On the Mutual Coefficient of Restitution in Two Car Collinear Collisions

/4/006 physics/06068 On the Mutual Coefficient of Restitution in Two Car Collinear Collisions Milan Batista University of Ljubljana, Faculty of Maritie Studies and Transportation Pot poorscakov 4, Slovenia, EU ilan.batista@fpp.edu (Updated Feb. 006) Abstract In the paper two car collinear collisions are discussed using Newton's law of echanics, conservation of energy and linear constitutive law connecting ipact force and crush. Two ways of calculating the utual restitution coefficient are given: one based on car asses and one based on car stiffness. A nuerical exaple of an actual test is provided.. Introduction For the odeling of the collinear car collision two ethods are usually used. The first is the so-called ipulse-oentu ethod based on classical Poisson ipact theory, which replaces the forces with the ipulses ([3], []). The second ethod treats a car as a deforable body; so the constitutive law connecting contact force with crush is necessary. For the copression phase of ipact the linear odel of force is usually adopted and the odels differ in the way the restitution phase of collision is treated ([7], [3], [4], [7]). The purpose of this paper is to extend the linear force odel discussed in [] to the collinear ipact of two cars. In the quoted article it is proposed that a car is characterized by its ass, stiffness and liit velocity for peranent crush. The latter properties can be established by a fixed barrier crush test. Also, the proposed restitution odel is siple: rebound velocity is constant. The question arises as to how these

/4/006 physics/06068 characteristics can be incorporated into the two car collision odel since it is well known that the utual coefficient of restitution is the characteristic of ipact; i.e., it is a two car syste and not the property of an individual car ([], [7]). To answer the above question, first the well-known theory of central ipact is specialized for collinear car collisions. The kinetic energy losses are then discussed and the restitution coefficient is related to the. The third section of the paper discusses two odels for calculating the utual restitution coefficient based on individual car characteristics. The last section is devoted to a description of the use of the present theory in accident reconstruction practice. The section ends with a nuerical exaple.. Two car collinear collision Consider a collinear ipact between two cars where collinear ipact refers to rear-end and head-on collisions. Before ipact the cars have velocities v and v respectively and after ipact they have velocities u and u (Figure ). Figure. The two car ipact: (a) pre-ipact velocities, (b) end of copression velocity, (c) post-ipact velocities

/4/006 3 physics/06068 In the collision phase the oveent of cars is governed by Newton's nd and 3rd laws (Figure ). On the basis of these laws equations of otion of the cars can be written as follows dv F dv dt = and dt = F () where and are the asses of the cars and F is contact force. Figure. Newton's 3rd law applied to collinear ipact of two cars Following Poisson's hypothesis ([6]), the ipact is divided into two phases: copression and restitution. In the copression phase the contact force F raises and the cars are defored. The copression phase terinates when the relative velocity of cars vanishes; i.e., when cars have equal velocity (Figure ). The copression phase () thus integrates the changes fro initial velocities to coon velocity u. This leads to the following syste of equations ( ) ( ) u v = P u v = P () c c where Pc τ c Fdt is copression ipulse and τ c copression tie. Fro () one 0 obtains the velocity after copression v + v u = + (3)

/4/006 4 physics/06068 and the copression ipulse ( ) Pc = v v + (4) In the restitution phase the elastic part of internal energy is released. Equations () are integrated fro u to the end velocities, which gives two equations for three unknowns ( ) ( ) u u = P u u = P (5) r r where Pr τ c Fdt is restitution ipulse and τ r is restitution tie. In order to solve 0 syste (5) for an unknown's post-ipact velocity and restitution ipulse the constitutive equation is needed. According to the Poisson hypothesis the restitution ipulse is proportional to copression ipulse P r = ep (6) c where e is the restitution coefficient. Because contact force is non-negative, so are copression and restitution ipulse. Fro (6) this iplies that e 0. Note. Instead of (6), one can use Newton's kineatical definition of restitution coefficient u e = v u v which is in the case of centric ipact without friction equivalent to Poisson s definition. However in the case of non-centric ipact with friction Newton's odel could lead to overall energy increase ([]).

/4/006 5 physics/06068 The total ipulse is P= Pc + Pr so by using (4) and (6) P= ( + e) Δv + (7) Solving (5) and (6) and taking into account (4) gives the well known forulas (see for exaple [3], []) for the cars post-ipact velocities ( + ) e = Δ = Δ u u e v v v + + ( + ) e = + Δ = + Δ u u e v v v + + (8) where Δ v= v v. The above equations can be used for calculation of post-ipact velocities if pre-ipact velocities are known, asses of cars are known and, in addition, the restitution coefficient is known. 3. Energy consideration At car ipact the kinetic energy is dissipated. Applying the principle of conservation of energy one obtains, after copression, ( + ) u v v + = +Δ E (9) where Δ E is axial kinetic energy lost (or axial energy absorbed by crush). By using (3) one has E Δ = + Δ v (0)

/4/006 6 physics/06068 Siilarly, by applying the principle of conservation of energy to the overall ipact process v v u u + = + +Δ E () one finds the well known forula for total kinetic energy lost (see for exaple []) Δ E = ( e ) v + Δ () Since, by the law of therodynaics, ΔE 0, it follows fro () that e. Now, fro (0) and () one has ( ) given by ([]) Δ E = e Δ E, so the utual restitution coefficient is ΔE e = = ΔE a ΔE ΔE 0 (3) where ΔE0 ΔE Δ E is the rebound energy. The forula obtained is the basis for relating the utual coefficient of restitution e with the restitution coefficients obtained for individual cars in the fixed barrier test. 4. The utual coefficient of restitution Let v T be a barrier test velocity of a first car and v T a barrier test velocity of a second car. Let these velocities be such that the axial kinetic energy lost can be written as v T v T Δ E = + (4) and in addition the rebound energy can be written as (see [9])

/4/006 7 physics/06068 e vt e vt Δ E0 = + (5) The utual restitution coefficient is therefore fro (3), (4) and (5), by using (0), e = e v + e v T T v T+ v T (6) For the odel of the barrier test proposed in [] the restitution coefficients of cars are e v in, 0 = vt v 0 and e = in, vt (7) where v 0 and v 0 are liited ipact velocities where all the crush is recoverable ([]). The task is now to deterine appropriate test velocities of cars which satisfy (4). 4. Model A - stiffness based utual restitution coefficient. Let v T be the barrier test velocity (or barrier equivalent velocity [8]) of the first car for the sae crush as in a two car ipact and v T the barrier test velocity for the sae crush for the second car. Then the test velocities for the sae crush ust satisfy relations ([], [8]) v T kδ = and v T kδ = (8) where k and k are stiffness of the cars and δ and δ are actual axial dynaics crush of the cars. Fro (8) one has

/4/006 8 physics/06068 v = k δ and k v = T δ (9) T On the other hand, fro (0), (4) and (8) it follows that kδ k δ Δ E = Δ v = + + (0) Defining overall axial crush δ δ + δ and taking into account the law of action and reaction kδ = kδ one obtains δ k δ δ k = = δ k+ k k+ k () Substituting () into (0) yields Δv kδ Δ E = = () where is syste ass and k is syste stiffness, given by kk k + k + k (3) Fro () one has δ = Δ v and therefore fro (9) the required test velocities are k (see also [8]) k k v = Δ v and v = Δ v (4) T T k k

/4/006 9 physics/06068 Substituting (4) into (4) leads to identity provides the required utual restitution coefficient = + and substituting it into (6) k k k e = ke k + ke + k (5) This equation for the calculation of e were published by various authors ([4],[5],[5]). Knowing the ass and stiffness of the cars and Δ v one can calculate test velocities fro (4), restitution of individual cars fro (7), the utual restitution coefficient fro (5) and post-ipact velocities fro (8). 4. Model B - ass based utual restitution coefficient. This odel does not include cars stiffness and it's based on (0) and (4) only. Equating (0) and (4) results in the equation Δ v = v + v (6) T T for two unknowns. To solve it one could set v = v v v = v v (7) T 0 T 0 where v 0 is a new unknown velocity. Substituting (7) into (4) one obtains after siplification ( ) ( ) v v + v v = 0 0 0, so v 0 v = + v + (8) This is in fact the velocity of the centre of the ass of colliding cars. Substituting (8) into (7) yields unknown test velocities

/4/006 0 physics/06068 v ( ) ( ) v v v v = v = T T + + (9) Note that in calculation of restitution coefficients (7) the absolute values of test velocities should be used. Substituting (9) into (6) gives the utual restitution coefficient e = e + e + (30) This forula was derived by different arguents of Howard et al ([9]) and is also quoted by Watts et al ([8]). 4.3 Copartent of the odels Coparing (4) and (5) one finds that test velocities of both odels are the sae if stiffness is proportional to the ass; i.e., k = k 0 and k = k 0 where k 0 is a constant. While the test velocities of the odels differ, the utual restitution coefficient differs only in the case when just one car is crushed peranently, since when vt v0 and vt v0 then both e = e = so by (5) or (30) it follows e = and when vt > v0 and vt > v0 then substituting (7) and appropriate test velocities into (5) or (30), and taking (0) into account, yields e = v + v Δv 0 0 (3)

/4/006 physics/06068 Note that (3) can not be used directly for calculating the utual restitution coefficient in advance since the classification of ipact--fully elastic, fully plastic or ixed-- depends on test velocities. At last the question arises as to which odel is ore physically justified. While Model A has a sound physical base connecting test velocities with crushes, Model B requires soe additional analysis. It turns out that it can be interpreted as follows. The copression ipulse (4), can be written by using (3) as Pc = Δ v. Using () one could define test velocities of individual cars as velocities resulting at the end of the copression phase in a fixed barrier test as the sae ipulse as in an actual two car collision; i.e., P = Δ v = v = v (3) c T T Fro this equation, test velocities given already by (9) result. Now by (6) restitution ipulse is Pr = epc = e Δ v, so by (5) and (3) one ust have e Δ v = ev = e v. But this can be fulfilled only in the special case when e T T = e, and consequently, by (30), when e= e. This consequence raises a doubt about Model B s adequacy for general use. 4.4 Exaples The above forulas were ipleented into the spreadsheet progra (Table ).As the exaple, a full scale test (test no. 7) reported by Cipriani et al ([6]) was executed. In this test the bullet car ade ipact with the rear of the target car at a velocity of 5 /s or 8 k/h. The ass of the cars and their stiffness was taken fro the report; however, the liit speed was taken to be 4 k/h for both cars ([]). The result of the calculation is shown in Table. The calculated velocity difference for the target car is 4.8 k/h, which differs fro that easured (3.9 /s or 4.0 k/h) by about 5%. The calculated velocity change for the bullet car is.3 k/h and the easured one was.9 /s or 0.4 k/h. The discrepancy is thus about 7%. If one takes the liit speed to be 3 k/h,

/4/006 physics/06068 then the calculated value of velocity change for the bullet car is 3.6 k/h, differing fro that easured by about %, and the calculated value of velocity change for the target car is 0.4, which actually atches the easured value.. Table. Spreadsheet progra for calculation of post-ipact velocities Full scale test 7 of Cipriani et al ([6]) Vehicle Vehicle ass kg 46 495 stiffness kn/ 886.07 564.687 liit velocity k/h 4 4 ipact velocity k/h 8 0 Delta V k/h 8.00 velocity after copression k/h 7.8 syste ass kg 648.7 syste stiffness kn/ 565.7 test velocity k/h 0.8 7.3 test restitution 0.37 0.56 restitution 0.45 post ipact velocity k/h 3.4.3 Delta V k/h 4.76 -.3 Maxial crush 0. 0.06 Residual crush 0.07 0.03 5. Accident Reconstruction In a real car accident the proble is not to deterine post-ipact velocities but usually the opposite; i.e., to calculate the pre-ipact velocities. For deterining pre-ipact velocities, however, the post-ipact velocities deterined fro skid-arks should be known. If only the peranent crushes of cars are known then only the velocity changes for individual cars in an accident can be calculated. If the characteristics of cars are known--i.e., ass, stiffness and liit velocity--then the proble is solved as follows. Let δ r be residual crush of the first vehicle. The axial crush, then, is ([])

/4/006 3 physics/06068 δ = δ + δ (33) r 0 where the recoverable part of crush is calculated as δ 0 = v0. The axial crush of k the second car can be calculated in the sae way or fro Newton s 3rd law as δ k = δ (34) k The axial energy lost at ipact is then calculated fro Δ E =Δ E +Δ E (35) kδ where Δ E = and fro (), kδ Δ E =. The pre-ipact velocity difference is thus, ΔE Δ v = (36) To calculate velocity changes of individual vehicles the first test velocities are calculated by (8) v ΔE ΔE = v = (37) T T Fro (7) the restitution coefficient for individual cars are calculated and fro (5) the utual coefficient of restitution. Fro (8) the velocity differences of individual cars at ipact are ( + e) ( + ) e Δ v = v u = Δv Δ v = v u = Δv + + (38)

/4/006 4 physics/06068 The above forulas were prograed into a spreadsheet progra (Table ). As the exaple, the car to car test described by Kerkhoff et al ([0]) is considered. In this test the test car (bullet) struck the rear of the stationary car (target) at a speed of 40.6 ph or 65 k/h. The actual easured Δ v was.6 ph or 36. k/h. As can be seen fro Table, the calculated value Δ v for the bullet car is 36. k/h; i.e., the discrepancy between actual and calculated value is 0.% and the calculated ipact velocity 64.4 k/h differs fro the actual by.3 %. Note that the deforation of the stationary car was not reported, so (34) is used for calculation of its axial dynaic crush. The liit speed for both cars was taken to be 4 k/h ([]). The discrepancy of calculated values in the previous case is so inial because the actual low ipact velocity tests were used for deterination of stiffness. If one used for the calculation the default values of CRASH stiffness and appropriate calculated liit velocity for class cars the discrepancy would increase. Thus, in this case the calculated velocity change of the bullet car is 38.5 k/h, which differs fro the actual change by about 6% and the calculated Δ v is 5. k/h, differing by about 0%. Table. Spreadsheet progra for calculation of velocity differences at ipact. Car to car test no by Kerkhoff et al ([0]) Vehicle Vehicle ass kg 00.44 0. Data stiffness kn/ 68.9 87.89 liit speed k/h 4.00 4.00 crush 0.6? recoverable crush 0.03 0.04 axial crush 0.9 0.36 syste ass kg 550.39 syste stiffness kn/ 574.65 ax energy lost kj 9.86 57.53 test velocity k/h 6.5 36.80 test restitution 0.5 0. restitution 0. Delta V k/h 36.09 64.5-36.06

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