Bubka physics

[pvph]

I did some easy phisict calculations.. I'm not sure if this is the right place to post this, but I've some technical questions aswell.
I wanted to see the correlation between energy in the vaulter at take off and at the top of the vault. I used data from Bubkas' jumps.
Weight - 80kg
V(last 5m) - 9.9m/s

I used this formula E=0.5mv^2 + mgh. E(kinetic) + E(potential) = E(total)
I calculated the energy in two places: Right before the take-off (at this point there is no potential energy, just kinetic), and at the top of the vault, here, the athlete pass the bar (here I assumed the the kinetic energy equaled "0")

So, here we go:
Energy at take-off (or right before take-off): 0.5x80kgx9.88m/s^2 + 0 = approx 4000J
Energy at the top: 0 + 80kgx9.81x615 = approx. 4850J

As you see there is 800J more energy when Bubka passed the bar (if there is some kinetic energy at the top of the jump, the difference between energy at the take-off and at the top will be even bigger). 800J is about 20% more energy. So my question is, where is this energy made? Is it during the recoil of the pole and when Bubka got inverted? The most logical answer, in my head, is that this energy is made by the athletes actions during the recoil. What do you think?

I hope you guys understood what I was trying to say.

[altius]

Take a look at Figure 7.7 on page 44 of Beginner to Bubka. Data from the 1987 World Champs in Rome suggested just over 4000 joules at take off and around 5600 when the pole was straight. The additional energy comes from Bubka continuously interacting with the pole from the instant of take off until he released it.

[vault3rb0y]

More specifically, while he is on the pole his transfers more energy into the pole by rotational kinetic energy, creating a moment around his top hand. Also, during the top part of his inversion he transfers energy from his muscles into moving his hips upwards in a free-hip to handstand action. These two sources of energy more than overcome the energy loss to friction and the forward motion necessary for clearance.

[VaultPurple]

I posted this in a thread on run speed vs. height. But it kind of goes along.

I know there is alway the rumor of Bubka and a 10.1 m/s vault. But does anyone have the documented measurements of this? What meet and what did he jump?

What I am trying to do is calculate the energy put into the vault, and then how much energy they added after they leave the ground.

If Bubka were to run at 10.1 m/s then he would put in just enough kinetic energy to raise his center of mass about 17', and since his center of mass is probably about 3' off the ground that puts the vault at just over 20' (ie. the world record).

But in the 1987 vault if he was only running 9.5 m/s then he would have put in enough kinetic energy from the run to raise his center of mass 15' which is just over a 18' vault. But since he jumped 19'2 in that vault. Which means he added about 260 Joules of energy to the vault.

One of the goals to this is, had he used the same technique at the top of the vault would he have jumped 6.45 meters (which is the equivalent of what you would get from adding the extra 260J to a 10.1 m/s vault). And I know some people argue he has jumped close to that height, so maybe he was able to, just the bar was not there.

And I know that we are not in a vacuum and there is loss of energy in the pole and things like that. But the data is still very close to the real numbers we get from Bubka's vaults. Since 10.1m/s comes out to 6.12m vault and the energy added from the technique in the 1987 vault comes out to about 6.45m vault, which is what some people argue he should have been able to jump if he was not raising his PR 1cm at a time.

But as a side not: in Jenn Surr's 4.92 vault, she only added about 56J of energy to her vault. This amount of energy raised her vault 9cm. Because of her weight being less than that of Bubka's she needs less energy to raise her vault height, so if she were to put in enough energy to raise her vault 33cm more than just her speed allows (as Bubka did) you would get a vault of about 5.16m.

Also Bubka's kinetic energy never reached 0 because he always landed deep in the pit. Had it reached 0 he would land in the box.

[ADTF Academy]

All this is good if the object is being shot from the ground. However, pole vaulters are not they are from an upright position. The true question is at what height does this all begin.

The original poster stated mgh or (80)(9.81)(6.15) this would be as if Bubka came from the ground and does not take into account COM at TO or Reach height.

Later someone stated a COM was at 3' so Bubka really only needed to rise 5.25 changing the entire thought of energy gained in vault.

(80)(9.81)(6.15) = 4827
(80)(9.81)(5.25) = 4120


However where is the energy actually being applied? The exact location changes based on models use, but I feel it is being applied through or near the hands. Please correct me if I am wrong but wouldn't that mean we need to calculate off reach at impact? If not please explain why starting height is lower than point of application. Lets assume 7'6" reach (probably higher)

(80)(9.81)(3.86) = 3030

9.6 m/s speed would be 1/2(80)(9.6)^2 = 3686

A loss of kinetic to Potential energy of 656 joules.

If we look at it from the ground than yes energy was created or even if we look at it from COM at takeoff energy was created. If we look at it from reach height, point of rotation and exactly how high we need to go than energy was not created energy was conserved. Unless I am naive I do believe it takes energy to bend a pole, rotate that pole and unbend it. Yes our swing and extension can add energy back into the system, but I do not feel until proven wrong that it can add enough energy to overcome the other factors. If anything it aids in minimizing energy loss thus energy conservation. This is in lines with the extension of the top arm at top. I have stated before try to do a 1 handed pushup upside down. How far can you raise your COM? Extension of arm at top is about conserving energy hence staying in line with the pole, not creating it.

Till we can actually get data on the amount of energy needed to bend and unbend a pole at x stiffness with y mass, many of these talks are hypothetical at best. Any data can be made to look great at proving a point till you are asked to show hard data to support.

[ADTF Academy]

here is another interesting point I thought about.


At what point in time exactly does the concepts switch from Kinetic to Potential?

As soon as we leave the ground
As soon as the pole stops bending
As soon as the pole starts unbending
As soon as your imagination states it does


Depending on your thoughts on this it will also have an effect on the equation and actually Potential energy needed or used to vault X height and thus the relationship between energy in and energy out changes. Personally I don't think we have ever scratched the surface of how complex the physics is in the vault. Anyone got the money to front a high end project cause till one is done we are guessing.

[VTechVaulter]

energy can not switch directly from kinetic to potential

as the athlete runs its pure kinetic = 1/2 m*v*v
as the pole begins to bend kinetic energy begins to decrease, stored potential in the pole begins to increase, and hopefully the vaulter is rising increasing the gravitation potential energy. when pole reaches full bend, there is a lot of stored potential, additional gravitation potential from being higher in the air, and the vaulter should still be moving so theres additional kinetic. as the pole unbends, the stored potential energy decrease, kinetic increases, and gravitation potential increases. At the peak of the vault, kinetic energy approaches zero, stored potential in the pole is zero, and gravitational potential = mgh which equals the original kinetic 1/2 m *v*v minus any losses due to inefficencies, plus internal energy added by the vaulter via leg swing, row, pull things of that nature. a moment before the vaulter hits the pit his potential approaches zero and kinetic nears its peak again, and then sitting the pit its all zero.

here is another interesting point I thought about.


At what point in time exactly does the concepts switch from Kinetic to Potential?

As soon as we leave the ground
As soon as the pole stops bending
As soon as the pole starts unbending
As soon as your imagination states it does


Depending on your thoughts on this it will also have an effect on the equation and actually Potential energy needed or used to vault X height and thus the relationship between energy in and energy out changes. Personally I don't think we have ever scratched the surface of how complex the physics is in the vault. Anyone got the money to front a high end project cause till one is done we are guessing.

[vault3rb0y]

Somewhat relevant:

http://www.personal.psu.edu/jmp575/blog ... ltpdf.html

I wrote this for a technical definition for an english project.

It's dumbed down but should still pretty sound, minus the additional energy input once off the ground, which was not mentioned.

[ADTF Academy]

Thank you Brian......


#1 question I have always wanted to know in pure data... How much energy does it take to bend a pole to x % when the pole is Y stiffness? Does the energy used get redistributed when the pole unbends or what % is lost from the bending to unbending process. These are only a few things that really have never been studied or at least I have never seen any data on.

What % of KE at takeoff to PE available on top is considered good? How does this link and match up with grip height. Higher the grip the more you can get away with, but this also must take into account energy available to rotate such a longer grip into the pits safely.

All excellent questions.... Excellent way to shed the light on the issue of the vault Brian.

[AVC Coach]

Interesting topic. Wouldn't kinetic energy switch to potential at the apex over the bar? It's like flipping a coin in the air. At what point does that coin quit accending and start decending? There has to be a precise moment when that coin stops and there's no kinetic energy, only potential.

[perezn03]

flipping a coin at the apex, where the v=0 at the highest point in it s parabolic path, the acceleration is -9.8, but its kinetic energy is 0. this is bc KE(translational) = 1/2m(v)^2. The potential energy is at its max as PE = mgh. So at the apex thats where the PE is the greatest, but to consider, at the apex if the coin is flipped, there is still a constant Kinetic energy (rotational) neglecting air resistance.

I have read a lot of this forum, i feel that one major thing not being taken into account is the moment of inertia of each individual point of the vault, therefore that would effect the energy. as for KE(rotational) = 1/2i(omega)^2. reflecting on that point that i is usually most equal to a value with a R (radius) squared which would be detrimental to the final KE as KE(trans)+KE(rot)+PE(pole) need to be accounted for.

Anyway just a thought.
Nick

[vault3rb0y]

Thank you Brian......


#1 question I have always wanted to know in pure data... How much energy does it take to bend a pole to x % when the pole is Y stiffness? Does the energy used get redistributed when the pole unbends or what % is lost from the bending to unbending process. These are only a few things that really have never been studied or at least I have never seen any data on.

How difficult could that be? We could put a force-meter between the connection of a flexing machine and the pole, and measure the pressure on recoil as you flex the pole. If we compare this to the force of the flexing machine and do this on different poles we should get a decent average of how much energy is conserved when a fixed amount is put in. I've never used a flexing machine before, but even if it wouldn't quite work on one, we could definitely create a machine to control the conditions and get accurate measurements.