Traffic Crash Reconstruction

1
Traffic Crash Reconstruction

• Derivation of Slide to Stop Formula

2
Definitions

• Ke Kinetic Energy
• W Work
• F Force
• M Mass
• g gravity (32.2fps2)

3
Definitions (cont.)

• S Speed (mph)
• v Velocity (fps)
• d Distance
• w Weight
• f Coefficient of Friction
• (Drag Factor)

4
Work Energy Theorem

• A vehicle in motion possess (translational)
  Kinetic Energy (Ke), which can be represented
  numerically by the following equation

5
Work Energy Theorem
A vehicle in motion possess (translational)
Kinetic Energy (Ke), which can be represented
numerically by the following equation
6
Stopping the Vehicle

• Work is required to bring that vehicle to a stop.
  Work is a product of force over distance OR force
  times distance, as represented in the following
  equation

7
Stopping the Vehicle

• Work is required to bring that vehicle to a stop.
  Work is a product of force over distance OR force
  times distance, as represented in the following
  equation

8
Newtons Third Law

• If we employ Newtons Third Law, we know that the
  work required to bring the vehicle to a stop must
  be equal to the kinetic energy of the vehicle, OR

9
Newtons Third Law

• If we employ Newtons Third Law, we know that the
  work required to bring the vehicle to a stop must
  be equal to the kinetic energy of the vehicle, OR

10
The Work Energy Theorem

• By substituting the values, (1/2Mv2 for Ke and Fd
  for W), we arrive at our work energy theorem

11
The Work Energy Theorem

• By substituting the values, (1/2Mv2 for Ke and Fd
  for W), we arrive at our work energy theorem

12
Friction

• Friction, the resistance of two surfaces in
  contact, is employed to bring the vehicle to a
  stop. In this case. the vehicles tires and the
  roadway surface. Friction is a product of force
  divided by weight......

13
Friction

• Friction, the resistance of two surfaces in
  contact, is employed to bring the vehicle to a
  stop. In this case. the vehicles tires and the
  roadway surface. Friction is a quotient of force
  divided by weight......

14
Friction

• That equation can be rewritten to show that Force
  is equal to friction times weight.........

15
Friction

• That equation can be rewritten to show that Force
  is equal to friction times weight.........
  F fw

16
Friction

• That equation can be rewritten to show that Force
  is equal to friction times weight.........
  F fw
• We then substitute fw for Force in our equation
  so that....

17
Friction

• That equation can be rewritten to show that Force
  is equal to friction times weight.........
  F fw
• We then substitute fw for Force in our equation
  so that....

18
Newtons Second Law

• Newtons Second Law tells us that Mass is equal
  to weight divided by gravity....

19
Newtons Second Law

• Newtons Second Law tells us that Mass is equal
  to weight divided by gravity....

20
Newtons Second Law

• Newtons Second Law tells us that Mass is equal
  to weight divided by gravity....

• w/g is substituted in the equation for Mass, so
  that

21
Multiplication Tables

• The equation is then multiplied through so that

22
Multiplication Tables

• The equation is then multiplied through so that

23
Losing the Weight

• Applying the associative properties of
  multiplication, we will divide both sides by w.

24
Losing the Weight

• Applying the associative properties of
  multiplication, we will divide both sides by w.
  As you can see, the weight does not fit into the
  
• equation!

25
The Gravity of the Situation

• Gravity can be represented numerically at 32.2
  feet per second per second or 32.2 fps2.
  Substituting this value into our equation it now
  reads.....

26
The Gravity of the Situation

• Gravity can be represented numerically at 32.2
  feet per second per second or 32.2 fps2.
  Substituting this value into our equation it now
  reads.....

27
The Gravity of the Situation

• By multiplying by 2, it shows

28
The Need For Speed

• Ultimately, we are trying to determine how fast
  the vehicle was going, so we need to isolate the
  velocity. This is done by dividing the equation
  by 64.4 so that

29
Terminal Velocity

• Since we are trying to solve for Speed (mph) we
  need to convert velocity (fps).
• S 1.466v

30
Terminal Velocity

• Since we are trying to solve for Speed (mph) we
  need to convert velocity (fps).
• S 1.466v
• Why 1.466? It is a numeric representation of feet
  in a mile to seconds in an hour.

31
Terminal Velocity

• Since we are trying to solve for Speed (mph) we
  need to convert velocity (fps).
• S 1.466v
• Why 1.466? It is a numeric representation of feet
  in a mile to seconds in an hour. SO

32
The Need For Speed

• 1.466S is then substituted into our equation for
  velocity so that.......
• (1.466S)2 64.4df

33
The Need For Speed

• 1.466S is then substituted into our equation for
  velocity so that.......
• (1.466S)2 64.4df

1.466 squared is 2.14 (We always truncate to the
second decimal point), so that
34
The Need For Speed

• 1.466S is then substituted into our equation for
  velocity so that.......
• (1.466S)2 64.4df

1.466 squared is 2.14 (We always truncate to the
second decimal point), so that 2.14S2 64.4df
35
The Need For Speed

• Both sides of the equation are then divided by
  2.14, again to help us isolate Speed ....

36
The Need For Speed

• Both sides of the equation are then divided by
  2.14, again to help us isolate Speed ....
• S2 30.04df

37
The Need For Speed

• Both sides of the equation are then divided by
  2.14, again to help us isolate Speed ....
• S2 30.04df
• ...and the right side of the equation is moved
  under the radical, so that

38
The Need For Speed

• Both sides of the equation are then divided by
  2.14, again to help us isolate Speed ....
• S2 30.04df
• ...and the right side of the equation is moved
  under the radical, so that

39
The Final Equation

• We drop the decimal points do that our working
  equation is

40
The Final Equation

• We drop the decimal points so that our working
  equation is