• Support PF! Buy your school textbooks, materials and every day products Here!

Osculating circle for 3D curve

  • Thread starter jks
  • Start date
  • #1
jks
2
0

Homework Statement


Given the space curve r(t)=<t^2,(2t^3)/3,t>, give the parametrized equation for the osculating circle along the trajectory at r(p).

I should note this is so I can give a plot of the circle in Winplot, which is a free grapher/plotter type program.

Homework Equations


From r(t), the Frenet frame is:
T(p)=<2p,2p^2,1>/(2p^2+1)
N(p)=<1-2p^2,2p,-2p>/(2p^2+1)
B(p)=<-2p,1,2p^2>/(2p^2+1)

the radius of curvature is c(p)=((2p^2+1)^2)/2
so the expression for centers of curvature is:

<p^2,(2p^3)/3,p> + <1-2p^2,2p,-2p>(2p+1)/2 = r(p) + N(p)c(p)


The Attempt at a Solution



Ok my first crack at this was to try and solve it as the intersection of the osculating sphere and osculating plane, which would give a circle. So the osculating plane's equation would be

(x-p^2)(-2p)/(2p^2+1) + (y-(2p^3)/3)/(2p^2+1) + (z-p)(2p^2)/(2p^2+1) = 0

then I chose to represent the sphere as a parametric surface:
defining x0 = (1-p^2)(2p^2+1)/2
y0 = (2p^3+6p)(2p^2+1)/6
z0 = (-p)(2p^2+1)/2
are the x, y, and z components of the center of curvature respectively.

x = c(p)cos(θ)sin(φ) + x0
y = c(p)sin(θ)sin(φ) + y0
z = c(p)cos(φ) + z0

so then I rearranged the plane's equation for x, y, and z separately, and set them equal to the parametric equation. I don't think I have to say that the equations were quite big, but I noticed that they were a linear system, so I solved it by using matrices. I'm going to stop right here because this approach didn't give me anything useful, problem #1 of which was that the parametric representation of the intersection would necessarily have been an expression of 2 parameters, giving a surface, while only 1 parameter is required for a curve, which is what I need.

I have a nagging feeling that I'm overlooking something very simple that could make this problem a lot easier. Might I be able to construct the circle, then rotate it so that it fits into place? I'm not too well versed on how to do 3d rotations.
 

Answers and Replies

  • #2
I have an idea. You already have the following quantities:
- The point r(p)
- The unit normal n(p)
- The radius of curvature c(p)
Then the center of the osculating circle is r(p)+c(p)N(p).
So now you have a center, radius and normal B(p) to the circle,
it shouldn't be hard if you find the right formulas...
 

Related Threads on Osculating circle for 3D curve

  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
2
Views
10K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
1
Views
5K
  • Last Post
Replies
4
Views
3K
Replies
3
Views
3K
Replies
4
Views
2K
Replies
1
Views
3K
Replies
3
Views
4K
  • Last Post
Replies
1
Views
476
Top