CODEWITHSUNDEEP
matlabsignal-processingcurve
Alessandro Togni
Alessandro Togni

Reputation: 885

How to calculate the point-by-point radius of curvature of a trajectory that is not a proper function

with Matlab i'm trying to calculate the "radius of curvature" signal of a trajectory obtained using GPS data projected to the local cartesian plane. The value of the signal onthe n-th point is the one of the osculating circle tangent to the trajectory on that point. By convention the signal amplitude has to be negative when related to a left turn and viceversa.

With trajectories having a proper function as graph i'm building the "sign" signal evaluating the numeric difference between the y coordinate of the center of the osculating circle:

for i=1:length(yCenter) -1
    aux=Y_m(closestIndex_head:closestIndex_tail );

    if yCenter(i) - aux(i) > 0
        sign(i)=-1;
    else
        sign(i)=+1;
        
    end
end

The above simple method works as long as the trajectory's graph is a proper function (for every x there is only one y).

The trajectory i'm working on is like that:

enter image description here

and the sign signal got some anomalies:

enter image description here

The sign seems to change within a turn.

I've tried to correct the sign using the sin of the angle between the tangent vector and the trajectory, the sign of the tangent of the angle and other similar stuff, but still i'm looking at some anomalies: enter image description here

I'm pretty sure that those anomalies came from the fact that the graph is not a proper function and that the solution lies on the angle of the tangent vector, but still something is missing.

Any advice will be really appreciated, thank you.

Alessandro

Upvotes: 1

Views: 3783

Answers (1)

SleuthEye
SleuthEye

Reputation: 14579

To track a 2D curve, you should be using an expression for the curvature that is appropriate for general parametrized 2D functions.

While implementing the equation from Wikipedia, you can use discrete differences to approximate the derivatives. Given the x and y coordinates, this could be implemented as follows:

% approximate 1st derivatives of x & y with discrete differences
dx  = 0.5*(x(3:end)-x(1:end-2))
dy  = 0.5*(y(3:end)-y(1:end-2))
dl  = sqrt(dx.^2 + dy.^2)
xp  = dx./dl
yp  = dy./dl
% approximate 2nd derivatives of x & y with discrete differences
xpp = (x(3:end)-2*x(2:end-1)+x(1:end-2))./(dl.^2)
ypp = (y(3:end)-2*y(2:end-1)+y(1:end-2))./(dl.^2)

% Compute the curvature
curvature = (xp.*ypp - yp.*xpp) ./ ((xp.^2 + yp.^2).^(1.5))

For demonstration purposes I've also constructed a synthetic test signal (which can be used to recreate the same conditions), but you can obviously use your own data instead:

z1 = linspace(2,1,N).*exp(1i*linspace(0.75*pi,-0.25*pi,N))
z2 = 2*exp(-1i*0.25*pi) + linspace(1,2,N)*exp(1i*linspace(0.75*pi,2.25*pi,N))
z = cat(1,z1,z2)
x = real(z)
y = imag(z)

Sample trajectory

With the corresponding curvature results:

Curvature plot

Upvotes: 3

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