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Steward Graded Stakes ![]() |
Its been a while since I looked at those formulas (there are three of them on display for comparison).
I'm not sure I like your formula though, but I forget exactly why. I think its because it over biases the E1-ness of the E2 figure. Looks acceptable for 6 furlong races, but what about an 8 furlong race: (98 * 6) - (92 * 4) / (6-4) (588 - 368) / 2 220/2 110 "e1-e2", or "turn time" 110 seems a bit high. -N |
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Graded Stakes |
I must have missed the original thread on this. Can someone point me towards it?
Thanks, Bill |
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Steward Graded Stakes ![]() |
quote: There may be a thread or two lingering around here somewhere - you can try the Forum's search facility. The info that Secretariat refers to can be found here: http://www.thorotech.com/freestuff/turntime.html -N |
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Graded Stakes |
Thanks Nathan
Bill |
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| <Secretariat>
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Interesting point Nathan in dealing with Pace figures rather than feet per second real times.
However, it seems that if a horse chugs along at a 92 pace for four furlongs, and picked it up to average a 98 for six furlongs, his turn time may very well have been 110. I understand a 100 looks like a giant move (which it is). Without being able to convert the Pace Figures to real time it is difficult to know exactly. I would be interested from mathematical genuises out there though why my suggested formula is not correct. |
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Steward Graded Stakes ![]() |
I'll have to revisit this when I can focus better.
Is your 6 furlong formula more simply: E1E2 = E2 + (E2-E1) In the example this would yield: 104 = 98 + (98-92) 104 = 98 + 6 To put it another way, the E1E2 would be calculated by subtracting E1 from E2, and then adding that back to E2. This is what BRIS recommends for determing "turn time" (though they don't say to add it back to E2). I find it acceptable in sprints, its the routes where E1 is four furlongs and E2 is six that bug me. -N |
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Graded Stakes![]() |
Secretariat and Nathan,
Looking at the your methods of calculating turn time both seem reasonable, but I still have a problem with the approach. Let me try to explain with an example. Looking at the first race from Ap on tomorrow's card, there is a match-up of two horses running in a race on June 4th (Fleet Boss & Biblical). Fleet Boss E1(80) E2(84) BL1 7.3 BL2 5.5 gained 1.8 lengths Biblical E1(86) E2(88) BL1 4.3 BL2 3.5 gained 0.8 lengths Clearly, Fleet Boss should have a higher turn time, since he gained more lengths between the 1st and 2nd call in the same race. But using the calculation just the opposite appears true: Fleet Boss 88 = 84 + (84-80) Biblical 90 = 88 + (88-86) Does this make sense, or am I missing something? Bill |
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| <Secretariat>
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I'm assuming the first call was 4F and second call 6F
Using my formula in above thread. Fleet Boss ((84*6)-(80*4))/(6-4)= 92 Bilbical ((88*6)-(86*4))/(6-4)=92 Equal turn time. A larger gain in lengths does not always mean faster turn time. For example, suppose horse A showed these times: 46-110 or a 24 turn time in actual speed Another horse B showed a 47-113 or a 26 in actual speed. Let's say horse B gained 5 lengths and horse A gained no lengths. Horse' B's turn time is now 25 and horse A's is still 24. In other words even though Horse B gained more lengths his turn time is slower. In Sartin/Brohamer turn time is simply second call-first call adjusted for lengths gained or lost during the calls. What makes it difficult is that this is based on actual running times or feet per second, and not Bris Pace Times. If Bris supplied a way of converting their Pace Figures to running speeds we could more accurately solve the issue. Unfortunately, they don't and most likely never will. It's a interesting subject though as Turn Time is according to Brohamer a very useful handicapping tool (the hidden fraction), and hence accurately computing it can be profitable in some situations. As you can see Nathan's formula is a bit different than mine as I had the two equal in TT. But since we're mixing apples and oranges here I'm content to disagree. I guess here is my viewpoint. Furlong 1234 Furlong 123456 80 84 Arithmatically, how much faster does a horse have to run in two furlongs to increase his pace speed by four points. My answer is a 92 ((E2*# of furlongs)-(E1*#of furlongs))/(E2 furlongs-E1 furlongs) Oh well. Was just interested in others comments. Thanks |
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Graded Stakes![]() |
You are correct the race was a route and the first call was 4f and the second 6f. And I totally agree that if Bris pace figures were better correlated to each other (and that includes all the calls even their late pace figures), your method of calculating turn time makes sense.
But because of concerns I have had regarding Bris figures for some time, it is apparent that looking at the race I mentioned certainly reflects inaccuracy in using their pace figures to calculate turntime. We are in full agreement with regard to beaten lengths, but like you mentioned looking at horses from the same race is totally different than different races with totally different internal fractions. I am sure you would agree, Fleet Boss should have a better turntime than Biblical based on the beaten length difference in the same race. Like you I enjoy sharing ideas and discussing how we can improve upon what we have. Thanks, Bill |
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