Bubka physics

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Re: Bubka physics

Unread postby perezn03 » Sat Mar 19, 2011 5:34 pm

also, keep in mind that whatever force is being applied, you need to take into account angle. Sure someone straight on can have a force = t. but your gonna want the individual X and Y components of the force and apply that to the bend of the pole because the way a flew machine works, it applies a force at both end of the pole horizonatally as the pole lays with the prebend up, therefore the bend occurs naturally due to the bend of the pole and the force applied directly above. When vaulting there are multiple forces that act on the pole at take off, not just two. The flex machine, and an actual vault are two very separate events.

a prior article about Flex machine viewtopic.php?f=8&t=19280

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Re: Bubka physics

Unread postby VTechVaulter » Wed Mar 30, 2011 9:48 am

AVC Coach wrote: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.


right there is a nanosecond there where vertical velocity does equal zero

So rather than look at percent of bend, you have to look at chord length.

Potential energy absorbed by the pole = 1/2 k*x*x where x is the amount of compression in the pole (amount chord shortened) and K is the stiffness of the spring. The difficult thing to comprehend here is the following. If we say a pole is 180lbs, that is NOT a stiffness. Stiffness is measured as a rate of force/distance, like a one dimensional pressure. So we would need to know that information first (im guessing decamouse has it). I dont know if this number will be anywhere close but lets say a pole has a stiffness of 10 lbs per inch, and you shorten the chord by 4 ft (48 inches), than the amount of energy absorbed by the pole would be

PE = 1/2 *10*48*48 = 11520 lbs *in. You could also then calculate the rate at which the vaulter is rising.

SInce you have a total energy of 1/2 m * running speed * running speed, and at that point you lose 11520 lbs*in. The remaining energy is all motion, 1/2mv^2 again. so you could calculate vertical velocity. This also proves theoretically why jumping up puts less stress on the pole. sorry about the weird units, this is why we should all go to metric.
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Re: Bubka physics

Unread postby Lax PV » Thu Apr 14, 2011 12:49 pm

perezn03 wrote: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.
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I like the mention of the rotational energy here (after all--much energy comes from the swing). I (moment of inertia) however, will not be constant. Mathematically, one can account for these changes if the data is present (I spoke with Dr. McGinnis about this at the summit a couple years ago, but life got in the way, never followed up on it, and haven't done anything with it yet). In response to ADTF's post, until force transducers are placed in the hands of the athlete, emg sensors put on the body and arguably, a box made out of force plates, the values are never going to precisely be known and even then (and this is one of my biggest cruxes with biomechanics) all we know is what happened--not what one SHOULD be doing. At that point, an optimization algorithm would have to be made to maximize different parameters... and at this point, you are looking at Ph.D. level research. :(

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Re: Bubka physics

Unread postby agapit » Sun May 01, 2011 11:47 pm

ADTF Academy wrote:
9.6 m/s speed would be 1/2(80)(9.6)^2 = 3686



I like this conversation! it goes away from all this driving chest and dropping shoulders stuff and gets into the real meat.

FYI the actuall takeoff speed is in the 8.5-8.7 m/s range due to losses in redirecting horizontal speed (impulse) into 16-17% takeoff angle before pole resistance, so the numbers of a kinetic energy are lower and more imphasis is on the second part of the jump i.e. inversion and pull-push.

By the way the recoiling pole (see original post) does not add energy to the system, it just gives back some stored energy after losses due to the friction.
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Re: Bubka physics

Unread postby yoboe » Wed May 04, 2011 11:12 pm

right, the pole only stores kinetic energy given to it from the run on the X plane, and when it recoils it puts the same amount of energy back into the person on the Y plane.

if this is correct i have no idea how it is physically possible to have more joules of engery at the top of a vault compared to the takeoff. energy is always conserved in a closed system - the system of a vaulter becomes closed as soon as the takeoff foot leaves the ground. you can't actually add energy to the pole while you're in the air, the thing that makes one vaulter go higher than another is how they use this energy efficiently. you are basically floating as you go over the bar, that is you have zero Y plane KE. in this state, every movement your body makes has an equal and opposite effect on another part. for instance if you are flaoting in space, and you move your arms down, your body will go up. with this in mind the goal is to have every body part not directly over the bar as low as possible so that the part of the body over the bar will be as high as possible.

i guess i just dont understand when people say that middle phases put energy into the pole, when really the amount of energy in the pole will remain the same the whole time? maybe someone can clear this up for me

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Re: Bubka physics

Unread postby altius » Thu May 05, 2011 1:44 am

"you can't actually add energy to the pole while you're in the air" - the question you have to ask is 'are you in the air when the pole is connected to the ground'?"
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Re: Bubka physics

Unread postby Pogo Stick » Thu May 05, 2011 1:58 am

yoboe wrote:i guess i just dont understand when people say that middle phases put energy into the pole, when really the amount of energy in the pole will remain the same the whole time? maybe someone can clear this up for me


Because:
altius wrote:the question you have to ask is 'are you in the air when the pole is connected to the ground'?"


And because your body (muscles) can add energy in the system. "Trained athletes can manage up to about 2.5 hp briefly..." (Wikipedia)
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Re: Bubka physics

Unread postby Decamouse » Thu May 05, 2011 7:06 am

Simple analogy - jump up and grap a high bar - you are not longer touching the ground - so how could you then do any movement or do swings faster and faster?? Ok that is a rigid or semi-rigid device attached to ground

So next what about rings - yep - they are a flexible attachment

With the pole - as long as it is in the box and you are hanging on to it (attached) - while it is a bit more complex system -- work can be done - energy generated - question is how much and can you use it - some can - most of us - that is one of the goals
Last edited by Decamouse on Thu May 05, 2011 6:45 pm, edited 1 time in total.
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Re: Bubka physics

Unread postby agapit » Thu May 05, 2011 5:21 pm

yoboe wrote:the system of a vaulter becomes closed as soon as the takeoff foot leaves the ground.

i guess i just dont understand when people say that middle phases put energy into the pole, when really the amount of energy in the pole will remain the same the whole time? maybe someone can clear this up for me


No the system is not closed. The pole has support on the ground and muscle work adds energy and you can measure it with a tenso box (platform) if you have one like in Colorado Springs.
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Re: Bubka physics

Unread postby agapit » Thu May 05, 2011 5:24 pm

yoboe wrote:the system of a vaulter becomes closed as soon as the takeoff foot leaves the ground.

i guess i just dont understand when people say that middle phases put energy into the pole, when really the amount of energy in the pole will remain the same the whole time? maybe someone can clear this up for me


No, the system is not closed. The pole has support on the ground and muscle work adds energy and you can measure it with a tenso box (platform) if you have one like in Colorado Springs.
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Re: Bubka physics

Unread postby altius » Thu May 05, 2011 6:57 pm

Was that it so important agapit -that you had to post it twice?????
Its what you learn after you know it all that counts. John Wooden

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Re: Bubka physics

Unread postby agapit » Thu May 05, 2011 9:45 pm

altius wrote:Was that it so important agapit -that you had to post it twice?????


Oops
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