deceleration in g (gravities) =
= ((velocity in miles per hour)^2) / ((stopping distance in feet) * 30)
where:
• The stopping distance is the distance over which the occupant decelerates from the initial speed until he or she stops. If the person is wearing a seat and shoulder belts this may be 2 or more feet. For an unrestrained passenger, the stopping distance may be an inch.
• 1 gravity = 32.17 feet per second squared.
• Miles per hour = 5280 feet per 3600 second = 1.46 feet per second
• Kinetic deceleration = 0.5 * ((velocity)^2)
• The 30 in denominator comes from 0.5 * (1.46^2) / 32.17
The deceleration of a restrained passenger is often in the order of 10 g's, while the unrestrained occupant may experience decelerations of over 200 g's.
Once the deceleration is known, the force can be determined from the occupant's body weight. The pressure on the body at contact points can be estimated from dividing the total surface area at the points of contact into this force.
