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A Time Independent Model of Box Safety for Stunt Motorcyclists

A Time Independent Model of Box Safety for Stunt Motorcyclists. Ivan Corwin Sheel Ganatra Nikita Rozenblyum Harvard University Analysis: Steve Tuckerman, Bloomsburg University. The Problem:.

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A Time Independent Model of Box Safety for Stunt Motorcyclists

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  1. A Time Independent Model of Box Safety for Stunt Motorcyclists Ivan Corwin Sheel Ganatra Nikita Rozenblyum Harvard University Analysis: Steve Tuckerman, Bloomsburg University

  2. The Problem: • You are the stunt coordinator for a movie. James Bond will be leaping an elephant on a motorcycle, and landing on a pile of cardboard boxes. Your job is to minimize the number of boxes required.

  3. Key Assumptions: • Temperature and weather are ideal, they do not affect the strength of the box. • The wind is negligible, because the combined weight of the motorcycle and the person is sufficiently large. ST: wind resistance is important, a model should indicate this.

  4. Key Assumptions • The ground on which the boxes are arranged is a rigid flat surface that can take any level of force. ST: It is likely that you would need less boxes if the boxes were placed on a more yielding surface. • All boxes are cubic, for greatest strength.

  5. Variable Assumptions • Tire rectangle length in machine direction: 7 cm front, 10 cm back. • Tire rectangle length in cross machine direction: 10 cm. ST: Would a wider tire require fewer boxes? • Mass of rider and motorcycle: 300 kg. ST: would a lighter motorcycle and lighter rider require fewer boxes?

  6. Variable Assumptions • Angle Variation away from y-axis leaving the ramp: π/36. • Ramp angle of elevation: π/6 • Initial Velocity: 1500 cm/s

  7. Unanswered Question • Is it an Indian elephant? • Or an African elephant? • Indian elephants reach a height of 2.7 meters, African elephants reach a height of 4 meters.

  8. How high will the motorcycle in the model go? • Initial velocity: 15 m/s • Angle of Elevation: π/6 → A (t) = < 0, -9.8 > → V (0) = 15 <cos π/6, sin π/6>

  9. → → V (t) = 15 <√3/2, ½> + A = <(15√3)/2, 15/2 – 9.8t> → R (t) = < ((15√3)/2)t, (15/2) t – 4.9t^2)> Maximum height occurs when 15/2 – 9.8 t = 0 So t = .765 seconds.

  10. Maximum height is 7.5t – 4.9t^2, t = .765, equals 2.9 meters. • Indian elephants reach a height of 2.7 meters, African elephants reach a height of 4 meters. • Clearly, the motorcycle in the model used by the Harvard group would not clear an African elephant, and also would likely not clear an Indian elephant, due to the broadness of an elephant.

  11. Not good for either the elephant or the stunt person

  12. Why is this important? • The initial assumptions of velocity and angle are not valid. Their assumptions were carried through the model to come up with specific recommendations for number and size of boxes, and the arrangement of the boxes.

  13. What methodology was used? • Cardboard boxes were modeled as springs. • The surface striking the box was a rectangular tire hitting the center of the top of the box. ST: This shows great control on the part of the stuntperson. • It was assumed that there would be no torque on the box. • Force in the “y” direction was analyzed, not force in the “x” direction.

  14. The tops of cardboard boxes can fail in three ways: puncturing, cracking, and buckling. Each method of failure absorbs energy.

  15. Force Up (Spring Action, Top of Box) • Fup = [2EcdLmd(1-(l-Lcd/(√((l-Lcd/2)^2+((x(t))^2) • + same factor for machine direction

  16. Cracking • XCd = √((Pcdw/Ecd)^2 + (Pcdw/Ecd)(l-Lcd)

  17. Speed of Motorcycle after hitting box At (x) = integral from 0 to x of Fnet ds = EcdLmd (x^2 –(l – Lcd)(√(l – Lmd/2)^2 + x^2) + (l – Lcd)^2 / 2 ) + EmdLcd(x^2 – (l – Lmd) (√((l – Lmd)/2)^2 + x^2) + (l – Lmd)^2/2 ) + mgx

  18. Fnet = total force • Ecd = 800,000 kg /(s^2 cm) • Lmd = 7 cm front, 10 cm back (tire radius, machine direction) • Small l = length of box edge • Lcd = 10 cm, (tire radius, cross direction) • Emd = 3,000,000 kg/ (s^2 cm)

  19. Effect on Energy Absorption by sides of Box • Acd–edge = ELmdx^2 cd-edge times (x^2f)/(3(l – Lcd)) + mgx cd-edge This calculation treats the sides as springs and utilizes Hooke’s law.

  20. Effect of Friction Vfy = √ (2 * ((1/2)(mv^2) – Δ Ay)/(m + mb)) Vfy is new horizontal speed due to friction Mb = mass of the boxes displaced M = mass of motorcycle and stuntperson

  21. Conclusion of Harvard students • The most efficient arrangement of boxes occurs in a pyramid shape, each box resting on 4 others, in order to dissipate the downward force. ST(assumes rider will land in center of top box and does not knock over boxes)

  22. Critique • The team did an excellent job modeling box failure • The team did an excellent job determining energy absorption per box, and in determining the ideal size of the boxes for their model.

  23. Critique • The rider is not likely to land in the center of the top box • The rider is likely to knock down many boxes, in the more likely event that they land on the face of the box • The rider must go higher than that indicated, either at a sharper angle or an increased speed. These modifications may affect the conclusions of the researchers.

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