

Title: Rolling Circle on Ellipse Post by chetangarg on Dec 18^{th}, 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 22^{nd}, 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 27^{th}, 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 27^{th}, 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 30^{th}, 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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