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Effect of weight on horse speed

Yes use all times not just winners(remember to use the corrected times adjusted for the weight and the track variant as already discussed how to do this)
15th percentile down the ladder
why 15th …just experience tells me this is as good as any….no correct answer….the median I believe would include too many slow times, 15th is a sensible representation and cuts out ant outliers at the top

I thought you just wanted to line up the times, if you are attempting to create usable ratings to adjust for future races, then it’s very difficult to give full advice…I don’t have the data you have.
Im guessing we are dealing with a country I know nothing of with just one racetrack, I don’t know where you obtain your track variant, the track variant could be useless. Creating an accurate track variant is essential.
In a days results if the track variant is 2,30 seconds is this the same variant for 6 furlong races and 7 furlong races, if so that makes no sense.
Does your data contain official ratings for the horses or any indication of class of runners in each race.
When you find out that a horse has run a corrected time of 1.13.5 and your standard is 1.14
You know this horse is 1/12 sec/furlong fast. You need to multiply this by a constant that will give your horse a rating that will then line up with kg so you can adjust for future weights in future races.
Again there is no correct answer…for U.K. I ended up with a constant of 126 to convert to a pounds friendly scale.
I would suggest you use 58 as the constant.
so in our example the horse is 1/12 sec/fur fast so 1/12 x 58 so the horse is 4.83 kg fast
So we could call this horse 104.83 speed rating
Do this for every horse, then adjust for kg carried in future races
The track variant will be vital, it will be a case of if your variant is accurate your ratings will be accurate.

Well, I suppose it's got to be the same everywhere.
Most of our horses are from UK, some are bred in Greece and a few French.
But as I said different types of surface may produce different effects (dirt-grass-artificial). Ours is dirt.

The track variant is a long story.
The best track variant is the one that finds most winners really, in the sense that we use the recent time figures only to predict the
winner of a race and forget all other things we usually take into account (such as jockey, early speed, days off ...).
I tried without any track variant and certain poor results obtained.
Then I tried to work out class par times. Better but not quite.
The one they publish in our press is the one doing better.
This is not the same for all distances of course and I don't know the precise method. It's a method used by DRF in the States I 'm
told (or adapted from it) but at this time I don't know more.
So let's accept that now (despite some grumbles I hear from other people).

The issue here is to compute the weight effect, just that.
I never use ratings. To compute probabilities I use raw times and also Hill's integral:


There are some ratings BHA produced for us. I don't fancy them a lot. It helps the jockey club to group horses better than before
but I don't fancy.

I 'm trying out something now along the lines you suggest. I will return after a while
 
I did various things:

There are 5809 horse histories as I said.
The number of sorties in each horse history varies, from 1 to about 200.
For each entry we are given total time, kilos and track variant.

From each horse history I exclude the first four sorties, because the horses tend to be immature.
From what remains I take only the entries in the distance I 'm interested in.

Now for those entries in which the weight was not the same I compare pairwise.
That gives me a set of gradients δΤ/δW.

Again I exclude the pairs that are distanced by more than a year, to eliminate the aging effect.

I compute the mean gradient for each horse history and the mean of the mean gradients over all horse histories in the distance.

The δt values of course cannot be attributed to change in kilograms only.
For each horse, the horse times vary because the form is not always the same, but we hope the number crunching will win the day and give us what we want.

The distances I have are all flat and range from 5f to 15f.
I found these values (sec/kg):

5f = 0.04
5.5f = 0.04
6f = 0.06
6.5f = 0.01
7f = 0.03
8f = 0.09
9f = 0.02
12f = 0.21

15f is missing - too few readings (horse histories) to make sense (only 17).
There is also 10.5f and that too is missing because of few readings (10).

This looks like a fail. They look disorderly.

I tried the same calculation with median values.
That is I computed the mean of the median gradients of each horse history.
The median is less susceptible to outliers so perhaps it would work.
But I found little improvement.

Lastly I did this:
I restricted the pairs of comparable performances to less than six months apart.
Also they had to be consecutive in the distance as well as chronologically consecutive (so form is the same as far as possible).
This restricted the number of readings a little and the results were:

5f = 0.05 (no of readings = 837)
5.5f = -0.06 (!) (no of readings = 69)
6f = 0.07 (no of readings = 2507)
6.5f = 0.17 (no of readings = 107)
7f = 0.06 (no of readings = 2322)
8f = 0.06 (no of readings = 531)
9f = 0.22 (no of readings = 195)
12f = 0.21 (no of readings = 54)

This looks more logical though not quite satisfactory and the discrepant ones are the 5.5f and 6.5f but with few readings in each case.

The evaluation of the standard time I don't think comes into play as we have no shipped horses.
In the UK there are many, perhaps 10 out of 10 in a 10 horse race, but here no.
 
Last edited:
I repeated my calculations.
This time I compare pairs of performances with the following restrictions:

a) disregard the first three sorties of every horse
b) the pairs of performances to be compared must n't be more than one year apart

The formula we are looking for is:

corrected time = official time + track variant + k . δw

where δw is change in weight (= weight in new race - weight carried in old race) and k are the elusive coefficients.

The criterion for k is minimum variance and to minimize the variance of the readings this time I used analytical expressions
(it's a quadratic form and the minimum can be derived using calculus).

The new results:

5 furlongs = 0.03 seconds / kg
5 1/2 furlongs =0.03 seconds / kg
6 furlongs = 0.03 seconds / kg
6 1/2 furlongs = 0.05 seconds / kg
7 furlongs = 0.04 seconds / kg
8 furlongs = 0.05 seconds / kg
8 1/2 furlongs = 0.05 seconds / kg
9 furlongs = 0.05 seconds / kg
10 1/2 furlongs = 0.13 seconds / kg
12 furlongs = 0.10 seconds / kg
15 furlongs = 0.15 seconds / kg


As you can see there is only one that looks a little jumpy. The 6 1/2 furlongs.
The 10 1/2 furlongs is also jumpy but in this case there are very few readings.

The racecourse standard times are not really needed.
You were confused because in the UK the horses move about the many racecourses of the country and so one needs to make adjustments
every time. But in my case there is only one race course and all the readings are from this race course (the Markopoulo park as the Greek race course is called).
Only in one occasion do I need standard time. The 5 furlongs because we have a fast track and a slow track for 5f. So I made the
necessary conversions for this case.

But why was I getting erratic results before ?
I tried to calculate the variance for values of k = 0, 0.01, 0.02, 0.03 ... and choose the minimum.
Did I make a mistake ?
Or do the math processors make mistakes when such long sums are involved ?

In the event it does look indeed like weight is overhyped.
 
Hi TheBluesBrother TheBluesBrother, very interesting stuff. I noted your divide into 200 rule, can you possibly explain the logic behind it? And do you make allowances for changes in going? One of the reasons I've become interested in this is because sometimes I am completely confused by Topspeed and his lbs per length calculations. Sometimes the difference between his scale and that of the RPR scale can be quite marked. To give an example look at Cardano in the Copper Horse Stakes at Ascot (15 June) won by Amtiyaz. The winner recorded an RPR of 104 and TS of 96. Cardano beaten 10 and 1/2 lengths scored an RPR of 93 and TS of 75. Difference in RPR is 11 and difference in TS is 19. Therefore RPR scale is 1.3 lbs/length and the TS scale is 2.2 lbs/length. TS scale also changes from course to course and also sometimes changes for the same course but different going conditions. Using your divide into 200 rule for the Ascot race the lbs/length would be 1.1.
I realise that the scale can be fluid and related to the time taken but 2.2lbs/length over 14 furlongs seems a lot to me. Can I ask how does the Ascot race compare to your own calculations for the race and do you think that TS scale is accurate given the time and conditions of the race?

Thanks in advance
Billy
 
Nellsman Nellsman

I used to have a formula for calculating different lbs per length using the standard times for every racecourse and distance until davejb davejb pointed out to me using a constant of 200 did the same thing.

Example: Ascot 5f RP standard time is 59.40s, 200/59.40s =3.367 (3.67 lbs per length)
do you make allowances for changes in going? One of the reasons I've become interested in this is because sometimes I am completely confused by Topspeed and his lbs per length calculations.
Yes, that is why we have going allowances, I also adjust for any rail movements'
I am completely confused by Topspeed and his lbs per length calculations. Sometimes the difference between his scale and that of the RPR scale
RPR ratings are not on the same scale as the Top Speed figures, the RPR's are a private handicap dealing with class, whereas the Top speed figures deal with how fast they ran.
There is a major flaw in how the Top speed figures are calculated, Dave Edwards uses the same static lbs per length scale for each and similar distances.

Can I ask how does the Ascot race compare to your own calculations for the race and do you think that TS scale is accurate given the time and conditions of the race?
Ascot.png

Note I used two going allowances on the day.

Mike.
 
TheBluesBrother TheBluesBrother Nellsman Nellsman
I feel it's only fair to point out that Mike's 200/x formula is very much his own work, to which my contribution was solely to point out that mathematically two of the formula steps could be replaced by the 200 calculation without altering the result...I think there was something like divide something by a thousand then multiply the result by five at some final point, why Mike picked the values he did (the thousand and the five, or whatever) to do the various bits with was very much all his own work.

Dave
 
Hi TheBluesBrother TheBluesBrother, thanks for the very informative reply. I worked out the lbs/length scale that Topspeed has used in each race that day and there is a quite large discrepancy when compared to yours in each and every race. Also I note the large discrepancy in SF's in the last race (Amtiyaz) with your sf of 65 and Topspeeds being 96. The time recorded in that race was 3m1.23s Topspeed has that as fast by 2.17 giving a standard time of 3m3.4s. Whereas you have the race slow by 0.78s giving a standard time of 3m0.45s. Are these numbers correct or have I miscalculated. The difference in standard times is 2.9s which would equate to about 14 lengths of a difference. Am I way off beam here or can there really be so big a differential between ratings?

Billy
 
Nellsman Nellsman

You cannot really compare Top Speed's methods to mine, he uses WFA and I don't, on the day I had the Amitiyaz race going allowance at +0.40s/f (good to firm) and TS had it 20lb lower at +0.20s/f, the rail movements added 11yds (subtract 0.85s).

The standard time you use is constant and does not change unless I know of an error and make an adjustment.

Mike.
 
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