Friday, November 23, 2012

Joe Theismann vs Lawrence Taylor: Fibula goes Snap


By Alex Girden

(Warning- not for the squeamish)

Basic info:
Lawrence Taylor=245 lbs=111 kg
Gary Reasons=234lb=106 kg
Joe Theismann=192 lbs=87 kg

Young’s Modulus of bone=15X10^9 N/m^2

What change in length occurred in the fibula of Joe Theismann?
F=E*(deltaL/L)*A

Force we will assume for now is just mtotal*g
Force= (111kg+106kg)*9.8=2120N
A=3.6X10^-4 m^2
Maximum compressive strength=170X10^6 N/m^2
Average length of fibula=40.50cm=.4050m
Change in Length of bone if would not snap= F=E*(deltaL/L)*A.
Delta L=1.6X10^-4m.

Force/Area=5.9X10^6 N/m^2.  Thus, our bone should not have snapped unless Lawrence Taylor added extra compressive force besides his weight.

170X10^6N/m^2*3.6X10^-4m^2=61,200 N.  -2120 N=59,100 N.  It appears that I need more information about this as there is no way Lawrence Taylor supplied that much force to Joe Theismann’s leg.  I believe we should be looking at sheer strength and not compressive strength, but there is no value for shear strength of the bone.  According to this result,
LT would have needed to add a minimum of 59,100 N to break Joe Theismann’s leg. 

Calculate the torque applied to Theismann’s leg.
T=Frsintheta.  Theta appeared to be 90 degrees, sin90=1.
Assume force was applied halfway from axis of rotation or hinge of fibula.L/2
Torque if just using weight=2120N*(.405/2)=5230N*m
Torque if use maximum force needed to break leg calculated from above:

T=61,200N*(.405/2)=12,400 N*m. 

Ultimately, a specific torque was applied to Joe Theismann’s leg as seen by the awkward angle in which the quarterback’s leg is caught underneath his body.  The weight of the linebackers creating a sheer or compressive force combined with the torque from the awkward rotation resulted in the infamous snap of Joe Theismann’s leg causing his fibula to break through his skin as seen in the video.  This injury ultimately ended his career. 

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