**Stiffness Coefficients -
Energy and Damage**

Copyright Ó George M.
Bonnett, JD 2001 All Rights Reserved

Politics
is not the only thing that makes for strange bedfellows, so do collisions. A wonderfully strange relationship exists between
momentum, energy, force, damage and speed as uniquely illustrated in dealing
with barrier collisions. Two of the
critters in the menagerie that exemplify this relationship are Crush Energy
Equivalent Speed (CEES) and Kinetic Energy Equivalent Speed^{1} (KEES),
which are both used in computing stiffness coefficients.

Now
before you think this matter is of little importance, remember that when
dealing with a change of speed from damage, you come face to face with this
relationship and the beasts therein.

In most instances when a testing organization crashes a vehicle, the vehicle is run into a non-deforming, immovable barrier of some type. These barriers run the gamut from a loading dock to the sophisticated devices used by the government. The barriers are usually for dealing with frontal impacts although there are some unusual contraptions designed for rear- and side-impact collisions.

In
an effort to make this as mathematically easy as possible, speed and velocity
(vector) will both be expressed in terms of **units,** which can be in any
system you desire. As a result, energy
will also be expressed in terms of units.
While SI units are the choice for some, others prefer a different unit of
measure. The purpose of this paper is
not to quibble about which is more correct, but to convey an idea. The unit of speed or velocity is the correct
unit for the energy formulae and for conversion to a unit of momentum. It is for the reader figure out what unit
this should be considering the instant application.

When a vehicle is run into a non-deforming, immovable barrier like a loading dock, the speed of the vehicle can be measured and the vehicle can be weighed so that we know how much energy the vehicle possessed at impact. At the higher speeds, there is usually very little bounce (read restitution) and it is often ignored, as it does not have much influence on the measurements. Armed with the weight and speed and a tape measure for determining the measurements of the damage, we can determine the stiffness coefficients for the particular vehicle.

Sometimes the vehicle is not run into the barrier. Sometimes the barrier, or impactor, is run into the vehicle. Knowing the weight and speed of the impactor, the kinetic energy the impactor brings into collision can be determined.

The KEES computations are complicated by the necessity of determining the speed of both the impactor and the struck vehicle post-impact. This post-impact information is necessary in order to obtain the energy of these two bodies post-impact so that it may be subtracted from the energy of the impactor pre-impact in order to determine the energy that went into damage.

What
happens when the post-impact speeds of the objects are not reported? It is out of this dilemma that KEES was
spawned. The KEES, or similar,
computation is necessary for rear and side impacts since the post-impact speed
is not reported.

The
KEES determination has at its core the following assumptions:

- The Impactor and the
Vehicle remain together after impact, and

- The Impactor and the
Vehicle have a common post-impact speed.

KEES is computed as
follows:

KEES
= Sqr (30 * (Energy before Impact – Energy after Impact)/Impactor Weight)

KEES
= Sqr (30*0.5/32.2*(IW*(IS*5280/3600)^{ 2}- ((IW+VW)*(PIS*5280/3600)^{
2}))/IW)

Where:

IW
= Impactor Weight IS = Impact
Speed VW = Vehicle Weight PIS = Post-impact Speed

And

Post-impact
Speed (PIS) = (Impact Speed * Impactor Weight) / (Impactor Weight + Vehicle
Weight)

KEES is that portion of the speed of the Impactor that
actually goes into damaging the vehicle.
This is then the closing speed necessary between the struck vehicle and
a non-deforming, infinite mass barrier to duplicate the damage on the vehicle.

While this is one
approach to the problem and might even be a solution, it is made more difficult
than necessary, as we shall soon see.

Three Collisions

We need to look at three
related but different collisions:

1. Vehicle 1 hits an immovable barrier.

2. Vehicle 2 hits Vehicle 3 in a head on
collision.

3. Vehicle 4 hits a stationary feather.

All of the vehicles are
identical and are all traveling at the same speed of 30 units.

Which
of these vehicles has the most damage? Which
of them has the least damage? How many
of them have identical damage? How does
this relate to KEES?

A collision begins at first contact between the vehicles. The collision ends when the transfer of energy between the vehicles ceases. This may be at maximum engagement if there is no restitution, or if there is restitution, it is at separation of the vehicles. This time frame is the total duration of the collision impulse. Everything after this is post-impact and has nothing to do with damage or with the change of velocity (deltaV) of either vehicle due to the collision impulse. Post-impact information can help us compute the speeds of the vehicles at impact, but has absolutely nothing to do with damage.

Instinctively
we “know” that Vehicle 4 has no damage.
Vehicle 1 hits the barrier and comes to an immediate stop. Vehicles 2 and 3 hit each other at twice
the closure speed between Vehicle 1 and the barrier. They also come to a complete stop at impact as the momentum of
Vehicle 2 cancels the momentum of Vehicle 3 as the momenta are equal but
opposite in direction. In collision 2
there is twice the energy at collision as in collision 1, but there are 2
vehicles that sustain damage. In
actuality, the first three vehicles all absorb identical amounts of damage
resulting in identical damage to each vehicle.

Why
was there no damage in collision 3? The
answer is “**mass**.” Not the mass
of the vehicle, it is identical to the mass of all of the other vehicles. The mass, or lack thereof, of the feather is
also critically important in the solution.
While this may seem too simple to be discussed, it proves a very
important principle. It illustrates,
and convincingly so, the principle of “**reduced mass**.” It is the reduced mass of the objects in
collision that determines the amount of damage sustained in the collision.

Reduced
mass **[M1*M2/(M1+M2)]** allows the mass of each of the objects to be
proportionally represented in the equation.
It is necessary to know the impact speeds of both objects as well as the
post-impact speeds of both objects in order to determine the energy going into
damage if KEES is used. This amount of
detail is unnecessary. If we can determine
the closing velocity of the objects, assuming there is no rotation or separation
velocity, we have all of the information necessary using the reduced mass
principle to determine the energy that results in damage. The ** reduced** mass must be
applied to the closing velocity vector (resulting from subtracting the two
impact velocity vectors) to determine the energy associated with the closing
velocity of the two objects.

In
central collisions, where the principal direction of force (PDOF) passes
through the center of mass of both vehicles, it is the actual mass of the
vehicles that is determinant. In
non-central collisions, it is the **effective** (the gamma value - working
through the lever arm) mass that is the determinant of the actual damage. The remainder of the ** damage energy**
goes into rotation. Effective mass
deals with the PDOF acting on the center of mass through a lever arm requiring
less force than acting directly on the center of mass.

How
does this relate to KEES? Well, at
least we now know that the damage in the collision is a related to the mass of
the objects and the closing speeds. Do we really need the post-impact data?

** **

The additional collisions consist of:

4. Vehicle 5 is traveling at 30 units and Vehicle 6 is stopped. Both of the vehicles in this rear-end collision are identical.

5. Vehicle 7 is traveling at 90 units and Vehicle 8 is traveling at 60 units. Both of the vehicles in this rear-end collision are identical to the vehicles in Collision 4.

How are these collisions similar and how are they different? Both collisions involve identical vehicles with identical closure rates of 30 units. If it were not for the post-impact information, we could not tell these collisions apart. Interesting!

** **

** **

**Proof of Identical Damage**

In Collision 4, using a mass of 1, we have a momentum value of 30 for Vehicle 5 and zero for Vehicle 6. With zero restitution, both vehicles will stick together after collision and each will have a momentum value of 15. Dividing the momentum by the mass, we get a speed of 15 units for both vehicles.

Using the kinetic energy formula (KE = 0.5 * Mass *
Velocity ^{2}) we have KE = 0.5 * 1 * 30^{2} or KE = 450 units
total into the collision. For each
vehicle out of collision the KE = 0.5 * 1 * 15^{2} or 112.5 units for a
total out-of-collision energy of 225 units.
This leaves 225 units of energy to be shared equally (identical
vehicles) as damage.

In Collision 5, using a mass of 1, we have a momentum value of 90 for Vehicle 7 and 60 for Vehicle 8. With zero restitution, both vehicles will stick together after collision and each will have a momentum value of 75. Dividing the momentum by the mass, we get a speed of 75 units for both vehicles.

Using the kinetic energy formulae (KE = 0.5 * Mass *
Velocity^{2}) we have KE = 0.5 * 1 * 90^{2} or KE = 4050 units
for Vehicle 7 into the collision. For
Vehicle 8, KE = 0.5 * 1 * 60^{2} or KE = 1800 units into the
collision. This gives a total energy
into the collision of 5850 units. For
each vehicle out of collision the KE = 0.5 * 1 * 75^{2} or 2812.5 units
for a total out of collision of 5625 units.
This leaves 225 units of energy to be shared equally (identical
vehicles) as damage, which is exactly what we have in Collision 4.

If we disregard the post-impact information, the collisions are identical as far as the damage is concerned. Unless we need to know an exact impact speed, we can disregard the post-impact data, as the collisions are identical as far as the energy for damage is concerned. Damage does not involve anything post-impact. If identical vehicles are involved, it is the closing speed of the vehicles that is determinant in the damage resulting from the collision. Interesting, most Interesting!

** **

**A Second Look**

In Collision 1, Vehicle 1 strikes the barrier at 30 units. However, what if the barrier was traveling at 30 units when it struck Vehicle 1, which was stationary?

As we saw in the first three collisions, it is the **reduced
mass** of the two objects in collision that determines the amount of damage
if the objects are non-deforming. In
collisions 4 and 5 we discovered that it is not the instantaneous speed of the
individual vehicles that determines the amount of damage but the closing
speed.

If both of these principles apply, then it does not
matter if the vehicle strikes the barrier or the barrier strikes the vehicle. It is the closing speed (**V _{c}**)
and the

**The energy absorbed by each object is directly related
to the stiffness coefficients of the objects and the volume of the damage.**

Since the only variables that go into determining the damage in a central (or nearly central) collision without restitution are the closing speeds of the objects and the reduced mass of the objects involved, why do we need the post-impact speed of anything in order to determine the energy that goes into damage? It is as important as the color of the vehicle in helping to determine the damage.

** **

** **

**Frame of Reference**

Are there still skeptics? Let us take two more looks at Collision 1:

1. In this look, we will be in the same reference frame as the barrier. We see the vehicle approaching the barrier at 30 units until collision. After collision, all motion ceases between the barrier and the vehicle.

2. In the next look, we will be in the same reference frame as the car. We now see the barrier approaching the car at 30 units until collision. After collision, all motion ceases between the barrier and the vehicle.

In these ** collisions, **which was the

We must remember that a definition of mass is the resistance of the object to a change in velocity. When the two objects in collision reach a common velocity, the vehicles have reached maximum engagement and there can be no further damage from the collision.

**Maximum Damage Energy**

The
maximum energy available for damage is equal to 0.5 times the **reduced mass**^{
}times the closing velocity squared.
Collisions 7 and 8 have a closing velocity of 30 units regardless of the
reference frame. With this formula, it
is not necessary to compute the total collision energy and subtract the energy
after collision to determine the damage.

Determination
of **reduced mass** involves dividing the product of the masses by the sum
of the masses. This fraction is then
used as the **reduced mass**. In
Collision 2, 4, and 5 the result is 1 * 1 divided by 1 + 1 or 1/2.

The
maximum energy available for damage using this method for Collisions 4 and 5
would be 0.5 times 1/2 times the closing velocity squared. With a closing velocity of 30, the maximum
energy available for damage is E_{Max} = 0.5 * (1/2) * 30^{2}
or 225 units. This is the same energy figure computed previously.

For
Collision 2 the energy of each vehicle coming into collision is KE = .5 * 1 *
30^{2} or 450 units. Since both
vehicles stop at impact, there is no post-impact energy. This means both
vehicles share a total of 900 units of energy that is available for damage.

With
a closing velocity of 30, using the reduced mass principle, the maximum energy
available for damage is E_{Max} = 0.5 * (1/2) * 60^{2} or 900 units.
This is the same energy figure computed using the other method.

What
happens in Collision 1 with the infinite mass, non-deforming barrier? In order to make the problem a little more
manageable, lets assign a mass of 1000 to the barrier since it is difficult to
add and multiply using Infinity as one of the numbers.

If
the frame of reference is the car and the barrier is moving at 30 units, the
total KE into collision is then KE = .5 * 1000 * 30^{2} or 450,000
units of energy. We then have to fuss
with all of the post-impact data and KEES formula to arrive at the Kinetic
Energy Equivalent Speed and then convert back to energy so that figure can be
subtracted from the total energy into the collision to get the energy that goes
into damage.

If
the frame of reference is the barrier and the car is moving at 30 units, the
total KE into collision is then KE = .5 * 1 * 30^{2} or 450 units of
energy. Since in this frame of
reference the car will come to a stop, there is no post-impact information to
deal with and there are a total of 450 units of energy going into damage. With
a closing velocity of 30, using the Reduced Mass formula, the maximum energy
available for damage is E_{Max} = 0.5 * (1000/1001) * 30^{2} =
450,000/1001 or 449.5504 units. This is not the same energy figure computed
using the other method.

Well,
what happened? It looks like there is a
problem with the Reduced Mass formula.
Well, not really. Remember we
assigned a mass of 1000 to the barrier.
The higher the number we assign the closer this number gets to 450. With a mass, less than Infinity, the barrier
and the car have a different **reduced mass**. They will remain together but both reach a common velocity that
is no longer zero and the total energy coming into collision goes into both
damage and translational motion of the car/barrier object. In other words, the barrier moves ever so
slightly and that takes some energy away from the total going into collision
with the remainder (449.5504) going into damage.

** **

** **

**Reduced Mass
Formula**

The closing velocity (V_{C}) is the magnitude
of the resultant of the vector subtraction of the velocity vectors of the
objects. It (V_{C}) is the
third side of the triangle, computed using the Law of Cosines.

V_{C} = |**V _{1}
- V_{2}**|

Momentum equation:

V_{1} * M_{1} + V_{2} * M_{2}
= V_{3} * M_{1} + V_{4} * M_{2}

The maximum possible deformation is reached when both
vehicles achieve a common velocity (**V _{common}**).

V_{3} = V_{4} = V

Substitute into the momentum equation:

V_{1} * M_{1} + V_{2} * M_{2}
= V * (M_{1} + M_{2})

Solve for V'**:**

V = (V_{1} * M_{1} + V_{2} * M_{2})
/ (M_{1} + M_{2})

An energy equation similar to the momentum equation
is:

Ke_{1} + Ke_{2} = E_{D }+ Ke_{3}
+ Ke_{4}

Ke_{1} and Ke_{2} are the kinetic
energies at first contact for the two vehicles, respectively, Ke_{3}
and Ke_{4} are the post-impact energies for the two vehicles and E_{D
}is the total energy due to deformation to both vehicles.

Kinetic energy formula:

Ke = .5 * M * V^{2}

** **

**Therefore:** .5 * M_{1} * V_{1}^{2}
+ .5 * M_{2} * V_{2}^{2} = E_{D }+ .5 * M_{3}
* V_{3}^{2} + .5 * M_{4} * V_{4}^{2}

Multiply by 2:

M_{1} * V_{1}^{2} + M_{2}
* V_{2}^{2} = 2 * E_{D }+ M_{3} * V_{3}^{2}
+ M_{4} * V_{4}^{2}

Substitute V for V_{3} and V_{4}:

M_{1} * V_{1}^{2} + M_{2}
* V_{2}^{2} = 2 * E_{D }+ M_{3} * V^{2}
+ M_{4} * V^{2}

M_{1} * V_{1}^{2} + M_{2}
* V_{2}^{2} = 2 * E_{D }+ V^{2} * (M_{3}
+ M_{4})

Substitute for V:

M_{1} * V_{1}^{2} + M_{2}
* V_{2}^{2} = 2 * E_{D }+ ((V_{1} * M_{1}
+ V_{2} * M_{2}) / (M_{1} + M_{2}))^{2}
* (M_{3} + M_{4})

Cancel:

M_{1} * V_{1}^{2} + M_{2}
* V_{2}^{2} = 2 * E_{D }+ (V_{1} * M_{1}
+ V_{2} * M_{2})^{2} / (M_{1} + M_{2})

Multiply by (M_{1} + M_{2}) and
cancel:

(M_{1} + M_{2}) * (M_{1} * V_{1}^{2}
+ M_{2} * V_{2}^{2}) = 2 * E_{D }+ (V_{1}
* M_{1} + V_{2} * M_{2})^{2}

Multiply:

M_{1}^{2} * V_{1}^{2}
+ M_{1} M_{2} * V_{1}^{2} + M_{1} M_{2}
* V_{2}^{2} + M_{2}^{2} * V_{2}^{2}
=

2 * E_{D }* (M_{1} + M_{2}) +
M_{1}^{2} * V_{1}^{2} + 2 * M_{1} * M_{2}
* V_{1}* V_{2} + M_{2}^{2} * V_{2}^{2}

Simplify by canceling like terms:

M_{1} M_{2} * V_{1}^{2}
+ M_{1} M_{2} * V_{2}^{2} = 2 * E_{D }* (M_{1} + M_{2})
+ 2 * M_{1} * M_{2} * V_{1}* V_{2}

Subtract 2 * M_{1} * M_{2} * V_{1}*
V_{2} from both sides:

M_{1} M_{2} * V_{1}^{2}
- 2 * M_{1} * M_{2} * V_{1}* V_{2} + M_{1}
M_{2} * V_{2}^{2}
= 2 * E_{D }* (M_{1} + M_{2})

Factor the left side of the equation:

M_{1} M_{2} * (V_{1}^{2}
- 2 * V_{1}* V_{2} + V_{2}^{2}) = 2 * E_{D }*
(M_{1} + M_{2})

M_{1} M_{2} * (V_{1} - V_{2})^{
2} = 2 * E_{D }* (M_{1} + M_{2})

Substitute V_{C} for** **(V_{1} - V_{2}):

M_{1} M_{2} * V_{C}^{ 2}
= 2 * E_{D }* (M_{1} + M_{2})

** **

Solve for E_{D:}

** **

E_{D} = .5 * M_{1} M_{2} * V_{C}^{
2} / (M_{1} + M_{2})

** **

This ** reduced
mass** formula can be used to find the maximum damage energy for a
collision. By substituting separation
velocity (

** **

**E _{MaximumDamage} = .5 * M_{1} M_{2}
* V_{C}^{ 2} / (M_{1} + M_{2})**

** **

**E _{SeparationTranslationalMotion} = .5 * M_{1}
M_{2} * V_{S}^{ 2} / (M_{1} + M_{2})**

**Reduced Mass
Formulae**

** **

The practical value of the formula for the Energy for Maximum
Damage and the formula for Energy of Separation Translational Motion is a
different matter entirely. Either, or
both, may play a significant role in the reconstruction.

The formula for the Energy for Maximum Damage, **E _{MD} = .5 M_{1}M_{2}V_{C}^{2}/(M_{1}+M_{2})**,
eliminates the need to subtract the zero restitution kinetic energy of the
vehicles departing the collision from the total kinetic energy of the vehicles
entering a collision in order to arrive at the maximum possible energy going
into deformation. In addition to the
mass of each of the vehicles involved, only the closing velocity is required
for the computation.

Substituting the separation velocity for the closing
velocity, the formula now computes the energy restored to the system in the
form of translational motion, **E _{R(STM)}
= .5 M_{1}M_{2}V_{S}^{2}/(M_{1}+M_{2})**.

The formulae can also work together. If **.5
M _{1}M_{2}V_{S}^{2}/(M_{1}+M_{2})**
is divided by

**Tsunami and
Other Waves**

** **

One of the contenders for the energy contained in the
pool is the viscoelastic wave that is generated within the object itself. This wave is similar in many respects to the
Tsunami or "harbor wave" generated by an earthquake or resulting from
landslides under the surface of a body of water. The tsunami, almost invisible until it reaches shallow water,
causes damage that is spatially far removed from the origin of the wave. As the viscoelastic wave reaches
structurally weaker sections, it may cause stress within the object. The
oscillation created by the wave causes the object to damage itself. This is
commonly referred to as induced damage and may be spatially removed from the
contact damage of the collision. If no
damage is caused this energy is radiated by the object just as the object
radiates sound wave energy.

** **

** **

**Reduced Mass
Applications**

The Reduced Mass set of formulae can be used to
determine the actual energy going into damage, which can be of great importance
to the Reconstructionist. It is an easy
method of determining the total energy in the damage pool, which must be
apportioned out to the different contenders.

**Energy:
Kinetic versus Potential **

** **

The
kinetic energy formula (KE = 0.5 * Mass * Velocity^{2}) and its variant
(E = Weight * friction coefficient * Displacement) differ from the potential
energy formula (PE = 0.5 * k * x^{2}) in one very important aspect -
Mass. This formula for Potential Energy
does not deal with mass.

Crush energy uses the CRASH3 model developed by
McHenry. This system is based on the ** spring
model** using Hooke's Law (F

Newton's
Third Law of Motion tells us that the forces must be equal and opposite and
must act on different bodies. Since the
forces are equal and opposite (F_{1} = F_{2}), and a
"spring model" is used to model the collision damage, it follows that
in the collision, the energy related to damage is potential energy and not
kinetic energy.

E_{1}
= 0.5 * k_{1} * x_{1}^{2} and E_{2} = 0.5 * k_{2}
* x_{2}^{2}. Therefore,
E_{1} / E_{2} **= **x_{1}** **/**
**x_{2}.

In
order to illustrate the **reduced****
mass** principle, two more collisions will be considered. The first involves two identical 18-wheelers
loaded to 80,000 pounds each. Attached
to the front of each is a giant spring.
Both springs are identical.
These springs can absorb all of the energy that would normally go into
damage. When the vehicles collide, both
springs compress absorbing all of the energy.
The springs compress equally, but this is to be expected, as both the
vehicles and the springs are identical.

In
the next collision, the second vehicle is a 1000-pound dune buggy. It has a spring attached to the front that
is identical to that of the 18-wheeler.
The closing speed is identical to the collision between the two
18-wheelers. How much damage do we
expect in this collision?

It
is obvious that if each of the vehicles is traveling at 30 units at impact, the
post-impact motion will be in the same direction as the pre-impact direction of
the 18-wheeler. This is the only
realistic way that momentum can be conserved in this collision, as momentum is
a result of the inertial qualities of the vehicles. Mass is really a measure of the inertia of an object. If the dune buggy has very little relative
mass, it must accelerate faster than the 18-wheeler when equal forces are
applied to both. If it takes less force
to accelerate it to the same speed in the same time frame, the springs will not
compress as much as they did in the collision involving the two
18-wheelers. To belabor a point, for
those non-believers, instead of a dune buggy, imagine again the feather. How much spring compression will the feather
generate?

As
can be seen from the collisions above, it is the **reduced mass** of the
vehicles in collision that accounts for the difference in the spring
compression. If a "spring
model" is used for damage, then it follows that it is the **reduced mass**
that is the "effective mass" in any collinear collision with 100
percent overlap or any central collision where all action passes through the
centers of mass of both vehicles.

However,
by using a spring model for damage, it is not necessary to consider mass in
order to determine how the energy is distributed between the vehicles. The mass is in effect factored out of the
distribution of the energy. It is
important to get rid of mass in the damage distribution model for a collision
in order to eliminate all of the qualifiers at the end of the last sentence in
the preceding paragraph.

Once
the collision is non-central (action does not pass through the center of mass),
the mass of the object or vehicle is not what is being displaced. It is the end of the moment arm that the
force is acting on, not the mass of the object. It is rotational inertia that the force is acting against, not
the inertial mass of the object.

The
proceeding section demonstrated how energy is shared using springs to
illustrate a point. Just as ice cream
comes in different flavors, springs come in different sizes. If we put a spring capable of resisting the
weight of the 18-wheeler by only compressing a fraction of an inch and place it
in front of the 18-wheeler while using a spring that will compress 100 times as
much in front of the dune buggy, and then re-stage the collision, the results
will be much different. The springs
are equivalent to the A and B stiffness values. They represent the resistance to deformation of the object. Since the non-deforming, infinite mass
barrier cannot be deformed, all of the energy available for damage goes into
damaging the vehicle. ** Energy is
only shared equally between identical vehicles.** If the resistance to deformation is not
identical, the energy is no longer shared equally for the same volume of
damage. The

When
the vehicle is struck on the side by a vehicle with less mass than the struck
vehicle, it is analogous to being struck by the feather. It will not cause the same amount of damage
as would have occurred had the vehicle been struck by an infinite mass,
non-deforming barrier. The damage is
based on the ** spring model** and the

Since Kinetic Energy Equivalent Speed computations result
in the same energy values obtained from using the reduced mass computations,
and the closing speed of the barrier is known, then KEES computations are
really nothing but an involved mathematical exercise. In order to distinguish between the KEES computations and the **reduced
mass** computations, the term Crush Energy Equivalent Speed (CEES) will be
applied to the latter.

It
is important to emphasize the *crush** *reference in the term
CEES. This term (CEES) refers to the
process of converting the potential energy of actual crush damage into a speed
for a vehicle of identical weight (mass) as the damaged vehicle using the
kinetic energy formulae.

** **

** **

**A Fundamental Error**

** **

An analysis of two more collisions is necessary. The striking vehicle in both collisions will have a speed of 30 units and a mass of 1.

6. In Collision 6, the vehicle strikes a non-deforming infinite mass barrier.

7. In Collision 7, the vehicle strikes a non-deforming object with a mass of 1.

These collisions are not identical.

An examination of both collisions using the conservation of momentum reveals that the striking vehicle has 30 units of momentum. Both of the non-deforming struck objects are stationary and therefore generate a momentum value of zero. As both objects are non-deforming, they will absorb no damage energy as a result of the collision. So far, both collisions are identical.

A closer examination, again using momentum, reveals a slightly different story. After collision, according to the law of Conservation of Momentum, we must have the same total momentum as before the collision.

In Collision 6, using the reduced mass formula, all of the energy brought into collision goes into damaging the striking vehicle, as an infinite mass object, by the very definition of mass, cannot move. If the non-deforming infinite mass barrier does not move, there is no momentum for that object in the post collision phase. If zero restitution is a given, then both objects must reach a common velocity of zero.

In Collision 7, using the reduced mass formula, only a portion of the energy brought into collision goes into damaging the striking vehicle. Some of the energy goes into accelerating the non-deforming object. If the non-deforming object moves, the energy used to accelerate this object must be subtracted from the total energy pool available for damage. Conservation of Momentum tells us that the two objects will reach a common speed. Again, with zero restitution, there is no force to separate the vehicles so they will remain at this common velocity. With both objects having a mass of 1, there is a total mass of 2 with a momentum value of 30 that gives a common velocity of 15 units. With both units traveling at 15 units, the energy (which must also be conserved) can be computed.

Using the Kinetic energy formula (Ke = 0.5 * M * V^2), the energy figure for each vehicle is 112.5 units for a total energy out of collision of 225 units.

The energy brought by the vehicle into each of the collisions using this same formula was 450 units.

In Collision 6, all of the energy (450 units) went into damage. In Collision 7, only 225 units of energy were available for damage.

Clearly, the vehicle in Collision 6 will sustain twice the damage as the vehicle in Collision 7. If this is true, should both vehicles have identical stiffness coefficients?

Assume that both vehicles have identical resistance to deformation. With this as a given, both vehicles will have identical deformation if they both strike an infinite-mass, non-deforming barrier and both vehicles will have identical deformation if they both strike a non-deforming impactor with a mass of 1. So where is the problem?

Wait, how can they have identical resistance to damage when one has twice the damage as the other and they were both closing at the same speed?

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**A Note for the Purists**

There is no such thing as an infinite mass barrier here on our little planet. It is a theoretical object, an object of our fantasy and a figment of our imagination. In reality, momentum is truly conserved, as the earth moves when our infinite mass object is struck, and this is where the momentum goes. It also requires energy to move the earth, but not as much as you might suspect, as it is proportional not to the speed, but to the square of the speed.

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**Where the Rubber Meets the Road**

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A problem arises when we try to compute the stiffness values from the information garnered from the test collision. The formula calls for the following variables:

1. The (average for average stiffness) crush depth,

2. The maximum speed without damage,

3. The width of the damage,

4. The weight of the vehicle, and

5. The speed of the vehicle.

This appears to be fairly straightforward data that is required. So where is the problem?

The fundamental flaw is very carefully hiding. The flaw, like the devil, is in the
details. The average crush depth is
relatively easy to compute even though many improperly perform this
computation. The maximum speed without
damage seems simple enough, as do the width of the damage and the weight of the
vehicle. This only leaves the speed of
the vehicle, and that can be measured very precisely. So, again, where is the ** Big Problem**, the

*Is the culprit the term speed? Do we want a speed, or is it a velocity that
is required? It must have a quantity,
but it also has a direction. Is it a
vector, or is it a scalar? That depends
on how you look at it. Not unlike light
that can be either a particle or a wave, depending upon which is being
sought. But this is not in and of
itself the answer.*

* *

Does the term speed mean **closing speed** and** **if
so, between which objects? Here is the
real problem! Speed does not
necessarily mean closing speed, as it refers to the impact speed ** with a non-deforming
infinite mass barrier **and needs to be adjusted accordingly. The term speed refers to that speed computed
using either KEES or CEES. Using the
term CEES has an advantage over KEES in that it helps to denote that the speed
obtained from the energy computation is equivalent to the closing speed between
the vehicle and the non-deforming infinite mass barrier required to produce
identical damage.

CEES, KEES and BES all generate a value for speed. Their computed values are identical, as we have seen. They have a distant cousin – absolute speed. It is only in a collision with an infinite mass (immovable) non-deforming barrier that absolute speed has the same value as the others. The absolute speed should never be used to determine stiffness coefficients in a test collision. With an immovable barrier, they are all identical in value, but even in this situation, absolute speed is not the correct input into the stiffness computations.

Using the absolute speed of the impactor, which is not an infinite mass object, introduces the fundamental flaw into the calculations. Energy is transformed into a change of speed for the impactor. This energy is no longer available for damage and therefore must be taken out of the pool of energy used to deform the test vehicle. Only the speed computed as a result of the energy of actual damage can be used in determining the stiffness coefficients of the test vehicle. Using the absolute speed of the impactor, especially one with a mass nearly identical to the test vehicle, results in abnormally high stiffness coefficients. When these stiffness coefficients are used in computations that employ the CRASH3 algorithms, abnormally high speeds are then generated.

Suppose a test is preformed with an impactor at 30 miles per hour that has a mass equal to that of the test vehicle. If the absolute speed (30 M/H) of the impactor is used to compute the stiffness coefficients, the resulting CRASH3 speed changes are almost 50% higher than those computed using the correct stiffness values.

How much of an effect these incorrect stiffness values and the correspondingly erroneous speed change computations have had is a matter of speculation. Certainly, the flawed computations have been used as evidence of either fault or negligence within the civil and criminal divisions of our legal system.

**Better the Devil You Know?**

** **

This identification of a serious problem poses another one that is potentially more serious. Now, where do we go from here? All of the stiffness computations made using absolute speeds involving impactors with a mass that was not several orders of magnitude greater than the mass of the vehicle being tested have this fundamental flaw, unless they were computed using either CEES, KEES or BEV as the Impact Speed in the computation.

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**A Cautionary Note**

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**CEES and KEES do not deal with Rotational Energy! **Usually minimal rotation is involved in
the tests, but any energy that does go into rotation must be subtracted from
the *Damage* energy pool before calculating any equivalent speed. This work does not deal with rotational
energy, as its inclusion would only detract from our central focus. This cautionary note is to serve only as a
reminder that rotational energy must also be removed from the pool of energy before
calculating a barrier equivalent speed that represents the crush damage to the
vehicle.

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** **

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**Conclusion**

The procedure for computing stiffness coefficients requires using the speed obtained from energy computations as the impact speed. The energy equivalent speed and the absolute speed are identical in value only in a collision with an infinite mass barrier.

Either by oversight or carelessness, in test collisions
with a movable barrier instead of an infinite mass barrier, the **absolute
speed** of the test vehicle has been used in the stiffness computations
instead of the **crush energy equivalent speed**. This work illustrates the correct procedures for determining the
stiffness coefficients of vehicles in collision as well as the relationship
between momentum, energy and force and their relationship to speed and damage.

The underlying problem of allotting all of the pre-impact energy of the vehicle (or impactor) to damage is not confined to the calculation of stiffness coefficients. Americans seem to be obsessed with the CRASH3 methodology for determining deltaV and accident severity. The European community seems to prefer a comparison between the damage severity of the accident vehicle and of exemplar photographs of damage where the speed of the vehicle was known or could be accurately determined.

The inherent problem still exists with this methodology
and has the potential of infecting any resulting calculations. It is a wolf hiding in sheep’s
clothing. If using the comparison for establishing
absolute speed, it is just as flawed as the stiffness calculations, and for
exactly same reason. The deformation of
a vehicle is not related to absolute speed – **EVER!** It does not matter if the deformation
occurred during a staged test crash, or a real world collision, the deformation
does not result from absolute speed.

Deformation results from a change of velocity or deltaV. In dealing with collisions involving vehicles, it is the change of velocity resulting from the collision impulse. A more precise statement would be that deformation results from the force of acceleration, or deltaV divided by deltaT where deltaT is the time interval for deltaV. It is not usually the deltaV of the fall with a large deltaT that causes injury; it is the deltaV of the stop with the small deltaT of the impact that can be fatal.

A comparison of photographs, where the first involves a collision with a bridge abutment and the second involves an identical vehicle striking a small stationary vehicle, will yield remarkably different deformation characteristics if the bullet vehicle was a large sedan traveling at identical speeds in both collisions. The deltaV in the first instance will be extremely close to the absolute speed of the vehicle, as the speed of the rotation of the earth will not be markedly changed. While close in value, or even identical in value, the absolute speed is not the deltaV. The deltaV in the second instance will be much less, as the smaller vehicle will be accelerated to a common velocity with the larger vehicle and they will both continue in motion after maximum engagement.

This is the reasoning behind the correct approach for stiffness coefficient determination and why photographs of damage should not be used to determine absolute speed.

An understanding of the basic principles involved will prevent errors that in the world of the accident reconstructionist can have serious civil and criminal consequences.

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**References:**

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1.
The first use of the term ** Kinetic Energy Equivalent
Speed** was, to the best of my knowledge, by Daniel W. Vomhof III to
identify a speed (mph) equivalent to the energy used in damaging a vehicle,
specifically in the NHTSA side- and rear-crash tests. The problem he found with using the more commonly referred to
term "Barrier Equivalent Speed" was that in most vehicle vs. vehicle
accidents, no "barrier" is involved. This term, he believes, more
correctly describes what is being computed when making speed computations from
"crush" and has been in the AutoStats® program since originally
offered for sale in 1991. Dan and I
agree in principle and on methodology; we disagree on terminology,
understandable in light of the fact that he lives in California and I live in
Florida.

2.
Glenn A. Burdick Ph.D. introduced me to the ** reduced mass**
principle. Glenn received his Ph.D. in
physics from Massachusetts Institute of Technology and is a Dean Emeritus of
Engineering and a Distinguished Professor of Engineering at the University of
South Florida. He is also a Full
Professor of both Transportation and Electrical Engineering.