Turn Deceleration

[Cratos]

Sometime ago there was a thread started about Randy Moss of the DRF quest to determine why and how horses slow down because of the turns in the racetrack layout.

I believe I have the answer and the correct approach to calculating turn deceleration and I have done it for the current meet at Saratoga.
The formulation is universal, but it must be adjusted for turn radius, banking angle, and surface.

The assumptions on the model at Saratoga’s 1 1/8 mile Main track are as follows:

• Average Saratoga horse weight is 1,084 pounds
• Average weight toted by the Saratoga horse is 119.62 pounds (average weight from Belmont Summer Meet)
• Coefficient of kinetic friction at Saratoga’s Main track is 0.9775304
• Saratoga’s Main track turn radius at tangent to the rail is 468.23 feet
• Distance around Saratoga’s Main track turn at tangent to the rail is 1,473 feet
• Each path width on Saratoga’s Main track turn is approximately 4 feet
• At Saratoga Main track, the closest running radius is 470.23 feet
• At Saratoga Main track, the shortest running turn arc is 1,477.27 feet
• At Saratoga Main track, g in path 1 range from 4.22 (farthest out) to 8.40 (the most inner)
• At 12.468 feet/one fifth second the horse’s side force in path 1 is 8461.90 Lbs.

Therefore two horses entering the turn at 12.578 per one-fifth second, the horse in path 1 at Saratoga on the Main track will decelerate to 12.428 feet per one-fifth second while the horse running in path 2 would decelerate to 12.432.

The decrease in g and the larger radius will allow the horse in path 2 to run a faster rate, but it will have a longer distance to travel (1,477.27 feet vs. 1,496.12 feet).

Converting this to speed around the turn for each horse we get 1,477.27/12.428 = 23.8 seconds whereas the horse in path 2 time would be 1.496.12/12.432 = 24.1 seconds.

The difference between the two, 24.1 – 23.8 = .30 seconds or 1.77 lengths at .167 seconds per length.

This can easily be proven by taking data from the result charts and comparing the actual data against the calculated data.

How helpful is the calculation? Its merit comes in two forms; one is that the so called “track variant” which is calculated by some by just using final race times is bogus. The reason being is that a horse final time will change based on turn position even if the surface is perfect? Secondly, the calculation is very helpful when a horse which went wide in its last outing and an assessment is now being made of what time it might run in today’s race.

Comments, please.

[Saratoga_Mike]

Sometime ago there was a thread started about Randy Moss of the DRF quest to determine why and how horses slow down because of the turns in the racetrack layout.

I believe I have the answer and the correct approach to calculating turn deceleration and I have done it for the current meet at Saratoga.
The formulation is universal, but it must be adjusted for turn radius, banking angle, and surface.

The assumptions on the model at Saratoga’s 1 1/8 mile Main track are as follows:

• Average Saratoga horse weight is 1,084 pounds
• Average weight toted by the Saratoga horse is 119.62 pounds (average weight from Belmont Summer Meet)
• Coefficient of kinetic friction at Saratoga’s Main track is 0.9775304
• Saratoga’s Main track turn radius at tangent to the rail is 468.23 feet
• Distance around Saratoga’s Main track turn at tangent to the rail is 1,473 feet
• Each path width on Saratoga’s Main track turn is approximately 4 feet
• At Saratoga Main track, the closest running radius is 470.23 feet
• At Saratoga Main track, the shortest running turn arc is 1,477.27 feet
• At Saratoga Main track, g in path 1 range from 4.22 (farthest out) to 8.40 (the most inner)
• At 12.468 feet/one fifth second the horse’s side force in path 1 is 8461.90 Lbs.

Therefore two horses entering the turn at 12.578 per one-fifth second, the horse in path 1 at Saratoga on the Main track will decelerate to 12.428 feet per one-fifth second while the horse running in path 2 would decelerate to 12.432.

The decrease in g and the larger radius will allow the horse in path 2 to run a faster rate, but it will have a longer distance to travel (1,477.27 feet vs. 1,496.12 feet).

Converting this to speed around the turn for each horse we get 1,477.27/12.428 = 23.8 seconds whereas the horse in path 2 time would be 1.496.12/12.432 = 24.1 seconds.

The difference between the two, 24.1 – 23.8 = .30 seconds or 1.77 lengths at .167 seconds per length.

This can easily be proven by taking data from the result charts and comparing the actual data against the calculated data.

How helpful is the calculation? Its merit comes in two forms; one is that the so called “track variant” which is calculated by some by just using final race times is bogus. The reason being is that a horse final time will change based on turn position even if the surface is perfect? Secondly, the calculation is very helpful when a horse which went wide in its last outing and an assessment is now being made of what time it might run in today’s race.

Comments, please.

Doesn't Ragozin & Co already try to adjust for this very factor (among others)? In any case, I like your quantitative approach.

[gm10]

Sometime ago there was a thread started about Randy Moss of the DRF quest to determine why and how horses slow down because of the turns in the racetrack layout.

I believe I have the answer and the correct approach to calculating turn deceleration and I have done it for the current meet at Saratoga.
The formulation is universal, but it must be adjusted for turn radius, banking angle, and surface.

The assumptions on the model at Saratoga’s 1 1/8 mile Main track are as follows:

• Average Saratoga horse weight is 1,084 pounds
• Average weight toted by the Saratoga horse is 119.62 pounds (average weight from Belmont Summer Meet)
• Coefficient of kinetic friction at Saratoga’s Main track is 0.9775304
• Saratoga’s Main track turn radius at tangent to the rail is 468.23 feet
• Distance around Saratoga’s Main track turn at tangent to the rail is 1,473 feet
• Each path width on Saratoga’s Main track turn is approximately 4 feet
• At Saratoga Main track, the closest running radius is 470.23 feet
• At Saratoga Main track, the shortest running turn arc is 1,477.27 feet
• At Saratoga Main track, g in path 1 range from 4.22 (farthest out) to 8.40 (the most inner)
• At 12.468 feet/one fifth second the horse’s side force in path 1 is 8461.90 Lbs.

Therefore two horses entering the turn at 12.578 per one-fifth second, the horse in path 1 at Saratoga on the Main track will decelerate to 12.428 feet per one-fifth second while the horse running in path 2 would decelerate to 12.432.

The decrease in g and the larger radius will allow the horse in path 2 to run a faster rate, but it will have a longer distance to travel (1,477.27 feet vs. 1,496.12 feet).

Converting this to speed around the turn for each horse we get 1,477.27/12.428 = 23.8 seconds whereas the horse in path 2 time would be 1.496.12/12.432 = 24.1 seconds.

The difference between the two, 24.1 – 23.8 = .30 seconds or 1.77 lengths at .167 seconds per length.

This can easily be proven by taking data from the result charts and comparing the actual data against the calculated data.

How helpful is the calculation? Its merit comes in two forms; one is that the so called “track variant” which is calculated by some by just using final race times is bogus. The reason being is that a horse final time will change based on turn position even if the surface is perfect? Secondly, the calculation is very helpful when a horse which went wide in its last outing and an assessment is now being made of what time it might run in today’s race.

Comments, please.

What does this mean? Does it have to do with banking?

• At Saratoga Main track, g in path 1 range from 4.22 (farthest out) to 8.40 (the most inner)

[michiken]

Are your calculations based on both horses running at constant velocity (no acceleration) when entering the turn? If both horses enter the turn running at the exact same velocity, then your idea may have merit.

If your premise is correct, then how do some horses manage to put in higher turn time fractions?

Personally, I think they decelerate because of logistics - they have to angle their legs and bodies to negotiate the turn. Have you ever tried to steer your car around a corner at the same speed you drive on the straight roads?

Can you be a good engineer and document the formulas you used?

Ken

[Cratos]

Yes, the banking angle plays a part, but The weight can be made more precise with the gravitational force, g. g is a determinant based on velocity squared.

[illinoisbred]

Am I following you correctly-would the chord measurement be 936.46?

[gm10]

Yes, the banking angle plays a part, but The weight can be made more precise with the gravitational force, g. g is a determinant based on velocity squared.

this is probably a matter of terminology but I am a bit confused
is centripetal force the same as what you call side force?

also, how do you know the average weight? as someone who understands physics much better than I do, you will know that the weight of the horse has a huge impact on the effect that the jockey and weight carried have
I am not disputing your average (although it seems a bit low)

finally, I don't agree with your DTV conclusion
these things come out in the wash - any projection or par method will not be relying on the winner being situated at the rail or X wide

[Cratos]

Are your calculations based on both horses running at constant velocity (no acceleration) when entering the turn? If both horses enter the turn running at the exact same velocity, then your idea may have merit.

If your premise is correct, then how do some horses manage to put in higher turn time fractions?

Personally, I think they decelerate because of logistics - they have to angle their legs and bodies to negotiate the turn. Have you ever tried to steer your car around a corner at the same speed you drive on the straight roads?

Can you be a good engineer and document the formulas you used?

Ken
Correct they must be entering the turn at constant velocity (no acceleration). I don’t disagree with your assumption of logistics, but I don’t know how to quantify the logistics of a race horse.

These are standard formulas for rate of turn and turn radius which I used and can be found in many physics textbooks. However you will either agree or disagree with the assumptions I made to derive the determinants.

[Cratos]

Am I following you correctly-would the chord measurement be 936.46?

I am not understanding your chord measurement of 936.46

[illinoisbred]

I am not understanding your chord measurement of 936.46
I guess I'm asking what the is distance from rail to rail at the 2 tangental points(beginning of turn/end of turn).

[michiken]

Another way to explain my loose term of the word 'logistics' would be:

When horses run in a straight line, their acceleration is linear => a = dv/dt.

As the horse goes around the turn, their acceleration becomes angular (http://www.tutor4physics.com/motioncircular.htm) => a = v2 / r.

Did your calcs consider the angular acceleration or are you calculating centripital force?

Illinois,

The chord you are describing would just be the diameter assuming a 180 degree turn. That diameter would increase as the distance away from the rail increases.

[Cratos]

Another way to explain my loose term of the word 'logistics' would be:

When horses run in a straight line, their acceleration is linear => a = dv/dt.

As the horse goes around the turn, their acceleration becomes angular (http://www.tutor4physics.com/motioncircular.htm) => a = v2 / r.

Did your calcs consider the angular acceleration or are you calculating centripital force?

Illinois,

The chord you are describing would just be the diameter assuming a 180 degree turn. That diameter would increase as the distance away from the rail increases.

I like your way of thinking and for a horse traveling at a speed v around a turn of radius r, the side force is mv2/r; and the frictional force available to prevent sliding is μm where µ is the coefficient of friction and g the gravitational force.

Therefore v2 = µgr

[Cardus]

I would think that the length of a horse's stride would figure into its ability to declerate and accelerate into and out of turns.

[Cratos]

this is probably a matter of terminology but I am a bit confused
is centripetal force the same as what you call side force?

also, how do you know the average weight? as someone who understands physics much better than I do, you will know that the weight of the horse has a huge impact on the effect that the jockey and weight carried have
I am not disputing your average (although it seems a bit low)

finally, I don't agree with your DTV conclusion
these things come out in the wash - any projection or par method will not be relying on the winner being situated at the rail or X wide

GM10, I apologize for not answering your post in a more timely manner, but Saratoga is time consuming and (fun).

However the first part of your post was answered in my reply in post #12, but the second part of your post where you disagreed with me about DTV I will say to you that if 2 hundredth of a second is the smallest winning time margin as estimated by the DRF, then a time margin of 3 tenths of a second cannot “come out in the wash” because statistically it becomes bias.

Another point that I would like to make is that there are far too many people making “figures” (and I am not speaking of you or anyone on this forum) who do not have the mathematical or statistical expertise to do so.

[Cratos]

I would think that the length of a horse's stride would figure into its ability to declerate and accelerate into and out of turns.

I believe that "Michiken" addressed those issues in post #4 of this thread.

[gm10]

GM10, I apologize for not answering your post in a more timely manner, but Saratoga is time consuming and (fun).

However the first part of your post was answered in my reply in post #12, but the second part of your post where you disagreed with me about DTV I will say to you that if 2 hundredth of a second is the smallest winning time margin as estimated by the DRF, then a time margin of 3 tenths of a second cannot “come out in the wash” because statistically it becomes bias.

Another point that I would like to make is that there are far too many people making “figures” (and I am not speaking of you or anyone on this forum) who do not have the mathematical or statistical expertise to do so.

I have thought about it since I posted last time and you are right.
On most occasions, it will actually not matter, because today's winners will be in 'random' lanes. Today's winners will be a representative sample of the full population of winners (from which I ultimately derive par times and DTV).

However, on days where there is a positive or negative rail bias, the DTV will be very wrong, for the reason that you mentioned.

[classhandicapper]

I believe most figure makers that adjust final time for ground loss (without consideration for the ability to run faster when racing wider) have calculated the ground loss to be approximately equal to 1 length per path. Perhaps they are using fewer feet between paths.

So if you consider the fact that being outside allows the horse to run faster, if anything the appropriate adjustment would less than 1 length.

[classhandicapper]

I should add that the reason Moss brought this issue up was because "I" pressed him to review the adjustments he makes to the 6F time of route races in order to compare them properly the 6F times sprinters are running.

Without a proper adjustment, it's more difficult to evaluate speedy turn backs.

[Native Texan III]

Personally, I think they decelerate because of logistics - they have to angle their legs and bodies to negotiate the turn. Have you ever tried to steer your car around a corner at the same speed you drive on the straight roads?


Ken

That is right Ken - it is more horse locomotion than elementary physics. A horse is not a ball on a string going in a circle. It turns itself, leaning over into the curve so it and the jockey weight provide the sideways force component - correct banking can do that without leaning. It places its legs so that they each place more in a straight line than in two straight lines. As the horse has to keep its back straight when pushing with its hind legs it crabs around the bend in a series of straight lines with its front legs turning into the curve. A short coupled horse can do that far easier than a long striding galloper which is why some Europeans of that type fail on the tight Breeders Cup tracks. Some horses lose confidence in that they may fall when leaning over and drop back so a jockey trying to make a move gets no response.

[classhandicapper]

I don't have the physics background to discuss this issue with you guys, but since I'm the one that initiated the Moss thread I can explain why he's doing it.

The idea is to equalize the fractions of a route race and 6F sprint for the impact of a turn to better evaluate turn backs and to make better comparisons between speed horses.

The problem is that part of the reason fractional times in routes tend to be slower is due to jockeys rating their horses in order to get the longer distance and part is that different parts of the race are around the turn (which is harder to run fast around)

If you could isolate the impact of the turn alone then you'd have a better basis for comparison. But it gets very complex because you also have to consider how much energy the horse is actually consuming when running at a given speed on the turn and on a straight.

He has very good numbers, but I'm not entire convinced he has it perfect and neither is he.

[Cratos]

That is right Ken - it is more horse locomotion than elementary physics. A horse is not a ball on a string going in a circle. It turns itself, leaning over into the curve so it and the jockey weight provide the sideways force component - correct banking can do that without leaning. It places its legs so that they each place more in a straight line than in two straight lines. As the horse has to keep its back straight when pushing with its hind legs it crabs around the bend in a series of straight lines with its front legs turning into the curve. A short coupled horse can do that far easier than a long striding galloper which is why some Europeans of that type fail on the tight Breeders Cup tracks. Some horses lose confidence in that they may fall when leaning over and drop back so a jockey trying to make a move gets no response.

What you just stated is elementary physics

[Robert Goren]

What make it even more complex is that some horses corner a lot better than others.

[Cratos]

I believe most figure makers that adjust final time for ground loss (without consideration for the ability to run faster when racing wider) have calculated the ground loss to be approximately equal to 1 length per path. Perhaps they are using fewer feet between paths.

So if you consider the fact that being outside allows the horse to run faster, if anything the appropriate adjustment would less than 1 length.

In North American thoroughbred racing there are 2 primary dirt track layouts, 1 mile and 1 1/8 mile circumferences. In addition there are Belmont at 1 ½ miles and Keeneland at 1 1/16 miles. There are probably some minor ovals at either 5/8 mile or 7/8 mile in circumference. A turf and poly surface requires a different coefficient of friction and is not used in this calculation.

However what distinguish each track beside its surface composition is its stretch length and that difference changes the turn radii.

I am not saying that you are wrong, but I would love to see the figure makers’ calculations that adjust final time for ground loss under the different track turn radii. Also correct me if I wrong, but did not Randy Moss come up with 1.5 lengths per path?

[Cratos]

What make it even more complex is that some horses corner a lot better than others.

You are correct, but we are speaking of average and not extremes; and I will grant you that there are some horses that corner better than others which says that horse cornering is stratified due to the horse's cornering efficiency.

[Cratos]

I don't have the physics background to discuss this issue with you guys, but since I'm the one that initiated the Moss thread I can explain why he's doing it.

The idea is to equalize the fractions of a route race and 6F sprint for the impact of a turn to better evaluate turn backs and to make better comparisons between speed horses.

The problem is that part of the reason fractional times in routes tend to be slower is due to jockeys rating their horses in order to get the longer distance and part is that different parts of the race are around the turn (which is harder to run fast around)

If you could isolate the impact of the turn alone then you'd have a better basis for comparison. But it gets very complex because you also have to consider how much energy the horse is actually consuming when running at a given speed on the turn and on a straight.

He has very good numbers, but I'm not entire convinced he has it perfect and neither is he.

Given the mass (weight) and the velocity (speed) the amount of kinetic energy consumed by the horse on the straightaway or the turn is easy to calculate

[Native Texan III]

What you just stated is elementary physics

If so, you have not understood a word.

[Cratos]

If so, you have not understood a word.

I am not the arguring type and I will defer to you that you are correct; I have not understood a word.

[classhandicapper]

In North American thoroughbred racing there are 2 primary dirt track layouts, 1 mile and 1 1/8 mile circumferences. In addition there are Belmont at 1 ½ miles and Keeneland at 1 1/16 miles. There are probably some minor ovals at either 5/8 mile or 7/8 mile in circumference. A turf and poly surface requires a different coefficient of friction and is not used in this calculation.

However what distinguish each track beside its surface composition is its stretch length and that difference changes the turn radii.

I am not saying that you are wrong, but I would love to see the figure makers’ calculations that adjust final time for ground loss under the different track turn radii. Also correct me if I wrong, but did not Randy Moss come up with 1.5 lengths per path?

I'm not sure what Randy uses, but I'm almost certain that Thorograph and Sheets use 1 length per path

[classhandicapper]

Given the mass (weight) and the velocity (speed) the amount of kinetic energy consumed by the horse on the straightaway or the turn is easy to calculate


I'd like to see some of those calculations translated into approximations of impact on final time (just the incremental energy portion).

[gm10]

Sometime ago there was a thread started about Randy Moss of the DRF quest to determine why and how horses slow down because of the turns in the racetrack layout.

I believe I have the answer and the correct approach to calculating turn deceleration and I have done it for the current meet at Saratoga.
The formulation is universal, but it must be adjusted for turn radius, banking angle, and surface.

The assumptions on the model at Saratoga’s 1 1/8 mile Main track are as follows:

• Average Saratoga horse weight is 1,084 pounds
• Average weight toted by the Saratoga horse is 119.62 pounds (average weight from Belmont Summer Meet)
• Coefficient of kinetic friction at Saratoga’s Main track is 0.9775304
• Saratoga’s Main track turn radius at tangent to the rail is 468.23 feet
• Distance around Saratoga’s Main track turn at tangent to the rail is 1,473 feet
• Each path width on Saratoga’s Main track turn is approximately 4 feet
• At Saratoga Main track, the closest running radius is 470.23 feet
• At Saratoga Main track, the shortest running turn arc is 1,477.27 feet
• At Saratoga Main track, g in path 1 range from 4.22 (farthest out) to 8.40 (the most inner)
• At 12.468 feet/one fifth second the horse’s side force in path 1 is 8461.90 Lbs.

Therefore two horses entering the turn at 12.578 per one-fifth second, the horse in path 1 at Saratoga on the Main track will decelerate to 12.428 feet per one-fifth second while the horse running in path 2 would decelerate to 12.432.

The decrease in g and the larger radius will allow the horse in path 2 to run a faster rate, but it will have a longer distance to travel (1,477.27 feet vs. 1,496.12 feet).

Converting this to speed around the turn for each horse we get 1,477.27/12.428 = 23.8 seconds whereas the horse in path 2 time would be 1.496.12/12.432 = 24.1 seconds.

The difference between the two, 24.1 – 23.8 = .30 seconds or 1.77 lengths at .167 seconds per length.

This can easily be proven by taking data from the result charts and comparing the actual data against the calculated data.

How helpful is the calculation? Its merit comes in two forms; one is that the so called “track variant” which is calculated by some by just using final race times is bogus. The reason being is that a horse final time will change based on turn position even if the surface is perfect? Secondly, the calculation is very helpful when a horse which went wide in its last outing and an assessment is now being made of what time it might run in today’s race.

Comments, please.

How did you come to such a precise coefficient of friction by the way?

[Cratos]

How did you come to such a precise coefficient of friction by the way?

I believe that μk which equals kinetic friction coefficient should be close to 1 and I made some assumptions about the normal force then inserted them into the formulas. The decimal precision is because I didn’t want rounding to make μk equal to 1.

[Cratos]

I'd like to see some of those calculations translated into approximations of impact on final time (just the incremental energy portion).

I think that those calculations can be made.

[gm10]

I believe that μk which equals kinetic friction coefficient should be close to 1 and I made some assumptions about the normal force then inserted them into the formulas. The decimal precision is because I didn’t want rounding to make μk equal to 1.

Ok thx.

Do you differentiate between dry and wet dirt?

I also still wonder about your value of horse weight.
I did some study on a high-quality data set not so long ago and the average value was quite a bit higher than the one you propose. Is it based on actual body weights of race horses?

[Cratos]

Ok thx.

Do you differentiate between dry and wet dirt?

I also still wonder about your value of horse weight.
I did some study on a high-quality data set not so long ago and the average value was quite a bit higher than the one you propose. Is it based on actual body weights of race horses?

The rule of thumb for a thoroughbred weight is 1,100 pounds. The data I collected averaged 1,084 pounds and that is what I used.

However horses like Point Given, Secretariat, Forego, and others weigh 1,200 pounds or more and then you have MTB, Bold Forbes, Birdstone and others weighing in at less than 1000 pounds.

All calculations are made on dry surfaces. Friction changes dramatically when liquid is applied.