Rotation of a set of points onto the X axis

Hi Guys,

I know this is a simple problem but I am just not to able to figure out how to do it. I dont know if the code I have written is correct. So my problem. I have a set of 4 points in a 2D plane. The points are linear and can be fit to a line in the form on y=m*x+b. Usually b=0. So these fitted line will pass through the origin. The points are in all quadrants. So the fitted lines look like they radite out of the origin. I want to rotate all these points onto the x axis. I am having trouble finding out the exact angle of rotation since this would depend on the quadrant the points/fitted line lie in.

Can you please tell me if my code is right. It is very crude.

My points are in matrix X. It is a 2 row 4 column matrix. Row 1 has X coordinates and row 2 has corresponding Y coordinates.

%Fit a line to the points
 poly=polyfit(X(1,1:4),X(2,1:4),1);
 theta=atan(poly(1));

%computation of angle of rotation based on quadrant trajectory %lies. If in first or fourth it is -theta and if in second or third %it is -(pi+theta)

sumx=sum(X(1,:));
sumy=sum(X(2,:));

if (sumx>0)
theta=-theta;
else
theta=-(pi+theta);
end

%computation of rotation matrix. rotation is about X axis
rot = [cos(theta) -sin(theta); sin(theta) cos(theta)];
Xrot=rot*X;

Your help is greatly appreciated.

Nancy

Answer 1

[Honglei Chen]

I'm not sure why you need to play with theta, unless you care about the orientation. If you just want the line to become horizontal, rotate an angle of -theta should serve the purpose.

The reason that it does not lie on the x axis is because the rotation matrix is centered on origin. Therefore, once rotated, you need to move it onto the x axis.

I modified your program a little bit and put it below. I just used the -theta and the main change is at the end, where I compensate for the displacement after I did rot*X

% Test data
X = [-2 -1 1 2;0 -1 -3 -4];

%--- Your program starts ---

%Fit a line to the points
poly=polyfit(X(1,1:4),X(2,1:4),1);
theta0=atan(poly(1));

theta=-theta0;

%computation of rotation matrix. rotation is about X axis
rot = [cos(theta) -sin(theta); sin(theta) cos(theta)];
Xrot=rot*X-poly(2)*cos(theta0);   % <- change here

% --- Your program ends ---

% plot
plot(X(1,:),X(2,:),Xrot(1,:),Xrot(2,:))

[Nancy]

I am playing with theta because the range of atan is only -pi/2 to pi/2. So if the points are in the second or third quadrant, wont that give a erroneous result.

I want the points to be oriented on the positive X axis.

[Nancy]

I mean I want the fitted line to lie on the positive X axis.

[Honglei Chen]

If the slope is positive, the line is always in quadrant I and III, and the theta is positive. You rotate -theta to make it horizontal.

If the slope is negative, the line is always in quadrant II and IV, and the theta is negative. Again you rotate -theta to make it horizontal.

So I don't see the need to play with theta. But I guess you mean to have the upper half of the line toward the positive x axis? In any case, your original code already make the line horizontal. All you need to do is to displace it on to the x axis. The change I made still applies, you just may need to use '+' instead of '-' when you rotate it with pi+theta.

[Nancy]

If the points are in quadrant II or III, rotation with a angle theta that lies between -pi/2 to pi/2 wont bring the line onto the X axis.

You can try running the code with X = [-4 -3 -2 0;4 3 2 0]; The rotated line lies on the negative X axis.

In case of the intercept poly(2) 99% of the times is 0.

I just want to know if my code is correct, it looks too crude. :)

Thanks for your help anyways.

[Honglei Chen]

If poly(2) is most of time 0, then your program should work already.

[Nancy]

My data set involves a lot of set of points. It seems to work ok. There are still some points that dont fall on the positive X axis. I want to know if my code is correct or if there is a more elegant way to do this.

[Honglei Chen]

Then why not show an example that gives you the unexpected result?