You Too Can Be james Bond 263 You too Can Be james bond Deng X 8 Xu W Zhang Zhenyu Southeast university Advisor: Chen enhui Abstract We divide the jump into three phases: flying through the air, punching through the stack, and landing on the ground. We construct four models to minimize the number and the cost of boxes In the Ideal Mechanical model, we consider the boxes'force on the motorcycle and stunt person as constant. In the Realistic Mechanical model, we focus on how the boxes support the motorcycle and stunt person, which includes three phases: elastic deformation, plastic deformation, and and crush-down deforma- tion. However, in the Ideal Air Box model, the internal air pressure of each box an' t be ignored. As a matter of fact, the boxes are unsealed, so we amend the Ideal Air Box model to develop a Realistic Air Box model. We discuss the strengths and weaknesses of each model We define a metric U, which is a function of the cost and the number of boxe By mathematical programming, we calculate the size and the number of the boxes In normal conditions, we that assume the safe speed is 5.42 m/s. Fora total weight of stunt person and motorcycle of 187 kg, we need 196 boxes of size 0.7m x07m x 0.5 m. We analyze the accuracy and sensitivity of the result to such factors as the total weight, the contact area, and the velocity. We also offer some important uggestions on how to pile up the boxes and how to change the shape of the Assumptions and Analysis about boxes and the pile of boxes All the boxes are the same size. The ratio of length to width has little effect on the compression strength of a cardboard box; so to simplify the problem, we assume a square cross section The UMAP Journal 24(3)(2003)263-280. 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
You Too Can Be James Bond 263 You Too Can Be James Bond Deng Xiaowei Xu Wei Zhang Zhenyu Southeast University Nanjing, China Advisor: Chen Enshui Abstract We divide the jump into three phases: flying through the air, punching through the stack, and landing on the ground. We construct four models to minimize the number and the cost of boxes. In the Ideal Mechanical model, we consider the boxes’ force on the motorcycle and stunt person as constant. In the Realistic Mechanical model, we focus on how the boxes support the motorcycle and stunt person, which includes three phases: elastic deformation, plastic deformation, and and crush-down deformation. However, in the Ideal Air Box model, the internal air pressure of each box can’t be ignored. As a matter of fact, the boxes are unsealed, so we amend the Ideal Air Box model to develop a Realistic Air Box model. We discuss the strengths and weaknesses of each model. We define a metric U, which is a function of the cost and the number of boxes. By mathematical programming, we calculate the size and the number of the boxes. In normal conditions, we that assume the safe speed is 5.42 m/s. For a total weight of stunt person and motorcycle of 187 kg, we need 196 boxes of size 0.7 m × 0.7 m × 0.5 m. We analyze the accuracy and sensitivity of the result to such factors as the total weight, the contact area, and the velocity. We also offer some important suggestions on how to pile up the boxes and how to change the shape of the boxes. Assumptions and Analysis About Boxes and the Pile of Boxes • All the boxes are the same size. The ratio of length to width has little effect on the compression strength of a cardboard box; so to simplify the problem, we assume a square cross section. The UMAP Journal 24 (3) (2003) 263–280. 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
264 The UMAP Journal 24.3 (2003) Table 1 Variables, parameters, and physical constant Notation Description Units h height of box zA perimeter around top of bo total surface area of box pressure in box(cylinder) PPkmm pressure in box at time t when it collapses pressure in box at time t, in atmospheres atm rate of air leaking from box m3/s volume of box(cylinder) volume of box in time interval i Pile length of the pile m width of the pile height of the pile L number of layers of boxes in the pile total number of boxes ost of boxes S upper surface area of pile Jump Elephant average height of the elephant Hm naximum height of the jump 4 m fraction of Hmax that a person can reach gle of the ramp launch speed /s safe speed at which to hit the ground mass of motor plus stunt person Kellicut formula F compressive strength of the box P comprehensive annular compressive strength of the pape corrugation constant circumference of the top surface of the box J box shape coefficient Fo maximum supporting force from the box F buffering force of the box N constant concerning the properties of paper Other distance that the cylinder is compressed linder displacement when box collapses ompression distance at which box collapses dt interval of time W )s, k1, k2 quantities related to cost 入h,入 weight factors T, functions to be optimized Constants acceleration due to gravity, at Earths surface standard atmospheric ressure at Earth's surface Pa
264 The UMAP Journal 24.3 (2003) Table 1. Variables, parameters, and physical constants. Notation Description Units Box h height of box m r length of side of box m Z perimeter around top of box m A0 total surface area of box m2 P pressure in box (cylinder) Pa Pt pressure in box at time t when it collapses Pa k pressure in box at time t, in atmospheres atm σ rate of air leaking from box m3/s σi rate of air leaking from box in time interval i m3/s V volume of box (cylinder) m3 Vi volume of box in time interval i m3 Pile Lpile length of the pile m Wpile width of the pile m Hpile height of the pile m L number of layers of boxes in the pile Num total number of boxes Cost cost of boxes S upper surface area of pile m2 Jump Helephant average height of the elephant m Hmax maximum height of the jump 4 m ν fraction of Hmax that a person can reach H0 height of the ramp m θ angle of the ramp v0 launch speed m/s vsafe safe speed at which to hit the ground m/s M mass of motor plus stunt person kg Kellicut formula P compressive strength of the box Px comprehensive annular compressive strength of the paper dx2 corrugation constant Z circumference of the top surface of the box m J box shape coefficient F0 maximum supporting force from the box N F buffering force of the box N b constant concerning the properties of paper Other x, z distance that the cylinder is compressed m xt cylinder displacement when box collapses m zm compression distance at which box collapses m dt interval of time s W work done J Ds, k1, k2 quantities related to cost λh, λc weight factors T, U functions to be optimized Constants g acceleration due to gravity, at Earth’s surface m/s2 P0 standard atmospheric pressure at Earth’s surface Pa
You Too Can Be james Bond 265 After the box has been crushed down to some extent, we ignore the support- ing force that it can still supply Considering the practical production and transport limitations, the size of the box should not be too large The boxes are piled together in the shape of a rectangular solid When the motorcycle impacts one layer, it has little effect on the next layer The layer below is considered to be rigid flat(its displacement is ignored) We ignore the weight of the boxes; they are much lighter than the person plus motorcycle About the Stunt Person and the motorcycle We ignore the resistance of the air to the horizontal velocity of the person and motorcycle. The friction is so little that it is negligible The stunt person has received professional training, is skilled, and is equipped nything allowable for The average weight of a stunt person is 70 kg We choose a certain type of motorcycle(e.g, Yamaha 2003 TT-R225), which weighs 259 Ib [Yamaha Motor Corp. 2003 About the elephant The elephant keeps calm during the jump We adopt the classic value of 3. 5 m for the height of the elephant [PBS Online n.d. about the weather The weather is fine for filming and jumping, including appropriate tem- perature and humidity. On a gusty day, the wind might make the person lose balance in the air about the Camera The most attractive moment is when the person is over the elephant and maximum height Hmax. We have to make sure that the boxes do not appear on camera, namely, we need Pile vHmax, where the coefficient v is best determined empirically In our model, we set v=0625
You Too Can Be James Bond 265 • After the box has been crushed down to some extent, we ignore the supporting force that it can still supply. • Considering the practical production and transport limitations, the size of the box should not be too large. • The boxes are piled together in the shape of a rectangular solid. • When the motorcycle impacts one layer, it has little effect on the next layer. The layer below is considered to be rigid flat (its displacement is ignored). • We ignore the weight of the boxes; they are much lighter than the person plus motorcycle. About the Stunt Person and the Motorcycle • We ignore the resistance of the air to the horizontal velocity of the person and motorcycle. The friction is so little that it is negligible. • The stunt person has received professional training, is skilled, and is equipped with anything allowable for protection. • The average weight of a stunt person is 70 kg. • We choose a certain type of motorcycle (e.g., Yamaha 2003 TT-R225), which weighs 259 lb [Yamaha Motor Corp. 2003]. About the Elephant • The elephant keeps calm during the jump. • We adopt the classic value of 3.5 m for the height of the elephant [PBS Online n.d.]. About the Weather The weather is fine for filming and jumping, including appropriate temperature and humidity. On a gusty day, the wind might make the person lose balance in the air. About the Camera The most attractive moment is when the person is over the elephant and at maximum height Hmax. We have to make sure that the boxes do not appear on camera, namely, we need Hpile ≤ νHmax, where the coefficient ν is best determined empirically. In our model, we set ν = 0.625
266 The UMAP Journal 24.3(2003) About the ramp for Jumping The ramp for jumping is a slope at angle e of length Slope, as determined by the height or horizontal distance for landin The Development of models When the stunt person begins to contact the cardboard boxes or the ground, he or she may suffer great shock. To absorb the momentum, the contact time must be extended We divide the whole process into three independent phases, that is 1. flying through the air, 2. punching through the stack, and 3. landing on the ground find out the maximum height in phase 1, the greatest speed of hitting the ground in phase 3, and how the person plus motorcycle interact with the boxes is based on the results of phases 1 and 3. Phases 1 and 3 are simple and we solve them first Flying through the air The stunt person leaves the ramp with initial speed vo at angle g to the horizontal at height Ho(Figure 1) Initial Person a The Height of Elepl A Pile of Cardboard Boxe gure 1. The jum Q In the air, the stunt person on the motorcycle is affected only by the constant eleration of gravity. Based on Newtons Second Law, we have (t)=(vo cos 8)t, t)=(vo sin 0)t -igt
266 The UMAP Journal 24.3 (2003) About the Ramp for Jumping The ramp for jumping is a slope at angle θ of length Lslope, as determined by the jump height or horizontal distance for landing. The Development of Models When the stunt person begins to contact the cardboard boxes or the ground, he or she may suffer great shock. To absorb the momentum, the contact time must be extended. We divide the whole process into three independent phases, that is: 1. flying through the air, 2. punching through the stack, and 3. landing on the ground. We find out the maximum height in phase 1, the greatest speed of hitting the ground in phase 3, and how the person plus motorcycle interact with the boxes is based on the results of phases 1 and 3. Phases 1 and 3 are simple and we solve them first. Flying through the Air The stunt person leaves the ramp with initial speed v0 at angle θ to the horizontal at height H0 (Figure 1). Figure 1. The jump. In the air, the stunt person on the motorcycle is affected only by the constant acceleration of gravity. Based on Newton’s Second Law, we have x(t)=(v0 cos θ)t, y(t)=(v0 sin θ)t − 1 2 gt2
You Too Can Be james Bond 267 where a(t)and y(t)are the horizontal and vertical displacements from the launch point after t seconds. The launch speed vo and the maximum height h g(Hmax-Ho) For an elephant of height 3.5 m, we take Hmax =4 m. For Ho=0.5 m and 0=30, we get vo=v2.9.8.3. 5/0.5=16.6 m/s With a 2 m-high box-pile the stunt person hits the pile with vertical speed 6.3 m/s and horizontal speed 14.3 m/s; the distance between the landing point and elephant is D=9.2m Would the landing be safe? To simplify the problem, we ignore the complex process when the person begins to touch the ground. We consider that there is a critical safe speed usafe If the speed hitting the ground is less than or equal to that speed the person would not be injured. The safe speed is related to the ground surface(hard, grassplot, mud, etc )and materials used(paper, rubber etc. ) Our simulation uses a typical valu 5.42m/ Is the Pile area Large Enough? The height Hile of the pile of boxes is related to the maximum height Hmax that the stunt person reaches and also to the vertical speed of hitting the boxes The greater Hmax, the greater Pile is required, with Pile- Lh, where L is the number of the layers of boxes and h is the height of a single box Would Lpile equal to the length of the person be enough? The answer is no When accelerating on the ramp, the stunt person cant make the initial jump speed exactly what we calculate. We think that 3-5 times the length of the person is needed The stunt person does not leave the ramp aligned exactly along the central axis and does not keep the motorcycle exactly along that axis after hitting the boxes. That there may be some horizontal movement means that Pile should be 24 times the length of the person n our simulation we let Lpile=6 m, Pile=4 m Boxes: How to Cushion the person Ideal mechanical model is destroying the eneral assumptions, we suppose that while the stunt person boxes of the current layer, boxes in lower layers are seldom affected and ke
You Too Can Be James Bond 267 where x(t) and y(t) are the horizontal and vertical displacements from the launch point after t seconds. The launch speed v0 and the maximum height Hmax are related by v0 sin θ = 2g(Hmax − H0). For an elephant of height 3.5 m, we take Hmax = 4 m. For H0 = 0.5 m and θ = 30◦, we get v0 = √2 · 9.8 · 3.5/0.5 = 16.6 m/s. With a 2 m-high box-pile, the stunt person hits the pile with vertical speed 6.3 m/s and horizontal speed 14.3 m/s; the distance between the landing point and elephant is D = 9.2 m. Would the Landing Be Safe? To simplify the problem, we ignore the complex process when the person begins to touch the ground. We consider that there is a critical safe speed vsafe. If the speed hitting the ground is less than or equal to that speed, the person would not be injured. The safe speed is related to the ground surface (hard, grassplot, mud, etc.) and materials used (paper, rubber etc.). Our simulation uses a typical value, vsafe = 5.42 m/s. Is the Pile Area Large Enough? The height Hpile of the pile of boxes is related to the maximum height Hmax that the stunt person reaches and also to the vertical speed of hitting the boxes. The greater Hmax, the greater Hpile is required, with Hpile = Lh, where L is the number of the layers of boxes and h is the height of a single box. Would Lpile equal to the length of the person be enough? The answer is no. When accelerating on the ramp, the stunt person can’t make the initial jump speed exactly what we calculate. We think that 3–5 times the length of the person is needed. The stunt person does not leave the ramp aligned exactly alomg the central axis and does not keep the motorcycle exactly along that axis after hitting the boxes. That there may be some horizontal movement means that Wpile should be 2–4 times the length of the person. In our simulation, we let Lpile = 6 m, Wpile = 4 m. Boxes: How to Cushion the Person Ideal Mechanical Model Based on our general assumptions, we suppose that while the stunt person is destroying the boxes of the current layer, boxes in lower layers are seldom affected and keep still
268 The UMAP Journal 24.3 (2003) To illustrate the process of collision with just one layer separately, we sup pose that the mutual effect of the stunt person and the box-pile is motion with a constant acceleration. During that, the stunt person plus motorcycle is sup- ported by a constant vertical force F--that is, we treat f as the average force during the whole process It can be proved that although the stunt person strikes the boxes of different layers at different initial velocities, the work consumed to fall through each box is the same(Appendix A). The number of layers L of the box-pile is determined by the formula Work×L where L is the smallest integer that satisfies this inequality. Strength and weaknesses This model is simple and efficient. However, we have ignored the detailed process and substituted constant work, though in fact the force changes with Realistic mechanical Model First we study the empirical deformed load curve of the cushion system 2000 g the boxes'deformation under a static load(Figure 2)[Yan and Yuan Deforation Figure 2. Deformation curve The compression process is divided into three phases, as shown in the figure Oa phase: Elastic deformation, according to Hooke law AB Phase: Plastic deformation. The compression grows more slowly and reaches the maximum BC phase: Crush-down deformation: After compression reaches the maxi- mum the rate of deformation starts to fall. The unrecoverable deformation goes on IncreasIng
268 The UMAP Journal 24.3 (2003) To illustrate the process of collision with just one layer separately, we suppose that the mutual effect of the stunt person and the box-pile is motion with a constant acceleration. During that, the stunt person plus motorcycle is supported by a constant vertical force F— that is, we treat F as the average force during the whole process. It can be proved that although the stunt person strikes the boxes of different layers at different initial velocities, the work consumed to fall through each box is the same (Appendix A). The number of layers L of the box-pile is determined by the formula Work × L − mgh ≥ 1 2mv2 0 − 1 2mv2 safe, where L is the smallest integer that satisfies this inequality. Strength and Weaknesses This model is simple and efficient. However, we have ignored the detailed process and substituted constant work, though in fact the force changes with time. Realistic Mechanical Model First we study the empirical deformed load curve of the cushion system, showing the boxes’ deformation under a static load (Figure 2) [Yan and Yuan 2000]. Figure 2. Deformation curve. The compression process is divided into three phases, as shown in the figure: • OA phase: Elastic deformation, according to Hooke Law. • AB phase: Plastic deformation. The compression grows more slowly and reaches the maximum. • BC phase: Crush-down deformation: After compression reaches the maximum, the rate of deformation starts to fall. The unrecoverable deformation goes on increasing
You Too Can Be james Bond 269 According to the Kellicut formula [Yan and Yuan 2000; Zhao et al. the com pressive strength of a box is d P=P Z/4 ZJ where P is the compressive strength of the box, Pr is the comprehensive annular compressive strength of the paper, d is the corrugation constant Z is the circumference of the top surface, and J is the box shape coefficient For the stunt person plus motorcycle, the maximum supporting force from the box is nearly 2/3 F0=D=P2)2J1p=b2-5, where b is a constant concerning the properties of paper We assimilate the static loading process to the dynamical process of getting impacted and obtain the following buffering force and its deformation graph gure Fo F ≤x≤b; h The model describes the mechanical capability of the box and offers an ap propriate curve of the relationship between buffering power and deformation We can measure the energy consumed by the crushing of boxes. One limitation is the Kellicut formula, which applies only to certain kinds of cardboard boxes Error may also occur in replacing the dynamical process with a static process Ideal air box model We consider the depleting energy consumed by the resistance of air in the process of compression. We divide the process into two phases Phase 1: Assume that the cardboard is closed(gas can' t escape). The pressure in the box rises from standard atmospheric pressure Po to Pt=k po(k atmo- sheres)at time t when the box ruptures. We consider that the impact is so
You Too Can Be James Bond 269 According to the Kellicut formula [Yan and Yuan 2000; Zhao et al.], the compressive strength of a box is P = Px dx2 Z/4 1/3 ZJ, where P is the compressive strength of the box, Px is the comprehensive annular compressive strength of the paper, dx2 is the corrugation constant, Z is the circumference of the top surface, and J is the box shape coefficient. For the stunt person plus motorcycle, the maximum supporting force from the box is nearly F0 = P s (Z/4)2 = Px dx2 Z/4 2/3 ZJ s (Z/4)2 = bZ−5/3s, where b is a constant concerning the properties of paper. We assimilate the static loading process to the dynamical process of getting impacted and obtain the following buffering force and its deformation graph (Figure 3). F = F0 a x, x ≥ a; F0, a ≤ x ≤ b; F0 exp −a(x − b) h , x ≥ b. The model describes the mechanical capability of the box and offers an appropriate curve of the relationship between buffering power and deformation. We can measure the energy consumed by the crushing of boxes. One limitation is the Kellicut formula, which applies only to certain kinds of cardboard boxes. Error may also occur in replacing the dynamical process with a static process. Ideal Air Box Model We consider the depleting energy consumed by the resistance of air in the process of compression. We divide the process into two phases: Phase 1: Assume that the cardboard is closed (gas can’t escape). The pressure in the box rises from standard atmospheric pressure P0 to Pt = kP0 (k atmospheres) at time t when the box ruptures. We consider that the impact is so
270 The UMAP Journal 24.3(2003) b Deformation Figure 3. Force-deformation figure quick that the gas in the cardboard doesnt exchange energy with environ- ment. So we can just deem it an adiabatic compression process. Using the First Law of Thermodynamics, we get PV= constant The proof is in Appendix B Phase 2: Under the effect of the impact and internal air pressure, cracks appear in the wall of the cardboard box. The internal air mixes with the air outside quickly falling to standard atmospheric pressure Po. We assume that the cardboard box is a rigid cylinder and the compressive face(top surface) of the box is a piston( Figure 4) We calculate the internal pressure when the piston shifts downward a dis- tance a from height h. Let s be the area of the top of the cylinder. According to(1), we have P·[hs]14=P·(h-x)s}4, h P= Po h The graph of P is shown in Figure 5 Consider Phase 1. According to(2), we have h Pt=kPo= Po h We solve for the displacement at time t k
270 The UMAP Journal 24.3 (2003) Figure 3. Force-deformation figure. quick that the gas in the cardboard doesn’t exchange energy with environment. So we can just deem it an adiabatic compression process. Using the First Law of Thermodynamics, we get P V 1.4 = constant. (1) The proof is in Appendix B. Phase 2: Under the effect of the impact and internal air pressure, cracks appear in the wall of the cardboard box. The internal air mixes with the air outside, quickly falling to standard atmospheric pressure P0. We assume that the cardboard box is a rigid cylinder and the compressive face (top surface) of the box is a piston (Figure 4). We calculate the internal pressure when the piston shifts downward a distance x from height h. Let s be the area of the top of the cylinder. According to (1), we have P0 · [hs] 1.4 = P · [(h − x)s] 1.4, P = P0 h h − x 1.4 . (2) The graph of P is shown in Figure 5. Consider Phase 1. According to (2), we have Pt = kP0 = P0 h h − x 1.4 . We solve for the displacement at time t: xt = h 1 − k−5/7 .
You Too Can Be james Bond 271 Air the air pressure Figure 4. The cylinder model Figure 5. Internal pressure as a function of the volume of compressed air. The compression work done in Phase 1 is Pdv=s h P Posh=k2 +k Consider Phase 2. The compression work of Phase 2 is W Po-po)dV so the total work done is given by(3) In this model, first we get the curve of internal pressure and deformation Based on the graph, we calculate the energy consumed by the resistance of gas However, considering the box as a rigid cylinder may be inaccurate. Another weakness is the airtightness of the box; actually, the internal gas leaks during the whole process of compression
You Too Can Be James Bond 271 Figure 4. The cylinder model. Figure 5. Internal pressure as a function of the volume of compressed air. The compression work done in Phase 1 is W1 = xt 0 P dV = s h(1−k−5/7) 0 P0 h h − x 1.4 − 1 ! dx = P0sh 5 2 k2/7 − 7 2 + k−5/7 . (3) Consider Phase 2. The compression work of Phase 2 is W2 = (P0 − P0) dV = 0, so the total work done is given by (3). In this model, first we get the curve of internal pressure and deformation. Based on the graph, we calculate the energy consumed by the resistance of gas. However, considering the box as a rigid cylinder may be inaccurate. Another weakness is the airtightness of the box; actually, the internal gas leaks during the whole process of compression.
272 The UMAP Journal 24.3 (2003) Realistic air box model The height of the box is less than 1 m, so the speed of the motorcycle and stunt person changes little when passing through one box height's distance hence, we think of it as constant during compression of a single box. The effect of the air in the crushing process is just like the air in a cylinder with a hole to leak air. In Figure 6, we have s is the area of the top of the box(top of the cylinder), h is the height of the box(cylinder) o is the air-leaking rate, z is the crushing distance from the top of the box(cylinder), and v is the speed of the motorcycle and stunt person Figure 6. The unsealed cylinder model. In the crushing process, the cardboard bo and larger. As a result, the air leaks more quickly. Generally, when a cardboard box is pressed to about half of its original height, the effect of air-leaking is so prominent that we cannot ignore it L We assume that the air-leaking rate changes with the crushing distance z forz∈(0,h).Letx=2/ h for a∈(0,1) The goal is to calculate the total work that the air does to the motorcycle and stunt person. It is necessary to calculate the pressure of the air in a box while the crushing is occurring. We divide the crushing process into N time periods; in each time period the pressure of the air can be calculated according to the universal gas law PV=nRT. We assume that temperature is constant, so PV In the first time period, we have PoVo= Pogo dt+(Vo- su dt)P1 P=Po(Vo-go dt
272 The UMAP Journal 24.3 (2003) Realistic Air Box Model The height of the box is less than 1 m, so the speed of the motorcycle and stunt person changes little when passing through one box height’s distance; hence, we think of it as constant during compression of a single box. The effect of the air in the crushing process is just like the air in a cylinder with a hole to leak air. In Figure 6, we have s is the area of the top of the box (top of the cylinder), h is the height of the box (cylinder), σ is the air-leaking rate, z is the crushing distance from the top of the box (cylinder), and v is the speed of the motorcycle and stunt person. Figure 6. The unsealed cylinder model. In the crushing process, the cardboard box’s deformation becomes larger and larger. As a result, the air leaks more quickly. Generally, when a cardboard box is pressed to about half of its original height, the effect of air-leaking is so prominent that we cannot ignore it. We assume that the air-leaking rate changes with the crushing distance z via σ(z)=2e4z/hs, for z ∈ (0, h). Let x = z/h for x ∈ (0, 1). The goal is to calculate the total work that the air does to the motorcycle and stunt person. It is necessary to calculate the pressure of the air in a box while the crushing is occurring. We divide the crushing process into N time periods; in each time period, the pressure of the air can be calculated according to the universal gas law P V = nRT. We assume that temperature is constant, so P V is constant. In the first time period, we have P0V0 = P0σ0 dt + (V0 − sv dt)P1, P1 = P0(V0 − σ0 dt V0 − sv dt
