Im in cali at the moment away from all my texts in florida and was wondering if anyone knew off the top of their head or could look up the equation for calculationg the force generated by a falling body at height “x” and weight “y”. i know its in supertraining so if anyone could do me a favor and post it i would be greatly appreciated.
Its been years since i have done physics so if i’m wrong I’m sorry but I’m sure Duxx can help us further…
I can’t find the equation you are talking about but if you are wondering what the relationship between hight of depth jump and energy they put into the ground on landing well it is directly proportinal to the hight of the drop because once you drop the only thing affecting you is gravity. Mass doesn’t make a difference.
velocity on impact should be = SQRT(2xgxh) if that is any use…
As for the energy well it is simply mass x g x height from surface (gravitational potential energy).
Perhaps you are asking for the Reactivity Coefficient.
In which case RC = Fmax / (tmax x W) = RFDmax / Weight (changed this due to typo before )
Fmax = max force
Tmax = time taken to reach max force
W = weight of object or body mass
To be honest, I already wrote a answer to this question, but I erase it, because I did the calculus for altitude landing (jump and stick) and I thought James was asking for Depth Jumps…
I am going to eat something in this bloody student restoran in campus, so for an 20min I will post the “equation”… Hold on
Before I start I must say that tc0710 did a error in this line
RC = Fmax / (tmax x W) = RFDmax x Weight
If RFDmax = Fmax / tmax, then RC = RFDmax / Weight
My Enoka, Siff and others are in another town so I will be writing from my head.
James, you cannot calculate the “force generation” if you don’t have force plate, because we are talking about the collision between athlete and ground, and the force developed during a time (peaks etc) is not easily calculated because it depends on various factors like shoes, ground, leg stiffnes etc.
Anyway, what you could calculate with the data you have is Potential Energy, Kinetic Energy, Linear Momentum at the ground contact, Ground Reaction Force (GRF) Impulse, and if you have ground contact time (which I doubt, because we are talking about Altitude Landings, so ground contact is “hardly” measured – you are in contact with the ground while you stand on it) you could calculate Average Force, Average Eccentric Power.
So, here we go…
Potential Energy [Joules] = m [kg] * heigh [m] * g [gravitational constant = 9,81] (eq. 1)
If we neglect air resistance and if we assume that starting velocity from where you jump is zero (it is if you drop from from box, but if you jump up or down things change a lot) we can say that, at the instant of ground contact,
Kinetic Energy = Potential Energy (eq. 2)
Because Kinetic Energy equals
Kinetic Energy [Joules] = m [kg] * velocity^2 [m/s] * 1/2 (eq. 3)
then the velocity at the ground contact is (as tc0710 wrote)
m [kg] * velocity^2 [m/s] * 1/2 = m [kg] * heigh [m] * g [gravitational constant] (eq. 4)
velocity^2 [m/s] = 2 * g [gravitational constant] * heigh [m] (eq. 5)
velocity [m/s] = SQR (2 * g [gravitational constant] * heigh [m]) (eq. 6)
From this you can calculate
Linear Momentum [kg m/s] = m [kg] * velocity [m/s] (eq. 7)
GRF Impulse should equals Linear Momentum (m*v = F * t), but you should also add a weigh of the athlete in the equation with a duration of stoping (time until the velocity is zero), but I don’t thing you got this time becasue it would ask for 3D analysis system, so try something with this equation…
Hope it helps!
thanks guys appreciate it im not looking to do a study so it doesnt matter the exact number i just need to know force in lbs expereinced when doing say and altitude drop.
As stated earlier, those forces depends on surface, landing techniques etc. Collision is very complex part of physics.
You could calculate average force, but for that you need decceletation time (time till stop)
i know there is a general equation for calculating the force in lbs of a falling body from some hight at some weight. forget the alititude drop just consider it a ball weighing “X” falling from a height “y”. i knew i should have brought those books…
even when considering athlete as a ball, it is imposible to use a formula for instantenuous force because, once again, we are talking about collisions and they are chaotical!
It is posible that average force is about 2,000 N while the force peak can easiliy be 10,000 N.
You must differentiate between instantenuous force (lim deltat-> 0) and average force (over some time)
I don’t believe that formula you search for exists… if you find it please post it, but I think you confused something…
peak force achived during the fall in lbs. im trying to figure peak muscle tension. im sure that im not confused ive seen the equation in both a hatfield book (where he was talking about impulse/inertia) and siff has a passage on the muscle tension generated by a falling body during an altitude drop and how it can be many times greater than muscle tension generated during a squat in the same postion even though for only a brief moment.
Where do you normally reside in florida?
yea central florida but im in cali for a couple months
lol why do you say that?
the equation i was looking for was power. just got back to cali and was able to check my books. power = work/time so for example if you drop a 10 lb weight 5 feet it takes .559 seconds to fall at a acceleration of 32 ft/sec^2. this works out to 89.4427 lbs of force upon impact. very useful for traiing purposes considering that jumping off a height of 3.2 m can yeild muscular force generation of over 20 times your bodyweight.
James, can you elaborate how you figured that out and give a few more examples. My knowledge of physics is lacking.
How the hell did you calculated this? From work you just jump to force… Please elabortae (in metric units). Thanks
np guys i got this from an old hatfield book. it was also briefly mentioned in supertraining under depth landings. sorry you wont be getting a metric conversion anytime soon us americans are stuck on our craptacular system.
“THE MATH BEHIND DROPPING TEN POUNDS ON YOUR TOE”
power = work/time …yes? ok good.
work = force * distance … yes? ok good.
time to fall 5 feet = .559 (sqrt((5*2)/(32ft/sec^2)))
work = 50
power = 50 /.559 = 89.4427 ft-lbs per sec
that is why the impusle inertial machine is so freaking amazing. no maladaptive stress, only adaptive stress. mmm sounds like a good buy.
I would suggest you taking a look at the book called “Life Science Physics”, and you should do it ASAP… take a look at any basical mechanic book!
If the 10kg object falls to your feet, from 1m,then
Force [N] = mass [kg] * g [9,81]
Force = 10 [kg] * 9,81 = (around) 100N
if it falls from 1m of heigh to 0m of heigh (ground or your foot), then potential energy at 1m will become kinetical energy at the bottom (if we negleck air resistance).
Kinetic Energy [Jouless] = Potential Energy [Jouless]
Kinetic Energy = deltaH [m] * mass [kg] * g [9,81]
Kinetic Energy = (1-0) [m] * 10 [kg] * 9,81 = 100 Joules
The power you calculated (Kinetic Energy / time to fall) is the AVERAGE POWER of gravity, not the power of impact!!!
The power of impact depends on the contact time… forces also depends on various stuff… you can only calculate Kinetic Energy at the impact, not the power of the impact…
you can calculate peak power and thats what i did
The peak power equals F * v, and the average equals Work/Time…
Thus, peak power of gravity at the instant of collision is
Gravity Power = F [N] * v[instant velocity at collision]
Average Gravity Power = Kinetic Energy [Jouless] / Time of falling
Play with this and you will see that they are not the same…
Anyway, this is GRAVITY POWER and NOT the power of the impact!!!