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Horse Racing calcuations 
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#1
Feb2314, 09:26 PM

P: 2

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: http://www.youtube.com/watch?v=a17sohcly10) 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 taskhence why I have come to Physics Forum for help! Contributions will be credited in all publications. Thanks in advance! burtonk 


#2
Feb2314, 10:33 PM

P: 211

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. 


#3
Feb2314, 10:53 PM

P: 2

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? 


#4
Feb2414, 09:30 AM

Sci Advisor
P: 3,282

Horse Racing calcuations



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