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riddles >> putnam exam (pure math) >> Rolling Circle on Ellipse
(Message started by: chetangarg on Dec 18th, 2018, 1:55am)

Title: Rolling Circle on Ellipse
Post by chetangarg on Dec 18th, 2018, 1:55am
If a circle or radius r is rolling around a fixed standard ellipse
x^2/a^2 + y^2/b^2 =1 .
Find the locus of the center of the circle.
Also can you provide the simulation of the path.

Title: Re: Rolling Circle on Ellipse
Post by Michael Dagg on Dec 22nd, 2018, 5:50pm
Are you sure it's the locus of the center you want
and not the locus of a boundary point on the circle?

Title: Re: Rolling Circle on Ellipse
Post by Grimbal on Dec 27th, 2018, 12:56pm
It should be something like this

x = a cos t + r b cos t/sqrt((b cos t)^2 + (a sin t)^2)
y = b sin t + r a sin t/sqrt((b cos t)^2 + (a sin t)^2)

where a, b are the 1/2 diameters and r is the circle radius.

Example: a=2 b=1 r=0.5
on https://www.wolframalpha.com
parametric plot ( 3 cos t + .5 (2 cos t)/sqrt((2 cos t)^2 + (3 sin t)^2) , 2 sin t + .5 (3 sin t)/sqrt((2 cos t)^2 + (3 sin t)^2) )


Title: Re: Rolling Circle on Ellipse
Post by Michael Dagg on Dec 27th, 2018, 4:15pm
I made a post here earlier and tried to attach an image but keep getting a server error and then couldn't save the post. I will not post an image this time.

The locus of the center is just another ellipse whose major and minor axes lengths are merely extended by r at their ends. That is, it's just a copy of the standard ellipse extended r units along a vector normal to the tangent at each (x,y) of the stardard ellipse. Just replace a, b in the standard equation with a+r, b+r to get this new ellipse.

There are various ways to write a polar representation of this new ellipse.

For the pair (R, theta) you can start with substituting

x = R(theta) cos(theta)
y = R(theta) sin(theta)

into the equation of the new ellipse and then solve for R(theta).

It can also be written using the eccentricity.

Same idea applies if the circle is inside the ellipse provided that the radius of curvature of the circle is less or equal to smallest radius of curvature of the ellipse.

Title: Re: Rolling Circle on Ellipse
Post by Grimbal on Dec 30th, 2018, 3:18pm
Adding a vector perpendicular to the tangent is what I did.

u = (a cos t, b sin t) is the path of the ellipse.
u' = (-a sin t, b cos t) is the derivative, giving a vector parallel to the tangent.
v = (b cos t, a sin t) is the same rotated 1/4 turn.  v is perpendicular to the ellipse at point u.

Resizing v to have length r and adding that to u gives the answer: u + r*v/|v|

Or in other words:
x = a cos t + r b cos t/sqrt((b cos t)^2 + (a sin t)^2)
y = b sin t + r a sin t/sqrt((b cos t)^2 + (a sin t)^2)



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