Thinking Outside the box 257 the motorcycle will ever exert on the flats and divide it by the FCT value for a single C-flute cardboard flat order to obtain the total number of sheets needed brief examination of a piece of cardboard demonstrates that bending it in direction parallel to the flutes is much easier than bending it in the direction pendicular to the flutes. Hence, it would be risky to orient all flats in the ne direction; if force is applied along a strip of the surface, all of the flats may sily give way and bend in succession. So, it would be wise to alternate the orientation of the flats in the stack. To make sure that we have the full strength in any direction, we use twice the number of flats calculated alternating the direction of the flutes of each flat as we build the stack. Then no matter how the motorcycle is oriented when it lands on the stack, at least the required strength exists in every direction Finally, we determine the overall height h1 of the structure as follows. The motorcycle accelerates downward at a rate of g from the peak of its flight to the top of the structure; this distance is ho -h1. It then decelerates constantly at a rate of g until it reaches the ground, over a distance h1. Since the motorcycle is at rest both at the apex of its flight and when it reaches the ground, the relation (ho-hi)g=h1g must hold. It follows then that hog 9+g Finding the desired deceleration To determine g, we need the height H of the jump. We discuss how to build the platform so that the rider does not experience a deceleration greater than for a 0.25-m jump We assume that most of the cushion of the landing is in the compression of the tires, which compress Ar =5 cm. The vertical velocity of the rider on impact is v2gH. We also make the approximation that the motorcycle experiences constant deceleration after its tires hit the ground. So the rider travels at an initial speed of v2g H and stops after 0.05 m. We determine the acceleration: Solving yields a= 20gH. That is, if the motorcyclist jumps to a height of 4 m, on landing the ground exerts a force on the motorcycle that feels like 80 times the force of gravity; if the jump were from 0. 25 m, this force would be only 5mg To simulate a 0.25-m fall, we should have the motorcyclist decelerate at a ratThinking Outside the Box 257 the motorcycle will ever exert on the flats and divide it by the FCT value for a single C-flute cardboard flat order to obtain the total number of sheets needed. A brief examination of a piece of cardboard demonstrates that bending it in the direction parallel to the flutes is much easier than bending it in the direction perpendicular to the flutes. Hence, it would be risky to orient all flats in the same direction; if force is applied along a strip of the surface, all of the flats may easily give way and bend in succession. So, it would be wise to alternate the orientation of the flats in the stack. To make sure that we have the full strength in any direction, we use twice the number of flats calculated, alternating the direction of the flutes of each flat as we build the stack. Then, no matter how the motorcycle is oriented when it lands on the stack, at least the required strength exists in every direction. Finally, we determine the overall height h1 of the structure as follows. The motorcycle accelerates downward at a rate of g from the peak of its flight to the top of the structure; this distance is h0 − h1. It then decelerates constantly at a rate of g until it reaches the ground, over a distance h1. Since the motorcycle is at rest both at the apex of its flight and when it reaches the ground, the relation (h0 − h1)g = h1g must hold. It follows, then, that h1 = h0g g + g . (4) Finding the Desired Deceleration To determine g , we need the height H of the jump. We discuss how to build the platform so that the rider does not experience a deceleration greater than for a 0.25-m jump. We assume that most of the cushion of the landing is in the compression of the tires, which compress ∆x = 5 cm. The vertical velocity of the rider on impact is √2gH. We also make the approximation that the motorcycle experiences constant deceleration after its tires hit the ground. So the rider travels at an initial speed of √2gH and stops after 0.05 m. We determine the acceleration: v2 0 + v2 f = 2ax, 2gH 2 +0=2a(0.05). Solving yields a = 20gH. That is, if the motorcyclist jumps to a height of 4 m, on landing the ground exerts a force on the motorcycle that feels like 80 times the force of gravity; if the jump were from 0.25 m, this force would be only 5mg. To simulate a 0.25-m fall, we should have the motorcyclist decelerate at a rate of 5g.