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Gerald.
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- March 22, 2010 at 16:57 #14508
The moderators will probably want to move this thread to the Systems section, but I’d like them to hold off for a week or two, just in case interested parties don’t notice this thread. I’m writing this up as I’m going along today, and I want to make sure I’m doing it the right way.
Why Ireland?
a) No snakes, just certain trainers.
b) Fewer horses, fewer races, less work.
c) Next time Darren asks me my opinion about an Irish horse, I might actually have one!
We are given these constants:-
E(0) = 2615.3 J/kg
T(E) = 18.5s = Anaerobic time constant
P(m) = 68.296 W/kg
T(p) = 15.054s = Aerobic time constant
C(R) = 3.728 J/kg/m = Cost of running
k = 0.002273Ws cubed / kg m cubed = Air resistanceT(0) = E(0)/P(m) = 2615.3/68.296 = 38.29s
Working out Power Ratings for yesterday’s meeting at the Curragh
STEP 1
Work out F(T) for the standard times at the distancesT(s) = Standard Time
Standard Time (Racing Post)
5f 59.3
6f 71
6f 71
8f 97
8f 97
7f 83.5
8f 97F(T(s)) = [T(0)/T(s)]*(1-EXP(-T(s)/T(E))) + 1 – [T(p)/T(s)]*(1-EXP(-T(s)/T(p)))
Okay, these are the figures I’ve got:-
F(T(s))
1.371 5f
1.318 6f
1.318 6f
1.238 8f
1.238 8f
1.274 7f
1.238 8fThese seem to be about right, as F(T) decreases from 1.4 to about 1.04 over race times, and the furthest race yesterday was only a mile.
STEP 2
Work out the Cost of Running the Standard Time carrying Standard weight, on standard going.I’ll do this in the next post.
March 22, 2010 at 17:39 #284891No moles either; is that bioenergetic? [what is bioenergetics?]
March 22, 2010 at 17:55 #284894STEP 2
Cost of Running the standard time, carrying standard weight, on standard going.
Equation 24, top of page 54.
C(R) = P(m)F(T(s))- K*[(D/T(s))^3] – 2*(1+ w(s)/W)[(D/T(s))^2]/T(s)
Divided by
(1 + w(s)/W)(D/T(s))D is in metres so is furlongs x 201.16
w(s) = standard weight = 9 stone = 126 lbs
W = weight of horse = 1100 lbsAnyway, these are the figures I’ve got
3.793 5f
3.681 6f
3.681 6f
3.668 8f
3.668 8f
3.645 7f
3.668 8fI’m not so sure as in Step 1 that I’ve got this right. It would help if I worked out what units these were. If it was something along the lines of Watts per kilogram, it sort of makes sense. Will go all out at 5f, so that is the one where the horse is working hardest. Interesting that 7f is easier than 6f or 8f though . . .
March 22, 2010 at 17:57 #284895I think Bioenergetics is something along the lines of
anaerobic supply + aerobic supply = cost of locomotion + air resistance + kinetic energy.
March 22, 2010 at 18:05 #284897No moles either
They’re all in Ballydoyle.
March 22, 2010 at 18:12 #284899This is a bit tedious.
I think
STEP 3
is to calculate F(T) for the actual times that were run.
March 22, 2010 at 18:24 #284900STEP 3
Calculating F(T) for each actual race time.
Use Equation 23 at the top of page 53. Alternatively, use the equation in STEP 1, replacing F(T(s)) with F(T).
Here are yesterday’s racetimes:-
66.71
80.01
78.05
109.69
108.46
95.9
112.9Hopefully, these are the correct corresponding values of F(T):-
1.335
1.285
1.292
1.211
1.213
1.240
1.205March 22, 2010 at 18:37 #284901STEP 4
Calculate the
Power Ratings
of the winners, assuming that there is no Going Allowance ( F(G) = 1 ) .
P = [1 + w/W)*F(G)*C(R)*D/T + k*[(D/T)^3] + 2*(1 + w/W)[(D/T)^2]/T ]
divided by
F(T)P is the power rating in Watts per kilogram.
w is the weight carried
W is the weight of the horse (1100 lbs).
F(G) is the Going Allowance, which for the moment will be set at 1.
C(R) is the Cost of Running, calculated in STEP 2.
D is the Distance of the race, in metres.
T is the time of the race, in seconds.
F(T) was calculated in STEP 3.Erm, I might be gone a while. In fact, drunk too much Sella, we’ll carry this on tomorrow.
March 22, 2010 at 19:31 #284917Moved it here Gerald.
March 22, 2010 at 20:32 #284945That’s a pity – Prufock, Cavelino
et al
will never find it here. I’ll just be talking to myself, and I won’t know whether I’m doing it the right way.
March 22, 2010 at 23:41 #284984Don’t worry, Gerald; as long as you’re doing it bioenergetically it’ll be fine.
March 23, 2010 at 15:23 #285078
edit:
operator missing
=[(1+((14*E2+F2)/1100))*R2*((201.16*B2)/C2)+0.002273*((201.16*B2)^3)+2*(1+(14*E2+F2)/1100)*((201.16*B2/C2)^2)/C2]/S2

Replaced [ & ] by ( )
Values for the seven races are
1731943.31
3110144.66
3094342.08
7822645.02
7807581.58
5116672.87
7860765.29Erm, a bit big. Should be in the region of 60. Forgot to divide D by T before I cubed it.
=((1+((14*E2+F2)/1100))*R2*((201.16*B2)/C2)+0.002273*((201.16*B2/C2)^3)+2*(1+(14*E2+F2)/1100)*((201.16*B2/C2)^2)/C2)/S2
Okay, these are the unadjusted Power Ratings of the winners, before a Going allowance is calculated:-
59.3811
58.7363
61.0067
59.5012
59.2287
57.6398
57.3567March 23, 2010 at 20:38 #285164I was planning to create speed ratings by assuming that the 5th in the 22 runner Lincolnshire (Race 5) had run to their form rating. However, Topspeed is indicating that they were 45 lbs or so slow in this race.
Instead, I shall assume that the 3rd race (a Handicap) was truly run, and that the 3rd horse ran a couple of pounds higher than its rating, and the fourth horse a couple of pounds lower.
I’ll keep that for tomorrow.
Before, I want to go on a little diversion.
STEP 5
Calculate the Power Ratings for a Standard Horse running in the same times as the winning times. Standard Horse carries 9 Stone?
Pretty much the same calculation as Step 4, except the weight w is replaced by 126, instead of the actual weight carried.
March 23, 2010 at 20:58 #285171Okay, these are the Power Ratings if the winner had been carrying 9 Stone:-
59.2504
59.1694
60.9179
59.0677
59.8422
57.9374
57.1459These are the differences from the Power Rating of the winner, together with the winners’ weight:-
0.1307 9-03
-0.4331 8-04
0.0888 9-02
0.4334 9-10
-0.6135 8-00
-0.2977 8-07
0.2108 9-05So it looks like a 1lb is about 0.043 of a Power Rating, well, at the Curragh on a soggy day, anyway. I just wanted to know a ballpark figure for the Power Rating scale.
March 23, 2010 at 21:24 #285179STEP 6
Getting a bit lost and confused, but what I’d like to calculate next is the Power Rating of the Standard Horse, carrying Standard Weight, running Standard Time on Standard Going.
Actually, this seems to be what the constant P(m) is.
P(m) = 68.296 W/kg
I’ll do the calculation for each race, tomorrow, just to make sure.
March 30, 2010 at 10:05 #286428Good stuff, Gerald, following with interest, although I havent bought the book yet. Awaiting your final figures with anticipation…
March 30, 2010 at 10:32 #286431Yeah, sorry Cav. I had a bad day last week when I found out that Kingsfort was out, I was supposed to do this, the form for a race on my betfair thread, plus I promised to write instructions for work on how to do my job. Was a bit much for me, and in the end I did nothing that day. (Can’t cope with stress.) Haven’t gotten back on the bicycle yet. Got the form to do for two races at Southwell tomorrow, and then I’m clear to return to this. Got next week off as holiday.

The problem comes from working out what the Going allowance should be on a particular day, especially if its the first day and you don’t have any collateral ratings!
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