You have summoned me with M A T H!

## MATH!

While that equation can give a good estimate for a short fall, Tahu wouldn’t be falling at a constant speed. He starts at zero speed, accelerates, and… well, we don’t know if he hits terminal velocity, but we’ll ignore air drag for now. So his speed is

x = x(0) +v(0)t +1/2 a t^2

v(0) is zero, and we’re going to treat x(0) as zero as well, so we need his acceleration. But wait, what’s the acceleration of Spherus Magna? According to the calculations done on this topic, it would be about 4.84 m/s^2.

So if he falls for 20 seconds, his distance is

1/2 * 4.84 * (20^2) = **968 meters**.

But what if we include DRAG! For a fall of 20 seconds, drag is going to have a noticeable impact and slow his fall. So let’s have a look at that.

The drag force on a body is 1/2(density of fluid)(velocity squared)(coefficient of drag)(frontal area). So how the heck do we get any of these values?

First off, density. You might thing it’s just the density of air, but this isn’t atmospheric air, it’s atmospheric air on another planet that was once part of a larger planet and who knows how that would work. What we’ll do is find the density of air on earth at the altitude where the surface of Aqua Magna would be, or an “altitude” of 1059 meters below Earth Sea level. This puts the density of air at 1.34kg/m^3.

Okay, what’s the drag coefficient? The coefficient of Drag for a falling person is 1 for belly down, .7 for head down. Tahu flips several times during this fall, so we’ll use the average, .85

And the frontal area? This is surprisingly hard to get reference data for, but the frontal area of a person is 5.5 ft^2 (.51 m^2). Toa are about 1.2x the height of an average person, and we’ll assume that their other measurements scale too, giving a frontal area of .84 m^2. But again, Tahu flips, so his average surface area will vary from .84 to .5184, so we’ll take the average of .6792.

But wait! Takua’s there too. So we’ll add in the surface area of .5184/2 (since Takua will only give surface area half the time), for a total of .93

putting it all together, we get:

F(drag) = (1.34)(kg/m^3)(v^2)(m^2/s^2)(.85)(.9384)(m^2)

F(drag) = (1.068)(v^2)

Now we can use Force equals Mass times Acceleration to–

Wait

Karzahni

We don’t know the mass of a Toa.

…Well, that won’t stop me! Let’s assume for one second that Tahu hit terminal velocity by the time of the scene you just measured, and we can actually *solve* for the mass of a Toa.

m * a = 1.068 v^2

m * 4.84 = 1.068 (21.9)^2

m=105 kg

Huh. If we were to assume that their mass scaled up in comparison to a human just like size, we’d get an average mass of 89 kg. Plus, Tahu has Takua with him. While there’s no canonical size for Matoran, the Toa Nuva sets are roughly 1.85x the size of the 03 Matoran, so the mass of a Matoran would be about 13kg, for a total of 102kg, which is shockingly close to 105kg. So at least we can assume that this Bohrok Swarm is somehow accurate so far.

putting it all together, we get:

force = weight - f(drag)

force = 105 * 4.84 - 1.068(v^2)

It was at this point that I stopped to wonder what I’m doing with my life.

So, the force is going to vary with velocity. That’s a headache to deal with. However, since we assumed that the velocity in that scene was terminal (or close enough to it), and since acceleration due to gravity is constant, we can take his average velocity, or 21.9/2 = 10.95. So we have an average force of

force = 105 * 4.84 - 1.068(10.95^2) = 380 newtons

F = m * a

380 = 105 * a

a=3.6

And now, back to where we started

x(t) = x(0) +v(0)t +1/2 a t^2

x(20) = 0 + 0t +1/2 * 3.6 * 20^2 = **720 meters**.

So there you have it. His fall was either 968 meters, or 720 meters.

As for his ascent, it would take either

968/1.46 = 663 seconds = 11 minutes

or

720/1.46 = 493 seconds = 8 minutes

On topic: The screw pins in the Legend Reborn. *Why do they spin?*