General equation of a circle in 3D?

Sandyscott

Hi,

I've been trying to deduce the general equation of a circle in 3D space, but without much luck.

Well, I'll jump in with what I've got so far.


[tex]P = (X_{P},Y_{P},Z_{P}) = \mbox{General point on diameter of circle circle.}[/tex]
[tex]C = (X_{C},Y_{C},Z_{C}) = \mbox{Centre of the circle.} [/tex]
[tex]\bar{d} = \left(\begin{array}{cc}x_{d}\\y_{d}\\z_{d}\end{array}\right) = \mbox{Perpendicular to the plane of the circle.} [/tex]
[tex]r = \mbox{radius of circle} [/tex]

These are the identities I've used to describe it:
[tex]\vec{PC}[/tex] is perperpendicular to [tex]\bar{d}[/tex] so
[tex]\vec{PC}.\bar{d}=0[/tex]

and,

[tex]|\vec{PC}| = r[/tex]

these expand (somewhat clumsily) to:
[tex]x_{d}(X_{C}-X_{P})+y_{d}(Y_{C}-Y_{P})+z_{d}(Z_{C}-Z_{P})=0 [/tex]
and
[tex]\sqrt{(X_{C}-X_{P})^2+(Y_{C}-Y_{P})^2+(Z_{C}-Z_{P})^2}=r[/tex]

Not sure what to do from here, or if I'm even barking up the right tree, any help would be much appreciated

Hootenanny

Hi there Sandyscott and welcome to PF,

The easiest way is to define the 3D circle using parametric equations.

Sandyscott

And how do I do that?

Bearing in mind that in the problem I want to have the circle at a funny angle.

arunbg

Usually the equation of a circle in 3D is given by the locus of points that satisfies the eqn of some sphere and a plane which cuts off the circular cross section required.
What are the given parameters to determine the equation in your question ?

Sandyscott

This isn't for a formal maths question*, so I don't have any specifc boundaries. I think ideally I'd like to have parametric description of the curve in terms of one parameter, with these input constants: the centre of the circle, the perpendicular vector and the radius.


*I want to determine the shape of a piece of wood mouth of a duct (circular) and a rectangular hole in a model hovercraft I'm building (I have no intention of describing the piece of wood, I just want to "take measurements" from my mathematical model)

Hootenanny

Using vectors, generally if t is the parameter then and point P on the circle is given by;

[tex]P = R\cos(t) \vec{u} + R\sin(t) \;\;\vec{n}\times\vec{u} + c[/tex]

Where u is a unit vector from the centre of the circle to any point on the circumference; R is the radius; n is a unit vector perpendicular to the plane and c is the centre of the circle.

Sandyscott

Thanks very much, that should be very useful.

Incidentally, is that derived from the intersection of a sphere and plane re: arunbg? (so i can have a go at understanding it myself ;-) )

Hootenanny

Yes it is derrived from the intersection of a sphere and a plane using vector geometry.

Sandyscott

Cool,

Thanks for the prompt responses, I've never had an answer as quickly in any forum.

Cheers

Hootenanny

My pleasure, if you need any help deriving the general form do not hesitate to post here again.

play

Why does the equation of a 3d circle need a unit vector perpendicular to the plane and a unit vector from the center of the circle to any point on the circumference?

Hootenanny

Because in three dimensions one needs both vectors to determine the vector that is tangential to the circumference at a given point and lies in the plane of the circle.

play

Oh, thanks :)

dan33

Hi,

How would one describe a circle in non-parametric form? I understand that a circle in 3d is the intersection of a sphere and a plane. Is it possible to have an equation describing the circle with only the following elements:
- coordinate of the center of the sphere (also a point on the plane), x_c, y_c, z_c
- components of the vector normal to the plane, n_x, n_y, n_z
- the radius of the sphere, r
- the locus of points of the circle which would of course be x, y, z

Thanks in advance.

Chris001

"Where u is a unit vector from the centre of the circle to any point on the circumference"

How exactly do you derive vector U? its more complicated than just Rcos(t),Rsin(t) because you have that third axis, right?

LCKurtz

I'm a little late to this thread, but as you have noticed, nobody has given an answer that contains only the three pieces of information posited by the original poster: The center, the radius, and a normal vector. Here's how to do that; the algebra gets messy in the general case but not bad for specific numbers.

Let [itex] N = \langle n_1,n_2,n_3\rangle[/itex] be the normal vector, [itex]\vec C = \langle a,b,c\rangle[/itex] be the position vector of the center, and r be the radius.

Let [itex]\vec u = N \times \vec i[/itex] and [itex]\vec v = N \times \vec u[/itex]. Normalize these two vectors to get [itex]\hat u[/itex] and [tex]\hat v[/itex] which are now orthogonal unit vectors in the plane of the circle. The parametric equation of the circle becomes:

[tex]\vec R(t) = \vec C + r\cos(t)\hat u + r\sin(t) \hat v[/tex]

[Edit] Additionally note that crossing N with i is to get a vector perpendicular to N. If N is parallel to i that won't work, but then just use j.

gofall

Could anyone please give the matrix form of the equation like:


with,


and [tex]A[/tex] the matrix.

Thanks in advance.