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The physics of falling and landing

  1. Jan 14, 2008 #1
    Hey there! I am from a community called Parkour. We are interested in mastering our environment and surroundings. One thing that we commonly face is whether or not training jumps from a height is reasonable and safe. Some people train and drill jumps from 10-20 feet. Others argue that it should never be done. To explore this, we wanted to explore the physics of the movement. I have done some preliminary calculations but I got stuck on how to convert this to something that can be compared to squatting a weight. The numbers I come up with, shown below, seem very high when compared to those of squatting a 100 kg barbell. Here is what i have so far, I hope someone here can help me...

    Equation List
    1. Force = Mass * Acceleration (F=ma)
    2. Velocity = Time * Acceleration (v = at or Vf - Vo = at)
    3. Momentum = Mass * Velocity (p = mv)
    4. Drop Distance = Initial Velocity * Time to Drop +One-Half * Acceleration * Time to Drop squared (dx = Vo*dt + (1/2)*a*dt^2)
    5. Avg Acceleration = Change in Velocity /Change in Time (a = dv/dt)
    6. Change in Velocity = Final Velocity -Initial Velocity (dv = Vf-Vo)
    7. Change in distance during compression = Average Velocity * Time to stop moving (dx = v*dt)

    For now lets assume that we are dropping like a formless ball. Cannonballing off of the top of a roof to neglect things like air resistance and the fact that your body will rotate on your center of mass.

    Let's get to the math.

    dx = Vo*dt + (1/2)*a*dt^2 --> Assume Vo is 0 since we are dropping from a height into free fall.
    dx = (1/2)*a*dt^2 -->Rearrange equation 5 to solve for dt. Plug that in to this equation..
    dx = (1/2)*a*dv^2/a^2 --> Cancel out the a's and plug in equation 6 which leaves you with this...
    dx = (1/2)*(Vf-Vo)^2/a --> Once again, Vo is 0, which leaves us with an equation where we have all known parameters and can find our final velocity such that..
    dx = 1/2*Vf^2/a --> Rearrange this to solve for Vf (so that we can solve for our momentum later)
    Vf = sqrt(2*dx*a) OR Vf = (2*dx*a)^.5 -->sqrt = square root. Now we can find out our momentum when we hit the floor as a cannon ball.

    Lets find out our velocities from different heights..

    Sample Calculation at 3 meters (~10 Feet)
    Vf = (2*3 meters * 9.8 meters/second squared)^.5 = (58.8 meters squared/seconds squared)^.5 = 7.67 meters/second.

    1 meter/~3 feet = 4.43 m/s
    1.5 meters/~5 Feet = 5.422 m/s
    3 Meters/~10 Feet = 7.67 m/s
    3.5 Meters/~11.5 feet = 8.28 m/s
    6 Meters/~20 Feet = 10.844 m/s
    6.5 Meters/~21 feet = 11.29 m/s

    Now we have some velocities that we will hit the ground with. But how does that translate into the collision/landing. Firstly, the collision will most likely be elastic, since we will not STICK to the earth, but rather bounce back away from it. We can assume that your sneakers or feet will compress by something like 1 millimeter, for calculations sake. From here we must find the time it will take our bodies to stop once we hit the floor (remember this will be pretty fast). We can do this with equation 7.

    dx = v*dt --> Solve for the time to stop moving
    dt = dx/v --> Plug in using known values.

    Sample Calculation using 3 meters/~10 feet
    dt = .001 meters/(7.67 meters/second)/2 -> we divide by 2 because we are taking the average of 0 m/s (rest) and 7.67 m/s (final velocity)
    dt = 0.00006520506637 s (6.52 * 10^-5 s or 65 microseconds) to stop moving.

    Note: 65 us = 65 microseconds

    1 meter/~3 feet = 112 us
    1.5 meters/~5 Feet = 92 us
    3 Meters/~10 Feet = 65 us
    3.5 Meters/~11.5 feet = 60 us
    6 Meters/~20 Feet = 46 us
    6.5 Meters/~21 feet = 44 us

    Now that we have these times to stop moving once we hit the ground, we need to figure out how much we need to accelerate against gravity in order to support our bodies upright from these heights.

    Using Equation 5, we can solve for our acceleration.
    a = v/t

    Sample Calculation using 3 meters/~10 feet
    a = (7.66 m/s)/65 microseconds) = 117600 m/s^2 or 12000 Gs (12000 times the acceleration due to gravity.

    1 meter/~3 feet = 39200 m/s^2 or 4000 G
    1.5 meters/~5 Feet = 58800 m/s^2 or 6000 G
    3 Meters/~10 Feet = 117600 m/s^2 or 12000 G
    3.5 Meters/~11.5 feet = 137200 m/s^2 or 14000 G
    6 Meters/~20 Feet = 235200 m/s^2 or 24000 G
    6.5 Meters/~21 feet = 254800 m/s^2 or 26000 G

    These calculations show that your acceleration will increase linearly, not exponentially. The velocity increases logarithmically, the time decreases slightly exponentially. The gravitational force increases linearly. Maybe I did something wrong...but it seems form these calculations that you can increase your jump hight linearly, or at a constant rate over time. Do note though, that even a 3 foot drop means you are going again 4000 times the acceleration due to gravity. Increasing from a 3 foot drop to a 10 foot drop increases your load threefold, according to this.

    Unfortunately, thats usually where I get stuck in calculating these things...i even asked some mechanical engineers these things and they got all befuddled too.

    From what I understand, a 100 kg (220 lb) mass, when resting on your shoulders in push press is exerting 100kg * 9.8 m/s^2 of force on your body downwards. This is 980 Newtons of force. This is really low compared to the Forces and accelerations that we examined during a fall and a 220 lb weight is no joke. To think that a 1 foot drop is harder to execute than a 220 lb squat is pretty absurd.
  2. jcsd
  3. Jan 14, 2008 #2


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    I think those g-forces are way off. The thing is, how well you land depends on how much you change your momentum over time. You will have a fixed change in momentum for each height, that you want to dissipate over the longest amount of time possible. That's why you don't stiffen your joints on landing, you make them as flexible as possible and let your muscles absorb/counteract the force.

    What you should do is calculate how much time someone can dissipate a force, you know that a smoothly done fall will result in very little impact felt (due to lots of distribution over time) and a badly executed jump will hurt because you haven't prepared your muscles for it.

    65 microseconds is far too small a time value, I think, considering that by careful movement you should be able to distribute the force over at least a second. Let's take a 1m fall, impact velocity is some 4.41m/s. If you have a person of mass, say 60kg, then their change in momentum is roughly -270kg.m/s. Over 65 microseconds, that would be 4153846N, or an obscene 7000g. Over 1s, it's barely 0.5g. A shock is usually not absorbed in just 65*10^-6s and if it were, we would be smashed into tiny little pieces.
  4. Jan 14, 2008 #3

    Ben Niehoff

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    Science Advisor
    Gold Member

    Even if you did stiffen your joints upon landing, your shoes would compress far more than a millimeter! At least a centimeter or two.

    However, in a real landing, you also bend your knees, and for higher jumps, you roll. The purpose of these is to

    1. Increase the distance over which your center of mass decelerates, and

    2. Increase the area of your body that makes contact with the ground, in order to distribute the force.

    (1) is the most important of the two. Your center of mass is about a meter high when standing...so if you flex and roll, you can give your body an entire meter over which to decelerate. This decreases the G-forces dramatically.

    (2) is partially accomplished by rolling. Anyone training in martial arts, acrobatics, or similar activity is taught to roll when they fall; the reason is that the force of impact can be spread out, and you can avoid breaking bones. Ideally, you want to tuck your head, and roll over your shoulder, thus distributing the impact over your back.
  5. Jan 14, 2008 #4
    Pretty much, from what I gather here, assuming there is no roll and its a straight down "suicide" drop, is that my main error in calculation was where I calculate the time of compression and distance of compression...is that right?

    When I compress my center of mass (around the hips) downward, we are talking about a drop of over a meter rather than a millimeter. Additionally, 65 microsec is much too short when u count that both the sneakers/feet will compress as well as the compression of the center of mass.

    In short, we are realistically looking at around a 1 second compression time over about 1 meter for a typical male. Recalculating with these changes means that the G forces will be severely reduced to under 1 G for most drops just a couple of feet high, correct?
  6. Jan 14, 2008 #5
    You should have multiplied by 2 here. The velocity while decelerating is on average
    half the maximum velocity. So the time will be 2 * distance / (final velocity).
    This means all your times are 4 times to short and your accelerations are 4 times too
  7. Jan 15, 2008 #6
    So, you're gonna land at about 20 mph. Would you step in front of a bus going 20 mph?
  8. Jan 15, 2008 #7
    There is no way these are correct. The average eye-blink is around 250ms (Google "blink your eye milliseconds")

    The time to land, bend your knees, and absorb the impact would be around this time.

    Your largest time of 112us is the time it takes for a 1200 m/s (2600mph) bullet to go about 5 inches.

  9. Jan 15, 2008 #8
    If you land heels first it hurts like hell.

    Somebody linked to some slow-motion videos on here recently. One showed the diiference between landing heels first (tremendous shock wave through the legs) and landing toes first - far more fluid. Unfortunately I can't find the link.
  10. Jan 16, 2008 #9
    Parkour is a lot more than basic physics. It involves some pretty complex biomechanics. Calculating this stuff isn't easy. It's much more simple to attach some equipment and actually measure what happens, but that is probably extremely expensive.
  11. Jan 16, 2008 #10


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    I agree.

    Doing this sort of stuff is easiest by empirical observation.

    You just can't sit down and calculate anything but a solution that'll work for a spherical human in a vacuum.
  12. Jan 16, 2008 #11
    You make me laugh.

  13. Jan 16, 2008 #12
    thanks guys, u provided me with a ton of information and a laugh or two :P
  14. Jan 16, 2008 #13
    Well I think you can get useful info from some basic maths/physics.

    20ft fall is about 6m
    and you fall at 1g

    after 20ft, (6 meters):
    In v^2 - u^2 = 2.a.s
    you get v^2 = 2.g.6= 12.g
    so v=squrt[12x9.8] = 10.84 m/s

    When you land you probably want your head to stop moving in around 1 meter so that it doesn't hit the floor.
    So in v^2-u^2=2.a.s again,
    you get -12g=2.a.1
    because u^2= 12.g
    and s=1 meter

    So a=-6g

    which is a rough figure for your head's average decelleration when you land from a 20ft jump.
  15. May 23, 2009 #14
    Sorry to reply to an old thread, but I found this with Google.

    My friend and I have come into possession of a rather accurate and small accelerometer, and want to do some experiments to find these numbers empirically.

    Any suggestions on what we should do? (where to attach the accelerometer, what types of landings to attempt, etc.)
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