286 The UMAP Journal 24.3(2003) Let r(t) be the displacement from the point of entry as a function of time. Since (0)=0, we have (t) and so the total distance traveled through the boxes is 1 △x=x(tf) Therefore, we arrive at: Given an impact velocity vo N 20 m/s and deceleration bounded by 5g, the stunt person requires at least 4 m to come to rest The energy that must be dissipated in the boxes is roughly equal to the kinetic energy that the motorcycle and stunt person enter with. ( Since the box rig should be only 3-4 m high, the potential energy is a much smaller fraction of the total energy. Thus, for vo= 20 m/s and a mass of 200 kg the change in energy is 40,000 J From(1), we calculate that the crush energy of a standard (30 inch)box is 633 J, so we need. 40, 000 /633 a 60 boxes. Trajectory Analysis and Cushion Location Cardboard boxes won t dissipate any energy unless the stunt person lands on them. It is therefore important to consider the trajectory, so we know where to place the box rig and what the uncertainty is in the landing location We calculate trajectories by solving the following differentialequation, where v is the speed, h is the drag coefficient, and i is the position: We used Matlab's ODE45 function to solve an equivalent system of first-order equations. We use an air drag coefficient of h=1.0 [Filippone 2003]. We see from Figure 5 that it would be unwise to ignore air resistance, since it alters the landing position by up to several meters. to le ith exactly the same initial velocity and angle every jump. We therefore need to allow for some uncertainty in the resulting trajectory and ensure that the cardboard cushion is large enough to support a wide range of possible landing locations The ramp angle o is constant, but the motorcycle might move slightly to one side as it leaves the ramp. Let 6 be the azimuthal angle between the ramp axis and the motorcycle's velocity vector. Ideally 6 should be zero, but small variations may occur. The other uncertain initial condition is the initial velocity In modeling possible trajectories, we assume the following uncertainties Initial velocity: vo Intended t 1 m/s
286 The UMAP Journal 24.3 (2003) Let x(t) be the displacement from the point of entry as a function of time. Since x(0) = 0, we have x(t) = v0t − 1 2 at2 and so the total distance traveled through the boxes is ∆x = x(tf ) = v2 0 a − 1 2 a v0 a 2 = v2 0 2a . Therefore, we arrive at: • Given an impact velocity v0 ≈ 20 m/s and deceleration bounded by 5g, the stunt person requires at least 4mto come to rest. The energy that must be dissipated in the boxes is roughly equal to the kinetic energy that the motorcycle and stunt person enter with. (Since the box rig should be only 3–4 m high, the potential energy is a much smaller fraction of the total energy.) Thus, for v0 = 20 m/s and a mass of 200 kg, the change in energy is 40,000 J. From (1), we calculate that the crush energy of a standard (30 inch)3 box is 633 J, so we need. 40, 000/633 ≈ 60 boxes. Trajectory Analysis and Cushion Location Cardboard boxes won’t dissipate any energy unless the stunt person lands on them. It is therefore important to consider the trajectory, so we know where to place the box rig and what the uncertainty is in the landing location. We calculate trajectories by solving the following differential equation, where v is the speed, k is the drag coefficient, and x is the position: (x) = −gzˆ − k m|v| 2vˆ We used Matlab’s ODE45 function to solve an equivalent system of first-order equations. We use an air drag coefficient of k = 1.0 [Filippone 2003]. We see from Figure 5 that it would be unwise to ignore air resistance, since it alters the landing position by up to several meters. It is unreasonable to expect the stunt person to leave the ramp with exactly the same initial velocity and angle every jump. We therefore need to allow for some uncertainty in the resulting trajectory and ensure that the cardboard cushion is large enough to support a wide range of possible landing locations. The ramp angle φ is constant, but the motorcycle might move slightly to one side as it leaves the ramp. Let θ be the azimuthal angle between the ramp axis and the motorcycle’s velocity vector. Ideally θ should be zero, but small variations may occur. The other uncertain initial condition is the initial velocity v0. In modeling possible trajectories, we assume the following uncertainties: • Initial velocity: v0 = vintended ± 1 m/s
Cardboard Comfortable When It Comes to Crashing 287 resistance Figure 3. Air resistance significantly changes the trajectory. Azimuthal angle:6=0°±2 We use this to identify the range of possible landing locations by plotting the trajectories that result from the worst possible launches(Figure 6 If the intended initial velocity is 22 m/s, the ramp angle is 200, and the mass of the rider plus motorcycle is 200 kg, then the distance variation is 2.5 m and the lateral variation is +1.5 m Impact simulation To evaluate the effectiveness of various box rig configurations, we con struct a numerical simulation of the motion of the stunt person and motorcycle through the box rig Assumptions The full physics of the box rig is far too complex to model accurately. We make the following assumptions to approximate and simplify the problem The problem is two dimensional. We restrict out attention to the plane of motion of the stunt person
Cardboard Comfortable When It Comes to Crashing 287 0 5 10 15 20 25 30 35 40 45 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 distance (m) height (m) Effect of Air Resistance on Trajectory no air resistance air resistance Figure 3. Air resistance significantly changes the trajectory. • Azimuthal angle: θ = 0◦ ± 2◦ We use this to identify the range of possible landing locations by plotting the trajectories that result from the worst possible launches (Figure 6). If the intended initial velocity is 22 m/s, the ramp angle is 20◦, and the mass of the rider plus motorcycle is 200 kg, then the distance variation is ± 2.5 m and the lateral variation is ± 1.5 m. Impact simulation To evaluate the effectiveness of various box rig configurations, we construct a numerical simulation of the motion of the stunt person and motorcycle through the box rig. Assumptions The full physics of the box rig is far too complex to model accurately. We make the following assumptions to approximate and simplify the problem. • The problem is two dimensional. We restrict out attention to the plane of motion of the stunt person