Adjusted Screen

[joefour]

Hey Guys (and Gals),

A question occurred to me which I have not been able to answer for myself using the search function and going back through past postings.

On the adjusted screen, concerning Final time only, is there a time value (in tenths?) that can be assigned to the beaten lengths? For instance,
a horses running line shows on the adjusted screen a projected 47.4, 111.8, and 139.6, final beaten lengths are 9.8. How many tenths (timewise) are those 9.8 final beaten lengths worth (hypothetically)? Or to put it another way, how much is each (final time) beaten length worth (hypothetically speaking) on the adjusted screen?

[Mark]

The Segments screen Compiled Beaten lengths at 1st, 2nd calls and finish do all these calculations for you. These are done from the adjusted numbers.

[joefour]

Thanks Mark,

But what I'm looking for is the time value (in tenths of a second) of any beaten lengths shown at the Final projected time shown on the adjusted screen. Is a beaten length after the final time only in RDSS's adjusted screen worth 1/10 sec. or 1/5 sec.?

In other words, following the above example, the final projected time of the leader of the race would be 139.6, and our horse is projected to come in 2nd by 9.8 lengths. What is his projected final time, 149.4 (if final beaten lengths are worth a .1 sec) or is it 159.2 (if final beaten lengths are worth a 1/5 sec.).

[Mark]

Joe,
Ted is adjusting the hell out of these numbers. He is applying ITV and DTV and then he is applying the value of the beaten lengths.
His algorithms I am sure are in feet per second. The value of a beaten length is different for how fast the horse is travelling at the time of the race. But for simplicity lets assume a beaten length is 8ft. If the winner runs a mile in 96 seconds than his average velocity is 5280/96 or 55 feet per second.
If your horse was 9.8 lengths behind and a length is 8ft he ran 5280 -(8*9.8) or 78.4ft so in 96 seconds he traveled 5280-78.4 = 5201.6/96 = 54.18 fps average velocity. Now convert to time: 5280/54.18 = 97.45 or 1:37.45.
This is at best an approximation but workable. To answer your question 8/54.19 = .147 seconds. But the faster the average velocity the shorter time it will take to cover 8 lengths. Also the distance of the pace segment will have a hell of alot of effect on the average velocity. In the first fraction you are
working with average velocity in the high 50s and 60s. The Average velocity of a final fraction is probably 51 -52fps.

[joefour]

Thanks Mark,

I completely forgot (another senior moment) the fact that a length at say-5 furlongs is quicker than a length at a longer distance like a mile. I'll have to calculate a chart at various distances to use as a rough guide for myself.