The Physics of Equestrian

Jumping

By Daryl MacLeod

3 part Jump

• The - ApproachIncludes the three strides before a fence and the 'takeoff' in front of the fence.

• The Jump- The air time as the horse rises in the air, experiences 'hangtime', and begins to descend back down to the ground.

• The Landing- Includes actual 'impact' of the horse hitting the ground, and the follow through as horse and rider continue on to the next fence.

Quick Course

The Approach

• The horse and rider possess only Kinetic Energy (if the ground is considered to be 'y-zero’). This kinetic energy is expressed

by: KE = 1/2mv2

The Jump

• Over the fence - Horse and rider reach maximum height and their velocity is reduced to zero; they possess only Potential Energy, PE. This Potential Energy is expressed by: PE = mgy

The Landing

• The horse returns to the ground, horse and rider are once again at 'y-zero' and possess only Kinetic Energy, KE, expressed

by: KEf = 1/2mvf2

What if….

If the fence is y= 5 1.52 m (5 ft).In calculation, assume the horse may jump up to six inches higher than the fence (depending on their perception) making y = 1.52m + 0.15m or y=1.52m

Due to non conservative forces, such as air resistance and heat, the potential energy reached at the top of the jump will only be about 80% of the kinetic energy present on approach.

1/2v2 = 0.8(9.8m/s2*1.52m) vlow = 6.1m/s vhigh = 6.4m/s

Projectile motion

Range equation: R = (vo2/g) (sin

2θ) Height equation: H = (vo

2 sin2θ)/(2g)

For this analysis we are going to determine the initial velocity at which the horse must leave the ground in order to get over a jump of a certain height and width.

With what initial velocity will a horse need to take off in order to

jump a 1.5m high fence?Assume the horse leaves the ground at 45°.

(angle would change depending on the different properties of the intended jump)

H = (vo2 sin2θ)/(2g)

vo= √((H*2g)/(sin2θ))

vo= √((1.5m*2*9.8m/s)/(sin245)vo= 7.7m/s

Horse power

The term horsepower was coined by James Watt, the inventor of the steam engine, when he needed a unit of power large enough to describe the output of his new invention. He worked in a coal mine and observed the work of the ponies that were used in the mine to haul coal. He saw that a mine pony could do about 22,000 foot-pounds of work in a minute. He then increased that number by 50% to make the measurement of horsepower (not ponypower!) at 33,000 foot-pounds of work in one minute. So, according to Watt, a horse can haul 330 pounds of coal 100 feet in one minute.

The End

http://wiki.croomphysics.com/index.php?title=The_Physics_Of_Horse_Jumping

http://www.unc.edu/~cgable/physics24.htm

Neat video-http://www.youtube.com/watch?

v=B9N0cAzoQVI