Collision pulse modeling: Closed form solutions for sinusoids exponentiated to integer powers

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Collision pulse modeling: Closed form solutions for sinusoids exponentiated to integer powers Jai Singh1 1Biomechanical Engineering Analysis & Research, Inc., Thousand Oaks, California, USA ---------------------------------------------------------------------***---------------------------------------------------------------------

Abstract - The subject work extends the mathematical

mortality risk for the occupants of the collision partner [24]. The scientific literature contains numerous studies that have focused on the relationship between vehicular collision partner crash pulse metrics and the potential for occupant injury [8, 25-28].

framework for the well-established use of half period and quarter period sine and sine squared functions for modeling collision pulses within the contextual fields of accident reconstruction, occupant kinematics and biomechanical engineering. The topical focus of the material presented herein is the derivation of the kinematic equations (i.e. acceleration, velocity and displacement) for the cases in which either a sine or cosine function, exponentiated to an arbitrary, positive, integer value, is used as the base function for modeling a segment of the underlying acceleration. The closed form analytic solutions presented also account for non-zero valued initial acceleration offsets and for the use of partial full period responses beyond, but inclusive of, the quarter and half period responses.

Generally, the event pulses associated with actual collisions contain multiple local extrema and therefore are modeled in only an approximate sense when a single analytic closed-form modeling equation is used. The scientific literate contains a number of studies that have evaluated the utility of exactly such an approach, using simple functions, for the kinematic response due to collision loading [9, 13, 26, 28-31]. One function that has been commonly used is the sine function, raised to either the first power or raised to the second power (i.e. haversine function). If x denotes the operand of the sine or haversine function, then both functions are zero valued when x = n, where n is an integer. Both functions have peak magnitude when x = 0.5n (again, where n is an integer, but excluding zero). The sign of the peak magnitude alternates between and any successive values of n and n + 1. Both functions are oscillatory. The period of the sinusoidal waveform, T, represents the time for one full cycle of the sinusoid. The period is related to the ordinary frequency, f (measured in units of Hz, i.e. cycles per second), by T = 1/f, and to the circular frequency,  (measured in units of Hz), of the sinusoid, by T = 2/.

The haversine function rectifies the magnitude of the sinusoid. For modeling the entirety of an event pulse, the half period response is typically used. Figure 1 depicts both functions shown over four periods and also over one half of a period.

Key Words: Collision pulse modelling, Accelerationtime history analytic modelling, Trigonometric modelling, Sine, Cosine

1. INTRODUCTION The synonymous terms crash pulse and collision pulse, as either explicitly defined or implicitly used in the scientific literature, refer to the shape and characteristics of the acceleration or deceleration (referred herein as acceleration) experienced by an object or structure when subjected to an impact [1-7]. The characterization of the shape of the acceleration-time history along with its other characteristics can be applied to any event in which an object or structure incurs an acceleration due to the event.

The term event pulse is thusly adopted to account for this extension.

Contextually, the literature contains evaluations of the event pulses for motor vehicles subjected to collisions [5-7, 8-12], motor vehicle seats [13], motor vehicle occupants [14-16] and coupled analyses involving multiple structures [6, 17-23].

One of the determinable characteristics of any event pulse is the area encompassed between the abscissa and the acceleration response curve. Mathematically, this area is the first time integral of the acceleration and represents the velocity change (delta-v or v) incurred by the object or structure due to the event. Within the context of motor vehicle collisions, the velocity change incurred by a vehicular collision partner, due to an impact, represents the most important factor when it comes to predicting the

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Figure1: Four period response for a sine wave (blue) and haversine wave (orange) shown on the left and the half period responses shown on the right.

The half period response of either the sine function or haversine function has certain desirable characteristics when it comes to modeling the entirety of an event pulse.

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The first is that the modeled acceleration response terminates at the same value as the initial acceleration (zero as shown in Figure 1). The second is that the magnitude of peak acceleration occurs at an interior temporal value (if the initial time value is t = 0 and the event duration is t, then the peak acceleration occurs at t = t/2). If the peak acceleration is denoted as Ap then the sine and haversine functions can be written as a(x) = Ap sin(x) and a(x) = Ap sin2(x), respectively. The operand of both functions, x, is a function of time (i.e. x = f(t)). When a single sine or haversine function is used to model the entire event pulse, x = (t)/(0.5T) = (2t)/T. If the event duration is denoted as t then the relationship between the event duration and the period of the sinusoid is T = 2t.

Therefore, x = t/t and the two modeled acceleration functions, operative over {t: 0 ≤ t ≤ t} are:  t   t  a  t   A p sin   a  t   A p sin 2    t   t 

case in which both collision partners exhibit a linear forcedeflection response for both the closure and separation phases [36]. The linear force-deflection response can be derived directly from a half period sine model (i.e. the half sine model).

Matching the results of a phenomenological model is not a requisite study-independent requirement. Nor is there a physical or modeling reason as to why the exponentiation of the sine function must be limited to either a value of one or two. The subject paper, specifically, focuses on the theoretical development of the analytic closed form solutions for the case in which the power of the exponentiation, n, for a sine based event pulse, is of an integer value {n: n is an integer greater than or equal to unity}. The case of n = 0 is quite simple, with the sin 0(x) reducing to unity for all values of x.

2. THEORY

(1)

From the introductory material presented, it is clear that the base function of interest can be written as f(x) = sinn(x), where n {n: 1 ≤ n, n is an integer }. Furthermore, f(x) = a(x) and because x is a function of time during the event pulse, f(x) = a(x) = a(t). The operative temporal boundary can be defined as being between and inclusive of the indexed times t = ti and t = tj where the index j = i + 1 (i.e. the next indexed location in a series of locations) and where {ti, tj: ti < tj}. The operative temporal duration of the event can be written as tji = tj – ti. The accelerations at time t= ti and t = tj are denoted as ai and aj.

The half sine acceleration response is also the response of the simplest phenomenological model (i.e. the half period response of a single degree of freedom (SDOF) simple harmonic oscillator). Such a model consists of a single lumped mass (m) and a single massless element that develops force as a linear function of deflection (where k is the constant slope of the relationship). The model can be equivalently characterized by defining the circular frequency for the system,  = (k/m)0.5. In the case of the simple harmonic oscillator, the kinematic responses at t = T/4 are of peak acceleration magnitude, zero velocity and peak displacement. The kinematic responses at t = T/2 are of zero acceleration, the negative of the initial velocity and zero displacement. This simple model may also be used for modeling the response of a vehicle subjected to a full-width engagement, collinear, axial, impact with a fixed, rigid, massive barrier (FRMB).

This approach to the symbolic terms in the problem may appear to be an unnecessary complication, on first glance.

However, if one considers even the simple problem of a two phase response (the closure and separation phases), the problem involves three critical points. These points are the start of the closure phase, the contemporaneous terminus of closure and start of separation and the terminus of the separation phase). The indexes can be numerically labeled as one, two and three, respectively.

The time at which the terminus of closure between the collision partners (i.e. for the case described, this is the time at which the velocity of the test vehicle is zero valued) can be used to segment the kinematic responses into a closure phase and separation phase response [29, 32-33]. This approach does not change the shape of the kinematic responses, but instead, segments them. The matching of all terminus of closure and terminus of separation phase kinematic responses is generally not achieved, when working with actual test data, secondary to the fact that the estimation of any single parameter between the set of phasic stiffnesses, phasic circular frequencies, the times at which closure or separation terminate, or the deflection at the terminus of closure, or separation analytically fixes the values of the other parameters [34]. The SDOF modeling approach underpins the most commonly utilized, closure phase only, residual damage based model [35] and has been extended to the

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The explicit representation of the operative equations for each phase can readily be avoided (a tedious task as the number of critical points increases) by the use of the sequential alphabetic subscripts coupled with the nature of the phasic response that is being modeled between each successive pair of indexed points. It should be readily apparent that t = ti will not always be zero valued.

Furthermore, a(ti) will not always be zero valued and a(ti) = ai may generally not be equal to a(t2) = a2. The base form of the sine function can readily be generalized. The initial value offsets of (ti, ai) can be accounted for by introducing a phase shift and an amplitude shift, respectively. a  t   ai  asin n  b  t  t i  

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In equation (2), the response amplitude is denoted symbolically (a), the period T is equal to 2b-1 and the phase shift is bti. The constant term b can be solved for by algebraically rearranging the definition of the period, b= (2)/T = /(0.5T). This relationship is based upon a full period response and can be generalized to a partial period response by changing the value of the constant term in the denominator. For the case when the constant term is in the numerator and is set equal to one, rather than two, the relationship is based in on the half period response. For the case when the constant term is set to 0.5 (in the numerator), rather than two, the relationship is based on a quarter period response. When the temporal domain of interest is tji (specifically, rather than T), the term b may be written as b = c/tji. Rewriting equation (2):  c  a  t   a i  a sin n   t  t i    t ji  

sin  x  

n

(3)

cos  x  



(10)

(11)

z  z 1 2

sin  x  

z  z 1 j2

(12)

n

 z  z 1  1 1 n cos n  x      n z  z  2 2   1 n k 1 n  k 1 n  n  z z   n  zk zk  n 2 k 0 2 k 0 1 n 2k  n  n z 2 k 0

(5)

Substitution for z, in the form to the right of the last equality, lead to the following result.

(6)

cos n  x  

1 n j 2k  n  x e 2 n k 0

(13)

Using Euler’s formula on this equation and employing the symmetry properties of the trigonometric functions leads to the following result.

1 n  cos   2k  n  x      2 n k  0  j sin   2k  n  x     1 n  cos   n  2k  x      n  2 k  0  j sin   n  2k  x    

cos n  x  

(7)

Summing equations (6) and (7) followed by algebraic rearrangement yields the following solution for the cosine function.

e jx  e  jx 2



For the case of cosn(x):

Where j is the complex unit (i.e. j is the square root of -1).

The cosine function is an even function and therefore cos(x) = cos(-x). The sine function is an odd function and therefore sin(-x) = -sin(x). Substitution for x, by -x, in equation (6) and employing the symmetry properties of the cosine and sine functions leads to the following:

cos  x  



Letting z = ejx and z-1 = e-jx, equations (8) and (9) can be rewritten in the following manner:

Euler’s formula defines the relationship between the complex exponential function and the trigonometric functions of sine and cosine. For all x in the reals:

 cos  x   jsin  x 

 n  k  0  k  n

n n!    k  k ! n  k !

2.1 Euler’s Formula

e



Where the binomial coefficient, representing the k th entry of the nth row in Pascal’s triangle, is defined as:

When c is equal to 0.5, in equation (4), the sine function has a value of one and a = aji. This represents the quarter period response. More generally:

 jx

 n  k  0  k  n

 a  b      a n  k b k      a k b n  k 



e jx  cos  x   jsin  x 

(9)

The binomial theorem states that the sum of two terms, a and b, raised to a power, n, can be expressed in the following form.

 c  a j  ai  a sin n  t j  t i   a ji  a sin n  c  (4)  t ji   

a  a ji sin  n  c 

e jx  e  jx j2

2.2 The binomial theorem

Equation (3),written as t = tj is:



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(14)

(8)

Subtracting equation (7) from equation (6) followed by algebraic rearrangement yields the following solution for the sine function.

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For the case of sinn(x):

cos n  x   2 1 n

n

n  z  z 1  1 sin  x    z  z 1    n  j2  j2    n

   

1

n

 n  k

1 n  k

 n  k

1 n  k

 n  k

n k

  k  z  z 

 j2  k 0   n

1

n

1

n

1

n

k n



 n 

  k   1

n k

 z 2k  n  

Substitution for z, in the form to the right of the last equality, leads to the following result. sin n  x  

1

n

 n 

  k   1

 j2  k 0   n

n k

e

j 2k  n  x

  

(15)

n

n

n

 n 

  k   1

n k

n



n

 j

n

n

n k

 j

n1

k 0

(17)

(18)

    (19)  sin   n  2k  x    

cos   n  2k  x    

Because of the symmetry of each even row (the rows under consideration are even because n + 1 is even when n is odd), the following holds for the binomial coefficients. n  n     k n k

The summation resulting from the employment of the binomial theorem, as shown in equation (14), entails the involvement of n +1 terms. Therefore, when n is odd, there will be an even number of terms under the summation operator. Correspondingly, when n is even, there will be an odd number of terms under the summation operator. Starting with case where n is odd, the total number of terms summed will be equal to 2k.

Because of symmetry this can also be viewed as being equal to the presence of pairwise terms for each k and -k {k: k is not equal to zero}. If q is equal to n – 2k, then these pairwise terms will be found in the form of cos(qx) and cos(-qx) = cos(qx) and additionally as -jsin(qx) and -jsin(qx) = jsin(qx). The summation, therefore, when n is an odd integer, eliminates the sine terms in equation (14).

The cosine terms, however, are pairwise additive. Thusly, when n is equal to an odd integer, equation (14) reduces to the following:

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n k

k 0

n

2.3 Form of cosn(x) based on exponential parity

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 

  n     k   1  1    sin  x   2   n       1    k 

Further reduction of equation (14) and equation (16) depends upon the parity (i.e. odd or even integer valuation) of the exponent, n.

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k 0

Just as was the case with the cosine function, when n is an odd integer, the summation involves an even number of terms. We thusly rewrite equation (16) as follows:

 cos   2k  n  x        j2  k  0    j sin   2k  n  x    (16) cos   n  2k  x     1 n  n  n k       1  j sin   n  2k  x    j2 n k  0  k    1



2.4 Form of sinn(x) based on exponential parity

Using Euler’s formula and employing the symmetry properties of the trigonometric functions (as was done for the previous case) leads to the following result. sin  x  

 n 

  k  cos   n  2k  x 

 n      cos  0   j sin  0     0.5n 1    cos n  x   n  0.5n  1 2  n   2    cos   n  2k  x     k 0  k       n      1   0.5n    n 0.5n  1  n 2     2    cos   n  2k  x     k 0  k     



 j2  k 0   n

  

  k  z  1  z 

 j2  k 0   n

1 n  k

0.5  n  1 

When n is even, the total number of terms summed will be equal to 2k + 1. The extra term occurs when k = 0.5n. As a result, the summation may be partitioned. The terms for k not equal to 0.5n, representing an even number of terms, reduces to the form shown by equation (17) but with the modification for the upper limit on the summation operator being 0.5n – 1.

  

  k  z  1   z 

 j2  k 0   n

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The operand of each trigonometric function is (n – 2k)x for k = 0 to k = 0.5 (n -1) and is equal to –(n-2k)x for k = 0.5 (n-1) + 1 to k = n. Because of the even symmetry in the cosine function, however, the corresponding pairwise terms have the same sign for the cosine function. The term (-1)n-k follows the segment of the repeating sequence of {-1, 1, -1, 1, …} over the domain of k = 0 to k = 0.5 (n -1) and the negative of this sequence over the domain of k = 0.5 (n-1) + 1 to k = n. As a result, the pairwise terms, consisting of the multiplication of the cosine term and (1)n-k, associated with each value of k and n – k have opposite signs. Because the binomial coefficient terms for each of these pairwise terms is the same (on a pairwise basis), the summation associated with the cosine terms is zero valued. Equation (19), for any odd integer, n, becomes:

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sin  x  

1

n

 j2 

n

 n 

k 0

 

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

  k   1   j sin   n  2k  x   (21)

n

n k



Using the definition of j as (-1)0.5 and reducing the terms that involve the exponentiation of (-1) leads to the following result. n

0.5  n  1 



  1 k 0

The generic variable x can be replaced with the following, which brings the subject formulation into the time domain.

0.5  n  1   k



x

  n     k   1    sin  x   2    n        1    k  n

n

0.5 n  k

k 0

n

k 0

   0.5  n  1   k  n   k  1   0.5  n  1       (28) a  t   ai  a  2 1 n        c n  2k   k 0   sin  t  ti        t ji     

   (24)  sin   n  2k  x      

cos   n  2k  x  

Substitution of equation (27) into equation (26) followed by substitution of the resultant into the form of equation (3) leads to the following solution for the modeled acceleration-time response for the case in which n is an even integer.

The previous finding of equation (20) also holds here, in regards to the binomial coefficient terms. For the summation term involving the sine function, when n is an even integer, the central term (i.e. at k = 0.5 n) is zero valued. The signs of the pairwise terms at {k, n – k} are valued {1, -1}, respectively, due to the odd symmetry of the sine function. The exponentiated -1 term, for each pairwise {k, n – k} are equally valued. The result of the multiplication of the three terms under the summation operator associated with the sine function is of equal magnitude and oppositely signed terms for all pairs {k, n – k} and a zero valued central term. The summation operation results in a zero valuation, resulting in the following relationship:

 n  n    2  0.5n          n      (29)  1 0.5 n  k    a  t   ai  a     1  n 0.5 n  1   k  2      c  n  2k    k 0   cos    t  t i      t ji      

Equations (28) and (29) are operative over the temporal domain of t {t: ti less than or equal to t and t less than or equal to tj}.

3.1 Segmented acceleration response for cosn(x) In certain cases, such as when the magnitude of acceleration is decreasing between the two indexed points, it is preferable to use a cosine function rather than a sine function. Subtracting the operand, x, from 0.5, and taking the cosine, yields the sine of the operand, x. Taking the sine of 0.5 – x results in the cosine of the operand, x.

Substitution for x, using the definition in equation (27), in equation (17), followed by writing the resultant in the form of equation (3), leads to the following solution for the modeled acceleration-time response for the case in which n is an odd integer.

n  n    0.5n  k sin n  x   2  n     1  cos   n  2k  x   (25) k  0  k  

Equation (25) reduces following the same approach as described for cosn(x), for n being an even integer. The result of this reduction is the following:

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(27)

Substitution of this relationship into equation (23) and substitution of the resultant into the form of equation (3) leads to the following solution for the modeled acceleration-time response for the case in which n is an odd integer.

 n  k  sin   n  2k  x   (23)   

0.5 n  k  0.5

c  t  ti  t ji

3.1 Segmented acceleration response for sinn(x)

When n is an even integer, the reduction starts by consolidating the terms in equation (19) that are exponentiated values of -1.

n

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n

The finding shown in equation (20), in regards to the binomial coefficient, is again noted. However, the sine function, being odd, results in the pairwise sine terms having the opposite sign. For example, for n = 3, the summation occurs over k = 0 to 3 and the corresponding values of n – 2k are (3, 1, -1, -3) respectively. For the last two terms, sin(-x) = -sin(x) and sin(-3x) = -sin(3x).

Accounting for this fact leads to the final form for the solution of sinn(x): sin n  x   2 1  n

n

3. COLLISION PULSE MODELLING

 n   0.5n  1.5  k sin  x   2     1   sin   n  2k  x   (22) k  0  k   n

 n   0.5n    0.5n  1   0.5n  k  n  2 1  n   1   k  cos   n  2k  x   k 0    

sin  x   2 n

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a t  aj

  n  1 n 0.5n  1  n   a i  a j  a  2  n  2     k 0  k    0.5n  

 1  n 0.5 n  1  n   c  n  2k     (30) a  2  t  t i       k   cos  t  k  0  ji       

a t  aj

a t  aj

  n           0.5n       (36) n    n   a ji  2        0.5n  1  k     2       k  0  cos  c  n  2k   t  t      i       t ji       

  n          n   0.5n    (31) a  2  0.5 n  1  n  c n  2k       2    cos   t  t i            k  0  k  t ji        Equations (30) and (31) are operative over the temporal domain of t {t: ti less than or equal to t and t less than or equal to tj}.

The terminus boundary condition requires that the acceleration has a value of aj at t = tj. Substitution of t = tj into equation (30) leads to the following result:

3.3 Boundary value assessment

  1 0.5 n  1  k    1 n 0.5 n  1    a ji 2  n     a k  0   sin  c  n  2k       k     

Each of the four equations for the acceleration must yield a response of ai at the temporal boundary of t = ti.

Substitution of t = ti into equation (28) results in the parenthetical term that multiplies a, being zero valued (the sine function is zero-valued when the operand is zero). This produces the expected result of ai = ai.

Substitution of t = ti into equation (29) leads to the following result:

(37)

The term on the left of the equality in equation (37) is not readily reducible. However, if that term is plotted as a function of positive, odd, integer values of n (from 1 to 99) and over a domain of c {c: c is greater than equal to zero and c is less than or equal to ten}, it can be seen (see Figure 2) that the ratio of the acceleration difference, aji, to the amplitude, a, for equation (37), is constant over the domain of positive, odd, integer values of n, and that it is cyclical with respect to c. For any value of n within this domain, the ratio is zero for integer values of c, 0.5 for c = (2k+1)/2 for k {k: k is an even integer greater than or equal to zero}, and -0.5 for k {k: k is an odd integer greater than or equal to one}.

(32)

The parenthetical term, for any arbitrary, positive, even, integer value of n, is zero valued. This leads to the expected results of ai = ai. Substitution of t = ti into equation (30) leads to the following result: 0.5 n  1  n  ai  a j  a  2 1 n     k 0  k   

(35)

The parenthetical term, for any arbitrary, positive, even, integer value of n, is unity. This leads to the result of a = aij. Equation (29) can thusly be rewritten as:

Following the same approach leads to the following solution for the modeled acceleration-time history for the case in which n is an even integer.

 n  n   2     0.5n    ai  ai  a  0.5n  1  n   2 1 n   1 0.5n  k      k 0   k    

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(33)

The parenthetical term for any arbitrary, positive, odd, integer value of n, is unity. This leads to the result that a = aij. Equation (30) can thusly be rewritten as: a t  aj

 2 1 n     (34) a ji  0.5 n  1  n   c  n  2k     t  t i        k  cos  t ji     k  0    Substitution of t = ti into equation (29) leads to the following result:

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Figure 2: Ratio of the acceleration difference aji to the amplitude, a, for the case of the sine function raised to a

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positive, odd, integer value. The response surface is shown as continuous solely for the purpose of aiding the visualization. In actuality, the response surface exists at discrete positive, odd integer, values of n.

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equation (39) and substituting the result into equation (29) leads to the following result. a  t   ai

   0.5 n  k  n   k  1      n  0.5 n  1       c n  2k  2            0.5n  k 0  t ji   cos        t  t    i   a ji      0.5 n  k  n    n  0.5 n  1  1  k         0.5n   2   k 0     cos  c n  2k        

Rearranging equation (37) and substituting the result into equation (30)leads to the following result: a  t   ai

  1 0.5 n  1  k    0.5 n  1        n   c  n  2k   t  t i      k  0   sin  t ji k     (38)   a ji 0.5  n  1   k    1    0.5  n  1         n  sin c n  2k   k 0      k          

(40)

For the cosine function where n is an odd integer, substitution of t = tj into equation (30) leads to the following result.

Substitution of t = tj into equation (29) leads to the following result.

0.5  n  1   n   a  2 1 n    cos  c  n  2k      0   k k  0    

 1 0.5n  k   0.5n  1 n  1 n   a ji n  2   2   n     a (39) cos  c  n  2k   k 0  0.5n     k    

(41)

The case where a = 0 is the trivial case and thusly not germane to the subject problem. Therefore, it is the parenthetical term that is zero valued. Clearly, 21-n, is not zero valued for positive, odd, integer values of n. Thusly, it is the summation that is zero valued. For any positive, odd, integer values of n, the solution for c is ±0.5 + 2k where k is within the integers. The solution of c = 0.5 is the quarter period response.

The term on the left of the equality is, again, not readily reducible. Following the same approach as for the previous case leads to the result shown in Figure 3.

For the case of the cosine function with n being an even integer, substitution of t = tj into equation (31) leads to the following result.

  n         n   0.5n   a2    0 0.5n  1  n      2    cos  c  n  2k       k 0  k         

Following the same approach used for the previous case, the inner parenthetical must be zero valued. This leads to the following:

Figure 3: Ratio of the acceleration difference aji to the amplitude, a, for the case of the sine function raised to a positive, even, integer value. The response surface is shown as continuous solely for the purpose of aiding the visualization. In actuality, the response surface exists at discrete positive, even integer, values of n.

  n  0.5n  1  n     2    cos  c  n  2k     0 k  0  k   0.5n  

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(43)

For any positive, even, integer value of n, the solution for equation (43) is c = ±0.5.

From Figure 3, it can readily be seen that the ratio of the acceleration difference, aji, to the amplitude, a, for equation (39), is constant over the domain of positive, integer values of n and that it is cyclical with respect to c.

For any value of n within this domain, the ratio is zero for integer values of c and unity for c = k/2 {k: k is an odd integer that is greater than or equal to zero}. Rearranging

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(42)

4. TIME DIFFERENTIATION 4.1 Velocity solutions For a given acceleration function, a(t), the velocity solution can be obtained by implicit integration of the acceleration.

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v  t   vi  a j  t  t i  

2 1 n t ji a ji

c  n   1 n  2k        0.5  n  1   k         k 0 sin  c  n  2k   t  t    i    t ji   

The constant of integration, C1, is typically solved for by using a point in time at which both the time and velocity are known. If the problem is an initial value problem, the point is at {ti, vi}. This leads to the following solution:

C 1  v i    a  t  dt    ti

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(45)

(48)

Finally, for the case when the acceleration is a cosine function raised to a positive, even, integer power, the velocity solution is:

For the case when the acceleration is a sine function raised to a positive, odd, integer power, the velocity solution is the following:

v  t   vi  a j  t  t i 

v  t   vi  ai  t  t i 

 n  n    2  0.5n   t  t i          n   1   k   n  2k     a ji  1  n    0.5 n  1  2 t ji    c  n  2k      c    k 0    t ji  sin       t  t      i   

  0.5  n  1   k  n  1     k   n  2k     1        0.5 n  1  1          c  n  2k     k  0      cos    (46)  t ji      t  t    t ji a ji       i    c   0.5  n  1   k  n    0.5  n  1   k    1           k 0    sin  c  n  2k     

(49)

When t = tj, the solution for the velocity, v(tj), must be equal to vj. This can be written as vj = vi + vji. For the case where the underlying acceleration is a sine function raised to a positive, odd, integer power, writing equation (46) at t = tj leads to the following result.

For the case when the acceleration is a sine function raised to a positive, even, integer power, the velocity solution is the following: v  t   vi  ai  t  t i 

v ji  a i t ji 

 n  n    2  0.5n   t  t i          n     0.5n  k 1   k   n  2k     1   1 n    0.5n  1   2 t ji    c  n  2k    c    k 0   t ji   sin    (47)     t  t    i    a ji    n   n 2         0.5n      0.5n  k  n    0.5n  1  1       1n   k 2       k 0  cos  c  n  2k     

a ji t ji c

   0.5  n  1   k  n   0.5 n  1  1  k         k  0  n  2k 1  1  cos  c  n  2k          0.5 n  1 n     0.5  n  1   k     1   k  sin  c  n  2k         k 0 

(50)

The denominator within the parenthetical is zero valued when c is an integer. For the case where the underlying acceleration is a sine function raised to a positive, even, integer power, writing equation (47) at t = tj leads to the following result. v ji  a i t ji  a ji t ji

  0.5n  k  n  1     n  2 0.5n  1  1   k   n  2k        0.5n   c    k 0   sin  c  n  2k     (51)     0.5n  k  n     n  0.5n  1  1  k      2      0.5n    k  0      cos  c  n  2k     

For the case when the acceleration is a cosine function raised to a positive, odd, integer power, the velocity solution is:

For the case when the underlying acceleration is a cosine function raised to a positive, odd, integer power, writing

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equation (48) at t = tj and using the previously derived result of a = aji leads to the following solution.

v ji  a j t ji  0.5 n  1

  k 0

u  t   ui  vi  t  t i  

c

 n   1    n  2k  sin  c  n  2k     k  

(52)

For the case where the underlying acceleration is a cosine function raised to a positive, even, integer power, writing equation (49) at t = tj and using the previously derived result of a = aji leads to the following solution.

v ji  a j t ji  a ji t ji   n  n  2 1 n   2     0.5n  c     0.5n  1  n         n  2k 1 sin  c  n  2k      k 0  k     

(53)

u  t   ui  vi  t  t i  

(54)

The constant of integration, C2, can be solved for by using a point where both the time value and displacement are known. For an initial value problem, this condition is {t i, ui}. For the case in which the acceleration is a sine function raised to a positive, odd, integer power, the displacement solution is given by the following.

2

t  t 

u  t   ui  vi  t  t i  

2

i

Impact Factor value: 8.315

(56)

aj 2

t  t 

2

i

aj 2

t  t 

2

i

 n 1  n   2 2     t  ti   0.5n       (58)  n   2   k   n  2k    a ji  1  n 2     2 t ji 0.5n  1    c n  2k        2 c 2   k  0        t ji  1  cos       t  t         i   The displacement at the terminus boundary point, tj, must be equal to uj. This displacement may also be written as uj = ui + uji. For the case in which the underlying acceleration is a sine function raised to a positive, odd, integer power:

For the case in which the underlying acceleration is a sine function raised to a positive, even, integer power, the displacement solution is given by the following.

|

2

i

For the case in which the underlying acceleration is a cosine function raised to a positive, even, integer power, the displacement solution is given by the following:

  0.5  n  1   k  n  1     k   n  2k     1          t ji   0.5 n  1   t  t i     c  n  2k        k  0    (55)  c  n  2k          t ji   sin            a ji t ji        t  t i    n c     0.5  n  1   k 0.5  n  1     1    k         k 0      sin  c  n  2k     

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t  t 

 n   2  k   n  2k      1 n 2  (57) 2 a ji t ji 0.5 n  1     c  n  2k       2 2    c k 0   t ji  1  cos     t  t       i 

The displacement solution, for a given velocity function, v(t), can be obtained by implicit integration.

u  t   ui  vi  t  t i  

2

For the case in which the underlying acceleration is a cosine function raised to a positive, odd, integer power, the displacement solution is given by the following:

4.2 Displacement solutions

ai

ai

  n   2  0.5  0.5n   t  t i           0.5 n  k  n  2   1 n  2k            k 2   0.5 n  1  2 t ji     c  n  2k       2 c 2     k  0   t ji   1  cos        t  t        i   a ji  n        n  0.5 n  1  1 0.5 n  k        k   2   k 0   0.5n   cos  c  n  2k     

2 1 n a ji t ji

u  t    v  t  dt  C2

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u ji  v i t ji 

ai 2

t ji 2 

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solutions were coded in a symbolic mathematics package (Mathematica v. 12.0; Wolfram Research, Inc.; Champaign, Illinois, USA). The exponential power was arbitrarily chosen as four for the even integer case and five for the odd integer case.

For the sine based acceleration functions, the initial value of {ti, ai} = {0.1, 5} along with tji = 0.07, aji = 20 and c = 0.19 (0.095 of a full period) were arbitrarily chosen. For the cosine based acceleration functions, the values of ti = 0.1, aj = 5, tji = 0.07, aji = 20 and c = 0.5 were chosen. The value of c, for the cosine case, was not arbitrary while the values of the other parameters were arbitrary. The results from the coded equations were compared against the results based upon the solutions from direct entry of each function (in the form of ai + a sinn(x) or aj + a cosn(x)). The results for the acceleration functions are shown in Figures 4 and 5.

a ji t ji 2 c

  0.5  n  1   k  n  1    0.5 n  1  1   k   n  2k              sin  c  n  2k    k 0   1    c  n  2k          n     0.5  n  1   k 0.5  n  1     1    k         k 0    sin  c  n  2k     

(59)

For the case in which the underlying acceleration is a sine function raised to a positive, even, integer power: u ji  v i t ji 

1 2

a i t ji 2  a ji t ji 2

   0.5n  k  n   k   1        0.5n  1 2  0.5  n   2    n  2k     0.5n  2 c 2     (60) k 0   1         cos  c  n  2k            0.5n  k  n     n  0.5n  1  1  k         0.5n   2   k 0      cos  c  n  2k     

Figure 4: Acceleration response for the sine functions.

The image on the left shows the overlay for the case where n = 5 (the results from equation 38 are shown in blue with the direct entry results shown in orange). The image on the right shows the overlay for the case where n = 4 (the results from equation 40 are shown in green with the direct entry results shown in yellow).

For the case in which the underlying acceleration is a cosine function raised to a positive, odd, integer power:

2 1 n a ji t ji 2 1 u ji  vi t ji  a j t ji 2  2 2 c 2 (61) 0.5  n  1  n   2      n  2k  1  cos  c  n  2k    k  0  k  



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

For the case in which the underlying acceleration is a cosine function raised to a positive, even, integer power: u ji  v i t ji 

aj

t ji  a ji t ji 2   n  1  n  2 1 n  2  0.5n   2 2    c  2

2

     n   2     n  2k     0.5 n  1  k        c     k  0  1  cos        n  2k       

Figure 5: Acceleration response for the cosine functions.

The image on the left shows the overlay for the case where n = 5 (the results from equation 34 are shown in blue with the direct entry results shown in orange). The image on the right shows the overlay for the case where n = 4 (the results from equation 36 are shown in green with the direct entry results shown in yellow).

(62)

5. SOLUTION VALIDATION

For the velocity comparisons, an arbitrary initial velocity, vi = 2, was used. The results for the velocity comparisons are shown in Figures 6 and 7.

The kinematic solutions for the sine based acceleration formulations as per equations (38) and (40), cosine based acceleration formulations as per equations (34) and (36), and the corresponding velocity and displacement

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Figure 6: Velocity response for the sine functions. The image on the left shows the overlay for the case where n = 5 (the results from equation 46 are shown in blue with the direct entry results shown in orange). The image on the right shows the overlay for the case where n = 4 (the results from equation 47 are shown in green with the direct entry results shown in yellow).

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Figure 9: Displacement response for the cosine functions.

The image on the left shows the overlay for the case where n = 5 (the results from equation 57 are shown in blue with the direct entry results shown in orange). The image on the right shows the overlay for the case where n = 4 (the results from equation 58 are shown in green with the direct entry results shown in yellow).

The solutions generated by the equations presented in the subject work, matched, exactly, the solutions based upon direct entry for the values of n that were considered.

6. DISCUSSION The closed form analytic solutions presented in the subject work are certainly more complex, in appearance, than those typically found in the extant scientific literature.

The basis for this increased complexity is multifactorial.

Firstly, the equations presented in the subject work account for non-zero valued initial acceleration offsets.

Not every analysis involves a situation in which the initial value of time is zero and the initial acceleration is zero. An excellent example of such a situation is that in which one is evaluating the separation phase of a collision while having time t = 0 being the time at which closure initiates.

Secondly, for the case in which the acceleration response is modeled using the sine function, the subject formulation allows for the consideration of fractional period responses beyond the standard half period or quarter period responses. Historically, the half period model has been used when modeling the entirety of an event and with the acceleration being symmetric in shape [29] whereas the quarter period model has been used when segregating the acceleration response (i.e. for a collision in regards to closure and separation) [34]. The half period model can also be used for a segregated response analysis. However, this segregation occurs in regards to the velocity response while leaving the symmetric shape of the acceleration response unchanged. While not addressed herein, the half period and quarter period models, as historically used, generally produce poor matches for the total set of segregated kinematic response parameters for data based upon even the simplest collision test configuration (i.e. full-width engagement, collinear, axial impact between a test vehicle and a FRMB). Thirdly and finally, the greatest source of complexity in regards to the forms of the subject equations is the inclusion of arbitrary, positive, integer exponentiation.

Equations (16) and (14) are valid

Figure 7: Velocity response for the cosine functions. The image on the left shows the overlay for the case where n = 5 (the results from equation 48 are shown in blue with the direct entry results shown in orange). The image on the right shows the overlay for the case where n = 4 (the results from equation 49 are shown in green with the direct entry results shown in yellow).

For the displacement comparisons, an arbitrary initial displacement of ui = 0.5 was used. The results for the displacement comparisons are shown in Figures 8 and 9.

Figure 8: Displacement response for the sine functions.

The image on the left shows the overlay for the case where n = 5 (the results from equation 55 are shown in blue with the direct entry results shown in orange). The image on the right shows the overlay for the case where n = 4 (the results from equation 56 are shown in green with the direct entry results shown in yellow).

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equations for sinn(x) and cosn(x) where {n: n is in the integers} and could have readily been used for the baseline acceleration functions. However, both equations are complex equations in the mathematical sense (i.e. they both contains terms associated with j = (-1)0.5). The separation of the responses based on odd and even integer valuation avoids having to deal with complex analysis but is at the expense of increasing the complexity in the appearance of the equations. Again, it should be noted that there is no underlying, uniform and ubiquitous requirement that the exponential order of the trigonometric functions employed should be limited to either a value of one or two.

work lies within the solution validation and is due to the use of singular parameter values for each case. This choice was made due to the fact that the solution validation was simply meant to show that the equations presented herein match the results of the solutions based upon the symbolic solutions for the case in which the exponential has a defined, ‘known’ value. The presence of derivational errors or typographical errors, both being a concern given the complex appearance of the equations, would have been readily manifest for any arbitrary odd or even valued integer of choice. While not shown here, the arbitrary changing of each exponential value to other integer values within each subset (i.e. odd or even) produced an equivalent result when compared with entered, symbolic, solutions of the form of both sinn(x) and cosn(x).

The presentation of the solutions based upon the use of the use of the cosine function might appear to be extraneous on first glance. However, such is not the case.

The cosine function correctly models the shape of the second quarter period response of a half period sinusoid.

This can readily be seen in Figure 10, which depicts a simple half period sine response, the corresponding initial quarter period sine response (overlayed) and the overlayed cosine and sine models for the corresponding final quarter period response.

REFERENCES

Figure 10: A simple half period sine function with an initial offset and return to the original ordinate value (left). The middle image shows the first quarter period sine response (green) overlayed on the half period response. The image on the right shows the cosine response (red) as well as the sine response (green), for the second quarter period, overlayed on the half period response.

The formulative nature of the subject work was strictly to derive the closed form analytic solutions for the kinematic responses of an event pulse modeled using sinusoids exponentiated to integer powers. In that regard, the equations shown, to the author’s knowledge are presented for the first time within the context of event pulse modeling. The exclusion of application examples, based upon the stated scope of the subject work, in the author’s view, does not represent an intrinsic limitation of the subject work. This choice was made due to space limitations and the necessity for the presentation, in a subsequent work, of the equations when the exponential power of the sinusoidal functions is not limited to integer powers. Example applications predicated upon real-world data do not involve an a priori basis of being limited, in the modeling sense, to sinusoidal functions that are raised to just integer powers. The intrinsic limitation of the subject

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Proceedings: International IRCOBI Conference, Munich, Germany, September 18-20, 2002.

[15] Evans NC, LM Furton and DA Cok (1992) Occupant energy management technique for restraint system analysis and design – theory and validation. SAE Technical paper no. 922082.

[27] Linder A, M Avery, M Krafft and A Kullgren (2003) Change of velocity and pulse characteristics in rear impact: real world and vehicle tests data.

Proceedings: 18th International Technical Conference on the Enhanced Safety of Vehicles, Nagoya, Japan, May 19-22, 2003. Paper No. 285.

[16] Grimes WD and FD Lee (2000) The effect of crash pulse shape on occupant simulations. SAE Technical paper no. 2000-01-0460.

[17] Katoh H and R Nakahama (1982) A study on the ridedown evaluation. Proceedings: 9th International Technical Conference on Experimental Safety Vehicles, Kyoto, Japan, November 1-4, 1982, pp. 190195.

[28] Siegmund GP, BE Heinrichs, DD Chimich, AL DeMarco and JR Brault (2005) The effect of collision pulse properties on seven proposed whiplash injury criteria. Accident Analysis & Prevention 37(2): 275285.

[18] Bonello KJ (1992) Occupant energy management technique for restraint system analysis and design – understanding the physics of the system. SAE Technical paper no. 922083.

[29] Varat MS and SE Husher (2003) Crash pulse modeling for vehicle safety research. Proceedings: 18th International Technical Conference on the Enhanced Safety of Vehicles, Nagoya, Japan, May 1922, 2003. Paper no. 501.

[19] Huang M, J Laya and M Loo (1995) A study on ridedown efficiency and occupant responses in high speed crash tests. SAE Paper No. 950656.

[30] Singh J and J Perry (2005) Comparative analysis of FMVSS 208 sled and dynamic deceleration pulse characteristics.

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[20] Wu J, GS Nusholtz and S Bilkhu (2002) Optimization of vehicle crash pulses in relative displacement

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Vehicles, Washington DC, USA, June 6-9, 2005. Paper no. 05-0028.

[31] Hynd D and C Willis (2007) Recommendations for a low-speed rear impact sled test pulse. Document No. 166, European Enhanced Vehicle Safety Committee (EEVC) WG20 Report.

[32] Singh J and J Perry (2007) Implication of symmetric sine-based collision pulse models on force-deflection characteristics. Collision 2(2): 104-113.

[33] Singh J and J Perry (2008) An alternative formulation for symmetric sine based collision pulse models.

Collision 3(1): 34-43.

[34] Singh J (2013) Further developments regarding the dynamic modeling of motor vehicle collision response using the SDOF approach. Collision 8(1): 10-31.

[35] McHenry RR (1975) A comparison of results obtained with different analytical techniques for reconstruction of highway accidents. SAE Technical paper no. 750893.

[36] Singh J (2013) Laplace domain analysis of the collinear vehicle-to-vehicle collision under linear force-deflection constraints coupled with net externally applied loads.

Proceedings: 23rd International Technical Conference on the Enhanced Safety of Vehicles, Seoul, South Korea, May 27-30, 2013. Paper No. 13-0016.

BIOGRAPHY

© 2025, IRJET

Mr. Singh is a private practice engineer with specialties in the fields of motor vehicle accident reconstruction and biomechanical engineering.

He holds a BS in mechanical engineering from the University of Illinois at UrbanaChampaign and a MS in biomedical engineering from the University of Southern California.

He has authored numerous peer reviewed conference proceedings, technical articles and scientific papers.

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