This is NOT homework (I don’t go to school)
Can we figure out how much potential energy is stored in a pole vault pole at maximum flex?
The vaulter runs, plants the tip of the pole and the pole flexes. At maximum flex, how much energy is “stored” in the pole?
It should be a factor of the vaulter weight and speed right? So if I converted that back into a figure of XX kg, how much would it be (assuming an 80kg weight of the vaulter, and a typical sprinting speed)?
As a follow up question, if we compared the technology of today to the technology of the first ash poles, how much of the increase in height can be attributed to an improvement in technology, and how much to an improvement in technique / strength?
Would it be fair to (for example) compare the % increase in high jump record in the same time frame to the % increase in pole vault over that same period and the difference attribute to technology?
I think this is an extremely tricky question because the stored energy in the pole goes from 0 (right before it gets planted in the ground) to some maximum during maximum pole flex back down to 0 (when the athlete releases the pole at the top of the jump).
So finally, the easiest method to find out precicely how much stored potential energy is stored in a pole vault is to photograph what’s the maximum angle deformation between the two pole tips, then run an iterative nonlinear (geometric) finite element analysis until you match the deformation of the photo, then add up the strain energy at the final step. Done.
That’s a great answer, but to employ the local vernacular - “I ain’t catching no balls with your england”
In other words, the mathematics, photo taking and the like is far far beyond my capabilities.
I have been able to research and find out that the pole vault record increased by 27% between 1960 and 2000 (vs 10+% for other jumpring sports), but I am still not sure how much “fling” is in a pole vault.
To put it another way - at maximum flex, how far would a pole vault pole be able to throw a tennis ball?
Probably the easiest way to do this is to calculate the energy required to fling the vaulter over the bar - after all, that’s the energy that comes out of the flexed pole.
So if a 80kg jumper is reaching a height of 6.1m how much energy is that?
Does the time it take make a difference?
I suspect that because all of the forward momentum goes into bar flex, the energy stored by the bar is some multiple of the vaulters weight - but I dunno any physics to figure this out.
Well I think moving 80kg to a height of 6.1 metres would take about 5000 joules of energy being transferred from the vaulter to the pole then back to the vaulter and finally dissipated on the landing mat. (10 joules being 1kg raised to 1m)
There would be losses along the way in terms of inefficiencies in the energy transfer (sound, heat, friction etc)
So let us imagine that it requires 25% more potential energy stored in the pole and you are looking at about 6250 joules (for the maximum amount that could be stored before breakage you may be looking at 8-10,000 joules)
But of course all these estimates are FPOOMA (figures pulled out of my ass)
Adding some more figures here - if my calculations are roughly correct that seems to suggest an equivalence to a 42hp electric motor (assuming the energy release from the pole takes place over 0.2 of a second) that actually seems quite weedy to throw a burly Russian 20 feet in the air but of course if that energy release takes a shorter time (which I suspect is the case) then the power rating climbs substantially.
Unsurprisingly I have never timed the flexing of a vaulting pole…anyone like to try?
That’s probably a fair approximation. For good measure, add in the weight of the pole that the jumper is carrying with them; that thing’s gotta weigh like fifteen pounds, right?
One source of inaccuracy in this method is that you never store 100% of the jumper’s kinetic energy in the pole. If that ever happened, the jumper would find himself completely stationary in the air, and that never actually happens; he slows down during the ascent, but never hits zero MPH.
You could estimate the gravitational potential energy of the jumper as he clears the bar, but this includes all of his initial kinetic energy (some of which was not actually stored in the flexing of the pole), and also includes energy he’s added during the ascent by flexing his legs and abs to get his lower body high enough to clear the bar.
I think the second best estimate is going to come from judging the jumper’s running speed just prior to the ascent, and calculating the associated kinetic energy - and then subtracting the kinetic energy associated with his speed at the instant of the pole’s greatest flex. This is before he really starts to raise his lower body, and so the difference in kinetic energy here must be stored in the flexing of the pole.
I think the best (but probably most difficult) estimate is going to come from The Niply Elder’s FEA approach.
If you had some method of estimating or measuring the distance the end of the pole has deflected axially, you could get your your old copy of Timeshenko’s “Theory of Elastic Stability” where you will find several methods of calculating the force required for this deflection. Once you have that, assume that the change of force from zero deflection to maximum deflection is a straightline function, and calculate the energy within the resulting triangular area. Come to think of it, Timeshenko will even tell you how much energy is stored in the pole, so you could check your calculation.
You need only the axial deflection of the pole, not the shape, as the shape of a uniform rod as it bends due to an axial force is a constant - in other words, all elastic rods have the same basic deflection pattern under those conditions.
I’d give you the formulas, but my copy of Timeshenko is burried in storage somewhere. I’ll bet you could find it via Google.
There! Did that help?
Not all of the vaulter’s energy comes from the energy stored in the pole, either-- Some stays as kinetic energy from the running/moving, and some comes from the vaulter’s muscles at the moment of jump. And a finite-element analysis assumes that you already know the properties of the pole. The best approach is probably to measure the maximum flex of the pole, then put the pole in a machine where you can compress it and measure the force at each distance of compression, until you get it to the amount of deformity you photographed.
One more thing. You can easily calculate the initial force required to start the buckling of the pole by using Euler’s buckling equation. This is:
Pcritical =Pi^2EI/L^2
where Pcritical is the load that first causes the pole to buckle, Pi = 3.14159, E is the modulus of elasticity of the material, I is the Moment of Inertia of the cross section of the pole, and L is the length of the pole. A lower bound of the potential energy can be found by the formula
Energy = Pcritical * dl
where dl is the axial deflection of the pole. In other words, if the force required to start the deflection is 200 pounds, and the pole deflects 2 feet, the potential energy stored in the pole is 400 ft-lbs.
As the load on the pole will need to increase to force the pole to deflect more than the initial buckling position, the energy will actually be somewhat more. For that increase you will need something like Timeshenko.
Note, in all the above, ^2 means square the number, and * represents the times symbol.
From Chronos
This is exactly right. Back in the days when I was designing fishing rods using graphite composites, I used this method to determine the modulus of elasticity of the composite. I will definitely spare you the details of this.