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Lumbar Spine Forces - Summary

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  • Lumbar Spine Forces - Summary

    Here are the responses along with the original posting. Thanks for the
    info. To help clear things up. The person would be walking and the
    brick was dropped on their head.


    I have been given an interesting problem to solve and was wondering what
    are the various way to go about solving it.

    A brick weighing 1 Kg falls 2 meters from a building and lands on the
    top of a mans head. What are the forces on the lumbar spine?

    Kath Bogie
    Shannon Dill

    Depends if he is standing or sitting.
    There are loads of references on biomechanics of the spine -
    of the lumbar spine.
    You could try searching Medline for them - also try;

    Schultz AB, Anderson GBJ.
    Analysis of loads on the lumbar spine.
    Spine 1981;6(1):76-82.

    Son K, Miller JAA, Schultz AB.
    The mechanical role of the trunk and lower extremities in a seated
    weight-moving task in the sagittal plane.
    J. Biomech. Eng. 1988;110:97-103.

    Dietrich M, Kedzior K, Zagrajek T.
    A biomechanical model of the human spinal system.
    Proc. Instn. Mech. Engrs. 1991;205:19-26.

    Shannon Dill

    Energy of the brick = m*g*h (mass * acceleration due to gravity *
    height fallen) = 1 * 10 * 2 = 20 J.

    This energy is transmitted to the man's head. How much is
    transmitted through the lumbar spine depends on how much is
    by permanent deformation of the hair, scalp and skull bones!! It
    depends on how much is transmitted through the abdomen and
    at the level of the lumbar spine.

    "Kenneth St. John"

    I would think you would use the acceleration of gravity on the brick to
    determine the velocity of the brick when it hits the man's head. You
    would then assume a distance that the man's head would be dented by the
    impact and use that and the velocity to determine the time period over
    which the brick decelerates. F=ma This would be the force on the
    cervical spine. After that, it would take quite a few assumptions to
    determine the force on the lumbar spine.

    Is this a homework assignment?

    Kenneth R. St. John Phone: 601-984-6199
    Assistant Professor Fax: 601-984-6087
    Orthopaedic Surgery and Rehabilitation
    University of Mississippi Medical Center
    2500 North State Street
    Jackson, MS 39216-4505

    George Davey

    This is actually a trick question. If we assume total elastic
    collisions you
    can get an answer but it will not be right. To obtain the correct
    you would have to determine the amount of force absorbed by the
    tissue surrounding each vertebre and the forces that are translated to a
    different axis. This is not equal to zero and will vary proportional to
    If you assume elastic collisions between vertebre then the force on each
    vertebra would be the same and would be equal to the weight of the
    brick 9.81N plus the force required to stop the brick which would vary
    according to the stopping distance (this would be the amount the body
    compresses in response to the brick stricking)
    As you can see this is a very complex problem and can only be solved if
    you know these biomechanical variables.
    As you know a vehicle airbag presents a similar problem when
    calculating the forces on the brain and in fact this is why these forces
    cannot be accurately estimated and must be determined by crash tests.
    Such variables as airbag deflation rate determine the stopping distance
    of the brain which is what ultimately determines whether you live or die
    in a car crash.
    One way to esimate this force is to drop a brick on someone's head from
    a smaller distance to estimate the lateral force absorbtion involved in
    stopping the brick which will determine the stopping distance and will
    affect the forces on the spine. I would guess the stopping distance is
    not small as the brick would break and I don't think the brick would
    If you have a forceplate you could drop the brick from 0.1 meters onto
    someone's head and measure the ground reaction forces of the feet (or
    butt is they are sitting down) . some of the initial vertical force of
    brick will be tranlated to a medial/latteral orforward/backward force.
    This experiment will tell you what percentage of the force is expended
    some other direction other than vertical. You can assume that nearly
    the vertical force component will be transfered through all the lumbar
    vertibre equally. This is an estimate that will eliminate the need for
    calculating a stopping distance because if you know the resulting force
    then the stopping distance can be calculated backwards. As the
    stopping distance decreases the force on the brick and also the body
    will increase.
    The experiment described here would only be a estimate for short brick
    drops. As you can see if the brick was dropped from 10 meters this
    experiment would not apply because you would have more lateral force
    absorbtion involved with the skull fracturing, etc. Also at some point
    neck would flex and this would disipate a lot of the energy laterally.
    Good luck and be careful if you decide to drop bricks on people's heads.
    You may even want to start at 0.05 meters. Once you collect the
    background data it should be good for all brick drops where no skull
    fracturing or significant neck flection occours.
    Just guessing I would say that the forces on the lumbar are going to be
    85% of what is seen by the atlas/axis bones but it might be closer to
    100% because otherwise African women would not be able to carry
    100 pound loads of water on their head without injuring their necks but
    maybe if they relaxed they would injure their necks.

    George Davey

    Paul Bourassa
    Shannon Dill

    the brick will reach a final velocity of sqrare root of
    6.3 m/s. The linear momentum would be 1kg*6.3m/s=6.2Kgm/s = 6.3 N.s.
    Assume that the force versus time has the shape of a half sine or else
    shape of a triangle. If the force rise to a maximum within 0.01s and
    back to zero in another 0.01s, the maximum force would then be 620N.
    check area of triangle = 0.5*Fmax*time = 0.5*620N*.02s = 620*.01=6.2Ns
    In this approximation g (9.81m/s2) was replaced by 10 m/s2. The largest
    approximation comes from the contact time and the shape of the impulse.
    Personnaly, I think that the damage to the brain would be much much more
    important than any posssible damage to the spine.


    From: (Garry Allison)
    Shannon Dill

    That would make an interesting PhD or two.. I suggest go onto the
    and ask for verification with something like...

    Has the person got their knees hyper extended?



    "M Swanepoel"

    Wow Shannon! This sounds like quite a tricky one, probably referred
    to you by a lawyer working on a building site case!? The best approach
    would be
    experimental - ideally one would drop a brick on the head of a cadaver
    propped up in a standing position, and have a piezoelectric force
    plate under the feet of the body.

    A feeble second best would be to calculate the energy stored in the
    brick, and then estimate a suitable stopping distance for the brick -
    a millimetre or two perhaps. Equate the work performed in stopping the
    motion of the brick, to its original potential energy, and you'll have
    an order of magnitude answer. Sadly the exact deceleration of the
    brick, and whether parts of it continue to conserve some of their
    kinetic energy after impact, mean that this is at best a guesstimate.

    The manner in which the impact occurs is critical. If a sharp corner
    of the brick cracked the skull, then deceleration would be more
    gradual than a "flat sided" impact - hence the lumbar spine would
    experience lower force. In general I would expect the severity of
    the head injury to be inversely related to the force experienced by
    the spine.

    Mark W Swanepoel
    School of Mechanical Engineering
    University of the Witwatersrand

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