A Lengthy Love Affair: III

Greetings MechWarriors,

Over the past few posts, I have been talking about my adoration for Highlander pilots and their unique maneuver. This time, I’d like to take a peek into what would happen in real life to a Light Mech caught in the path of a dropping Highlander! There are a number of simplifying assumptions that I will make to ball-park useful numbers. This will include the values for acceleration due to gravity, wind resistance, the reality of three dimensional vectors, etc, etc. Although these assumptions remove us from the “real world” and are somewhat idealized, they will allow us to fill in as many gaps as we can to come to a conclusion. And, let’s be real here, there are some details that we just can’t get from a work of fiction!

So, let’s start by determining, as best we can, what the jump capacity of a Highlander is. A 90m jump capacity is what is listed on sarna.net, but a 20m jump capacity is listed in MWO. We’ll split the difference and assume the jump capacity of the average Highlander is 55m. Next, we need to determine the height of a Light Mech for us to crush. An Atlas is referenced as being 13-15m tall. A Jenner, being an average Light, would stand just under half as tall. So, let’s assume that a Jenner is 6m tall.

So, what this tells us is that a Highlander will be able to produce around a DY of 55m. At the peak of its ascent, the Highlander’s +/-V_y and +a_y will be zero, as it’s no longer traveling upward. Given the height of a Jenner, the Highlander will only have 49m to fall before it strikes the Jenner. Now, we need to determine how fast the Highlander will be traveling when it strikes that puny Jenner. To do this, we will assume that the acceleration due to gravity is what it is at sea-level on Earth, ~9.81m/s². For simplicity, we will assume that the Highlander travels all 49m without experiencing positive acceleration or external forces. Given the change in position and the acceleration during motion, we can determine the amount of time it would take the Highlander to strike the Jenner.

DY = a₀ Dt²/2 + V_Y0Dt

This equation states: The change in position is equal to the initial acceleration multiplied by the (change in time)² all divided by two PLUS the initial velocity multiplied by the change in time. Because our Highlander is not moving when it comes to the peak of its jump, the initial velocity is zero. The initial acceleration is not zero because the Highlander’s acceleration has simply changed from a net positive to a net negative, or it is always experiencing gravity. Simplified, we’re left with:

Dt = sqrt(2DY/a₀)

Plugging in our known values, we are able to determine that the Highlander’s descent will last 3.35 seconds! Now, that’s moving pretty quick! Next, let’s determine the velocity at the time of impact.

DV_y  = a₀Dt + V_Y0

Again, our initial velocity is zero, so, we’re able to simplify the equation once more.

DV_y  = a₀Dt = (9.81m/s²)*(3.35s) = 32.85 m/s

There, we have now determined how fast an average Highlander would be travelling when it strikes a Jenner after a full jump in Earth equivalent gravity. How exciting! But, you may be wondering where we are to go from here if we want to find the amount of force a Highlander would impart on a Jenner. Next time, we’ll continue our look into the details of that very question! Anyone that is interested in reading ahead will find what they’re looking for from the concept of momentum and impulse!

Good luck and good hunting, pilots.

Sources:
http://www.sarna.net
http://mwomercs.com/forums/topic/4688-mwo-battlemech-scale/