using simultaneous equations to solve inline collisions · inline collisions solved using...

Inline Collisions Solved Using

Simultaneous Equations

Nathan Shigemura

Traffic Safety Group, LLC Name

Inline Collisions Solved Using Simultaneous

Equations

Institute of Police Technology and Management’s

2019 Symposium on Traffic Safety

© 2015 N. Shigemura, J. Daily, R. Strickland 1

Presented By:

Nathan Shigemura

Traffic Safety Group, LLC

IPTM Adjunct Faculty

Institute of Police Technology and Management’sSymposium on Traffic Safety

Orlando, FLJune 3 - 6, 2019

© 2015 N. Shigemura, J. Daily, R. Strickland1

� When reconstructing a traffic crash, one of the aspics that is usually determined is the speeds of the colliding vehicles.

� A tool commonly used to find impact speeds is the principle of Conservation of Linear Momentum (COLM).

� Using COLM to find impact speeds in collision with relatively widely spaced approach angles generally yields good results.


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� However, when analyzing inline collisions with COLM the results may not be as definitive as desired.

� At times, all that can be obtained is a wide range of impact speeds for each vehicle.

� Another tool used to analyze traffic crashes is the principle of Conservation of Energy (COE).

� However, as with COLM, the results may be as less than desired.


� The two methods commonly used to find impact speeds of colliding vehicles:

� Conservation of Linear Momentum

� Conservation of Energy

1 1 2 2 1 3 2 4m v m v m v m v+ = +

1 2 3 4 otherE E E E E+ = + +


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� Both COLM and COE share the same difficulty of being one equation containing two unknowns (the impact speeds) which inherently prohibits the solving of either equation unless one of the unknowns is available.

� However, by utilizing the mathematical procedure of simultaneous equations the two unknowns (impact speeds) can be found using the COLM and COE equations.


� This is successful because there are now two equations and two unknowns.

� During this session we will discuss COLM, COE and the techniques to solve simultaneous equations and use those techniques to solve an inline collision.


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� Conservation of Linear Momentum states that the of the total momentum (vector sum) of two colliding bodies after the collision is the same as the total momentum (vector sum) of two colliding bodies before the collision.

Initial momentum = Final momentum


= the pre-collision momentum of Unit #1,

= the pre-collision momentum of Unit #2,

= the post-collision momentum of Unit #1,

= the post-collision momentum of Unit #2.

1 1 2 2 1 3 2 4m v m v m v m v+ = +

1 1m v

1 3m v

2 2m v

2 4m v


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Since:x

x

wm

g=

1 1 2 2 1 3 2 4

1 2 1 21 2 3 4

1 1 2 2 1 3 2 4

m v m v m v m v

w w w wv v v v

g g g g

w v w v w v w v

+ = +

+ = +

+ = +


� Since momentum is a vector quantity we must also consider direction.

x-direction components:

1 1 2 2 1 3 2 4cos cos cos cosw v w v w v w vα ψ θ φ+ = +

y-direction components:

1 1 2 2 1 3 2 4sin sin sin sinw v w v w v w vα ψ θ φ+ = +

α = the approach angle of Unit #1,

ψ = the approach angle of Unit #2,

θ = the departure angle of Unit #1,

φ = the departure angle of Unit #2.

Where :


w1 & w2 = the weights of Units #1 & #2 respectively,

v1 & v2 = the impact speeds of Units #1 & #2 respectively,

v3 & v4 = the departure speeds of Units #1 & #2 respectively,

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Since the approach vector of Unit #1 is commonly placed on the x-axis (but is not a requirement) the approach angle of Unit #1, α, is 0o, and cos α = 1, and sin α = 0. Thus,

1 1 2 2 1 3 2 4

1 1 2 2 1 3 2 4

1 1 2 2 1 3 2 4

cos cos cos cos

(1) cos cos cos

cos cos cos

w v w v w v w v

w v w v w v w v

w v w v w v w v

α ψ θ φ

ψ θ φ

ψ θ φ

+ = +

+ = +

+ = +

x-direction components:

1 1 2 2 1 3 2 4

1 1 2 2 1 3 2 4

2 2 1 3 2 4

sin sin sin sin

(0) sin sin sin

sin sin sin

w v w v w v w v

w v w v w v w v

w v w v w v

α ψ θ φ

ψ θ φ

ψ θ φ

+ = +

+ = +

= +

y-direction components:


� Solving the y-direction component equation for v2

yields:

� Solving the x-direction component equation for v1

yields:

� Note that v2 must be solved for first since it appears as a variable in the v1 equation.

1 3 42

2

sin sin

sin sin

w v vv

w

θ φ

ψ ψ= +

2 4 2 21 3

1 1

cos coscos

w v w vv v

w w

φ ψθ= + −


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� COLM works well with planar collisions.� However, with in-line collisions both

approach and departure angles are in the x-direction.

� This causes all sine values (y-direction components) to equal 0 which eliminates the v2 equation and prevents v2 from being calculated.

� Therefore, the impact speed of one of the vehicles must be known and assigned as v2which will then allow the impact speed for v1to be calculated.


� A two-vehicle, head-on collision occurs killing both drivers.

� Unit #1 is a brown 1986 Chrysler LeBaron GTS, 4-door passenger car with a weight of 2601 pounds. Unit #2 is a white 1988 Ford Escort GL, 4-door passenger car with a weight of 2222 pounds.

� The LeBaron, was proceeding southbound on a two-lane US highway, crossed over the centerline, and struck the Escort, which was proceeding northbound. The collision occurred in the northbound lane.

� Witness statements indicate that the crossover action of the LeBaron was gradual, not jerky or sudden.


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� No evidence of pre-impact evasive action was found at the scene or noted by witnesses.

� The crash occurred in the early afternoon and the weather was clear and dry.

� The impact halted the forward progress of the Escort and drove it rearward approximately 10 feet 9 inches. The Escort rotated clockwise from the impact.

� The LeBaron traveled approximately 19 feet after impact. Both vehicles’ front wheels were locked due to damage after impact.

� The roadway was asphalt with a coefficient of friction of 0.72.


� LeBaron is vehicle #1 with approach at 0 degrees.

� Escort is vehicle #2 with approach at 180 degrees.

� Departure angle for both vehicles is 0 degrees.

� Departure speed for LeBaron, v3, is calculated at 15.69 mph.

� Departure speed for Escort, v4, is calculated at 11.80 mph.


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� A range of impact speeds was chosen for the Escort: v2 = 40 mph - 80 mph.

� A corresponding impact speed was then calculated for each impact speed value of the Escort.

Escort impact speed (mph)

LeBaron impact speed (mph)

40 59.94

45 64.21

50 68.48

55 72.75

60 77.02

65 81.29

70 85.57

75 89.94

80 94.11


� As can be seen, the calculated range of impact speeds for the LeBaron is extremely wide and provides little insight to the crash.

� The most that can be said is that if the Escort was traveling x-mph at impact then the LeBaron was traveling y-mph at impact.

� Thus, without knowing the pre-impact speed of one of the vehicles we cannot determine a definitive pre-impact speed of the other.


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� Since COLM isn’t providing very usable results, let’s explore a different method.

� Conservation of Energy.


� Conservation of Energy (COE) is a physics principle that states:� The total energy is constant.� Energy can be neither created nor destroyed, but can

be transformed from one form to another.

� Thus, when two bodies collide, the total energy leaving the collision is equal to the total energy entering the collision.

Total pre-collision kinetic energy = Total post-collision energy


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� In an elastic collision� No damage is sustained by the colliding bodies.

� Most of the energy will leave the collision in the form of motion.

� A small amount of energy will leave in other forms such as heat and sound.

1 2 3 4 otherE E E E E+ = + +

E1 = the pre-collision kinetic energy of Unit #1,


E3 = the post-collision kinetic energy of Unit #1,


Eother = other post-collision energy such as heat and sound .

Where :


� In an inelastic collision

� Damage is sustained by the colliding bodies.

� Some of the energy will leave the collision in the form of motion.

� Some of the energy will be used in damaging the bodies.

� A small amount of energy will leave in other forms such as heat and sound.

1 2 3 4 1 2crush crush otherE E E E E E E+ = + + + +





Ecrush1 = the energy it took to damage Unit #1,

Ecrush2 = the energy it took to damage Unit #2,

Eother = other post-collision energy such as heat and sound .

Where :

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� Since the vehicles are moving, the energy E, is kinetic energy.

� The kinetic energy that is used to do the work that damages the vehicles is large compared to any other form of energy dissipated in the collision itself. Therefore Eother can be ignored.

� We can substitute the kinetic energy equation into equation :

� Resulting in:

1 2 3 4 1 2crush crush otherE E E E E E E+ = + + + +

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2Ke mv=

2 2 2 2 2 2

1 2 3 4 1 2

1 1 1 1 1 1

2 2 2 2 2 2crush crush

mv mv mv mv mv mv+ = + + +

24© 2015 N. Shigemura, J. Daily, R. Strickland

Since:x

x

wm

g=

2 2 2 2 2 2

1 2 3 4 1 2

2 2 2 2 2 21 2 1 2 1 21 2 3 4 1 2

1 1 1 1 1 1

2 2 2 2 2 2

1 1 1 1 1 1

2 2 2 2 2 2

crush crush

crush crush

mv mv mv mv mv mv

w w w w w wv v v v v v

g g g g g g

+ = + + +

+ = + + +

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Can now cancel common terms and obtain::

2 2 2 2 2 2

1 1 2 2 1 3 2 4 1 1 2 2crush crushw v w v w v w v w v w v+ = + + +

Where : w1 & w2 = the weights of Units #1 & #2 respectively,

v1 & v2 = the impact speeds of Units #1 & #2 respectively,

v3 & v4 = the departure speeds of Units #1 & #2 respectively,

vcrush1 & vcrush2 = the energy-equivalent crush speeds of Units #1 & #2 respectively


� We will continue with our crash example using the LeBaron and the Escort.

� Crush measurements (“C” measurements) were taken of both vehicles.

� These measurements, along with the stiffness values for each vehicle, will allow damage analyses to be made that will determine the damage energy for each vehicle and the subsequent energy-equivalent crush speed, vcrush-x.

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� For the LeBaron:


C1 3 ft 9 in 45 in

C2 3 ft 9 in 45 in

C3 4 ft 7 in 55 in

C4 5 ft 5 in 65 in

Stiffness values

A = 356 lb/inB = 34 lb/in2

G = 1874 lb

L = 69 in damage width

α = 0o angle between the collision force and a line normal to damage face

� Substitute the values into:

� Which yields:

� Divide by 12 to get ft-lb:


( ) ( ) ( )2 2 2 2 2

1 2 3 4 1 2 3 4 1 2 2 3 3 46 2 2 2 2 1 tan6 3

L BE G A C C C C C C C C C C C C C C α

= + + + + + + + + + + + +

E = 4,580,900 in-lb

E = 381,740 ft-lb

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� Calculate the EBS for the LeBaron:


E = 381,740 ft-lb

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30

66.31 mph

crush

Kev

w=

=

� Do the same for the Escort:


C1 3 ft 7 in 43 in

C2 4 ft 0 in 48 in

Stiffness values

A = 373 lb/inB = 38 lb/in2

G = 1849 lb

L = 66 in damage width

α = 0o angle between the collision force and a line normal to damage face

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� Substitute the values into:

� Which yields:

� Divide by 12 to get ft-lb:


( ) ( ) ( )2 2 2

1 2 1 2 1 2 1 tan2 6

A BE L G C C C C C C α

= + + + + + +

E = 3,839,491 in-lb

E = 319,957 ft-lb

� Calculate the EBS for the Escort:


E = 319,957 ft-lb

2

30

65.68 mph

crush

Kev

w=

=

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� Substituting values for weights, departure speeds and crush EBS into the COE equation yields:

� We see we have one equation with two unknowns and thus cannot solve for speed without additional information.


2 2 2 2 2 2

1 1 2 2 1 3 2 4 1 1 2 2crush crushw v w v w v w v w v w v+ = + + +

2 2

1 22601 2222 21,971,699v v+ =

� Doing the same with the COLM equation yields:

� Again we have one equation with two unknowns and cannot solve for speed without additional information.


1 22601 2222 67,029v v− =

1 1 2 2 1 3 2 4cos cos cos cosw v w v w v w vα ψ θ φ+ = +

Recall the vehicles collided head-on which means the direction of Unit #2 is opposite that of Unit #1. This means ψ = 180 o and cos ψ = -1. Thus the second term is negative.

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� However, looking at both the COE and COLM equations together it can be seen that they are a pair of simultaneous equations.

� There are now two equations and two unknowns (the same unknowns) therefore the pair of values for v1 and v2

which make both equations true can be found.

� This will be accomplished by using the mathematical technique of simultaneous equations.


1 22601 2222 67,029v v− =

2 2

1 22601 2222 21,971,699v v+ = COE

COLM

� Let’s take time to look at the mathematical technique of solving simultaneous equations.

� Simultaneous equations are two equations which apply to the same situation at the same time

� Let’s say we have a situation in which the following two equations apply.

� In other words, both equations are true (valid) for the given situation.


2x + 14y = 28

9x + 7y = 18

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� The two unknowns are x and y.

� Simultaneous equations can be solved one of two ways:� Substitution method

� Elimination method


� In the Substitution Method one of the equations is solved (rewritten) for one of the unknowns.

� The resulting expression for that unknown is then substituted into the other equation which makes one equation with one unknown.

� The numerical value of that second unknown can now be found.

� Recall, the two equations in our example are:


2x + 14y = 28

9x + 7y = 18

Eq. 1

Eq. 2

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� So in our example, Eq. 1 is solved for y:


2 14 28

14 28 2

28 2

14

x y

y x

xy

+ =

= −

−=

� Now take that expression for y and substitute it into Eq. 2 and solve for x:


28 2

14

xy

−=

9 7 18

28 29 7 18

14

9 14 18

9 4

8 4

0.50

x y

xx

x x

x x

x

x

+ =

− + =

+ − =

− =

=

=

Now a single equation with one unknown.

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� Now that x has been found it can be used in Eq. 1 to find y:

� Thus x = 0.50 and y = 1.92.© 2015 N. Shigemura, J. Daily, R. Strickland

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( )

2 14 28

2 0.50 14 28

1 14 28

14 27

27

14

1.92

x y

y

y

y

y

y

+ =

+ =

+ =

=

=

=

� In the Elimination Method the two equations are either added or subtracted to eliminate one of the unknowns.

� The resulting expression will be one equation with one unknown.

� The numerical value of that unknown can now be found.

� Many times, one of the equations has to be multiplied or divided through by a number so that the coefficient of one of the unknowns is the same as the coefficient of that same unknown in the other equation.


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� Recall, the two equations in our example are:

� Multiply Eq. 2 by the number 2 to make the coefficient of y equal to the coefficient of y in Eq.1.


2x + 14y = 28

9x + 7y = 18

Eq. 1

Eq. 2

2x + 14y = 28

2(9x + 7y) = 2(18)

Eq. 1

Eq. 2

� The two equations now look like:

� Subtract Eq. 3 from Eq. 1 to eliminate the unknown y:


2x + 14y = 28

18x + 14y = 36

Eq. 1

Eq. 3

2x + 14y = 28

-(18x + 14y = 36)

Eq. 1

Eq. 3

-16x = -8 Eq. 4

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� Eq. 4 can now be solved for x:

� Substituting this value for x into either of the original equations (Eq. 1 or Eq. 2) yields y = 1.92.

� Thus x = 0.50 and y = 1.92 and are the same as those found by the Substitution Method.

� Now, back to the crash example!


-16x = -8 Eq. 4

x = 0.50

� Recall, the simultaneous equations pair is:

� We’ll solve this pair using the Substitution Method.

� Begin by dividing both equations by 2601 to reduce each equation:


1 22601 2222 67,029v v− =

2 2

1 22601 2222 21,971,699v v+ = COE

COLM

1 20.854 25.77v v− =

2 2

1 20.854 8447.40v v+ = COE

COLM

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� Solve the COLM equation for v1:


1 2

1 2

0.854 25.77

25.77 0.854

v v

v v

− =

= +

COLM

� Substitute that expression for v1 into the COE equation:

� The final equation is a quadratic equation and thus can be solved using the Quadratic Formula.


( )

( ) ( )

2 2

1 2

2 2

2 2

2

2 2 2

2 2

2 2 2 2

2

2 2

2

2 2

0.854 8447.40

25.77 0.854 0.854 8447.40

25.77 0.854 25.77 0.854 0.854 8447.40

0.729 22 22 664.09 0.854 8447.40

1.58 44 664.09 8447.40

1.58 44 7783.31 0

v v

v v

v v v

v v v v

v v

v v

+ =

+ + =

+ + + =

+ + + + =

+ + =

+ − =

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2

2 21.58 44 7783.31 0v v+ − =

Where : x = v2,

a = 1.58,

b = 44

c = -7783.31

2 4

2

b b acx

a

− ± −=

Quadratic Formula


x = v2,

a = 1.58,

b = 44

c = -7783.31

( )( )

( )

2

2

2

4

2

44 44 4 1.58 7783.31

2 1.58

44 1936 49190.51

3.16

44 51126.51

3.16

44 226.11

3.16

b b acx

a

v

− ± −=

− ± −=

− ± +=

− ±=

− ±=

2

182.11

3.16

57.62 mph

v =

=

Choose the “+” branch which will yield a positive answer

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� Substitute that expression for v2 into the COLM equation:

� Thus v1, the impact speed of the LeBaron was 74.97 mph and v2, the impact speed of the Escort was 57.62 mph .

( )1 2

1

1

1

0.854 25.77

0.854 57.62 25.77

49.20 25.77

74.97 mph

v v

v

v

v

− =

− =

− =

=

COLM


� We can check our answers against the chart calculated earlier:

Escort impact speed (mph) LeBaron impact speed (mph)

40 59.94

45 64.21

50 68.48

55 72.75

60 77.02

65 81.29

70 85.57

75 89.94

80 94.11

74.97 mph57.62 mph

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Graphical solution:

Energy Balance Equation Momentum Equation


The solution by superimposing one graph over the other.


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The positive answers are in the first quadrant.


� We saw how in some cases, such as inline collisions, COLM or COE alone may not yield very meaningful answers for impact speeds.

� However, we saw that, by utilizing both concepts of COLM and COE to create simultaneous equations and then applying the mathematical technique of solving simultaneous equations by the substitution method we were able to determine meaningful numerical values for impact speed.

� By having an open mind and utilizing sound scientific and mathematical procedures the traffic crash investigator can solve many more cases.


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