222 The UMAP Journal 24.3 (2003) directions(perpendicular to the force), creating a strain in the walls. For a given strain E, the potential energy density v stored in the walls of the box is 号Ye2+O(∈) where Y is the Youngs modulus for boxboard. For this calculation the effects of longitudinal (along the direction of the force) contractions in the box walls are negligible compared to the transverse expansion a stress s is created in the box walls as a result of the battle bety walls pressure to expand and its resistance to expansion, S=Y∈+O(e2) Only a small strain(e 1)is necessary to cause the box to buckle, so we neglect higher-order terms. There is a point where the box stops expanding and begins to give way to the increasing stress placed on it. When the stress in the box walls reaches its tensile strength Ts, it loses its structural integrity and the box bursts as it continues to buckle. Thus we can set a limit on the maximum strain that the box walls can endure Ts ∈max The typical tensile strength of cardboard is on the order of a few MPa, while the Young s modulus is on the order of a few GPa. Thus, we may assume that E<l. The maximum energy density allowed in the walls of the box is now 1T2 Umax 2Y The total energy stored in the walls just before the box bursts is where v=(lateral surface area)x(thickness)is the volume of cardboard in the box walls. If we assume that the force to deform the box is constant over the deformation distance, then by conservation of energy, we have △Hdt2△Hyy, where△ h is the ch ange In neight o he box over the course of the buckling process. SInce due dU the change in velocity of the falling stunt person due to buckling the box is 12y dt2ms△HY222 The UMAP Journal 24.3 (2003) directions (perpendicular to the force), creating a strain in the walls. For a given strain , the potential energy density v stored in the walls of the box is v = 1 2Y 2 + O( 4), where Y is the Young’s modulus for boxboard. For this calculation, the effects of longitudinal (along the direction of the force) contractions in the box walls are negligible compared to the transverse expansion. A stress S is created in the box walls as a result of the battle between the wall’s pressure to expand and its resistance to expansion, S = Y + O( 2). Only a small strain ( 1) is necessary to cause the box to buckle, so we neglect higher-order terms. There is a point where the box stops expanding and begins to give way to the increasing stress placed on it. When the stress in the box walls reaches its tensile strength TS, it loses its structural integrity and the box bursts as it continues to buckle. Thus, we can set a limit on the maximum strain that the box walls can endure, max = TS Y . The typical tensile strength of cardboard is on the order of a few MPa, while the Young’s modulus is on the order of a few GPa. Thus, we may assume that 1. The maximum energy density allowed in the walls of the box is now vmax = 1 2Y 2 max = 1 2 T2 S Y . The total energy stored in the walls just before the box bursts is Vmax = 1 2 T2 S Y V, where V = (lateral surface area) × (thickness) is the volume of cardboard in the box walls. If we assume that the force to deform the box is constant over the deformation distance, then by conservation of energy, we have dUs dt + dVs dt = −dUbox dt = −dUbox dt dz dt = −∆Ubox ∆H dz dt = us 2∆H T2 S Y V, where ∆H is the change in height of the box over the course of the buckling process. Since msus dus dt = dUs dt , the change in velocity of the falling stunt person due to buckling the box is dus dt = 1 2ms∆H T2 S Y V. (1)