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```The Georgia Police Academy
Traffic Accident Reconstruction Level V
Background and Purpose
The National Highway Traffic Safety
automobiles sold in this country meet crash
worthiness standards. Each automobile
manufacturer must provide new automobiles for
the purpose of crash testing so NHTSA can verify
that the standards are being met. When crash tests
are done all the data is collected and maintained.
Since the tests are done in a controlled
environment the vehicle’s speed and resulting
crush damage are known.
This testing has allowed engineers to develop a
mathematical model that can be used to measure a
vehicle’s speed based on the amount of crush
damage it receives in a collision. Although its
application is limited, this information does
provide traffic accident investigators with another
tool that may be used to calculate speed. Traffic
Accident Reconstruction Level V, introduces
students to the proper techniques for making
accurate speed estimates from crush damage.
Critical issues such as work, energy, the
conservation of energy, equation application
and the energy equation derivation are
complete this course should have the
knowledge and confidence to accurately
estimate speed from vehicle crush in some
instances.
Terminal Performance Objective
Given a simulated traffic accident, students
will evaluate vehicle crush data to
accurately calculate vehicle impact speeds.
Enabling Objectives
• Demonstrate the ability to use work, force and
distance to calculate energy.
• Demonstrate one method for calculating elastic
potential energy when force constant and distance
are known.
• Define the Law of Conservation of Energy.
• Demonstrate one method for calculating energy of
a vehicle that skids over numerous surfaces, when
weight, drag factor and distance are known.
Enabling Objectives (con’t)
• Demonstrate two methods for calculating
vehicle speed when the amount of crush and
after impact vehicle dynamics are known.
• Define the terms dynamic collapse and
restitution as they apply to vehicle damage.
Work
Technically, work is done when a force acts
on an object through a distance. Work is
also a measure of the effect force has on
changing an object. The amount of change
or work done is reflected in changes in the
object's velocity, position, size, and shape.
Work can be described mathematically as
the product of the force (F) and the distance
(d) through which it acts, provided the force
and distance covered are in the same
direction. This may be written in equation
form:
W = Fd
Remember from past courses that the unit
for force is pounds (lb) and the unit for
distance is feet (ft). W = Fd
F = force (lb)
d = distance (ft)
W = work (ft-lb)
Example
Assume a force of 300 lbs is required to
push a box 20 feet. Calculate the amount of
work done.
W = Fd
W = (300 lbs) (20 ft)
W = 6000 ft-lbs
If the force and distance are not in the same
direction, then the equation is rewritten.
W = Fd Cos 1
Example
Assume the force is equal to 300 lbs, the
angle is 45 degrees, and the distance is 20
feet. Calculate the amount of work done.
W = Fd Cos 1
W = (300) (20) (Cos 45E)
W = 4242 ft-lbs
Work is the product of the magnitudes of force and
distance. Therefore, work is a scalar quantity.
Being a scalar quantity, work has only magnitude
and sign (positive or negative), but has no
direction associated with it. Work is positive if
1 is greater than or equal to 0 degrees but less
than 90 degrees. Work is negative if 1 is greater
than 90 degrees but less than or equal to 180
degrees.
An accelerating vehicle is an example of
positive work. The force accelerating the
vehicle is in the direction the vehicle is
traveling. In that case 1(angle) is 0 degrees.
d
Direction of Travel
Example
Assume the force acting on the vehicle in
the previous slide is 1000 lbs, the distance
the force acts through is 100 ft, and the
angle is 0 degrees. Calculate the amount of
work done.
W = Fd Cos 1
W = (1000) (100) (Cos OE)
W = 100,000 ft-lbs
Now suppose the vehicle is decelerating. In
this case the work has a negative sign since the
force is opposite the direction of travel. In this
case the 1(angle) is 180 degrees. Assume the
force acting on this vehicle is 2000 lbs through
a distance of 50 feet.
d
Direction of Travel
Friction
W = Fd Cos 1
W = (2000) (50) (Cos 180E)
W = -100,000 ft-lbs
Energy
When work is done, energy is transferred
between different objects. The amount of
energy possessed by an object is an indication
of its ability to do work. Work and energy are
proportional. The more energy an object has,
the more work it can perform.
Equations can be derived to calculate the amount
of energy transferred between objects when work
is done under a variety of conditions. These
derivations use the positive sign if the object's
energy is increased and negative if it is decreased
when work is done. This agrees with the previous
examples given of the accelerating and
decelerating vehicles. The accelerating vehicle is
gaining energy and increasing its ability to do
work. The decelerating vehicle is losing energy
and decreasing its ability to do work.
Energy can be grouped into three general
categories:
• Rest energy
• Kinetic energy
• Potential energy
1. Rest energy is energy an object possesses
due to its mass. This form of energy is
related to Einstein's theory of relativity and
will not be discussed.
2. Kinetic energy (KE) is energy an object
possesses due to its motion. A vehicle in
motion has more energy than a vehicle at
rest. When a vehicle accelerates, it gains
velocity and also increases its energy. The
energy of a moving body is equal to:
• KE = ½mv²
• KE = Kinetic energy (ft-lbs)
• M = Mass (weight divided by the
acceleration of gravity)
• v = Velocity (fps)
The amount of work required to accelerate
the vehicle from rest to a given velocity can
be calculated using this equation. The
energy calculated is the energy gained by
the vehicle and it is kinetic energy. In the
equation mass and weight are related by the
acceleration due to gravity. Considering
this relationship the equation could also be
written as follows:
KE = wv²/2g
KE = Kinetic energy(ft-lbs)
w = Weight (lbs)
v = Velocity (fps)
g = Gravity (32.2 ft/sec²)
Example
Assume a 3,000 lb vehicle is accelerated
from rest to a velocity of 40 fps. Calculate
the amount of work done and energy
gained.
KE = wv²/2g
KE = (3000) (40)²/(2)(32.2)
KE = 4,800,000/64.4
KE = 74,534 ft-lbs
The same amount of work would be required to
bring the vehicle to rest. The same amount of
energy that was originally gained, would have to
be dissipated. Energy is dissipated when it is
converted into thermal (heat) energy. This
happens as a result of the friction between the tires
and road surface, tires and brake system, or
through rolling resistance in coasting to a stop.
In the derivation of the speed from skid
mark equation we learned that the following
equation can be used to calculate the
amount of kinetic energy that must be
converted to another form in order to bring
a vehicle to a stop:
W = wfd
W = Work (ft-lbs)
w = weight (lbs)
f = drag factor
d = distance (ft)
Example
Assume a 3000 lb vehicle from the previous
example skidded to a stop with a drag factor
of .71 from 40 fps, it would slide
approximately 35 ft. Calculate the amount
of work done.
W = wfd
W = (3000)(.71)(35)
W = 74,550 ft-lbs
The following two equations can be used to
calculate speed estimates from skid marks.
•
W = wfd
•
KE = wv²/2g
The first equation is used to find the amount
of kinetic energy the vehicle dissipated
while skidding. The second equation,
rearranged to solve for velocity, uses the
energy calculated to find the velocity of the
vehicle.
• KE = wv²/2g
• Multiply both sides by 2g
• 2g * KE = wv²/2g * 2g/1
•
•
•
•
Cancel leaving:
2g * Ke = wv²
Divide weight from both sides:
2g * KE/w = wv²/w
•
•
•
•
Cancel leaving
v² = 2g * KE/w
Take the square root of both sides leaving
v = %(2g)(KE)/w
Since the quantity of one form of energy is
the same as the quantity of any other form
of energy, the equation can be written:
v = %(2g)(E)/w
Example
From the previous example, the amount of
kinetic energy that was dissipated while
skidding was approximately 74,534 ft-lbs
and the vehicle weighed 3000 lbs.
Calculate the velocity.
v = %(2g)(E)/w
v = %(2)(32.2)(74,534)/3000
v = %4,800,000/3000
v = %1600
v = 40 fps
Kinetic energy is a scalar quantity. The
kinetic energy of a given mass depends on
the magnitude of its velocity and not on the
direction of travel. Any change in kinetic
energy depends on the sign of the work
energy is increased. A negative sign
indicates the kinetic energy is decreased.
Acceleration rate is not a factor in kinetic
energy. The mass and velocity of a body
govern the amount of kinetic energy the
body has. The rate at which the body
reaches that velocity does not affect the
kinetic energy.
3. Potential energy (PE) is energy an object
possesses due to its position. Two forms of
potential energy will be covered in this
course.
• Gravitational potential energy (PEh)
• Elastic potential energy (PEk)
• Gravitational potential energy is the energy
an object possesses due to its position above
some reference plane, usually the surface of
the earth.
Position
1
Position
2
Horizontal Motion
W
h1
W
h2
Reference Line
When the box moves from position 1 to position 2 there is no change
in the vertical distance. PEh1 = PEh2
The boxes in the overhead have a
downward gravitational force which is their
weight (w). The distance moved is the
height (h) of the center of mass above its
original position. These variables can be
substituted into the equation for work:
W = Fd
W = wh
This is the equation for gravitational potential
energy:
PEh = wh
PEh = gravitational potential energy (ft-lbs)
w = weight (lbs)
h = vertical distance (ft)
Example
Assume a vehicle weighing 3000 lbs has its
center of mass raised 10 ft above the
ground. Calculate the amount of
gravitational potential energy the vehicle
has.
PEh = wh
PEh= (3000) (10)
PEh = 30,000 ft-lbs
Calculate how many feet per second a 3000
lb vehicle would be traveling to have
30,000 ft-lbs of
v = %(2g) (E)/w
v = %(2) (32.2) (30,000)/3000
v = %1,932,000/3000
v = %644
v = 25.37 fps
The change in gravitational potential energy depends
only on the difference between the initial and final
heights of the object, not on the path taken to reach the
final height. The vehicle moves up and down while
going from position 1 to position 2. The change in
gravitational potential energy is only dependent upon the
difference between h1 and h2. Peh = w(h2-h1)
Position 2
Position 1
h2-h1
h2
Reference Plane
h1
• Elastic potential energy is energy an object
possesses because of its shape. A spring
which is compressed or stretched has elastic
potential energy because of its shape. The
amount of elastic potential energy it has
depends upon the amount of work done to
compress or stretch the spring.
A force is required to compress a spring from it
free length. The force required is not constant. It
must increase as the spring is compressed more
and more. The amount of force is directly
proportional to the amount the spring is
compressed. This relationship is known as
Hooke's Law. The proportionality constant for
Hooke's Law depends on the stiffness of the spring
and is called the force constant (k). A stiff spring
will have a larger force constant than a spring that
can be easily stretched.
Free Length
X
Compressed
A spring is compressed a certain distance, X from its
free length. The force required can be calculated by the
equation F = kx
F = kx
F = force (lbs)
k = force constant (lbs/ft)
x = distance compressed (ft)
In the overhead the distance, x, is zero
before the spring is compressed. When the
spring is compressed a certain distance, x,
the force is F = kx
The average force would be calculated by
using the following equation.
Favg = (0 + kx)/2 or Favg = ½kx
The work equation is W = Fd
Therefore, the work done to compress the spring is
W = Favg x
Remember F = ½kx, so ½kx can be substituted for
F, leaving W = (½kx) (x) or
W = ½kx²
W = work (ft-lbs)
k = force constant (lb/ft)
x = distance compressed/stretched
This is also equal to the amount of elastic
potential energy possessed by the spring.
PEk = ½kx²
Example
Assume a spring with a force constant of
10,000 lbs/ft is compressed 2 ft from its free
length. Calculate the elastic potential
energy possessed by the spring.
PEk = ½kx²
PEk = ½ (10,000) (2)²
PEk = 20,000 ft-lbs
Remember this equation applies equally for
a spring that is stretched and to one that is
compressed. The elastic potential energy is
equal to one-half the product of the force
constant times the square of the distance
compressed/stretched from its free length.
If the distance is doubled, the energy
increases by a factor of 4.
Conservation of Energy
When work is done energy is converted
from one form into another. We are able to
apply energy to accident situations because
it is a conserved quantity. The total energy
remains the same, but it may change forms.
The law of conservation of energy states
that when work is done and energy is
converted from one form into another, no
energy is created and no energy is
destroyed. As accident investigators, this
concept can be another tool which may
enable us to determine accident vehicle
speed.
First we must account for the work done
during the accident sequence. Once the
total energy dissipated has been calculated it
can then be converted into velocity. The
total energy must be calculated by using an
energy balance equation.
ET = E1 + E2 + E3.....Ei
ET = initial energy possessed by the vehicle
E1 = first amount of energy dissipated
E2 = second amount of energy dissipated
E3 = third amount of energy dissipated
Ei = last amount of energy dissipated or left
unchanged
Example
Assume a vehicle weighing 3000 lbs, was
initially traveling at 60 fps and then skidded
73.75 ft with a drag factor of .50, and its
velocity after skidding was 35 fps.
Substitute these values into the energy
balance equation and calculate.
ET = E1 + E2
ET = wv²/2g + wfd
ET = (3000) (35²)/64.4 + (3000) (.50)
(73.75)
ET = 57,065.217 + 110,625
ET = 167,690.22 ft-lbs
Now considering our energy balance equation this
total amount of energy should equal our initial
amount of energy (ET). The initial velocity was
60 fps and the weight is 3,000 lbs.
KE = wv²/2g
KE = (3000) (60²)/64.4
KE = 167,701.86 ft-lbs
As you can see, the two are close enough to be
considered equal.
Example
Assume a vehicle that weighs 4,000 lbs skids for
100 ft on a surface having a .80 drag factor, and
then continued to slide an additional 60 ft on a
second surface having a .40 drag factor. At the
end of the 60 ft skid on the second surface the
vehicle's brakes were released and it was still
going 30 fps. Calculate the total amount of energy
possessed by the vehicle.
ET = E1 + E2 + E3
ET =
4,000*.80*100+4,000*.40*60+4,000*(30²)/
64.4
ET = 320,000 + 96,000 + 55,900
ET = 471,900 ft-lbs
This total represents the initial amount of energy
the vehicle possessed when it first began to slide.
Since this is kinetic energy, the velocity of the
vehicle can be found as follows:
v = %2gE/w
v = %(2) (32.2) (471,900)/4000
v = %7597.59
v = 87.16 fps
Example
A vehicle weighing 3,500 lbs skids to a stop
in 100 feet while traveling up a 4% grade.
The drag factor is .60. Calculate the
vehicle's initial kinetic energy.
W = E1 = wfd
E1 = (3500) (.60) (100)
E1 = 210,000 ft-lbs
A vehicle traveling 100 feet up a 4% grade
will rise approximately 4 ft. The
gravitational potential energy gained by the
vehicle due to this height change can be
found by using the gravitational potential
energy equation.PEh = E2 = wh
E2 = (3,500) (4)
E2 = 14,000 ft-lbs
The energy balance equation can be written
to calculate the amount of kinetic energy the
vehicle initially possessed.
KE = ET = E1 + E2
ET = (210,000) + (14,000)
ET = 224,000 ft-lbs
The initial velocity can now be found.
v = %2gET/w
v = %(2) (32.2) (224,000)/3500
v = %4121.6
v = 64.19 fps
If the same vehicle was sliding down the
be the sum of the initial kinetic energy and
gravitational potential energy. If the vehicle
stopped in 100 ft, the energy dissipated
would have to equal the sum of the initial
kinetic energy and gravitational potential
energy:
KEi + PEh = E1
Rearrange the equation to solve for KEi,
leaving
KEi = E1 PEh
KEi = (210,000) - (14,000)
KEi = 196,000 ft-lbs
The velocity of the vehicle when it first
began to slide would be calculated as
follows:
v = %2gET/w
v = %(2) (32.2) (196,000)/(3500)
v = %3606.4
v = 60.05 fps
The previous example demonstrates that a
vehicle will require less initial velocity
(kinetic energy) to slide down a grade than
to slide the same distance up the grade.
In some instances energy might be
dissipated in subtle ways that we are unable
to measure. In these cases it may be
difficult to account for all the energy in an
energy balance equation.
Speed Estimates From Vehicle
Damage
Energy is dissipated in ways other than sliding.
Suppose a vehicle weighing 3,000 lbs is traveling
40 fps when it collides with a solid (nondeforming) immovable barrier and comes to rest.
E1 = wv²/2g
E1 = (3,000) (40²)/64.4
E1 = 4,800,000/64.4
E1 = 74,534 ft-lbs
This vehicle would possess 74,534 ft-lbs of
kinetic energy due to its weight and
velocity. This energy is dissipated in the
collision with the barrier, and the result is
damage to the vehicle. The quantity of
energy used to do damage to a vehicle must
be determined in order to estimate speed
based on that damage.
After a collision the vehicle is permanently
damaged. The crush measurements taken
on a damaged vehicle show the resulting
damage to the vehicle. In the collision itself
the vehicle will have slightly more damage
than is revealed after the collision. This
damage is referred to as dynamic collapse.
A vehicle has some restitution in a collision.
Restitution means the vehicle tries to return
to its original shape. The effect of
restitution is insignificant in nearly all
severe collisions. In low-velocity impacts,
restitution may have to be considered.
The definition of work indicates that the amount
of energy dissipated is related to the force and the
distance through which it acts. The distance is
determined by making crush measurements. Once
this has been done, the speed can be calculated
provided information exists regarding the amount
of force required to produce the damage. This
quantity of force is related to the dynamic force
deflection characteristics, or crush resistance
(stiffness), of the vehicle structure being crushed.
The dynamic force-deflection
characteristics of a vehicle are determined
by tests, primarily barrier impact tests. In
barrier impact tests it is assumed that all the
energy is used in doing damage to the
vehicle. The velocity at which the vehicle
strikes the barrier is an indication of the
amount of kinetic energy (crush energy)
available to damage the vehicle.
The assumed relationship between a barrier test
vehicle and a similar accident vehicle is that an
equal quantity of energy must be used to damage
the latter the same as the barrier test vehicle. For
example, suppose a 1990 Chevrolet Impala is used
in a barrier impact test and reveals an average of
15 inches of crush from a 30 mph impact. You
investigate and accident in which a 1990
Chevrolet Impala strikes a bridge abutment.
You measure the crush damage and find that
it averages 15 inches. In this scenario it
would be safe to assume the accident
vehicle struck the bridge at a speed of 30
mph. The barrier impact speeds are known
as the equivalent barrier speed (EBS) or
barrier equivalent velocity (BEV).
Once the EBS for the accident vehicle has
been determined, the energy dissipated in
doing damage can be calculated. This
energy along with energy dissipated in other
ways, can be used in an energy balance
equation to calculate the vehicle's initial
velocity.
Example
Assume a 3,000 lb vehicle (vehicle 1) skids with a
drag factor of .70 for 60 ft, then strikes a parked
vehicle (vehicle 2) weighing 3,500 lbs in the front.
Barrier impact tests run at 50 fps using the same
type of vehicle as the striking vehicle (1) and tests
run at 30 fps using the same type of vehicle as the
struck vehicle (2) resulted in similar damage
patterns as reflected by the accident vehicles.
After impact, the two vehicles remained
locked together and slide as a unit for 60 ft,
with a drag factor of .50. In this twovehicle system, the striking vehicle has all
the initial energy. The amount of energy
initially possessed by the striking vehicle is
calculated by quantifying the amount of
work done in sliding and doing damage.
Energy (work) dissipated by vehicle 1 while
skidding, before impact:
E1 = wfd
E1 = (3,000) (.70) (60)
E1 = 126,000 ft-lbs
Energy (work) dissipated in doing damage
to vehicle 1:
E2 = w1v1²/2g
E2 = (3,000) (50²)/64.4
E2 = 7,500,000/64.4
E2 = 116,459 ft-lbs
Energy (work) dissipated in doing damage
to vehicle 2:
E3 = w2v2²/2g
E3 = (3,500) (30²)/64.4
E3 = 3,150,000/64.4
E3 = 48,913 ft-lbs
Energy (work) dissipated by vehicles 1 and
2 while sliding after impact:
E4 = (w1 + w2)fd
E4 = (3,000 + 3,500) (.50) (60)
E4 = (6,500) (.50) (60)
E4 = 195,000 ft-lbs
Total amount of energy possessed by the
striking vehicle (1) when it first began to
skid:
ET = E1 + E2 + E3 + E4
ET = 126,000 + 116,459 + 48,913 +
195,000
ET = 486,372 ft-lbs
This amount of energy can be used to
determine the velocity of the striking
vehicle when it first began to skid:
v = %2g ET/w1
v = %(2) (32.2) (486,372)/3,000
v = %31,322,357/3,000
v = %10,440.786
v = 102.18 fps
It is obvious, from this example, that the
key to this method of estimating speed is
assigning an appropriate EBS value for the
amount of energy required for damage. To
accomplish this we could go out and find
like vehicles and conduct crash tests until
the actual accident scenario is reproduced.
This however, is impractical.
On the other hand, we can compare our accident
data to crash tests which have already been
performed. Yet, even with this method the
possibilities are infinite due to the number of
different damage profile configurations. The best
way would be to have a general mathematical
model. This model would relate energy to damage
(crush) for as many damage profile configurations
as possible.
For years crash tests have been conducted which
examine the dynamics of automobile collisions.
The results of these tests indicate that the
deceleration force on an automobile in a head-on
collision was directly proportional to the
deformation of the vehicle. The studies also
proved that vehicles behave like linear plastic
springs, in other words they dissipate energy
rather than store it (the damage remains after the
force of the collision is removed).
Earlier in the course you were given the
equation for the elastic potential energy of a
compressed spring.
PEK = ½kx²
This equation also applies to a linear plastic
original shape.
Vehicles behave like linear plastic springs when they are crushed,
dissipating energy. This is modeled by a spring at its free length
attached to the vehicle.
The vehicle makes contact with a barrier and the spring begins to
compress, dissipating kinetic energy and slowing the vehicle.
The vehicle stops when all the Ke has been dissipated by compressing
the spring.
If the vehicle in the previous slide was
traveling 30 fps and weighed 3,000 lbs, it
would initially possess approximately
41,900 ft-lbs of kinetic energy. To bring the
vehicle to a stop, the spring would have to
deform enough to dissipate 41,900 ft-lbs of
kinetic energy. In equation form this is
written:
½mv² = ½kx²
The actual amount of deformation would depend
on the spring's stiffness. This concept can be used
to estimate the amount of energy dissipated by
damage if the permanent deformation and the
vehicle's stiffness are known.
The process of calculating an accident
vehicle's speed from damage begins by
understanding the variables involved. First
of all you have to determine the A and B
stiffness coefficients for the accident
vehicle. These stiffness coefficients reflect
the force deflection characteristics
(stiffness) of the vehicle structure.
The A and B coefficients can be estimated
from fixed barrier impact tests by plotting
permanent crush against the impact speed.
Remember, the deceleration force of an
automobile is directly proportional to the
crush. One of the variables for the A and B
coefficients, b1, must be solved first. The
following equation will solve for b1.
b1 = (v - bo)/C
b1 = change in impact speed to change in crush
(unknown)
v = impact velocity (mph not fps) (crash test
vehicle)
bo = 3 mph to 8 mph (the maximum barrier impact
velocity resulting in no permanent damage. The
lower number will give the lower speed in the
energy dissipated doing damage equation)
C = average crush (inches) (crash test vehicle)
NOTE:
Both (v) and bo must be multiplied by 17.6
(12)(22)/15 = 17.6 (converts to in/sec.).
The A coefficient can be determined from
the following equation.
A = (w)(bo)(b1) / (g)(W)
A = A coefficient (unknown)
w = vehicle weight (lbs) (crash test vehicle)
bo = substituted from the b1 equation
b1 = solved from the b1 equation
g = 386.4 in/sec²
W = crush width (inches) (crash test
vehicle)
The B coefficient can be determined from
the following equation.
B = (w)(b1²) / (g)(W)
B = B coefficient (unknown)
w = vehicle weight (lbs) (crash test vehicle)
b1 = change in impact speed to change in
crush, substituted from b1 equation
g = 386.4 (in/sec²)
W = crush width (inches) (crash test
vehicle)
Your handouts will give you the vehicle
weight of the test vehicle. Crush width is
given in some of the crash test results.
When it is not given the Crash III chart will
be used to determine A and B coefficients.
Remember that all the variables given so far
reflect information from the crash test
vehicle.
Once the A and B coefficients have been
determined, they can be substituted into the
equation for energy dissipated in doing
damage. There are three equations. They
are based on the number of crush
measurements ("C" measurements) taken
from the accident vehicle.
If only two crush measurements are taken
the following equation is used.
E=W[G+A/2(C1+C2)+B/6(C1²+C2²+
C1C2)](1+TAN²1)
If four crush measurements are taken the
following equation is used.
E=W/6[6G+A(C1+2C2+2C3+C4)+B/3(C1²
+2C2²+2C3²+C4²+C1C2+C2C3+C3C4)](1
+TAN²1)
If six crush measurements are taken the
following equation is used.
E=
W/5[5G+A/2(C1+2C2+2C3+2C4+2C5+C6)
+
B/6(C1²+2C2²+2C3²+2C4²+2C5²+C6²+C1C
2
+C2C3+C3C4+C4C5+C5C6)](1+TAN²1)
E = energy dissipated (in-lbs) (divide end
result by 12 to convert to ft-lbs)
G = A²/2B (lbs)
W = width of crush region (inches)
(accident vehicle)
A = the maximum force per inch of damage
width which will not cause permanent
damage (lb/inch)
B = the spring stiffness per inch of damage width
(lb/inch²)
1 = the angle between a line perpendicular to the
vehicle's damaged surface and the force direction
It is recommended that this factor not exceed 2.
The tangent squared of any angle above a 45E
angle added to 1 will exceed 2.
C = crush measurements of damage
Assume the PDOF angle is 50
degrees.
Force Direction 1
d1
d2
C’s
d1<d2
Tan 50 = 1.19
1+ Tan² 50
1+ 1.19²
1+1.42
2.42 Since this exceeds 2 it would
be an indication that the force
came from the side and not the
front. Therefore, the angle should
be measured from the side.
Example
Assume we are investigating an accident involving
a 1982 Chevrolet Camero, 2 door which strikes a
measurements are made of the accident vehicle:
C1 = 29.4 inches; C2 = 30.4 inches; C3 = 30.6
inches; C4 = 30.2 inches; C5 = 29.3 inches; C6 =
27.5 inches. The width of the crush is 71 inches.
Example (continued)
Calculate the impact speed of the accident
vehicle. From our handouts we find that
impact tests done on a 1982 Camero were
done from a speed of 35.4 mph, weight
3428 lbs, crush width 62.4 inches, and
average crush 24.57 inches, use 5 mph for
bo.
Solve for b1 first.
b1 = (v - bo) / C
b1 = (35.4)(17.6) - (5)(17.6) / 24.57
b1 = 623.04 - 88 / 24.57
b1 = 535.04 / 24.57
b1 = 21.77
Solve for the A coefficient.
A = (w)(bo)(b1) / (g)(W)
A = (3428)(5)(17.6)(21.77) / (386.4)(62.4)
A = 6567225.3 / 24111.36
A = 272.24 lb/in.
Solve for the B coefficient
B = (w)(b1²) / (g)(W)
B = (3428)(21.77²) / (386.4)(62.4)
B = (3428)(473.93) / 24111.36
B = 1624642 / 24111.36
B = 67.38 lb/in.
Solve for G:
G = A² / 2B
G = 272.24² / (2)(67.38)
G = 74114.618 / 134.76
G = 549.97 lbs
Now we can solve the energy dissipated by
damage equation. Since we have six crush
measurements taken from the accident vehicle
we will use the following equation.
E=W/5[5G+A/2(C1+2C2+2C3+2C4+2C5+
C6)+B/6(C1²+2C2²+2C3²+2C4²+2C5²+C6²
+C1C2+ C2C3+C3C4+C4C5+C5C6)](1 +
TAN²1)
E=71.0/5{(5)(549)+272/2[29.4+(2)(30.4)+(
2)(30.6)+(2)(30.2)+(2)(29.3)+27.5]+67/6[2
9.4²+(2)(30.4²)+(2)(30.6²)+(2)(30.2²)+(2)(2
9.3²)+27.5²+(29.4)(30.4)+(30.4)(30.6)+(30.
6)(30.2)+(30.2)(29.3)+(29.3)(27.5)]}{1 +
0}
E=14.2{[(2745)+(136)(29.4+60.8+61.2+60.
4+58.6+27.5)+11.16(864.36+1848.32+1872
.72+1824.08+1716.98+756.25+893.76+930.
24+924.12+884.86+805.75)]}(1)
E=14.2[(2745)+(136)(297.9)+(11.16)(1332
1.44)] (1)
E=14.2(2745+40514.4+148667.2)(1)
E=14.2(191926.6)(1)
E=2,725,358.7 in-lbs/12 = 227,113.22 ft-lbs
EBS = v = %2gE/w
v = %(2)(32.2)(227,113.22)/3428
v = %146,626,091.79
v = %4266.65
v = 65.31 fps (15/22) = 44.53 mph
Equation limitations
Remember that A and B coefficients are
established from barrier tests of many vehicles. If
the accident vehicle is not listed in the barrier test
information, then it must be assigned to a group.
That group's A and B coefficients must then be
used. This may or may not be accurate.
A different uniform stiffness is assumed for the
entire front and rear widths and the entire lengths
of the sides. Frame or uni-body structure
involvement is an important factor in front and
rear collisions. If the uni-body structure is
involved, the stiffness coefficients may be
accurate. If only sheet metal is involved the
stiffness coefficient may be too high. This applies
to the sides of the vehicle also. There are "hard"
spots at the wheel areas and "soft" spots between
the wheels in the passenger area.
The frame and frame-type structure provide
the major stiffness to the vehicle. It is likely
to be more accurate to weigh the crush
measurement in favor of the deformation in
the frame or uni-body structure area. Be
careful when there is under-ride or override.
Damage can be difficult to measure. Refer
to the S.A.E. handout on measuring
protocol. Do not use the equation for
rollover or sideswipe accidents. There is no
stiffness data for large trucks.
Alternative Method For
Calculating Speed From Barrier
Test Data.
Some of the NHTSA crush data you have
been provided with does not list the width
of the damage sustained by the test vehicle.
Without knowing the width you would have
to refer to the Crash III chart for the A & B
coefficients. There is an equation which
will give the desired results and is much
simpler to use.
From our earlier example, a 1982 Chevrolet
Camero test vehicle sustained 24.57 inches
of average crush from a speed of 35.4 mph.
The 1982 Chevrolet Camero accident
vehicle sustained an average crush of 29.56
inches. We computed a speed of 44.56 mph
using the long equation. Now we will use
the same information in the short formula.
Ratio = Test speed - speed without
sustaining damage/average crush from test
From our example:
Ratio = (35.4 - 5)/24.57
Ratio = 30.4/24.57
Ratio = 1.23
Once the ratio has been established it can be
substituted into the following equation which will
give the equivalent barrier speed.
Speed = Ratio (Cavg) + speed without sustaining
damage
From our example:
Speed = 1.23 (29.56) + 5
Speed = 41.35 mph (speed of the accident vehicle)
– From the long formula a speed of 44.56 mph
was computed. This is a difference of just over
3 miles an hour. Remember, the ratio for each
case needs to be determined based on crash test
results. In this case the ratio was 1.23. 1.23 is
not a constant for every situation.
Equation limitations
This equation assumes a straight linear
relationship between the amount of vehicle crush
and the velocity of that vehicle required to create
the designated amount of crush. A limitation of
the linear function is its lack of input for exact
crush width. This hampers ability to compensate
for collisions involving deformation to only a
portion of the vehicle end.
The linear function assumes damage across
the entire end width of the vehicle. This can
be compensated for by assigning some of
your C1 through C6 measurements the
value of zero. These zeros would be used
for calculation of Cavg.
C4 C5 C6
Note: C4, C5
and C6 equal
zero inches.
Vehicle Crush Profile
Cavg = C1+C2+C3+0+0+0
6
Derivation of the Kinetic Energy
Equation
The work done on an object by a force is
defined as the product of the magnitude of
the force and the distance that the object
moves in the direction of the force:
Work = Fd (1)
The concept of work is that work represents
the amount of energy transformed from one
form to another. Therefore, work and
energy have different appearances for the
same thing and are both measured in foot
pounds.
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