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Help with Logarithm Manipulation for an Athletics question

  1. Dec 7, 2015 #1
    I am interested in track & have found through some hard work that for 800 that 1'42.00 is same value as for 3'44.00 for 1600
    ( note : for 1600m not 1 mile )

    It is a Log relationship but I can't quite settle on what the value for equivalence should be for 1500m.

    I have a provisional figure of 3'28.25, but I might be a tenth or so out.

    I'd appreciate on any thoughts on "best" log manipulation of "known" 1'42.00 for 800 & 3'44.00 for 1600 to give an exact as possible figure for 1500.
  2. jcsd
  3. Dec 7, 2015 #2


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    Your question is unclear. log(800)=1'42.00 or log(1'42.00)=800? Does 1'42.00 mean 1 minute and 42 seconds?
  4. Dec 7, 2015 #3


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    By hard work do you mean you just took the world record values for each of those races?

    How have you decided from two figures alone that it's a log relationship? Why not a linear relationship? A hyperbolic relationship? Any of the other infinite possibilities?

    If you believe it must follow a log curve, then you simply need to solve these equations simultaneously to find the values of a,b:

    [tex]a\log(800)+b = 102[/tex]
    [tex]a\log(1600)+b = 224[/tex]

    So you can then calculate the value of [itex]a\log(1500)+b[/itex].
  5. Dec 7, 2015 #4
    1'42 = 102 secs
    Last edited by a moderator: Dec 8, 2015
  6. Dec 7, 2015 #5

    There is no world record for 1600m


    I know it is a Log relationship from years of research/calculation

    Then try some to link 1'42.00 for 800 with 3'44.00 for 1600

    Thank you

    I normally end up with 3'28.25 with my alternative attempts at solution, but I believe that may be 1 or 2 tenths out

    What did you get with your method ?
  7. Dec 7, 2015 #6


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    Alright, I'll take your word for it.

    There are infinitely many ways to link two points on a graph. Most of them will not even be close to representing an accurate or even an approximate model of the real life situation though.

    Well I was hoping that you'd check it yourself to see that my model can't be correct.

    For 1500m it produces a result of 3'52.60, but extending it out to the extremes is where it goes very wrong. It gives the marathon (42.195km) a time of 13'20 and the 100m sprint has a negative time.
  8. Dec 7, 2015 #7


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    The method suggested above yields 3'32.64 which I think is too high. (12sec on the last 100 of 1600 is usually not achievable.) My calculation gives me an estimation of 3'28.14 (ln 102/800 ≈ -2.06 and ln 224/1600 ≈ -1.966; so I estimated ln x/1500 ≈ -1.975 which yields x ≈ 3'28.14).
  9. Dec 7, 2015 #8
    I'm interested only in 1500.

    Don't bother with extreme distances above/below.

    The answer is somewhere in 3'28-low region +/- tenths.

    It is refining precision...
  10. Dec 7, 2015 #9


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    I find 3'28.14 fits pretty well on your target. (I have only been calculating 1500. What I didn't know was how the "energy rate" declines. You started with an exponent -2.06 (on 100m) and ended with an exponent -1.966 on 1600. So to estimate -1.975 on 1500 seemed a good guess to me.)
  11. Dec 7, 2015 #10
    I don't have much confidence/interest in your answer.

    I want as much precision possible.

    I have no interest in "energy rate".

    Just 2 parameters with what 1500 is...
  12. Dec 8, 2015 #11


    Staff: Mentor

    @eldrick, ow about a little more civility from you? The question you have asked is ambiguous, and the members posting here are just trying to help. Show them more courtesy, and a lot less attitude...
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