Understanding 3D circle parameterization

  • #1
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Understanding 3D circle parameterization

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  • #2
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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.
 
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  • #3
PeroK
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:welcome:

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##.
 
  • #4
user1003
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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
 
  • #5
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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.
 
  • #6
user1003
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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
 
  • #7
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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}|##
 

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