## 1513 Estimation of a Collision Coefficient of Restitution from Fixed Barrier Data

When a vehicle impacts a fixed barrier, it is presumed that the fixed barrier does not crush or otherwise absorb any significant amount of kinetic energy from the vehicle. The fixed barrier pushes back against the vehicle during impact with a force equal to that being applied by the vehicle. In essence, since no work can be done on the fixed barrier, the vehicle does work on itself.

Earlier, Equation (xxv) was derived to show the relationship between energy dissipated by the collision and the coefficient of restitution. For convenience, the equation is shown again below.

Eloss = I(1 - e2)/2][(mAmB)/(mA + mB)][vA1 - vbJ2 (xxv)

If it is assumed that a fixed barrier essentially has an "infinite" or immovable mass, and the wall has no initial speed prior to the collision, then the equation reduces to the following:

When e = 0, a fully plastic collision, all of the initial kinetic energy is dissipated as work in crushing the vehicle. When e = 1.0, none of the initial kinetic energy is dissipated as work; all of it goes into the rebound of the vehicle.

While not exactly contributing to the point at hand, the following aside may be interesting to the reader. By substituting Equation (xviii) into (lix), a function for collision energy loss against a fixed-barrier collision can be derived that is solely dependent upon initial speed:

where s = velocity of Vehicle "A" given in mph and e = 2.718 etc.

An alternate version of the above equation where velocity is given in ft/sec is as follows:

However, getting back to the point of this section, if it is assumed that crush caused by collision with the fixed barrier is proportional to the change in kinetic energy of the vehicle, then:

where U = work done in crushing the vehicle, sA = amount of crush in the vehicle, F = the average force applied to the vehicle during collision with the fixed barrier, and %vA2 = vA22 - vA12.

When two vehicles collide head on, the total work done in crushing is as follows:

U = (1/2)mA(%vA2) + (1/2)mB(%vB2) = sA ■ F + sB ■ F. (lxiii)

The force "F" applied between the two vehicles during collision is the same in magnitude but opposite in sign.

Now, if in the above equation, Vehicle "B" is sitting still with respect to Vehicle "A" (i.e., %vB2 = 0), then the crush damage on both vehicles would be due to the action of Vehicle "A" alone. As can be seen in comparing Equations (lxii) and (lxiii), this means that the crush on Vehicle "A" in colliding with Vehicle "B" would be about half that of Vehicle "A" colliding with a fixed barrier at the same speed.

where vAeq = impact velocity of "A" with a fixed barrier and vA1 = impact velocity of "A" with Vehicle "B," which is sitting still.

Thus, the coefficient of restitution for the above collision situation can be estimated by converting "(0.707)vA1" into mph units, and substituting into Equation (xviii).

In a similar fashion, equivalent fixed-barrier impact velocities can be derived for other situations. For example, if the two vehicles have about the same mass and collide with the same velocity head on, then the equivalent fixed-barrier impact velocity is equal to the actual impact velocity.

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