Crush and Restitution

Institute of Police Technology and Management’s

24th Annual Special Problems in Traffic Crash Reconstruction

Crush Analysis with Under-rides and theCoefficient of Restitution

Jeremy Daily∗ Russell Strickland† John Daily‡

27 April 2006

Abstract

In this paper, a detailed discussion of the damage momentum technique is presented

which involves an interpretation of the damage profiles, understanding the origin of the

stiffness coefficients, and derivation of ∆v. Also, a technique for estimating the energy

dissipated during deformation based on residual crush measurements is explained in the

context of the damage momentum solution. A staged crash test from the Special Problems

2005 conference is used as an example to validate the technique. Also, a discussion of the

coefficient of restitution is given with the derivation of its relationship with crush energy. Fi-

nally, a discussion of the misapplication of damage energy techniques is outlined for trailer

underride collisions.

There were four crash tests were conducted at Special Problems 2005 in which passenger

vehicles were run into a stationary tractor-trailer unit. Two of the impacts were collinear

into the back of the trailer, while the other two were at right angles to the trailer tandems.

Analysis for the collinear impacts was limited to standard COLM techniques, while the

side impacts were analyzed by means of rotational mechanics. Damage crush profiles were

recorded for later use with a damage-energy technique.

In this paper, we will examine the previous impacts using a damage-energy technique.

Furthermore, comparison of the damage-energy solution to recorded pre-crash speed mea-

surements will validate the technique for the side impacts. Finally, the inability to apply a

damage momentum solution to trailer under-ride collisions will be explained.

∗Jackson Hole Scientific Investigations, Inc., 7845 Timber Hill Dr, Huber Heights, OH 45424, (937) 235-5693,[email protected]

†Fairfield City Police Department, 5320 Pleasant Ave., Fairfield, OH 45014, (513) 325-8703, [email protected]‡Jackson Hole Scientific Investigations, Inc., P.O. Box 2206, Jackson, WY 83001, (307) 733-4559,

[email protected]

1

Contents

Contents

1. Introduction 4

1.1. Recap of Crash Tests from 2005 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.2. Review of Conservation of Linear Momentum . . . . . . . . . . . . . . . . . . . . 7

2. Damage Momentum Analysis 8

2.1. Crush Energy and Equivalent Barrier Speed (EBS) . . . . . . . . . . . . . . . . . . 8

2.2. A Planar (Two Dimensional) Impact Model . . . . . . . . . . . . . . . . . . . . . . 9

2.3. Relationship to ∆v . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

3. Determining Crush Energy 16

3.1. The General Energy Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

3.2. CRASH III Deformation Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

3.2.1. Determining Stiffness Coefficients . . . . . . . . . . . . . . . . . . . . . . . 18

3.2.2. Crash Test Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

3.2.3. Determining ∆vtest . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

3.2.4. Determine Campbell Model Coefficients . . . . . . . . . . . . . . . . . . . 20

3.2.5. Determining A, B, and G Values . . . . . . . . . . . . . . . . . . . . . . . . 22

3.3. Determining Damage Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

3.4. Determining the Location of the Damage Centroid . . . . . . . . . . . . . . . . . 25

3.4.1. Longitudinal Location of the Centroid in the Damage Area . . . . . . . . . 26

3.4.2. Lateral Location of the Centroid in the Damage Area . . . . . . . . . . . . 26

3.4.3. Locating the Damage Centroid with Respect to the Local Axis of the Vehicle 26

3.5. Crush Energy Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

4. Relationship between Crush Energy and the Coefficient of R estitution 28

4.1. Physics of an Impact . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

4.2. Taxonomy of Impacts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

4.2.1. Nature of Elasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

4.2.2. Relative Velocities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

4.2.3. Orientation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

4.3. Understanding the Coefficient of Restitution . . . . . . . . . . . . . . . . . . . . . 32

4.3.1. Kinematic Definition of Restitution . . . . . . . . . . . . . . . . . . . . . . . 32

4.3.2. Kinetic Definition of Restitution . . . . . . . . . . . . . . . . . . . . . . . . 33

4.3.3. Energetic Definition of Restitution . . . . . . . . . . . . . . . . . . . . . . . 36

4.4. Computing Restitution based on Damage Energy . . . . . . . . . . . . . . . . . . 39

4.5. Computing Coefficient of Restitution Based on Crash Test Data . . . . . . . . . . 41

4.6. Concluding Remarks on Restitution . . . . . . . . . . . . . . . . . . . . . . . . . . 43

2 c©2006 Jeremy Daily, Russell Strickland, and John Daily

Contents

5. Analysis of Underride Collisions 43

5.1. 1989 Plymouth Voyager Van into the rear of the tractor-trailer . . . . . . . . . . . 43

5.2. 1994 Jeep Cherokee into the rear of the tractor-trailer . . . . . . . . . . . . . . . . 45

5.3. Under-ride Analysis Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

6. Summary and Conclusions 47

A. Analysis of the Nissan Crash 50

B. Analysis of the Plymouth Van Crash 51

C. Analysis of the Jeep Crash 52

Copyright Information

The following material contains excerpts from Fundamentals of Traffic Crash Reconstruction, an

IPTM publication. It is copyrighted and has been reprinted with permission of the authors for

use in IPTM training programs.

c©2006 Jeremy Daily, Russell Strickland, and John Daily 3

1. Introduction

Figure 1: Scene photo of the mini van into the rear of the trailer.

1. Introduction

1.1. Recap of Crash Tests from 2005

There were four tests performed in two days. Engineers from MacInnis Engineering used a tow

cable system to pull the bullet vehicles into the trailer. The tow cable was fed through the center

of the rear duals for two crashes and along the right and left side for the other two crashes.

The tractor trailer was common to all crashes. The trailer was a Pine 48 ft box van with sliding

tandems. The tractor was a 2004 Mack single axle day cab (VIN: 1M1AE02YX4N00138).

Rear Under-ride Crash #1 A 1989 Plymouth Voyager SE (VIN: 2P4FH4531KR174080) was

pulled into a stationary tractor trailer, Figure 1. It was pulled into the left rear of the box van

trailer and penetrated to the rear tires of the trailer. The tractor-trailer combination had its

spring brakes applied and was pushed forward a small amount due to the impact.

Rear Under-ride Crash #2 A 1994 Jeep Cherokee (VIN: 1J4FJ28S6RL225912) was run into the

right rear of the box van trailer in the same manner as crash #1.

Rear Dual Axle Crash #3 A 1992 Nissan Sentra (VIN: 1N4EB32A7NC734928) was pulled

by cable into the rear duals of the box van trailer. The trailer rotated and the tractor remained


1.1. Recap of Crash Tests from 2005

Figure 2: Scene photo of the Jeep into the rear of the trailer.

Table 1: Comparison of speed estimates using reconstruction techniques to measured speedsin miles per hour.

Bullet Test Vehicle Analysis Type Lower Bound Upper Bound Actual Speed

Jeep In-line Momentum 38 49 37

Voyager In-line Momentum 30 52 39

Nissan Rotational Mechanics 34 41 39

Honda Rotational Mechanics 27 32 31

stationary. The impulse of the collision rocked the tractor trailer but did not tip it over.

Rear Dual Axle Crash #4 A 1993 Honda Accord (VIN:1HGCB7693PA090661) was pulled

into the right rear duals of the box van trailer. The trailer rocked and the tractor drive axles

moved about 1 inch.

Table 1 shows a comparison of the reconstructed speeds with the actual speeds measured

by a RADAR system. The two under-ride collisions were problematic when trying to predict

impact speeds using momentum due to the high mass ratio.

The impact analysis using rotational mechanics concepts proved to be accurate. Moreover,

using the middle values of the ranges yielded answers within a couple miles per hour. The

analysis presented herein is valid and the evidence can be easily gathered at the scene, provided


1. Introduction

Figure 3: Scene photo of the Nissan into the rear of the trailer. Notice the two units did notstick together.

Figure 4: Scene photo of the Honda into the rear of the trailer. Notice the lack of damage tothe rear duals of the trailer.


1.2. Review of Conservation of Linear Momentum

the on-scene investigator is trained to look for evidence under the trailer.

In this paper, we are going to analyze in detail the damage-momentum solution for the im-

pact speed of the Honda. The other three crashes are analyzed using WinCRASH, a commercial

implementation of the CRASH3 computer program.

1.2. Review of Conservation of Linear Momentum

Conservation of Linear Momentum (COLM) techniques are a proven tool in any reconstruc-

tionists toolbox. There are times, however, where using COLM is not useful (e.g., high mass

ratios, insufficient evidence and so forth). A derivation of the momentum equation is provided

in Section 4.3.1.

• A collision is considered to be a closed system. That is, the only forces acting on the

vehicles during the collision phase are the collision forces themselves. Ground frictional

forces are neglected. This may not be valid for low speed collisions or for collisions with

high mass ratios.

• Collisions are inelastic, that is, kinetic energy is not conserved.

• Linear momentum is conserved in the collision system. In other words, system momen-

tum before equals momentum after.

• Collisions may be collinear or angled. Collisions are also either central or non-central.

• Approach and departure angles are based upon the impact circle. See Section ?? for fur-

ther review.

• Departure speeds are based upon calculations using the Work–Energy Theorem. The

analysis chosen is based upon the type of post-impact trajectory (spin, etc.)

• Change of velocity vectors (∆v) may be calculated in a straight-forward manner using the

vector geometry with the law of cosines.

• The direction of the ∆v vectors may be calculated using the Law of Sines. This gives us

the Principal Direction of Force (PDOF).

• If both ∆v vectors are calculated using the Law of Cosines, then Newton’s Third Law may

be used to check the results of the calculation.

• A simple geometry check may be used to see if the ∆v vectors are actually opposite in

direction.


2. Damage Momentum Analysis

We see this analysis is not dependent on the amount of vehicle damage and the energy causing

the damage. Neither is it dependent on pre- or post-impact vehicle rotations, except in choosing

correct trajectory type.

There are some impact configurations or conditions for which conservation of linear mo-

mentum is not the sole analysis. These include, but are not limited to, barrier impacts, moving

barrier impacts, fixed object collisions, and in-line collisions. The in-line collisions can be either

head-on or in the same direction. We will examine some of these impact types in this chapter.


2.1. Crush Energy and Equivalent Barrier Speed (EBS)

The equivalent barrier speed (EBS), sometimes referred to as the barrier equivalent velocity (BEV),

is determined by calculating the amount of energy that the vehicle dissipates during a crash.

The speed is determined from

v =

√

2KE

mor S =

√

30KE

w(1)

where m is the mass (slugs or kilograms) of the vehicle and KE is the kinetic energy (ft-lb or

joules) of the vehicle immediately before it was used to crush the vehicle. Therefore, there

is no post-impact velocity and the vehicle comes to a complete stop. The reason it is called an

“equivalent speed” is because it may not be an actual speed. If there is any post-impact velocity,

then the EBS and the actual speed will be different.

There are some instances in which the EBS will be different than the ∆v as outlined in refer-

ence [1]:

1. The EBS will be higher than the ∆v of a vehicle if the vehicle strikes an object that is rigid

and movable.

2. The EBS will be lower than the ∆v of a vehicle if the vehicle strikes an object that is soft

and massive. An example of this would be a vehicle running into a snow bank.

3. The EBS and ∆v will be the same whenever a vehicle strikes an object whose stiffness is

proportional to the weight ratio of the object to the vehicle.

In this section, we will know the equivalent barrier speed and the amount of crush energy. The

coefficient of restitution will be assumed away for this section.


2.2. A Planar (Two Dimensional) Impact Model


Often, the collision force between two vehicles passes through one or both vehicles offset from

the center of mass. When this situation occurs, we will see the vehicle upon which this force

acts to both move in the direction of the force and also to rotate about its center of mass. A

schematic is illustrated in Figure 5.

Let us now relate this general collision model to our system of two eccentric, in-line vehicles

shown in Figure 6.

When looking at Figure 6, we would intuitively expect the acceleration of the center of mass

of each vehicle to be different from each other, simply because the vehicles will tend to rotate

away from the collision. In the same way, we would expect the acceleration at the centroid

of the damage areas to be larger than the acceleration of the center of mass of each vehicle,

respectively.

Let us look at the governing equations for vehicle #1. Vehicle #2 may be analyzed in a similar

way.

F1 = M1a1 (Newton’s Second Law) (2)

τ1 = I1α1 (Newton’s Second Law for rotation) (3)

Recall I = moment of inertia = mk2, where k = radius of gyration. So,

τ1 = m1k21α1 (4)

Torque is also defined as the product of a force and a lever arm:

τ1 = F1h1 (5)

Now consider the following relationship that equates the accelerations:

ac = a1 + h1α1 (6)

From rotational mechanics a = rα and r = h, so:

ac − a1 = h1α1 (7)

Solving for α1:

α1 =ac − a1

h1(8)



a

F

h

τ

α

F is the collision force in lbs (N).

a is the translational acceleration of the center of mass, in-line with the direction of force. Theunits are ft/sec2 or m/sec2.

h is the lever arm upon which the force acts (ft or m).

τ is the torque about the center of mass. Torque is the cross product of F and h. Units are lb-ftor N-m.

α is the angular acceleration caused by τ about the center of mass. Basic units are rad/sec2 forboth systems of measure.

Figure 5: The general case of a non-central collision. This figure shows the overall collisionforce acting on a vehicle.



M1

M2

aca1 a2

Fh2

τ2, α2

h1

τ1, α1

M1 and M2 are the respective vehicle masses.

F is the total collision force.

a1 and a2 are the respective accelerations of the CM of each vehicle. In an offset collision, theseprobably will not be the same for each vehicle.

ac is the common acceleration of the crush zone of the respective vehicles. Therefore, the cen-troids of the damage areas must reach a common velocity since collision times are identi-cal.

h1 and h2 are the lever arms upon which force F acts.

τ1 and τ2 are the respective torques acting about the mass centers.

α1 and α2 are the respective angular accelerations of each vehicle about their centers of mass.

Figure 6: A schematic of a general non-central, in-line collision.



Now, the torques in Eqs. (4) and (5) are the same, so:

F1h1 = m1k21α1 (9)

Substitute for α1:

F1h1 = m1k21

[ac − a1

h1

]

(10)

Divide both sides by m1 and multiply by h1:

F1

m1h2

1 = k21(ac − a1) (11)

Recall from Newton’s Second Law that F1 = m1a1 or a1 = F1m1

. Substitute this value into Eq. (11):

a1h21 = k2

1(ac − a1)

a1h21 = k2

1ac − k21a1

a1h21 + a1k2

1 = k21ac

a1(k21 + h2

1) = k21ac

a1 =

(k2

1

k21 + h2

1

)

ac (12)

We will define the effective (dynamic) mass ratio as:

γ =k2

k2 + h2(13)

Substituting the definition of an effective mass ratio into Eq. (12) gives the simple proportion:

a1 = γ1ac (14)

Equation (14) shows us the acceleration of the CM of vehicle #1 is a proportion, γ1, of the

acceleration of the centroid of the damage area. We may also see that γ will always be less

than or equal to 1 because the denominator of Eq. (13) will never be less than the numerator. If

γ = 1, then we have a central collision without rotation. This is a proof of our intuition.

2.3. Relationship to ∆v

Consider the following general relationship:

a =∆v

∆t(15)



For vehicle #1:

a1 =∆v1

∆t(16)

ac =∆vc

∆t(17)

So substituting Eqs. (16) and (17) into Eq. (14) yields:

∆v1

∆t= γ1

∆vc

∆t(18)

The ∆t terms are common in the denominators on both sides and will cancel:

∆v1 = γ1∆vc (19)

In offset collisions, the acceleration ac and velocity change ∆vc of the damage centroid will

always be larger than the velocity change and acceleration of the center of mass.

Using an energy and momentum based approach [2, 3, 4], formulas for the change in velocity

for an eccentric impact can be determined as:

∆v1 =

√√√√

2γ1Ecrush

m1

(

1 + γ1m1γ2m2

) or ∆v1 =

√√√√

2gγ1Ecrush

w1

(

1 + γ1w1γ2w2

) (20)

We calculate ∆v2 knowing the change in momentum from one vehicle has to be equal and

opposite the change of momentum in the other.

m1∆v1 + m2∆v2 = 0

∆v2 = −m1

m2∆v1 (21)

Newton’s Third Law is satisfied because the impulse vectors, and thus the ∆v vectors, are

opposite in direction.

Example 1 Recall from 2005 that the 1993 Honda Accord crashed into the rear dual axles of a station-

ary semi-trailer. There was no permanent damage done to the trailer and the Honda absorbed 81,200

ft-lb (110,100 J) of crush energy. The trailer rotated around its kingpin and had a moment of inertia of

421,470 lb-ft-sec2 (571,513 kg-m2). The distance from the kingpin to the point of impact is 36.4 ft (11.09

m). The Honda weighs 2900 lb (1315 kg) and the trailer weighs 13,425 lb (6090 kg). Determine the

impact speed of the Honda.



Solution

Since there is no permanent damage to the trailer wheels, there is no contribution to the total

crush energy from the trailer. Therefore:

Ecrush = EHonda + Etrailer = EHonda + 0

We also know that the trailer has no initial lateral acceleration. As such, the ∆v of the trailer

is its actual post-impact velocity. To determine the change in velocity of both vehicles, we will

use Eq. (20). Equation (20) requires the effective mass ratios of each of the vehicles. Since the

Honda hit the center of the rear duals, the collision force acts through the Honda’s center of

mass, so its effective mass ratio is 1. On the other hand, the effective mass ratio of the trailer

must be computed from Eq. (13). The square of the radius of gyration (k2) of the trailer was

determined to be 1010.90 ft2 (93.83 m2). Therefore, the effective mass ratio of the trailer is:

US SI

γ2 =k2

2

k22 + h2

2

=1010.9

1010.9 + 36.42

= 0.433

γ2 =k2

2

k22 + h2

2

=93.83

93.83 + 11.092

= 0.433

Notice that both ratios are the same because a ratio has no dimension. Let us use subscript 1

for the Honda and subscript 2 for the trailer. Thus, for the Honda:

∆v1 =

√√√√

2gγ1Ecrush

w1

(

1 + γ1w1γ2w2

)

=

√√√√

2(32.2)(1)(81, 200)

2900(

1 + 1(2900)0.433(13,425)

)

= 34.68 ft/s

∆v1 =

√√√√

2γ1Ecrush

m1

(

1 + γ1m1γ2m2

)

=

√√√√

2(1)(110, 100)

1315(

1 + 1(1315)0.433(6090)

)

= 10.57 m/s

In a similar fashion, we can determine the ∆v for the center of mass of the trailer:



∆v2

∆vc

27.3 ft (8.32 m)

36.4 ft (11.09 m)

Figure 7: The ∆v of the damage centroid is different than the ∆v of the center of mass of thetrailer.

∆v2 =

√√√√

2gγ2Ecrush

w2

(

1 + γ2w2

γ1w1

)

=

√√√√

2(32.2)(0.433)(81, 200)

13, 425(

1 + 0.433(13,425)1(2900)

)

= 7.49 ft/s

∆v2 =

√√√√

2γ2Ecrush

m2

(

1 + γ2m2

γ1m1

)

=

√√√√

2(0.433)(110, 100)

6090(

1 + 0.433(6090)1(1315)

)

= 2.28 m/s

Since Eq. (20) was developed by assuming the vehicles remain in contact after the collision,

the coefficient of restitution is zero. In order for the Honda to experience the ∆v computed in

this example, the impact speed must be equal to the change in velocity of the Honda plus the

post-impact velocity. This post-impact velocity is the ∆v of the damage centroid of the trailer,

∆vc, because it was initially at rest.

Computing the change in velocity of the damage centroid of the trailer involves using the

concept of similar triangles. Consider Figure 7, which shows the geometric relationship of the

damage centroid, the center of mass, and the point of rotation. We can consider the trailer to

be a rigid body rotating about its kingpin, so the relative velocities of any point on the trailer is

proportional to its distance from the kingpin. This fact allows us to use the property of equal

ratios to determine a relationship between ∆v2 and ∆vc.


3. Determining Crush Energy

US SI

∆vc

∆v2=

36.4

27.3= 1.333

∆vc = 1.333∆v2

= 1.333(7.49) = 9.98 ft/s

∆vc

∆v2=

11.09

8.32= 1.333

∆vc = 1.333∆v2

= 1.333(2.28) = 3.04 m/s

US SI

v1 = ∆v1 + ∆vc

= 34.68 + 9.98

= 44.66 ft/s

S = 30.46 mph

v1 = ∆v1 + ∆vc

= 10.57 + 3.04

= 13.61 m/s

S = 48.99 kph

The actual speed measured of the Honda at impact for this staged crash was close to 31 mph

(50 kph).

The accuracy of a damage momentum analysis is not always guaranteed, because the energy

absorbed in the crash by crushing the vehicle is empirically based. In other words, the tech-

niques used to determine the energy in a crash are not based completely on physics, but rather

a curve fit to crash test data. The curve-fitting technique is the only tractable way to obtain

energy values for vehicle crashes and the details will be discussed in the next section.


The acronym CRASH stands for Computer Reconstruction of Automobile Speeds on the High-

way. The CRASH3 algorithm is implemented in many commercial software packages such as

EDCRASH, WinCRASH, and m-CRASH. The computer programs provide great tools to per-

form analysis; however, comprehension of the physics and models implemented in the com-

puter code enable the user to make a more sound interpretation of the evidence and results of

the computer analysis. CRASH has two parts: the trajectory calculator and the damage algo-

rithm. This chapter explains some of the aspects of the damage algorithm.


3.1. The General Energy Model

L

x

Narrow widthrectangles

AD = Area of Damage

Figure 8: A large rectangle that represents the damage area can be broken into a series of smallrectangles. Here, L is the damage width, AD is the area of the damage projection,and x is the depth of crush.

3.1. The General Energy Model

Consider Figure 8, in which we have taken a rectangular damage profile and have segmented

it into smaller rectangles that are 1 unit wide. Each of these rectangles is x units in depth. We

may now define A and B stiffness coefficients for each unit of width where:

The units of A are lb/inch (N/m),

The units of B are lb/in2 (N/m2),

The units of G are lb (N), and the value of G is equal toA2

2B.

Now consider our energy equation:

E = Ax +Bx2

2+

A2

2B(22)

This equation was used to determine the energy per unit width for a rectangular damage pro-

file. There, the unit width was the entire front of the vehicle. The A and B values were also

for the entire width of the vehicle. Here, since we divided the entire rectangular damage area

into smaller rectangles one unit wide (e.g., 1 inch), Eq. (22) can be used to calculate the damage

energy for each of the narrow rectangles. Correspondingly, the A and B values are for one of

the narrow rectangular strips. If we want to calculate the total damage energy, ET , we will have

to multiply E by L.

ET = EL = AxL +Bx2L

2+

A2L

2B(23)

The product of L and x is the area of the large rectangle, which is the area of the damage. If we

use the variable AD to denote this area of damage, then Eq. (23) becomes:

ET = A(AD) +Bx

2(AD) +

A2L

2B(24)



Factoring out the area term gives:

ET =

(

A +Bx

2

)

AD +A2L

2B(25)

The centroid of a rectangle, denoted as x, in relationship to its length, x, isx

2. Thus, we can

write Eq. (25) in terms of the centroid of the damage area:

ET = (A + Bx) AD +A2L

2B(26)

3.2. CRASH III Deformation Model

To calculate the total energy it took to damage (crush) a vehicle, three things are needed:

Stiffness coefficients. These are the A, B, and G values and come from crash test data. Ways

to calculate these values are discussed in Section 3.2.1.

Area of damage. This is the AD value. In essence, we integrate over the damage profile to get

the damage area in square inches (or square meters). There are several ways to do this

and we will explore the trapezoid rule in Section 3.3.

Depth of the centroid of the damage area. This is the x value. Calculating this value is dis-

cussed in Section 3.4.1.

The following discussions and the equations developed are for a collision in which the PDOF

angle is 90◦to the damage face. The situation in which the PDOF is not 90◦ is beyond the scope

of this paper.

3.2.1. Determining Stiffness Coefficients

The A stiffness coefficient represents the amount of force that the vehicle can sustain before it

begins to permanently deform. The B stiffness coefficient represents the amount of force needed

to permanently deform the vehicle structure. The G value is the area of the force-deflection

triangle to the left of the abscissa (y-axis) and is related to A and B by the relationship

G =A2

2B(27)

3.2.2. Crash Test Data

Most vehicle crashes are conducted under the New Car Assessment Program (NCAP), which

requires a 35-mph (56 kph) full-frontal barrier crash test. Other crash tests include: angled



frontal-fixed barrier collisions, movable barrier collisions (both deformable and non-deformable),

narrow object impact tests, and side-movable deformable barrier tests. The remaining discus-

sion on crash test data will deal only with full-frontal barrier tests.

When we obtain crash test data, we will get the average crush of the vehicle. We will call this

Cavg. This assumes a uniform crush profile. Procedures for dealing with a non-uniform crush

profile are discussed in Ref. [5].

Test reports are available to the public through the National Highway Traffic Safety Adminis-

tration web site, http://www-nrd.nhtsa.dot.gov/. These reports may have a data sheet

entitled “Accident Investigation Division Data” that will reveal:

• The approach speed of the vehicle in kph: ∆vtest

• The test mass of the vehicle in kg: mtest

• Six vehicle crush depths in mm: C1, C2, C3, C4, C5, and C6

• Width of the damage profile in mm: Ltest

Unit conversions will be required to work in US units. Dimensions must agree in all equations.

3.2.3. Determining ∆vtest

There are different ways in which people have defined ∆vtest when using crush energy formu-

las. One method is to use the total ∆v, which is the actual change in velocity experienced by

the center of mass of the vehicle. It is defined as:

∆v = v1 − v3

where v1 is the approach velocity and v3 is the rebound velocity. Using the definition of the

coefficient of restitution gives:

∆v = v1(1 + ǫ)

The equations for ∆v are based on the energy, Ecrush , dissipated during the collision. These

equations also assume no restitution. As such, the simplest way of computing stiffness values

is to ignore the effects of restitution. This means that ∆vtest = vapproach .

The CRASH III Manual (Ref. [4]) presents the idea of using an “effective” energy that will

account for the restitution of the crash test in the computation of the ∆v values by introducing

the “effective” energy term

Eeff = EA(1 + ǫ)2



where EA is the kinetic energy of the approaching vehicle. This is misleading because the actual

energy dissipated in crushing the vehicle is:

Ecrush =1

2mv2

A(1 − ǫ2)

= EA(1 − ǫ2)

Therefore, it is the recommendation of the authors to either ignore restitution when computing

crush energy or include the effects of restitution on total ∆v calculation separate from the dam-

age energy calculation. Some authors may advocate determining stiffness coefficient based on

the total ∆v. Again, this is ill-advised because using the total ∆v will give crush energy values

that are erroneously high unless restitution is near zero. For example, if ǫ = 0.1, then Eeff will

be 21% higher than EA, whereas Ecrush is only 1% lower than EA. Therefore, ignoring the resti-

tution for barrier impact tests at 30 or 35 mph will result in small errors. It will also result in

a consistent definition of ∆vtest, namely, the change in speed before rebound. Therefore, for a

fixed barrier test, use the actual approach speed to determine the stiffness values.

3.2.4. Determine Campbell Model Coefficients

Campbell noted that, for early 1970’s full-sized General Motors vehicles, the impact speed and

the depth of crush followed a straight line similar to the line shown in Figure 9.

The model for relating frontal barrier impact speed to crush damage takes a linear form:

v = b0 + b1C (28)

where v is the impact speed,

C is the residual crush,

b0 is the zero crush speed (intercept) in units of speed, and

b1 is the slope in units of speed per length (i.e. mph/in).

When a vehicle crashes into an object, energy is expended. The majority of the energy lost in

a collision is due to the plastic (permanent) deformation of the vehicle(s) and/or the objects

involved in the crash. Quantifying this energy is difficult since there are many different mech-

anisms of dissipating the energy. However, a model based on crash tests can approximate the

amount of energy dissipated. This is called an empirical model because it is based on obser-

vations rather than physical principles. The linear model shown in Figure 9 is an empirical

model. Since energy is dependent on mass as well as velocity, a “standard” weight was used

for diagrams similar to Figure 9.

This technique can be used if only a single crash test is available. We will outline a simple

procedure to extract the empirical crush coefficients. These coefficients are used to determine



0

10

20

30

40

50

60

0 10 20 30 40 50 60

Residual Crush, inches

Imp

act

Sp

eed

,mil

esp

erh

ou

r

b

b

b

b

bbbbb

b

bbbbbb

b

b

bb

b

b0

1

b1

Figure 9: Impact speed plotted against the measured residual crush for a frontal-fixed barrierimpact. The slope of the line is b1 and the intercept is b0. This plot resembles Fig. 1from Campbell’s 1974 paper [6].

the energy dissipated for an irregular crush profile.

The first step is to define a “no-damage” speed for our test vehicle, bo. Typical values for the

speed at which no residual crush exists are 5 mph (8 kph) for front and rear impacts in which

the vehicle is protected by bumpers. On the other hand, side impact no-damage speeds are

typically 2 mph (3.2 kph).

When using the US system of measurement, our A and B values will be in units of lb/in

and lb/in2, respectively, so we will need to express all of our no-damage speeds and ∆v values

in inches/second. In the same way, acceleration will be expressed in inches/second2. Thus, 5

mph = 88 in/sec, 2 mph = 35 in/sec, and 32.2 ft/sec2 = 386 in/sec2.

Similarly, when using the SI system of measurement, our A and B values will be in units

of newton/meter and newton/meter2, respectively, so we will need to express all of our no-

damage speeds and ∆v values in meters/second. In the same way, acceleration will be ex-

pressed in meters/second2. Thus, 8 kph = 2.22 m/sec, and 3.2 kph = 0.89 m/sec.

Let us now determine the slope of the line that would intersect the y-axis (speed) at bo and

the crash data point (Cavg, ∆vtest). This slope is b1:

b1 =∆vtest − bo

Cavg(29)

This equation assumes uniform crush depth. Vehicles with significant taper on the front or



Crush

Impact Speed

bo

∆v

Cavg

(Cavg, ∆v)

Figure 10: Impact speed-crush graph. Point (Cavg, ∆v) is on the graph and can be used tocalculate the slope, b1, of the graph.

some otherwise irregular crush profile need different crush averaging. Neptune provides a

formula to determine stiffness coefficients based on irregular crush profiles in Ref. [5].

3.2.5. Determining A, B, and G Values

Once we have bo and b1, we can calculate the A, B, and G stiffness coefficients. These equations

take the Campbell impact speed-crush data (Figure 9) and convert it to force-crush data (Figure

11).

The A stiffness coefficient can be determined with the following equation:

A =mTbob1

Ltest(30)

where mT is the mass of the test vehicle with instrumentation. This is usually given in kilograms

and needs to be in slugs for the US system (1 kg = 0.0685 slugs).

The B stiffness coefficient can be determined with the following equation:

B =mTb2

1

Ltest(31)

The G value is a straight-forward calculation based upon the triangle geometry.

G =A2

2B(32)

3.3. Determining Damage Area

The initial models for the CRASH III program approximated irregular damage profiles by di-

viding the damage area into equally spaced trapezoids. Each trapezoid was bounded on the


3.3. Determining Damage Area

O x = residual crush

F

F =A + Bx

B

G

A

Figure 11: The geometric relationship between A, B, and G.

ends by the damage face and the undamaged collision face profile. The trapezoids are bounded

on the sides by the crush measurements, Cn. Two, four, or six crush measurements are taken and

must be equally spaced. This results in one, three, or five crush zones, as seen in Figure 12.

The damage area is computed by summing the areas of all the zones, so for Figure 12:

AD = A1 + A2 + A3 + A4 + A5

After some geometry and algebra, we get the damage area for six equally spaced crush mea-

surements:

AD =L

10(C1 + 2C2 + 2C3 + 2C4 + 2C5 + C6) (33)

Example 2 Determine the stiffness coefficients for a 1993 Honda Accord in SI units based on the

NHTSA crash test 1875 performed by Calspan Corp.

This crash test was at 56.3 kph (assume no rebound velocity) and the test weight was 1579

kg. The overall length of the damage region is Ltest = 1460mm. The crush depth dimensions

reported are:

Location C1 C2 C3 C4 C5 C6

Depth (mm) 423 481 522 523 483 376

The mean crush depth is calculated using the area determined in Eq. (33):

Cavg =AD

L

=�L(423 + 2(481) + 2(522) + 2(523)+ 2(483) + 376)

10�L

= 482 mm = 0.482 m



deformedvehicle

L

x

− +

y−

+

C1C2

C3

C4C5

C6zone 1 zone 2 zone 3 zone 4 zone 5

Figure 12: An irregular damage profile of width L pictured here is broken up into five trape-zoidal zones by taking six equally spaced crush measurements.


3.4. Determining the Location of the Damage Centroid

Employing Eq. (29), converting 56.3 kph to 15.64 m/s, and assuming that b0 = 2.22 m/s gives

the value for b1:

b1 =∆vtest − bo

Cavg

=15.64 − 2.22

0.482

= 27.84

This value is used in Eq. (30) to determine the A coefficient:

A =mTbob1

Ltest

=1579 kg (2.22 6m

s )(27.84 1s )

1.46 6 m

= 66, 842kg

s2

If we multiply the units for A by m/m, we would get units ofkg-msec2 ( 1

m ), which are units of

newtons/meter. The value for the B coefficient is:

B =mTb2

1

Ltest

=1579 kg (27.84 1

6s )2

1.46 m

= 838, 239N

m2

The value for the G coefficient is:

G =A2

2B

=

(66, 842 N

m

)2

2(838, 239 N

m2

)

= 2665 N

3.4. Determining the Location of the Damage Centroid

The damage centroid is the point within the damage profile at which the collision force acts.

It can be thought of as the center of mass of the damage area. If the CRASH III model is to



work for two-vehicle collisions, the damage centroids must reach a common velocity. There-

fore, sideswipe collisions cannot be modeled with CRASH III. This constraint is not placed on aIMPORTANT!

COLM solution! For a COLM solution, there is no requirement that the damage centroids reach

a common velocity.

In the interest of brevity, the results only are presented in this section. A detailed derivation

can be found in Ref. [2].

3.4.1. Longitudinal Location of the Centroid in the Damage A rea

x =C2

1 + 2C22 + 2C2

3 + 2C24 + 2C2

5 + C26 + C1C2 + C2C3 + C3C4 + C4C5 + C5C6

3(C1 + 2C2 + 2C3 + 2C4 + 2C5 + C6)(34)

3.4.2. Lateral Location of the Centroid in the Damage Area

Having determined the longitudinal location of the centroid, x, we will determine the lateral

location of the centroid, y, which pinpoints the centroid of the damage area. We will need this

location to determine where the PDOF is acting.

y =L

30

(−13C1 − 18C2 − 6C3 + 6C4 + 18C5 + 13C6

C1 + 2C2 + 2C3 + 2C4 + 2C5 + C6

)

(35)

3.4.3. Locating the Damage Centroid with Respect to the Loca l Axis of the Vehicle

Once the centroid of the damage area has been located within the damage area, we can locate

the centroid with respect to the local x-y axis of the vehicle. The local axis has its origin at

the center of mass (CM) of the vehicle. Positive x is forward of the CM. Positive y is to the

passenger side of the CM. The x-y location of the centroid is an ordered pair with + or − signs

used depending on where it is with respect to the local axis.

In the x-direction

The value for x locates the depth of the centroid from the damage face of the vehicle. From

vehicle specification databases (such as Expert Autostats R©) we can determine various mea-

surements such as, front overhang, wheelbase, front bumper to front axle, center of mass to

front axle, etc. Using these measurements, as necessary, along with x, the location of the cen-

troid can be located with respect to the local vehicle axis. It may be helpful to sketch a picture

showing these measurements to assist in locating the centroid and to help remember the sign

of the location.


3.5. Crush Energy Equations

In the y-direction

The value for y laterally locates the centroid from the center of the damage area (half of the

measured damage width). From vehicle specification databases, we can determine various

measurements such as vehicle width, front overhang, wheelbase, front bumper to front axle,

center of mass to front axle, etc. The center of mass of the vehicle is generally located at physi-

cally half the vehicle width.

When measuring the damage area of a vehicle, the center of the damage area (half the dam-

age width) is located with respect to the center of mass of the vehicle and is called D. Since y is

referenced to the center of the damage width, knowing D allows us to locate the centroid with

respect to the local vehicle axis. It may be helpful to sketch a picture showing these measure-

ments to assist in locating the centroid and to help remember the sign of the location.

3.5. Crush Energy Equations

In the previous sections, we have developed equations for damage area and for damage cen-

troid depth. Now we substitute these values into Eq. (26) on page 18:

ET = (A + Bx) AD +A2L

2B

The result for six evenly spaced crush measurements (five crush zones) is:

ET =L

5

(A

2(C1 + 2C2 + 2C3 + 2C4 + 2C5 + C6)

+B

6

[C2

1 + 2C22 + 2C2

3 + 2C24 + 2C2

5 + C26 + C1C2 + C2C3 + C3C4 + C4C5 + C5C6

]

+5A2

2B

)

(36)

Example 3 Determine the crush energy of a 1993 Honda Accord given the following equally spaced

crush measurements across a front width of L = 63.75 in:

Location C1 C2 C3 C4 C5 C6

Depth (in) 3.3 11.0 14.78 15.0 14.0 7.0

The A stiffness coefficient was determined from the data in Example 2 to be 396 lb/in and the

B coefficient is 129 lb/in2. This is a six-crush measurement that can be cumbersome by hand.


4. Relationship between Crush Energy and the Coefficient of Restitution

Therefore, let us introduce the temporary variables:

x1 = C1 + 2C2 + 2C3 + 2C4 + 2C5 + C6

x2 = C21 + 2C2

2 + 2C23 + 2C2

4 + 2C25 + C2

6

x3 = C1C2 + C2C3 + C3C4 + C4C5 + C5C6

So from Eq. (36) we get:

ET =L

5

(Ax1

2+

B(x2 + x3)

6+

5A2

2B

)

(37)

Solving for the x values gives:

x1 = 3.3 + 2(11.0) + 2(14.78)+ 2(15.0) + 2(14.0) + 7.0 = 119.86 in

x2 = 3.32 + 2(11.02) + 2(14.782) + 2(15.02) + 2(14.02) + 7.02 = 1580.78 in2

x3 = 3.3(11.0) + 11.0(14.78) + 14.78(15.0)+ 15.0(14.0) + 14.0(7.0) = 728.58 in2

Substituting these values into Eq. (37) yields:

ET =63.75

5

(396(119.86)

2+

129(1580.78 + 728.58)

6+

5(396)2

2(129)

)

= 12.75(23, 732.28 + 49, 651.24 + 3039.07)

= 974, 390 in-lb

= 81, 200 ft-lb

4. Relationship between Crush Energy and the Coefficient of

Restitution

4.1. Physics of an Impact

An impact happens when two objects interact with large forces over a short period of time.

We refer to this interaction as a collision. This causes large impulsive forces which typically

deform the objects. The moment two objects come together is called incidence. Immediately

after incidence the compression phase begins where the colliding objects deform and absorb ki-

netic energy. This compression phase has a finite duration and is terminated when the dynamic

deformation reaches a maximum. Following the maximum deformation, a period of restitution

occurs where the object may rebound. During the rebound phase some (not all) of the stored en-

ergy is turned back into kinetic energy as the objects depart with some relative velocity. We also


4.1. Physics of an Impact

(a) Location of the vehicle at incidence, t = 0.

Fixed Barrier

v = vin

(b) Location of the vehicle at maximum crush, t = tc .

Fixed Barrier

v = 0

(c) Location of the vehicle at separation, t = ts .

Fixed Barrier

v = vout

Co

mp

ression

Reb

ou

nd

Figure 13: The different phases of an impact between a vehicle and a solid fixed barrier.

define the collision as taking place with no displacement with respect to an inertial (ground)

reference frame. Hence, we do not have to consider any gain or loss of potential energy during

the collision phase.

Figure 13 shows the two phases of a collision of a vehicle into a fixed barrier. In Fig. 13a, the

vehicle is just touching the barrier and has some kinetic energy. The contact generates a force

that acts over some distance to the point of maximum crush. The collision force acting through

the distance is the work done in the crash. All of the energy used to do the work associated

with deformation comes from the initial kinetic energy.



4.2. Taxonomy of Impacts

4.2.1. Nature of Elasticity

The discussion in this section is based upon classical mechanics, which have a foundation in

Newton’s three laws of motion as well as the work done by Huygens and others just before

Newton’s time. We use classical mechanics for crash reconstruction because, in most cases, the

results we obtain are reasonable and may be done with relatively simple math. The variables

involved are simple to define and are measurable by means of testing. The core of classic

impact analysis is the impulse-momentum model. There are other impact models that are based

upon elastic wave propagation, contact mechanics, and plastic deformation or hydrodynamic

modeling which are far beyond the scope of this paper.

We may place collisions into two general categories: an elastic or inelastic collision:

An elastic collision is one in which kinetic energy is conserved. As such, an elastic collision

is a conservative system, meaning a system in which kinetic energy is conserved. Real

world systems are never completely elastic, as kinetic energy is never completely saved.

The work done to deform the objects is therefore called reversible work. In elastic collisions,

the colliding objects bounce off of each other.

An inelastic collision is one in which kinetic energy is not conserved. This does not mean that

all of the kinetic energy of the system goes away in the collision. In fact, usually it does

not, as the colliding objects will move away with some velocity after the impact. The

amount of kinetic energy that may be lost in a collision is consistent within the bounds

set by conservation of linear momentum. In a completely inelastic collision, the kinetic

energy required to do the work to deform the objects is transformed into other forms of

energy, such as heat. As such, the work done to deform the colliding objects is called

irreversible work, because we cannot get it back. A characteristic of inelastic collisions is

the two objects tend to stick together.

Real world collisions usually fall somewhere between being elastic and inelastic. A measure

of the elasticity of a collision is the coefficient of restitution, which, by Newton’s definition, is a

ratio of the relative velocity of approach to the relative velocity of recession. The coefficient

of restitution may vary between zero and one. A value of one indicates a completely elastic

collision, while a value of zero indicates a completely inelastic collision. For example, if we

drop a steel ball bearing on a hard steel plate, it will bounce quite a bit. We would say that

collision has a relatively high coefficient of restitution. If we take a soft lead ball and drop it on

the same plate, it will not bounce quite as much. Hence, its coefficient of restitution is lower

than that of the steel ball, but still has some value. If we replace the lead ball with a soft lump

of clay, then we will see an inelastic collision, as the clay will hit and stick to the surface. As


4.2. Taxonomy of Impacts

the coefficient of restitution becomes closer to zero, more of the system kinetic energy is being

dissipated into other forms of energy, primarily heat.

We also consider that the only forces acting on the bodies during the collision are the im-

pulsive (collision) forces themselves. As such, for either one or two dimensional collisions, we

ignore the effects of any ground frictional forces. This is an example of an unconstrained impact.

In the real world, this assumption may not always be valid and depends on several things, such

as relative vehicle masses, presence of significant ground forces, or other constraints that may

affect vehicle motion.

4.2.2. Relative Velocities

Consider for a moment the steel plate and ball mentioned above. How might the relative ve-

locity at impact affect the impact behavior?

Let us think about what happens if we drop the steel ball on the plate from a height of 14

feet. This will result in an impact velocity of about 30 fps. If we examine this ball after the

impact, we probably will not be able to discern and permanent deformation in the ball. If we

do the same thing with the lead ball, then we may see a small flat spot on it, but still no great

deformation.

Now, let us fire the balls in turn out of an air gun at 300 fps. The steel ball will still bounce off

the plate, but there may well be some measurable deformation in it. It’s coefficient of restitu-

tion, in the sense of classical mechanics, will likely be less because of this permanent deforma-

tion. The lead ball, being softer (less internal strength), will probably be flattened by the impact

and may have little or no bounce at all.

As a final example, we will replace the air gun with a high velocity rifle that is capable of

launching the balls at 3000 fps. In this case, the steel ball will likely penetrate through the

plate, resulting in large plastic flows of both the ball and the plate. This is the beginning of

hydrodynamic behavior where the extreme stresses make the solid behave like a fluid. In a similar

way, the lead ball may also penetrate the plate, even though it is much softer than the plate.

These high velocity impacts are called ballistic impacts, and we may not use classical impact

mechanics to determine impact behavior.

As we may see with our thought experiment, the coefficient of restitution is a function not

only of the material properties of the impacting bodies, but also on the relative velocity at

impact. We will discuss this further in Section 4.3.

4.2.3. Orientation

There are generally two different types of impacts: collinear and oblique:

A collinear impact occurs when the direction of travel coincides with the direction of force.



Head-on or rear-end collisions are examples of collinear impacts. Also, a collinear impact

occurs whenever one vehicle or object is stationary.

An oblique impact occurs when the line of force is not coincident with the direction of travel.

The oblique impact is the general case of any planar (two dimensional) impact which

means a central impact is a special case of an oblique impact.

In traffic crash reconstruction, we may also categorize collisions as being either central or non-

central. A central collision is one in which the impulse force (PDOF) passes through the center

of mass of the vehicle (object). A non-central collision means the impulse force does not go

through the center of mass. In a central collision for both vehicles, we would expect the post-

impact velocity (both magnitude and direction) to be similar and for the vehicles to move away

from the collision with little or no rotation. These constraints are not placed on a non-central

collision.

4.3. Understanding the Coefficient of Restitution

The coefficient of restitution provides the analyst with a technique of dealing with energy losses

in a collision. It is fairly simple to use with in-line collisions but it fails to provide all the miss-

ing pieces to a planar impact problem. There are three basis for the definition of the coefficient

of restitution: kinematic, kinetic, and energy. These definitions will be presented and the rela-

tionship between them will be explained.

4.3.1. Kinematic Definition of Restitution

Sir Issac Newton provided the first definition of restitution with a formula based on the relative

velocities of each object:

ǫ =v2,out − v1,out

v1,in − v2,in(38)

where v refers to the magnitude of the velocity normal to the impact plane. For in-line colli-

sions, these velocities are the actual velocity magnitudes. If the collision is oblique, then the

velocity used in Eq. (38) is the component of the velocity vector normal (perpendicular) to the

plane of impact.

While Newton was correct with most everything he wrote, he mistakenly presumed the coef-

ficient of restitution was only a material property. It has been since shown that restitution also

depends on the relative velocities themselves. Also, application of Eq. 38 for eccentric impacts

may lead to an apparent increase in energy which is physically inadmissible.



4.3.2. Kinetic Definition of Restitution

In the early 1800’s Poisson developed a kinetic definition of the coefficient of restitution. This

definition says the coefficient of restitution is the ratio of the magnitude of the normal rebound

impulse to the magnitude of the normal deformation impulse perpendicular to the plane of

contact:

ǫ =(F∆t)rebound

(F∆t)de f orm(39)

We can show that this definition is consistent Newton’s definition. When only collision forces

are significant, Newton’s Second Law gives us the definition of impulse as a change in momen-

tum:

F∆t = m∆v

where the bold fonts indicate vector quantities that have both magnitude and direction. The

overall change in velocity follows the kinematic definition:

∆v = vout − vin

At the point of maximum deformation, the contact location reaches some unknown velocity,

vc. This velocity may be zero if an object strikes a solid barrier. The peak deformation is also

achieved at a unique time sometime between when the collision started and when the objects

are no longer interacting. Using this point in time, we can break up the change in velocity into

the sum of two distinct events:

∆v = vout − vc︸︷︷︸

rebound

+ vc − vin︸︷︷︸

de f orm

Knowing that mass is conserved in a crash, we can get the following relationship:

F∆t = mvout − mvc︸︷︷︸

rebound

+ mvc − mvin︸︷︷︸

de f orm

and

F∆t = (F∆t)rebound + (F∆t)de f orm

which says the total impulse is the vector sum of the deformation impulse and the rebound

(restitution) impulse. In all physical cases, the deformation impulse is larger than the restitution

impulse. For a central collision, the rebound impulse will be both smaller in magnitude and

opposite in direction to the deformation impulse which makes the total impulse smaller than



the deformation impulse. The following relationships should also be noted:

(F∆t)rebound = mvout − mvc (40)

(F∆t)de f orm = mvc − mvin (41)

The rebound and deformation impulses act along the line of impact which is perpendicular to

the plane of impact. During an actual collision, the line of impact may change so the choice for

which line of impact to use is based on the overall effect of the impulse. Since collision times

are short, the line of impact is fairly consistent. The line of impact is also referred to as the

Principal Direction of Force (PDOF).

Notice that Eqs. (40) and (41) contain an unknown velocity vector vc. This unknown will be

determined by using Newton’s Third Law which says that impulses act equally and opposite

when two objects interact. Therefore, if we have two objects, #1 and #2, then during an impact:

F1∆t = −F2∆t

Since the time scale is common to both objects, we can express Newton’s Third Law for both

phases of the crash as:

(F1∆t)rebound = −(F2∆t)rebound

(F1∆t)de f orm = −(F2∆t)de f orm

For the restitution phase, Newton’s Third Law can be rewritten as:

m1v1,out − m1v1,c = −(m2v2,out − m2v2,c) (42)

and for the deformation phase:

m1v1,c − m1v1,in = −(m2v2,c − m2v2,in) (43)

Notice that the collision velocities of each object do not have be the same. Solving for the

collision velocity at maximum engagement of object #1 from Eq. (42):

v1,c =m1v1,out + m2v2,out − m2v2,c

m1(44)

and solving for v1,c in Eq. (43) is:

v1,c =m1v1,in + m2v2,in − m2v2,c

m1(45)

Setting Eq. (44) equal to Eq. (45) gives the Conservation of Linear Momentum Equation that



says the total momentum in is equal to the total momentum out (even if the velocities v1,c and

v2,c are different):

m1v1,in + m2v2,in = m1v1,out + m2v2,out

When dealing with two dimensions, or planar impact mechanics, the kinetic definition of

restitution only characterizes the collision along the line of line of impact. In introductory

dynamics textbooks, the assumption is made that the impact is frictionless for oblique collisions

[7, 8]. Obviously, real impacts (car crashes) can contain friction when dealing with oblique

impacts. Including the effect of tangential impulses from friction has been addressed by various

authors [9, 10, 11].

Since the definition from Eq. (39) only relates the impulse vectors that are normal to the

impact plane, the force and velocity vectors must be expressed in terms of their components:

F = (Fn, Ft) and v = (vn, vt) (46)

where the subscript n refers to the normal direction and the subscript t corresponds to the tan-

gent direction. The statement earlier about the velocities at the point of contact being different

at maximum compression can now be further qualified. In the direction normal to the contact

plane, the points of contact of both bodies must reach a common velocity in the normal direc-

tion to prevent interpenetration. The velocity components of the contact points in the direction

tangent to the impact plane can be different if sliding exists.

Considering only the vector components in the normal direction we can substitute Eqs. (40)

and (41) into Eq. (39):

ǫ =mvn,out − mvn,c

mvn,c − mvn,in(47)

where n indicates the vector component normal to the impact plane. From hereon, the normal

component of velocity is implied. Since mass is a scalar, it can be factored out and canceled:

ǫ =vout − vc

vc − vin(48)

Since, in the normal direction, the velocities at maximum compression are the same v1,c =

v2,c = vc we can simplify Eqs. (44) and (45):

vc =m1v1,out + m2v2,out

m1 + m2(49)

and

vc =m1v1,in + m2v2,in

m1 + m2(50)

Substituting Eq. (50) into the numerator of Eq. (48) and Eq. (49) into the denominator of Eq. (48)



for object #1 gives:

ǫ =v1,out −

m1v1,out +m2v2,out

m1+m2

m1v1,in+m2v2,in

m1+m2− v1,in

ǫ = ��m1v1,out + m2v1,out −��m1v1,out − m2v2,out

��m1v1,in + m2v2,in −��m1v1,in − m2v1,in

Factoring out and canceling m2 gives the kinematic definition of restitution from Section 4.3.1:

ǫ =v1,out − v2,out

v2,in − v1,in

4.3.3. Energetic Definition of Restitution

In 2000, W. J. Stronge [9] published a book where he defined the square of the coefficient of

restitution based on the work done by the normal forces in the collision. This is known as

Stronge’s Hypothesis and is stated mathematically as:

ǫ2 = −Wrebound

Wde f orm(51)

where Wrebound = W1,rebound + W2,rebound is the sum of the work done by both normal impulsive

forces during the rebound phase of the collision. Similarly, Wde f orm = W1,de f orm + W2,de f orm is

the sum of the work done by both normal impulsive forces during the deformation (or com-

pression) phase of the collision. This definition requires the colliding objects to be deformable.

This, however, is not a limitation because every real object is deformable to some extent.

Energy is the ability to do work and the work done by the force normal to the impact plane is

equal to the kinetic energy of the relative velocities normal to the plane. The diagram in Fig. 14

may be helpful in understanding the relationship between work, energy, impulse, velocity,

force and time. The work done by the deformation force is determined as the scalar product of

force and displacement:

Wde f orm =∫ xmax

0F(x) dx (52)

which can be though of graphically as the area under the force displacement curve. If we use a

simple linear spring model where F(x) = kx, then

Wde f orm =1

2kde f orm x2

max (53)

Similarly, the work done by the rebound force is

Wrebound =∫ xresid

xmax

F(x) dx (54)



But xmax is greater than xresid , so we can swap the limits of integration:

Wrebound = −

∫ xmax

xresid

F(x) dx (55)

and for a spring:

Wrebound = −1

2krebound (x2

max − x2resid) (56)

This represents the area under the curve in Fig. 14d.

The work done can be related to the impulse using a relationship developed by Poisson:

W =∫

Fv dt (57)

= vavg

∫

F dt (58)

= vavg F∆t (59)

which says the work done is equal to the velocity multiplied by the impulse. Incorporating this

relationship into the energetic definition of the coefficient of restitution gives:

ǫ2 = −v1,avg,rebound (F∆t)rebound + v2,avg,rebound (F∆t)rebound

v1,avg,de f orm (F∆t)de f orm + v2,avg,de f orm (F∆t)de f orm(60)

We also know that vavg = aavg∆t and aavg = Fm . The normal component of the rebound velocity

is opposite in direction of the deformation so the negative signs cancel. Making the appropriate

substitutions:

ǫ2 =F∆tm1

(F∆t)rebound + F∆tm2

(F∆t)rebound

F∆tm1

(F∆t)de f orm + F∆tm2

(F∆t)de f orm

(61)

ǫ2 =��(

1m1

+ 1m2

)

(F∆t)2rebound

��(

1m1

+ 1m2

)

(F∆t)2de f orm

(62)

Taking the square root of both sides renders the kinetic definition of restitution:

ǫ =(F∆t)rebound

(F∆t)de f orm(63)

The energetic definition of restitution has recently gained popularity as it prevents a solution

that violates the conservation of energy while maintaining the classic definitions.



time

Force

tmax0 tsep

(F∆t)de f (F∆t)reb

(a) General representation of force as afunction of time. The areas correspond tothe deformation impulse and the reboundimpulse.

time

displacement

tmax0 tsep

xres

xmax

(b) General representation of displacementas a function of time.

time

Impulse

tmax0 tsep

(c) Impulse is always increasing

displacement

Force

xmax0 xres

Ecrush −Wreb

(d) The area under the force vs. deflectionlines represent the work done by the colli-sion force.

Figure 14: Illustrations of the relationship between work, energy, impulse, velocity, force, andtime.


4.4. Computing Restitution based on Damage Energy

4.4. Computing Restitution based on Damage Energy

In this section, a relationship between the magnitude of the incoming normal velocities, dam-

age (crush) energy, and the coefficient of restitution is given based on Stronge’s Hypothesis. We

begin by performing an energy balance:

1

2m1v2

1 +1

2m2v2

2︸︷︷︸

KEin

=1

2m1v2

3 +1

2m2v2

4︸︷︷︸

KEout

+Ecrush (64)

In an earth fixed coordinate system, the point of contact may have some velocity. This velocity

is unknown but special in that it is the common velocity of the damage centroids for both the

objects in the collision. This common velocity is called vc. The energy balance be written with

respect to the common velocity. If this is done, then the work done during deformation can be

related to the kinetic energy in reference to the common velocity, vc. Therefore, Eq. (64) can be

written as:

1

2m1(v1 − vc)

2 +1

2m2(v2 − vc)

2

︸︷︷︸

KEin

=1

2m1(v3 − vc)

2 +1

2m2(v4 − vc)

2

︸︷︷︸

KEout

+Ecrush (65)

Notice that the energy balance gives a value for the residual damage energy as:

Ecrush = KEin − KEout (66)

This is not the maximum crush energy as some of the maximum energy absorbed in the crush-

ing object is returned as kinetic energy out. To do this, some work has to be done by the deform-

ing object. This is the work done by the collision force during rebound. Since the deformation

during rebound is opposite in direction to the deformation of crush, the sign on the work must

be negative. The magnitude of the work done by the rebounding force is represented by the

area under the grey triangle in Fig. 14. Mathematically, the work done by the rebounding force

is related the the kinetic energy as:

Wrebound = −(KEin − Ecrush ) (67)

Also, the ability for the compressive force to do work comes from the kinetic energy relative to

the common velocity. Using this statement and Eq. (65) in the energetic coefficient of restitution



gives:

ǫ2 =KEin − Ecrush

KEin

=KEin

KEin−

Ecrush

KEin

= 1 −Ecrush

KEin

= 1 −Ecrush

12 m1(v1 − vc)2 + 1

2 m2(v2 − vc)2(68)

It is necessary to determine the common velocity based on impulse-momentum concepts. Dur-

ing the compression phase, the changes in velocity are ∆v1 = v1 − vc for object #1 and ∆v2 =

v2 − vc for object #2. Also, Newton’s Third Law says that the compression impulses are equal

and opposite:

F1∆t = −F2∆t (69)

The concept of impulse and momentum (a variant of Newton’s Second Law) gives the relation-

ships:

F1∆t = m1(v1 − vc) (70)

and

F2∆t = m2(v2 − vc) (71)

Using Eqs. (69)-(71) gives an expression for vc:

vc =m1v1 + m2v2

m1 + m2(72)

This equation is still valid for incredibly large masses (rigid barriers). If m1 → ∞, then vc = v1.

Similarly, if m2 → ∞, then vc = v2. The difference between v1 and vc is:

v1 − vc =��m1v1 + m2v1 −��m1v1 − m2v2

m1 + m2(73)

Likewise,

v2 − vc =m1v2 +��m2v2 + m1v1 −��m2v2

m1 + m2(74)

Substituting these differences into Eq. (68) and simplifying gives the result:

ǫ2 = 1 −2Ecrush (m1 + m2)

m1m2(v1 − v2)2(75)


4.5. Computing Coefficient of Restitution Based on Crash Test Data

It should be noted that Eq. (75) can be expressed as:

ǫ2 = 1 −Ecrush

KEin(76)

where

KEin =1

2

(m1m2

m1 + m2

)

(v1 − v2)2 (77)

is the maximum amount of energy available to crush the vehicles. The derivation of this equa-

tion can also be found in Ref. [12]. For real values of ǫ the right hand side of Eq. (75) must be

positive. Also, for physically plausible results, the RHS of Eq. (75) must be less than 1. Given

these constraints are satisfied, we can take the square root to get:

ǫ =

√

1 −2Ecrush (m1 + m2)

m1m2(v1 − v2)2(78)

The above equation gives a value for the coefficient of restitution based on the estimation of

the damage energy and the solution of the incoming velocities. It is important to note that the

velocities are vector quantities and we must use vector subtraction to determine the relative

velocities. This means that if the velocities are the same (both magnitude and direction), then

the denominator is zero and no solution exists when performing the division of Eq. (78). This

corresponds to the physical fact that a collision cannot occur if both objects have the same

velocity.

The primary use of Eq. (78) for a sanity check on any solution based on a damage momentum

technique. Since the coefficient of restitution has physical meaning and has been tested and

reported in the literature, determination of ǫ should render a typical value if both the crush

energy and the impact velocities were computed correctly. With measurable crush profiles

values of ǫ typically less than 0.30. Also, ǫ should be greater than zero if the vehicles did not

stick together. Typical values are between 0.05 and 0.15. Higher relative impact velocities give

lower coefficients of restitution.

4.5. Computing Coefficient of Restitution Based on Crash Tes t Data

In a frontal barrier crash test, the impact is collinear and the kinematic definition of restitu-

tion proves to be the easiest to compute the coefficient of restitution. The incoming velocity is

known (usually 35 mph) and the exit velocity must be read from the data and graphs in the

crash report. For example, we’ll use report number CAL-93-N10 for the 1993 Honda Accord.

In that report, there are nine accelerometer locations measuring acceleration in the x direction

(along the direction of travel) listed on page 3-11. In Appendix B of the CALSPAN report, the

acceleration, velocity and displacement curves are shown. Every acceleration trace should start



Table 2: Data extracted from Calspan report # CAL-93-N10 for a full frontal barrier crash testof a 1993 Honda Accord. The entrance speed was 56 kph.

Accelerometer Location Exit Velocity (km/h) Coef. of Rest.

1 Left Rear Seat –4 0.071

2 Right Rear Seat –4 0.071

3 Top of Engine Accelerometer Destroyed N/A

4 Bottom of Engine 0 N/A

5 Right Front Brake 35 N/A

6 Left Front Brake –30 N/A

7 Instrument Panel –10 0.18

8 Left Rear Seat 8 N/A

9 Right Rear Seat –3 0.054

Table 3: Mean and standard deviation of the coefficient of restitution for vehicle to barrierimpacts for different types of vehicles at both 30 mph and 35 mph taken from Ref. [14].

VehicleType

Passenger(30mph)

Passenger(35mph)

Pickup(30mph)

Pickup(35mph)

SUV(30mph)

SUV(35mph)

Van(30mph)

Van(35mph)

Avg. ǫ 0.139 0.152 0.105 0.160 0.135 0.146 0.131 0.143

Std. Dev. 0.045 0.028 0.023 0.036 0.058 0.026 0.044 0.041

and end at zero. If it does not, then either the accelerometers was knocked off axis or was not

calibrated. Each velocity trace should also start and end with the same velocity and the final

velocity should be negative (indicating rebound).

Table 2 shows the data gathered from the report. It can be seen that the crash test only

produced four physically plausible data point to determine the coefficient of restitution. The

four results have a large relative spread but an overall small value for restitution. Keep in mind,

the restitution values computed are for a vehicle striking a solid barrier at 35 mph.

Justification of using the coefficient of restitution from crash test to actual crashes is not based

on physical principle, but rather an understanding that the variation of the coefficient of resti-

tution from the crash test most likely encompasses a particular crash in question. Recent re-

searchers have developed techniques to determine a composite coefficient of restitution based

on physical models [13].

If a crash test is not available or the coefficient of restitution is desired for different speeds

and vehicles, then a literature search is necessary to justify the coefficient of restitution. A

detailed paper by Monson and Germane [14] contains both results of the mean and standard

deviation for coefficient of restitution based on crash test data. Some results from Ref. [14] are

shown in Table 3. Another detailed report of the coefficient of restitution based on test results

was written by Prasad in Ref. [15].


4.6. Concluding Remarks on Restitution

4.6. Concluding Remarks on Restitution

In this section, we have shown three definitions of the coefficient of restitution. All three defini-

tions give the same results for in-line collisions. The energetic definition reflects reality the best

as it prevents the violation of the conservation of energy. In eccentric impacts with friction (i.e.,

cars stick together or slide in the impact zone) the coefficient of restitution only applies to the

direction normal (perpendicular) to the contact plane. Analyzing a collision with a significant

impulse in the tangent (sliding) direction requires the introduction of additional coefficients

and is beyond the scope of this paper.1

The empirical problem of high data scatter plagues the use of coefficient of restitution as well

as stiffness values. One argument for using the coefficient of restitution is that it has a defined

lower bound (zero) that has physical meaning. Furthermore, extensive studies in the literature

can be found on restitution, for example, see Ref. [14]. Techniques have also been published on

combining crush energy and planar impact mechanics (e.g. [16, 17]).

Finally, the evaluation of the coefficient of restitution provides a check on a completed recon-

struction.

5. Analysis of Underride Collisions

In this section we will examine the two under-ride collisions to see if we can apply a damage

energy technique. We will develop the damage profile and see if we can use it to determine

the impact speeds for the collisions. The damage-momentum solution was determined using

WinCRASH and the detailed results are contained in the Appendix.

5.1. 1989 Plymouth Voyager Van into the rear of the tractor-t railer

At incidence, the van had a speed of 39 mph while the tractor-trailer was sitting still with the

spring brakes engaged. Post collision measurements show a damage profile at two different

levels: the bumper and the roof. Both levels were measured (in inches) with the following

equally spaced crush measurements across a width of L = 57.7 inches:

Location C1 C2 C3 C4

Bumper 0 10 15 20

Roof 2 8 11 30

The A stiffness coefficient was determined to be 284.2 lb/in and the B stiffness coefficient is

72 lb/in2. If we use only the bumper crush measurements as our damage profile we get the

1A detailed discussion of collisions with friction can be found in Refs. [9, 10, 11].



following WinCRASH results:

Ecrush = 47, 500 ft-lb

∆v = 19.6 mph

which is far below the actual impact speed of 39 mph. This underestimation is expected because

we are ignoring all the damage above the bumper. Tumbas and Smith [18] propose that we not

only measure the crush at the bumper but also at the level where maximum crush occurs. They

propose that we use the deepest measurement at each measurement station unless there is a

difference of 5 inches or more between the two measurements. In this case we average the two

measurements and use the average in calculations. When we conduct the analysis using this

measurement protocol we get the following WinCRASH results:


∆v = 21.2 mph

which still under estimates the speed.

Why are the ∆v estimates from CRASH III erroneous in the case of an underride? The fol-

lowing list gives some explanations:

1. Frictional forces from the tractor-trailer are impulsive in the case of an underride collision.

The first reason for this is that the tractor-trailer unit is significantly heavier than the bullet

vehicle and duration of the crash is longer. Since impulse is F∆t both longer collision

times and larger frictional forces push the formerly neglected impulse into relevance.

2. The collision impulse acts to rotate the rear of the bullet vehicle down and to the outside.

The 2-D limitation of our analysis completely ignores this downward impulse. Further-

more, the bullet vehicle is constrained by the earth which imparts an impulsive constrain-

ing force back on the vehicle. Again, we do not have a way of quantifying the impulsive

ground forces during the collision.

3. The stiffness values are empirically derived from NHTSA crash tests. These crash test

then indicate stiffness for the particular configuration (e.g., frontal barrier), not the stiff-

ness of the softer upper portion of the vehicle.

4. The measurement protocol is not well understood and differences in measuring render

different results.

5. The energy absorbed by the ICC bumper of the truck was completely ignored.

The coefficient of restitution can be calculated based on Eq. (78) given that m1 = 2900/32.2 =

90.06 slugs and m2 = 27, 850/32.2 = 864.9 slugs. The erroneous approach speed was given in


5.2. 1994 Jeep Cherokee into the rear of the tractor-trailer

WinCRASH as 25.6 mph for the bullet van. The coefficient of restitution is:

ǫ =

√


m1m2(v1 − v2)2

=

√

1 −2(55, 308)(90.06 + 864.9)

90.06(864.9)[25.6(1.466)− 0]2

= 0.19

This value for restitution is not unreasonable for a vehicle to vehicle crash with post collision

separation. However, since these vehicle remained in contact, a coefficient of restitution closer

to zero would indicate closer agreement with he physical evidence. Computing a “reason-

able” coefficient of restitution does not guarantee the validity of the solution using damage

momentum because the closing speeds were based on the damage energy to begin with. If the

estimated damage energy is incorrect, then the speed estimates will also be incorrect– even if

the ratio of crush energy and incoming kinetic energy are consistent. The real power of check-

ing a solution is using the determination of the coefficient of restitution to assess the accuracy

of the crush energy based on an independent solution for the incoming speeds. If we use the

known actual impact speed of 39 mph, then the coefficient of restitution is determined as:

ǫ =

√


m1m2(v1 − v2)2

=

√

1 −2(55, 308)(90.06 + 864.9)

90.06(864.9)[39(1.466)− 0]2

= 0.768

which is obviously way to high. Therefore, the estimated crush energy was too low.

5.2. 1994 Jeep Cherokee into the rear of the tractor-trailer

Post collision measurements show a damage profile at three different levels. Those levels are

the bumper, the hood, and the roof level. All three levels were measured (in inches) with the

following equally spaced crush measurements across a width L = 50.5 inches:

Location C1 C2 C3 C4

Bumper 6 7 7 0

Hood 56 48 45 34

Roof 8 7 12 1



The A stiffness coefficient was determined to be 358.8 lb/in and the B stiffness coefficient

is 114.2 lb/in2. If we use the bumper height measurements we get the following WinCRASH

results:


∆v = 12 mph

which is far below the actual impact speed of 37 mph. If we follow the Tumbas and Smith

protocol and average the deepest crush with the bumper we get the following WinCRASH

results:


∆v = 39 mph

The WinCRASH analysis gives a closing speed of 49 mph which overestimates the speed of the

Jeep by about 10 mph or 27%. Due to the shape of the Jeep, the hood was engaged with the

rails of the truck from the incidence until separation and represented the most amount of crush

damage. Since the hood is soft compared to the frame, the energy energy estimated based on

the hood will be high.

5.3. Under-ride Analysis Conclusions

Let’s examine some of the results. The ∆v values listed are a function of post-impact velocity,

effective (dynamic) mass ratio, and the damage energy.

If we look at the results together we see that we have an average error of about 30%. One

vehicle result is 33% below the actual speed and one vehicle result is 27% above the actual

speed. One reason for this is the difference in the height and length of the front of the vehicles.

This affected how much above the bumper interaction occurred before the bumper struck a

solid part of the trailer. Another problem that we face is that although we are able to measure

the crush and mathematically arrive at an answer for the under-ride collisions we are actually

measuring the crush horizontally and using stiffness coefficients that were determined with

crashes that have horizontal impulse vectors (forces parallel to the roadway) and trying to

apply them to crashes where the impulse vectors are not parallel to the roadway. The impulse

vectors are actually trying to drive the colliding vehicles into the roadway. This non-horizontal

force is evident in the crash video and the post collision photos where we can actually see the

vehicles have been crushed downward as well as to the rear. Hence, both of these collisions

are constrained by the ground and we have not quantified any ground forces that may be

acting on the bullet vehicle. The angle of the impulse vector in these under-ride collisions is

affected by many factors. Some of these factors are vehicle shapes and sizes as well as trailer

characteristics. This makes any attempt to use a damage momentum solution for this type of


underride collision unreliable when using current techniques.

6. Summary and Conclusions

We began by discussing the Crash III impact and energy models, defining such terms as crush

energy, effective mass ratio, stiffness coefficients and the like. We have seen the purpose of the

Crash III model is to ultimately calculate the magnitude of the ∆v vector. We have seen how to

calculate the Campbell model coefficients, A, B, & G, which are based upon the deformation of

an ideal, inelastic (irreversible) spring. We have seen that damage energy may be determined

by the planar damage area, the depth of the centroid of the damage area, and the A, B, & G

coefficients. We have seen how to locate the centroid and damage area mathematically.

In this paper, we have discussed classical impact models based upon Newton’s Laws of Mo-

tion. We have examined the idea of coefficient of restitution in three different ways. The first

definition of coefficient of restitution is the kinematic model, based upon the ratio of the relative

velocities of approach divided by the relative velocities of rescission. This is Newton’s formula-

tion. Next, we looked at the kinetic model of coefficient of restitution, as postulated by Poisson

in the 1800’s. This kinetic model essentially defines the coefficient of restitution as the ratio of

the magnitude of the rebound impulse divided by the magnitude of the deformation impulse.

Lastly, we examined an energetic definition of impulse as formulated by Stronge. Stronge’s

Hypothesis says, in essence, that the coefficient of restitution squared is the ratio of the work

done in rebound divided by the work done to deform the colliding objects. Since work is a

force acting through a distance, the objects must be deformable to some degree. Most all real

world objects are deformable, if only a small amount, so the method has utility for many im-

pact configurations. We have also seen how the coefficient of restitution is a function not only

of material properties, as postulated by Newton, but is also a function of the relative impact

velocity.

We have also seen, for oblique impacts, that the coefficient of restitution is defined normal

to the impact plane. In oblique impacts, there may also be a tangent frictional force present,

which may need additional terms outside the coefficient of restitution. We have not quantified

any tangential (sliding) forces, but have referred the reader to several references where this

topic is discussed.

We have seen how to determine the maximum amount of kinetic energy that may be lost in

a collision. We have presented a coefficient of restitution model that may be used to determine

the coefficient of restitution using the damage energy calculated from the Crash III method.

We have seen how this coefficient of restitution may be used as a sanity check on any damage-

momentum solution. We have presented test data showing reasonable values for coefficient of

restitution for real world crashes.


References

Finally, we have re-examined the crash tests from SP 2005. We see the Crash III model gives

good results for the impacts that occurred at right angles into the trailer tandems. We have

further shown that treating the tractor-trailer as a movable barrier and using the measurement

protocol from Tumbas and Smith will result in inconsistent results and should not be used for

speed reconstruction. Because the trailer was treated as a movable barrier, there was no damage

energy associated with it. In reality, there was work done to deform some of the trailer struc-

ture and this work was not accounted for. In addition, the bullet vehicles in the two underride

crashes were also constrained by the ground, and these increased ground forces were not quan-

tified. As of this writing, there are no tested methods for dealing with either the ground force

from this constraint nor the work required to deform the trailer structure.

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References

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[15] A. K. Prasad, “Coefficient of restitution of vehicle structures and its use in estimating the

total δv in automobile collisions,” ASME AMD: Crashworthiness and Occupant Protection in

Transportation Systems, vol. 126, pp. 217–246, 1991.

[16] J. F. Kerkoff, M. S. Varat, S. E. Husher, A. M. Busenga, and K. Hamilton, “An investigation

into vehicle frontal impact stiffness, BEV and repeated testing for reconstruction,” SAE

Technical Paper Series, no. 930899, 1993.

[17] R. M. Brach, D. F. Rudny, and D. W. Sallmann, “Comparison of tire friction test methodolo-

gies used in accident reconstruction,” in Accident Reconstruction: Technology and Animation

VIII SP-1319, no. 980367, pp. 239–248, Society of Automotive Engineers, Warrendale, PA,

February 1998.

[18] N. S. Tumbas and R. A. Smith, “Measuring protocol for quantifying vehicle damage from

and energy basis point of view,” SAE Technical Paper Series, no. 880072, 1988.


A. Analysis of the Nissan Crash

The results for the ∆v of the Nissan and the total damage energy are valid. The closing speed

results are not valid. The program is not flexible enough to impose constraints on the motion

so the rotation around the kingpin is not modeled properly. Instead, we used a movable barrier

with a weight of of γ2w2 = 0.433(13, 425) = 5812 lbs. This rendered accurate results only for

the ∆v of the Nissan.

File Name : Project Name : Desc :

NISSAN.SLM Nissan Crash IPTM 2005 Crash Test

Date : 4/11/2006 File Number :

Licensed to : Russell H Strickland Organization : Fairfield Police Serial Number : SW200-183-144-114499

WinCrash Project Report

Vehicle 1 Vehicle 2

NISSAN NEWCAR2

Vehicle Description 1992 Nissan Sentra 2dr

Weight and Car Data Page: 2





VEHICLE 1 NISSANDefault Type Sub CompactCollision Surface FrontWeight 2408 lbRadius of Gyration squared 2951 in2

Vehicle DimensionsLoa : 170.0 in

Roh : 38.0 in WB : 96.0 in Foh : 36.0 in

Wtb : 891 lb Wta : 1517 lb

b : 60.5 in a : 35.5 in

Yw : 67.2 in

Xr : 98.5 in Xf : 71.5 in

VEHICLE 2 NEWCAR2Default Type Moveable BarrierCollision Surface FrontWeight 5813 lbRadius of Gyration squared 4024 in2


Roh : 30.0 in WB : 120.0 in Foh : 30.0 in

Wtb : 2616 lb Wta : 3197 lb

b : 66.0 in a : 54.0 in

Yw : 100.0 in

Xr : 96.0 in Xf : 84.0 in

Damage Data Page: 3





DAMAGE DATA NISSAN NEWCAR2

Profile Standard Standard

Damage Width, L 66.0 in 0.0 inProfile Offset, D 0.0 in 0.0 inDamage Offset, LR -33.0 in 0.0 in

Pdof 0.0 deg 0.0 deg

Force Location Crush Centroid User DefinedXp 59.5 in 84.0 inYp 0.6 in 0.0 in

Stiffness, A 188.0 lb/in 9999999 lb/inStiffness, B 41.0 lb/in2 9999999 lb/in2

Number of Coefficients 6 0

Damage Dimensions C L C L

1 22.0 in 0.0 in2 21.5 in 13.2 in3 24.6 in 26.4 in4 26.6 in 39.6 in5 24.5 in 52.8 in6 22.0 in 66.0 in

Damage Results Page: 4





DAMAGE RESULTS NISSAN NEWCAR2

Delta V 28.3 mph +/- 1.1 mph 11.7 mph +/- 0.4 mph EBS 33.7 mph +/- 1.3 mph 0.0 mph +/- 0.0 mph Force 76925 lb +/- 6570 lb 0 lb +/- 0 lbDelta Omega 6 deg/sec +/- 101 deg/sec 0 deg/sec +/- 43 deg/secMoment Arm 0.6 in +/- 10.4 in 0.0 in +/- 14.7 inMagnification Factor 1.00 +/- 0.00 0.00 +/- 0.00Mass Ratio 1.00 +/- 0.00 1.00 +/- 0.00Damage Energy 91423 ft-lb +/- 6864 ft-lb 0 ft-lb +/- 0 ft-lb

COMMON RESULTS

Impulse, IMP 3110.7 lb-sec +/- 116.9 lb-secCollision Time, Dt 0.081 sec +/- 0.008 secTotal Energy, Et 91423 ft-lb +/- 6864 ft-lbClosing Speed, Vd 40.1 mph +/- 1.2 mph

Project Warnings Page: 5





Current Warning Flag Settings

Vehicle Warnings NISSAN NEWCAR2

No vehicle warnings were noted

Project Warnings

No Project warnings were noted

Confidence levelsA Stiffness Value 10.0 %

B Stiffness Value 10.0 %

Pdof Error 10.0 deg

Distance Error 10.0 %

Lockup Sensitivity 0.1

Separation Heading Error 5.0 deg

Approach Heading Error 5.0 deg

B. Analysis of the Plymouth Van Crash


VAN.SLM voyager1




Vehicle 1 Vehicle 2

VOYAGER NEWCAR2

Vehicle Description 1989 plyouth voyager



VAN.SLM voyager1



VEHICLE 1 VOYAGERDefault Type VanCollision Surface FrontWeight 3046 lbRadius of Gyration squared 3187 in2


Roh : 40.8 in WB : 112.2 in Foh : 28.8 in

Wtb : 1249 lb Wta : 1797 lb

b : 66.2 in a : 46.0 in

Yw : 72.0 in

Xr : 107.0 in Xf : 74.8 in



Roh : 30.0 in WB : 120.0 in Foh : 30.0 in

Wtb : 12533 lb Wta : 15318 lb

b : 66.0 in a : 54.0 in

Yw : 100.0 in

Xr : 96.0 in Xf : 84.0 in

Damage Data Page: 3


VAN.SLM voyager1



DAMAGE DATA VOYAGER NEWCAR2








1 0.0 in 0.0 in2 10.0 in 19.2 in3 15.0 in 38.5 in4 20.0 in 57.7 in



VAN.SLM voyager1



DAMAGE RESULTS VOYAGER NEWCAR2

Delta V 19.6 mph 2.1 mph EBS 20.6 mph 0.0 mph Force 64875 lb 0 lbDelta Omega 111 deg/sec 0 deg/secMoment Arm 17.8 in 0.0 inMagnification Factor 1.00 0.00Mass Ratio 0.91 1.00Damage Energy 47500 ft-lb 0 ft-lb

COMMON RESULTS

Impulse, IMP 2728.1 lb-secCollision Time, Dt 0.084 secTotal Energy, Et 47500 ft-lbClosing Speed, Vd 23.7 mph



VAN.SLM voyager1




Vehicle Warnings VOYAGER NEWCAR2


Project Warnings




Pdof Error 10.0 deg






VAN2.SLM voyager1




Vehicle 1 Vehicle 2

VOYAGER NEWCAR2

Vehicle Description 1989 plyouth voyager



VAN2.SLM voyager1



VEHICLE 1 VOYAGERDefault Type VanCollision Surface FrontWeight 3046 lbRadius of Gyration squared 3187 in2


Roh : 40.8 in WB : 112.2 in Foh : 28.8 in

Wtb : 1249 lb Wta : 1797 lb

b : 66.2 in a : 46.0 in

Yw : 72.0 in

Xr : 107.0 in Xf : 74.8 in



Roh : 30.0 in WB : 120.0 in Foh : 30.0 in

Wtb : 12533 lb Wta : 15318 lb

b : 66.0 in a : 54.0 in

Yw : 100.0 in

Xr : 96.0 in Xf : 84.0 in

Damage Data Page: 3


VAN2.SLM voyager1



DAMAGE DATA VOYAGER NEWCAR2











VAN2.SLM voyager1



DAMAGE RESULTS VOYAGER NEWCAR2

Delta V 21.2 mph 2.3 mph EBS 22.2 mph 0.0 mph Force 69710 lb 0 lbDelta Omega 120 deg/sec 0 deg/secMoment Arm 17.9 in 0.0 inMagnification Factor 1.00 0.00Mass Ratio 0.91 1.00Damage Energy 55308 ft-lb 0 ft-lb

COMMON RESULTS




VAN2.SLM voyager1




Vehicle Warnings VOYAGER NEWCAR2


Project Warnings




Pdof Error 10.0 deg





C. Analysis of the Jeep Crash


JEEP2.SLM jeep bumper jeep




Vehicle 1 Vehicle 2

JEEP NEWCAR2

Vehicle Description 1994 jeep cherokee






VEHICLE 1 JEEPDefault Type IntermediateCollision Surface FrontWeight 3350 lbRadius of Gyration squared 2626 in2


Roh : 36.2 in WB : 101.6 in Foh : 27.6 in

Wtb : 1397 lb Wta : 1953 lb

b : 59.2 in a : 42.4 in

Yw : 70.5 in

Xr : 95.4 in Xf : 70.0 in



Roh : 30.0 in WB : 120.0 in Foh : 30.0 in

Wtb : 12533 lb Wta : 15318 lb

b : 66.0 in a : 54.0 in

Yw : 100.0 in

Xr : 96.0 in Xf : 84.0 in

Damage Data Page: 3





DAMAGE DATA JEEP NEWCAR2


Damage Width, L 50.5 in 0.0 inProfile Offset, D -12.8 in 0.0 inDamage Offset, LR -38.0 in 0.0 in


Force Location Crush Centroid User DefinedXp 56.8 in 84.0 inYp -14.7 in 0.0 in










DAMAGE RESULTS JEEP NEWCAR2

Delta V 39.0 mph 4.7 mph EBS 41.1 mph 0.0 mph Force 167117 lb 0 lbDelta Omega -221 deg/sec 0 deg/secMoment Arm -14.7 in 0.0 inMagnification Factor 1.00 0.00Mass Ratio 0.92 1.00Damage Energy 204920 ft-lb 0 ft-lb

COMMON RESULTS








Vehicle Warnings JEEP NEWCAR2


Project Warnings




Pdof Error 10.0 deg










Vehicle 1 Vehicle 2

JEEP NEWCAR2

Vehicle Description 1994 jeep cherokee






VEHICLE 1 JEEPDefault Type IntermediateCollision Surface FrontWeight 3350 lbRadius of Gyration squared 2626 in2


Roh : 36.2 in WB : 101.6 in Foh : 27.6 in

Wtb : 1397 lb Wta : 1953 lb

b : 59.2 in a : 42.4 in

Yw : 70.5 in

Xr : 95.4 in Xf : 70.0 in



Roh : 30.0 in WB : 120.0 in Foh : 30.0 in

Wtb : 12533 lb Wta : 15318 lb

b : 66.0 in a : 54.0 in

Yw : 100.0 in

Xr : 96.0 in Xf : 84.0 in

Damage Data Page: 3





DAMAGE DATA JEEP NEWCAR2


Damage Width, L 50.5 in 0.0 inProfile Offset, D -12.8 in 0.0 inDamage Offset, LR -38.0 in 0.0 in


Force Location Crush Centroid User DefinedXp 66.8 in 84.0 inYp -16.3 in 0.0 in










DAMAGE RESULTS JEEP NEWCAR2

Delta V 12.0 mph 1.4 mph EBS 12.6 mph 0.0 mph Force 50820 lb 0 lbDelta Omega -75 deg/sec 0 deg/secMoment Arm -16.3 in 0.0 inMagnification Factor 1.00 0.00Mass Ratio 0.91 1.00Damage Energy 19569 ft-lb 0 ft-lb

COMMON RESULTS








Vehicle Warnings JEEP NEWCAR2


Project Warnings




Pdof Error 10.0 deg





Documents

Crush and Restitution