Conference rusure::math

        Yes. :-)
        
        Dan

I figured that out but can't come up with a good method of finding
integers m and n such that m/n = i/o.  Any suggestions?

tom_p

    Re .0, .2:
    
    If you divide o by i, you'll get some fraction, yes?  Like 4/3 or 12/8,
    or whatever?  Just reduce that fraction to simplest form (divide
    numerator and denominator by their greatest common divisor).  That
    gives you p/q.  p is the number of times the small circle's tangent
    point touches the larger circle before returning to its starting point. 
    q is the number of times the small circle touches the big circle's
    original tangent point before the original tangents come together
    again.  I think p-q is the number of revolutions the small circle makes
    as seen by a fixed observer.
    
    
    				-- edp

That works ok if o and i are both integers.

What I've been able to come up with so far is multiplying both o and i by
10^n where n is the smallest number which will make both results
integers.  I then remove common factors resulting in p and q as in .-1.
Any better ideas?

If o/i = p/q and p and q are integers derived as above, it looks like p
is the number of full revolutions made by the small circle, q is the
number of times the small circle passes the origin on the large circle
and p+q is the number of times the origin of the small circle touches the
circumference of the large circle.

tom_p

        Let the inner circle first touch the outer circle at "3 o'clock".
        Let P be that point on the inner circler and Q be that
        point on the outer circle.
        
        The circumference of the outer circle is 2 pi o, so the
        inner circle must roll a distance of 2 pi o before it is
        again tangent at point Q.  Likewise the inner circle will
        be tangent at Q if and only if it has rolled a distance
        of 2 n pi o for some integer n.
        
        The circumference of the inner circle is 2 pi i, so the
        inner circle must roll a distance of 2 pi i before point
        P on the inner circle is again oriented at 3 o'clock. 
        Likewise the inner circle will have P oriented at 3 o'clock
        if and only if it has rolled a distance of 2 m pi i for
        some integer m.
        
        If you want both the inner circle to be tangent at Q and
        have point P be the point of tangency (i.e., for P to be
        at 3 o'clock when the inner circle is tangent at Q) then
        you must have 2 n pi o = 2 m pi i or o/i = m/n.
        
        So if o/i is irrational, you will never have both again
        once you start rolling.  If o/i is rational and equal to
        m/n then you can have both after m complete rotations
        (point P moving from 3 o'clock to 3 o'clock) of the inner
        circle and n complete rotations of the inner circle
        around the outer circle (the inner circle moving from Q
        to Q tangency).  This is true for any n and m such that
        o/i = m/n, and is true first at the starting point (sort
        of like saying m = n = 0) and next when m/n is o/i
        reduced to lowest terms.
        
        By the way, if the ratio o/i is irrational then the
        position along the outer circle at which P is oriented at
        3 o'clock do come arbitrarily close to point Q as you
        continue rolling.  In fact they are dense in the outer
        circle.
        
        Dan

    Re .4:
    
    If you have o and i in decimal form, then multiplying them by 10^n and
    then dividing by the common divisor is fine.  This presumes that o and
    i are EXACT when in decimal form, not just approximations.  If these
    are only approximations, the answer will be incorrect, at least if you
    want to know exactly when the circles will be back as they were.
    
    
    				-- edp

        What about a ball in a torus ?

	Alain