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Alignments of runners

  1. Nov 14, 2009 #1
    If I have n runners on a circular tracks at different speeds [itex]s_i[/itex], will they always meet up arbitrarily close together in a group?

    So does there always exist a time t such that
    [tex]
    \forall i,\varepsilon>0 ,\exists t: [t\cdot s_i]<\varepsilon
    [/tex]
    where the bracket denote the fractional (non-integer) part.

    And if that time always exists, what is the combination of runner speeds such that it takes them longest time to meet up again?
     
  2. jcsd
  3. Nov 14, 2009 #2
    Lets make this into a simple problem using whole numbers to specify the number of revolutions per minute A1,A2,A3,...A(n) all "n" whole numbers being coprime to each other. If they all start up on the same radius, at a time t they will indeed all be radially aligned again. Can you prove this?
     
  4. Nov 14, 2009 #3
    All rationals is easy to prove I guess, so I'm more worried about algebraic independent numbers (if thats the correct term). And it's easy to show that 3 runners will meet.
    But I'm not sure about more of them, since it might take so long before this 3 group meets the 4th runner, so that the 3 group might split by that time.
    I found http://eom.springer.de/k/k055910.htm but that doesn't help either?
     
  5. Nov 15, 2009 #4
    Here is my take on the problem. I hope I'm not wrong bigtime. : ) -- Edit P.S.: oh well, this was the "easy" part that Gerenuk referred to.

    Let [itex]T_i = 2 \pi R / s_i[/itex] be the period (time to complete one revolution) of each runner.

    Then you look for positive integers [itex]n_i[/itex], such that all [itex]n_i T_i[/itex] are equal.

    Since some [itex]T_i[/itex] can be rational and some irrational, this won't be generally possible. But you can always find rational approximations to your [itex]T_i[/itex] (f.i. by a bisection method), as close to the originals as you wish.

    So assume now that all [itex]T_i[/itex] are expressed as the reduced fraction [itex]a_i / b_i[/itex], with [itex]a_i, b_i[/itex] coprime integers.

    Let [itex]L[/itex] be the least common multiple of all [itex]b_i[/itex]. The integer [itex]c_i = a_i L / b_i[/itex] is the numerator of each fraction, when expressed with [itex]L[/itex] as the (common) denominator. Let [itex]C[/itex] be the lcm of all [itex]c_i[/itex]; now each [itex]n_i[/itex] will be [itex]C / c_i[/itex]. As we have taken the least common multiple on each step, there are no smaller integers satisfying this condition.
     
  6. Nov 15, 2009 #5
    As you noticed there is a tricky part :smile:

    The rational approximation might need such long times to meet up, so that the small deviation from rationals isn't small anymore.
     
  7. Nov 15, 2009 #6
    Oh, I was wrong. That guy Kronecker did actually prove it and the link I mentioned contains the statement.
     
  8. Nov 16, 2009 #7
    I am pretty much lost in that topology link, but I do think there is a problem with my argument. Looking for an *exact* match after n_i revolutions each, given rational approximations of the periods, is exceptionally impractical. What you were asking is if the runners *approximately* meet after some revolutions, which is a completely different question. My apologies for the pointless ornamentation. : )
     
  9. Nov 16, 2009 #8
    Your result for rational numbers solves the problem. An exact match is of course also an approximate match.
    Yet, for irrational numbers this proof doesn't work. In the problem I pose I'm looking for an "arbitrary close" match (not exact though), but it's not that trivial to show that this occurs. You can of course approximate all number by rational, but this still existing little deviation might be blown up, once the number is multiplied by a large number of revolutions.
     
  10. Nov 16, 2009 #9
    Not really. Two runners with periods 333/1000 and 667/1000 will meet at 222111 turns of the first, while what you probably wanted to know is that they nearly meet every other turn at the beginning.

    (Edit: sorry, I meant 667 turns of the first, not 222111. But you get the idea.)
     
  11. Nov 16, 2009 #10
    You were right in the first place with your derivation.
    I actually have "no time" limit. I wanted to know if they ever meet within any given distance.
     
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