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  • eldrick
    replied
    you've posted 1/2 dozen nonsense posts ( & you can't even work out relevance of absolute speed & relative speed ?! ), fiddling with stats & running around like a headless chicken

    when advised to modify the stats to come up with more meaningful numbers, you stick your head in the sand

    taking a cursory look at the numbers, the coefficient variation is likely to be almost identical for shot & discus indicating wind-effects, outliers, etc cancel out when best throws are considered, leaving hammer as an anomaly

    however, looking at 3rd best throw analysis, the coefficient of variation for all 3 is nearly identical, indicating that you may have screwed up your initial hammer calculation with best throw or if they are correct, some anomaly does exist

    Leave a comment:


  • Cottonshirt
    replied
    eldrick,

    I daresay there are people who find your overly bombastic approach to imparting information, your rude, witless ramblings and your arrogant and supercilious manner either amusing or helpful. I am not one of them.

    I daresay there are people who need to look petty statistical formulae up on Wikipedia because they are unable to remember what they are, didn't understand them in the first place and don't have the wits to correctly apply them. I am not one of those people either.

    I daresay there are people who, having grown tired of the ignorant ramblings of the sad, lonely morons who inhabit some internet message boards discover how to use the "hide posts from this member" facility. I am one of those people. Good bye.

    Martin

    Leave a comment:


  • eldrick
    replied
    you clearly haven't a clue

    here's wiki take on it - read it & learn :

    The coefficient of variation is useful because the standard deviation of data must always be understood in the context of the mean of the data. The coefficient of variation is a dimensionless number. So when comparing between data sets with different units or wildly different means, one should use the coefficient of variation for comparison instead of the standard deviation.

    Leave a comment:


  • Cottonshirt
    replied
    Originally posted by eldrick
    look up coefficient of variation
    I don't have to look it up. I am perfectly familiar with the coefficient of variation, thank you. It is a standardised measure useful when the values in a batch are measured on a ratio scale. In athletics, we measure using a tape measure which is an absolute scale, sometimes referred to as an interval scale because the interval between any two adjacent marks is the same along the length of the scale. A ratio scale might apply if you were measuring something logarithmically; earthquakes for example.

    Please do go and see if you can find an earthquake to measure.


    Martin

    Leave a comment:


  • eldrick
    replied
    Originally posted by Cottonshirt
    Please put the chalk down and let the grown ups play with the black board
    quoting gsce formulas is the extent of your talents ?!

    listen up kid, you are clueless on the matter

    look up coefficient of variation

    use your brain ( whatever there is ) & then learn to apply it...

    Leave a comment:


  • Powell
    replied
    Originally posted by Cottonshirt
    I wonder if this is possibly because it is not aerodynamic at all whereas the discus is at least a little bit like a flying wing. It might not be much but a spinning discus can generate a little lift (in the right wind) but a hammer cannot.
    It's a very good point. Actually, in the discus, the wind effect is much more than 'a little bit' significant. Just compare the results that the Americans (and some others) achieve in the California wind tunnels to what they do in enclosed stadiums. The right wind adds meters, not just inches, to your distance.

    Leave a comment:


  • Cottonshirt
    replied
    Considering your point that an athlete's best throw is less likely to be representative, I repeated the calculation but this time I used each athlete's 3rd best throw of all time. This meant that I had fewer athletes data to play with and have been limited to only 50 throwers, but this should be a sufficient sample.

    Top 50 - Based on 3rd best throw.
    1. SP 0.39m.....34[/*:m:uyx7eoj6]
    2. DT 1.22m.....32[/*:m:uyx7eoj6]
    3. HT 1.37m.....35[/*:m:uyx7eoj6]

    Notice how the SD for both the SP and DT have come down by over 20% (SP = -22%, DT = -25%) indicating that a thrower's third-best is likely to be nearer the mean, more "average". This would seem to indicate that their best was exceptional, what we earlier referred to as an outlier.

    But for the hammer the SD has got bigger (HT = +17%). It is also worth noting that for the HT the mean went up (81.59 versus 81.87) when considering 3rd best throws. The only way I can interpret this is that it means their 3rd best throw is closer to their best than in the other two events. Or looking at it from the other direction, their best is less likely to be an outlier because it is closer to their 3rd best throw than in either the discus or shot.

    This tells me that the "lucky long outlier" is less likely in the hammer (at the level of the top throwers in the world) than in the other two throws we looked at. In other words, this again reinforces what Powell said; the hammer is highly dependent on good technique, less dependent on luck. I wonder if this is possibly because it is not aerodynamic at all whereas the discus is at least a little bit like a flying wing. It might not be much but a spinning discus can generate a little lift (in the right wind) but a hammer cannot.


    Martin

    Leave a comment:


  • Cottonshirt
    replied
    Originally posted by eldrick
    thank you imagy, & while you're at it, you can tell this idiot he needs to divide the standard deviations he's got by the mean of the population in order to get viable comparisons
    Where x is any value in the batch, and x is the mean of those values, and where n is the number of values in the batch. And, where for the sake of demonstration we shall call E the Sigma notation symbol. Then standard deviation is calculated as...


    ..........((E(x-x)^1/2)/n)^1/2

    In some cases it is appropriate to use n-1 rather than n. There is no need to divide by the "mean of the population" which is in any event a meaningless term for a non-existent value.

    Please put the chalk down and let the grown ups play with the black board.


    Martin

    Leave a comment:


  • eldrick
    replied
    Originally posted by Powell
    I disagree. Eldrick's recent post estimating Henry Rono's potential at distances he competed in based on made-up PBs in events he never contested was more ridiculous than this.
    coming from a guy who was barely out of diapers when rono was running in his pomp, i'll stick to my findings, based also on 1st hand observation - i had the good fortune to see his race against ovett in london & had him sign my programme

    Leave a comment:


  • eldrick
    replied
    thank you imagy

    & while you're at it, you can tell this idiot he needs to divide the standard deviations he's got by the mean of the population in order to get viable comparisons

    Leave a comment:


  • imaginative
    replied
    Originally posted by Cottonshirt
    Originally posted by imaginative
    Let us say that, in the
    language of statistics, the hammer-throw has a larger standard
    deviation than the shot-put (in my impression so far; have not
    crunched the numbers, though).
    Well now I have crunched the numbers and they don't really tell the
    story you assumed they would.
    Thank you for taking the trouble. It is always nice to get numbers on
    things. In particular, I suspect that I might have been fooled by the
    absolute size of the numbers. Even knowing that the hammer usually
    goes four times the distance of the shot, it is easy to view 2 "hammer
    meters" as a greater improvement than 0.5 "shot meters".

    I took the top 90 throwers all-time in each event, SP, HT, and DT (I
    didn't do the JT because they have changed the model twice and it got
    too complicated to allow for the differences).
    It might (and I am not sure about this) be better to look at each
    individual. If you look at the PB of the top-90 (which is my
    impression) this will likely yield a misleading SD because the PBs
    will all be above-to-noticeably-above the expectation value.

    (I do not think that this would have a large effect on the
    conclusions, however.)

    Using what I consider to be the standard approach to calculating SD
    (using n rather than n-1 as my divisor) I got the following numbers.
    The columns are, from the left, event, standard deviation, number of
    men within 1-SD of the mean.
    1. SP 0.50m.....65[/*:m:2yyomk6b]
    2. DT 1.63m.....65[/*:m:2yyomk6b]
    3. HT 1.17m.....64[/*:m:2yyomk6b]


    So, not only does HT not have a larger standard deviation than the
    other throws, but the event with the largest standard deviation does
    not have more men in that interval than the others.
    The constant number of athletes was to be expected (based on how the
    standard deviation works), and is an extra validation that the
    calculations are good.


    Notice how distances in the shot are approx. 1/4 of those in the other
    throws but SD is approx. 1/3. What this tells us is that (for those
    men in the top 90 in the world) you don't have to be that much better
    than average in the shot to be considered really good, but you have to
    work damn hard at the hammer and discus to achieve the same. This
    supports an earlier comment (from Powell I seem to recall) that the
    hammer is highly technical and depends on a lot of coaching know-how.
    Ironically, one of the reasons why I expected the hammer to be prone
    to outliers---the more technical an event, the harder it is to hit the
    perfect throw/jump/whatnot, and the greater the reward when one does.
    On second thought, this is only partially true, and the SD would
    indeed likely shrink with a greater technical component.

    I'm glad you suggested this, it's told me a lot more about the throws
    than I thought it would.
    I, in turn, am glad you went through the effort---for the very same
    reason.

    Leave a comment:


  • imaginative
    replied
    Originally posted by Cottonshirt

    So what exactly was all this magical hand-waving pseudo-BS-mathematics for?

    I too strongly doubt whether Eldrick's calculation above is helpful;
    however:

    There are common, semi-accepted approximate deltas for converting
    non-synthetic to synthetic. The 1s is unlikely to have been made up
    for this thread.

    The speed improvement is roughly 1/45 (of the original speed). You
    calculate the absolute, not the relative, speed difference.

    The 1/45 is then used to calculate a (1 1/45)^3/2 ~ 1.034 conversion
    factor.

    This is applied to the previous 55.92m to yield roughly two meters
    more---which, under the many assumptions made, would be a correct
    calculation of the ``surface part'' of the 9m over-all improvement.

    Leave a comment:


  • Cottonshirt
    replied
    Originally posted by imaginative
    Let us say that, in the language of statistics, the hammer-throw has a larger standard deviation than the shot-put (in my impression so far; have not crunched the numbers, though).
    Well now I have crunched the numbers and they don't really tell the story you assumed they would.

    I took the top 90 throwers all-time in each event, SP, HT, and DT (I didn't do the JT because they have changed the model twice and it got too complicated to allow for the differences).

    Using what I consider to be the standard approach to calculating SD (using n rather than n-1 as my divisor) I got the following numbers. The columns are, from the left, event, standard deviation, number of men within 1-SD of the mean.
    1. SP 0.50m.....65 [/*:m:1tf3xt7f]
    2. DT 1.63m.....65[/*:m:1tf3xt7f]
    3. HT 1.17m.....64[/*:m:1tf3xt7f]


    So, not only does HT not have a larger standard deviation than the other throws, but the event with the largest standard deviation does not have more men in that interval than the others.

    Notice how distances in the shot are approx. 1/4 of those in the other throws but SD is approx. 1/3. What this tells us is that (for those men in the top 90 in the world) you don't have to be that much better than average in the shot to be considered really good, but you have to work damn hard at the hammer and discus to achieve the same. This supports an earlier comment (from Powell I seem to recall) that the hammer is highly technical and depends on a lot of coaching know-how.

    I'm glad you suggested this, it's told me a lot more about the throws than I thought it would.


    Martin

    Leave a comment:


  • Powell
    replied
    Originally posted by Cottonshirt
    It would be difficult to find a more ridiculous example of pseudo-mathematics.
    I disagree. Eldrick's recent post estimating Henry Rono's potential at distances he competed in based on made-up PBs in events he never contested was more ridiculous than this.

    Leave a comment:


  • Cottonshirt
    replied
    Originally posted by eldrick
    ... using same ballpark conversion of 1s/400m improvement ... then improvement is ~ 1/45 in speed in the circle
    It would be difficult to find a more ridiculous example of pseudo-mathematics.

    You start by assuming that the speed improvement gained by the change over to concrete circles is 1s/400m.

    You then assume an initial time of 45s for 400m, and you incorporate this second assumption into a Mickey Mouse cartoon calculation that you have not shown but which somehow magically results in a speed improvement of 1/45. (no units ?)

    If you really want to calculate it then you should proceed along lines something like this:

    Formula for speed.............Speed = Distance / Time

    Initial Time per 400m.................45s

    Therefore, initial speed...............400/45 = (80/9)m/s


    After improving by 1s per 400m

    Time per 400m..........................44s

    New improved speed.....................400/44 = (100/11)m/s

    Improvement is difference between the two

    so we have............................(100/11) - (80/9) = (20/99)m/s

    Which is approximately nine times more than your (guess) figure.

    Not that it means anything, because you just pulled your 1s/400m out of thin air anyway, and that time of 45s per 400m has about as much to do with hammer throwing as I do with the Large Hadron Collider.

    You know the real funny part... having gone to all the trouble of convincing absolutely no one that you know what you're talking about, you didn't even use this nonsensical figure in anything. The big denouement to your post was the improvement in the 1950's was 9m, and you got that from mikli's figures by simply subtracting 55.92 from 64.77 = 8.85. A ten year-old child could have done that; in fact my ten year-old niece did do it!

    So what exactly was all this magical hand-waving pseudo-BS-mathematics for?


    Martin

    Leave a comment:

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