Rotate Normal Around Tangent

I want to rotate a normal vector around a tangent vector to create a circle. I have not been able to find anything to do this. Does such a function exist? Or how would I generate one?

Thanks.

[Matt J]

Clarify what this is supposed to do. What data are you given and in what form? What will the output data be, and in what form?

[Paul Huter]

I have a 3D curve, and I have calculated normal and tangent vectors at each point on the curve based on a velocity (I am using Frenet-frame here). I want to be able to draw a 3D circle that is normal to the curve at some points (not all...there are a lot of points...). I figured the best way would be to rotate the normal vector about the tangent vector.

Answer 1

[Matt J]

I'm assuming you're choosing the radius, R, of this circle. Then if T and N are the tangent and normal vectors at point P on the curve (all in column vector form):

n=1000;
theta=linspace(0,2*pi,n+1); 
  theta(end)=[];

refcircle = [R*cos(theta);R*sin(theta);zeros(1,n);ones(1,n)] ;

T=T/norm(T);
N=N/norm(N);
E=cross(N,T);

A=[0 0 0; R 0 0; 0 R 0].';
B=[P,P+R*N,P+R*E];

params=absor(A,B);  %get this function from FEX

C = params.M*refcircle; %circle points at 3D curve

plot3(C(1,:), C(2,:), C(3,:)) %plot the circle

The above uses ABSOR, available here

http://www.mathworks.com/matlabcentral/fileexchange/26186-absolute-orientation-horns-method

[Paul Huter]

Two things: * What is "P"? Is this supposed to be T? * Is there any way to do this without ABSOR? I have a problem with getting code from home to work, so if I could get a solution that does not use this function (which is rather large...), I would appreciate that

I appreciate the help - this looks like a good solution, aside from the big external function.

Thanks!

[Matt J]

As I mentioned in the preamble to the code "T and N are the tangent and normal vectors at point P on the curve". Here's an alternative sans-ABSOR

n=1000;
theta=linspace(0,2*pi,n+1); 
  theta(end)=[];
refcircle = [R*cos(theta);R*sin(theta)] ;

T=T/norm(T);
N=N/norm(N);

C=bsxfun(@plus, [N,T]*refcircle, P);

plot3(C(1,:), C(2,:), C(3,:)) %plot the circle

[Paul Huter]

Whoops, I read the reply, then forgot about it when reading the code. I appreciate the help.

[Paul Huter]

There seems to be a problem when I multiply [N, T] by refcircle. refcircle as 1000 columns and [N, T] may be two columns. How do I fix this?

[Matt J]

There's nothing wrong with refcircle having more than 2 columns. Since this is matrix multiplication, it's only critical that refcircle have 2 rows. I've tested all versions of the routine before giving them to you and they run with no MATLAB errors on my end.

One small issue, though, is that I did have the circle tangent to the curve instead of normal to it. The version below should fix that.

    n=1000;
    theta=linspace(0,2*pi,n+1); 
      theta(end)=[];
    refcircle = [R*cos(theta);R*sin(theta)] ;
    T=T/norm(T);
    N=N/norm(N);
    E=cross(N,T);

    C=bsxfun(@plus, [N,E]*refcircle, P);
    plot3(C(1,:), C(2,:), C(3,:)) %plot the circle

[Paul Huter]

Alright, I got it. However, when I plot, I end up with a bunch of lines (which are normal to the curve). Maybe they are long thin ovals, but they are certainly not circles. Again, any help is appreciated.

[Matt J]

Clarify whether the plot you're talking about is from the code as I gave it to you, or the result of you adapting/inserting it into your larger problem. If the latter, I'd have to see what you did.

However, when I run it in isolation with the sample data P,T,N,R data below, I definitely get a plot of a circle floating in 3D space. Verify first that you can reproduce this.

P=[1;1;1];
T=[1;1;1];
N=[-1;2;-1];
R=3;

    n=1000;
      theta=linspace(0,2*pi,n+1); 
        theta(end)=[];
      refcircle = [R*cos(theta);R*sin(theta)] ;
      T=T/norm(T);
      N=N/norm(N);
      E=cross(N,T);
      C=bsxfun(@plus, [N,E]*refcircle, P);
      plot3(C(1,:), C(2,:), C(3,:)) %plot the circle

[Paul Huter]

Turns out the way I was calculating my tangent/normal vectors was doing them as row-vectors, not columns. Looking at the way you did it (with columns) got it to work.

Thanks for your help.

Answer 2

[Muthu Annamalai]

Paul, You need to find the points of a 2D rotation transform using the equations, for example affine transformation http://en.wikipedia.org/wiki/Rotation_(mathematics), and then you may visualize it using plot() commands. HTH, -Muthu

[Paul Huter]

As stated above, I am calculating Frenet-frame vectors, which I have (in the past) used to generate a rotation matrix (3D). Is this same rotation matrix going to allow me to draw a circle perpendicular to the curve (i.e. by multiplying the x, y, z points of a circle by the rotation matrix)?

Answer 3

[Image Analyst]

Sounds like the streamtube() function. Could that be used? Or maybe morphological dilation, imdilate(). For morphological dilation, imagine a sphere whose center is tracing out your 3D curve. The dilated volumetric image is the volume swept out by that sphere as it travels along your curve.