Wikipedia:Reference desk/Archives/Mathematics/2012 May 8

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May 8[edit]

Transitioning from one arc to another and finding the radii of curvature and the joining point[edit]

I want to construct two connecting arcs that have the same tangent at where they join. One has a constant radius of curvature and the other ends at a given point. To draw an example in your head:

  1. A circle with a radius of 1 has its center at (-4,0).
  2. Point A is at (0,2).
  3. The first arc starts at point A and is drawn in a counter-clockwise direction to where it is tangent to the circle.

What is the radius of curvature of the arc and where is it tangent to the circle? --Melab±1 20:33, 8 May 2012 (UTC)[reply]

I don't think you've added enough constraints. You could just draw a line segment from that point tangent to the circle, in which case the radius of curvature of a line is infinite (actually there would be two possible line segments, one tangent near the top of the circle, one tangent near the bottom).
Or, you could add the constraint of having the arc match the curvature of the circle at the tangent point, in which case it would have the same radius as the circle there. I believe this arc would have a degree of 2, if drawn using 3 constraints (point A, and the tangent and curvature constraints at the circle) so would be a quadratic arc. StuRat (talk) 21:41, 8 May 2012 (UTC)[reply]
It could be a conic section or it could be something else. —Tamfang (talk) 00:43, 10 May 2012 (UTC)[reply]
A clothoid (or Cornu spiral or Euler spiral) is commonly used in road and railway design as a Track transition curve to join straight parts or circular arcs of a track. Also the 3rd degree curve (described by the equation of the form y = ax3 + bx2 + cx) is used for this purpose. --CiaPan (talk) 07:41, 9 May 2012 (UTC)[reply]
But this problem isn't about connecting a line segment to a circle, it's about connecting a point to a circle. StuRat (talk) 16:41, 9 May 2012 (UTC)[reply]
If we read the given problem as as requiring the solution curve to osculate the given circle, it allows a family of clothoid solutions with one degree of freedom: where the transition curve meets the circle, or the tangent angle at the given point. (I recently got interested in clothoids for font design!) —Tamfang (talk) 00:43, 10 May 2012 (UTC)[reply]

Let A be a given point outside a given circle, B the circle's center point and r the circle's radius. We're seeking a circular arc through A, which is internally tangent to the given circle and has radius R. Let C be the arc's centre and D an arc and circle meeting point. Then |CA| = |CD| = R and |CB| = R−r = |CA|−r. That means the arc's centre C is always closer by r to B than to A, so C is a point of one branch of a hyperbola of foci A,B, the arc's radius is R=|CA| and the tangency point D is found by intersecting the given circle with a |CB| line as the intersection which is not between B and C. The minimum R is (|AB|+r)/2.
The other branch of the hyperbola is a locus of centers of arcs through A which are externally tangent to the given circle.
If point A is inside the circle, then points C form an ellipse with foci A,B.
In the degenerate case of A being a point of the circle, the set of C's is a straight line AB and the arc sought is any circle tangent (internally or externally) to the given circle at A. --CiaPan (talk) 07:40, 11 May 2012 (UTC)[reply]