# Quantifying Equestrian Show Jumping: A Large Context Problem for Physics Students

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3 Jumping Fences The following descriptions of the approach, takeoff, flight and landing during a jump are adapted from a comprehensive piece by Sheila Schils, a well-known expert in equine rehabilitation, entitled Biomechanics of Jumping. 2 The Approach and the Push-Off The Flight and the Landing Figure 2 Fence jumping: approach, push-off, flight and landing Approach The horse must reach the fence at an even, steady gait (usually a canter motion) so that she can focus on the best spot for takeoff. See Figure 2a. The horse reaches forward and down with her neck in order to lower the front legs and her CG. The front legs are propped or strutted out in front of the body. This relatively sudden braking action allows momentum to carry the hind legs further under the body of the horse than would otherwise be possible. Takeoff See Figures 2b, 2c and 2d. As she finishes the last whole stride before the jump, the horse begins to shift her weight backward by raising her head, shortening her neck and lifting her shoulders. Her neck continues to shorten to help move the weight backward. This also helps to stop the normal forward movement of the canter. As the weight moves backward, the hind legs compress or coil. With the maximum amount of flexion in the hind joints, the horse can then create the maximum 46 ASEJ, Volume 43, Number 1, June 2013

4 push against the ground to propel herself up and forward. The horse has the most effective takeoff when her hip joint is placed vertically above the hoof. Flight See Figures 2d, 2e and 2f. The horse s hind legs reach maximum extension after they leave the ground, and her front legs are curled tight against her body. Her knees lift and bend to curl the legs up, the tighter the better, to reduce the chance of her hitting the fence with her front legs. To bend her knees and lift her forehand (the front part of the horse s body), the scapula (shoulder) rotates upward and forward. During the flight, the CG follows an approximate parabolic trajectory. Landing See Figures 2f and 2g. To slow the forward momentum and reduce the force of impact, the horse swings her neck and head up as her forelegs reach toward the ground. The horse s nonleading front leg lands first. When her leading front leg lands, both legs push against the ground in an upward and backward direction. The hindquarters rotate underneath the trunk and reach toward the ground as the forehand moves forward and out of the way of the hindquarters. Jumping Fences: Physical Principles Approach and Takeoff I am referring, for analysis, to one of the 1.6 m fences used in the Grand Prix. Derly approached this high fence with a speed of about 6.0 m/s. Her speed was reduced by a shorter stride to about 4.0 m/s just before she anchored her hind legs. We assume that the horizontal speed of 4.0 m/s does not change significantly during the 0.2 s push-off. In Figure 1, Derly s front legs are lifted and her hind legs stop moving for about 0.2 s. This is called the stance phase. The front legs are coiled so that the body of the horse, with reference to the horizontal, can be as high as 45, just before push-off. The hind legs uncoil, and at the point of leaving the ground, the angle for the trajectory is about 40. During this stance phase, the body moves about 0.8 m (4.0 m/s 0.2 s). At the moment of takeoff, Derly s CG is about 20.0 cm along the line of her body, in front of Lamaze s right foot. Knowing the angle at takeoff and the horizontal velocity, we can easily determine the instantaneous vertical velocity at the beginning, the height, the range and the time of flight for the trajectory. The Kinematics of the Jump See Figure 3. v y = 3.4 m/s Parabola v x = 4.0 m/s V = 5.2 m/s 5.2 m/s 40 V x = 4.0 m/s Height= 0.58m 40 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx Range = 2.8m Time of flight = 0.68s Figure 3 The motion of the CG of horse and rider over a 1.6 m fence ASEJ, Volume 43, Number 1, June

5 Derly pushes off when the angle to the horizontal is about 40, and the CG at this moment happens to be at about the same height as the fence. The horizontal velocity (v x ) is constant at about 4.0 m/s. (We will assume that the value of g, the gravitational attraction, is g = 10.0 m/s 2.) The vertical velocity (v y ) at the moment of liftoff is given by v y = v x tan 40. (1) Therefore, v y = 4.0 tan 40 = 3.4 m/s and the time t to reach height h is obtained from or v y = gt (2) t = v y /g = 3.4/10.0 = 0.34 s. The total time for the trajectory is 2t, or 0.68 s. Since v y 2 = 2gh, (3) the height reached by the CG is h = /20.0 = 0.58 m. The total height from the ground to the CG is = 2.18 m. The range R of the CG is given by or R CG = 2v x t (4) R = = 2.72 m. The tangential velocity v T at the point of liftoff is or v T = (v x 2 + v y2 ) 1/2 v T = ( ) 1/2 = 5.2 m/s. Remember that the distance from the spot where the hind leg hoofs take off and the horizontal distance to the CG (that is, the start of the trajectory) is about 1.0 m. If the jump is perfectly symmetrical (and it seldom is), the distance from the base of the fence to the point of contact, in this case, is about 2.72 m. The total range R is R = = 4.8 m. It should be noted that the location of Derly s CG (see Figure 1) does change somewhat during the flight as the horse s body configuration changes (due to the movement of the neck and the leg during the flight). Therefore, the trajectory is only an approximate parabola, as indicated in Figure 4. The Dynamics of the Jump The motion during the push-off stage that takes about 0.2 s is fairly complicated. Looking at Figure 2, it is clear that the force F produced by the hind legs is almost vertical. However, it misses the CG by a small distance d, which increases during this short time of contact. At the start of the push, Derly s body is along an elevation of about 45 (see Figure 5). As the hind legs push forward (remember that the hoofs are CG Parabola: R = 2.80 m H= 0.58m Time for flight: 0.68 s Fence: 1.60m m m m m xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx Total range= 4.80 m Figure 4 The kinematics of jumping a 1.6 m fence 48 ASEJ, Volume 43, Number 1, June 2013

6 stationary during this brief period), the body moves about 0.8 m forward. The product of df is a torque, which causes Derly to rotate clockwise for about 0.2 s. When the hoofs leave the ground, the CG has moved about 0.8 m horizontally and Derly is moving with an initial vertical velocity of 3.4 m/s and a constant horizontal velocity of 4.0 m/s essentially along a parabolic trajectory. The resultant initial velocity, then, must be 5.2 m/s, as shown earlier. When contact with the ground has ended, the direction of the body of the horse is about 40 with the horizontal. CG F (average) = mg + mδv y /t F (Total) = N 0.80m d F (average) = mδv y /t = 570x3.4 / 0.20 = Take-off angle: N t = 0.20 s Initial v y = 3.4 m/s angle: 45 mg = 5700 N Hind hoofs Vx = 4.0 m/s ( Note: The CG moves about 0.80m during the contact time of 0.20 s Force (N) Average force 0.20 Time (seconds) xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx Figure 5 Forces acting during takeoff (0.2 s) ASEJ, Volume 43, Number 1, June

9 Finally, it should be mentioned that just as in the case of the takeoff force acting behind the CG of the horse, producing a clockwise rotation, the contact force produced by the front legs on landing acts in front of the CG and produces a counter-clockwise rotation. Indeed, if the angle of descent is large enough, the horse will rotate clockwise, which results in a dangerous somersault, with a potential of severe injury to both rider and horse. Water Jumping If equestrian jumping is like jumping hurdles, then water jumping is similar to the long jump. Figure 8 shows an ideal water jump. The width of the jump at the Grand Prix at Spruce Meadows was 4.2 m. So Lamaze and Derly had to make sure the jump was at least 5.0 m long (see Figure 9). The angle of elevation at takeoff was about 25 and the approach speed about 7.5 m/s, because for a range of trajectory to be 5.0 m, at an angle of 25, the horizontal velocity must be 7.5 m/s. Following the same reasoning as before and using equations 1, 2, 3 and 4, 5.0 = v x t = 7.5t. Therefore, t = 0.7 s. Since tan 25 = v y /v x, v y = 3.5 m/s, the height h of the trajectory is h = v y2 /20 = 0.61 m. The contact time for the takeoff is also about 0.2 s and, therefore, the vertical force necessary for the trajectory is given by F y = mg + m(3.5/0.2) = 570( ) = 15,600 N, very much the same as for the 1.6 m fence. On landing, the vertical force F y, as for the hind legs on takeoff, is about 15,600 N. As before, the horse is reducing her speed, this time to about 5.0 m/s, from 7.5 m/s. Therefore, the horizontal force is F x = 570(2.5/0.2) = 7,100 N. The total force then is F = (15, ,100 2 ) ½ = 17,000 N, a little larger than the force required for the fence jumping. These large forces acting on the horses, even if only for a short time, are stressful for them. Riders are concerned about their horses and make sure that they are healthy, both physically and emotionally. Horses are examined by veterinarians before each competition. Serious accidents in Grand Prix jumping, unlike in steeple chasing or racing, are rare. CG Parabola Range: 4.2 m Height: 0.61m 1.50 m Time of flight: 0.70 s 1.0m. Range: 6.2m xxxxxxxxxxxxxxxxxxxxxxxxxx 1.0m Water barrier Figure 8 Kinematics of jumping the water barrier 52 ASEJ, Volume 43, Number 1, June 2013

11 Notes I would like to thank my wife, Ann, for the sketches in Figure 2 and her helpful suggestions for improving the article See (accessed May 7, 2013). 3. Unfortunately, since the time of writing, Eric Lamaze has sold Derly and is concentrating on developing his young stallion Wang Chung M2S. He has had some good wins lately. We wish them a successful future. References Meershoek, L S, H C Schamhardt, L Roepstorff and C Johnston Forelimb Tendon Loading During Jump Landings and the Influence of Fence Height. Equine Veterinary Journal 33, no S33 (April): Moghaddam, M S, and N Khosravi A New Simple Biomechanical Method for Investigating Horses Jumping Kinematics. In XXV International Symposium on Biomechanics in Sports 2007, Ouro Preto Brazil, Also available at (accessed May 7, 2013). Stinner, A O Physics and the Bionic Man. The Physics Teacher 18, no 5 (May): (Reprinted in New Scientist, December 18 25, 1980, ) 54 ASEJ, Volume 43, Number 1, June 2013

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