HOW MANY Lengths per Second - Horse Racing Forum - PaceAdvantage.Com

Robert Fischer · 03-20-2007, 02:30 PM

I have seen 4.5 (.22seconds/length) , 5(.20 s/l), and 6(.167 s/l) being used.

cj · 03-20-2007, 02:41 PM

There is no exact answer. All horses are not the same length, and when tired horses cover less ground in a second than they do when fresh.

If a horse is 9 feet long, and a 6f race is run in :22, :45, and 1:10, what is his lengths per second?

1320 feet in 22 seconds is 60 feet per second, or 6.67 lengths per second for the 9 foot long horse.

2640 feet in 45 seconds is 58.67 feet per second, or 6.52 lengths per second for the 9 foot long horse. However, at that point, he covered the 2nd 1/4 in 23 seconds, or 57.39 fps, which equates to 6.38 lengths per second for the 9 foot long horse.

3960 feet in 1:10 is 56.57 fps, or 6.29 lengths per second. Broken down by last quarter, which was 25 seconds, is 5.87 lengths per second.

Which one do you want to use?

(Of course, that is only if you believe the horse is 9 feet, which most aren't in my opinion)

Greyfox · 03-20-2007, 02:55 PM

I know certain handicappers, particularly those in the Sartin camp, want dead on accuracy. But will it matter? All horses in a computer program are going to be treated as constant . If you call them 9, 10, or 11 feet long, and the number of lengths that that travel per second 4, 5 or 6, you'll still be applying the same equation to all of the horses. Even if you absurdly assumed a horse to be 20 feet long and running 7 lengths a second, this is all going to take place in ratio and the outcome will be a number, that when converted to a percentage, will be the same as if you were dead on the money.

shanta · 03-20-2007, 03:26 PM

Quote:

Originally Posted by Greyfox

I know certain handicappers, particularly those in the Sartin camp, want dead on accuracy.

Grey,
Actually at the end Howard created formulas that brought the top 2-4 contenders closer together in "look" to allow for what he termed "wagercapping" based on VALUE instead of "handicapping" based on the best horse camp.

Richie

Robert Fischer · 03-20-2007, 03:27 PM

Quote:

Originally Posted by cj

3960 feet in 1:10 is 56.57 fps, or 6.29 lengths per second. Broken down by last quarter, which was 25 seconds, is 5.87 lengths per second.

This gives me a pretty good guess at what i want to do.

greyfox - good point about using the ratio

shanta · 03-20-2007, 03:29 PM

Quote:

Originally Posted by Greyfox

IAll horses in a computer program are going to be treated as constant . If you call them 9, 10, or 11 feet long, and the number of lengths that that travel per second 4, 5 or 6, you'll still be applying the same equation to all of the horses. Even if you absurdly assumed a horse to be 20 feet long and running 7 lengths a second, this is all going to take place in ratio and the outcome will be a number, that when converted to a percentage, will be the same as if you were dead on the money.

Sartin's last software changed the value of beaten lenghts according to distance and the particular fraction in the race.

This is how quite often a horse that was beaten by another in the previous race rated better when matched up TODAY.

Richie

Greyfox · 03-20-2007, 03:46 PM

Quote:

Originally Posted by shanta

Sartin's last software changed the value of beaten lenghts according to distance and the particular fraction in the race.

Richie, undoubtedly Sartin did that, ultimately.
I won't get into why he did that, but Dick Schmidt would know.
Nevertheless, even if you give a horse 6 lengths/ second in the first fraction,
5.4 lengths per second inthe second fraction, and 4.8 lengths per second in the third fraction, you're still performing the same operation as a constant across all horses in a given race.
If you are coming up with a result of f.p.s. and total energy based on f.ps.
there might be some very minor changes to accuracy of the number versus reality. But, simply assuming that horses were 10 feet long and travelling 5 lengths per second across all three fractions and then putting it into percentages will give you the same outcome. I know that you'll disagree but some of his earlier simpler programs were every bit as good as his "progressive" improved ones.

shanta · 03-20-2007, 03:58 PM

Quote:

Originally Posted by Greyfox

[b]Richie, I know that you'll disagree but some of his earlier simpler programs were every bit as good as his "progressive" improved ones.

I would NOT disagree with that at all Grey. I think you are on the money man!
Richie

bobphilo · 03-20-2007, 04:01 PM

It all depends on how fast the horse is running and how exact you want to be. One thing that is certain is that the traditional notions that 1 length = 10 feet and 1 length = 1/5 second are wrong. The average Thoroughbred length is closer to 8 feet. Clydesdales are 10 feet. Thoroughbreds cover more than 5 lengths per second. Such a velocity is more like trotting horse speed. Charles Carrol covers the subject in some detail in his “Handicapping Speed”.

His method of getting a precise beaten lengths - seconds conversion involves determining the velocity in FPS based for the race time, or even more precise, the fractional split, and to use this velocity in converting beaten lengths to seconds.

A “quick and dirty” method that I have found to be almost as accurate is to multiply beaten lengths by .15 to find the time behind in seconds.

Bob

ranchwest · 03-20-2007, 04:37 PM

Quote:

Originally Posted by Greyfox

Richie, undoubtedly Sartin did that, ultimately.
I won't get into why he did that, but Dick Schmidt would know.
Nevertheless, even if you give a horse 6 lengths/ second in the first fraction,
5.4 lengths per second inthe second fraction, and 4.8 lengths per second in the third fraction, you're still performing the same operation as a constant across all horses in a given race.
If you are coming up with a result of f.p.s. and total energy based on f.ps.
there might be some very minor changes to accuracy of the number versus reality. But, simply assuming that horses were 10 feet long and travelling 5 lengths per second across all three fractions and then putting it into percentages will give you the same outcome. I know that you'll disagree but some of his earlier simpler programs were every bit as good as his "progressive" improved ones.

You were right when you were talking about the feet in a length being somewhat arbitrary. One reason is that I suspect you'll get a different real value of a length in different charts. You'd have to know who made up the chart to have any idea of the value, so it just makes sense to settle on a figure, such as 9 feet. And, as you pointed out, most people are likely going to use some constant value, so it doesn't make a huge difference what the constant is. I figured up one time what seems to be a reasonable figure and used it, don't recall off the top of my head what it was. It will make some difference, though, because the lead horse is always at zero.

However, the seconds to a length definitely does vary from fraction to fraction and race to race, as CJ was explaining.

The reason Sartin's results were similar probably has more to do with pace being an inexact science (as far as exact values of split times) to begin with more than the appropriateness of methodologies in determining speed. IMHO, there's a lot more to pace handicapping than just calculating spilt times. You have to choose appropriate running lines, evaluate need to lead, etc.

pressman · 03-20-2007, 05:23 PM

The office of animal husbandry uses the following equation:
5/3 times height in inches
the standard horse being between 15-17 hands they use 16
16 hands x 4 inches (a hand)=64 inchesx5/3= 320 inches /3=107 " which is just shy of 9 feet(rounding up)

the_fat_man · 03-20-2007, 05:27 PM

Quote:

Originally Posted by Greyfox

I know certain handicappers, particularly those in the Sartin camp, want dead on accuracy. But will it matter? All horses in a computer program are going to be treated as constant . If you call them 9, 10, or 11 feet long, and the number of lengths that that travel per second 4, 5 or 6, you'll still be applying the same equation to all of the horses. Even if you absurdly assumed a horse to be 20 feet long and running 7 lengths a second, this is all going to take place in ratio and the outcome will be a number, that when converted to a percentage, will be the same as if you were dead on the money.

Exactly right --- as long as your conversions are constant, the numbers gods have no problem

bobphilo · 03-20-2007, 05:32 PM

Quote:

Originally Posted by pressman

The office of animal husbandry uses the following equation:
5/3 times height in inches
the standard horse being between 15-17 hands they use 16
16 hands x 4 inches (a hand)=64 inchesx5/3= 320 inches /3=107 " which is just shy of 9 feet(rounding up)

Thanks Pressman,

This is the best estimate not based on conjecture that I'v seen. I've been using 8 feet feet but it seems that 9 is a better estimate and I will adjust accordingly.

Bob

ranchwest · 03-20-2007, 05:32 PM

Quote:

Originally Posted by pressman

The office of animal husbandry uses the following equation:
5/3 times height in inches
the standard horse being between 15-17 hands they use 16
16 hands x 4 inches (a hand)=64 inchesx5/3= 320 inches /3=107 " which is just shy of 9 feet(rounding up)

That's good, but when you're trying to handicap I think you have to evaluate the length as it is used in practice. You have to look at enough figures to zero in on a reasonable value to use. You have to determine what value for a length makes the data work with reasonable consistency.

ranchwest · 03-20-2007, 05:35 PM

Quote:

Originally Posted by the_fat_man

Exactly right --- as long as your conversions are constant, the numbers gods have no problem

Well, as I pointed out the lead horse is always at zero. So, if you use 8 feet then the horse a length back is 8 feet back or if you use 9 feet then the same horse is 9 feet back, but the lead horse is still at the constant zero. So, it does make a difference in relation to the lead. It makes no difference among trailers (non-leaders).