Analysis of Four Staged Crashes of Passenger Vehicles into a Semi

Institute of Police Technology and Management

Analysis of Four Staged Crashes of

Passenger Vehicles into a

Semi-Trailer

Jeremy Daily

Russell Strickland

Special Problems in Traffic Crash Reconstruction

25-29 April 2005

1 Introduction

1.1 Purpose

The purpose of this report is to show the details of traffic crash reconstruction tech-

niques as applied to crashes between commercial vehicles and passenger vehicles.

There were four test conducted: two offset under-ride collisions and two crashes

into the rear duals of a stationary semitrailer. These tests were instrumented which

enables verification of the reconstruction. However, the authors did not use the in-

strumented data until after the reconstruction was finished. This removed the temp-

tation to “fudge” numbers to make the reconstruction fit the data. Furthermore, the

1

1 Introduction

use of some assumptions can be verified because we have known speeds from radar

trace of the bullet vehicle going into the target.

1.2 Testing Procedure

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 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.

2 c©2005 Jeremy Daily and Russell Strickland


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

c©2005 Jeremy Daily and Russell Strickland 3

1 Introduction

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



Figure 3: Scene photo of the Nissan into the rear of the trailer. Notice the two unitsdid not stick together.


1 Introduction

Figure 4: Scene photo of the Honda into the rear of the trailer. Notice the lack ofdamage to the rear duals of the trailer.


In all the tests, the a cable broke at the junction between the steel cable and the

polypropylene ski rope. The impulse required to snap the rope has no measurable

effect on the collision.

The next section presents a review of the physics used in the reconstruction of the

aforementioned collisions. A list of all the evidence should provide guidance for the

investigating accident reconstructionist. As with any collision analysis, the evidence

is the single most important factor in determining the success of the outcome of the

reconstruction. Following the reconstruction, the instrumented data is presented to

validate some of the assumptions we make during a crash reconstruction.

2 Reconstruction of Staged Crashes

2.1 Evidence Required

2.1.1 Drag Factor For Tractor Trailer

The surface of the test site was level. Skid tests with passenger vehicles give a drag

factor of 0.75 (±0.05). Since the tractor trailer has commercial vehicle tires, the drag

factor is reduced by a factor of 0.80 to compute a drag factor of

fCV = n fcar = 0.80(0.75) = 0.60

where n is the commercial vehicle tire factor.

This drag factor for the truck was verified by performing instrumented skid tests

from 30 mph with the same tractor trailer unit on the same surface of the crash tests.

The drag factor was computed using

f =S2

30d

Where d is the skid distance that includes the shadow marks. Speed was estimated

from the speedometer (undesirable) or integrated from an accelerometer trace when

available. The data are summarized in Table 1. A prudent range to use in the drag

factor for the rear tires would be from 0.55 to 0.65 based on these tests.



Table 1: Drag factor tests for the tractor-trailer rig.

Run Number Speed (mph) Distance (ft) Computed f Measured f ABS

1 30 50 0.60 NA Yes2 30 52 0.57 NA Yes3 27.5 45 0.56 0.523 No4 31.5 51.5 0.64 0.472 No

Note: The stopping was violent while riding in the cab and the deceleration pulsewas very noisy

Table 2: The weights of all the bullet vehicles. The bottom two weight are providedto show the difference between the published curb weight an the as testedweights. The DOT scales were not available for the first two vehicles.

Vehicle Weight (lbs) Source

Voyager 3046 Expert AutoStats (curb weight)Sentra 2408 Expert AutoStats (curb weight)Accord 2900 DOT portable scales

Cherokee 3350 DOT portable scales

Accord 2778 Expert AutoStats (curb weight)Cherokee 3554 Expert AutoStats (curb weight)

2.1.2 Vehicle Weights

The bullet vehicle weights are shown in Table 2.

The tractor trailer was measured using portable scales from the Department of Trans-

portation (DOT). The raw data are summarized in Table 3. Only four scales were

available so the weight of the trailer had to be determined by resting the last axle on

a “dummy” surface of similar thickness.

2.1.3 At Scene Measurements

The most difficult piece of information to obtain from the scene is the displacement

of the tractor trailer. In the case of an in-line collision, the markings from the tires

will be very faint. If the wheels rotate, (which they may) then speckling may exist

on the tire which indicates how much it rolled. Keep in mind this speckling is short

lived and is gone after the tire rolls one complete revolution.



Table 3: Tractor trailer weights in pounds. Notice the weight distribution of thetrailer from left to right is not centered.

Tractor Trailer Combination

Axle Number 1 2 3 4 Total

Right Side 4500 4400 2950 2200 14,050Left Side 4500 4900 2500 1900 13,800

Total 9000 9300 5450 4100 27,850

Trailer Weight Only

Axle Number Landing 3 4 Total

Right Side 3300 2750 Dummy 8100Left Side 1100 2400 Dummy 5325

Total 4400 5150 3875 13,425

The total weight on axle #4 was determined by subtracting a proportional amountfrom the combination configuration because only four scales were available. Theproportionality was determined as

ratio =4100

5450= .75

Since the landing gear takes more load than the tractor, the weight on the rear axlesis reduced when the landing gear is lowered. The weight lost on axle #3 was 300pounds, thus the weight lost on axle # 4 would be 300(0.75) = 225.



Figure 5: This photo illustrates the markings left by the rear duals on the pavement.The mark existed under the trailer and gives an amount the rear axles weredisplaced.



For our crash test, the wheel positions were marked before and after the collision.

This enables us to accurately record the actual tire displacements of the tractor-trailer.

In the side collisions, the best evidence will be under the trailer as created by the

opposite duals as shown in Figure 5. The same types of evidence should exist for

smaller scale collisions, such as a motorcycle into a pickup. The initial and final

positions of the trailer are important for the reconstruction.

2.1.4 Physical Properties of the Trailer

We need to use the weights, along with the physical dimensions, to determine the

location of the center of mass. The height of the center of mass is not relevant to our

analysis. All calculations assume planar motion with no slope in relation to gravity.

Center of Mass of the Trailer The center of mass of any object is computed using

the formula

dCM =∑

ni=1 diwi

wTotal

where n is the number of weight forces

dCM is the distance the center of mass is from from a reference point

di is the perpendicular distance the ith force (wi) is acting from the refer-

ence point

wTotal is the total weight

The kingpin location was the reference point. The weight measurements are shown

in Table 3 and the dimensions are shown in Figure 6. The distance the center of mass

of the trailer is located behind the kingpin is calculated as:

d =9(4400) + 34.5(5150) + 38.5(3875)

13, 425

= 27.3 feet

The center of mass is not in the center of the trailer as one may expect. The lateral

location of the center of mass is determined in reference to the center of the left duals



Figure 6: A side view of the trailer used in the crash tests.

as shown in Figure 7. The formula to determine the distance the center of mass is

from the left duals is

dL = tw

(wR

wT

)

where tw = dL + dR is the track width. Substituting the data from Table 3 along with

a 68 inch track width for the trailer gives:

dL = 68

(5325

8100

)

= 44.7 inches

Yaw Moment of Inertia There are few ways of determining the mass moment of

inertia. The first, and easiest, is to look up the value in a database. The following

information is from Fancher, P.S., et al., DOT HS 807 125, “A Factbook of the Mechanical

Properties of the Components for Single Unit and Articulated Heavy Trucks”, December

1986:

IG = 110, 739 lb-ft-sec2

for an empty 48 ft box van trailer that weighs 13,800 lbs. The subscript G indicates

the moment of inertia assumes rotation about the center of gravity.

If the location of rotation is different from the center of mass, then, by the parallel axis



Figure 7: A rear view of the trailer used to determine its center of mass.



Figure 8: A top view of the trailer used to determine its center of mass.

theorem, the yaw moment of inertia changes to

I = IG + mh2

where I is the yaw moment of inertia about the point of rotation

IG is the mass moment of inertia about the center of mass

m is the mass of the object

h is the distance from the center of mass to the point of rotation

The center of mass of the trailer is slightly off center, but not enough to make a dif-

ference when computing the value of h using the Pythagorean Theorem. The value

of h, as seen in Figure 8, is 27.3 feet. So, by the parallel axis theorem,

I = IG +

(wT

g

)

h2

= 110, 739 +

(13, 425

32.2

)

27.32

= 421, 470 lb-ft-sec2

The next way to determine the mass moment of inertia is by assuming simple shapes

with uniform density. The individual objects each have a simple formula for the yaw

moment of inertia and the yaw moment of inertia for the composite structure can be

determined by using the parallel axis theorem. This technique requires practice and

time to become comfortable with the calculations.

The most reliable way (and expensive) it to measure the dynamic response of the


2.2 Under-ride Analysis Using Momentum

Fc1Fc2

Unit 2 Unit 1

m1a1m2a2

Fg Positive Direction

Figure 9: A free body diagram of a vehicle #1 striking vehicle #2. The forces in-clude the ground force, collision forces and inertial forces. Let positive bedefined as to the right. Unit 1 is left bound and has a negative direction.

vehicle on a rotating table. These types of tests are conducted by large laboratories

and the results are usually published.


2.2.1 Free Body Diagram

An in-line momentum analysis begins with a proper construction of a free body dia-

gram. This diagram should represent the forces present during the event (crash). A

simple free body diagram of a vehicle striking another vehicle is shown in Figure 9.

2.2.2 Derivation of Governing Equations

Let’s break down the free body diagram for each vehicle shown in Figure 9. The

stationary vehicle is denoted as #2 and the striking vehicle is denoted as #1. The

dynamic equation of motion for vehicle #1 is determined by summing the forces

during a collision according to Newton’s Second Law:

0 = Fc1 − m1a1 (1)

where Fc is the collision force. We will assume no other external forces are acting on

Unit 1. Similarly, sum the forces for Unit 2:

0 = Fg + m2a2 − Fc2 (2)



where Fg is the ground friction force.

We know from Newton’s Third Law that the collision forces are equal and opposite:

−Fc1 = Fc2 (3)

Solving eq. (1) for Fc1 and eq. (2) for Fc2 gives:

Fc1 = m1a1 (4)

Fc2 = Fg + m2a2 (5)

Substituting eq. (4) and (5) into eq. (3) leads to the following equation:

−m1a1 = m2a2 + Fg (6)

where Unit #1 is the bullet car

Unit #2 is the Tractor Trailer Combination

Fg is the ground force forces from the braked tractor-trailer

Recall from physics that acceleration is a change in velocity divided by a change in

time:

a =∆v

∆t(7)

Substitute eq. (7) into eq. (6) to get:

−m∆v1

∆t= m2

∆v2

∆t+ Fg (8)

Multiply through by ∆t to get:

−m1∆v1 = m2∆v2 + Fg∆t (9)

Our definition of a positive ∆v is as follows:

∆v1 = v1 − v3

∆v2 = v2 − v4



where v3 is the post impact velocity of Unit 1

v1 is the pre impact velocity of Unit 1

v4 is the post impact velocity of Unit 2

v2 is the pre impact velocity of Unit 2

Substitute these values into eq. (9):

−m1(v1 − v3) = m2(v2 − v4) + Fg∆t

−m1v1 + m1v3 = m2v2 − m2v4 + Fg∆t

Multiply through by −1:

m1v1 − m1v3 = −m2v2 + m2v4 − Fg∆t

Move all the pre impact momentum values of the left hand side of the equation:

m1v1 + m2v2︸︷︷︸

Pin

= m2v4 + m1v3︸︷︷︸

Pout

− Fg∆t︸︷︷︸

Impulse

(10)

Which says the momentum before the collision is equal to the momentum after the

collision less what is lost due to the impulse.

If there is no significant impulse, which is often the case, then we have the principle

of the Conservation of Linear Momentum. Notice the conservation of linear momen-

tum is not a law of physics, it is based on the laws of physics.

The impulse from ground force is determined using the following equation:

Fg∆t = w2n f ∆t (11)

where f is the drag factor

w2 is the weight of vehicle 2

∆t is the duration of the crash

n is the braking efficiency of the vehicle.

Typically ∆t values are on the order of 100 to 150 milliseconds. This is an assumption

to be examined with the instrumented crashes.



The post impact velocities are determined by the work-energy theorem which re-

duces to:

S =√

64.4d f (fps) (12)

for the US system and

S =√

18.2d f (mps) (13)

for the SI system. The drag factor is determined as

f = µn + m (14)

where µ is the coefficient of friction, n is the braking efficiency and m is the slope of

the surface.

Keep in mind that direction does matter when assigning values to v3 and v4. Also,

since we are combining momentum and impulse, special care must be taken to use

consistent units. This is why we aren’t using S =√

30d f

2.2.3 Application to Under-ride Crashes

Since the tractor-trailer was not being operated, its spring brakes were set. This

means all but the front axles are being braked and the braking efficiency can be de-

termined as

n =27, 850 − 9000

27, 850= 0.677

We’ll assume the spring brakes apply maximum force and the drag factor for the

tractor trailer is between 0.55 and 0.65. This means the ground force of the tractor

trailer combination is

Fg = wn f



Low f High f

Fg = 27, 850(0.677)(0.55)

= 10, 370 lb

Fg = 27, 850(0.677)(0.65)

= 12, 255 lb

The post impact velocities are equal because the vehicles stuck together. Therefore:

v3 = v4

The magnitude of their velocity is determined from eq. (12) and the direction is

assigned according to the desired coordinate system. For this analysis the positive

direction is the same as in Figure 9.1

Since the tractor-trailer was sitting still, v2 = 0. We can incorporate these simplifica-

tions into eq. (10) and solve for v1:

v1 =(m2 + m1)v3 − Fg∆t

m1(15)

Convert mass to weight by substituting m = wg and simplifying:

v1 =(w2 + w1)v3 − Fgg∆t

w1(16)

Predicted Speed for the Jeep

The mass ratio of the trailer to the Jeep is

mass ratio =27, 850

3350= 8.3

1This choice was intentional to emphasize to the reader the importance of keeping track of direction(minus signs).



The Jeep displaced the tractor-trailer 1.5 ft. Hence, the post impact velocity can be

computed using a low and high drag factor:

Low f High f

v3 = −

√

64.4(1.5)(0.677)(0.55)

= −6.00 fps

v3 = −

√

64.4(1.5)(0.677)(0.65)

= −6.52 fps

The post impact velocity is negative (left bound) according to our positive sign con-

vention.2

Substitute the appropriate values for the Jeep striking the tractor trailer:

Low f , No ∆t:

v1 =(27, 850 + 3350)(−6.00)− 10, 370(32.2)(0)

3350

v1 = −55.88 fps

S1 = 38.1 mph

High f , High ∆t:

v1 =(27, 850 + 3350)(−6.52)− 12, 255(32.2)(0.100)

3350

v1 = −72.5 fps

S1 = 49.4 mph

2Accelerometer data suggest v3 = 4.7 fps is a better number for the post impact speed of the tractor-trailer.



Both speeds are in the negative direction. Our predicted speeds based on ranging

the drag factor for the truck and the magnitude of the external impulse are between

38 and 49 mph.3

Predicted Speed for the Minivan

The mass ratio of the minivan to the tractor-trailer is:

mass ratio =27, 850

3046= 9.14

For this case, we assume there is some uncertainty concerning the post impact dis-

tance moved by the tractor trailer. From photographic evidence and witness state-

ments we assume a post impact distance traveled between 0.8 ft an 1.4 ft. When

using these values for distance we get a post impact range as define below:

Low f , Low d High f , High d

v3 = −√

64.4n f d

= −

√

64.4(0.677)(.55)(0.8)

= −4.37 fps

v3 = −√

64.4n f d

= −

√

64.4(0.677)(.65)(1.4)

= −6.30 fps

Substitute the appropriate values for the minivan striking the tractor trailer:

Low f , Low d, and no Impulse:

3The collision time from the accelerometer traces indicate an impact time duration of around 300msec for an underride collision.



v1 =(27, 850 + 3046)(−4.37)

3046

v1 = −44.32 fps

S1 = 30.01 mph in the negative direction

High f , High d, and High ∆t:

v1 =(27, 850 + 3046(−6.30) + 12, 255(32.2)(0.100)

3046

v1 = −76.85 fps

S1 = 52.40 mph in the negative direction

Both speeds are in the negative direction. Our predicted speeds based on ranging

the drag factor for the truck, the presence of ground forces and the duration of the

crash are between 30 and 52 mph.

2.2.4 Sensitivity Analysis of In-line Momentum

The analysis of the in-line momentum cases with the high mass ratios reveal that the

small range in exit speed, v3, lead to a large change in the impact speed ,v1.

The sensitivity of an output to a change to an input is determined by its first deriva-

tive. The derivatives of eq. (16) with respect to all of the input variables are as

follows:dv1

dw1=

v3

w1−

(w1 + w2)v3 − FGg∆t

w21

dv1

dw2=

v3

w1

dv1

dv3=

(w1 + w2)

w1

dv1

dFg= −

g∆t

w1


2.3 Eccentric Collision Analysis Using Rotational Mechanics

dv1

d∆t= −

gFg

w1

These formulas provide a tool to analytically determine how sensitive each input

variable is to the resulting output (v1). Substitution of the values used in the analysis

herein would reveal that the v1 determination is sensitive to ∆t, w1 and v3. Therefore,

any uncertainty in these values is amplified in the determination of the impact speed.

This is why there is a large range for v1 for a normal range of valid drag factors.

2.3 Eccentric Collision Analysis Using Rotational Mechani cs

When a bullet vehicle strikes a semi-trailer at near right angles and rotates the trailer,

we can determine the ∆v of the bullet vehicle by using the concepts of rotational

mechanics. Let us begin with Newton’s Second Law for rotation:

τ = Iα (17)

where τ (tau) is the torque applied to a body (equivalent to force)

I is the effective mass moment of inertia of the body (equivalent to mass)

α is the angular acceleration in radians per second per second (equivalent

to acceleration).

The radian is a unit of angular measure and there are 180/π = 57.29 degrees per

radian.

Newton’s Third Law for rotation says that every applied torque is matched by an

equal but opposite torque. Using these two laws, we can derive the principle of the

conservation of angular momentum.

Consider a collision between a ball and a board as shown in Figure 10 on the follow-

ing page. When the ball strikes the board, it gives the board a linear impulse. This

linear impulse acts at a distance away from a fulcrum (pivot point) and becomes an

angular impulse, τ∆t which acts to rotate an object. Since this board is free to rotate,

it accelerates according to Newton’s Second Law for rotation. The ball under goes a

change in velocity and its initial impact speed can be determined using the concept

of the coefficient of restitution. Let’s see how this is done mathematically.



bPivot Pointτ = Iα

Fc1

Fc2

h

F = ma

Figure 10: A ball strikes a board with an initial velocity and creates a torque aboutthe pivot point.

2.3.1 Derivation of ∆v from Rotational Mechanics Concepts

Begin by separating the individual bodies from the system and analysing their free

body diagram. For the ball, the relation holds that

Fc1 = ma (18)

which says the force imparted on the ball by the board is equal to the inertial force

of the ball.

The moments (or torques) about the pivot point are equal to the angular inertia. The

torque is equal to the product of the force the ball imparts on the board and the length

of the lever arm used by the applied force:

Iα = hFc2 (19)

which means:

Fc2 =Iα

h



By Newton’s Third Law:

−Fc1 = Fc2 (20)

Making the substitution of eqs. (18) and (19) into eq. (20) gives:

−ma =Iα

h(21)

Substitute a = ∆v∆t into eq. (21) to get:

−m∆v

∆t=

Iα

h(22)

An analogous substitution can be made for the angular acceleration:

α =∆ω

∆t(23)

where ω (omega) is the angular velocity in radians per second. Continue by substi-

tuting the value for α into eq. (22):

−m∆v

∆t=

I∆ω

h∆t(24)

Multiplying both sides of the equation by h∆t yields:

−mh∆v = I∆ω (25)

The definition of ∆ω is similar to ∆v:

∆ω = ωi − ω f (26)

where ω f is the post impact angular velocity of the board

ωi is the pre impact angular velocity of the board. In this case, wi = 0.

Since the initial angular velocity is zero, we will let ω f = ω so ∆ω = −ω. Continue

by substituting this simplification into eq. (25):

−mh∆v = −Iω



Multiply though by −1 to get:

mh∆v = Iω

Now solve for ∆v:

∆v =Iω

mh(27)

This result says we can determine the speed change of the ball if we can determine

the angular velocity of the board after the impact. We will extend this same concept

to vehicle hitting the side of a trailer.

2.3.2 The Work–Energy Theorem for Rotation

The work–energy theorem says that the work done by a torque comes from kinetic

energy:

KE = Work (28)

The kinetic energy of a rotating object is:

KE =1

2Iω2 (29)

The work done by a torque is:

Work = τθ (30)

where τ is the torque and θ (theta) is the rotational displacement in radians. Also the

definition of torque is:

τ = Fh (31)

where F is a force and h is the lever arm for the force. If we substitute eq. (31) into

eq. (30) and invoke eq. (28), then we get:

1

2Iω2 = Fhθ (32)

Solving for ω gives:

ω =

√

2Fhθ

I(33)



2.3.3 Application to Side Impact Collisions

The evidence of the scene indicates the tractor did not move during the collision. This

means the kingpin can be considered rigid. As a result, the rotation of the trailer oc-

curs around the kingpin and all measurements are referenced to this point as shown

in Figure 6. Recall from Section 2.1.4 the mass moment of inertia of the trailer about

its kingpin is 421,470 ft-lb-sec2. The weight on the rear dual axles is 9550 lbs.

The force used in eq. (33) comes from the friction force of the rear duals (axles 3

and 4). The lever arm for this frictional force is h = 36.4 ft. The tires are sliding

perpendicular to their rolling direction so they are producing the maximum available

force. We can use a lower friction coefficient as an upper value of µL = 0.55 and

µH = 0.65 which give us two distinct values for Fh:

FLh = µLw3,4h

= 0.55(9550)(36.4)

= 191, 191 lb-ft

FHh = µHw3,4h

= 0.65(9550)(36.4)

= 225, 953 lb-ft

All values determined so far are specific to only the trailer.

Nissan Sentra Collision The mass of the Nissan Sentra is computed from the def-

inition of weight:

m =w

g=

2408

32.2= 74.78 slugs

The angle of rotation is determined using the definition of tangent:

θ = tan−1

(rise

run

)



Figure 11: The pre impact and post impact scene diagrams superimposed to get theangle of rotation for the Nissan crash.

For the Nissan Crash, the distance the front duals moved was 39 inches (3.25 ft) and

the distance from the kingpin to the front duals is 34.5 ft. Using these values gives

the rotation of the trailer as:

θ = tan−1

(3.25

34.5

)

= 5.38◦

The angular values must be in radians, thus:

θ = 5.38( π

180

)

= 0.0939 rad

Substitute the appropriate values into eq. (33):

ωL =

√

2(191, 191)(0.0939)

421, 470= 0.292 rad/s

ωH =

√

2(225, 953)(0.0939)

421, 470= 0.317 rad/s

These values are now used in eq. (27) to compute the ∆v of the Nissan:

∆vL =421, 470(0.292)

36.4(74.78)= 45.21 ft/s



Figure 12: The pre impact and post impact scene diagrams superimposed to get theangle of rotation for the Honda crash.

∆vH =421, 470(0.317)

36.4(74.78)= 49.08 ft/s

Honda Accord Collision The mass of the Accord is computed from the definition

of weight:

m =w

g=

2900

32.2= 90.01 slugs

For the Honda Crash, the distance the front duals moved was 31 inches (2.58 ft) and

the distance from the kingpin to the front duals is 34.5 ft. Using these values gives

the rotation of the trailer as:

θ = tan−1

(2.58

34.5

)

= 4.27◦

Also, the angle of rotation as measured from the scale drawing in Figure 12 gives

4.27 degrees, which agrees with our hand calculations. The angular values must be

in radians, thus:

θ = 4.27( π

180

)

= 0.0746 rad

Substitute the appropriate values into eq. (33):

ωL =

√

2(191, 191)(0.0746)

421, 470= 0.260 rad/s



ωH =

√

2(225, 953)(0.0746)

421, 470= 0.283 rad/s

These values are now used in eq. (27) to compute the ∆v of the Honda:

∆vL =421, 470(0.260)

36.4(90.01)= 33.48 ft/s

∆vH =421, 470(0.283)

36.4(90.01)= 36.40 ft/s

2.3.4 Computing Impact Speed from ∆v

In order to compute the impact speed from ∆v we must define the coefficient of

restitution as:

ǫ =v4 − v3

v1 − v2(34)

The coefficient of restitution is the ratio of the relative velocity after the collision to

the relative velocity before the collision. It has a value from 0 to 1. For some low

speed impacts, the coefficient of restitution can be as high as 0.3. As the speeds get

larger, the coefficient of restitution drops. We will assume a value (on the high side)

of 0.1. A higher value for the coefficient of restitution leads to a lower impact speed.

We also have a definition of ∆v that is:

∆v = v1 − v3

or

v3 = v1 − ∆v

Let v2 = 0 and substitute the expression for v3 into eq. (34):

ǫ =v4 − (v1 − ∆v)

v1 − 0



Solve for v1:

ǫv1 = v4 − v1 + ∆v

v1 + ǫv1 = v4 + ∆v

v1(1 + ǫ) = v4 + ∆v

v1 =v4 + ∆v

1 + ǫ(35)

The value of v4 is determined from the kinematics of rotating bodies:

v4 = rω (36)

where r is the distance between the centroid of the collision and the point of rotation.

The variable of ω describes the post impact angular velocity and is determined using

eq. (33). Substitute eq. (36) into eq. (35) to get:

v1 =rω + ∆v

1 + ǫ(37)

Impact Speed of the Nissan The value for r is the distance between the kingpin

and the center of the rear duals, 36.4 feet. We can see from eq. (37) that any restitution

decreases the impact velocity. Thus to get the high end of the range of impact speeds,

use no restitution and the high values of ω and ∆v:

v1H =36.4(0.317) + 49.08

1 + 0

= 60.61 ft/s

S1H = 41.33 mph

We know that some restitution exists, but not much. Let’s assume ǫ = 0.1. Also, we

will use the low values for ∆v and ω:


3 Crash Data Analysis and Conclusions

v1L =36.4(0.292) + 45.21

1 + 0.1

= 50.76 ft/s

S1L = 34.61 mph

Our reconstructed impact speed range for the Nissan is between 34 and 41 mph.

Impact Speed of the Honda The same impact area and trailer was hit by the

Honda. Again, to get the high end of the range of impact speeds, use no restitu-

tion and the high values of ω and ∆v:

v1H =36.4(0.283) + 36.40

1 + 0

= 46.70 ft/s

S1H = 31.81 mph

We know that some restitution exists, but not much. Let’s assume ǫ = 0.1. Also, we

will use the low values for ∆v and ω:

v1L =36.4(0.260) + 33.48

1 + 0.1

= 39.04 ft/s

S1L = 26.63 mph

Our reconstructed impact speed range for the Honda is between 27 and 32 mph.


Table 4 shows a comparison of the reconstructed speeds with the actual speeds mea-

sured by a RADAR system. The two under-ride collisions reveal problems when


Table 4: Comparison of speed estimates using reconstruction techniques to mea-sured speeds in miles per hour.

Bullet Test Vehicle Lower Bound Upper Bound Actual Impact Speed

Jeep 38 49 37Voyager 30 52 39Nissan 34 41 39Honda 27 32 31

trying to predict impact speeds using momentum concepts when the vehicle mass

ratios are high. Even though inclusion of external impulses is physically more cor-

rect, it tends to make the speed estimates less conservative (higher). Furthermore,

the impulse forces are difficult to determine, especially when little evidence exists.

It is not recommended to use a momentum analysis in court when high mass ratios

exist.

The impact analysis using rotational mechanics concepts proved to be accurate. More-

over, 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 gath-

ered at the scene, provided the on-scene investigator is trained to look for evidence

under the trailer.

Another assumption we were able to analyze using instrumentation was the dura-

tion of the crash. Some approximate values are shown in Table 5. The time of the

impact in an under-ride collision is about three times larger than a “normal” crash.

This means the average acceleration would be lower for a given ∆v which would

tend to indicate that passengers in the back seat could possibly survive such crashes

(if they wear seat belts). Furthermore, the effect of external forces become more pro-

nounced since ∆t is larger.

Detailed analysis of the high speed video could reveal parameters such as the coeffi-

cient of restitution, the duration of the crash, and the mechanisms of force generation

during a crash. These values would be helpful in further understanding the physics

of crashes.

Crush measurements were taken of the cars involved in the side impact crashes.

The analysis of the ∆v using crush is left as an exercise for the reader. The crush



Table 5: Measure of collision times in milliseconds.

Bullet Test Vehicle Internal Vericom

Jeep 300Voyager 320Nissan 100Honda 110

measurements are included in the appendix.

Acknowledgements

We would like to thank David King and Jonathan Lawrence from MacInnis Engineer-

ing for their successful execution of a well thought out testing plan. Thanks also to

all the people helping conduct the crash testing who include, but are not limited to:

David Brill, Walter Dobson, John Daily, Nate Shigemura, Bill Wright, Wade Bartlett,

Tony Becker, the mapping team (PSFM) and the high speed photography crew.


Figure 13: The crush measurements for the Nissan Sentra



Figure 14: The crush measurements for the Honda Accord


Documents

Analysis of Four Staged Crashes of Passenger Vehicles into a Semi