Technically, Power is equal to the integral of Fdv. If force is constant, then P=Fv. However, if the force is not constant, then the answer will be different. Since impact is not instantaneous but rather best modeled by a delta function (I think…) where t does approach 0, the fact that a is not constant could affect the equation and thus power absorption and production.
Additionally, since an athlete does not “bounce” of the ground, but rather absorbs the impact and will spend as much as a full second on the ground, the bands might continue to provide some downward force while the athlete is on the ground.
Again, I’m just throwing out some half-assed theory. I really don’t have any practical experience to draw from.
I think the best experiment to verify this would be to measure time spent on the ground by an athlete. Since Power is the change in energy with respect to time (rate of change of energy), the amount of kinetic energy at impact should be the same. That is, put an athlete on a box with bands and then on a higher box without bands so that the intial potential energies are the same. If the athlete spends less time on the ground with the AMT setup, delta t will be less and therefore power will be greater. Producing a greater power absorption stimulus with the same intitial kinetic energy is the whole point of doing rate work, correct?
Any thoughts on that?
[quote]Donzi wrote:
jtrinsey wrote:
To keep with the calculus comments, I’ll throw something out there that could be an added reason for AMT jumps. If acceleration is constant (as with the higher box jumps), than the rate of change of acceleration (3rd deriv) = 0. With bands the acceleration is not constant, so the rate of acceleration is different.
So, during the AMT jump, acceleration is greater than gravity.
Since maximal force (F=ma) is different than maximal power output,
P= F*V(at impact)
P= ma*V
Whether you use additional acceleration or you use additional height, power at impact is equivalent. If the bands are released at impact then a=gravity at impact and the only other variable is velocity. If velocities are the same then power should be as well.
perhaps having a different rate of change of acceleration might have an effect on power absorption and thus power production?
Thats possible as far as the CNS is concerned but externally there would be no difference in power at impact.
I think its a good observation about the possible effects of +g’s on power output and training effects but it needs to be pinned down further.
Bruce Lee’s 1" punch is very powerful. He is able to generate tremendous acceleration for his fist which results in a high terminal velocity. He also is able to continue delivering power because the mass of his fist is too small for him to generate max power but upon impact he is accelerating a greater load with smaller joint angles and is able to administer additional power.
In a depth jump, your bodyweight is your load and joint angle is equivalent between AMT and regular drop. Typically, max power is thought to be generated between roughly 50-70% of a 1RM load for a given movement. Meaning that F and V are maximized in this load range. An athlete may be able to generate more power if extra load is utilized, ie: weighted vest(mass), bands are used(acceleration), or a taller drop height (velocity) since P=FV=(MA)V
As I understand it though, in depth jumps its more a matter of muscle stiffness allowing elastic tendencies of tendons and connective tissue to translate downward force into upward force rather than muscle actions ability to accelerate or decellerate a load.
Just some thoughts…
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