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Radius of Curvature

Date: 07/23/2003 at 16:10:43
From: Jim Fife
Subject: Ellipses

I have several ellipses whose major and minor "diameters" I know, but 
I have no information about their foci.  I need to know a formula in 
terms of those diameters or their radii to find the minimum radius 
of curvature at the pointy end of the ellipse, so that, for example, 
I could know what diameter milling cutter could be used to mill an 
elliptical shaped pocket into a piece of metal (too big a tool won't 
fit all the way into the most tightly curved part of the pocket).

The math gets tough (I'm an engineer, but not really a mathematician), 
and the math books generally define ellipses in terms of being the 
locus of points the sum of whose distances from two foci are equal, 
and go from there.

Date: 07/23/2003 at 16:46:02
From: Doctor Douglas
Subject: Re: Ellipses

Hi Jim,

Thanks for writing to the Math Forum.

The major diameter is sometimes called the major axis. Let this have 
length 2*a. Let the minor diameter (minor axis) have length 2*b.  We 
often say that a is the "semimajor axis" and that b is the
"semiminor axis."  Then the eccentricity of the ellipse is 

  e = sqrt(a^2 - b^2)/a

This should be a number between 0 and 1. The distance from the 
center to the foci is c = a*e = sqrt(a^2 - b^2). 

Now, this tells you where the foci are--they both lie on the major 
axis, at a distance of c from the center of the ellipse. But if you 
are trying to calculate the radius of curvature at the pointy end 
(where the major axis intersects the ellipse), you can work directly 
from the formula for the ellipse:

  x^2   y^2
  --- + --- = 1            this assumes that the coordinate system
  a^2   b^2                has the origin at the ellipse's center.

We need the radius of curvature at (x,y) = (a,0).  This is actually
a question that is found using calculus:

                             [(x')^2 + (y')^2]^(3/2)
     radius of curvature R = -----------------------
                                  x'y" - y'x"

where the x and y coordinates can be parameterized as

  x(t)  =  a cos(t), y(t)  = b sin(t)
  x'(t) = -a sin(t), y'(t) = b cos(t)
  x"(t) = -a cos(t), y"(t) = -b sin(t)

and plugging these into the expression for R gives us

       [a^2 sin^2(t) + b^2 cos^2(t)]^(3/2)
  R =  -----------------------------------
             ab[sin^2(t) + cos^2(t)]

The point (x,y) = (a,0) occurs when t=0, so we plug t=0 into this 
expression to find the maximum possible radius of your cutting tool:

           [0 + (b^2)*1]^(3/2)   b^3
  R(a,0) = ------------------- = --- = b^2/a
                  a*b*1          a*b

Does this expression make sense? You can see that if b/a is small 
(i.e., the ellipse is very squashed), then the radius of curvature is 
b*(b/a), so that it is smaller than the semiminor axis b. And if b=a, 
then the ellipse is actually a circle, and it has radius of curvature 
equal to a, as required.

I hope this helps.  Please write back if you have more questions
about this.

- Doctor Douglas, The Math Forum 
Associated Topics:
College Calculus
College Conic Sections/Circles
High School Calculus
High School Conic Sections/Circles

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