Re: RC: Up Hill or Down

[David LeBlanc <dleblanc@mindspring.com>]

At 06:10 PM 1/9/00 EST, Tivers@aol.com wrote:
>In a message dated 1/9/00 2:59:05 PM Pacific Standard Time, 
>dleblanc@mindspring.com writes:
>
><< Actually, if you look at the static load on a horse standing on a hill
> (aligned with the slope), you'll find that the proportion of the load on
> the front is going to be a function of where the animal's center of gravity
> is located.  If the horse lowers the back-end, it will carry more weight on
> the rear legs. >

>Pard, standing on the flat the horse bears 60% of its weight on the
forelegs. 
>You're telling me that going down hill at a trot is going to move the center 
>of gravity rearward far enough that the hindlegs take the burden? Please
show 
>me a picture of this happening so I can send it to the Smithsonian. 

The assumption that the center of gravity will change is based on the fact
that the saddle and rider both move back and forth depending on the slope
of the back of the horse.  Tell you what - let's make it really simple -
we'll take the rider out of the picture.  Let's just consider the horse.

I doubt the Smithsonian would be interested - I'd thought all this would be
intuitively obvious to the casual observer (horse sense tells you it's true
<g>), but apparently not.  So I'll work out the math for you in a moment.

> >Also, if the horse brings the back legs up under itself,
> this also increases the loading on the rear legs.  Any time the load on
> something isn't perfectly distributed, the support loading is going to be a
> function of 2 things - the first is that the sum of the loads on the
> supports must be equal to the total load (else it falls or flies), and that
> the distance to the support from the center of gravity times the load on
> the support must equal out, or the object will spin about the center of
> gravity.  That's the picture from a statics engineering standpoint, and the
> statics of a horse going downhill with a rider on it is anything but simple>

>You'ved greatly oversimplified it here--in the wrong direction, defying 
>gravity and the laws of physics. 

Nope.  I'm very well acquainted with gravity and the laws of physics.  If
you're really interested I could work the same problem assuming the horse
were on Mars, or perhaps even a space station.  Gets a bit tricky if we
assume that we're on a really fast-moving spaceship near a black hole, but
I doubt we'll be holding any endurance rides in space any time very soon.

Let me give you a really simple lesson in statics.  If you push harder on
the bottom of something than gravity pulls down, it goes up.  If you don't
push hard enough, it goes down.  If you assume it isn't moving, then the
two have to equal out.  This is called a force balance.  So our first
equation is:

Ff + Fh = W (load on the front legs, plus load on the hind legs = weight)

Next thing to consider is leverage, or an engineering student would know
this as moment balance.  If these don't equal out, whatever it is spins
around the center.  We'd prefer our horses not spin around the center,
since they'd either fall on their heads or flip over if they did.  So let's
assume these are in balance, too.  This gives us:

Ff*X = Fh*Y 

where X and Y are the distances from each end of the horse (or where the
legs hook up to the body) to the center of gravity.

Now you assert that the horse carries 60% of the weight on the front legs
standing flat on level ground.  I'm just going to accept that as true, and
that means that X works out to 2/5 the length of the horse, which is a
useful number and simplifies the next step.

Now if you tilt the ground, and the horse's body stays parallel to the
ground, then none of this changes - you still have the same distribution,
until you get the poor horse stood completely up on one end.  I'm also
assuming that the horses hooves dig a bit into the soil and the force acts
straight up against gravity, not perpendicular to the surface.

So now let's assume that the horse does like Heidi says it does, and like
I've seen mine do - it brings the hind legs up under itself, brings its
butt down a bit, and levels out its back.  Assume we're on a 30 degree
slope, and that the angle between the line of the horse's hind legs is at
60 degrees to the line of the back.  After applying a bit of high school
geometry to the problem, now we get:

Ff + Fh = W 

and

Ff*X = Fh*Y 

but now, 

X = 2/5Hl (2 fifths the length of the horse)
and
Y = (3/5 - sqrt(3)/4)Hl (3 fifths minus square root of 3 divided by 4) *
length

(draw it out if you're confused)

So now applying some basic algebra, we work all this out to find that:
Ff = .295W
Fh = .705W

Which is a bit different than the horse on level ground, which gave us:
Ff = .6W
Fh = .4W

So you see that under this situation, the bulk of the load has actually
shifted to the hind legs and the hind legs have contracted (better to carry
the load with).

Plus, if you've ridden a good horse down many hills (one of our's is really
good at this), you can see exactly what it is doing.  The horse is also
going to try to keep the rider off its withers if it can, and leveling out
its back helps on this account, too.  Also makes sense from the standpoint
of being able to control its motion, since if there's any slipping that's
going to happen, it is most likely to occur first in the rear, and the
horse still retains its ability to control where it is going from side to
side using the front legs, and if the rear legs come completely out from
under it, it falls on its butt, not its face.  I'd a lot rather fall on my
butt than my face, and I think the horse would agree with me on this.

Note that these numbers could change a little if I actually went and
measured a couple of horses to get real numbers for leg length and distance
from hip to shoulder, but the principle is the same, and the overall result
is going to be pretty close.

>No need to delve deeply into the religions of math--accelerometer data is 
>already available in the horse.

As you can see, the math is pretty trivial - a good high school student
who's had either a good geometry course or trig ought to be able to deal
with it.  I've taught harder problems to algebra students at an art school
(which surprisingly was a really fun job).
 
>> Incidently, not really understanding the dynamics of a system is why we
> build bridges with a 'safety factor' of 10 - what it means is that we
> really don't understand how the bridge really works, so if we add a lot of
> extra steel and concrete, we hope it won't fall down. <g>>

>The existentialist bridge builder. 

No, I just know that anytime you figure something out to 3 decimal places,
and then multiply by 10, there's some bullshit involved somewhere.  You run
into the same thing all over the place in engineering when we don't quite
understand how something works.  When you _do_ understand how something
works, you can usually build it more cheaply and reliably.  Otherwise,
you're just screwing around in the dark hoping that you're either not being
incredibly wasteful or that something might blow up in your face.  Point is
that even with the static load distribution shifting, what's a lot more
likely to be critical are the stresses on some joint somewhere - which is
dynamics, and is a lot harder problem.


David LeBlanc
dleblanc@mindspring.com


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