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Shannon Dill
07-21-1998, 05:58 AM
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.

Shannon

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?

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From:
Kath Bogie
To:
Shannon Dill


Depends if he is standing or sitting.
There are loads of references on biomechanics of the spine -
particularly
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.

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From:
Andrew_Pinder@hsl.gov.uk
To:
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
absorbed
by permanent deformation of the hair, scalp and skull bones!! It
also
depends on how much is transmitted through the abdomen and
musculature
at the level of the lumbar spine.

Andrew_Pinder@hsl.gov.uk
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From:
"Kenneth St. John"
To:
sm.dill@baylordallas.edu


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
E-mail: kstjohn@sod.umsmed.edu
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From:
George Davey
To:
sm.dill@baylordallas.edu


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
answer
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
the
force.
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
break.
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
the
brick will be tranlated to a medial/latteral orforward/backward force.
This experiment will tell you what percentage of the force is expended
in
some other direction other than vertical. You can assume that nearly
all
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
the
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
gdavey@stu.uomhs.edu
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From:
Paul Bourassa
To:
Shannon Dill


the brick will reach a final velocity of sqrare root of
2*g*h=sqr(2*10*2)=
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
the
shape of a triangle. If the force rise to a maximum within 0.01s and
falls
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.
Paul

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From:
iallison@info.curtin.edu.au (Garry Allison)
To:
Shannon Dill

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

Has the person got their knees hyper extended?

cheers
GTA.

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From:
"M Swanepoel"
To:
sm.dill@baylordallas.edu


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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