Undergrad Defining a generalized coordinate system

Click For Summary
A coordinate system is being developed to represent points in space based on a parametric curve, defined by coordinates (s, ρ, ω). Here, ρ represents the shortest distance from a point not on the curve to the curve itself, while s is the distance along the curve from a starting point. The unit tangent vector to the curve, denoted as ẑ(t), helps define a plane where the point lies on a circle of radius ρ. The direction ω needs to be orthogonal to ẑ(t) and should align with cylindrical coordinates when the curve is a straight line in the z-direction. The challenge lies in ensuring that the ω direction adapts appropriately as the curve changes, maintaining the desired properties.
dipole
Messages
553
Reaction score
151
(Note that the title of this thread might be incorrect - I'm just drawing on the vocabulary people use when discussing Lagrangian Mechanics...)

Hi, I'm trying to set up a coordinate system to represent points in space where one of the coordinates is the distance along a parametric curve, one is the shortest distance from a point to the curve, and one is an angle defined relative to some direction.

I have some curve in space defined by:

x = x(t)
y = y(t)
z = z(t)

and I want to define a coordinate system (s, \rho, \omega) relative to this curve.

\rho is defined by the shortest distance from a point p' = (x',y',z') (I will use primes to denote points not on the curve) in space to the curve (assume this is unique).

s is defined by the distance along the curve, starting from some initial point \left (x(t_0), y(t_0), z(t_0) \right) to the point \left (x(t), y(t), z(t) \right) such that

d \left (x', y', z', x(t), y(t), z(t) \right ) = \rho

where d is the euclidean distance.

In other words, say I have some point in space p' = (x',y',z'), then \rho is the length of the smallest line segment between p' and some point on the curve p(t) = (x(t), y(t), z(t)). I want to represent the vector from p(t) to p' by an orthogonal coordinate system that is attached to the curve.

Let \hat{s}(t) be the unit tangent vector to the curve at the point p(t). This defines a plane where the point p' lies on a circle of radius \rho in the plane. The only other thing I need to uniquely define the point p' in this plane is some reference direction to measure the angle at which the point p' lies on the circle. Call this direction \hat{\omega}(t).

Here's the kicker though - I want \hat{\omega}(t) to be defined in such a way that when the curve is a straight line, and in the z-direction, that the coordinate system becomes plain old cylindiral coordinates, and \hat{\omega}(t) = \hat{x} . However, when the curve is not a straight line, then \hat{\omega} should rotate with the curve in such a way that it is always orthogonal to \hat{s}(t), and locally if \hat{s}(t) \simeq \hat{z} then \hat{\omega}(t) \simeq \hat{x}.

My question is how to define \hat{\omega}(t) in such a way that satisfies these constraints.

Does this make sense? I hope at least the idea of what I want to do is clear. I further hope this problem is not ill-posed. Any help would be appreciated and please let me know if something is unclear.
 
Mathematics news on Phys.org
Hi, yes I remembered that as I was writing this, but I'm still not sure how to enforce that the Frenet-Serret frame becomes normal cylindrical coordinates in the limit that the curvature goes to zero (assuming that any straight line will be in the z-direction).
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 10 ·
Replies
10
Views
1K
  • · Replies 1 ·
Replies
1
Views
3K