 2-D X vector from Ellipse_plot.m as an input to EllipseDirectFit.m (Link 3). This function returns the coefficients for the ellipse function ax^2+bxy+cy^2+dx+dy+f=0. Using these coefficients I calculate the angle between the x-axis and the major axis of the ellipse. This angle, along with the center and major/minor axes are passed into ellipseMatrix.m (Link 4), which returns a filled matrix. Unfortunately, the matrix appears to be out of rotation from what I want. Here is the portion of my code:
N = 20; %Number of grid points in ellipse
ellipsepoints = clusterpoints(clusterpoints(:,1)==i,2:3)';
[A,C] = minVolEllipse(ellipsepoints,0.001,N);
%%%%%%%%%%%%%%
%
%Adapted from:
% Ellipse_plot.m
% Nima Moshtagh
% nima#seas.upenn.edu
% University of Pennsylvania
% Feb 1, 2007
% Updated: Feb 3, 2007
%%%%%%%%%%%%%%
%
%
% "singular value decomposition" to extract the orientation and the
% axes of the ellipsoid
%------------------------------------
[U D V] = svd(A);
%
% get the major and minor axes
%------------------------------------
a = 1/sqrt(D(1,1))
b = 1/sqrt(D(2,2))
%theta values
theta = [0:1/N:2*pi+1/N];
%
% Parametric equation of the ellipse
%----------------------------------------
state(1,:) = a*cos(theta);
state(2,:) = b*sin(theta);
%
% Coordinate transform
%----------------------------------------
X = V * state;
X(1,:) = X(1,:) + C(1);
X(2,:) = X(2,:) + C(2);
% Output: Elip_Eq = [a b c d e f]' is the vector of algebraic
% parameters of the fitting ellipse:
Elip_Eq = EllipseDirectFit(X')
% http://mathworld.wolfram.com/Ellipse.html gives the equation for finding the angle theta (teta).
% The coefficients from EllipseDirectFit are rescaled to match what is expected in the wolfram link.
Elip_Eq(2) = Elip_Eq(2)/2;
Elip_Eq(4) = Elip_Eq(4)/2;
Elip_Eq(5) = Elip_Eq(5)/2;
if Elip_Eq(2)==0
if Elip_Eq(1) < Elip_Eq(3)
teta = 0;
else
teta = (1/2)*pi;
endif
else
tetap = (1/2)*acot((Elip_Eq(1)-Elip_Eq(3))/(Elip_Eq(2)));
if Elip_Eq(1) < Elip_Eq(3)
teta = tetap;
else
teta = (pi/2)+tetap;
endif
endif
blank_mask = zeros([height width]);
if teta < 0
teta = pi+teta;
endif
%I may need to switch a and b, depending on which is larger (so that the fist is the major axis)
filled_mask1 = ellipseMatrix(C(2),C(1),b,a,teta,blank_mask,1);
EDIT 2:
As a response to the suggestion from #BenVoigt, I have written a for-loop solution to the problem, here:
N = 20; %Number of grid points in ellipse
ellipsepoints = clusterpoints(clusterpoints(:,1)==i,2:3)';
[A,C] = minVolEllipse(ellipsepoints,0.001,N);
filled_mask = zeros([height width]);
for y=0:1:height
for x=0:1:width
point = ([x;y]-C)'*A*([x;y]-C);
if point < 1
filled_mask(y,x) = 1;
endif
endfor
endfor
Although this is technically a solution to the problem, I am interested in a non-iterative solution. I'm running this script over many large images, and need it to be very fast and parallel.
EDIT 3:
Thanks #mathematical.coffee for this solution:
[X,Y] = meshgrid(0:width,0:height);
fill_mask=arrayfun(#(x,y) ([x;y]-C)'*A*([x;y]-C),X,Y) < 1;
However, I believe there is yet a better way to do this. Here is a for-loop implementation that I did that runs faster than both above attempts:
ellip_mask = zeros([height width]);
[U D V] = svd(A);
a = 1/sqrt(D(1,1));
b = 1/sqrt(D(2,2));
maxab = ceil(max(a,b));
xstart = round(max(C(1)-maxab,1));
xend = round(min(C(1)+maxab,width));
ystart = round(max(C(2)-maxab,1));
yend = round(min(C(2)+maxab,height));
for y = ystart:1:yend
for x = xstart:1:xend
point = ([x;y]-C)'*A*([x;y]-C);
if point < 1
ellip_mask(y,x) = 1;
endif
endfor
endfor
Is there a way to accomplish this goal (the total image size is still [width height]) without this for-loop? The reason this is faster is because I don't have to iterate over the entire image to determine if my point is within the ellipse. Instead, I can simply iterate over a square region that is the length of the center +/- the largest principle axis.
Expanding the matrix multiply, which is an elliptic norm, gives a fairly simply vectorized expression:
[X,Y] = meshgrid(0:width,0:height);
X = X - C(1);
Y = Y - C(2);
fill_mask = X.^2 * A(1,1) + X.*Y * (A(1,2) + A(2,1)) + Y.^2 * A(2,2) < 1;
This is what I intended by my original comment.