speed of the bar tip. Also, since m2 is very small, the equations predict that the speeds of the spinner robot will almost remain unchanged, which makes sense since very little energy was transferred to the debris. The other limit case is the spinner hitting a very heavy arena wall. The wall is so much heavier than the robot that we can assume that m2 → ∞ and I2 → ∞, resulting in M2 → ∞ and therefore the system effective mass is M ≅ M1, the effective mass of the spinner robot. This will result in the maximum impact that the spinner can deliver, J = M1⋅(1+e)⋅vtip. This value is twice the impact that would be delivered to an opponent with M2 = M1. This is why it is a much tougher test to hit an arena wall than an opponent with similar mass. And, of course, the equations will tell that the speeds of the arena after the impact will be approximately zero.
6.4.3. Impact Energy Before the impact, we’ll assume that the attacking robot (such as a spinner) will have an energy Eb stored in its weapon. For the spinner impact problem presented above, Eb = Ib⋅ωb2 / 2. The impact usually lasts only a few milliseconds, but it can be divided into two phases: the deformation and the restitution phases. In the deformation phase, a portion Ed of the stored energy Eb is used to deform both robots (such as bending the spinner bar or compressing the opponent’s armor), while the remaining portion Ev is used to change the speeds of both robots and weapon. It is not difficult to prove using the presented equations that Ed = M⋅vtip2 / 2 for the tip spinner impact problem. Interestingly, this would be the deformation energy Ed that a mass M with a speed vtip would generate if hitting a very heavy wall, as pictured to the right. So, the higher the effective mass M, the higher the Ed. We’ll see later in this chapter how an attacking robot can manage to maximize M to increase the inflicted damage to the wall opponent. After the deformation reaches its peak, the restitution phase starts. A portion Ek of the deformation energy Ed was stored as elastic deformation, which is then retrieved during the restitution phase to change even more the speeds of both robots. The remaining portion Ec of Ed (where Ed = Ek + Ec) is the dissipated energy, transformed into permanent deformations, fractures, vibration, noise, as well as damped by the robot structure and shock mounts. We can show that, for an impact with COR equal to e, Ek = Ed⋅e2 and Ec = Ed⋅(1−e2). So, a perfectly elastic impact (e = 1) would have no dissipated energy (Ec = 0), and a perfectly inelastic impact (e = 0) would dissipate all its deformation energy (Ec = Ed). Note that inelastic impact does not mean that the entire energy of the system (which originally is Eb) is dissipated, it only means that the portion Ed is completely dissipated. In summary, Eb = Ev + Ed = Ev + Ek + Ec, where the energy (Ev + Ek) will account for the changes in linear and angular speeds of the robots and weapon, and Ec will be dissipated.
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