# Impact Force

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 * 28 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 parameters as above).

- 60 mph = 96.6 km/h

### 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 _{ g } h / s (5b) *

The 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 *

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