I Understanding 3D circle parameterization

[user1003]

TL;DR Summary

Understanding 3D circle parameterization

Hey
I saw this post: https://www.physicsforums.com/threads/general-equation-of-a-circle-in-3d.123168/ trying to understand 3D circle parameterization.
I saw the formula given by Hootenanny
But I didn't understand it and would like some help.
What are u and n and how can I find/calculate them?

thanks!

 

[FactChecker]

1) The third term, $\overrightarrow{c}$ just locates the center.
2) The first term, $R\cos(t)\overrightarrow{u}$ is like the u vector is the X-axis of a simple circle of radius $R$ in the XY plane.
3) The second term, $R\sin(t)\overrightarrow{n}\times\overrightarrow{u}$ is like the Y-axis of a simple circle of radius $R$ in the XY plane.
( $\overrightarrow{n}\times\overrightarrow{u}$ is a unit vector at right angles to both $n$ and $u$. Since $n$ is normal to the plane of the circle, that puts $\overrightarrow{n}\times\overrightarrow{u}$ in the plane of the circle at right angles to $u$.)

How you can calculate $\overrightarrow{n}$ and $\overrightarrow{u}$ would depend on what information you have in the problem you are working on.

 

[PeroK]

I think that $\vec u$ is any unit vector, and $\vec n$ is any vector perpendicular to $\vec u$. Any choice defines a circle. For any given circle, you have a choice of $\vec u$ and two options for $\vec n$.

 

[user1003]

1) The third term, $\overrightarrow{c}$ just locates the center.
2) The first term, $R\cos(t)\overrightarrow{u}$ is like the u vector is the X-axis of a simple circle of radius $R$ in the XY plane.
3) The third term, $R\sin(t)\overrightarrow{n}\times\overrightarrow{u}$ is like the Y-axis of a simple circle of radius $R$ in the XY plane.
( $\overrightarrow{n}\times\overrightarrow{u}$ is a unit vector at right angles to both $n$ and $u$. Since $n$ is normal to the plane of the circle, that puts $\overrightarrow{n}\times\overrightarrow{u}$ in the plane of the circle at right angles to $u$.)

How you can calculate $\overrightarrow{n}$ and $\overrightarrow{u}$ would depend on what information you have in the problem you are working on.

Thanks!
I have the normal of the circle's plane, center point and radius

 

[FactChecker]

Thanks!
I have the normal of the circle's plane, center point and radius

I had to correct a mistake for the third line. It is the second term, not the third.

 

[user1003]

1) The third term, $\overrightarrow{c}$ just locates the center.
2) The first term, $R\cos(t)\overrightarrow{u}$ is like the u vector is the X-axis of a simple circle of radius $R$ in the XY plane.
3) The second term, $R\sin(t)\overrightarrow{n}\times\overrightarrow{u}$ is like the Y-axis of a simple circle of radius $R$ in the XY plane.
( $\overrightarrow{n}\times\overrightarrow{u}$ is a unit vector at right angles to both $n$ and $u$. Since $n$ is normal to the plane of the circle, that puts $\overrightarrow{n}\times\overrightarrow{u}$ in the plane of the circle at right angles to $u$.)

How you can calculate $\overrightarrow{n}$ and $\overrightarrow{u}$ would depend on what information you have in the problem you are working on.

Hey

so my question is how do I get the vector u if I have the normal of the circle's plane, center point and radius?
thanks

 

[FactChecker]

Hey

so my question is how do I get the vector u if I have the normal of the circle's plane, center point and radius?
thanks

Start with any vector at right angles to the normal vector $\overrightarrow{n} = (n_x,n_y,n_z)$,
The general equation for a non-zero vector, $\overrightarrow{v} = (v_x,v_y,v_z) \ne (0,0,0)$, at right angles is to say that the dot product with $\overrightarrow{n}$ is 0.
$0 = n_x v_x + n_y v_y + n_z v_z$.
We know that $\overrightarrow{n}$ is not the zero vector. Assuming that $n_z\ne 0$, we can set $v_x=1, v_y=1, v_z = (-n_x - n_y)/n_z$.
That will give a vector in the plane at right angles to $\overrightarrow{n}$.
Then we want to normalize $\overrightarrow{v}$ to a unit length to get $\overrightarrow{u}$:
$\overrightarrow{u} = \overrightarrow{v}/|\overrightarrow{v}|$