Adjusting times

[whodoyoulike]

Originally Posted by Cratos

...
Example:

Belmont’s average winning dirt 6F = 4000.5 feet in 71.62 seconds or an average speed of 55.85 feet per second.

Belmont’s average winning dirt 1M = 5340 feet in 97.30 seconds or an average speed of 54.17 feet per second.

The data shows that a horse at Belmont running on the dirt surface runs an average of 3% faster at 6F than at the 1M distance. ...

Interesting, we've researched and calculated using different methods and from different time periods and come up with similar results.

Average winning dirt 6F = 3960 feet in 70.89 seconds or an average speed of 55.86 feet per second.

Average winning dirt 1M = 5280 feet in 97.76 seconds or an average speed of 54.01 feet per second.

Btw, I calculate your 1M = 5340 feet in 97.30 seconds or an average speed of 54.88 feet per second.

This is one of the things I look for on this forum which is comparisons to my methods as a sanity check.

[AndyC]

Originally Posted by whodoyoulike

....This is one of the things I look for on this forum which is comparisons to my methods as a sanity check.

If you are doing sanity checks based on things that happen on this forum it might be time to check yourself in.

[steveb]

Originally Posted by whodoyoulike

Interesting, we've researched and calculated using different methods and from different time periods and come up with similar results.

Average winning dirt 6F = 3960 feet in 70.89 seconds or an average speed of 55.86 feet per second.

Average winning dirt 1M = 5280 feet in 97.76 seconds or an average speed of 54.01 feet per second.

Btw, I calculate your 1M = 5340 feet in 97.30 seconds or an average speed of 54.88 feet per second.

This is one of the things I look for on this forum which is comparisons to my methods as a sanity check.

forgive me if i have misinterpreted, but i can't see what you have figured?
they just look like answers that everybody that knows basic math would get?

[traynor]

Originally Posted by Cratos

Wrong, there aren’t any assumptions made except that the Trakus data is correct.

This is basic physics and applies to any object moving across a surface in space.

What it shows is that you cannot develop relevant speed from Equibase data because by choice Equibase doesn’t publish distance traveled by the horse during the race.

Great! I thought you might be saying the values generated had some significance other than being the averaged result of a number of instances. Thank you for clarifying.

[whodoyoulike]

Originally Posted by steveb

forgive me if i have misinterpreted, but i can't see what you have figured?
they just look like answers that everybody that knows basic math would get?

I wasn't referring to the math but, rather despite using different methods and time periods we've both derived similar racing times for the two different distances which is a sanity check at least for me.

[steveb]

Originally Posted by whodoyoulike

I wasn't referring to the math but, rather despite using different methods and time periods we've both derived similar racing times for the two different distances which is a sanity check at least for me.

thanks for reply.
different people, different thoughts i guess, but yours and cratos's times are not relatively very close, as far as i can tell.

assuming cratos figures equated to 100, then yours equate to 110 and 95 respectively.
that is a big difference to me.
if 71.62 is to 70.89 then 97.3 is to 96.3(not 97.76)

i don't think the actual times matter, just the relationships.
but to me there is a huge difference between you and cratos, although i would have no idea, which, if either, are correct.

[whodoyoulike]

Originally Posted by AndyC

If you are doing sanity checks based on things that happen on this forum it might be time to check yourself in.

I know what you're getting at but, when I came up with my data and methods I really didn't have any supporting data other than that it worked most of the time. Hell, after all these years I'm still wanting to verify my results and methods with others without giving out my algorithms. When I used to get paid for analyzing problems I always built in check figures to ensure I wasn't on the wrong track (not a pun).

[whodoyoulike]

Originally Posted by steveb

thanks for reply.
different people, different thoughts i guess, but yours and cratos's times are not relatively very close, as far as i can tell.

assuming cratos figures equated to 100, then yours equate to 110 and 95 respectively.
that is a big difference to me.
if 71.62 is to 70.89 then 97.3 is to 96.3(not 97.76)

i don't think the actual times matter, just the relationships.
but to me there is a huge difference between you and cratos, although i would have no idea, which, if either, are correct.

I use ranges in my estimations so to me his numbers are in an acceptable ballpark range with mine. I've never estimated a horse to run a 8f in 95 seconds to actually expect him to run a 95 because the race's fractional paces usually determines final times.

[steveb]

Originally Posted by whodoyoulike

I use ranges in my estimations so to me his numbers are in an acceptable ballpark range with mine. I've never estimated a horse to run a 8f in 95 seconds to actually expect him to run a 95 because the race's fractional paces usually determines final times.

yes, but that is neither here nor there.
it has nothing to do with what you would expect, it is simply about relationships, nothing more.
i can cop that they are fine and in an acceptable range to you.
to me(and accepting cratos was right) then i would be investigating what to me is huge disparity.

[whodoyoulike]

Originally Posted by steveb

... it is simply about relationships, nothing more.
...

I think I'm in agreement with you if we are discussing the thread's topic. The time adjustments of horses changing different distances is all relative to their prior paces in the different distances.

[ebcorde]

The solutions I;ve read are measuring the Horse against the avg time by other Horses at the Track.

why not calculate the Horse's slope from his longest race(s)?

was that too easy for ya?

[steveb]

Originally Posted by whodoyoulike

I think I'm in agreement with you if we are discussing the thread's topic. The time adjustments of horses changing different distances is all relative to their prior paces in the different distances.

i think i should have kept my thoughts to myself about you and cratos's time similarity!

as far as the thread topic is concerned, although it is very very easy to equate times/fractions at different distances, it is not very smart to think that a horse can do the same at either distance.
all i would be doing is trying to find the optimal way to pace a race over this distance at this track.

optimal for ME, is when the horse runs the same number for every fraction and overall race.

but just because a horse can run 6 furlongs optimally, it's unlikely that it could do the same over 7 or 8 furlongs.

if i had numbers of 60, 60, 60, 60(or whatever the same) for fractions and final time over 6 furlongs, then there is no way i would expect the horse to run those same numbers over any other distance.
despite the fact that 60,60,60,60 would mean the same thing regardless of the distance.

and of course optimal rarely happens.

[whodoyoulike]

Originally Posted by steveb

...
but just because a horse can run 6 furlongs optimally, it's unlikely that it could do the same over 7 or 8 furlongs.

if i had numbers of 60, 60, 60, 60(or whatever the same) for fractions and final time over 6 furlongs, then there is no way i would expect the horse to run those same numbers over any other distance.
despite the fact that 60,60,60,60 would mean the same thing regardless of the distance.

and of course optimal rarely happens.

Again, I think I'm stating something similar to yours. I didn't mean that just because a horse ran 6f in 70.89, he would or should run 8f in 97.76 seconds.

Which is the reason I wrote:

... The time adjustments of horses changing different distances is all relative to their prior paces in the different distances.

[Cratos]

Originally Posted by ebcorde

The solutions I;ve read are measuring the Horse against the avg time by other Horses at the Track.

why not calculate the Horse's slope from his longest race(s)?

was that too easy for ya?

You are correct when you speak of an individual horse, but in the aggregate it is time relationships.

[Cratos]

Originally Posted by whodoyoulike

Interesting, we've researched and calculated using different methods and from different time periods and come up with similar results.

Average winning dirt 6F = 3960 feet in 70.89 seconds or an average speed of 55.86 feet per second.

Average winning dirt 1M = 5280 feet in 97.76 seconds or an average speed of 54.01 feet per second.

Btw, I calculate your 1M = 5340 feet in 97.30 seconds or an average speed of 54.88 feet per second.

This is one of the things I look for on this forum which is comparisons to my methods as a sanity check.

There is a quite argument in physics that deceleration doesn’t exist; it states that objects in motion are always in acceleration relative to their stating point, just at a slower rate of motion.

I don’t know if that assertion is true or false, but what I do know and understand is that an object slowing down would’ve gone faster at its previous point of travel than its current point.

Therefore given the 6F example of 4000.5 feet traveled in 71.62 seconds at 55.85 FT/sec, it will adjust back up the race curve to 70.15 seconds at 3960 feet, the exact 6F length.

Consequently, the 1M example of 5340 feet traveled in 97.30 seconds at 54.17 FT/sec, it will adjust back up the race curve to 96.38 seconds at 5280 feet, the exact 1M length.

The value of these adjustments is to bring all of the horse’s past races under consideration for today’s race to the same time distance which should be the distance of today’s race; this should be done before any other adjustments are applied.