It was one of those little hills that you hate. The ones where you top out the hill just in time to see the stop sign a few yards away at the bottom. Apparently not seeing the "Stop Ahead" sign, the driver of a 1998 Ford Expedition crested the hill moving into the left lane of the two-lane concrete road to pass another car. After changing lanes there was little time or opportunity for the driver to avoid hitting a 2000 Ford Escort stopped at the stop sign in the same lane on the other side of the intersection. The collision was dead center, head-on, and there was no significant under-ride or over-ride. The two young men in the Expedition suffered minor abrasions, bruises, and sore necks. The man driving the Escort was lucky enough to survive with a few broken bones and a whiplash injury.

Crash test data is used to determine the stiffness properties of a vehicle. Using a common linear model and ignoring bumper effects, the relationship between maximum impact force F and the depth of crush D is:

F = KD

K is the stiffness of the vehicle. Energy considerations determine the relationship between D and speed V,

D = (m/K)^{½} V

Where "m" (m^{x} = 5,140 lb/g) is the mass. To determine the speed of the Expedition at impact, it would be tempting to measure the crush D^{x} (12.7 inches), and solve the equation above for the speed V^{x}. Using K´^{x} = 6,488 lb/in, gives a speed of V^{x} = 15.9 mph. Called the equivalent barrier speed (EBS) this is the result corresponding to D^{x} when this vehicle hits a rigid fixed barrier.

Unfortunately, this is not the impact speed in this case. When hitting a rigid barrier, most of the kinetic energy (KE) is expended in crushing the Expedition. When hitting the Escort, some KE is also expended in crushing the Escort. Therefore, if the crush of the Expedition is D^{x} = 12.7 inch, we expect the impact speed to be larger than the EBS.

The impact speed can easily be determined by equating the total momentum before impact to that after impact. Since the two vehicles stick together in a plastic collision, we have

(m^{x})V^{x1} = (m^{x} +m^{s})V^{2}

where "m^{s}" is the mass of the Escort, and V^{2} is the common speed after impact. All wheels on both vehicles were locked at impact and left skidmarks 21.1 ft long. The post-impact KE (and V^{2}) can be calculated from the friction work. A friction coefficient of 0.7 gives V^{2} = 21.1 mph, and V^{x1} =
32 mph.

It is interesting that the speed of impact that produces D^{x} = 12.7 inches is approximately twice the EBS (15.9 mph). Thus, our erroneous calculation would have been in error by about 100%.

An energy analysis can also be used to determine V^{x1} by equating the initial KE to the energy consumed in crushing the two vehicles plus the KE just after impact. Thus,

0.5(m^{x})V^{2x}1 = 0.5F/(K^{2x}) + 0.5F/(K^{2s}) + 0.5(m^{x} + m^{s})V^{2}2

"F" is the same magnitude on both vehicles, but in opposite directions. Crash tests and vehicle data indicate that 'm^{s}' = 2663 lb/g, and K^{s} = 8649 lb/in. K^{x} = 5483 lb/in and is smaller than K´^{x} because the full width of the Expedition is not involved. The momentum equation can be solved for V^{x1} in terms of V^{2}, so that the energy equation can be solved for F in terms of V^{2}. This gives F = 69,512 lbs, which can be used to determine the 'g' forces on both vehicles.

The Expedition experiences 13.5 g's compared to 26.1 g's for the Escort. Similarly, the change in velocity is 10.9 mph for the Expedition compared to 21.1 mph for the Escort. These indicate a much more severe environment for the occupants of the Escort, explaining the difference in injuries. Even though the Escort is a little stiffer than the Expedition, the mass difference is large enough that the bigger vehicle still wins.