Radius of Curvature of Ellipse
Let us learn a few basic facts about Ellipse
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).
An Ellipse can be visualized as a Conic Section
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).
An Ellipse can be visualized as a Conic Section
While the equations of the Ellipse is given as shown below
While the equations of the Ellipse is given as shown below
In these ( h, k ) is the center of the Ellipse. For the ellipse a > b
While if b > a then the calculations are shown below
While if b > a then the calculations are shown below
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 point y 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"
Or it can be written as shown below
This is actually a question that is found using calculus:
[(x')^2 + (y')^2]^(3/2)
radius of curvature R = ----------------------------------
x'y" - y'x"
Or it can be written as shown below
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
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.
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
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.
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