I think I can illustrate just how true that is by putting some numbers to it.
Watching replays of yesterday's race I noticed that while Arrogate was visibly sent from the gate by Mike Smith - both he and California Chrome were within a length of each other position-wise through all of the clubhouse turn, the length of the backstretch, and they they were still almost on even terms up until about the mid point of the far turn.
The chart confirms this.
But the overhead or aerial replay shows that Arrogate raced against the rail through the first turn, for the length of the backstretch, and for about the first half of the second turn - before Mike Smith tipped him out so he could make his move.
The same aerial replay shows that California Chrome raced about five wide on the first turn and about four wide on the second turn.
So if Arrogate raced against the rail up through the mid point of the second turn where Mike Smith tipped him out so he could make his move - and if California Chrome raced 5 wide on the first turn and 4 wide up through the mid point of the second turn:
I have California Chrome traveling about 69.47 feet further than Arrogate did up to that point in the race.
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Racing closest to the rail while on a straight-away or racing wide while on a straight-away has minimal impact on distance traveled. (So I'll ignore that in this post for the sake of simplicity.)
But racing wide on a turn DOES have an impact on distance traveled.
And this is something that can be calculated.
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A racetrack (Gulfstream Park included) is nothing more than two straight-aways (the backstretch and the stretch) connected by two half circles (the clubhouse turn and the far turn.)
The formula for the circumference of a circle is: 2 pi r
Where pi (
Google it ) is approximately 3.1416 and r is the radius of the circle
If each turn is one half of a circle, then the distance traveled by a horse while navigating one turn on a race track is 1 pi r.
My guesstimate in this post for the radius of the Gulfstream Park dirt course turns is about 545 feet. (The actual radius might be different than that and if someone out there has that info please chime in and post it.)
My guesstimate in this post for the width of a horse (or the width of each path taken up by a horse as it navigates a turn) is about 3.25 feet. (Note that this is a guesstimate and if someone out there has a more accurate number than the 3.25 feet I am using in this post feel free to chime in an post it.)
Each path that a horse races wide on a turn adds about 3.25 feet to the radius of the turn.
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Based on the above, Arrogate traveled about 1712.17 feet while navigating the clubhouse turn - calculated as follows:
dist = pi times r
or
1712.17 = (3.1416) x (545)
At the same time California Chrome traveled about 1763.22 feet while navigating the clubhouse turn - calculated as follows:
dist = pi times r
or
dist = (3.1416) x (545 + (5 x 3.25))
or
dist = (3.1416) x (545 + (16.25))
or
1763.22 = (3.1416) x (561.25)
Note that the above calculation uses a turn radius of 561.25 feet instead of 545 feet (to adjust for California Chrome being 5 paths wide.)
Based on the above, I have California Chrome traveling about 51.05 feet further than Arrogate while navigating the clubhouse turn.
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Now let's do the first half of the far turn (that's about when Mike Smith tipped Arrogate off the rail to make his move.)
Since we are only talking about half a turn here, we can use the following formula to get distance traveled:
dist = (pi times r) divided by 2
For Arrogate that works out to 856.09 feet calculated as follows:
dist = (3.1416 x 545) / 2
or
858.09 = (3.1416 x 545) / 2
For California Chrome that works out to 876.51 feet calculated as follows:
dist = (pi times r) divided by 2
or
dist = (3.1416) x (545 + (4 x 3.25)) / 2
or
876.51 = (3.1416) x (545 + (13)) / 2
Based on the above, I have California Chrome traveling about 69.47 feet further than Arrogate did up until about the mid point of the far turn where Mike Smith tipped Arrogate out so he could make his move.
How significant is that?
I think that's VERY significant.
Both horses were within a length of each other at that point.
But California Chrome was forced to expend extra energy (relative to Arrogate) because he was forced 5 wide on the first turn and 4 wide on the second turn travelling 69.47 feet further than Arrogate did up to that point in the race.
Ignoring for second my belief system which says that racing wide on the first turn while being asked for speed is one of the more costly things energy-wise a rider can do to a horse...
I contend the following:
When Mike Smith tipped Arrogate out to make his move he was sitting on a fresher horse (one with far more in the tank) than the horse Victor Espinoza was sitting on - in large part because Espinoza and California Chrome were forced to travel 69.47 feet further than Smith and Arrogate.
If both horses were at the top of their respective games - and if both horses were about equal in ability going into the race - then at the point in the race where Mike Smith tipped Arrogate off the rail to make his move - because of the energy exerted by both horses up to that point in the race:
Mike Smith's horse should pull away from Victor Espinoza's horse.
I also contend that Gulfstream stats for the far outside posts at the 9 furlong distance are what they are for a reason.
-jp
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