wu :: forums « wu :: forums - Rolling Circle on Ellipse » Welcome, Guest. Please Login or Register. Feb 22nd, 2024, 8:46am RIDDLES SITE WRITE MATH! Home Help Search Members Login Register
 wu :: forums    riddles    putnam exam (pure math) (Moderators: Grimbal, towr, Eigenray, Icarus, william wu, SMQ)    Rolling Circle on Ellipse « No topic | Next topic »
 Pages: 1 Reply Notify of replies Send Topic Print
 Author Topic: Rolling Circle on Ellipse  (Read 606 times)
chetangarg
Newbie

Gender:
Posts: 30
 Rolling Circle on Ellipse   « on: Dec 18th, 2018, 1:55am » Quote Modify

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.
 IP Logged
Michael Dagg
Senior Riddler

Gender:
Posts: 500
 Re: Rolling Circle on Ellipse   « Reply #1 on: Dec 22nd, 2018, 5:50pm » Quote Modify

Are you sure it's the locus of the center you want
and not the locus of a boundary point on the circle?
 IP Logged

Regards,
Michael Dagg
Grimbal
wu::riddles Moderator
Uberpuzzler

Gender:
Posts: 7526
 Re: Rolling Circle on Ellipse   « Reply #2 on: Dec 27th, 2018, 12:56pm » Quote Modify

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) )

 « Last Edit: Dec 27th, 2018, 12:57pm by Grimbal » IP Logged
Michael Dagg
Senior Riddler

Gender:
Posts: 500
 Re: Rolling Circle on Ellipse   « Reply #3 on: Dec 27th, 2018, 4:15pm » Quote Modify

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.

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.

 « Last Edit: Dec 27th, 2018, 4:23pm by Michael Dagg » IP Logged

Regards,
Michael Dagg
Grimbal
wu::riddles Moderator
Uberpuzzler

Gender:
Posts: 7526
 Re: Rolling Circle on Ellipse   « Reply #4 on: Dec 30th, 2018, 3:18pm » Quote Modify

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)
 « Last Edit: Dec 30th, 2018, 3:19pm by Grimbal » IP Logged
 Pages: 1 Reply Notify of replies Send Topic Print

 Forum Jump: ----------------------------- riddles -----------------------------  - easy   - medium   - hard   - what am i   - what happened   - microsoft   - cs => putnam exam (pure math)   - suggestions, help, and FAQ   - general problem-solving / chatting / whatever ----------------------------- general -----------------------------  - guestbook   - truth   - complex analysis   - wanted   - psychology   - chinese « No topic | Next topic »