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Horse Racing calcuations

  1. Feb 23, 2014 #1
    Hi folks,

    I'm new here and I am looking for some help with calculations for a horse racing article that I am writing.

    I don't specifically know how to tackle the problem; which is "Does one length* (see glossary below) equal the industry standard 0.2 seconds, and does a change in the time it takes to complete a race create any variances in the given time of one length.

    I'd specifically like to know if one length equals 0.2 seconds in comparison to races that have historically been getting faster throughout the years.

    Things to consider:

    **Length: The distance from a horses nose to its tail, which is about 8 feet (2.4384 meters). It is also distance between horses in a race.

    Extras:

    -Each race is 1 mile in length and is started at full speed (see example video: )
    -The standard race time in 1960 was 2 minutes 30 seconds for a one mile harness racing race.**
    -The standard race time in 1985 was 2 minutes 0 seconds.*
    -The standard race time in 2014 was 1 minutes 50 seconds.*
    *Estimates only.
    **Harness racing is less popular form of horse racing (eg, Secretariat, Seabiscuit, etc), but I'll use it as a basis because each race is started at full speed.**

    Let's assume that acceleration is 0m/s as horses start at full speed and continue at the same speed throughout a race. This isn't realistic, but will likely help make calculations easier.

    The understanding is that one length equals 0.2 seconds and then that assumption is used to calculate the time it takes for horses to finish each race. If the race was completed in 2 minutes 0 seconds and the 2nd place horse finished 5 lengths behind the winner, it is expected that the horse who was second finished the race in 2 minutes 1 second.

    Please let me know if you need any other information. My background is in journalism and marketing, so I'm obviously out of my league with this task--hence why I have come to Physics Forum for help!

    Contributions will be credited in all publications.

    Thanks in advance!

    -burtonk
     
    Last edited by a moderator: Sep 25, 2014
  2. jcsd
  3. Feb 23, 2014 #2
    In 1960:
    5280f / 150s = 35.2f/s
    8/35.2 = .2272727… seconds per horse length

    In 1985:
    5280f / 120s = 44 f/s
    8/44 = .18181818… seconds per horse length

    In 2014:
    5280f / 110s = 48 f/s
    8/48 = .1666666… seconds per horse length

    1 mile is 5280 ft,
    length of horse is 8 ft, so
    5280 / 8 = 660 horse lengths around the track.

    For your standard of 8 ft horse going at a rate of .2 sec per length gives
    8/.2 = 40. That’s 40 ft per second, which would be 132 seconds to go around the track (2 minutes and 12 seconds)
    5280 / 40 = 132

    I didn’t accelerate the horse from 0 speed, and as you already understand, this is average distance over time. The horses might be going slower or faster than average at the finish line, so .2 seconds per length at the finish line can mean different overall total track time, depending on the speed around the rest of the track.

    Hope that gives you enough to answer your original questions (not looking for credits). If you still have questions ask some more.
     
    Last edited: Feb 24, 2014
  4. Feb 23, 2014 #3
    Thanks Mike!

    This looks great to me.

    Much appreciated!


    Obviously using a constant speed is unrealistic, but would there be any other issues that you could foresee when drawing a conclusion to this data?
     
  5. Feb 24, 2014 #4

    Stephen Tashi

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    Science Advisor

    If you write about a subject with associations to gambling, many of your readers will know about probability and statistics. They will expect some statistical analysis.
     
  6. May 8, 2016 #5
    The horses are timed with a photo-timer, basically a strip of film that is run through the camera at the same speed as the horses are running
    (pretty near) takes a continuous picture of about an eighth of an inch wide aperture that is right on the finish line. First it takes a picture of the
    nostrils, lips, ears, neck, withers, saddle, etc. until the horse has completely gone through this "window." The film runs at the speed of the horses
    in order that the picture will fix in focus on the film, that is why any part of the horse that is temporarily stationary in relation to the camera will be
    blurred, whereas the moving horse will be in focus. The time is stamped automatically right on the film, and it is easy to figure just how long it takes
    for the complete horse (1 length) to pass through this aperture. Also, it must be noted that the speed of these horses varies dramatically in different
    parts of the races, and also from one horse to another. There are so many factors influencing the outcome of any horse race that I personally
    consider the nit-picking of hundredths of a second, and maybe even fifths to be unnecessary worry. That does not apply to fifths of bourbon.
    (dogplayer)
     
  7. May 9, 2016 #6
    Any of the figures MikeGomez calculated would come to 0.2 when rounded to the nearest tenth, so saying .2 second per horse-length is correct as long as you understand that 0.2 is not the same thing as 0.20
     
  8. May 9, 2016 #7

    A.T.

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    Gold Member

    Then the horses need longer tails and 0.2s will still be fine.
     
  9. May 9, 2016 #8

    rcgldr

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    Homework Helper

    Unlike harness racing, the speed of the more popular form of horse racing hasn't changed much in 50 years or more. The Kentucky derby times have been around 2:01 to 2:05 for the last 50 years, with only Secretariat running a 1:59.40 back in 1973. Factors include the horses are running to win, not to set records, it's a once a year event, and track conditions change from year to year. Wiki article:

    http://en.wikipedia.org/wiki/Kentucky_Derby#Winners
     
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