Stopping distance g force calculator driver#
A crash which stops the car and driver must take away all its kinetic energy, and the work-energy principle then dictates that a shorter stopping distance increases the impact force. The work done to stop the driver is equal to the average impact force on the driver times the distance traveled in stopping.
The stopping distance is estimated to be about one fifth of that with a seatbelt, causing the average impact force to be about five times as great.
With no seatbelt to stop the driver with the car, the driver flies free until stopped suddenly by impact on the steering column, windshield, etc. Either a stretching or non-stretching seat belt reduces the impact force compared to no seatbelt. If the belt stretched 0.5 ft in the examplecar crash scenario, it would reduce the deceleration to 20 g's and the average impact force to 3200 lb compared to 30 g's and 4800 lb for a non-stretching seatbelt. Seatbelt variationsĪ moderate amount of stretch in a seatbelt harness can extend the stopping distance and reduce the average impact force on the driver compared to a non-stretching harness. A moderate amount of stretch in the seatbelts will reduce the average impact force. For the examplecar crash scenario the stopping distance is one foot, the force on a 160 lb driver is about 4800 lb or 2.4 tons, and the deceleration about 30 g's. A crash which stops the car and driver must take away all its kinetic energy, and the work-energy principle then dictates that a longer stopping distance decreases the impact force. The task of the seatbelt is to stop you with the car so that your stopping distance is probably 4 or 5 times greater than if you had no seatbelt.
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