Is it possible to estimate how much further Brother Derek ran than Barbaro?Someone sent me an email saying that they said 3-400 yards on TVG--Jett
As it says in the introductory materials on this site, a path on a turn is a length at the finish. 300 yards is nonsense.
That sounded like an awful lot.Maybe the person who sent the email got the info wrong.I\'m not sure exactly what paths Brother Derek and Barbaro took around the turns,but probably between 6 and 8 lengths difference.--Jett
Jett,
300 yards is more like 90 lenghts.
Mike
TGJB Wrote:
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> As it says in the introductory materials on this
> site, a path on a turn is a length at the finish.
> 300 yards is nonsense.
So the advantage of a path would be about eight feet... it that your distance approximation?
EZ-- you probably think that\'s a simple question, right? You forget what industry we\'re in.
As I\'ve posted before, a length in this business is a measure of time, not distance, and there is no official designation of how much time. The different teletimer companies use different formulas, and won\'t reveal what they are, although Equibase assures me they are close.
We eyeballed it a long time ago (before we knew how complicated the question is) as roughly 10 feet, and a path as somewhere around 3 feet or a little more, and using pi times radius, worked out it was approximately a length. (I actually came up with that way back at Ragozin\'s, and found out he was using almost exactly that, something like .96). As a practical matter, the ground info is limited in accuracy (trackman don\'t tell you the horse was in the 2.9 path), so it\'s never going to be super accurate, but it\'s pretty close.
TGJB Wrote:
-------------------------------------------------------
> EZ-- you probably think that\'s a simple question,
> right? You forget what industry we\'re in.
>
> As I\'ve posted before, a length in this business
> is a measure of time, not distance, and there is
> no official designation of how much time. The
> different teletimer companies use different
> formulas, and won\'t reveal what they are, although
> Equibase assures me they are close.
>
> We eyeballed it a long time ago (before we knew
> how complicated the question is) as roughly 10
> feet, and a path as somewhere around 3 feet or a
> little more, and using pi times diameter, worked
> out it was approximately a length. (I actually
> came up with that way back at Ragozin\'s, and found
> out he was using almost exactly that, something
> like .96). As a practical matter, the ground info
> is limited in accuracy (trackman don\'t tell you
> the horse was in the 2.9 path), so it\'s never
> going to be super accurate, but it\'s pretty close.
I was thinking more in aboslute terms with respect to the TVG Brother Derek & Barbaro question... we had a similar problem at work today figuring something out for the path of a mobile scientific instrument and related it to the distance a horse travels around the track (really, an ellipse). Assuming the path along rail is exactly one mile, like CD, how much farther in feet did a horse 2w 2w run than the horse 1w 1w.
It is in fact, a very complex problem as measuring the perimeter or cirumference of an ellipse (the shape the horse is tracing as it runs) is a high level calculus problem if you want an exact answer. For approximations, you\'d need to measure the two radii from the center of the ellipse. This can be difficult to figure for racetracks, as we all know, not all one mile tracks are created equal.... this is easy to show by the distance of the homestretch.
The approximation equation is pretty straightforward, but not sure how you\'d measure a and b
P ~ pi*(2(a2+b2) - (a-b)2/2)^1/2
I\'ll play with Google Earth and some GIS tomorrow to get the Churchill Downs info and maybe we can get an approximate answer for the difference in distance run by BD than Barbaro.
EZ-- most racetracks list the length of turns and stretch. But seems to me that the difference in distance is always going to be pi times the additional radius (distance out from the rail), or diameter around 2 turns,no?
Leaving office now.
Rubin Boxer\'s \"Engineering Analysis Of Thoroughbred Racing\" explains in detail how to compute the extra distance traveled by horses running outside the rail position. To compute the extra distance a horse runs from the outside post:
Multiply the distance, in feet, to the turn from the starting gate by itself.
Multiply the distance, in feet, from the outside post to the rail by itself.
Add the two together and take the square root of the result.
Example:
4f. to the turn = 2640\' x 2640 = 6969600
75\' to the rail = 75\' x 75 = 5625
6969600 + 5625 = 6975225
SqRt of 6975225 = 2641.07\'
Position from the rail around a turn
1 = .94 lengths
2 = 1.88 lengths
3 = 2.83 lengths
4 = 3.77 lengths
5 = 4.71 lengths