# Impact Force

## Impact forces acts on falling objects hitting ground, crashing cars and similar

The dynamic kinetic energy of a moving object, like a falling ball or a driving car, can be expressed as

*E = 1/2 m v ^{2} (1)*

*where *

*E = kinetic (dynamic) energy (J, ft lb)*

*m = mass of the object (kg, slugs)*

*v = velocity of the object (m/s, ft/s)*

In an impact - like a car crash - the work made by the impact force slowing down an moving object over a distance by **deforming the crumple zone** can be expressed as

*W = F _{avg} s (2)*

*where *

*W = work done (J, ft lb)*

*F _{avg} = average impact force during deformation (N, lb_{f})*

*s = deformation distance, crumple zone (m, ft)*

When a crumple zone deforms in a car crash the average impact force is designed to be as constant as possible.

In an impact where the object is **not deformed** - the work made by the impact force slowing down the moving object equals to the work done by a spring force - and can be expressed as

*W = 1/2 F _{max} s *

* = 1/2 k s ^{2} (2b)*

*where *

*W = work done (J, ft lb)*

*F _{max} = maximum force at the end of the deformation (N, lb_{f})*

*k = spring constant *

*s = deformation distance (m, ft)*

In a car crash the dynamic energy is converted to work and equation 1 and 2 can be combined to

*F _{avg}* s = 1/2 m v

^{2}(3)

The average impact force can be calculated as

*F _{avg}* = 1/2 m v

^{2 }/ s (3b)

The deformation slow-down distance can be calculated as

*s = 1/2 m v ^{2 }/ F_{avg} (3c)*

**Note!** - The deformation slow-down distance is very important and the key to limit the forces acting on passengers in a car crash.

### Example - Car Crash

A car with a mass of *2000 kg* drives with speed *60 km/h (16.7 m/s)* before it crashes into a massive concrete wall. The front of the car impacts *0.5 m* (the deformation distance).

The impact force can be calculated as

*F _{max}* = 1/2 (2000 kg) (16.7 m/s)

^{2 }/ (0.5 m)

* = 558 kN*

Note that the gravitation force (weight) acting on the car is only

*F _{w} = m g *

* = (2000 kg) (9.81 m/s ^{2})*

* = 19.6 kN*

The impact creates a force *28* times gravity!!

A person sitting inside the car with seat belts on will de-accelerate with a force *283 times gravity*. Note that the National Highway Traffic Safety Administration (NHTSA) states that "the maximum chest acceleration shall not exceed ** 60 times gravity** for time periods longer than

*3 milliseconds*". For a car crash with

*90 km/h (25 m/s)*the de-acceleration will be 64

*times gravity*(same as parameters above).

### Impact Force from a Falling Object

The dynamic energy in a falling object at the impact moment when it hits the ground can be calculated as

*E = F _{weight} h*

* = m a _{g} h (4)*

*where *

*F _{weight} = force due to gravity - or weight (N, lb_{f})*

*a _{g} = acceleration of gravity (9.81 m/s^{2}, 32.17405 ft/s^{2})*

*h = falling height (m)*

If the dynamic energy from the fall is converted to impact work - equation 2 and 4 can be combined to

* F_{avg} s* = m a

_{g}h (5)

The impact force can be expressed as

*F _{avg}* =

*m a*/ s (5b)

_{g}hThe deformation slow-down distance can be expressed as

*s = m a_{g} h / F_{avg} (5c)*

### Example - a Falling Car

The same car as above falls from height* 14.2 m *and crashes on the crumple zone with the front down on a massive concrete tarmac. The front impacts *0.5 m* (slow down distance) as above. The impact force can be calculated as

*F _{avg}* = (2000 kg) (9.81 m/s

^{2}) (14.2 m) / (0.5 m)

* = 558 kN*

**Note!** - a car crash in *90 km/h (25 m/s)* compares to a fall from * 32 m*!!

### Example - a Person falling from a Table

A person with weight (gravitational force) of *200 lbs (lb _{f})* falls from a

*4 feet*high table.

The energy of the falling body when it hits the ground can be calculated using *(4)* as

*E = (200 lb _{f}) (4 ft)*

* = 800 ft lb*

The impact on a human body can be difficult to determine since it depends on how the body hits the ground - which part of the body, the angle of the body and/or if hands are used to protect the body and so on.

For this example we use an impact distance of *3/4 inch (0.0625 ft) *to calculate the impact force:

*F _{avg}* = (800 ft lb) /

*(0.0625 ft)*

* = 12800 lb _{f}*

In metric units - person with weight *90 kg,* falling distance *1.2 m* and impact distance *2 cm*:

*E = (90 kg) (9.81 m/s ^{2}) (1.2 m)*

* = 1059 J*

*F _{avg}* = (1059 J) / (0.02 m)

* = 53 kN*