Why not use a 25 foot pole?

[DanCraig]

I don't get it--seems to me that the longer pole you use--the higher up you will go. You just have to hang on for dear life and let go at the right time. For instance, if you want to set a WR of 22 feet, then use, say, a 25 foot pole. That should get you over. You could work up to jumping over a 100 feet this way. I don't see what the big deal is--but I have never vaulted so maybe I am missing something. But it seems simple enough to me. Bigger pole=higher heights. I hope this helps someone, but it doesn't take a rocket scientist to firure it out.

[vaultguru6]

wow....never thought of that..............

[Decamouse]

It is not rocket science - but how fast you run and how well you transfer the energy dictates how high you can hold - if you held a 20' pole of appropriate stiffness - I would be surprised if you could get to 60 degress - getting vertical - or 90 to ground - yeah - maybe when we get a guy that is 6'9" and can run 10.2 - that is the pure physics of it -

[crash]

I don't see what the big deal is--but I have never vaulted so maybe I am missing something. .

yeah, your missing something!

[xtremevaulter]

It all has to do with Physics. I think it would be impossible for someone to use a 25 ft. Pole They could not get enough speed to even start getting upside down.

[Oldcoach]

Don't think about the method (long pole, short pole, big spring whatever)
it is strickly about energy conversion not the path or method to convert the energy.
A simple model 100% energy conversion

Change in Kinetic energy = Change in gravitational potential energy

1/2 M Vsquared = Mgh

or h (the change in vertical height)= 1/2 Vsquared /g

so for a guy that runs 10m/sec

h= 1/2 (100)/9.8 ( g is the acceleration due to gravity -9.8 m/sec/sec)

h = 5.10 meters-- this is the change in the vaulters CG

so if the vaulter 's cg is 1.1 m above the ground at take off

then the height the vaulters body can be raised is 5.10 +1.1= 6.20 meters above the ground. The World record is 6.15 meters.

Sorry for the garbled looking equations

[Decamouse]

wow OldCoach - great answer -eng wise as well - the 6.20 theory max does need one last thing added - the push off energy - if and when it is timed up - the pole is back to a relatively straight column you can probably generate a few cm more of potential - not me yet - getting of the pole early gets you zilch - give me a call sometime

[Oldcoach]

Decamouse,

You are correct that additional energy may be avaible in the pushoff potentially raising the height achieved. My simple model is what differentiates engineers from academics and scientists--- engineers are so lazy they only develope a theory that adequately satisfies the task at hand.
I'll give you a call sometime soon
Bob Johnson

[skola28]

Your explaination of the maximum polevault height doesn't really work here, but it definately shows the ignorance of the "longer pole=higher vaults" mentality. There are a 'few' more energies besides the run (one of the kinetic), and change in height (one of the potential). As stated the push off height increases the total height, but the vaulter swinging actually creates Rotational Kinetic energy. You also have to add in the impulse energy from the vaulter's 'jumping' off from the ground. I realize this was already stated, but this is assuming 100% conversion of all these energies into linear vertical kinetic energy (Upward speed). So none of this really takes into account the work done by the pole. The 'engineering' formula that should be used is this:

KE1 + PE1 + WORK DONE BY THE SYSTEM= KE2 + PE2
and we set KE2 to be 0 b/c we want the maximum height.

For KE (TOTAL KINETIC ENERGY):
1/2 MASS*Velocity^2 + 1/2 * I * Omega^2
where I is the Rotational Moment of Inertia of the vaulter which is
1/12 MASS * height of vaulters shoulders (sort of)

PE (TOTAL POTENTIAL ENERGY):
MASS*Gravitational Acceleration*Change in height+ 1/2 K(s1-s2)
where K is the 'poles' stiffness (relative to the flex number)

And I'm not even going to get started on how to figure out the total work done b/c there are many different parts which do work, and therefore many integrals which need to be solved. This is the actual engineering equation which would be used, the one which was mentioned earlier is simply the specific equation which satisfies most people but hardly an engineer! (Trust me, I know!)

If you REALLY wanted to get technical you'd have to deal with aerodynamics too! Anyway, I hope to one day figure out the real maximum height of the jump based on 3 variables:
Ground Speed, Push off Force(Impulse), Jumping force(Impulse). Where K (pole stiffness) is modified until the jump goes completely vertical. But that sounds like a lot of work. (Excuse the pun)

J. R. Skola

[Decamouse]

Don't think K is what you actual need then either - since if K is a spring constant the pole does not behave like a linear spring - flex number as currently derived is more beam theory which is also not technically correct - you would need to relate it to elastic column buckling theory -

[skola28]

Don't think K is what you actual need then either - since if K is a spring constant the pole does not behave like a linear spring - flex number as currently derived is more beam theory which is also not technically correct - you would need to relate it to elastic column buckling theory -

Just so you know, K doesn't have to be linear , so it really actually would be K, and yes it would be proportional to the flex number.

[Russ]

Boy, for a math-challenged person like me, these answers are terrific. These answers are also incredibly intelligent - which is all the more remarkable, given the original question that was asked.


[/img]