Fly with Confidence 299 Fly with Confidence Hu yuxiao Zhou anlu Zhejiang University Hangzhou, china Advisor: Tan Zhiyi Abstract of e develop a model to design a pile of cardboard boxes to cushion the fall stunt motorcycle; the kinetic energy of the motorcycle is consumed through breaking down the boxes We ignore the scattering effect of the boxes and begin by analyzing a single box. The energy to break it down has three components: the upper surfaces, the side surfaces, and the vertical edges. When big boxes are used, the upper surface provides the main supporting force; when small ones are used, the vertical edges play a great role We extend our analysis to the pile of boxes. Taking into account the maximum impulse that a person can bear, along with the camera effect and cost concerns, we determine the size of a box We conceive several stacking strategies and analyze their feasibility. We in- corporate their strengths into our final strategy. We also examine several modifi cations to the box to improve its cushioning effect To validate our model, we apply it to different cases and get some encouraging results Assumptions The stunt person and the motorcycle are taken as a system, which we refer to as the motorcycle system or for brevity as the motorcycle. We ignore relative movement and interaction between them The motorcycle system is a uniform-density block. We consider only the move- ment of its mass center, so we consider the motorcycle system as a mass particle The cardboard boxes are all made of the same material, single wall s-1 cardboard 4.5 mm thick [ Corrugated fiberboard..nd I The UMAP Journal 24(3)(2003)299-316. Copyright 2003 by COMAP, Inc. Allrights reserved Permission to make digital or hard copies of part or all of this work for personal or classroom use is granted without fee provided that copies are not made or distributed for profit or commercial dvantage and that copies bear this notice. Abstracting with credit is permitted, but copyrights for components of this work owned by others than COMAP must be honored. To to republish, to post on servers, or to redistribute to lists requires prior permission from COMAP
Fly With Confidence 299 Fly With Confidence Hu Yuxiao Hua Zheng Zhou Enlu Zhejiang University Hangzhou, China Advisor: Tan Zhiyi Abstract We develop a model to design a pile of cardboard boxes to cushion the fall of a stunt motorcycle; the kinetic energy of the motorcycle is consumed through breaking down the boxes. We ignore the scattering effect of the boxes and begin by analyzing a single box. The energy to break it down has three components: the upper surfaces, the side surfaces, and the vertical edges. When big boxes are used, the upper surface provides the main supporting force; when small ones are used, the vertical edges play a great role. We extend our analysis to the pile of boxes. Taking into account the maximum impulse that a person can bear, along with the camera effect and cost concerns, we determine the size of a box. We conceive several stacking strategies and analyze their feasibility. We incorporate their strengths into our final strategy. We also examine several modifi- cations to the box to improve its cushioning effect. To validate our model, we apply it to different cases and get some encouraging results. Assumptions • The stunt person and the motorcycle are taken as a system, which we refer to as the motorcycle system or for brevity as the motorcycle. We ignore relative movement and interaction between them. • The motorcycle system is a uniform-density block. We consider only the movement of its mass center, so we consider the motorcycle system as a mass particle. • The cardboard boxes are all made of the same material, single wall S-1 cardboard 4.5 mm thick [Corrugated fiberboard ... n.d.] The UMAP Journal 24 (3) (2003) 299–316. c Copyright 2003 by COMAP, Inc. All rights reserved. Permission to make digital or hard copies of part or all of this work for personal or classroom use is granted without fee provided that copies are not made or distributed for profit or commercial advantage and that copies bear this notice. Abstracting with credit is permitted, but copyrights for components of this work owned by others than COMAP must be honored. To copy otherwise, to republish, to post on servers, or to redistribute to lists requires prior permission from COMAP
300 The UMAP Journal 24.3 (2003) The cardboard box is cubic and isotropic Cost is proportional to the total surface area of cardboard Symbols and Terms Table 1 Symbols Motorcycle parameters mean acceleration during the landing b drag coefficient for the motorcycle plus rider e kinetic energy of the motorcycle when it hits the pile of boxes H height of the stage from which the motorcycle leaves m total mass of the motorcycle system s cross-sectional area of the motorcycle system, which we assume is 1.5 am standard deviation of vo standard deviation of the direction of vo angle between motorcycle direction and the horizontal initial projection speed of the motorcycle peed of the motorcycle when it lands on the pile Box parameters Ebox energy that a single box can absorb during its breakdown dge length of the box Ppole pressure needed to break down the pole of a box sure needed to break down the side of a box coefficient for transfer of pressure from side to pole P density of energy absorption(DEA)of a box Vbox original volume of a single box Pile parameters height of the pile Stotal total combined surface area of all boxes in the pile combined volume of all boxes in the pile Cardboard parameters bursting strength of the cardboard dgewise crush resistance of the cardboard Rider parameter amax the maximum acceleration that a person can bear Pole: the vertical edge of the box Edge: the horizontal edge of the box Corner: the intersection of poles and edges Bursting strength: the maximum pressure on the cardboard before it bursts
300 The UMAP Journal 24.3 (2003) • The cardboard box is cubic and isotropic. • Cost is proportional to the total surface area of cardboard. Symbols and Terms Table 1. Symbols. Motorcycle parameters a¯ mean acceleration during the landing b drag coefficient for the motorcycle plus rider E kinetic energy of the motorcycle when it hits the pile of boxes H height of the stage from which the motorcycle leaves m total mass of the motorcycle system S cross-sectional area of the motorcycle system, which we assume is 1.5 m2 σm standard deviation of v0 σd standard deviation of the direction of v0 θ angle between motorcycle direction and the horizontal v0 initial projection speed of the motorcycle v speed of the motorcycle when it lands on the pile Box parameters Ebox energy that a single box can absorb during its breakdown edge length of the box ppole pressure needed to break down the pole of a box pside pressure needed to break down the side of a box τ coefficient for transfer of pressure from side to pole ρ density of energy absorption (DEA) of a box Vbox original volume of a single box Pile parameters h height of the pile Stotal total combined surface area of all boxes in the pile Vpile combined volume of all boxes in the pile Cardboard parameters bs bursting strength of the cardboard es edgewise crush resistance of the cardboard Rider parameter amax the maximum acceleration that a person can bear. Pole: the vertical edge of the box Edge: the horizontal edge of the box Corner: the intersection of poles and edges Bursting strength: the maximum pressure on the cardboard before it bursts
Fly with Confidence 301 upper surtace ide surface corner Figure 1. Illustration of terminology Edgewise crush resistance: the maximum force on unit length of the edge be- fore the side surface crushes Piercing Strength: the energy of an awl piercing the cardboard Problem analysis Our primary goal is to protect the stunt person. Having ensured this, we should minimize the height of the pile(to get a good film effect) and minimize the total superficial area of boxes(to lower the cost) From the analysis of the jump, we get the pile area and kinetic energy of the motorcycle system. Then we consider the cushion process. Since the maximal impulse a person can bear is 12 kN [UIAA Safety Regulations n d. the problem is to ensure that the force exerted on the person stays within the range during 8 Consider the motorcycle landing in a pile of cardboard boxes. The boxes provide a supporting force to decelerate it. We examine the cushioning effect of both big boxes and small boxes. In our modeling, we focus on energy and reduce the problem to analyzing the energy to break down the boxes. We search for the relation between this energy and the size and number of the boxes. We then improve the cushioning by changing stacking approaches and modifying the box Model design The ump Local assumptions The speed v and direction of the motorcycle are random variables that are normally distributed
Fly With Confidence 301 Figure 1. Illustration of terminology. Edgewise crush resistance: the maximum force on unit length of the edge before the side surface crushes Piercing Strength: the energy of an awl piercing the cardboard Problem Analysis Our primary goal is to protect the stunt person. Having ensured this, we should minimize the height of the pile (to get a good film effect) and minimize the total superficial area of boxes (to lower the cost). From the analysis of the jump, we get the pile area and kinetic energy of the motorcycle system. Then we consider the cushion process. Since the maximal impulse a person can bear is 12 kN [UIAA Safety Regulations n.d.], the problem is to ensure that the force exerted on the person stays within the range during the cushioning. Consider the motorcycle landing in a pile of cardboard boxes. The boxes provide a supporting force to decelerate it. We examine the cushioning effect of both big boxes and small boxes. In our modeling, we focus on energy and reduce the problem to analyzing the energy to break down the boxes. We search for the relation between this energy and the size and number of the boxes. We then improve the cushioning by changing stacking approaches and modifying the box. Model Design The Jump Local Assumptions • The speed v and direction of the motorcycle are random variables that are normally distributed
302 The UMAP Journal 24.3(2003) The standard deviation(om)of the magnitude of vo is 0.05v The standard deviation(od of the direction of vo is 5 The stunt is considered safe if the probability of landing on top of the box ile is more than 99.7% Local variable 8: angle between velocity and horizontal direction h: nt of sta m: mass of motorcycle system. We first examine the path that the motorcycle follows. Taking the air resis- tance into account, we get two differential equations dux= m dt dt where Ux to cos A y Uo sin e d ty.(2) s, The drag coefficient b can be obtained by computing the mass and terminal eed of a skydiver [Halliday et al. 2001. For v=60 m/s and m= 200 kg have b≈30N/(m/s) A typical elephant is 3 m high and 5 m long, with a trunk 2 m long [Estes 1999]. To jump it over safely, we assume that the stage is higher than th elephants trunk reach, which is 5 m Solving equations (1)-(2)with Matlab, we get a quasiparabola. Air resis- tance makes a difference of no more than 5%, so we neglect it. With height difference H and initial speed vo, neglecting air resistance, the landing point is voV2H/g from the projecting point Determine the area of the pile The area of the box pile must be large enough for the actor to land on, that is, the upper face of the pile must capture more that 99. 7% of falls.(Very few may crush into the side face which is still quite safe To meet this criterion the surface must extend to cover six standard deviations (three on each side of the mean) of both the projection speed vo and its direction[ Sheng et al. 2001] We calculate the landing point in the combinations of vo +30m for the speed d +30d for deviation from straight. The resulting length is 3.03 m and the
302 The UMAP Journal 24.3 (2003) • The standard deviation (σm) of the magnitude of v0 is 0.05v0. • The standard deviation (σd) of the direction of v0 is 5◦. • The stunt is considered safe if the probability of landing on top of the box pile is more than 99.7%. Local Variables • v0: initial speed, • θ: angle between velocity and horizontal direction, • H: height of stage, • m: mass of motorcycle system. We first examine the path that the motorcycle follows. Taking the air resistance into account, we get two differential equations dvx dt = −bvx m , dvy dt = bvy m − g, (1) where vx = v0 cos θ, vy = v0 sin θ, and dx dt = vx, dh dt = vy. (2) The drag coefficient b can be obtained by computing the mass and terminal speed of a skydiver [Halliday et al. 2001]. For v = 60 m/s and m = 200 kg, we have b ≈ 30 N/(m/s). A typical elephant is 3 m high and 5 m long, with a trunk 2 m long [Estes 1999]. To jump it over safely, we assume that the stage is higher than the elephant’s trunk reach, which is 5 m. Solving equations (1)–(2) with Matlab, we get a quasiparabola. Air resistance makes a difference of no more than 5%, so we neglect it. With height difference H and initial speed v0, neglecting air resistance, the landing point is v0 2H/g from the projecting point. Determine the Area of the Pile The area of the box pile must be large enough for the actor to land on, that is, the upper face of the pile must capture more that 99.7% of falls. (Very few may crush into the side face, which is still quite safe.) To meet this criterion, the surface must extend to cover six standard deviations (three on each side of the mean) of both the projection speed v0 and its direction [Sheng et al. 2001]. We calculate the landing point in the combinations of v0 ±3σm for the speed and ±3σd for deviation from straight. The resulting length is 3.03 m and the
Fly with Confidence 303 width is 5.23 m. Taking the size of the motorcycle into account, the length needs to be approximately 4.5 m and the width 6 m. Since the motorcycle has a high horizontal speed, the box pile must in fact be longer to cushion the horizontal motion The Cushioning Process Definitions Big box: The cross-sectional area is much larger than that of the motorcycle so the motorcycle interacts with only one box when hitting the cushion we ignore the deformation and resistance of edges and poles. Thus, we need to consider only the interaction between motorcycle and the upper surface of that box Small box: The cross-sectional area is smaller than or comparable to that of mo- torcycle, so the motorcycle interacts with a number of them simultaneously. In this situation, the edges and poles play a great role in cushioning Since big and small boxes have different interactions, we analyze the two situations separately and compare their cushioning effect to determine the box Size Analysis of Big Box The motorcycle has considerable velocity when it hits the upper surface of the box, exerting a force on the upper surface. Because the corrugated card board is to some extent elastic, it first stretches a little. After some elongation it goes into the plastic region and finally ruptures. This process is too compli- cated for us to calculate the total energy that results in the final rupture, so we reduce it to the following extreme situation The area of the upper surface of the cardboard is infinitely large and the motorcycle can be taken as a point compared with the cardboard. The piercing strength of the corrugated paper is 4.9][Corrugated fiberboard .. n.d. ] The total energy of the motorcycle is E = mu2/ 2; since the order of magnitude of v is 10 m/s, the energy is approximately 104 J. Thus, the kinetic energy of the motorcycle is about 10 times the energy needed to pierce the cardboard. S the cardboard is easily pierced and provides little cushioning Although we examine the extreme situation, we can safely reach the follow- ing conclusion: The bigger the box, the easier for the motorcycle to penetrate the upper surface. In fact, as the cardboard becomes smaller, the motorcy cle cannot be taken as a point again, so the force distributes over the surface, making it more difficult to penetrate the upper surface When the box becomes even smaller, the edges and poles of the box provide great support. But the cost increases at the same time
Fly With Confidence 303 width is 5.23 m. Taking the size of the motorcycle into account, the length needs to be approximately 4.5 m and the width 6 m. Since the motorcycle has a high horizontal speed, the box pile must in fact be longer to cushion the horizontal motion. The Cushioning Process Definitions Big box: The cross-sectional area is much larger than that of the motorcycle, so the motorcycle interacts with only one box when hitting the cushion. We ignore the deformation and resistance of edges and poles. Thus, we need to consider only the interaction between motorcycle and the upper surface of that box. Small box: The cross-sectional area is smaller than or comparable to that of motorcycle, so the motorcycle interacts with a number of them simultaneously. In this situation, the edges and poles play a great role in cushioning. Since big and small boxes have different interactions, we analyze the two situations separately and compare their cushioning effect to determine the box size. Analysis of Big Box The motorcycle has considerable velocity when it hits the upper surface of the box, exerting a force on the upper surface. Because the corrugated cardboard is to some extent elastic, it first stretches a little. After some elongation, it goes into the plastic region and finally ruptures. This process is too complicated for us to calculate the total energy that results in the final rupture, so we reduce it to the following extreme situation. The area of the upper surface of the cardboard is infinitely large and the motorcycle can be taken as a point compared with the cardboard. The piercing strength of the corrugated paper is 4.9 J [Corrugated fiberboard ... n.d.]. The total energy of the motorcycle is E = mv2/2; since the order of magnitude of v is 10 m/s, the energy is approximately 104 J. Thus, the kinetic energy of the motorcycle is about 103 times the energy needed to pierce the cardboard. So the cardboard is easily pierced and provides little cushioning. Although we examine the extreme situation, we can safely reach the following conclusion: The bigger the box, the easier for the motorcycle to penetrate the upper surface. In fact, as the cardboard becomes smaller, the motorcycle cannot be taken as a point again, so the force distributes over the surface, making it more difficult to penetrate the upper surface. When the box becomes even smaller, the edges and poles of the box provide great support. But the cost increases at the same time
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
306 The UMAP Journal 24.3(2003) this pressure is exerted across an area S and through a distance h, for total work Ppole Sh. Equating this work to absorbed energy E=pSh, we have pSh =e= kppole sh and substituting Pole=Tes/e,we get Te together with the previous namax 2a Substituting kg, amax=100 m/s, v=13 m/s, k=05,e。=4×103N/m,S=1.5m2,andr=1.5, we C=0.225m,h=0.845m Number of boxes The landing area, 4.5 m by 6 m, needs to be extended to take the horizontal motion into account. The maximum length h that the motorcycle system can penetrate is sufficient for cushioning horizontal motion, so the pile should be (4.5+h)x6x h(meters)in dimension. From this fact, we can calculate the number of boxes needed in the pile From(7), we have mv2 Thus, we know the dimensions of the pile 45+h M Numerically, we get and the total number of boxes needed is N= 4x 27 x 24=2592. with total surface area Stotal=6a2N=6×0.2252×2592≈780
306 The UMAP Journal 24.3 (2003) this pressure is exerted across an area S and through a distance h, for total work kppoleSh. Equating this work to absorbed energy E = ρSh, we have ρSh = E = kppoleSh, so ρ = mamax S = kppole; and substituting ppole = τes/, we get = kτes mamax S, together with the previous h = v2 2amax . (7) Substituting m = 200 kg, amax = 100 m/s, v = 13 m/s, k = 0.5, es = 4 × 103 N/m, S = 1.5 m2, and τ = 1.5, we get = 0.225 m, h = 0.845 m. Number of Boxes The landing area, 4.5 m by 6 m, needs to be extended to take the horizontal motion into account. The maximum length h that the motorcycle system can penetrate is sufficient for cushioning horizontal motion, so the pile should be (4.5 + h) × 6 × h (meters) in dimension. From this fact, we can calculate the number of boxes needed in the pile. From (7), we have h = mv2 2kτesS . Thus, we know the dimensions of the pile: Nh = # h a $ = # mv2 2kτesS $ = 4, Nw = # 6 $ , Nl = # 4.5 + h $ . (8) Numerically, we get Nh = 4, Nw = 27, Nl = 24; and the total number of boxes needed is N = 4 × 27 × 24 = 2592, with total surface area Stotal = 6a2N = 6 × 0.2252 × 2592 ≈ 780 m2
Fly with Confidence 30 ext, we ana lyze the change in cost if we alter the edge length of the boxes As an approximation, we use 6×(4.5+h)108+24h108+12 We have calculated for minimum h and e. If we increase l, that is, use bigger boxes, we need fewer boxes but the total cost increases since 12mea 6a2N=6·108+ 648+540C. The number of layers is 4 regardless of edge length; but for increased e, th pile is lengthened to ensure that the motorcycle will not burst out of the pile due to the reduction of dea In conclusion, smaller boxes lower the pile and are cost efficient; from com- putation, we should choose the minimum size 22.5 cm on a side, and need 2592 boxes Stacking Strategy In the above discussion, we assume that we stack the boxes regularly stacked (no overlapping). We examine several other stacking strategies Pyramid stack yramid stacking stacks fewer boxes on top and more boxes at the bottom When a stress is exerted on the pile, it is divided into the normal stress and shearing force along the slopes so that the downward stress diverge ohnson 1985. Furthermore, a pyramidal stack is more stable than a regular stack Mixed stack Mixed stacking is to stack boxes of different sizes in the pile; a common practice is to lay big boxes on the top and small boxes at the bottom For a regular stack, we considered the cushioning a motion with constant acceleration because we assumed that the supporting force provided by boxes is constant. However, this is not the case; generally speaking, the force is larger in the first few seconds, so decreasing the supporting force in the beginning is good for cushioning We have shown that big boxes provide less support than small boxes. So the mixed stack can be characterized as softer on the top and stiffer at the bottom In fact, this kind of pile is similar to a sponge cushion, which is often used in stunt filming, high jump, pole vault, etc. In addition, cardboard is superior to a sponge cushion in this situation; a sponge cushion is too soft, so the motorcycle may lose bala ance
Fly With Confidence 307 Next, we analyze the change in cost if we alter the edge length of the boxes. As an approximation, we use N = Vpile Vbox = Nh · 6 × (4.5 + h) 2 = 108 + 24h 2 = 108 + 12mv2 kτesS 2 . We have calculated for minimum h and . If we increase , that is, use bigger boxes, we need fewer boxes but the total cost increases, since Stotal = 6a2N = 6 · 108 + 12mv2 kτesS ≈ 648 + 540. The number of layers is 4 regardless of edge length; but for increased , the pile is lengthened to ensure that the motorcycle will not burst out of the pile due to the reduction of DEA. In conclusion, smaller boxes lower the pile and are cost efficient; from computation, we should choose the minimum size 22.5 cm on a side, and need 2592 boxes. Stacking Strategy In the above discussion, we assume that we stack the boxes regularly stacked (no overlapping). We examine several other stacking strategies. Pyramid Stack Pyramid stacking stacks fewer boxes on top and more boxes at the bottom. When a stress is exerted on the pile, it is divided into the normal stress and shearing force along the slopes so that the downward stress diverge [Johnson 1985]. Furthermore, a pyramidal stack is more stable than a regular stack. Mixed Stack Mixed stacking is to stack boxes of different sizes in the pile; a common practice is to lay big boxes on the top and small boxes at the bottom. For a regular stack, we considered the cushioning a motion with constant acceleration because we assumed that the supporting force provided by boxes is constant. However, this is not the case; generally speaking, the force is larger in the first few seconds, so decreasing the supporting force in the beginning is good for cushioning. We have shown that big boxes provide less support than small boxes. So the mixed stack can be characterized as softer on the top and stiffer at the bottom. In fact, this kind of pile is similar to a sponge cushion, which is often used in stunt filming, high jump, pole vault, etc. In addition, cardboard is superior to a sponge cushion in this situation; a sponge cushion is too soft, so the motorcycle may lose balance.
308 The UMAP Journal 24.3(2003) Sparse stack ot arse stacking reserves a space c between adjacent boxes. Because each absorbs a constant amount of energy, the spaces decrease the density of energy absorption p. Thus, given the initial kinetic energy the height of the ile must increase to compensate for the decrease in p. Since pSh= E from(6), we have 知===m=(+) So, the cushioning distance h is proportional to the square of (1+c/e) sparse stack saves some material at the cross section but increase the cushie distance. With no change in base area, for rectangular stacking the surface area is constant, while in a pyramid stacking it decreases Crossed stack face d 01 B Crossed stacking is to lay the upper boxes on the intersection of the lower boxes, as shown in Figure 3. There are two kinds of interactions on the surface vertex-to-face: the interaction between pole of the upper box and the surface of the lower box. Because the pole is much stronger than the edge and surface, the pole will not deform but the upper surface may break down edge-to-edge: the interaction between edge of the upper box and the per pendicular edge of the lower box. The two edges both bend over. Figure 4 shows the deformation of the boxes compare the pile height of the two approaches ter than regular stacking, we To determine whether crossed stacking is be
308 The UMAP Journal 24.3 (2003) Sparse Stack Sparse stacking reserves a space c between adjacent boxes. Because each box absorbs a constant amount of energy, the spaces decrease the density of energy absorption ρ. Thus, given the initial kinetic energy, the height of the pile must increase to compensate for the decrease in ρ. Since ρSh = E from (6), we have hnew hold = ρold ρnew = 1/2 1/( + c)2 = 1 + c 2 . So, the cushioning distance h is proportional to the square of (1 + c/). The sparse stack saves some material at the cross section but increase the cushioning distance. With no change in base area, for rectangular stacking the surface area is constant, while in a pyramid stacking it decreases. Crossed Stack Figure 3. Crossed stacking: side and top views. Crossed stacking is to lay the upper boxes on the intersection of the lower boxes, as shown in Figure 3. There are two kinds of interactions on the surface: • vertex-to-face: the interaction between pole of the upper box and the surface of the lower box. Because the pole is much stronger than the edge and surface, the pole will not deform but the upper surface may break down. • edge-to-edge: the interaction between edge of the upper box and the perpendicular edge of the lower box. The two edges both bend over. Figure 4 shows the deformation of the boxes. To determine whether crossed stacking is better than regular stacking, we compare the pile height of the two approaches