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For the car crash scenario where a car stops in 1 foot from a speed of 30 mi/hr, what is the force on the driver? Assume a 160 lb (mass = 5 slugs) driver.
If firmly held in non-stretching seatbelt harness: Stopping distance 1 ft.
Deceleration = 967 ft/s2 = 294 m/s2 = 30 g's
Force = 4813 lb = 21412 N = 2.4 tons
Non-stretching seatbelt
If not wearing seatbelt, stopping distance determined by nature of collision with windshield, steering column, etc. : stopping distance 0.2 ft.
Deceleration = 4836 ft/s2 = 1474 m/s2 = 150 g's
Force = 24068 lb = 107059 N = 12 tons!!
No seatbelt!
If seat belt harness stretches, increasing stopping distance by 50%: 1.5 ft.
Deceleration = 645ft/s2 = 197 m/s2 = 20 g's
Force = 3209 lb = 14274 N = 1.6 tons
Stretching seatbelt
These calculated numbers assume constant deceleration, and are therefore an estimate of the average force of impact.
google出来的一个大概的计算. 按这个计算 160lb的司机 就是72kg, 30mi/h 就是50km/h, 就算是可拉伸的安全带也要14kn力量...
那个toyota的恐怕不够了.. |
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