Determine positions of projected points onto an ellipse

Hi,

I have a set of data points (blue dots below) and I woule like to project them onto an ellipse (with known a, b, x0, y0, and radian or tilt).

I wonder how I can determine the position of each projected points on the ellipse. Any advices/code/peudo code would be lovely. Thank you.

[Matt J]

Duplicate of,

https://www.mathworks.com/matlabcentral/answers/1873937-projecting-a-dataset-on-an-ellipse?s_tid=srchtitle

Answer 1

[Matt J]

Use the ellipseprj function below. It requires the download of trustregprob from the file exchange

https://www.mathworks.com/matlabcentral/fileexchange/53191-quadratic-minimization-with-norm-constraint

ab = [2 1];  %ellipse radii
xy0 = [1 1]; %ellipse center
rad=5; R = [cos(rad), -sin(rad); sin(rad),cos(rad)]; %ellipse tilt

xysamp = randn(2,5); %random points to be projected onto the ellipse

xyproj = ellipseprj(  R'*diag(1./ab.^2)*R  , xy0, xysamp); %the projected points

%%%Visualize
t = linspace(0,2*pi,50)';
xyc = xy0 + ab.*[cos(t), sin(t)]*R; %points for plotting the ellipse
plot(xyc(:,1),xyc(:,2),'b-')
axis equal
grid on

hold on
plot(xysamp(1,:),xysamp(2,:),'ro') %plot random points
plot(xyproj(1,:),xyproj(2,:),'ks') %plot their projections
line([xysamp(1,:);xyproj(1,:)],[xysamp(2,:);xyproj(2,:)],'color','g')
hold off

function Xp=ellipseprj(Q,xc,X0)
%Projects given 2D points onto 2D ellipse
%
%     Xp=ellipseprj(Q,xc,X0)
%
%IN:
%
% Q,xc: Ellipse equation matrices. Q is 2x2 and xc is 2x1 such that
%       ellipse equation is (y-xc).'*Q*(y-xc)=1
%   X0: 2xN matrix of points to be projected
%
%OUT:
%
%  Xp: 2xN matrix of projected points


N=size(X0,2);
xc=xc(:);

[Rt,D]=eig(Q);
Rt=real(Rt); D=real(D);
iD=diag(1./diag(D));
iDsqrt=sqrt(iD);


b=-iDsqrt*Rt.'*bsxfun(@minus,xc,X0);

Yp=nan(size(X0));

for i=1:N
    
    Yp(:,i)=trustregprob(iD,b(:,i),1);
    
    
end

Xp=bsxfun(@plus, Rt*iDsqrt*Yp,xc);
end

[Salad Box]

Absolutely brilliant. Thank you for your code, demo. Very clear and easy to implement.

[Salad Box]

Thank you Matt for your answer again. Both yours and John D's methods provides the same set of data for the projection points which verifies each others method. There is just something I couldn't explain.

I asked the same thing to John which I am posting this same question here hoping I can also get some advices from you.

Taking my 9th data point as an example,

A is where my 9th data point is, and B is the projection generated on the ellipse using your code. Purely based on the look, it seems that the distance between A and B is not the shortest. I would expect the projection is at C where A to C seems more perpendicular to the ellipse. I am not sure why this happens and how to explain this.

A = [0.2460  0.5098];
% after apply your code, I got the output of B
B = [0.2505  0.5134];

[John D'Errico]

@Salad Box very often it can LOOK like a line is not the line of closest approach, if you have not got the axes correct. You NEED to have the two axes with the same units. Otherwise, things will look distorted. For example

t = linspace(0,2*pi);
x = cos(t);
y = sin(t);
plot(x,y,'b-')

Now you should know absolutely that the curve I drew was a perfect circle. But it doe not look like a circle! It appears to be an ellipse. Next, draw a ray, protruding from the center of the circle at 45 degrees.

hold on
plot([0 1],[0,1],'r-')
hold off

Now look carefully at the plot. If you dod, you will come to the conclusion that the red line is NOT perpendicular to the curve. But since the red line is a radial line emanating fro mthe center of the circle, do you agree the line MUST have been perpendicular?

Now, I'll replot the circle and the radial line. But I'll add an extra command.

plot(x,y,'b-',[0 1],[0,1],'r-')
axis equal

Do you see that now the line is seen to be clearly perpendicular? As well, the circle actually LOOKS like a circle!

You need to be careful when you look at a plot like that, and come to conclusions based on the wrong axis scaling.

[Matt J]

@Salad Box, In addition to what John said, it is a pretty basic calculus exercise to compute the tangent to the ellipse at B, and to verify numerically that the tangent is orthogonal to (A-B). I encourage doing that in addition to any visual tests.

[Salad Box]

Thank you John and Matt, especially to John, now that explains everything. My axis was distorted I must admit. Well spotted!

Thank you Matt. Your adivce on checking the tangent was very valuable as well providing another way of checking the perpendicularity.

Great answers. Thank you.:)

Answer 2

[John D'Errico]

Simple. Just download my distance2curve utility, found on the file exchange.

For some example data, I'm not sure what you meant by radian as a variable.

t = linspace(0,2*pi,50)';
ab = [2 1];
xy0 = [1 1];
rotmat = @(R) [cos(R), -sin(R);sin(R),cos(R)];
xyc = xy0 + ab.*[cos(t), sin(t)]*rotmat(5);

plot(xyc(:,1),xyc(:,2),'b-')
axis equal
grid on

Now for a given set of points,

xy0 = randn(10,2);
xyproj = distance2curve(xyc,xy0);
hold on
plot(xy0(:,1),xy0(:,2),'ro')
plot(xyproj(:,1),xyproj(:,2),'rs')
line([xy0(:,1),xyproj(:,1)]',[xy0(:,2),xyproj(:,2)]','color','g')

Each point is projected to the nearest point on the ellipse.

distance2curve is found for free download from the file exchange, here:

https://www.mathworks.com/matlabcentral/fileexchange/34869-distance2curve?s_tid=ta_fx_results

[Salad Box]

Thank you for your answer. As you can see from my original plot, the ellipse is rotated/tilted. Radian is another way to describe the rotation instead of using the angle.

If I use theta = 5 (radian), then I use cos(theta) and sin(theta).

If I use theta = 30 (degree/angle), then I use cosd(theta) and sind(theta).

[Salad Box]

Thank you for your answer again. In YOUR example, the code seems doing a perfect job. When I applied your code to my data set, the projection is slightly different to what I expected.

I use my 9th data point as an example.

A is where my 9th data point is, and B is the projection generated on the ellipse using your code. Purely based on the look, it seems that the distance between A and B is not the shortest. I would expect the projection is at C where A to C seems more perpendicular to the ellipse. I am not sure why this happens and how to explain this.

A = [0.2460  0.5098];

% after apply your code, I got the output of B
B = [0.2505  0.5132];

[Torsten]

Among the t-values you supply, the code chooses the point of xyc that gives the shortest distance to your point.

So choosing a finer t-grid gives a better result.

Change

t = linspace(0,2*pi,50)';

to

t = linspace(0,2*pi,5000)';

e.g.

[Salad Box]

Thank you for your answer Torsten. I did what you have suggested to modify t to have 5000 intervals. It hasn't changed results. The projected point B still has the same x and y values as when t has 50 intervals. I am not sure whether t is the key to this.

Answer 3

[Torsten]

a = 2;
b = 1;             % main axes
x0 = 1;
y0 = 1;            % translation
alpha = 5;         % rotation angle of ellipse in degrees
px = randn(1,10);
py = randn(1,10);  % Sample points to be projected
syms theta xp yp xp_given yp_given
% Generate ellipse equation dependent on theta
x = a*cos(theta);
y = b*sin(theta);
% Solve for theta that minimizes the distance
f = (x-xp)^2 + (y-yp)^2;
df = diff(f,theta);
sol_theta = solve(df==0,theta,'MaxDegree',4)

sol_theta = 

% Transform given points (xp_given,yp_given) to coordinate system of the centered ellipse 
% and project on it
P_slash = [cosd(-alpha) -sind(-alpha);sind(-alpha) cosd(-alpha)]*[xp_given - x0 ; yp_given - y0];
sol_theta = subs(sol_theta,[xp yp],[P_slash(1) P_slash(2)])

sol_theta = 

% Transform projected points back to given ellipse
Px_proj = cosd(alpha)*subs(x,theta,sol_theta) - sind(alpha)*subs(y,theta,sol_theta) + x0;
Py_proj = sind(alpha)*subs(x,theta,sol_theta) + cosd(alpha)*subs(y,theta,sol_theta) + y0;

% Numerical example
for i = 1:numel(px)
  px_proj = double(subs(Px_proj,[xp_given yp_given],[px(i) py(i)]));
  py_proj = double(subs(Py_proj,[xp_given yp_given],[px(i) py(i)]));
  idx = abs(imag(px_proj))<1e-6 & abs(imag(py_proj))<1e-6;
  px_proj = px_proj(idx);
  py_proj = py_proj(idx);
  d = sqrt((px(i)-px_proj).^2 + (py(i)-py_proj).^2);
  [dist(i),index] = min(d);
  pxp(i) = px_proj(index);
  pyp(i) = py_proj(index);
end

%Graphical presentation
thetanum = 0:2*pi/100:2*pi;
xx = a*cos(thetanum);
yy = b*sin(thetanum);
xxx = xx*cosd(alpha) - yy*sind(alpha);
yyy = xx*sind(alpha) + yy*cosd(alpha);
xxx = xxx + x0;
yyy = yyy + y0;
plot(xxx,yyy)
hold on
arrayfun(@(i)plot([px(i), real(pxp(i))],[py(i), real(pyp(i))]),1:numel(px))
axis equal
grid on

[Salad Box]

Thank you for your answer. I am just wondering what theta, x_given and y_given are.

should I replace x_give and y_given with my own data points?

I assume x and y are real numbers, the thing is that my data set contains 85 data points, and in the case of the ellipse because I used theta = 0:0.01:2*pi, it actually generated 629 points to draw the ellipse.

So I am wondering how I can use

f = (x-x_given)^2 + (y-y_given)^2;

if the number of data points are not even matching (x_given and y_given are 85; x and y are 629)?

[Torsten]

px and py should be your data points.

You must admit 0 <= theta < 2*pi to determine the optimal projection point on the ellipse, but the graphical part is only for illustration - it's not a necessary part of the code.

Inputs are - as said -

a = 2;
b = 1;             % main axes
x0 = 1;
y0 = 1;            % translation
alpha = 5;         % rotation angle of ellipse in degrees
px = randn(1,10);
py = randn(1,10);  % Sample points to be projected

and outputs are pxp and pyp - the x-and y-coordinates of the points on the ellipse nearest to the (px /py) - and maybe (if it's of interest) the array "dist" which contains the Euclidean distance between the given data points and their counterparts on the ellipse.