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pole vaulting

 
 
aman
 
Reply Sat 23 Feb, 2008 03:10 am
What is the maximal height a pole vaulter could theoretically jump? Would it ever be possible that a pole vaulter passes the height of 10 metre?
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Type: Discussion • Score: 0 • Views: 1,465 • Replies: 5
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Tigershark
 
  1  
Reply Sat 23 Feb, 2008 03:19 am
Impossible.

Current WR is only 6metres something. 6.11 I think.

It would be impossible to get enough lift.
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Miller
 
  1  
Reply Sat 23 Feb, 2008 09:01 am
Increased height can be achieved by improved athleticism of the vaulter

It can also be achieved by altering the composition of the pole so as to increase the amount of stored strain energy within the pole.

Thus...it is possible.
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raprap
 
  1  
Reply Sat 23 Feb, 2008 09:21 am
Depends on how fast the jumper can sprint---classical physics says that potential energy is equal to kinetic energy (PE=KE)
so
mgh=1/2mv^2
and
h=1/2v^2/g
if the sprint to the box is on the order of say 9 m/s (~20 mph) and the technique is flawless then h= 4.1 m
10 m/s (22 mph)---h= 5.1 m
12 m/s (26 mph)---h=7.3 m
consequently the speed of the sprint to the box, assuming flawless technique, is the deciding factor.

Rap
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Miller
 
  1  
Reply Sat 23 Feb, 2008 09:25 am
Quote:
Essentially pole-vaulting involves the conversion of the kinetic energy of the running athlete to the potential energy of the jump using strain energy stored in the pole (the energy stored in elastic deformation). Consider an athlete of mass m = 80 kg running at a speed v = 10 m s­1, who has kinetic energy of 1/2mv2 = 4000 J. If this energy is converted with 100% efficiency into potential energy mgh, where g is the acceleration due to gravity and h is the height jumped, then the athlete can climb a height of 4000/mg, or just over 5 m. However, in reality, most pole-vaulters can jump heights of nearly 6 m. So where does the extra energy required to propel the athlete to these greater heights come from?

It turns out that the extra energy comes from the athleticism of the vaulter bending the pole. Energy is stored in the pole as it is bent or strained by the athlete's muscles, and returned to the vaulter as the pole recoils. The strain energy comes from the work done by the muscles of the athlete as he or she takes off, carrying out work on the pole as it is bent. The maximum strain energy of the pole is ms2/2rE, where s is the maximum or "failure" stress on the outside of the pole, r is its density and E is its Young's modulus (i.e. its "stiffness").

Bamboo has a relatively low Young's modulus and density, and a moderate failure stress. Glass fibre also has a low Young's modulus and density ­ but a much higher failure stress than bamboo. In fact, the maximum strain energy that can be stored in a glass-fibre pole before it breaks is about 2500 J, compared with just 100 J for bamboo. One consequence of this is that glass-fibre poles can be bent through much larger angles before breaking, which is why athletes can use them to jump gymnastically over the bar.

If we assume that the efficiency of a glass-fibre pole is 50%, an extra 2500 J of stored energy would be enough to get the athlete's centre of mass, feet first, over the bar. In other words, our athlete would have a kinetic energy of 4000 J plus a strain energy from the pole of 1250 J, giving a total energy of 5250 J. If all this is converted into potential energy, the athlete would climb a height of 5250/80g ~ 6.5 metres.

The advances in pole-vaulting performance are certainly not as good now as they were in the 1960s. That has not, however, stopped researchers from searching for further improvements, which have included introducing carbon fibres to make the pole still stronger and lighter (figure 2c), and allowing the pole to vary in thickness along its length. The latter innovation stemmed from work done in 1996 by Stuart Burgess, then at the Department of Engineering at Cambridge University. He demonstrated theoretically and experimentally that, during a jump, the greatest bending moments ­ and hence the greatest stresses ­ are at the middle of a pole, while the lowest stresses are at the ends. In other words, the ends of a pole can be made narrower ­ and hence lighter ­ without compromising the pole's performance. Burgess therefore optimized the thickness of a pole so that it tapered towards its end, giving a mass saving. Also, the bending moment ­ and hence the failure stress ­ increases with the stiffness of the pole for a given section of the pole and radius of bend. The ideal pole therefore has a low stiffness, a low mass and a high failure stress.

Clearly, pole-vaulting is an example of a sport in which technology has been used to improve athletic performance. As the Olympic winning heights in the discipline level off, it will be interesting to see if our ingenuity can provide another technological leap to allow pole-vaulters to jump even higher


www.neovault.com
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raprap
 
  1  
Reply Sat 23 Feb, 2008 09:40 am
Simple calc a 10 m vault would require a sprint speed at the box of 14 m/s (>31 mph).

Note the simple calc assumes the configuration of the center of vaulters mass doesn't change (technique)----assuming the runner is 2 m tall vaulter with a center of mass at 1 meter, and the vaulters c of m is at 10 m when they cross the bar, the change is 9m. so the sprint speed required for a 10 m vault would be sqrt(2*9.8m/s^2*9m)=13.3 m/s (29.7 mph). A world class sprinter in the 100m covers this distance in about 9.1 s (11 m/s). Usually in the 100, the sprinters speed maximizes before the finish line, so it might be possible for 13 m/s to be reached. However with steroids, HGH, amphetamines and genetic engineering anything is possible.

Rap
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