A Lengthy Love Affair: IV

Greetings MechWarriors,

Over the past few posts, I have been talking about the "Highlander Burial" and the physics involved. Today, we'll be bringing this subject to its smashing conclusion! We'll look at the forces involved in the impact between a Highlander and a filthy Jenner. This analysis involves the concept of impulse and the conservation of momentum. There are a few simplifying assumptions that we will once again make in order to get some useful numbers.

As a re-cap, let's look at what we know:

Highlander velocity right before impact: 32.85 m/s

Weight of the Highlander: 90 tons

Weight of the Jenner: 35 tons

Height of the Jenner: 6m

Let's start by converting the weights of each Mech into kilograms, the standard measure of mass of the metric system. One metric ton is equal to 1000kg, so the weights become easy to convert.

Weight of the Highlander: 90,000 kg

Weight of the Jenner: 35,000 kg

Now, there are a couple of ways that we can look at this:

1. This is a perfectly inelastic collision, the coefficient of restitution is 0, and the Jenner and Highlander become one after collision. (Conservation of momentum)
2. The collision completely crumples the Jenner and there is very little resistive force exerted on the Highlander. (Impulse)
3. The calculation of the Highlanders kinetic energy directly before impact will tell us how "hard" it hits the Jenner. (Energy)

I think we'll look at number 3 and then compare it to number 2. If the kinetic energy calculated in number 3 is far lower than the impulse required in number 2, we can get an idea of what will happen to the Jenner! Alright, let's do it.

So, the kinetic energy of the Highlander is equal to one half the mass multiplied by the square of the velocity. We could calculate the potential energy the Highlander has at the crest of it's jump, but we already know everything we need!

KE_H = ½ m_HV_H²

KE_H = ½(90000kg)*(32.85m/s)²

KE_H = 4.856E⁷  kg m²/s² or 48560 kJ! (Nowhere near enough to power the DeLorean…)

But, what we have is a unit of energy, not force! How do we get the amount of force from that? Energy is a unit of work per a unit of distance. Because the kinetic energy will be turned into “work” as the Highlander continues to the ground (through the Jenner), we will assume that the energy is dissipated over the 6m height of the Jenner.

F_I = KE_H/6m = 8,093,418.75 kg m/s² or 8.09E⁶ Newtons

Now, that’s a lot of force!

So, let’s calculate the force needed to completely crumple the Jenner. Impulse force is defined as the mass of an object multiplied by its change in velocity divided by the time taken for impact. The time for impact can be found by finding how long it would take the Highlander to travel 6m, the height of the Jenner.

6m / 32.85m/s = .183s

Given that impulse is:

F_i = m DV/Dt = (90000kg)*(32.85 m/s) / .183s = 1.62E⁷ N

Comparing this to the force derived from the Highlanders kinetic energy, we find that the kinetic energy is only half the required value!

So, in conclusion, we find that the Highlander lacks the energy required to completely squash that little Jenner! This, of course, doesn’t mean that the Jenner isn’t in for a world of hurt. The force of impact would put enormous strain on the legs, joints, and cockpit components and would certainly lead to failure. Even without taking into consideration the structural stability of the Jenner, it would be safe to assume that it would not fair very well. Let’s assume that the Jenner’s structure was able to absorb the impact; that still says nothing about the ground supporting the Jenner! The ground could certainly give way before the Jenner.

There we have it, a much much much deeper look into this little fanciful situation than was ever called for!

Over this coming week, I hope to post about some House Steiner technology and the early release of the King Crab in MWO!

Good luck and good hunting, pilots.