304 The UMAP Journal 24.3(2003) Small Box model down oe the r face Figure 2. Energy to break down a bo Energy to break down a box. The breakdown of a box consists of three pro- breakdown of the upper surface, breakdown of the side surfaces, and breakdown of the poles After all three components break down, the box is completely damaged and cannot provide any cushioning. The the total energy required to break down a Ebox= Upper Eside eps After some analysis(see Appendix), we find that Upper and Eside are rather small compared withe Ebox≈ Epole We cannot find any data for calculating Epole, so we make a rough estimate ur analogy is to steel, for which we have data. We obtain the relationship between the maximum pressure to break the pole and the side Pole= TPs where Pole is the breakdown pressure for the pole, Pside is the breakdown pressure for a side surface, and T is the transfer coefficient The breakdown pressure for a side surface is inversely proportional to the length e of a side, so with edgewise crush resistance of the cardboard es we hay
304 The UMAP Journal 24.3 (2003) Small Box Model Figure 2. Energy to break down a box. Energy to break down a box. The breakdown of a box consists of three processes: • breakdown of the upper surface, • breakdown of the side surfaces, and • breakdown of the poles. After all three components break down, the box is completely damaged and cannot provide any cushioning. The the total energy required to break down a box is Ebox = Eupper + Eside + Epole. After some analysis (see Appendix), we find that Eupper and Eside are rather small compared withEpole, so Ebox ≈ Epole. (3) We cannot find any data for calculating Epole, so we make a rough estimate. Our analogy is to steel, for which we have data. We obtain the relationship between the maximum pressure to break the pole and the side: ppole = τ pside, where ppole is the breakdown pressure for the pole, pside is the breakdown pressure for a side surface, and τ is the transfer coefficient. The breakdown pressure for a side surface is inversely proportional to the length of a side, so with edgewise crush resistance of the cardboard es we have pside = es . (4)
Fly with Confidence 305 Height of the pile. The motorcycle lands in a pile with an initial velocity and ultimately decelerates to zero, trapped in this pile. During that process, the force exerted on the motorcycle must be smaller than the maximum force that a person can bear; otherwise, the stunt person would be injured. Since 12 kN is the threshold, we consider 6 kN the safety bound. Thus, a 60 kg person can bear a maximum acceleration of amax =6000/ 60= 100 m/s. We want the mean acceleration to be smaller than this: a maxi we use mean acceleration because the cushion process has approximately constant deceleration. Thus, using kinematics, we obtain a Thus, we let the pile height h be u/ 2amax, so that the motorcycle just touches the ground when it stops. In terms of the kinetic energy e= mu/2 of the motorcycle, we have E (5) namax Size of boxes To see how a box cushions the motion of the motorcycle, we define the density of energy absorption(DEA)of a box as where ebox is the energy that the box can absorb during its breakdown and vbox is the original volume of the box. This density reflects the average cushioning ability of the box, and p can be thought of as the proportion of energy that is absorbed In a homogenous pile, all the boxes have the same DEA. The total energy that the pile absorbs is pVpile for the collapsed boxes. The height of the stack of boxes is h and the cross-sectional area of ones collapsed by the motorcycle is S, E=pVpile-pSh=os.E namax from(5). Cancelling the Es, we get Ne assume that the work done in breaking down a single box is proportional to Pole, with proportionality coefficient h. In breaking down the pile of boxes
Fly With Confidence 305 Height of the pile. The motorcycle lands in a pile with an initial velocity and ultimately decelerates to zero, trapped in this pile. During that process, the force exerted on the motorcycle must be smaller than the maximum force that a person can bear; otherwise, the stunt person would be injured. Since 12 kN is the threshold, we consider 6 kN the safety bound. Thus, a 60 kg person can bear a maximum acceleration of amax = 6000/60 = 100 m/s2. We want the mean acceleration to be smaller than this: a¯ ≤ amax; we use mean acceleration because the cushion process has approximately constant deceleration. Thus, using kinematics, we obtain a¯ = v2 2h ≤ amax, or h ≥ v2 2amax . Thus, we let the pile height h be v2/2amax, so that the motorcycle just touches the ground when it stops. In terms of the kinetic energy E = mv2/2 of the motorcycle, we have h = E mamax . (5) Size of Boxes To see how a box cushions the motion of the motorcycle, we define the density of energy absorption (DEA) of a box as ρ = Ebox Vbox , where Ebox is the energy that the box can absorb during its breakdown and Vbox is the original volume of the box. This density reflects the average cushioning ability of the box, and ρ can be thought of as the proportion of energy that is absorbed. In a homogenous pile, all the boxes have the same DEA. The total energy that the pile absorbs is ρVpile for the collapsed boxes. The height of the stack of boxes is h and the cross-sectional area of ones collapsed by the motorcycle is S, so E = ρVpile = ρSh = ρS · E mamax (6) from (5). Cancelling the Es, we get ρ = mamax S . We assume that the work done in breaking down a single box is proportional to ppole, with proportionality coefficient k. In breaking down the pile of boxes