[AusRace] Lord's Prayer 2

[Tony Moffat]

Mr Lord was a horse breeder, a farmer in the Western District, a football
club stalwart it seemed and would rather go to VFL than the races any day,
son, he told me.

See:

http://ausrace.com/pipermail/racing_ausrace.com/2019-July/000752.html


He reasoned that a horse having run a 1600 metres, with 57 kgs, from barrier
9, and finishing at 9 lengths from the winner had earnt that 9 and it didn't
seem to matter if the 9 was say 4.25 lengths, it was the 9 that was used
subsequently.

Now it runs over 1400 metres, with 59, from barrier 3 what will it earn in
this race.

 All of the elements in the form race constitute that 4.25 - he used finish
position and lengths lost in his calculations.

(a)He took the finish position (less 1 - why, because there are six horse
lengths between 1st and 7th, between the nose of the winner and the nose of
7th there are six lengths. It didn't matter to him that there may be 2, or
12 lengths, the numerical value is what he used - perhaps capped at 9
though).

(b)He reasoned that each barrier width out from the rail cost a runner a
10th of a length - a runner in 10 was disadvantaged 1 length according to
his math - and subsequent studies have shown him to be correct. Barrier 9
lost .9 (point 9 of a length) Barrier 14 lost 1.4 lengths.

(c)He divided the length of the race by the length of the horse
(standardised). He used 9 feet. This is 2.75 metres. 1600/2.75 = 581.8

(d) Weight carried. 57 kgs/1.5 = 38 *2.75 = 104.5. This was then the
handicappers value for a length loss over a mile - 1.5. The 2.75 is the
length of the horse, standardised. So the weight is divided by 1.5 kg to
ascertain the weight loss/penalty and then this value is multiplied by 2.75
to get a final value for that runner. He was indicating that 57 kgs causes a
loss/penalty of 104.5 lengths.

All of that was summed to get a 'horse value'. 

So 9 + (barrier 3 *.10 = .3) +(race distance 1600/2.75 = 581.8) + weight
penalty 104.5 = 9+.3+581.8+104.5 = 695.6 which he reasoned was the from
value for that run.

The upcoming race was dealt with similarly (without the inclusion of the
finishing penalty - the 9 in the example above)
Barrier 3*.1 = .3 + length 1400/2.75 = 509.9 + weight (59/1.5)*2.75 = 29.5.
Sum .3+509.9+29.5 = 539.7

How he dealt with the differing values or if the differences between the two
values is an indicator of form strength he did not divulge but it is easy to
see and deal with.

Another way - perhaps

Cheers

Tony


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