Safe 219 Safe landings Chad t kishimoto Justin C. Kao Jeffrey A. edlund California Institute of Technology Pasadena, California 91125 Advisor: Darryl H. Yor Abstract We examine the physical principles of stunt crash pads made of corrugated cardboard boxes and build a mathematical model to describe them. The model leads to a computer simulation of a stunt person impacting a box catcher. Together, the simulation and model allow us to predict the behavior of box catchers from physical parameters and hence design them for maximum safety and minimum We present two case studies of box-catcher design, a motorcyclist landing after jumping over an elephant and David Blaine's televised Vertigo stunt. These demonstrate the ability of our model to handle both high-speed impacts and large weights. For each case, we calculate two possible box-catcher designs, showing the effects of varying design parameters. Air resistance is the dominant force with high impact speeds, while box buckling provides greater resistance at low speeds We also discuss other box-catcher variations Basic concept Requirements A falling stunt person has a large amount of kinetic energy To land safely, most of it must be absorbed by a catcher before the performer hits the ground. Therefore, following Newtons Second Law, the catcher must exert a force to decelerate the performer, dt The UMAP Journal 24(3)(2003)219-232. 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
Safe Landings 219 Safe Landings Chad T. Kishimoto Justin C. Kao Jeffrey A. Edlund California Institute of Technology Pasadena, California 91125 Advisor: Darryl H. Yong Abstract We examine the physical principles of stunt crash pads made of corrugated cardboard boxes and build a mathematical model to describe them. The model leads to a computer simulation of a stunt person impacting a box catcher. Together, the simulation and model allow us to predict the behavior of box catchers from physical parameters and hence design them for maximum safety and minimum cost. We present two case studies of box-catcher design, a motorcyclist landing after jumping over an elephant and David Blaine’s televised Vertigo stunt. These demonstrate the ability of our model to handle both high-speed impacts and large weights. For each case, we calculate two possible box-catcher designs, showing the effects of varying design parameters. Air resistance is the dominant force with high impact speeds, while box buckling provides greater resistance at low speeds. We also discuss other box-catcher variations. Basic Concept Requirements A falling stunt person has a large amount of kinetic energy, Us = 1 2 msu2 s . To land safely, most of it must be absorbed by a catcher before the performer hits the ground. Therefore, following Newton’s Second Law, the catcher must exert a force to decelerate the performer, F = m dus dt . The UMAP Journal 24 (3) (2003) 219–232. 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
220 The UMAP Journal 24.3(2003) The total energy absorbed should equal the performer's initial energy(kinetic and potential) Fda =Uso vso However, the catcher itself cannot exert too large a force or it would be no better than having the performer hit the ground in the first place. We therefore set a maximum force, Fmax. The smaller we make F, the larger(and more expensive) the box catcher has to be. Therefore, to save both money and life, we would like to have F≈(1-6)Fmax, where d is a safety margin(0<8<1) The box catcher a box catcher consists of many corrugated cardboard boxes, stacked in yers, possibly with modifications such as ropes to keep the boxes together or inserted sheets of cardboard to add stability and distribute forces. When the stunt person falls into the box catcher, the impact crushes boxes beneath. As box collapses, not only does the cardboard get torn and crumpled, but the air inside is forced out, providing a force resisting the fall that is significant but not too large. As the performer passes though the layers, each layer takes away some kinetic energy Modeling the Cardboard Box We examine in detail the processes involved when a stunt person vertically impacts a single cardboard box. This analysis allows us to predict the effect of varying box parameters(shape, size, etc. on the amount of energy absorbed by the box Assumptions: Sequence of Events Although the impact involve ly complex interactions--between the performer's posture, the structure of the box, the air inside the box, the support of the box, the angle and location of impact, and other details--modeling thin- shell buckling and turbulent compressible flow is neither cost-effective for a movie production nor practical for a paper of this nature. We therefore assume and describe separately the following sequence of events in the impact 1. A force is applied to the top of the cardboard box
220 The UMAP Journal 24.3 (2003) The total energy absorbed should equal the performer’s initial energy (kinetic and potential), F dz = Us0 + Vs0 . However, the catcher itself cannot exert too large a force or it would be no better than having the performer hit the ground in the first place. We therefore set a maximum force, Fmax. The smaller we make F, the larger (and more expensive) the box catcher has to be. Therefore, to save both money and life, we would like to have F ≈ (1 − δ)Fmax, where δ is a safety margin (0 <δ< 1). The Box Catcher A box catcher consists of many corrugated cardboard boxes, stacked in layers, possibly with modifications such as ropes to keep the boxes together or inserted sheets of cardboard to add stability and distribute forces. When the stunt person falls into the box catcher, the impact crushes boxes beneath. As a box collapses, not only does the cardboard get torn and crumpled, but the air inside is forced out, providing a force resisting the fall that is significant but not too large. As the performer passes though the layers, each layer takes away some kinetic energy. Modeling the Cardboard Box We examine in detail the processes involved when a stunt person vertically impacts a single cardboard box. This analysis allows us to predict the effect of varying box parameters (shape, size, etc.) on the amount of energy absorbed by the box. Assumptions: Sequence of Events Although the impact involves many complex interactions—between the performer’s posture, the structure of the box, the air inside the box, the support of the box, the angle and location of impact, and other details—modeling thinshell buckling and turbulent compressible flow is neither cost-effective for a movie production nor practical for a paper of this nature. We therefore assume and describe separately the following sequence of events in the impact. 1. A force is applied to the top of the cardboard box
Safe landin 2. The force causes the sides of the box to buckle and lose structural integrity 3. Air is pushed out of the box as the box is crushed 4. The box is fully flattened We now consider the physical processes at play in each of these stages Table 1 Nomenclature Property Symbol Units Potential energy density J/m Youngs modulus Strain in box walls none Stress in box walls P Tensile strength Ts P Total kinetic energy Total potential energ. Volume of cardboard in box walls Distance scale over which buckling is significant AH m Width of box Thickness of box walls m Height of box of the top face of the be Proportion of top face through which air escapes Velocity of stunt person m/s an velocity of expelled air Mass of stur Density of air Acceleration due to gravity, g=9.8m/s2 Stage 1: Force Applied A force F(t) is applied uniformly over the top surface. The walls of the box expand slightly, allowing the box to compress longitudinally (along the direction of the force). While this applied force is small enough (less than the force necessary to cause buckling), the box absorbs little of the force and transmits to the ground (or next layer of the box catcher) a large fraction of the applied force. The applied force increases until it is on the order of the bucklin orce(F(t)N FB). At this point, the box begins to buckle Stage 2: Box Buckles The walls crumple, the box tears, and the top of the box is pressed down Once the box has lost structural integrity, although the action of deforming the box may present some resistance the force that the box itself can withstand is greatly diminished Consider the pristine box being deformed by a force applied to its upper face. To counteract this force, the walls of the box expand in the transverse
Safe Landings 221 2. The force causes the sides of the box to buckle and lose structural integrity. 3. Air is pushed out of the box as the box is crushed. 4. The box is fully flattened. We now consider the physical processes at play in each of these stages. Table 1. Nomenclature. Property Symbol Units Potential energy density v J/m3 Young’s modulus Y Pa Strain in box walls none Stress in box walls S Pa Tensile strength TS Pa Total kinetic energy U J Total potential energy V J Volume of cardboard in box walls V m3 Distance scale over which buckling is significant ∆H m Width of box w m Thickness of box walls τ m Height of box m Surface area of the top face of the box A m2 Proportion of top face through which air escapes α — Velocity of stunt person us m/s Mean velocity of expelled air ua m/s Mass of stunt person (and vehicle, if any) m kg Density of air ρ kg/m3 Acceleration due to gravity, g = 9.8 m/s2 g m/s2 Stage 1: Force Applied A force F(t) is applied uniformly over the top surface. The walls of the box expand slightly, allowing the box to compress longitudinally (along the direction of the force). While this applied force is small enough (less than the force necessary to cause buckling), the box absorbs little of the force and transmits to the ground (or next layer of the box catcher) a large fraction of the applied force. The applied force increases until it is on the order of the buckling force (F(t) ∼ FB). At this point, the box begins to buckle. Stage 2: Box Buckles The walls crumple, the box tears, and the top of the box is pressed down. Once the box has lost structural integrity, although the action of deforming the box may present some resistance, the force that the box itself can withstand is greatly diminished. Consider the pristine box being deformed by a force applied to its upper face. To counteract this force, the walls of the box expand in the transverse
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△HY
222 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)
Safe landin To determine AH, we assume an average force Fa FB, where FB is the buckling force of the cardboard walls. The Appendix shows that the buckle force is related to the Young s modulus and other physical parameters bp Ing F 12g2 In this case, V=cWr for a square box. Then AH can be estimated by FB△H T △H=2C For a typical box, AH is on the order of a few centimeters Stage 3: Box is Crushed Without structural integrity, the box is crushed by the applied force. Hov ever, in crushing the box, the air inside must be pushed out Let a be the surface area of the top of the box. We make the ad hoc as- sumption that when the box buckled, it is torn such that air can escape through an area a A. Provided a N O(1), we can assume incompressible flow, since the area of opening in the box is of the same order of magnitude as the area being pushed in. By conservation of mass, we obtain the the velocity of the air moving out of the box in terms of a and the velocity of the stunt person SA= ua(aa) Using conservation of energy, we equate the change in energy of the stunt person and the air leaving the box, dos dve dua dva\ The potential energy of the air does not change significantly as it is ejected, so the energy equation simplifies to dos dvs dU The energy gain of air outside the box is due to air carrying kinetic energy out of the boX dUa ldma u2=2(padua)u2 dt
Safe Landings 223 To determine ∆H, we assume an average force F ≈ FB, where FB is the buckling force of the cardboard walls. The Appendix shows that the buckling force is related to the Young’s modulus and other physical parameters by FB = Y π2 12 wτ 3 2 . (2) In this case, V = 4wτ for a square box. Then ∆H can be estimated by FB∆H = 1 2 T2 S Y V, Y π2 12 wτ 3 2 ∆H = 2wτ T2 S Y , ∆H = 24 π2 TS Y 2 τ 2 . For a typical box, ∆H is on the order of a few centimeters. Stage 3: Box is Crushed Without structural integrity, the box is crushed by the applied force. However, in crushing the box, the air inside must be pushed out. Let A be the surface area of the top of the box. We make the ad hoc assumption that when the box buckled, it is torn such that air can escape through an area αA. Provided α ∼ O(1), we can assume incompressible flow, since the area of opening in the box is of the same order of magnitude as the area being pushed in. By conservation of mass, we obtain the the velocity of the air moving out of the box in terms of α and the velocity of the stunt person: −usA = ua(αA) =⇒ ua = −us/α. Using conservation of energy, we equate the change in energy of the stunt person and the air leaving the box, dUs dt + dVs dt = − dUa dt + dVa dt . The potential energy of the air does not change significantly as it is ejected, so the energy equation simplifies to dUs dt + dVs dt = −dUa dt . (3) The energy gain of air outside the box is due to air carrying kinetic energy out of the box: dUa dt = 1 2 dma dt u2 a = 1 2 (ραAua) u2 a , (4)
224 The UMAP Journal 24.3 (2003) while the energy loss of the stunt person is from deceleration and falling: dt msu du dos dve t gus (5) Combining(3-5), substituting for ua, and rearranging, we obtain (6) Stage 4: Box Is Flattened Once the cardboard box is fully compressed and all the air is pushed out we assume that the box no longer has any effect on the stunt person. ummar In stages 1 and 4, the box is essentially inert--little energy goes into it Therefore in our mathematical description, we ignore these and concentrate on stages 2 and 3. Stage 2 occurs in the top AH of box deformation, while stage 3 occurs in the remainder of the deformation. Combining (1)and( 6), we get 2ms△HY ,|z|<|△Hi; 1 0A 2ma2-9,||21△H, where z is measured from the top of the box Modeling the Box Catcher Cardboard Properties We assume that each cardboard box is made of corrugated cardboard with uniform physical properties, using the data shown in Table 2 Table 2 Properties of corrugated cardboard [ Bever 1986 Property Symbol Value Tensile strength Ts 12.2MP Thickness 5 mm Youngs modulus 1.3 GPa
224 The UMAP Journal 24.3 (2003) while the energy loss of the stunt person is from deceleration and falling: dUs dt + dVs dt = msus dus dt + msgus. (5) Combining (3–5), substituting for ua, and rearranging, we obtain dus dt = 1 2 ρA ms u2 s α2 − g. (6) Stage 4: Box Is Flattened Once the cardboard box is fully compressed and all the air is pushed out, we assume that the box no longer has any effect on the stunt person. Summary In stages 1 and 4, the box is essentially inert—little energy goes into it. Therefore in our mathematical description, we ignore these and concentrate on stages 2 and 3. Stage 2 occurs in the top ∆H of box deformation, while stage 3 occurs in the remainder of the deformation. Combining (1) and (6), we get dus dt = 1 2ms∆H T2 S Y V, |z| < |∆H|; 1 2 ρA ms u2 s α2 − g, |z|≥|∆H|, (7) where z is measured from the top of the box. Modeling the Box Catcher Cardboard Properties We assume that each cardboard box is made of corrugated cardboard with uniform physical properties, using the data shown in Table 2. Table 2. Properties of corrugated cardboard [Bever 1986]. Property Symbol Value Tensile strength TS 12.2 MPa Thickness τ 5 mm Young’s modulus Y 1.3 GPa
Safe landings 225 Assumptions e Layers of boxes are comprised of identical boxes laid side-by-side over the entire area of the box catcher The box catcher is large enough compared to the stunt person that edge effects are negligible The boxes are held together so that there is no relative horizontal velocity between them Loose cardboard is placed between layers of cardboard boxes, so that any force transmitted through the top layer of boxes is well distributed to lower boxes, ensuring that only one layer is crushed at a time. In other words, we treat each layer of boxes independently and the box catcher as a sequence of Equations of Motion Since each layer is independent, the equations of motion for the box catcher look similar to the equation of motion for a single box(7). The equations depend on the dimensions of the boxes in each level, so we solve numerically for the motion of the stunt person. At each level of the box catcher, the performer impacts several boxes(approximately)at once. The number of boxes that act to area to the surface area of the top face of a cardboard box. Thus, we altera p decelerate the stunt person is the ratio As /A of the performers cross-sectiona equations of motion by this ratio, getting 1T2,A 2m△ByD <|△H (8) ≥|H where z is the vertical distance measured from the top of the stack, ztop is the value of z at the top of the current box, and As is the cross sectional area of the stunt person Now, given a suggested stack of boxes we can integrate the equations to see whether or not the stack successfully stops the falling stunt person, and if so, where in the stack
Safe Landings 225 Assumptions • Layers of boxes are comprised of identical boxes laid side-by-side over the entire area of the box catcher. • The box catcher is large enough compared to the stunt person that edge effects are negligible. • The boxes are held together so that there is no relative horizontal velocity between them. • Loose cardboard is placed between layers of cardboard boxes, so that any force transmitted through the top layer of boxes is well distributed to lower boxes, ensuring that only one layer is crushed at a time. In other words, we treat each layer of boxes independently and the box catcher as a sequence of layers. Equations of Motion Since each layer is independent, the equations of motion for the box catcher look similar to the equation of motion for a single box (7). The equations depend on the dimensions of the boxes in each level, so we solve numerically for the motion of the stunt person. At each level of the box catcher, the performer impacts several boxes (approximately) at once. The number of boxes that act to decelerate the stunt person is the ratio As/A of the performer’s cross-sectional area to the surface area of the top face of a cardboard box. Thus, we alter the equations of motion by this ratio, getting dus dt = 1 2ms∆H T2 S Y V As A |z − ztop| < |∆H|; 1 2 ρAs ms u2 s α2 − g, |z − ztop|≥|∆H|, (8) where z is the vertical distance measured from the top of the stack, ztop is the value of z at the top of the current box, and As is the cross sectional area of the stunt person. Now, given a suggested stack of boxes, we can integrate the equations of motion to see whether or not the stack successfully stops the falling stunt person, and if so, where in the stack
226 The UMAP Journal 24.3(2003) Analysis Our model allows us to predict a stunt persons fall given parameters for the box catcher. Now we would like to find analytic results to guide box-catcher Using the equations of motion( 8), we determine the force that the stunt person feels falling through the box catcher: 1 IS, As F=m dJ2△yA,.|2-al<△ 42-m39,|2-2opl≥△H We want to make this force large enough to stop but not ha arm the performer Therefore, we demand that F≤(1-6)Fmax However, we wish to find solutions that both minimize cost(fewest boxes) and conform to spatial constraints(we don't want the box catcher to be taller than the obstacle that the stunt person is jumping over). Thus, it is in our best interest to maximize the force applied to the performer subject to(9) We are faced with two independent equations with three unknowns; we solve for two of them in terms of the third. With the simplifying assumption that the top face of each box is a square(A=w), we can solve for the optimal dimensions of the box given the stunt persons impact velocity Y? w As 12g2A Fmax(1-8)=Thresh pA._m where i2 is the stunt persons velocity after causing the box to buckle(but before expelling all the air). So we have Define ?, the maximum number of gees of acceleration felt by the stunt person, thresh (1-6) max 2(-23+1)2m) (10) (11) 12m39C2
226 The UMAP Journal 24.3 (2003) Analysis Our model allows us to predict a stunt person’s fall given parameters for the box catcher. Now we would like to find analytic results to guide box-catcher design. Using the equations of motion (8), we determine the force that the stunt person feels falling through the box catcher: F = ms dus dt = 1 2∆H T2 S Y V As A , |z − ztop| < |∆H|; 1 2 ρAs u2 s α2 − msg, |z − ztop|≥|∆H|. We want to make this force large enough to stop but not harm the performer. Therefore, we demand that F ≤ (1 − δ)Fmax. (9) However, we wish to find solutions that both minimize cost (fewest boxes) and conform to spatial constraints (we don’t want the box catcher to be taller than the obstacle that the stunt person is jumping over). Thus, it is in our best interest to maximize the force applied to the performer subject to (9). We are faced with two independent equations with three unknowns; we solve for two of them in terms of the third. With the simplifying assumption that the top face of each box is a square (A = w2), we can solve for the optimal dimensions of the box given the stunt person’s impact velocity: π2 12 Y τ 3 w 2 As A = Fmax(1 − δ) = Fthresh, 1 2 ρ α2 Asu˜2 − msg = Fthresh, where u˜2 is the stunt person’s velocity after causing the box to buckle (but before expelling all the air). So we have u˜2 = u2 o − 4τ T2 S msY w As A . Define γ, the maximum number of gees of acceleration felt by the stunt person, by Fthresh = (1 − δ)Fmax = γmsg. We get 3 = π2 48γ Y TS 2 τ 2 g u2 0 − 2(γ + 1)α2 msg ρAs (10) and A(w) w = w = π2 12γ AsY msg τ 3 2 . (11)
Safe landings 227 Thus, we could write a routine that would integrate the equations of motion of the stunt person falling through each level of boxes and for each layer use the e incon ning velocity to calculate the optimal dimensions for the next layer of boxes. This would yield a box-catcher structure that would safely stop the stunt person in the fewest levels of boxes, minimizing the cost of the box catcher. General solutions The most difficult aspect of finding a general solution is the need to know the speed of the stunt person at impact, which depends on the height of the box catcher. Given sufficient computing time, we could solve the equations with condition that the height above the ground at impact is the height of the box catcher. However, this is unnecessary for a paper of this scope Instead we use the model to shed light on the qualitative aspects of building a box catcher. Equation (10)tells us that for higher speeds of thestunt person, we need taller boxes ( larger e )to keep the force on the stunt person at its maximum llowable level. This means that it is necessary to place the tallest boxes at the top of the stack, followed by shorter ones, for both cost effectiveness and safety. Inspecting (11) shows that the optimal width of the box is inversely proportional to the height (w x (-). Thus, the box catcher should have the thinnest boxes on top and the widest ones on the bottom It follows that the optimal box catcher has short, wide boxes at the bottom and tall, narrow boxes at the te Furthermore, the equations of motion(8)contain u- in the air expulsion stage but no velocity-dependent term in the buckling stage. Hence, for high impact speeds, air expulsion provides the dominant deceleration, while for smaller impacts, box buckling is the more important stage Results Motorcycle stunt A stunt person on a motorcycle is preparing to jump over an elephant for the filming of a blockbuster movie. What should be going through a stunt coordinator's mind? The average height of an African elephant is 4m; the mass of a top-of-the- line motorcycle is 230 kg Encyclopedia Britannica Online 2003] We assume that the stunt person clears the elephant by about 2 m, so the impact velocity is 8 m/s. Choosing the parameters (=3 m and w =2m,a 3 m-tall homogeneous box catcher stops the stunt person and motorcycle after they fall about 2.33 m(Figure 1)
Safe Landings 227 Thus, we could write a routine that would integrate the equations of motion of the stunt person falling through each level of boxes and for each layer use the incoming velocity to calculate the optimal dimensions for the next layer of boxes. This would yield a box-catcher structure that would safely stop the stunt person in the fewest levels of boxes, minimizing the cost of the box catcher. General Solutions The most difficult aspect of finding a general solution is the need to know the speed of the stunt person at impact, which depends on the height of the box catcher. Given sufficient computing time, we could solve the equations with condition that the height above the ground at impact is the height of the box catcher. However, this is unnecessary for a paper of this scope. Instead, we use the model to shed light on the qualitative aspects of building a box catcher. Equation(10)tells us that for higher speeds of the stunt person, we need taller boxes (larger ) to keep the force on the stunt person at its maximum allowable level. This means that it is necessary to place the tallest boxes at the top of the stack, followed by shorter ones, for both cost effectiveness and safety. Inspecting (11) shows that the optimal width of the box is inversely proportional to the height (w ∝ −2). Thus, the box catcher should have the thinnest boxes on top and the widest ones on the bottom. It follows that the optimal box catcher has short, wide boxes at the bottom and tall, narrow boxes at the top. Furthermore, the equations of motion (8) contain u2 in the air expulsion stage but no velocity-dependent term in the buckling stage. Hence, for high impact speeds, air expulsion provides the dominant deceleration, while for smaller impacts, box buckling is the more important stage. Results Motorcycle Stunt A stunt person on a motorcycle is preparing to jump over an elephant for the filming of a blockbuster movie. What should be going through a stunt coordinator’s mind? The average height of an African elephant is 4 m; the mass of a top-of-theline motorcycle is 230 kg [Encyclopedia Britannica Online 2003]. We assume that the stunt person clears the elephant by about 2 m, so the impact velocity is 8 m/s. Choosing the parameters = 1 3 m and w = 1 2 m, a 3 m-tall homogeneous box catcher stops the stunt person and motorcycle after they fall about 2.33 m (Figure 1)
228 The UMAP Journal 24.3 (2003) E88g308品 0.5 2.5 0.2 0.4 0.6 0.8 velocity E 2 6 0.6 time(s) Figure 1. Simulation of box catcher stopping stunt person and motorcycle However, we can do better. We take our previous solution, remove one layer of boxes, and change the top layer to boxes with height 6 m and width 3 m(for a y 1.3). Weve removed a row without compromising the safety of the stunt person. In this way, we can make changes until we arrive at an optimal arrangement one with the fewest boxes necessary David blaine's vertigo On May 22, 2002, David Blaine leaped from a ten-story pole onto a cardboard box catcher below. We consider his fall of more than 25 m onto approximately 4 m of boxes. His impact velocity was approximately 36.5 m/s(using the approximation uf=v2gh, which is valid since the terminal velocity for a person falling in air is about 60 m/s[Resnick et al. 1992D) Viewing his fall in the TV special [ David Blaine's vertigo 2002],we estimate the boxes to have height e=3 m and width w=3 m, with 12 layers of boxes in
228 The UMAP Journal 24.3 (2003) 0 0.2 0.4 0.6 0.8 2. 5 2 1. 5 1 0. 5 0 position distance below top of box catcher (m) time (s) 0 0.2 0.4 0.6 0.8 8 6 4 2 0 2 velocity velocity (m/s) time (s) Figure 1. Simulation of box catcher stopping stunt person and motorcycle. However, we can do better. We take our previous solution, remove one layer of boxes, and change the top layer to boxes with height 1 6 m and width 2 3 m (for a γ ∼ 1.3). We’ve removed a row without compromising the safety of the stunt person. In this way, we can make changes until we arrive at an optimal arrangement, one with the fewest boxes necessary. David Blaine’s Vertigo On May 22, 2002, David Blaine leaped from a ten-story pole onto a cardboard box catcher below. We consider his fall of more than 25 m onto approximately 4 m of boxes. His impact velocity was approximately 36.5 m/s (using the approximation vf = √2gh, which is valid since the terminal velocity for a person falling in air is about 60 m/s [Resnick et al. 1992]). Viewing his fall in the TV special [David Blaine’s Vertigo 2002], we estimate the boxes to have height = 1 3 m and width w = 2 3 m, with 12 layers of boxes in