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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,jeremy@jhscientific.com
†Fairfield City Police Department, 5320 Pleasant Ave., Fairfield, OH 45014, (513) 325-8703, russell-gina@fuse.net‡Jackson Hole Scientific Investigations, Inc., P.O. Box 2206, Jackson, WY 83001, (307) 733-4559,
john@jhscientific.com
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
4 c©2006 Jeremy Daily, Russell Strickland, and John Daily
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
c©2006 Jeremy Daily, Russell Strickland, and John Daily 5
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.
6 c©2006 Jeremy Daily, Russell Strickland, and John Daily
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.
c©2006 Jeremy Daily, Russell Strickland, and John Daily 7
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. Damage Momentum Analysis
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.
8 c©2006 Jeremy Daily, Russell Strickland, and John Daily
2.2. A Planar (Two Dimensional) Impact Model
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)
c©2006 Jeremy Daily, Russell Strickland, and John Daily 9
2. Damage Momentum Analysis
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.
10 c©2006 Jeremy Daily, Russell Strickland, and John Daily
2.2. A Planar (Two Dimensional) Impact Model
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.
c©2006 Jeremy Daily, Russell Strickland, and John Daily 11
2. Damage Momentum Analysis
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)
12 c©2006 Jeremy Daily, Russell Strickland, and John Daily
2.3. Relationship to ∆v
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.
c©2006 Jeremy Daily, Russell Strickland, and John Daily 13
2. Damage Momentum Analysis
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:
14 c©2006 Jeremy Daily, Russell Strickland, and John Daily
2.3. Relationship to ∆v
∆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.
c©2006 Jeremy Daily, Russell Strickland, and John Daily 15
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.
3. Determining Crush Energy
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.
16 c©2006 Jeremy Daily, Russell Strickland, and John Daily
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)
c©2006 Jeremy Daily, Russell Strickland, and John Daily 17
3. Determining Crush Energy
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
18 c©2006 Jeremy Daily, Russell Strickland, and John Daily
3.2. CRASH III Deformation Model
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
c©2006 Jeremy Daily, Russell Strickland, and John Daily 19
3. Determining Crush Energy
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
20 c©2006 Jeremy Daily, Russell Strickland, and John Daily
3.2. CRASH III Deformation Model
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
c©2006 Jeremy Daily, Russell Strickland, and John Daily 21
3. Determining Crush Energy
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
22 c©2006 Jeremy Daily, Russell Strickland, and John Daily
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
c©2006 Jeremy Daily, Russell Strickland, and John Daily 23
3. Determining Crush Energy
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.
24 c©2006 Jeremy Daily, Russell Strickland, and John Daily
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
c©2006 Jeremy Daily, Russell Strickland, and John Daily 25
3. Determining Crush Energy
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.
26 c©2006 Jeremy Daily, Russell Strickland, and John Daily
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.
c©2006 Jeremy Daily, Russell Strickland, and John Daily 27
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
28 c©2006 Jeremy Daily, Russell Strickland, and John Daily
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.
c©2006 Jeremy Daily, Russell Strickland, and John Daily 29
4. Relationship between Crush Energy and the Coefficient of Restitution
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
30 c©2006 Jeremy Daily, Russell Strickland, and John Daily
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.
c©2006 Jeremy Daily, Russell Strickland, and John Daily 31
4. Relationship between Crush Energy and the Coefficient of Restitution
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.
32 c©2006 Jeremy Daily, Russell Strickland, and John Daily
4.3. Understanding the Coefficient of Restitution
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
c©2006 Jeremy Daily, Russell Strickland, and John Daily 33
4. Relationship between Crush Energy and the Coefficient of Restitution
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
34 c©2006 Jeremy Daily, Russell Strickland, and John Daily
4.3. Understanding the Coefficient of Restitution
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)
c©2006 Jeremy Daily, Russell Strickland, and John Daily 35
4. Relationship between Crush Energy and the Coefficient of Restitution
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)
36 c©2006 Jeremy Daily, Russell Strickland, and John Daily
4.3. Understanding the Coefficient of Restitution
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.
c©2006 Jeremy Daily, Russell Strickland, and John Daily 37
4. Relationship between Crush Energy and the Coefficient of Restitution
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.
38 c©2006 Jeremy Daily, Russell Strickland, and John Daily
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
c©2006 Jeremy Daily, Russell Strickland, and John Daily 39
4. Relationship between Crush Energy and the 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)
40 c©2006 Jeremy Daily, Russell Strickland, and John Daily
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
c©2006 Jeremy Daily, Russell Strickland, and John Daily 41
4. Relationship between Crush Energy and the Coefficient of Restitution
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].
42 c©2006 Jeremy Daily, Russell Strickland, and John Daily
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].
c©2006 Jeremy Daily, Russell Strickland, and John Daily 43
5. Analysis of Underride Collisions
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:
Ecrush = 55, 308 ft-lb
∆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
44 c©2006 Jeremy Daily, Russell Strickland, and John Daily
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 =
√
1 −2Ecrush (m1 + m2)
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 =
√
1 −2Ecrush (m1 + m2)
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
c©2006 Jeremy Daily, Russell Strickland, and John Daily 45
5. Analysis of Underride Collisions
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:
Ecrush = 19, 569 ft-lb
∆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:
Ecrush = 204, 920 ft-lb
∆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
46 c©2006 Jeremy Daily, Russell Strickland, and John Daily
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.
c©2006 Jeremy Daily, Russell Strickland, and John Daily 47
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.
References
[1] P. V. Hight, D. B. Kent-Koop, and R. A. Hight, “Barrier equivalent velocity, delta v and
CRASH3 stiffness in automobile collisions,” SAE Technical Paper Series, no. 850437, 1985.
[2] J. G. Daily, N. Shigemura, and J. S. Daily, Fundamentals of Traffic Crash Reconstruction. Jack-
sonville, Florida: Institute of Police Technololgy and Managment, University of North
Florida, 2006.
[3] B. F. Schmidt, W. R. Haight, T. J. Szabo, and J. B. Welcher, “System-based energy and mo-
mentum analysis of collisions,” in Accident Reconstruction: Technology and Animation VIII
SP-1319, no. 980026, Society of Automotive Engineers, Warrendale, PA, February 1998.
[4] “CRASH3 Technical Manual.” National Highway Traffic Safety Administration, Washing-
ton D.C. Accessed on 2 Aug 2005 at
http://www-nass.nhtsa.dot.gov/NASS/MANUALS/Crash3Man.pdf.
[5] J. A. Neptune, G. Y. Blair, and J. E. Flynn, “A method for quantifying vehicle crush stiffness
coefficients,” SAE Technical Paper Series, no. 920607, 1992.
[6] K. L. Campbell, “Energy basis for collision severity,” SAE Technical Paper Series, no. 740565,
1974.
[7] F. P. Beer and E. R. J. Jr., Vector Mechanics for Engineers: Dynamics. New York: McGraw-Hill,
5th ed., 1988.
[8] R. C. Hibbeler, Engineering Mechanics: Dynamics. Upper Saddle River, NJ: Prentice Hall,
10th ed., 2004.
[9] W. J. Stronge, Impact Mechanics. Cambridge: Cambridge University Press, 2000.
48 c©2006 Jeremy Daily, Russell Strickland, and John Daily
References
[10] R. M. Brach, “Rigid body collisions,” Transactions of the ASME: Journal of Applied Mechanics,
vol. 53, pp. 133–138, 1989.
[11] J. B. Keller, “Impact with friction,” Transactions of the ASME: Journal of Applied Mechanics,
vol. 53, pp. 1–4, 1986.
[12] J. Daily, Fundamentals of Traffic Accident Reconstruction. Jacksonville, Florida: Institute of
Police Technololgy and Managment, University of North Florida, 1988.
[13] J. Coaplen, W. J. Stronge, and B. Ravani, “Work equivalent composite coefficient of resti-
tution,” International Journal of Impact Engineering, vol. 30, pp. 581–591, 2004.
[14] K. L. Monson and G. J. Germane, “Determination and mechanisms of motor vehicle struc-
tural restitution from crash test data,” in Accident Reconstruction: Technology and Animation
IX SP-1407, no. 1999-01-0097, Society of Automotive Engineers, Warrendale, PA, March
1999.
[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.
c©2006 Jeremy Daily, Russell Strickland, and John Daily 49
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
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
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
Vehicle DimensionsLoa : 180.0 in
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
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
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
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
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
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
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
File Name : Project Name : Desc :
VAN.SLM voyager1
Date : 3/28/2006 File Number :
Licensed to : Russell H Strickland Organization : Fairfield Police Serial Number : SW200-183-144-114499
WinCrash Project Report
Vehicle 1 Vehicle 2
VOYAGER NEWCAR2
Vehicle Description 1989 plyouth voyager
Weight and Car Data Page: 2
File Name : Project Name : Desc :
VAN.SLM voyager1
Date : 3/28/2006 File Number :
Licensed to : Russell H Strickland Organization : Fairfield Police Serial Number : SW200-183-144-114499
VEHICLE 1 VOYAGERDefault Type VanCollision Surface FrontWeight 3046 lbRadius of Gyration squared 3187 in2
Vehicle DimensionsLoa : 181.8 in
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
VEHICLE 2 NEWCAR2Default Type Moveable BarrierCollision Surface FrontWeight 27850 lbRadius of Gyration squared 4024 in2
Vehicle DimensionsLoa : 180.0 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
File Name : Project Name : Desc :
VAN.SLM voyager1
Date : 3/28/2006 File Number :
Licensed to : Russell H Strickland Organization : Fairfield Police Serial Number : SW200-183-144-114499
DAMAGE DATA VOYAGER NEWCAR2
Profile Standard Standard
Damage Width, L 57.7 in 0.0 inProfile Offset, D 10.0 in 0.0 inDamage Offset, LR -18.9 in 0.0 in
Pdof 0.0 deg 0.0 deg
Force Location Crush Centroid User DefinedXp 67.6 in 84.0 inYp 17.8 in 0.0 in
Stiffness, A 284.2 lb/in 9999999 lb/inStiffness, B 72.0 lb/in2 9999999 lb/in2
Number of Coefficients 4 0
Damage Dimensions C L C L
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
Damage Results Page: 4
File Name : Project Name : Desc :
VAN.SLM voyager1
Date : 3/28/2006 File Number :
Licensed to : Russell H Strickland Organization : Fairfield Police Serial Number : SW200-183-144-114499
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
Project Warnings Page: 5
File Name : Project Name : Desc :
VAN.SLM voyager1
Date : 3/28/2006 File Number :
Licensed to : Russell H Strickland Organization : Fairfield Police Serial Number : SW200-183-144-114499
Current Warning Flag Settings
Vehicle Warnings VOYAGER 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
File Name : Project Name : Desc :
VAN2.SLM voyager1
Date : 3/28/2006 File Number :
Licensed to : Russell H Strickland Organization : Fairfield Police Serial Number : SW200-183-144-114499
WinCrash Project Report
Vehicle 1 Vehicle 2
VOYAGER NEWCAR2
Vehicle Description 1989 plyouth voyager
Weight and Car Data Page: 2
File Name : Project Name : Desc :
VAN2.SLM voyager1
Date : 3/28/2006 File Number :
Licensed to : Russell H Strickland Organization : Fairfield Police Serial Number : SW200-183-144-114499
VEHICLE 1 VOYAGERDefault Type VanCollision Surface FrontWeight 3046 lbRadius of Gyration squared 3187 in2
Vehicle DimensionsLoa : 181.8 in
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
VEHICLE 2 NEWCAR2Default Type Moveable BarrierCollision Surface FrontWeight 27850 lbRadius of Gyration squared 4024 in2
Vehicle DimensionsLoa : 180.0 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
File Name : Project Name : Desc :
VAN2.SLM voyager1
Date : 3/28/2006 File Number :
Licensed to : Russell H Strickland Organization : Fairfield Police Serial Number : SW200-183-144-114499
DAMAGE DATA VOYAGER NEWCAR2
Profile Standard Standard
Damage Width, L 57.7 in 0.0 inProfile Offset, D 10.0 in 0.0 inDamage Offset, LR -18.9 in 0.0 in
Pdof 0.0 deg 0.0 deg
Force Location Crush Centroid User DefinedXp 66.9 in 84.0 inYp 17.9 in 0.0 in
Stiffness, A 284.2 lb/in 9999999 lb/inStiffness, B 72.0 lb/in2 9999999 lb/in2
Number of Coefficients 4 0
Damage Dimensions C L C L
1 2.0 in 0.0 in2 10.0 in 19.2 in3 15.0 in 38.5 in4 25.0 in 57.7 in
Damage Results Page: 4
File Name : Project Name : Desc :
VAN2.SLM voyager1
Date : 3/28/2006 File Number :
Licensed to : Russell H Strickland Organization : Fairfield Police Serial Number : SW200-183-144-114499
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
Impulse, IMP 2941.5 lb-secCollision Time, Dt 0.084 secTotal Energy, Et 55308 ft-lbClosing Speed, Vd 25.6 mph
Project Warnings Page: 5
File Name : Project Name : Desc :
VAN2.SLM voyager1
Date : 3/28/2006 File Number :
Licensed to : Russell H Strickland Organization : Fairfield Police Serial Number : SW200-183-144-114499
Current Warning Flag Settings
Vehicle Warnings VOYAGER 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
C. Analysis of the Jeep Crash
File Name : Project Name : Desc :
JEEP2.SLM jeep bumper jeep
Date : 3/28/2006 File Number :
Licensed to : Russell H Strickland Organization : Fairfield Police Serial Number : SW200-183-144-114499
WinCrash Project Report
Vehicle 1 Vehicle 2
JEEP NEWCAR2
Vehicle Description 1994 jeep cherokee
Weight and Car Data Page: 2
File Name : Project Name : Desc :
JEEP2.SLM jeep bumper jeep
Date : 3/28/2006 File Number :
Licensed to : Russell H Strickland Organization : Fairfield Police Serial Number : SW200-183-144-114499
VEHICLE 1 JEEPDefault Type IntermediateCollision Surface FrontWeight 3350 lbRadius of Gyration squared 2626 in2
Vehicle DimensionsLoa : 165.4 in
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
VEHICLE 2 NEWCAR2Default Type Moveable BarrierCollision Surface FrontWeight 27850 lbRadius of Gyration squared 4024 in2
Vehicle DimensionsLoa : 180.0 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
File Name : Project Name : Desc :
JEEP2.SLM jeep bumper jeep
Date : 3/28/2006 File Number :
Licensed to : Russell H Strickland Organization : Fairfield Police Serial Number : SW200-183-144-114499
DAMAGE DATA JEEP NEWCAR2
Profile Standard Standard
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
Pdof 0.0 deg 0.0 deg
Force Location Crush Centroid User DefinedXp 56.8 in 84.0 inYp -14.7 in 0.0 in
Stiffness, A 358.8 lb/in 9999999 lb/inStiffness, B 114.2 lb/in2 9999999 lb/in2
Number of Coefficients 4 0
Damage Dimensions C L C L
1 31.0 in 0.0 in2 27.5 in 16.8 in3 26.0 in 33.7 in4 17.0 in 50.5 in
Damage Results Page: 4
File Name : Project Name : Desc :
JEEP2.SLM jeep bumper jeep
Date : 3/28/2006 File Number :
Licensed to : Russell H Strickland Organization : Fairfield Police Serial Number : SW200-183-144-114499
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
Impulse, IMP 5956.3 lb-secCollision Time, Dt 0.071 secTotal Energy, Et 204920 ft-lbClosing Speed, Vd 46.9 mph
Project Warnings Page: 5
File Name : Project Name : Desc :
JEEP2.SLM jeep bumper jeep
Date : 3/28/2006 File Number :
Licensed to : Russell H Strickland Organization : Fairfield Police Serial Number : SW200-183-144-114499
Current Warning Flag Settings
Vehicle Warnings JEEP 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
File Name : Project Name : Desc :
JEEP1.SLM jeep bumper jeep
Date : 3/28/2006 File Number :
Licensed to : Russell H Strickland Organization : Fairfield Police Serial Number : SW200-183-144-114499
WinCrash Project Report
Vehicle 1 Vehicle 2
JEEP NEWCAR2
Vehicle Description 1994 jeep cherokee
Weight and Car Data Page: 2
File Name : Project Name : Desc :
JEEP1.SLM jeep bumper jeep
Date : 3/28/2006 File Number :
Licensed to : Russell H Strickland Organization : Fairfield Police Serial Number : SW200-183-144-114499
VEHICLE 1 JEEPDefault Type IntermediateCollision Surface FrontWeight 3350 lbRadius of Gyration squared 2626 in2
Vehicle DimensionsLoa : 165.4 in
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
VEHICLE 2 NEWCAR2Default Type Moveable BarrierCollision Surface FrontWeight 27850 lbRadius of Gyration squared 4024 in2
Vehicle DimensionsLoa : 180.0 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
File Name : Project Name : Desc :
JEEP1.SLM jeep bumper jeep
Date : 3/28/2006 File Number :
Licensed to : Russell H Strickland Organization : Fairfield Police Serial Number : SW200-183-144-114499
DAMAGE DATA JEEP NEWCAR2
Profile Standard Standard
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
Pdof 0.0 deg 0.0 deg
Force Location Crush Centroid User DefinedXp 66.8 in 84.0 inYp -16.3 in 0.0 in
Stiffness, A 358.8 lb/in 9999999 lb/inStiffness, B 114.2 lb/in2 9999999 lb/in2
Number of Coefficients 4 0
Damage Dimensions C L C L
1 6.0 in 0.0 in2 7.0 in 16.8 in3 7.0 in 33.7 in4 0.0 in 50.5 in
Damage Results Page: 4
File Name : Project Name : Desc :
JEEP1.SLM jeep bumper jeep
Date : 3/28/2006 File Number :
Licensed to : Russell H Strickland Organization : Fairfield Police Serial Number : SW200-183-144-114499
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
Impulse, IMP 1827.0 lb-secCollision Time, Dt 0.072 secTotal Energy, Et 19569 ft-lbClosing Speed, Vd 14.6 mph
Project Warnings Page: 5
File Name : Project Name : Desc :
JEEP1.SLM jeep bumper jeep
Date : 3/28/2006 File Number :
Licensed to : Russell H Strickland Organization : Fairfield Police Serial Number : SW200-183-144-114499
Current Warning Flag Settings
Vehicle Warnings JEEP 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
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