# Horse racing ... > wet grass vs. dry grass

by rolifantje
Tags: grass, horse, racing
 P: 8 Hi. I am a big horse racing fan, and while I am OK with maths, I don't have the knowledge in physics to explain the following well-known horse racing phenomena. When a horse race is run on grass that is wet/sticky, then a) It will take more time to cover the distance of the race than when the grass is dry (=firm ground). This is of course nothing new. Running through the mud is harder than running on a paved road. But how do you express this in terms of physics/mathematics? b) Horses that are able to 'quicken' at the end of a race on firm ground, usually fail to do so when there is some moist in the ground. If a horse needs to make up, say, 5 lengths in the final 200 meters of the race, why is this more difficult to do on a sticky surface? If horse A has 10% excess energy wrt horse B, does this convert into a difference (speed A - speed B) that depends on the actual speed they are running at (=lower when it's wet)? Maybe friction has something to do with it? A sticky surface will suck the hoofs into the ground. Any feedback will be greatly appreciated. I have been trying to explain this for a few years, but I just don't know enough about physics to come up with an answer myself.
 P: 748 No its because when the grass is wet the dirt is also wet and it is softer. When the horse pushes against a softer surface some of the energy goes into deforming the ground, and less work is done on the horse.
P: 8
Hi Curl,

For part (2), allow me to rephrase.

Scenario 1 - Firm ground
------------------------

Horse 1 is able to produce energy E1 which allows him to run at speed v1.
Horse 2 is able to produce energy E2 which allows him to run at speed v2.

One way of comparing the two horses is v2/v1. The higher above 1 this value is, the more distance there will be at the end between horse 1 and 2 (with horse 2 in front).

Scenario 2 - Soft ground
------------------------

Horse 1 is able to produce the same energy E1 which allows him to run at a new speed v'1.
Horse 2 is able to produce the same energy E2 which allows him to run at a new speed v'2.

We now compare the horses by looking at v'2/v'1.

In my experience v'2/v'1 is significantly larger than v2/v1. Can this be proven by using the laws of physics?

 Quote by Curl No its because when the grass is wet the dirt is also wet and it is softer. When the horse pushes against a softer surface some of the energy goes into deforming the ground, and less work is done on the horse.

PF Patron
P: 10,092

## Horse racing ... > wet grass vs. dry grass

This sort of situation involves complicated, non-linear, behaviour of the ground and the horse's physiology. How much effective propulsive energy you get with a hoof will depend on how the energy is stored in the tendons (potentially a huge proportion of energy-return, I believe) and how much gets lost in deforming the ground.

There must be an upward force to support the weight of the horse and a backward force (on the ground) for propulsion. This means the horse must be pushing at different angles, for different speeds and conditions.

All the above, and more, imply that nothing is simply proportional to anything else so it would actually be really surprising if you found that your ratios were the same. Actually, I can't see, from your post, which is the higher speed, v or vdash, which makes it difficult to decide which way the discrepancy goes and find some sort of an explanation.

There is a real difficulty in relating these sorts of physiological systems to the basic "laws of Physics" because there are so many variables involved which are so difficult to quantify. There was a similar thread, not long ago, involving weightlifting which just went on and on because people wouldn't accept this.
P: 8

I actually remember reading about the energy being stored in the horse's tendons a while ago.

I know that I am making a lot of assumptions, but for the sake of it, just say that horse 1 has 'available energy' E1 and horse 2 has E2 available. Suppose that both are equally efficient at transforming that energy into running speed. Then, if E2 > E1, the speed of horse 2 (=v2), will be higher than the speed of horse 1 (=v1).

Suppose the ground is firm.
Also suppose v1 = 15 m/sec, v2 = 15.15 m/sec

Then v2/v1 = 1.01.

After 60 seconds of racing, horse 2 will be 9 meters in front of horse 1.

If you repeat the same exercise on soft ground, making the same assumptions (horses have energy E1, E2, are equally efficient at converting this into speed), you will get results of the following nature.

v1 = 14 m/s
v2 = 14.28 m/s
v2/v1 = 1.02

So horse 2 will be slower than before, and will also be 16.8 meters ahead of horse 1, which is a lot more than before.

I understand that some horses are better suited to soft ground than others, but the average difference in running times of horses in 'wet' races is just a lot bigger. It happens time after time, there must be some reason for it.

So to sum up, I understand now why a) happens, but I would really like to see physics applied so that b) can be proven. I know it is the case from the data I have, I just want to see it explained within a theoretical framework. I tried proving it with the equations for kinetic energy and friction, assuming different friction coefficients for dry/wet, but couldn't really make it work.

 Quote by sophiecentaur This sort of situation involves complicated, non-linear, behaviour of the ground and the horse's physiology. How much effective propulsive energy you get with a hoof will depend on how the energy is stored in the tendons (potentially a huge proportion of energy-return, I believe) and how much gets lost in deforming the ground. There must be an upward force to support the weight of the horse and a backward force (on the ground) for propulsion. This means the horse must be pushing at different angles, for different speeds and conditions. All the above, and more, imply that nothing is simply proportional to anything else so it would actually be really surprising if you found that your ratios were the same. Actually, I can't see, from your post, which is the higher speed, v or vdash, which makes it difficult to decide which way the discrepancy goes and find some sort of an explanation. There is a real difficulty in relating these sorts of physiological systems to the basic "laws of Physics" because there are so many variables involved which are so difficult to quantify. There was a similar thread, not long ago, involving weightlifting which just went on and on because people wouldn't accept this.
 PF Patron Sci Advisor P: 10,092 This has a horribly deja vu feeling about it, I'm afraid. 1. You can't apply 'efficiency', as defined in Physics, to horses and people. In terms of work done, running on the flat has zero efficiency - the only energy that could possibly emerge would be the Kinetic Energy of the horse at the end of the race. Once it is up to speed, there is no KE gain. 2. Your quoted figures would need to be from actual measurements (I take it they are estimates) and made under controlled conditions. Without that, your ratios are not valid - so it can take you no further. Look at the thread about weight lifting that was locked a few days ago. It was fruitless from beginning to end because all we were ever supplied with was strings of numbers and wrongly used Science terms. Please believe me. This can't go anywhere without real data from experiments on turf, tendons and real horses. see http://www.physicsforums.com/showthread.php?t=472526
P: 8
OK, this is data for the one mile distance at Ascot race course in England.
The data is ordered by 'softness' of the ground. 'Good/Firm' is dry, whereas 'Heavy' is very sticky.

The data is for 'handicaps', which are designed to give each horse in the race an equal chance. v2/1v is the ratio of the speed of the winner to the speed of the 5th placed horse in the same race.

Ground	       Win Time (secs) v2/v1
Good/Firm	      100.96	1.011
Good	              101.62	1.012
Good/Soft	      103.23	1.012
Soft	              106.57	1.015
Heavy	              111.02	1.023
As you can see, the softer the ground, the longer it takes to run the distance. That was part a).

As you can also see from the final column, the 'difference-in-speed ratio' increases as the ground gets softer. That is part b).

To come back to a). If you throw a tennis ball on a concrete road it will take a certain amount of bounces, and a number of seconds before it has traveled 50 meters.
If you throw it over a muddy field, it will take more bounces and more seconds to travel the same 50 meters. I presume this is because the ball is heavier now, and maybe there is also more friction with the surface when it bounces.

However, what will happen if you use a bouncy rubber ball? Assume there would be no extra weight because of the mud sticking to the ball. So would it still take more bounces and travel at a lower speed?

 Quote by sophiecentaur This has a horribly deja vu feeling about it, I'm afraid. 1. You can't apply 'efficiency', as defined in Physics, to horses and people. In terms of work done, running on the flat has zero efficiency - the only energy that could possibly emerge would be the Kinetic Energy of the horse at the end of the race. Once it is up to speed, there is no KE gain. 2. Your quoted figures would need to be from actual measurements (I take it they are estimates) and made under controlled conditions. Without that, your ratios are not valid - so it can take you no further. Look at the thread about weight lifting that was locked a few days ago. It was fruitless from beginning to end because all we were ever supplied with was strings of numbers and wrongly used Science terms. Please believe me. This can't go anywhere without real data from experiments on turf, tendons and real horses. see http://www.physicsforums.com/showthread.php?t=472526
 PF Patron Sci Advisor P: 10,092 Those figures are encouraging. The difference is small but looks significant. This may have legs. I shall have to think on it for a while.
P: 8
 Quote by sophiecentaur Those figures are encouraging. The difference is small but looks significant. This may have legs. I shall have to think on it for a while.
That is much appreciated!
 PF Patron Sci Advisor P: 10,092 I wrote a whole contribution on this thread but it seems to have dropped off the end, somewhere due to something I must have done! I have a habit of breaking off near the end of a post and then coming back and shutting the computer down without looking and posting. Stupid boy. Anyway, here are some thoughts. The only situation that I can analyse is when the horse is accelerating. What goes on when it is at its maximum speed is far too hard because most of the losses are in the horse moving its legs as fast as poss and keeping itself off the ground. It is difficult to assign values and variables to this. However, it is easier to consider what is happening when the horse starts off. In particular, consider the very first stride. On solid ground, the available energy from the legs is 'all' transferred into Kinetic Energy. On soft ground, the energy goes into KE plus the work done in distorting the surface of the ground as the horse sinks. A more powerful horse will finish its first stride at a higher speed than a less powerful one. On softer ground, all other things being equal, both horses will need the same amount of energy to deform the ground so the more powerful horse will have proportionally more energy to transfer to KE. If you call the available energy from the legs Ef and Es (for the fast and slow horse) and the Work done whilst they are sinking into the ground is Ws (The weight W times the distance s that the hoof 'sinks'. So: KE (fast) = Ef - Ws and KE (slow) = Es - Ws KE is mv2/2 so the velocity is proportional to √(KE) So the ratio of v(fast)/v(slow) =√( (Ef-Ws)/(Es-Ws)) As Ws gets bigger and bigger, this ratio gets bigger and bigger. The same argument will apply to subsequent strides as the speed builds up, but is harder to write in simple terms. However, it indicates that the acceleration of the more powerful horse will be EVEN better under soft conditions. In the limit, you could imagine the ground so boggy that the slower horse has no surplus energy at all for running and uses all its energy in climbing back out of the hole it's made. The faster horse would still have enough energy available to stagger forward slowly. This is what the recorded figures show (qualitatively) but, to tie in with the actual figures I think i need to know some of the following. Weight of a typical racehorse. How long they spend in the acceleration phase. How deep the hooves sink into 'soft' ground. This analysis is very simplified because it implies that the horse with the quickest getaway would necessarily win the race. This isn't true and would make races very boring .
P: 8
 Quote by sophiecentaur I wrote a whole contribution on this thread but it seems to have dropped off the end, somewhere due to something I must have done! I have a habit of breaking off near the end of a post and then coming back and shutting the computer down without looking and posting. Stupid boy. Anyway, here are some thoughts. The only situation that I can analyse is when the horse is accelerating. What goes on when it is at its maximum speed is far too hard because most of the losses are in the horse moving its legs as fast as poss and keeping itself off the ground. It is difficult to assign values and variables to this. However, it is easier to consider what is happening when the horse starts off. In particular, consider the very first stride. On solid ground, the available energy from the legs is 'all' transferred into Kinetic Energy. On soft ground, the energy goes into KE plus the work done in distorting the surface of the ground as the horse sinks. A more powerful horse will finish its first stride at a higher speed than a less powerful one. On softer ground, all other things being equal, both horses will need the same amount of energy to deform the ground so the more powerful horse will have proportionally more energy to transfer to KE. If you call the available energy from the legs Ef and Es (for the fast and slow horse) and the Work done whilst they are sinking into the ground is Ws (The weight W times the distance s that the hoof 'sinks'. So: KE (fast) = Ef - Ws and KE (slow) = Es - Ws KE is mv2/2 so the velocity is proportional to √(KE) So the ratio of v(fast)/v(slow) =√( (Ef-Ws)/(Es-Ws)) As Ws gets bigger and bigger, this ratio gets bigger and bigger. The same argument will apply to subsequent strides as the speed builds up, but is harder to write in simple terms. However, it indicates that the acceleration of the more powerful horse will be EVEN better under soft conditions. In the limit, you could imagine the ground so boggy that the slower horse has no surplus energy at all for running and uses all its energy in climbing back out of the hole it's made. The faster horse would still have enough energy available to stagger forward slowly. This is what the recorded figures show (qualitatively) but, to tie in with the actual figures I think i need to know some of the following. Weight of a typical racehorse. How long they spend in the acceleration phase. How deep the hooves sink into 'soft' ground. This analysis is very simplified because it implies that the horse with the quickest getaway would necessarily win the race. This isn't true and would make races very boring .
Hi,

thanks a lot for your response.

Would it be reasonable to generalize your reasoning from 'first stride' to every phase of the race where a horse is trying to accelerate (usually at the start, but also towards the finish)?

In that case it makes a lot of sense to me, because it confirms another 'fact', one that I hadn't mentioned before. Namely, horses find a lot harder to 'kick on' late in the race, when the ground is soft.