co-axial impact theory - angelfire · 2002-07-05 · co-axial impact theory special problems y2k...

CO-AXIAL IMPACT THEORY LECTURE

J. Duff, P.Eng., Ph.D. For-Eng Consultants Ltd.

Introduction

This paper is an extended version of the first half of a paper on the came subject published last year and authored by Duff and Mikulcik [1999]. This paper first describes how the Conservation of Momentum equation is derived directly from Newton's Laws of Motion. It also shows that it is possible to solve for two unknowns in the Conservation of Momentum equation for Co-Axial Impacts by the addition of a second equation. Two examples of possible second equations are given—one being the Coefficient of Restitution condition, and the other being the Conservation of Energy equation.

In addition to the two sets of extra equations, the paper introduces a novel method of solving for any of the two unknowns in a co-axial accident investigation by expressing the collision impulse in four separate ways. The four expressions are combined in the Impact Identities Equation.

In order to implement the Impact Identities Equation it is necessary to appreciate the concept of the Coefficient of Restitution. This concept is explored fully, and its relevance to motor vehicle collisions is exposed. Its similarity to the Coefficient of Friction is described and an example of its use is also provided.

Similarly, the equations relating to the loss of kinetic energy are discussed, and a potential pitfall is highlighted. Some remarks are directed to the modes and distribution of the kinetic energy loss, but the paper stops short of discussing crush characteristics.

The paper covers the theory of impacts in a unique way, and introduces three factors that determine the severity of a co-axial impact. It is hoped that if the reader can grasp the significance of these three factors, it will lead to a clearer understanding of the effect of an impact between two bodies, and some commonly held misconceptions will be dispelled.

CO-AXIAL COLLISION THEORY WITHOUT EXTERNAL FORCES

Consider a body of mass M moving at velocity V approaching a second body. The second body has a mass m and is moving in the same co-axial direction at velocity v. In this format, all symbols dealing with the first body will be denoted by capital letters, i.e., M, V, and all symbols

Special ProblemY2K, Come Crash With Us IPTM Lecture May, 2000

Co-Axial Impact Theory Special Problems Y2K IPTM Lecture, May 2000

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dealing with the second body will be denoted by lower case letters, i.e., m, v. This is a similar format to that followed by Macmillan (1983).

If we arbitrarily choose the body denoted by capital letters to be the “bullet” and the other body as the “target”, it follows that in order for the bodies to collide V > v. In contrast to Macmillan’s approach, in which each body is assigned a positive motion as it advances towards the impact, in this theory motion to the right will be deemed positive for both bodies. If m were travelling in the opposite direction it would have a value of –v. However, continuing with Macmillan’s format, the subscript 1 will be used to indicate conditions at the start of impact, and the subscript 2 will represent conditions at the end of impact. Therefore at the start of impact, mass M is moving at velocity V1 and mass m is moving at velocity v1.

To help follow this theory we shall assign numerical values to each of the variables as they are introduced. For example, let V1 = 10 m/s and v1 = 1 m/s.

Concentrating on mass M for the moment, we can say that it is "A body in motion that will remain in motion without acceleration until acted upon by an external force". That description is Newton's First Law of Motion and therefore we see that mass M comes under the jurisdiction of Newton's First Law.

Mass M

V1

Force Fc

`

Time

Force

Fc

t1 t2

1(a) 1(b)

Figure 1 Force-Time Diagram from a Collision The "external force" is felt when the two bodies collide. A force Fc will be generated that will oppose the motion of the mass M, as shown in Figure 1(a). This force will vary in magnitude throughout the period of the impact. It will have a form somewhat like that shown in Figure 1(b).


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At this juncture, we do not need to know the exact shape of force, Fc. Neither do we need to know the duration of the impact, ∆t = t2 - t1. However, we will be interested in the area under the curve as shown in the figure.

Newton’s Second Law of Motion states that "The acceleration of the body will be directly proportional to the magnitude of the force, and inversely proportional to the magnitude of the mass". That statement can be cross-multiplied to define the general relationship, force equals mass times acceleration, i.e.,

F ma m vt

F t m v

= =

∴ =

∆∆

∆ ∆ (1)

When this general relationship is applied to a collision, force F becomes the variable Fc, which will be in effect for the duration of the collision, ∆t, i.e., from t1 to t2. The left side of the above equation, F∆t, is given by the area under the curve in Figure 1(b). The area can be expressed mathematically as:

F dt MdV M V Ict

t

V

V

c1

2

1

2z z= = ∆ = (2)

where Ic is known as the impulse, and has units of newton seconds, or kgm/s. ∆V is the change of velocity experienced by mass M due to the impulse of the impact. With the sign convention used in this lecture, it is defined as:

∆V V V= −1 2 (3)

If we decide that the final velocity, V2 = 2 m/s, we get ∆V = 10 - 2 = 8 m/s.

Combining equations (2) and (3), and re-arranging we get:

: MV MV I

V V IM

c

c

2 1

2 1

= −

= − (4)

Turning now to the other object, we know that mass m will experience exactly the same force Fc, but in the opposite direction during the collision, as per Newton’s Third Law of Motion, "Every action has an equal and opposite reaction". And of course the force will be felt for exactly the same time, i.e., from t1 to t2.


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Therefore: F dt mdv m v Ict

t

v

v

c1

2

1

2z z= = ∆ = (5)

where ∆v v v= −2 1 (6)

Notice that the order of the velocities is reversed when compared with equation (3) in keeping with the fact that the impulse is opposite in direction. Again, if we decide that the final velocity, v2 = 5 m/s, we get ∆v = 5 - 1 = 4 m/s.

Combining equations (5) and (6), and re-arranging we get

mv mv I

v v Im

c

c

2 1

2 1

= +

= + (7)

Equations (2) and (5) may be combined to form the equalities:

F dt I M V m vct

t

c1

2z = = =∆ ∆ (8)

Equation (8) contains the well-known expression M∆V = m∆v, registering the inverse relationship between mass and velocity change in a collision. The expression, M∆V = m∆v, is valid for the two body collision in our model. If other external forces, i.e., road forces, were being felt during the collision, an additional term would be necessary to account for these forces.

If we decide that M has a mass of 1000 kgs, the M∆V = m∆v expression requires that m = 2000 kilograms.

M∆V = m∆v can also be expanded into the expression for the Conservation of Momentum:

M V m vM V V m v v

MV MV mv mvMV mv MV mv

∆ ∆=

− = −

− = −+ = +

1 2 2 1

1 2 2

1 1 2

b g b1

2

g (9)

Numerically we get:

1000*10 + 2000*1 = 12,000 = 1000*2 + 2000*5.

What we have done is, we have used all three of Newton's Laws of Motion to prove that momentum is conserved in a collision. Again, if there were other external forces acting on the


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bodies during the collision, the Conservation of Momentum expression would require additional terms to account for these forces.

THE THREE FACTORS INVOLVED IN CO-AXIAL IMPACTS WITHOUT EXTERNAL FORCES

So far we have shown that the collision impulse, Ic, generates a change of momentum in each body, M∆V and m∆v. In essence we can determine the magnitude, or severity, of the collision by focusing on the outcome, i.e., the value of M∆V or m∆v. But what if we were conducting an experiment in which we were going to cause the two bodies to collide? Is it possible to predict the magnitude of the impulse, or the severity of the collision, beforehand? What are the factors that determine the magnitude of a collision?

First we have to introduce a phenomenon associated with colliding bodies. From experimentation and observation it has been noted that colliding bodies generally move apart at the conclusion of an impact. We shall allow for that possibility in our theory. It is common practice in the reconstruction industry to assume that, in most cases, there is negligible rebound. If a variable is included in our equations that accounts for rebound, it is always permissible to assign that variable a value of zero if there is no rebound in a particular case. This is similar in principle to assigning a velocity of zero to a stationary car.

Let the approach velocity and departing velocity between the two bodies be denoted by vc and vd respectively, so that:

v V vv v V

c

d

= −= −

1 1

2 2

(10)

Note that v2 > V2 in order that the bodies move apart. This rebound is accommodated by the Coefficient of Restitution. Let us designate the Coefficient of Restitution by the symbol, ε.

Therefore: ε = =−−

vv

v VV v

d

c

2

1 1

2 (11)

In practice ε varies from zero, where there is no rebound, to unity, where there is full rebound. Zero rebound is usually referred to as a perfectly plastic collision and full rebound is deemed to be a perfectly elastic collision.

In our numerical example we stated that V1 = 10 m/s, v1 = 1 m/s, V2 = 2 m/s, and v2 = 5 m/s. Therefore the Coefficient of Restitution for that example would be:


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ε = −−

FHG

IKJ =

−−

FHG

IKJ = =

v VV v

2 2

1 1

5 210 1

39

0 333.

Equations (4) (7) and (11) may be combined as follows to produce the impulse expression:

From (11), we get

from (4) and (7) v

leading to

1

v V V vIm

V IM

V v

Im M

V v V v

I M mMm

V v

I MmM m

V v

c c

c

c

c

2 2 1 1

1 1 1

1 1 1 1

1 1

1 1

1 1

1

1

− = −

+ − + = −

+FHG

IKJ = − + −

+FHG

IKJ = + −

= ++

FHG

IKJ −

ε

ε

ε

ε

ε

b gb g

b g b g

b gb g

b g b g

(12)

from which the equalities of statement (8) may be expanded to:

F dt I M V m v m v

m MmM m

V v

ct

t

c c

c

1

2 1

1 1

z = = = = +

=+

= −

∆ ∆ εb gwhere

and, from (10) vc

c

(13)

The expression, (1+ε)mcvc, groups together the three factors that contribute to a co-axial collision without external forces. They are:

1. Approach Velocity, embodied by the expression vc.

2. Mass Effect, embodied by the expression mc.

3. Structural Aspects, embodied by the expression (1+ε).

In effect, these three factors state that the intensity of the collision, i.e., the impulse generated by the impact, is governed solely by the velocity of approach between the objects, the effective mass of the objects, and the structural composition of the objects in the region on and behind the contact surfaces. If there were other external forces involved in the collision, a fourth factor would be added to the mix.


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The primary purpose of this lecture is to instill a deep understanding of these factors, so that their contribution to the outcome of a collision can be fully appreciated. We shall describe each one in turn:

Approach Velocity

vc is the velocity at which the two bodies approach each other. This factor points out that the intensity of the collision is not related to the individual velocity of either body relative to the ground. It is directly related to their relative velocity.

In other words, a collision in which a stationary vehicle is struck in the rear by another vehicle advancing at 20 kilometres/hour, will have the same affect as one in which the first vehicle is driving forward at 10 kilometres/hour and the other is travelling in reverse at the same speed. It is also identical to the situation where the first vehicle is travelling at 120 kilometres/hour and strikes the second vehicle that is travelling in the same direction at 100 kilometres/hour.

The first and third scenarios described above might be thought of as a low speed and a high-speed collision respectively, and the overall consequences of the accident are likely to be quite different. But the danger from the high-speed collision is due to the aftermath of the impact, particularly so if the drivers lose control. The damage caused by the actual vehicle-to-vehicle contacts will be identical.

Mass Effect

mc is called the effective mass. It is a more difficult concept to visualize. This factor shows that the intensity of the collision is not defined directly by either of the masses. Rather, the intensity is related to the combination of the masses expressed in the form of the mc equation. The units of mc are kilograms if the masses are measured in kilograms, or lbs if the masses are measured in lbs.

In co-axial theory, mc is a valuable concept in that it can be used to analyze impacts varying from bodies of equal masses, to impacts where one body is deemed to be of infinite mass, e.g., test barriers, bridge abutments, etc. As an example of how mc varies as one mass gets larger and larger consider the following:

Let M = 1000 kilograms and m = 1000 kilograms.

mc =×+

= × = × =1000 10001000 1000

10002000

1000 12

1000 500b g kgs

In this case, mc = M/2 = m/2, i.e., half of the "lighter" mass.


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Let M = 2000 kilograms and m = 1000 kilograms, or M = 1000 kilograms and m = 2000 kilograms, as it makes no difference which mass is the larger.

mc =×+

= × = × =2000 10002000 1000

20003000

1000 23

1000 667b g kgs

In this case, mc = 2/3 × M, or 2/3 × m, i.e., 2/3 of the lighter mass.

Increasing the heavier mass to 5000 kilograms while keeping the lighter one at 1000 kilograms produces

mc = 5/6 × the lighter mass = 833 kgs

At a mass ratio of 10:1 we get

mc = 10/11 × the lighter mass = 909 kgs

It can be seen that as the heavier mass approaches infinity, mc approaches 1000 kilograms, i.e., the lighter mass. If one of the masses equals infinity, i.e., m = ∝, then mc = (M × ∝)/(M + ∝) ≅ M.

Therefore we observe that mc varies from half of the lighter mass, when both masses are essentially equal, to the value of the lighter mass when the other is essentially infinite. It follows that the severity of the collision, as defined by the impulse, is governed by half of the lighter mass, then is increased by as much as a factor of two as the other mass varies from that of the lighter mass up to infinity.

Structural Aspects

The structural factor deals with the generation of forces as the colliding surfaces deform. The expression (1+ε) is just one manifestation of the structural factor. Other manifestations deal with the shape of the two objects, and the extent to which energy is consumed in the form of temporary or permanent deformation. Object shape becomes a consideration in oblique collisions, but can be ignored in simple co-axial theory where two flat surfaces approach each other squarely.

To demonstrate how the structural aspects play a part in influencing the outcome of a collision, consider dropping two balls onto a flat surface from the same height. Let the balls be of equal mass, and if the contact surface is firmly attached to the earth, we can say that each ball is impacting an infinite mass. Therefore mc will be equal to the mass of the ball and will be identical in each case.


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Because the balls are dropped from the same height, they will have the same impact velocity if we neglect air friction. Therefore the value of vc will also be identical in each case.

Now if one of the balls is made of material similar to that of a squash ball, and the other similar to a golf ball, we can expect that the "golf" ball will bounce higher than the "squash" ball. Therefore the Coefficient of Restitution will be greater in the case of the "golf" ball. It follows from equation (13) that the "golf" ball will have the higher Impulse and the higher ∆V. That demonstration shows how the structural aspects play a part in the outcome of a collision.

Numerical Examples The effects of varying mc and ε

Table 1 shows the effect of a bullet vehicle weighing 1000 kilograms, travelling at a velocity of 10 m/s, and striking a stationary object of various masses. Two values of the Coefficient of Restitution have been chosen. The ∆V for each object is calculated from the relationships in equation (13).

V1 = 10 m/s v1 = 0 m/s ∴vc = 10 – 0 = 10 m/s

ε = 0 ε = 0.5

M, kg 1000 1000 1000 1000 1000 1000

m, kg 1000 2000 infinite 1000 2000 infinite

mc, kg 500 667 1000 500 667 1000

Ic, kgm/s 5000 6667 10000 7500 10000 15000

∆V, m/s 5 6.67 10 7.5 10 15

∆v, m/s 5 3.33 0 7.5 5 0

V2, m/s 5 3.33 0 2.5 0 -5

v2, m/s 5 3.33 0 7.5 5 0

Table 1 Sample calculations for co-axial collisions without external forces

In the first column we see that mc has a value of half the "lighter" mass and there is an impulse of 5000 kgm/s. Because the masses are equal, the ∆V's are equal, and because ε = 0 the final velocities are equal. In the second column, we note an increase in mc with a corresponding increase in the impulse. The ∆V's are inversely proportional to the masses. In this case, though, mass M experiences a greater ∆V because it was involved in a collision with a more massive


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object than in the first column. Again the final velocities are equal because ε = 0. When M strikes the infinite mass in the third column, we note that mc = M and the impulse has doubled in comparison to the first column. Of course, all the velocity change has been sustained by mass M.

Changing ε from zero to 0.5 has no influence on mc, but has resulted in a 50% increase in the value for Ic. That is because of the term (1+ε) in equation (13). Notice that the ∆V's have also increased by 50% in all cases and would have been increased by 100% if ε had been given a value of unity.

Understanding the Coefficient of Restitution In any collision is it virtually impossible to calculate a value for the Coefficient of Restitution, ε, from First Principles. We cannot look at the materials that will be in contact with each other, assess their structures, and from those insights calculate the Coefficient of Restitution will be 0.25, for example. Likewise, it is virtually impossible to calculate the duration of the impact, ∆t.

We see that even if the impact velocities and masses are the same, if the collisions produce different Coefficients of Restitution then the post-impact velocities will be different. However, in all cases Conservation of Momentum will be honored.

The Coefficient of Restitution is analogous to the Coefficient of Friction, µ, in many ways. Because they are coefficients, both are non-dimensional. They also range in value from zero to unity depending on the circumstances. You should also notice that the value of the coefficient depends on the interplay between the two contacting bodies. It is never correct to refer to one body as having a Coefficient of Friction or a Coefficient of Restitution of some value. The body only has that coefficient because of its interaction with the second body. Therefore both bodies should be identified when referring to either coefficient. Also similar to the Coefficient of Friction, the Coefficient of Restitution is usually measured by experiment.

In the investigation of vehicle collisions, it is generally accepted that the Coefficient of Restitution tends to zero as the collision intensity is increased, say where vc > 50 kilometres/hour. However, it is well established that for Low Speed Impacts, i.e., from 5 kilometres/hour < vc < 15 kilometres/hour, the Coefficient of Restitution can have a value in the range of 0.2 to 0.6, and below 5 kilometres/hour, it can reach up to 0.9, [Emori, Horiguchi, 1990; Szabo, Welcher, 1992; Howard, Bomar, Bare, 1993; and Siegmund, Bailey, King, 1994].

Figure 2 is a typical example of Coefficient of Restitution experimental data that has been taken from the Siegmund, Bailey, King paper. This paper provides results from thousands of experiments conducted by the authors into the effects of Low Speed Co-Axial collisions. Although there is some scatter in the results—which may be indicative of the difficulty of obtaining consistent velocity measurements—the barrier test results clearly show a descending curve with increased ∆V. The vehicle-to-vehicle tests show even more scatter, but that may be


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due both to the difficulties in measurement, as well as the variations in the impact characteristics of a variety of vehicles.

Figure 2 Restitution Experiments—Frontal Collisions with a 1986 Chevrolet Cavalier

Problem Solving Typical question

The type of question that is posed for co-axial impacts in which the Coefficient of Restitution is deemed to be zero is

"Car M collides with car m which is stationary. The two vehicles stick together after impact. We can calculate the post-impact velocity of the two vehicles from skid-to-stop information. We know the masses of the two vehicles. The question is to find the impact velocity of vehicle M.

The Conservation of Momentum gives:

MV mv MV mv1 1 2 2+ = +


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This equation has four velocities and two masses. In the example, the masses are known and the fact that the two vehicles stick together means that:

V v2 2 v= =

Therefore the Conservation of Momentum equation becomes:

MV mv M m v1 1+ = +b g

leaving us with three velocity variables, V v . v1 1, and

From the skid-to-stop information we can find the value for v and therefore arrive at a value for the right side of the Conservation of Momentum equation. However, we are still left with the two variables, V1 and v1 on the left side of the equation. Luckily the input data informs us that v1 = 0. Therefore, it follows that V1 can be calculated.

In many cases, it is possible to declare that v1 = 0 because the vehicle was parked, and there was nobody inside. But what would happen if we could not be sure that v1 was stationary? Perhaps the driver of M said, "The reason I hit m was because the driver was driving backwards, and I could not stop in time". With the finding that the two vehicles stuck together after impact—which implies that the Coefficient of Restitution, ε = 0—a solution for V1 and v1 cannot be found. (I shall show you why later).

However, in most accidents the vehicles do not stick together and therefore V , and hence, ε ≠ 0. Let us assume that there is enough skid-to-stop information to work out values for V

v2 ≠ 2

2

2 and v2 independently. (Note that a reality check will require that v because M must remain behind m).

V2 ≥

With known values of V2 and v2, the right side of the Conservation of Momentum equation can again be calculated, just the same as if we had used a value for v . We are still left with the variables V1 and v1 on the other side of the equation. But now we will allow ourselves another piece of information—one that is independent of Newton's Laws of Motion as seen in the Conservation of Momentum equation. The new information says that "Experimental tests have shown that the Coefficient of Restitution for this impact would be, (say), 1/3". That statement introduces a second equation and therefore allows V1 and v1 to be calculated because now we have two equations and two unknowns.

Reverting to our previous example, let M = 1000 kilogram and m = 2000 kgm, and the calculated values for V2 and v2 are 2 and 5 m/s respectively. Given that ε = 1/3, the Coefficient of Restitution equation gives:


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ε = FHG

IKJ =

−−

FHG

IKJ =

−−

FHG

IKJ = −

− =

= +

13

5 2 3

99

2 2

1 1 1 1 1 1

1 1

1 1

v VV v V v V v

V vV v

b gb g

m / s

The Conservation of Momentum equation gave us:

1000 2000 1000 2 2000 5 12 0002 12

1 1

1 1

× + × = × + × =+ =

V vV v

, m / s

Substituting 9+v1 = V1, gives:

9 2 123 12 9 3

19 1 10

1 1

1

1

1

+ + == − === + =

v vvvV

Therefore m / sand m / s

We could also get this result from the Impulse Equalities. Given that we know M and m, we can get:

m MmM mc = +

FHG

IKJ =

×+

FHG

IKJ = ×

1000 20001000 2000

23

1000 kg

and from the Coefficient of Restitution equation we get:

ε = FHG

IKJ =

−−

FHG

IKJ =

−−

FHG

IKJ = −

− = =

13

5 2 3

9

2 2

1 1 1 1 1 1

1 1

v VV v V v V v

V v vc

b gb g m / s

Leading to

I m vc c c= +

= FHG

IKJ × × =

1

43

23

1000 9 8000

εb g Newton seconds


14

I M V I m vV V v v

V vV v

c c= =

= − = −

= − = −= =

∆ ∆

8000 1000 8000 20008 2 4 5

10 1

1 2 2 1

1 1

1 1

b g b

m / s m / s

g

which are the same answers as before.

What we have demonstrated is the fact that two unknowns can be solved in Co-Axial Impact theory if a second equation can be found. The second equation must be totally independent of the Newton's Laws of Motion/Conservation of Momentum equation. The Coefficient of Restitution equation is such a case. (There is another equation that can be used, as will also be shown in this study).

It was mentioned earlier that, a solution cannot be found for the two unknowns if ε = 0. This condition says:

ε = =−−

FHG

IKJ0 2 2

1 1

v VV v

We can see—both mathematically and logically—that for ε = 0, v2 must equal V2.

Therefore 0 = 0/(V1 - v1) which means that (V1 - v1) is indeterminate. (In the previous calculation, it had a value of 9). In effect, the second equation does not yield an answer that can be combined with the momentum equation.

Applicability of the Coefficient of Restitution Having established a theoretical basis for the use of the Coefficient of Restitution, we should now consider its practicality. We find that it is of limited application. It is generally accepted that the Coefficient of Restitution tends towards zero at the "normal' impact range, rather than the special type of accidents classified as Low Speed Impacts. (We shall suggest that the term Low Speed Impact might apply to collisions where the approach velocity is less than 15 kilometres/hour, say under 10 mph.)

Rearranging the Coefficient of Restitution equation, we can write:

v V vv V

c = − =−

1 12 2b g b gε


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This equation indicates that as ε approaches zero—as is the case for "normal" or high-speed impacts—the approach velocity, vc, becomes supersensitive. If ε is halved from, say 0.1 to 0.05, the approach velocity will be doubled. For this reason, accidents in which ε would be less than 0.2 are probably not suitable for this type of analysis.

In the Low Speed Impact, or Very Low Speed Impact category—say, where vc could be under 5 kilometres/hour—a Coefficient of Restitution value from 0.2 to 0.6 is more common. In these cases, the Coefficient of Restitution equation could be used. However at such low impact speeds, there is usually insufficient physical evidence to permit a reliable calculation of the post-impact speeds of the two vehicles. So you have to choose your accident if you are going to use the Coefficient of Restitution condition.

Case Study of Coefficient of Restitution Usage Consider the case where a bus of 10350 kilograms strikes the rear of a car of 1250 kilograms propelling it into the rear of a lead car of 1720 kilograms. It is known that both cars were stopped at a set of traffic lights at their moments of impact. From at-scene data it is also possible to calculate that the post-impact velocity of the bus was 3 m/s.

Therefore for the first collision, we know:

Vm

V v V

m MmM m

I m v M V

V V

V VV

c

c c c

2

1 1 1

1 1

1 1

1

3 010350 1250

10350 125010350 1250

111530 4

1

1 0 4 11153 10350

14 11153 10350 315614 31050

=

V2

== =

= − =

=+

FHG

IKJ

=×+

FHG

IKJ

==

= + =

= + × × = × −

= × × = × −

= −

m / s v m / sM kg kg

v

kgAssume

= 10350V

1

c

1

..

. .

. ..

εεb g

b g b gb g

∆


16

Therefore m / s

AlsoTherefore

m / sAlso

m / s m / s

8788 6 31050353

15614 35355117551171250

4 41

353 3 4 41 0053 4 41

1

1

2 1

2

1 2 2 1

..

. ...

.

. .

. .

VV

I mm v v

v

V V V v v v

c

==

= × =

− =

=

== − = −= − = −= =

∆

∆ ∆

b gv

Notice that the car has bounced off the bus at a velocity of 4.41 m/s, which is faster than the impact velocity of the bus, which was only 3.53 m/s.

If we assume that there was no decrease in velocity as the car lurched forward into the stationary lead car, we get for the second collision:

Vm

V v

m MmM m

I m v M V

VV

c

c c c

1

1 1

2

2

4 41 01250 1720

4 41 04 41

1250 17201250 1720

723 90 4

1

1 0 4 723 9 4 41 1250 4 414469 4 5512 5 1250

= == =

= −= −=

=+

FHG

IKJ

=×+

FHG

IKJ

==

= + =

= + × × = × −

= −

.

.

.

..

. . . .

. .

m / s v m / sM kg kg

v

m / s

kgAgain assuming

=

1

c

εεb g

b g b∆

g


17

Therefore m / s

AlsoTherefore

m / sAlso

m / s m / s

1250 10431083

4469 44469 44469 41720

2 60

4 41 083 2 60 0358 2 60

2

2

2 1

2

1 2 2 1

VV

I mm v v

v

V V V v v v

c

==

= =

− =

=

== − = −= − = −= =

..

...

.

. . .. .

∆

∆ ∆

b gv

Although the choices for the Coefficient of Restitution for each of the collisions might be subject to some dispute, we can see that the "sandwich" car sustained a rear impact causing a ∆V of 4.41 m/s followed by a frontal impact generating a ∆V of 3.58 m/s. By contrast, the lead car sustained a single rear impact, which generated a ∆V of only 2.6 m/s. Modifications to the input variables had little effect on the fundamental finding—that the lead car got off lightly in comparison to the "sandwich" car.

It was the injury claim by the occupant in the lead car that instigated the civil action. The Defense advanced the argument that the uninjured occupant of the "sandwich" car provided a better test of the propensity for injury. She was a relatively frail old lady compared to the strapping male youth in the lead car.

ENERGY CONSUMED IN CO-AXIAL COLLISIONS WITHOUT EXTERNAL FORCES

The preceding analysis has dealt only with the Conservation of Momentum via Newton’s Laws of Motion, combined with the Coefficient of Restitution condition. Nothing has been said about the degradation of the Kinetic Energy of the colliding bodies into crush, strain, heat, sound etc. In the investigation of collisions between vehicles, it is the crush, or permanent deformation, that is often the only obvious sign of kinetic energy degradation. In the analysis of Low Speed Impacts, where it is unlikely that there are marks on the roadway, the deformation of the vehicles is usually the only clue to the intensity of the collision, and to the propensity for injury to the occupants of the vehicles.

The equation for the Conservation of Energy may be written:

12 1

2 12 1

2 12 2

2 12 2

2MV mv MV mv E e+ = + + + (14)


18

in which the capital letters refer to one object and the lower case letters refer to the other object as before, and the subscripts 1 and 2 deal with the conditions before and after the collision. In this simplified version of the theory, no allowance has been made for rotation of the bodies. The terms E and e refer to the kinetic energy lost by each of the objects. In vehicle collisions this energy is usually in the form of crush, but it may also represent energy dissipated through internal vibrations, viscous dampening, etc.

Adding the Conservation of Energy equation (14) to the Impulse Statement, equation (13), and the Coefficient of Restitution Condition, equation (11), yields the following expression after many pages of algebraic manipulation:

E m v

m v m vm v m v

c c c

c c c c

c c c d

= −

= −

= −

12

2 2

12

2 12

2 2

12

2 12

2

1 ε

ε

c h (15)

Where Ec = E + e, and all the other symbols, ε, mc, vc and vd, have been previously defined.

Continuing with the algebraic manipulation, we can insert equation (15) into the equalities of equation (13) to add the additional equality:

F dt I M V m v m v E mct

t

c c cc c

1

2 1 1 21 2z = = = = + = +−

∆ ∆ ε εε

b g b g c h (16)

Equation (16), which we shall refer to as the Impact Identities Equation, expresses all the effects of a collision without external forces in one set of equalities. It is important to note that the loss of kinetic energy, Ec, is distributed between the two bodies, usually in the form of crush. Knowing the kinetic energy loss, it is possible to find each body’s change of velocity. Many authors call Ec the crush energy, [McHenry, 1975; and Hight, Lent-Koop, Hight, 1985]. This expression is not strictly true—although it is predominantly true—because the energy loss that occurs in other ways is usually small compared with that dissipated in crush.

Reverting back to equation (15), we can see that it contains the same three factors that contribute to a co-axial collision without external forces, albeit in a slightly different form:

1. Approach Velocity, embodied by the expression vc2, instead of vc.

2. Mass Effect, embodied by the expression ½mc, instead of mc.

3. Structural Aspects, embodied by the expression (1-ε2), instead of (1+ε).


19

Equation (15) states that the kinetic energy lost in any collision will be zero, when ε = 1, i.e., when there is total rebound in a perfectly elastic collision. The maximum loss of kinetic energy occurs when ε = 0, i.e., in a perfectly plastic collision. The maximum kinetic energy loss is given by the expression ½mcvc

2.

Recalling that in a "high-speed" collision where the value of the Coefficient of Restitution could be in the order of 0.1 or less, the term (1-ε2) ≅ (1-.01).represents 99% of the value for the case where ε = 0. Therefore if it can be shown that the Coefficient of Restitution was that low in a "high-speed" collision accident, the omission of the Coefficient of Restitution term would be acceptable, as it results in a negligible error—much less than the error associated with the crush measurements.

Examples of Kinetic Energy Loss Numerical Example

Given m / s m / s m / s m / s

kg kgWe also know v m / s

kg

c

V vV vM m

V v

v VV v

m MmM mc

1 1

2 2

1 1

2 2

1 1

10 12 51000 2000

9

13

23

1000

= == == =

= − =

=−−

FHG

IKJ =

=+

FHG

IKJ = ×

ε

Conservation of Energy gives12

giving Joules

MV mv MV mv E

E

E

c

c

c

12

12

22

22

2 2 2

12

12

12

12

1000 10 12

2000 1 12

1000 2 12

2000 5

24 000

+ = + +

× × + × × = × × + × × +

= ,

2


20

Alternatively E

Joules

c = −

= × ×FHG

IKJ × × − RST

UVWFHG

IKJ

=

12

1

12

23

1000 9 1 13

24 000

2 2

22

m vc c εc h

,

Impact into an Infinite Barrier

Consider the situation where a vehicle collides head-on into an infinite barrier. Let the vehicle have mass, M, and an impact velocity, V1. Because the barrier is infinite, it will have mass, m = ∝, meaning that mc = M. Also because the barrier is infinite it will be stationary, and its impact velocity, v1 = 0. Therefore vc = V1.

For simplicity, let us assume that there is no rebound, so that ε = 0, and V2 = v2 = 0. Therefore there was be no output kinetic energy. As v1 = 0, the total input kinetic energy will be 1

2 12MV .

But and

Therefore

where

M m V v

MV E m v

c c

c c c

≡ ≡

= ≡ −

=

1

12 21

212

1

0

ε

ε

c h2

In this case, we find that all the input kinetic energy is dissipated in the impact.

Equal Mass Impact

If, instead of striking an infinite barrier the vehicle of mass, M, strikes a stationary vehicle of equal mass, m. Because M = m, mc = ½M. Because m is stationary, its impact velocity, v1 = 0. Therefore vc = V1, as before. Again for simplicity, let us assume that there is no rebound. Therefore ε = 0 so that V2 = v2. However, in this case, V2 = v2 will not equal zero.

The kinetic energy dissipated in the impact is given by:

E m v M V Mc c c= ≡ VFHG

IKJ = RST

UVW12

12 2

12

12

212

12

which is just 50% of the kinetic energy dissipated in the infinite barrier collision. The comparison between the two collisions illustrates the influence of the Effective Mass. When the Effective Mass was equal to the lighter mass, all the input kinetic energy was consumed in the collision. When the Effective Mass was half the lighter mass, only 50% of the input kinetic


21

energy was consumed in the collision, the other 50% was retained as post-impact kinetic energy, i.e., from the Conservation of Momentum we get:

MV MV mvM mV v v

MV Mv

v V

1 2

2 2

1

1

0

212

+ 2= +== ==

=

whereLetso that

Therefore

The post-impact kinetic energy is therefore:

post - impact kinetic energy = 12

M m v

M V

MV

+

= RSTUVW

= RSTUVW

b g

b g

2

1

2

12

12

2 12

12

12

DISTRIBUTION OF KINETIC ENERGY LOSS

The loss of kinetic energy expression given in equation (15) is the sum of the crush energy—or other form of degradation of kinetic energy—in each of the two bodies involved in the impact. Unfortunately, it is not uncommon to find collision analyses in which the crush energy of only one of the bodies is used to determine the ∆V experienced by that body. That calculation is incorrect unless it can be shown that the other body was infinitely rigid—a concrete barrier for example—and therefore did not consume any kinetic energy.

If two vehicles have experienced crush in a collision, but only one is available for inspection, it is possible to estimate the kinetic energy loss in each vehicle—at least theoretically—if the crush characteristics are known for each vehicle. The method relies on the equalization of forces, as per Newton's Third Law of Motion. The analysis of crush characteristics, the equalization of forces, the shape of the collision force curve, and the duration of the impact, are all topics for further study. The essential point is that the kinetic energy consumption by both vehicles must be taken into account when using the Conservation of Energy concept to determine ∆V.

In the previous example of a vehicle collision into an infinite barrier—in which we deemed the Coefficient of Restitution was zero—it was shown that all of the input kinetic energy was


22

consumed in the collision. Unless they have been equipped with energy absorbing surfaces—honey-comb material, or surrounded by old tires, for example—most barriers consume negligible energy and are said to be infinitely rigid as well as infinitely massive. The NHTSA 30 mph frontal crash tests are an example, but not the IIHS offset impact tests.

If we now add the condition that the previous example was a vehicle crash into an infinite rigid barrier, we know that 100% of the input kinetic energy will be absorbed as crush in the vehicle—neglecting other forms of energy loss. In the other example in which the vehicle collided with a stationary vehicle of equal mass, we have shown that because of the Mass Effect, only 50% of the input kinetic energy will be consumed in crush.

If we now introduce the further assumption that the crush characteristics are identical for each vehicle, it follows that the 50% of input kinetic energy will be consumed equally between the two vehicles. Therefore the 'bullet" vehicle will only absorb 25% of its input kinetic energy in the identical vehicle-to-vehicle collision, in comparison to the 100% absorption in the barrier collision. Depending on the form of the crush characteristics for the vehicle, the crush depth should be approximately half as much in the vehicle-to-vehicle case compared to the barrier case.

Combining Momentum and Energy for Solving Co-Axial Impacts In a recent Special Problems forum, [Shigemura 1998], a method for solving the impact velocities of two vehicles involved in a head-on collision was presented. The method required that both post-impact velocities could be determined from scene information, and in addition the kinetic energy loss due to crush could also be calculated.

Because the Conservation of Momentum equation leaves the two impact velocities as unknowns, it is necessary to add a second equation to obtain a solution. The second equation is provided by the Conservation of Energy statement. An alternate method for solving the same problem is to use the Co-Axial Theory described in this lecture.

Illustrated below are the two techniques for solving co-axial collisions using both Momentum and Energy. In our example, we will assume that we know that the vehicles traveled from left to right post-impact. We also know from measurements and calculations:

M mV vE e

= == == =

1500 10003 512 000 18 000

2 2

kg kg m / s m / s

Joules Joules, ,

Simultaneous Equation using Quadratic Equation—Shigemura Technique

Conservation of Momentum gives:


23

MV mv MV mvV v

V v

v

1 1 2 2

1 1

1 1

1

1500 1000 1500 3 1000 5950095001500

10001500

6 3333 0 6667

+ = ++ = × + ×

=

= −

= −

kgm / s

Therefore

m / s. .

Conservation of Energy gives:

12

12

12

12

750 500 750 3 500 5 12 000 18 00049 25049 250

750500750

656667 0 6667

12

12

22

22

12

12 2 2

12

12

12

MV mv MV mv E e

V v

V v

v

+ = + + +

+ = × + × + +=

= −

= −

, ,,,

. .

Joules

Therefore

But from the Conservation of Momentum we write:

V v

v v12

12

1 12

6 3333 0 6667

401111 8 4444 0 4444

= −

= − +

. .

. . .

b g

Combining the two expressions for V12 leads to:

401111 8 4444 0 4444 65 6667 0 666711111 8 4444 255556 0

1 12

12

12

1

. . . . .. . .

− + = −

− − =

v vv vTherefore

v

This is known as a Quadratic Equation. The solution for a Quadratic Equation of the form:

aX bX c X b b aa

22

0 42

+ + = =− ± −is c

Inserting X vabc

=== −= −

1

111118 4444255556

...


24

leads to

or

m / s m / s

v

v v

1

2

1 1

8 4444 8 4444 4 11111 2555562 11111

8 4444 713079 11357932 2222

8 4444 184 88722 2222

8 4444 1359732 2222

22 04172 2222

515292 2222

9 9188 2 3188

=± − × × −

×

=± +

=±

=±

= =

=

−

= −

. . . ..

. . ..

. ..

. ..

..

.

.. .

b g

The two solutions for the impact velocity for the vehicle on the right tell us that either it was travelling at 9.9188 m/s from left to right, or it was travelling at 2.3188 m/s from right to left.

Inserting both values into the equation for V1 gives:

V vV V

1 1

1 1

6 3333 0 66676 3333 0 6667 9 9188 6 3333 0 6667 2 3188

0 2796 7 8792

= −= − × = + ×= − =

. .. . . . . .

. .Therefore or

m / s m / s

The reality of the collision dictates that the vehicle on the left could not be travelling towards the left, i.e., a negative velocity, and then travel to the right post-impact. Therefore the final solution must be:

V1 7 8792 2 3188= = −. . m / s and v m / s1

Co-Axial Impact Theory—Duff Method

From the at-scene measurements and calculations, we can write:

m MmM mv VE e

c = +=

×+

=

= − = − == + = + =

1500 10001500 1000

600

5 3 212 000 18 000 30 000

2 2

kg

v mE Joules

d

c , , ,/ s

The kinetic energy loss equation can be written:


25

E m v

m v vv

m v m v

c c c

c cd

c

c c c d

= −

= −FHG

IKJ

= −

12

1

12

1

12

12

2 2

22

2

2 2

εc h

Therefore

giving

m / s

30 000 12

600 12

600 2

300 1200300 31 200

31 200300

101980

2 2

2

2

,

,

,

.

= × × − × ×

= −

=

=

= ±

v

vv

v

c

c

c

c

Only the positive value of vc makes sense. Knowing vc and vd we can find the Coefficient of Restitution, ε, as well as the Collision Impulse, Ic.

ε

ε

= = =

= +

= + × ×

=

vv

I m v

d

c

c c c

2101980

01961

1

1 01961 600 1019807318 70

..

. ..

b gb g

Newton seconds

But

m / s m / s

m / s m / s

∆ ∆

∆ ∆

V IM

v Im

V V v vV V V v v v

c c= =

= =

= == − = −= + = −= + = −= =

7318 701500

7318 701000

4 8791 7 3187

3 4 8791 5 7 31877 8791 2 3187

1 2 2 1

1 2 1 2

. .

. .

. .. .−


26

Concluding Remarks The Conservation of Momentum equation is derived directly from Newton's Laws of Motion.

For co-axial impacts there is only one Conservation of Momentum equation with four velocity variables. If values for two of the variables can be determined, it is necessary to introduce another equation of different origin in order to find values for the two unknown variables.

Defining a Coefficient of Restitution produces a second equation that can be combined with the Conservation of Momentum equation to solve co-axial impact problems.

The Coefficient of Restitution condition has limited practical usage because of its sensitivity to the approach speed in low-speed impacts, and its tendency towards a zero value in high-speed impacts.

The effect of the Coefficient of Restitution can be neglected in making high-speed impact crush measurements because it introduces an error of approximately 1% which is insignificant in comparison to the inaccuracies inherent in measuring crush.

Given a fixed approach velocity and specific masses for two vehicles involved in a co-axial collision, the outcome will be influenced by the material and structure of the colliding vehicles. Depending upon the materials and structure, there could be a range of post-impact velocities. No matter what the post-impact velocities turn out to be, they will always be consistent with the Conservation of Momentum principle.

Given a fixed approach velocity and specific masses for two vehicles involved in a co-axial collision, the greater the crush that results from the collision, the greater will be the ∆V's experienced by the vehicles.

The Conservation of Energy equation is another example of an equation that can be used in combination with the Conservation of Momentum equation to solve for two unknowns in co-axial impact accidents.

In co-axial collisions, there is zero kinetic energy loss when the Coefficient of Restitution is unity.

In co-axial collisions, the maximum possible loss of kinetic energy occurs when the Coefficient of Restitution is equal to zero.

The maximum possible loss of kinetic energy is not the same as the total input kinetic energy except in the case where the effective mass is equal to the lesser mass, i.e., the other object has an infinite mass.


27

The kinetic energy that is lost in a co-axial collision can be distributed in any proportion from 0 to 100% in either of the two bodies involved in the collision. It is necessary to add together the amount of kinetic energy loss is each object in order to find the intensity of the collision, and the ∆V's experienced by each object.

The intensity of a collision is given by the Impulse, which is the combination of the collision forces and the duration of the impact. It is a vector quantity with a direction equal to the PDOF. Its units are newton-seconds, or kilogram-metres/second in S.I, and lbf-seconds or lbm-ft/second in Imperial units.


28

References

Emori RI, Horiguchi J; Whiplash in Low Speed Vehicle Collisions; SAE 900562; Sp-807; pp 103-108; 1990

Duff J, Mikulcik EC, Proceedings of the Canadian Multidisciplinary Road Safety Conference XI, 1999.

Hight PV, Lent-Koop DB, Hight RA; Barrier Equivalent Velocity, Delta V and CRASH3 Stiffness in Automobile Collisions; SAE 850437; 1985

Howard RP, Bomar J, Bare C: Vehicle Restitution Response in Low Velocity Collisions; SAE 931842; 1993

Macmillan RH; Dynamics of Vehicle Collisions; Editor-in-Chief Dorgham; Inderscience Enterprises Ltd.; Proceedings of the International Association of Vehicle Design; Special Publication SP5.1983; ISBN 0907776078; 1983

McHenry RR; A Comparison of Results Obtained with Different Analytical Techniques for Reconstruction of Highway Accidents; SAE 750893; SAE Transactions Volume 84; 1975

Shigemura N, Special Problems lecture, IPTM, 1998, and Daily J, Shigemura N, Fundamentals of Applied Physics for Traffic Accident Investigations.

Siegmund GP, Bailey MN, King DJ; Characteristics of Specific Automobile Bumpers in Low-Velocity Impacts; SAE 940916; Accident Reconstruction: Technology and Animation IV; SP-1030; pp333-370; 1994

Szabo TJ, Welcher J; Dynamics of Low Speed Crash Tests with Energy Absorbing Bumpers; SAE 921573; 1992

co-axial impact theory - angelfire · 2002-07-05 · co-axial impact theory special problems y2k...

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