Help: Force on landing of 650kg Horse

[Vet]

Dear All,

As a relatively senior individual my physics days are long behind me. As a veterinarian I am trying to calculate the following problem

The force on landing of a 650kg horse having jumped 1,5m showjump. At time of take off the horse was traveling 10km/hr. Distance from take off to point of landing is 3,5m.

I am looking for an approximative answer?

Many thanks and an explication might help fire up my grey matter

 

[CraigD]

Approximate answer: 6400 N

CraigD, AMInstP
www.cymek.com

 

[Vet]

Many thanks Craig

 

[Doc Al]

Approximate answer: 6400 N

On what do you base this answer?

 

[CraigD]

On what do you base this answer?

If we approximate the horse as a cylinder...F=ma, m=650 kg a=9.81 m/s/s

Now remember a few things, the force will be higher because of the horse is falling for some period before it hits the ground, which I just ignored for speed of approximation. Also that is spread out over 4 legs, assuming that all 4 legs hit the ground at the same time.

If you want a more detailed evaluation, remember that the force seen by the front legs, which usually hit the ground first, will be different than the back legs that hit at a later time.

Sorry if my approximation is to general, but he just asked for a quick approximation.

CraigD, AMInstP
www.cymek.com

 

[Doc Al]

If we approximate the horse as a cylinder...F=ma, m=650 kg a=9.81 m/s/s

All you did was give the weight of the horse! That's the force the ground exerts on the horse when there's no impact. (What does a cylinder have to do with anything?)

Now remember a few things, the force will be higher because of the horse is falling for some period before it hits the ground, which I just ignored for speed of approximation.

But the fact that the horse is falling is the entire problem--you can't just ignore it.

The real answer is that there's no simple way to calculate the force of impact for a falling horse (or anything else). The force of impact depends on how quickly (over how short a distance) the object is brought to rest. Slamming something down on concrete will create a different impact than landing on a soft mat. It depends on the nature of the impacting bodies--ground and horse--and how elastic they are. I'm sure some engineers can slap together a model, but not without some empirical data.

 

[Archduke]

We can use a basic model, with a few approximations, to hopefully get a rough and ready figure of the right order of magnitude.

We know from Newton that $$F=\frac{dp}{dt}$$. Now, assuming that the horse just clears the 1.5m fence, and all the gravitational potential energy is converted into kinetic energy, then the speed of the horse just before it hits the ground is:

$$\frac{mv^2}{2}=mgh \\<br /> <br /> \rightarrow v^2 = \sqrt{2gh}$$

So the Horse's vertical momentum just before it hits the ground is:

$$p_ {i} = m\sqrt{2gh}$$

And as the final vertical velocity is zero, then p_f = 0

Therefore:

$$dp = p_ {f}-p_ {i} = -m\sqrt{2gh}$$

So:

$$F= \frac{-m\sqrt{2gh}{dt}$$

Now, I did a little experiment! I jumped off my bed a few times, and noticed that I bent my legs as I hit the ground, so the time to change my momentum is longer (so the force is smaller). I would estimate it to be of the order of a couple seconds. So using this approximation and plugging in the rest of the numbers:

F = 2000N

A better answer can be get by perhaps timing the amount of time the horse, as I (hopefully!) weigh considerable less than a horse, and I jumped form about half a metre. But unless I've done something wrong, that value should be approximately correct.

 

[CraigD]

Doc Al, the cylinder is an old physics joke.

Archduke, you are correct, except that the 2000 N needs to be added to the weight of the horse to get the force on the entire system. Which means I was within 25% and for an approximation that isn't too bad. But as your experiment shows there are a lot of variables still not inlcuded in the calculation.

Really this seems to be more of a question of the impact on the internal orgins of the horse, which would add even more variables into the question.

CraigD, AMInstP
www.cymek.com

 

[Doc Al]

Doc Al, the cylinder is an old physics joke.

I think you meant... "Assuming a spherical horse..."

 

[siddharth]

As Doc Al said, there's no way to exactly calculate the answer without knowing how the horse comes to rest.

The time $\Delta t$ which the horse takes to come to rest will influence your answer greatly. Put some numbers and check it out.

 

[CraigD]

I think you meant... "Assuming a spherical horse..."

I always heard it as "If we assume the cow is cylindrical" but yea the same idea

This really is an interesting problem, and nicely illustrates the complex nature of real world systems vs approximate models.

CraigD, AMInstP
www.cymek.com

 

[say_physics04]

not enough info on this problem if you're trying to find the actual approximate force

 

[vanesch]

Now, I did a little experiment! I jumped off my bed a few times, and noticed that I bent my legs as I hit the ground, so the time to change my momentum is longer (so the force is smaller). I would estimate it to be of the order of a couple seconds.

I think the time is much shorter. Probably a better estimation is the *distance* over which there can be deceleration, and assume the deceleration constant. For instance, consider that the horse can bend its legs over, say, 1 meter height (probably too much). This means, that at an initial velocity of v0, it has to come to v = 0 over 1 meter, with an average deceleration of a during a time of deceleration T. So a = v0/T, and s = 1m = 1/2 a T^2 = 1/2 v0 T ; in other words, T = 2 s/v0.

Now, if the horse is falling from 1.5 meter, then it has a fall time of 1.5 = 1/2 9.81 t^2 => t = 0.55 seconds, and hence a velocity v0 of 0.55 * 9.81 = 5.4 m/s.

If the "braking distance" over which the horse "brakes" is 1 meter (it bends 1m through its legs), then, using our T = 2 s / v0, we have T = 0.36 seconds.
So we see that our "braking time" is only 0.36 seconds, and not "a few seconds". If the horse has a smaller braking distance (say, 50 cm), this will only be about 0.18 seconds.

Back to the force: the deceleration is then v0/T = 5.4/0.36 = 15 m/s^2.

So we see that the force is about 150% its weight. On top of that, the weight must be added because gravity doesn't stop during the braking, so the final force on the legs is about 2.5 times the static weight of the horse.

 

[CraigD]

So we see that the force is about 150% its weight. On top of that, the weight must be added because gravity doesn't stop during the braking, so the final force on the legs is about 2.5 times the static weight of the horse.

The 2.5 times might hold up on the legs, but the legs act as a shock absorber so that the actual impact on the entire system should not come close to this amount.

Also there is still the problem that not all four legs hit the ground at the same time so the impact timing is different at the two ends, throwing more wrenches into the calculation.

CraigD, AMInstP
www.cymek.com

 

[vanesch]

The 2.5 times might hold up on the legs, but the legs act as a shock absorber so that the actual impact on the entire system should not come close to this amount.

Well, it is simply that the entire body of the horse needs to decelerate from 5.4 m/s to 0 in about 0.36 seconds. So no matter how, a force corresponding to this, plus the normal force of the weight of the body must be applied to the entire body of the horse. The "shock absorber" is nothing else but the 1meter of deceleration room, which amounts to allowing for 0.36 seconds of deceleration.

Also there is still the problem that not all four legs hit the ground at the same time so the impact timing is different at the two ends, throwing more wrenches into the calculation.

As long as we assume that the body of the horse will "bend through" only about 1 meter, this doesn't matter much. One will simply have a re-distribution of the total force over the 4 legs as a function of time.

 

[david1701]

I remember one of the first things done with photography was a time lapse of a galloping horse, this might give some info on the way the hooves act on impact, there won't be as much force as on landing but it might help

 

[Worzo]

Take out the horizontal motion of the horse - assume that the forces on its legs as a result of this are the same whether it is jumping or not.

So the problem is reduced to simply a horse falling onto its front two legs, since the vertical motion of the horse has stopped by the time the rear legs hit the ground.

Firstly, recognise that even though it is a 1.5m jump, a horse's legs are already a good 1.5m tall! So the total centre of mass of the horse is not raised the height of the jump. The legs move fowards and upwards to clear the jump, and the lower leg flexes in order to tuck under the horse while it is going over. From watching a YouTube video just now, I'd say say the horse raises its centre of mass about 0.5m during a jump.

Secondly, a typical showjumping horse weighs less than 500kg = 5000N.

So, let's calculate the vertical speed of the horse as it hits the ground from about 0.5m: v = sqrt(2gh) = sqrt(16) = ~3m/s. This gives the horse a downwards vertical momentum of 1500Ns.

Also from my YouTubing, I'd say the frontlegs absorb this vertical momentum in less than 0.5s.

So, F = p/t ---> F = 3000N.

This is the average resultant force that must be applied over that time in order to bring the horse to zero vertical motion again. I would probably estimate it's just over half of this actually (~2000N?) since not all the momentum of the horse is absorbed by the front legs: as the front legs apply force, they will begin to rotate the horse about its centre of mass, thus retaining some momentum to be absorbed by the hind legs.

Also, we have calculated the resultant force required to slow the vertical motion. Given that maybe half the weight of the horse is over the front legs, we must add this (2500N) to the force required from the muscles. So this brings my estimate up to about 4500N from the frontlegs.

 

[Britsenigma]

Sorry for digging up this old topic.

Something to add is, nearly all horses actually land on 1 foot when jumping, then the 2nd leg comes down shortly after, I'm interested in the force on that 1 leg?

Watch this video:

(first slow moton vid I found on youtube, 1/2 way through they slow it down, and you can see the 1 foot landing)

This is why horses suffer physical wear and tear doing large jumps...as the force on ligaments is tremendous, if you watch the top class show jumping on tv, the force is so much their fetlock actually flexes and touches the ground as the 2nd foot lands.

So how much force we talking on that 1 leg, and can you make sense of these figures from this website:

"The big difference between jumping and other types of equine activities is the force of gravity that the horse must overcome during take off, and then reduce upon landing. The force on the horse’s foot upon landing from a 2-foot jump is about 3156 pounds, and landing from a four-foot fence places about 4509 pounds on one foot."

http://www.equinew.com/jumping.htm

Is that accurate, based on the previous maths above?