IRJET- Analytic Evaluation of the Head Injury Criterion (HIC) within the Framework of Constrained Optimization Theory

International Research JournalofEngineeringand Technology (IRJET) e-ISSN: 2395-0056
Volume: 07 Issue: 01 | Jan 2020 www.irjet.net p-ISSN: 2395-0072
© 2020, IRJET | Impact Factor value: 7.34 | ISO 9001:2008 Certified Journal | Page 2052
Analytic Evaluation of the Head Injury Criterion (HIC) within the
Framework of Constrained Optimization Theory
Jai Singh1
1Biomechanical Engineering Analysis & Research, Inc., 2060-D Avenida de los Arboles, No. 487, Thousand Oaks,
California, United States of America
---------------------------------------------------------------------***----------------------------------------------------------------------
Abstract - The Head Injury Criterion (HIC) is a weighted
impulse function, based upon the measured or calculated
resultant acceleration, a(t), at the head center of mass, which
predicts the probability of closed head injury when coupled
with an appropriate statistical model. Previous published
work has elucidated a number of important findings in
regards to determining the temporal segment, over the
operative domain of the resultant acceleration, which
maximizes the HIC when a(t) is an analytic closed-form
function and when there is no constraint on the length of the
temporal segment. Current formulations of the HIC, however,
are predicated upon using a temporal windowing function
that delimits the maximum durationoverwhichthefunctionis
calculated. Presented in the subject work, within the
framework of constrained optimization theory, are the
solutions for determining the initiating andterminusvaluesof
the windowing function, for internal domain points, for the
cases in which a(t) is defined by a single function over its
domain and when any internal domain point is based upon
two functions with differing analytic closed-form
representations. The general solutions are crystalized by
considering example application cases.
Key Words: Biomechanics, Closed head injury, Brain
injury, Head Injury Criterion, Constrained optimization
1. INTRODUCTION
The Head Injury Criterion (HIC) has been the primary
metric employed by United States (US) National Highway
Traffic Safety Administration (NHTSA), in regards to
predicting the probability of closed head injury, for
approximately half a century. The historical development of
the HIC traces to the severity index (SI), which in turn,
represents a log-linear fit to the Wayne State Tolerance
Curve (WSTC). The shortcomings of the WSTC, which was
developed for frontal headimpacts[1-6]havebeenreviewed
elsewhere [7-8]. The underlying issues with the SI [9] were
detailed in the initial development of the HIC [10]. Despite
the historical developmental shortcomings of the HIC, the
current instantiation of the same has been shown to
correlate well with the development, or lack thereof, of
concussive head injury [11]. Mathematically, this current
instantiation is expressed as the following:
   
2
1
2.5
t
2 1
2 1 t
max
1
HIC t t a t dt
t t
  
    
    
 (1)
In equation (1), a(t) is the resultant head center of mass
translational acceleration (i.e. the magnitude of the triaxial
translational acceleration at the center of mass of the head).
The temporal values of t = t1 and t = t2 {t1, t2: t1 < t2} are
within the temporal domain of a(t). The function is
maximized when t1 is equal to some value, 1, and t2 is equal
to some value, 2, within the domain of a(t) and with {1, 2:
1 < 2}. The function a(t) is expressed in terms of Gs
whereas the temporal variables are expressed in terms of
seconds. The current instantiation of the HIC also involvesa
limitation on themaximumdurationover whichequation (1)
is calculated. In other words, denoting d as d = t2 – t1, the
value of d is such that d ≤ dmax. Notations such as HIC15 or
HIC36 explicitly denote the maximum length of the temporal
window, in terms of milliseconds, in the subscript. These
formulations are referred to as clipped HIC formulations.
Chou an Nyquist, shortly after the introduction of the HIC
to the biomechanical engineering corpus, presented a set of
conclusions based upon the analytic evaluation of the
unclipped function [12]. These conclusions included the
finding that the critical values of {t1, t2},denotedaboveas{1,
2}, for the nontrivial case of 1 ≠ 2, occurred where a(1) =
a(2) and that the average acceleration of a(t) between the
two critical points was equal to fivethirdsoftheacceleration
at either critical point. The authors also presented closed-
form analytic solutions for the cases in which a(t) was
modeled using a half sine, triangular, trapezoidal or square
pulse shape. The unclipped nature of the formulations
developed by those authors can readily be shown by
example. For the case of a(t) modeled using a half sine pulse
shape, initiating at t = 0 and terminating at t = T, the authors
determined that the critical values were 1 = 0.1651T and 2
= 0.8349T. The critical duration,beingthedifferenceofthese
two values, is 0.6698T. The dependence on T directly
indicates that the evaluation was unclipped.
To the knowledge of the subject author, the analytic
evaluation of the clipped HIC function has not been
addressed within the scientific literature. The objective of
the author, by means of the subject work,isthedevelopment
of the analytic relationships for the case of the clipped HIC
function and the presentation of the results obtained by

applying these relationships to certain closed form pulse
shapes.
2. THEORY
2.1 Constrained optimization
The formal requirements for optimizing a function f(x),
subject to the equality constraints hi(x) = 0 {i: i = 1, 2, …, p}
and inequality constraints gj(x) {j: j = 1, 2, …, q} are
ensconced in the first order necessary and second order
sufficient Karush-Kuhn-Tucker (KKT) conditions [13]. The
constraint relationships are appended to f(x) by means of
Lagrange multipliers. The resulting Lagrangian function is:
        
p q
2
i i j j j
i 1 j 1
, , , f h g s
 
      x s x x xL   (2)
Where s is a vector of slack variables such that:
  2
j jg s 0 j  x (3)
For the case in which the optimization is a maximization
problem, the inequality constraints gj(x)areintheformgj(x)
≥ 0 j. For the maximization problem, when the jth
inequality constraint is active, sj
2 = 0 and j ≥ 0. When the jth
inequality constraint is inactive, j = 0 and sj
2 ≤ 0. The
solution vector x =  that maximizes f(x) subject to h(x) and
g(x), occurs when the derivative of the Lagrangian with
respect to all of its independent variables is zero-valued.
The first order necessary KKT optimality conditions consist
of the four, coupled, vector equations that express this first
derivative test.
 
     
p q
i i j j
i 1 j 1
, , ,
f h g 0
 

        

 
x s
x x x
x
L  
(4)
 
  2
j j
, , ,
g s 0 j

   

x s
x
L  

(5)
 
 i
, , ,
h 0 i

  

x s
x
L  

(6)
 
j j
, , ,
2 s 0 j

   

x s
s
L  
(7)
Equation (7) contains the switching conditions (i.e.
complementary slackness). The secondordersufficient KKT
conditions, corresponding to the second derivative test, can
be expressed as the following:
   
 
p
2 2
i i
T ni 1
2
j j
j active
f h
0 ,
g


 
     
     
   
 


z z z z 0
 

(8)
Where the vector z denotes any non-zero real valued
vector and 2 denotes the Hessian of its operand. It is
important to note that equation (8) is evaluated at the KKT
points (i.e. at x = ) and only for those inequality constraints
that are active. This equation is valid if the parenthetical isa
positive definite matrix. The feasiblechanges, z,mustsatisfy
the following gradient conditions for the constraints.
   
T T
i jh 0 g 0   z z  (9)
2.2 Clipped HIC as a constrained problem
Equation (1) can be rewritten by first separating the two
multiplicative terms that are exponentiatedtothe2.5power
and reducing the resultant two exponentiated terms
involving t2 – t1. Furthermore, the term t1 is replaced with ta
and the term t2 is replaced with ta + d, where d is the
temporal length of the windowing function. The resultant is
the following:
       
a
a
2.5
t d
1.5
a a
t
f f t ,d HIC t ,d d a t dt


 
    
 
 
x (10)
For the case in which the resultant acceleration, a(t), is
either an analytic closed-form function, or modeled as such,
it is useful to define N + 1 {N: N , N ≥ 1} transition points.
These points, collectively, represent the points at which the
known resultant acceleration undergoes a salientchange. In
the simplest case, that being one in which a(t) is monotonic
function, N = 1. The first transition point occurs at t = t0 and
with corresponding resultant acceleration a(t0) = a0, which
represents the start of the function. The second transition
point occurs at t = t1 and with corresponding resultant
acceleration a(t1) = a1, which represents the end of the
function. When a(t) is either a composite function of two
temporally segmented functions or when the form of the
function is explicitly based on a single value internal to the
domain, N = 2. The internal point for suchcasesisdenoted as
{t1, a(t1)} and the terminus point is denoted as {t2, a(t2)}. It
can thus readily be seen that for N ≥ 1, the resultant
acceleration function contains N - 1 internal transition
points. There is no upper limit on the number of transition
points, however, as N becomes arbitrarily large, theutilityof
using closed form analytic solutions decreases when
compared with simple numerical evaluation.
Equation (10) contains a number of implicit constraints,
which do not require reiteration in the formoftheinequality
constraints found within g(x). The first implicit constraint
arises from the fact that a(t) is the resultant acceleration.
Therefore, a(t) ≥ 0 t {t: t0 ≤ t ≤ tN}. The second implicit
constraint arises from the ordering of the limits of

integration. Because ta + d > ta, d > 0 and d-1.5 > 0. When
coupled with the first implicit constraint, the exponentiated
integral term in equation (10) is either positive or zero
valued t. As a consequence of these implicit constraints,
HIC(ta, d) ≥ 0 over the temporal domain.
In regards to the explicit constraints, thereare noequality
constraints and therefore h(x) is zero-valued and can be
dropped from the formulation. Three inequality constraints
are considered herein. The first is the relationship between
the temporal length of the windowing function that clips the
HIC function, with respect to a maximal specified length
(dmax). The inequality of dmax – d ≥ 0 leads to the following:
   2 2
1 a 1 max 1g t ,d s d d s 0     (11)
When this constraint is active, s1
2 = 0, 1 ≥ 0 and d = dmax.
When this constraint is inactive, s1
2 ≤ 0, 1 = 0 and the value
of d is unconstrained. The second constraint relatesthefirst
critical point to the first transition point. The inequality ofta
– t0 ≥ 0 leads to the following:
   2 2
2 a 2 a 0 2g t ,d s t t s 0     (12)
2 = 0, 2 ≥ 0 and ta = t0.
2 ≤ 0, 2 = 0. Even with ta
being unconstrained with respect to t0 when this constraint
is not active, solutions that involve ta < t0 can be excluded on
an implicit basis. The third constraint relates the second
critical point to the last transition point. The inequality of tN
– ta – d ≥ 0 leads to the following:
   2 2
3 a 3 N a 3g t ,d s t t d s 0      (13)
2 = 0, 3 ≥ 0 and ta + d = tN.
2 ≤ 0, 3 = 0. Even with ta
+ d being unconstrained with respect to tN when this
constraint is not active, solutions that involve tN < ta + d can
be excluded on an implicit basis.
There are a number of salient points in regards to the
subject formulation of the inequality constraints that
comprise g(x). Generally, for a maximization or
minimization problem, the solution approach requires
checking the boundary points (i.e. t = t0 and t = tN) in
addition to checking the internal domain (i.e. t0 < t < tN) for
the value(s) of the domain that optimize the function. When
N = 1, checking the first boundary point is achieved by
setting the second constraint as active and checking the
second boundary point is achieved by setting the third
constraint as active. The case when both the second and
third constraints are active is only valid, for N = 1, when t1 –
t0 = t10 = d. The determination of the necessity of
performing such a check can be determined a priori since
a(t) is considered as known. Secondly, whenitisknownthat
t10 < dmax, the first constraint can be rewritten by replacing
dmax with t10. Thirdly, when thefirstandsecondconstraints
are active for N = 1, ta = t0 and d = dmax. This is a check for the
first boundary point and with the clipping constraint set to
active. Similarly, when the first and third constraints are
active for N = 1, ta = tN – dmax and d = dmax. This is a check for
the second boundary point and with the clipping constraint
set to active. Finally, the number of constraints, generally,
when N > 1, must be increased in number. As an example,
consider the case of N = 2. The resultant acceleration, a(t),is
comprised of a function a1(t) that is active over the domain
{t: t0 ≤ t ≤ t1} and a second function a2(t) that is active over
the domain {t: t1 ≤ t ≤ t2}. Each segment, again, generally,
requires checking both boundary points and the internal
domain. The check of t = t1 as a second critical point solely
involves a1(t) whereas checking the same point as a first
critical point solely involves a2 (t). These checks involve the
inclusion of additional constraints in the form of the second
and third constraint. Additionally, one must check for the
case where ta < t1 < ta + d. Therefore, for N = 2, there are
seven checks involved. Generalizing this for N > 1, there are
4N – 1 checks involved. Furthermore, on a simple
combinatorial basis, there exist 23 = 8 combinations of the
three inequality constraints, per segment. The total number
of salient combinations can readily be reduced since the
form of a(t) is known.
2.3 Partial derivatives of the HIC function
The implementation of the KKT conditions requiresbeing
able to differentiate the HIC function. It is instructive to first
consider the case where a(t) is a single function. To simplify
the form of the solutions, the shorthand notation of a(ta) =aa
and a(ta + d) = ad are employed. Also, the difference ad – aa is
denoted as ada. Finally, the integration of a(t) over t = ta
and t = ta + d is denoted as:
   
a
a
t d
1
t
a t dt f t

 (14)
Taking the partial derivative of the HIC function with
respect to either ta or d is complicated by the fact that ta
appears in both the lower and upper limits of integration
and that d appears in the upper limit of integration.
Differentiation under the integration is employed by using
Leibniz’s rule. The context specific keys to implementing
Leibniz’s rule are:
   1 da 1 d
a
f t a f t a
t d
 
  
 
(15)
The partial derivatives of the HIC function with respect to
ta and d can therefore be written as:
 
  
1.5a 1.5
da 1
a
HIC t ,d
2.5d a f t
t

 

(16)

 
  
  
1.5 da 1.5
1
1
1
5
aHIC t ,d
31.5d f t
d
d f t


 
  
    
 
(17)
The four second partial derivatives oftheHICfunctionare:
 
  
 da
2 1
0.5a 1.5 a
12
a 2
da
a
f t
HIC t ,d 5 t
d f t
2t 3
a
2

 
   
  
 
 
(18)
 
  
 
 
da
12
0.5a 1.5
1
d
a
da 1
1
a
f t
dHIC t ,d 5
d f t a3t d 2
a
d f t2


 
       
     
  
(19)
 
  
 
 
d
12
a0.5a 1.5
1
da
da 1
1
a5
f t
3 tHIC t ,d 3
d f t
ad t 2 5
a
d f t2


 
       
     
  
(20)
 
  
 
 
d
12
0.5a 1.5
212
d
1
1
a
f t
dHIC t ,d 5
d f t
a2d 3
d f t2


 
   
  
    
  
(21)
When N ≥ 1, there exist N – 1 temporal domain delimited
acceleration functions. The evaluation of KKT points within
each acceleration function, inclusive of the boundary points
of the same, involves the first and second order partial
derivatives of the forms presented above. Additionally, as
noted previously, a check is required across each internal
transition point. The evaluation across the kth internal
boundary point {k: 1 ≤ k ≤ N – 1} requires a check across the
kth and k+1st acceleration functions. The former is denoted
as ak(t) and the latter is denoted as am(t). The HIC function
can then be defined as:
 
 
 
k
a
a
k
2.5t
k
t1.5
a t d
m
t
a t dt
HIC t ,d d
a t dt


 
 
 
  
 
 
 


(22)
The following notation is used to simplify the
presentation: ak(ta) = aka, am(ta + d) = amd and amd – aka =
amdka. Following the form of equation(14),the presentation
may be simplified by the introduction of:
     
k a
a k
t t d
k m km
t t
a t dt a t dt f t

   (23)
The partial derivatives of fkm(t) with respect tota andwith
respect to d can therefore be written as:
     km mdka km md
a
f t a f t a
t d
 
  
 
(24)
The first partial derivatives of the HIC function for this
case are:
 
  
1.5a 1.5
mdka km
a
HIC t ,d 5
a d f t
t 2

 

(25)
 
  
 
md
1.5a 1.5
km
1
km
5
a
HIC t ,d 2d f t
3d
d f t
2


 
 
  
  
 
 
(26)
The second partial derivatives of theHICfunctionthiscase
are:
 
  
 
2
2 mdka
0.5a 1.5
km2
mdkaa
km
a
3
a
HIC t ,d 5 2
d f t
a2t f t
t

 
  
 
  
  
(27)
 
  
 
  
2
0.5a 1.5
km
a
mdka
km
md
0.5mdka 1
km
HIC t ,d 5
d f t
t d 2
a
f t
d
a3
a
2 d f t



 
 
 
  
  
  
  
  
(28)
 
  
 
 
0.5a 1.5
km
a
md
km
a
md
mdka 1
km
HIC t ,d 5
d f t
d t 2
a
f t
t
a3
a
d f t2



 
 
 
  
  
     
  
(29)
 
  
 
 
md
km2
0.5a 1.5
2km2
md
1
km
a
f t
dHIC t ,d 5
d f t
a2d 3
d f t2


 
   
  
    
  
(30)

2.4 First order KKT conditions for the HIC
For the case in which a(t) is defined by a single analytic
function, the Lagrangian function as per equation (2) is:
      
3
2
a a j j a j
j 1
t ,d, , HIC t ,d g t ,d s

   sL  (31)
The coupled vector equations for the first order KKT
conditions are:
 
  
1.5a 1.5
da 1 2 3
a
t ,d, ,
2.5d a f t 0
t

    

sL 
(32)
 
  
  
1.5a 1.5
1
d
1 3
1
1
t ,d, ,
1.5d f t
d
5
a
3 0
d f t



 

 
 
   
 
 
sL 
(33)
2
max 1
2
a 0 2
2
N a 3
d d s 0
t t s 0
t t d s 0
  
  
   
(34)
1 1 2 2 3 32 s 0 2 s 0 2 s 0      (35)
For the case in which a(t) is comprised of N-1 {N: N > 1}
temporal domain delimited acceleration functions, the first
order KKT equations for each function involves evaluating
equations (32-35) as written for the kth acceleration
function. In this regard, the term k is added to each
subscript in the definitions and equations presented for the
case of a = a(t). For the evaluation across the kth transition
point, the second and third constraint conditions can be
dropped. The resulting Lagrangian is:
      2
a a 1 1 a 1t ,d, , HIC t ,d g t ,d s  sL  (36)
The four coupled equations for the first order KKT
conditions are:
 
  
1.5a 1.5
mdka km
a
t ,d, ,
2.5 a d f t 0
t

  

sL 
(37)
 
  
 
1.5a 1.5
km
md
1
1
km
t ,d, ,
1.5d f t
d
5
a
3 0
d f t



 

 
    
 
 
sL 
(38)
2
max 1d d s 0   (39)
1 12 s 0  (40)
2.5 Second order KKT conditions for the HIC
The Hessian in equation (8), for the HIC function, is of the
form:
 
   
   
2 2
a a
2
aa2
a 2 2
a a
2
a
HIC , HIC ,
t dt
HIC ,
HIC , HIC ,
d t d
      
 
     
      
 
    
(41)
Where a and  are the KKT point values of ta and d,
respectively. The first order gradients of the inequality
constraints are:
   
 
1 a 2 a
3 a
0 1
g , g ,
1 0
1
g ,
1
   
             
 
      
(42)
The Hessian for each of the inequality constraints is
[0]2x2. The second order gradients of the inequality
constraints are all zero valued. Equation (8), therefore,
reduces to:
 T 2 n
aHIC , 0 ,      z z z z 0 (43)
Expanding this equation leads to the following result:
 
   
 
2
a2
1 2
a
2 2
a a
1 2
a a
2
a2
2 2
HIC ,
z
t
HIC , HIC ,
z z
d t t d
HIC ,
z 0
d
  


      
       
  


(44)
When only the first inequality constraint is active, z = {z1,
0}T. When only the second inequality constraintisactive, z=
{0, z2}T. The first case results in the reduction of equation
(44) solely to the quadratic term in z1
2. The second case
results in the reduction of equation (44) solely to the
quadratic term in z2
2. When only the third inequality is
active, z = {z1, -z1}. For the case in which the HIC function is
based upon a single acceleration function, equation (44)
becomes:

  
 
  
2da
1
a
da d
1 1 2
a0.51.5
1
2d
2
2
da 1
1
d 1 2
a
z
t
a a
f t z z
d t5
d f t 0a2 z
d
a z3
a d f t z2


   
   
   
            
       
  
 
   
     
(45)
For the case in which the HIC is a function of two resultant
acceleration functions, equation (44) becomes:
  
 
  
2mdka
1
a
mdka
km 1 2
md
0.51.5
km a
2md
2
2
mdka 1
1
md km 2
a
z
t
a
d
f t z z
a
5
d f t 0t
2
a
z
d
a z3
a d f t z2


   
    
   
        
   
       
  
  
  
 
   
      
(46)
In both of these equations, the term that multiplies the
parenthetical is positive. Inside of each parenthetical, the
second additive term is positive given that it is a squared
term. When either z1 (second inequality constraint as the
sole active constraint) or z2 (first inequalityconstraintasthe
sole active constraint) arezerovalued,thefirstadditiveterm
in the parenthetical is positive for any real non-zero valued
z2 or z1, respectively.
3. APPLICATIONS
3.1 HIC based on singular a(t)
The coupled vector equations found under equations (32-
35) are the operative first order KKT equations for the case
when the HIC function is based upon a single acceleration
function a(t). As noted previously, there are eight cases that
are generated by the two unknowns and thethreeinequality
constraints. These eight cases can be evaluated for any
singular function a(t) without further specifying the form of
a(t) in an a priori manner. These cases along with the
solutions generated from the switching conditions, are
listed in Table 1.
Table -1: Eight constrained cases for singular a(t)
Case
Active gj(ta, d) Solutions from
sj
2
1 2 3 ta d
1
2 X dmax
3 X t0
4 X tN - ta
5 X X t0 dmax
6 X X tN -
dmax
dmax
7 X X t0 tN – t0
8 X X X t0 dmax
The second case, in Table 1, is of specific interest. The
detailed derivation for this case is presented below and the
results for all eight cases are shown in Table 2. For the
second case, in which only the first constraint is active, s1
2 =
0, 2 = 0 and 3 = 0. From the first equation under (34), d =
dmax. The second and third equations under (34) can be
directly solved, respectively, for s2
2 and s3
2.
2 2
2 0 a 3 a max Ns t t s t d t     (47)
A valid solution for this problem is one for which s2
2 ≤ 0
and s3
2 ≤ 0. The remaining unknowns are ta and 1.
Substitution of 2 = 3 = 0 and d = dmax into equation (33)
and solving for 1 leads to the following solution.
     
1.51.5 1
1 max 1 d max 1
5
1.5d f t a d f t
3
  
   
 
(48)
Because dmax is implicitly greater than zero and the
integral that defines f1(t) is greater than zero, the
parenthetical must be non-negative. Such a case occurs
when the following holds:
 
a max
a
t d
1
d max
t
3
a d a t dt
5


  (49)
For these equations, ad is equal to a(ta + dmax).
Substitution of 2 = 3 = 0 and d = dmax into equation (32)
leads to the following result:
  
1.51.5
da 12.5d a f t 0
  (50)
Because d and f1(t) are non-zero valued, solution to this
equation requires ada to be zero valued.
   a max d a aa t d a a a t    (51)
Therefore, for this case, the solution for ta is that which
results in aa being equal to ad and for which either value is

greater than or equal to three fifths of the average
acceleration between aa and ad.
Table -2: Solutions from i for singular a(t)
Case Solutions and requirements from i
1
d aa a and  
a
a
t d
1
a
t
3
a d a t dt
5


 
2
d aa a and  
a max
a
t d
1
d max
t
3
a d a t dt
5


 
3
d aa a and  
0
0
t d
1
d
t
3
a d a t dt
5


 
4
d Na a and    
N
a
t
1
a N a
t
3
a t t a t dt
5

  
5
d 0a a and  
0 max
0
t d
1
d max
t
3
a d a t dt
5


 
6
d aa a and  
N
N max
t
1
a max
t d
3
a d a t dt
5


 
7
   
N
0
t
1
a N 0
t
3
a t t a t dt
5

   and
   
N
0
t
1
d N 0
t
3
a t t a t dt
5

  
8 max N 0d t t 
3.2 HIC based on multiple a(t)
This section focuses on the evaluation of those situations
in which the form of the underlying acceleration functions
changes at an internal transitionpointandspecificallyonthe
evaluation of the case where ta is within the domain of ak(t)
and ta + d is in the domain of am(t) where m = k +1.
Generally, this evaluation is required for all internal
transition points k and m. As notedbefore,theseevaluations
are in addition to the evaluations that are done within the
domain of each function. Only the first constraint is
considered, herein, as potentially being active. When this
constraint is inactive, 1 = 0 and by equation (39), s1
2 = d –
dmax. A valid solution is one where s1
2 ≤ 0. From equation
(37), for the equation to be zero-valued, the term amdka
must be zero valued. Therefore:
   mdka m a max k aa 0 a t d a t     (52)
Because 1 is equal to zero, the equating of equations (37)
and (38) results in the following:
    
1.51.5 1
km ka km
5 3
d f t a d f t 0
2 2
  
  
 
(53)
For this equation to hold, the parenthetical must be zero-
valued. This results in the following solution.
     
k a
a k
t t d
1
k a k m
t t
3
a t d a t dt a t dt
5


 
  
 
 
  (54)
Therefore, for the unconstrained case, the solution that
maximizes the HIC function across the internal transition
point, tk, occurs at the values of {ta, d} where theacceleration
ak(ta) is equal to am(ta + d) and where the acceleration at
each point is equal to three fifths of the average acceleration
between the two points. This was the solution derived
previously [12]. When the first constraint is active, s1
2 = 0
and d = dmax. The requirement given by equation (52)
remains valid for the constrained case. The solution for the
Lagrange multiplier is obtained from equation (38).
    
1.51.5 1
1 max km md max km
5
1.5d f t a d f t
3
  
   
 
(55)
In order to meet the requirement that 1 ≥ 0, the
parenthetical must be greater than or equal to zero. As a
result:
 
 
 
k
a
a max
k
t
k
t1
m a max max t d
m
t
a t dt
3
a t d d
5
a t dt


 
 
 
   
 
 
 


(56)
Therefore, for the constrained case, the solution that
maximizes the HIC function across the internal transition
point, tk, occurs at d = dmax, ak(ta) = am(ta + dmax) and when
both point acceleration values have a minimum value equal
to three fifths of the average acceleration between the two
points.
3.3 HIC based on specific singular a(t)
The simplest acceleration function that falls within this
categorical descriptor is the ideal square wave, a(t) = Ap
where Ap is a constant value that is greater than zero over
the domain {t: t0 ≤ t ≤ t1}. The average value of this function
over any finite value of the operative domain is 0.5Ap. This
means that three fifths of the average acceleration over any
finite value of the operative domain is 0.3Ap. The point
accelerations for aa = ad = Ap for any two discrete points over
the domain. From Table 1, cases 1, 3 and 4 are excluded
based on form. The remaining cases, however, cover any
value of ta limited to {ta: t0 ≤ t ≤ t1 – d}. This is the expected
result.
The next level of complexity for this case are those
functions that are monotonic. For a function of time, the
function is monotonic increasing over its domain if the sign
of the first time derivative is positive over the domain.

Similarly, for a function of time, a function is monotonic
decreasing over its domain if the sign of the first time
derivative is negative over the domain. The solutions for
both cases are obvious by informal inspection. For a
monotonic increasing function, ta = tN – dmax andd=dmax. For
a monotonic decreasing function, ta = t0 and d = dmax.
Formally, for the family of monotonic increasing functions,
cases 1-3 and 5 can readily be excluded based on the
relationship between aa and ad. For the family of monotonic
decreasing functions, cases 1-2, 4 and 6 can be excluded for
the same reason. Case 8 is valid for monotonic increasing
and decreasing functions when dmax = t1 – t0.
Finally, one may consider the situation in which a(t) is
modeled as a symmetric half sine pulse shape. There are
three transition points that are of relevance for this function.
The first andlast, t = t0 and t = t2, respectively denotethetime
at which a(t) initiates and terminates. The second transition
point, t = t1, denotes both the time at which a(t) reaches its
maximum value (i.e. t = t1 = tm) and denotes the time value
about which a(t) exhibits symmetry. Because of this
symmetry, for this formulation, t10 = t21 = 0.5t20 and a0 =
a2. If T denotes the full period of the sinusoid, then t20 =
0.5T. The form of the acceleration function is:
   
 
0 10 0
0 10 0
20
2
a t a a sin t t
T
a a sin t t
t
 
    
 
 
    
 
(57)
The case in which only the first constraint is active (i.e.
case 2 of Table 1) is the sole case of interest. This particular
case requires that aa = ad, which for this specific function
means:
 
 
0 10 a 0
0 10 a max 0
2
a a sin t t
T
2
a a sin t d t
T
 
    
 
 
    
 
(58)
Subtractinga0 from both sides of the equalityand dividing
both sides by a10 leads to the following result:
   a 0 a max 0
2 2
sin t t sin t d t
T T
    
      
   
(59)
Because of the periodic nature of the sine function, the
solution for ta is of the form:
max
a 0 1 1
dT
t t c c
4 2
     (60)
Letting c1 = 0, using the (+) form of the quarter periodand
substituting T = 0.25t10 followed by simplification leads to
the following, expected, solution.
max
a 1
d
t t
2
  (61)
At t = td = t1 + 0.5dmax, a(t) is:
max
d 0 10
10
d
a a a cos
4 t
 
    
 
(62)
Three fifths of the average acceleration between the two
points is:
 1 10 10 max
max 0
max 10
12 t a d3 3
d a t dt a sin
5 5 5 d 4 t
    
   
  
 (63)
For the sign of the Lagrange multiplier to be correct,
thereby producingavalidKKTpoint,thefollowingmusthold:
0 max
10 10
max max max
10 10 10
a d
2
a 4 t
d d d
3sin 5 cos
4 t 4 t 4 t
 
 
  
       
     
       
(64)
The ratio of dmax/t10 will be valued between 0 and 1. At
the limit of 0, equation (64) reduces to 0 = 0. At the limit of
1, the right side the equation reduces to 2-0.5(3 – 1.25) ~ -
0.655481. The left side of the equation, however, is positive
valued and the inequality holds. The result regarding the
location of ta and td do not change for the case of the
exponentiated sine function (e.g. haversine or sin2 orforany
sinp where p is real-valued and greater than or equal to
unity) whereas the form of equations (63-64) will change.
3.4 HIC based on specific multiple a(t)
The simplest case where multiple acceleration functions
are apt is the triangular pulse shape. The temporal domain
of this function contains three transition points with t = t1
denoting the point of C0 continuity between the first linear
function, a1(t), and the second linear function, a2(t). The
acceleration functions are:
     
     
1 0 1 0 0 1
2 1 2 1 1 2
a t a k t t t : t t t
a t a k t t t : t t t
    
    
(65)
The slope of the first linear region is positive while the
slope of the second liner region is negative. Both slopes can
be written in terms of the boundary values for eachfunction.
10 21
1 2
10 21
a a
k k
t t
 
 
 
(66)
The triangular pulse is fully symmetric when k1 = -k2 and
t10 = t21. The evaluation, limited to an interval thatspanst

= t1, for d = dmax, requires determining ta for a1(ta) = a2(ta +
dmax).
 10 1 0 2 max 1
a
1 2
12 01
1 max
0 12 1 20 2 01
a k t k d t
t
k k
a t
t d
a t a t a t
   


 
 
    
(67)
For the Lagrange multiplier as per equation (56):
  10 12
2 d 2d 1 max
10 21 12 10
a a
a t a a d
a t a t
 
  
    
(68)
 1
max
10 12
1 max
10 21 12 10
3
d a t dt
5
a a3 3
a d
5 10 a t a t


 

    

(69)
Multiple formulations exist by which the rise phase
function, a1(t), and the fall phase function, a2(t), can both be
modeled using trigonometric functions. The simplest
formulation is for the case in which both phases are quarter
period responses such that t10 = 0.25T1 and t21 = 0.25T2.
The operative equations for such a formulation are:
     1 0 10 0 0 1
10
a t a a sin t t t : t t t
2 t
 
      
 
(70)
     2 2 12 1 1 2
21
a t a a cos t t t : t t t
2 t
 
      
 
(71)
When t = t1, the operand of the sine function in equation
(70) becomes 0.5, which leads to the maximum positive
value of the sine function of unity. The result is that a1(t1) =
a0 + (a1 – a0) = a1. When t = t1, the operand of the cosine
function in equation (71) becomes 0, which leads to the
maximum positive value of the cosine function of unity. The
result is that a2(t1) = a2 + (a1 – a2) = a1. Therefore, for this
case, the time at which the maximum of the resultant
acceleration occurs is t = tm = t1. Because each segment is a
quarter phase response, a0 = a2. The requirement that a1(ta)
= a2(ta + dmax), for the second case in Table 1, leads to the
following:
 
 
0 10 a 0
10
2 12 a max 1
21
a a sin t t
2 t
a a cos t d t
2 t
 
    
 
 
    
 
(72)
Because a0 = a2, a10 = a12. This leads to the following
solution:
10
a 1 max
20
t
t t d
t

 

(73)
This solution reduces to the expected result of ta = t1 –
0.5dmax when t10 = 0.5t20. Based on the solution shown in
equation (73):
10
d a max 1 max
20
t
t t d t d 1
t
 
     
 
(74)
If  = t10/t20, where  such that {: 0 <  < 1}, then
t21/t20 = 1 – . The requirement for the Lagrange
multiplier, given by equation (56), becomes:
0 max 20 max
10 20 max 20
a d t d5
cos sin 0
3 a 6 2 t d 2 t
      
     
     
(75)
The ratio of dmax/t20 is such that 0 < dmax/t20 ≤ 1. The
limit of the summation of the last two terms, as the ratio
approaches zero is /3. The summation of the last two
terms is zero valued when the ratio is approximately
2.10558-1. When the ratio is unity, the summation of the
last two terms is -1. It can therefore be stated that the
constraint on the Lagrange multiplier is conditionally met.
However, for most realistic applications, the ratio of
dmax/t20 is well below unity value and the requirements of
the constraint will be met.
The final case forconsiderationisthatofsituationinwhich
a(t) is modeled as a trapezoidal function, which in turn is a
special case of the use of three domain delimited linear
functions. The trapezoidal function is characterized by a
linear rise phase, a constant valued plateau and a linear fall
phase. The four transition points result in three temporal
domains consisting of {t: t0 ≤ t ≤ t1}, {t: t1 ≤ t ≤ t2} and {t: t2 ≤ t
≤ t3}. The acceleration functions corresponding to each
temporal domain are a(t) = a0 + k1 (t – t0), a(t) = a1 = a2 and
a(t) = a2 + k2 (t – t2). When t21 = dmax, then ta = t1 and ta +
dmax = t2. When t21 > dmax then t1 ≤ ta ≤ t2 – dmax. When t21
< dmax, the relevant checks are of the acceleration function
across t1, the acceleration function across both t1 and t2 and
the acceleration function across t2.
 
  
1
a
1
1
2.5t
0 1 0
t1.5
a t d
1
t
a k t t dt
HIC t ,d d
a dt


 
   
 
  
 
 
 


(76)

 
  
  
1
a
2
1
a
2
2.5
t
0 1 0
t
t
1.5
a 1
t
t d
1 2 2
t
a k t t dt
HIC t ,d d a dt
a k t t dt


 
   
 
 
   
 
 
  
 
 



(77)
    
a2
a 2
2.5
t dt
1.5
a 1 1 2 2
t t
HIC t ,d d a dt a k t t dt


 
    
 
 
  (78)
The solution for ta for equation (76)andequation(78)can
be determined based upon equation (52). For the case of
equation (76), ta = t1. However, when t21 < dmax then t2 = ta
+ dmax and ta = t21 – dmax. The solution for ta for equation
(78) is ta = t2 – d. However, when t21< dmax, ta = t1. For the
case of equation (77), with d = dmax, the solution for t = ta is:
 10 1 0 2 max 2
a
12
a k t k d t
t
k
   


(79)
The solution for ta is more heavily weighted towards the
linear function that has the smaller slope magnitude.
4. DISCUSSION
The quantification of the maximum HIC value, based on
anthropometric test device head center of mass resultant
acceleration calculations from triaxial translational
acceleration measurements, serves as a key metric when it
comes to passing regulatory compliance requirements for
the sale of production motor vehicles to thegeneral publicin
the United States. The quantification of the samemetricalso
serves as an important predictor, in a forensic sense, whenit
comes to predicting the potential for closed head injury
when based upon a base excitation (i.e. vehicular collision
partner collision severity from an accident reconstruction
evaluation) coupled with an occupant kinematics analysis.
From the mathematical perspective, the importance of
constraining the temporal duration of the calculation can
readily be seen by considering the HIC equation when a(t)is
a 1 G square wave and when the duration is 1 second. The
result is a HIC value of 1000. The problematic nature of HIC
calculations for which the temporal duration is
unconstrained is the same problem that was present for the
SI. HIC calculations, correspondingly, are conducted with a
maximum limit on the temporal window length. Thislimit is
generally either 15 milliseconds or 36 milliseconds [14].
Injury assessment reference values are typically presented
in conjunction with the length of the temporal window. For
example, a HIC15 value of 700 represents the IARV for the
HIC function, with a maximal temporal window length of 15
milliseconds, andwhichrepresentsa fivepercentprobability
of a severe head injury with severe being codedasa 4+score
on the Abbreviated Injury Scale [15]. For thecaseofdiscrete
resultant accelerationdata,theunoptimizedapproachwould
consist of calculating the HIC value over the interval {ti, ti +
nt},where t is the sampling interval and n {n: 1 ≤ n ≤
dmax/t, n }.
The computational requirements for the case of discrete
resultant acceleration data aresubstantivelymitigatedwhen
the acceleration is known or modeled as an analytical closed
form function. The findings of previous published research
work [12] are seminal in regards to showing that the HIC
function is maximized, for the unconstrained case, over the
interval t = t1 = 1 and t = t2 = 2, when a(1) = a(2) and when
the magnitude of the resultant acceleration at either point is
equal to three fifths of the average acceleration between the
points. Functionally, the evaluationoftheHICfunctionatthe
boundary points of t = t0 and t = tN is requisite, in the general
sense, when a(t0) falls within the domain of monotonically
decreasing function or when a(tN) falls within the domain of
a monotonically increasing function. Excluding the square
wave and the trapezoidal acceleration cases, the other
acceleration functions of interest are readilycharacterizable
as being concave. The determination of concavity can
readily be made in a priori manner due to the fact that the
acceleration functions are specified. When concavity is an
apt descriptor of the underlying resultant acceleration
function, the determination of the domain that maximizes
the HIC function reduces to the evaluation of points internal
to the temporal domain of the resultant acceleration
function. When the temporal domain of the resultant
acceleration function exceeds a maximum window or
clipping length, the optimization problem, in this case being
a maximization problem, reduces to the evaluation of case 2
as per Table 1. For this case, there is a single switching
condition and a single Lagrange multiplier. The former
encodes whether or not the constraint is active while the
latter encodes the inequality constraint of dmax – d ≥ 0, as an
additive term to the HIC function, thereby generating the
Lagrangian for the problem. The Lagrangian, in turn,
represents the converted form of the constrained problem
into one for which the first derivative test of the
unconstrained problem still applies. Thusly, it comes as no
surprise that the finding of a(a) = a(a + dmax) still holds
when the sole constraint for the case is active (the first
partial derivative of the Lagrangian with respecttota seesno
contribution from the constraint). The evaluation of the
problem with the constraint as active, which represents the
objective of the subject work, does lead to the following
additional results. The first is the expected result of d = dmax.
The second is that the equally valued resultant acceleration
at t = a and t = a + dmax is greater than or equal to three
fifths of the average acceleration between the points. These
solutions hold for concave resultant acceleration functions,
irrespective of the complexity of the same (implicit in this
statement that is that there is a one-to-one correspondence
between any value of t within the domain and a(t)).

The specific forms of the resultant acceleration function,
both as singular functions and as composite temporally
domain limited functions, were chosen primarily because of
commonality of usage. Clearly,thesetoffunctionspresented
is a subset of the totality of functions that could be
considered. Furthermore, for the composite case, one may
readily use functions from different families (e.g. a quarter
sine initial phase followed by a subsequent linear phase).
With the first inequality constraint set to active, the
requirement of equivalenceinthemagnitudeoftheresultant
accelerations for each function pair that generates an
internal transition point remains. Furthermore, the
requirement imposed by the Lagrange multiplier remains.
The specific form of this requirement varies as a function of
the form of each equation that is involved in generating a
transition point. For the case when only the first constraint
is active, the requirements of the second order KKT are met
for a valid first order KKT point.
5. CONCLUSIONS
The determination of the interval that maximizes the HIC
function associated with an underlying resultant head
acceleration, a(t), when the maximal length of the interval,
itself, is subject to constraint, was evaluated using the
standard first order necessary and second order sufficient
KKT conditions foroptimality. Themathematicalframework
enshrined within the KKT formulation for optimization
provided a formal mechanism for evaluating the underlying
the contextual problem of constrained maximization of the
HIC function. For the case of internal points, with respect to
the temporal domain of a(t), it was shown that when a(t) is a
singular function, that the window length d = dmax, that the
start of the maximal interval occurs at a(ta) = a(ta + dmax) and
that the Lagrange multiplier has a valid sign when the
resultant acceleration at either boundaryofmaximalinterval
is greater than or equal to three fifths of the average
acceleration between and inclusive of the maximal interval
temporal locations. For the case of internal points, with
respect to the temporal domain of a(t), it was shown that
when a(t) is acompositefunction, that the windowlengthd=
dmax, that the start of the maximal interval occurs at ak(ta) =
am(ta + dmax) and that the Lagrange multiplier has a valid sign
when the resultant acceleration at either boundary of
maximal interval is greater than or equal to three fifthsofthe
average acceleration between and inclusive of the maximal
interval temporal locations. These findings fill a
longstanding gap in the research literature in regards to
presenting the formally derived solutionsfortheconstrained
optimization of the HIC problem.
ACKNOWLEDGEMENT
The author wishes to acknowledge Dr. John Perry, who
served as the author’s thesis advisor and whose patient
guidance fostered the, as of now, longstandinginterestinthe
field of biomechanical engineering.
REFERENCES
[1] E. Gurdjian and H. Shawan, “Management of skull
fracture involving the frontal sinus,” Annals of Surgery,
vol. 95(1), 1932, pp. 27-32.
[2] H. Lissner, M. Lebow and F. Evans, “Experimental
studies on the relation between intracranial pressure
changes in man,” Surgery, Gynecology & Obstetrics, vol.
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[3] L. Patrick, H. Lissner and E. Gurdjian,“Survival bydesign
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Lissner, “Evaluation of the protective characteristic of
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[5] V. Hodgson and L. Thomas, “Effect of long-duration
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[8] J. Newman, “Head injury criteria in automotive crash
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[9] C. Gadd, “Use of a weighted-impulse criterion for
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Waeckerle, “Concussion in professional football:
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[12] C. Chou and G. Nyquist, “Analytical studies of the Head
Injury Criterion (HIC),” SocietyofAutomotiveEngineers
Technical Paper No. 740082, 1974.
[13] M. Bhatti, Practical Optimization Methods with
Mathematica Applications, New York: Springer-Verlag,
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[14] P. Prasad and H. Mertz, “The position of the United
States delegation to the ISO Working Group 6 on the use
of HIC in the automotive environment,” Society of
1985.
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children and adults in frontal and rear collisions,”
Society of Automotive Engineers Technical Paper No.
973318, 1997.

BIOGRAPHY
Mr. Singh is a private practice
engineer with specialties in the
field of motor vehicle accident
reconstruction and biomechanical
engineering. He holds a BS in
mechanical engineering from the
University of Illinois at Urbana-
Champaign and a MS in biomedical
engineering from the University of
Southern California. He has also
authored numerous other peer
reviewed conference proceeding,
technical and scientific papers.

IRJET- Analytic Evaluation of the Head Injury Criterion (HIC) within the Framework of Constrained Optimization Theory

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IRJET- Analytic Evaluation of the Head Injury Criterion (HIC) within the Framework of Constrained Optimization Theory