Calling Gareth (and others)

[Desert Orchid]

Re speed figures...

I wasn't sure which thread it was on but I found the articles I'd written for the Udate all those years ago (only 12 or 13, as it turns out!)

I quote one paragraph:

...I use my own version of the formula used by the Raceform team. The assumption [Raceform's] is that if a length equals 3lbs at five furlongs, and a horse runs at an average rate of six lengths per second at that distance, then one second equals 18lbs. For one second to equal one pound, a race would need to 90 furlongs long. To find the number of pounds per second (not per length) for any race distance, divide 90 by the number of furlongs in the race.

The square-bracketed insert is my own today. I used the info I'd got from Raceform (it might have been published in the Handicap/Update at some point prior to this if it isn't in DD's book).

I still divide 90 by the race distance then by 5 to get the number of pounds per length. I tend to round the lbs/length up to 4 at 5f when working out form ratings in close finishes but I use cumulative beaten distances as I go further back through the field. I find this more accurate.

I also tend to reduce it to 3lbs and sometimes two - occasionally less - the softer the ground seems, according to other figures.

Edit: Just noticed aother paragraph:

David Dickinson kindly wrote me a detailed explanation of how the new Speed Figures were calculated and I was impressed enough to simply let Raceform's computer deal with that side of things.

So it would have been a letter DD had sent in response to something I'd enquired about.

 

[Guest_]

Phew....when the thread title I got worried again!

 

[Desert Orchid]

Had you going, though, didn't I

 

[Harbinger]

Thanks DO. I'd be very interested to read the full set of articles, but if you don't have them in digital format don't worry (transcribing ain't much fun!).

I'll try to dig that David Dickinson book out later.

 

[EC1]

I hope its not that David Dickinson..that orange bloke :blink:

 

[Dices]

Re speed figures...

I wasn't sure which thread it was on but I found the articles I'd written for the Udate all those years ago (only 12 or 13, as it turns out!)

Hi everyone, how can i find the way to calculate "speed figures" like you are doing?
Your posts about the subject are difficult to follow, are there anywhere a basic guide posted by anyone of you?
Thanks

 

[Harbinger]

Right, the rationale from David Dickinson's "How To Compile Your Own Handicap" is:

A 5f race is 3300 feet long. The average weight of a horse and jockey combined is 1100lbs. So, over 5f, 1lb is equivalent to 3 feet. The average length of a horse is 9ft. So, over 5f, 3lb is equivalent to 1 length. From this, the following formula is derived:

Furlongs in race X Pounds per length = 15

or

Pounds per length = 15 / Furlongs in race


I've got a couple of problems with this.

First problem I have is the average weight quoted. In Hong Kong, where the weight of the horse is declared two days before the race, the average weight of the horse without the jockey is around 1100lbs.

I'm also not sure about the average length of a galloping horse being 9ft. I'm sure I've seen this challeneged somewhere (one of Mordin's books, possibly?).

If we correct the above for weight (assuming 9st / 126lbs), we get:

1226 lbs = 3300 ft
=> 1 lb = 2.69 ft,
=> 0.37 lb = 1 ft
=> 3.34 lbs = 9 ft = 1 length

Which would suggest a formula closer to:

Pounds per length = 17 / Furlongs in race

If the average length of a horse is closer to 10ft, then we would get something a bit over the formula:

Pounds per length = 18 / Furlongs in race

 

[Desert Orchid]

That's spooky.

This is my original text, written before DD published that book.

I came to the conclusion that I was going to have to work things out for myself. With the help of a pocket calculator I came up with the following:

Assuming a horse is nine feet long, there are 367 lengths in 5 furlongs.

Assuming an average flat horse weighs about 1000lbs (based on general reading in the racing papers) and carries 126lbs (nine stones), if we divide 1126 (total weight to be moved) by 367 (the number of lengths in the distance over which the weight has to be moved) the answer is 3.07. Even allowing for different weights of horses and minor tweaks for the actual measurement of a length, this is pretty close to the long-held view of the Jockey club about the relationship between the factors.

In the same way we can work out the value in pounds (to the nearest decimal point) of one length at these distances:

6f - 2.6; 7f - 2.2; 8f - 1.9; 10f - 1.5; 12f - 1.3; 16f - 1.0

Hmmm.