# How tall can a humanoid creature be without alerting the rest of the planet with earthquakes?

Planet is earth sized. Fantasy medieval setting that takes place on Continent A, which is on the other side of the planet from Continent B - about as far as South America is from Asia. Continent B has a very very tall giant lumbering around on it, so big that it basically wrecks the landscape of Continent B wherever it goes. Presumably, at a certain height the giant would cause such enormous tremors that everyone everywhere on the planet would hear it walking around - I don't want this. But I do want the giant to be as big as possible without being a nuisance to everyone else on the planet. A global tremor when it jumps up and down would be okay, but there's already so much going on in this world I don't want the distraction of a constant background rumble from this giant just going about its everyday business. How big can I get away with?

Update: don't worry about the structural integrity of the giant. The giant is magic, but the planet's geology is not*. Assume human-proportional mass for a given height.

Update: I want the giant to be as large and (locally) devastating as possible without causing global devastation. "Local" devastation can be as great as rendering all of Continent B uninhabitable, but if this isn't possible without causing a few raised eyebrows on Continent A, then no dice.

*except to the extent that all geology (and geologists!) is magical ;)

• I'm upvoting this just because the math gets more complicated the more I think about it. – Frostfyre Jul 16 '18 at 16:45
• I will, however, point out that if the giant's movements can be detected at the antipole, then half the planet is uninhabitable due to constant earthquakes/tsunamis of a size not seen on Earth.. – Frostfyre Jul 16 '18 at 16:48
• Wow this is complex, the "step energy" is going to go up as a really weird function of height since at a given average density a taller biped will weight more but also their step will gain more gravitational potential energy due to picking up their feet farther and... I'm too asleep to even work out the full equation structure. – Ash Jul 16 '18 at 17:22
• If you get to the point that there are measurable global tremors then the structure (e.g., bones, tendons, etc.) of the humanoid creature would have HUGE problems. Not to mention that above a certain size (think elephant or giraffe) "normal" animals will have issues with circulation, breathing and so many other things. Which reminds me: What do you get when you cross a kangaroo with an elephant? Potholes all over Australia. But I digress. – manassehkatz Jul 16 '18 at 17:31
• Agree with @manassehkatz - so much of the biophysics needs to be handwaved here, fudging the geophysics isn't going to make things any worse. Make the giant as big as you want and as loud as you want; people aren't reading fantasy stories with giants in them for the scientific realism. – Nuclear Wang Jul 16 '18 at 18:17

A real answer is a calculus problem that requires many different iffy variables that will be difficult to determine. So I am just going to ballpark it instead. In a lot of places, I have rounded or just taken sweeping approximations. It should hopefully be within an order of magnitude of the right answer.

However, I think getting a realistic answer to this question is almost impossible.

https://science.howstuffworks.com/environmental/energy/energy-hurricane-volcano-earthquake3.htm

A magnitude 4.0 earthquake is only equivalent to about 6 tons of TNT explosives.

https://en.wikipedia.org/wiki/TNT_equivalent

The "ton of TNT" is a unit of energy defined by that convention to be 4.184 gigajoules

So steps that produce $6 \times 4.184 \Rightarrow 25.1 \text{ gigajoules of energy}$ are an upper limit.

How do we calculate the energy imparted by a step? I am not really sure. This is not as easy as jumping as the question of how much of your mass is really behind each step is a difficult one to answer. Also and how fast are the giant's feet moving?

I am going to say that about 1/3 of the giant's mass is behind each step, and to step it simply lifts its foot and lets gravity pull it down to the ground. I will use potential energy to determine energy since it is a simpler calculation than kinetic energy (which is what it becomes when the giant steps).

$$\text{PE}_{\text{grav}} = \text{m} \times {g} \times{h}$$

So at $25.1 \text{ gigajoules} = \text{m} \times 9.8 \times \text{h}$, we have the mass of the giant's leg/part of body falling and the height from which this mass (presumably the leg) falls.

Assuming that it follows a human-like anatomy: The average man has a height of 167.2 cm. The average man has a mass of 70kg. The average volume of a human is 95 liters.

Height is related to the volume of mass: 1.67m has a ratio to the cube root of this mass, $70\text{kg} \Rightarrow 4.12$ untiless ratio ($\sqrt[3]{70}$).

$$\text{Ratio in terms of height} \Rightarrow m = 15.03 \frac{kg}{m^3}\times h^3$$ Where m is fraction of gian't body weight (1/3 of total body weight).

So when:

$25 \text{ GJ} = \frac{15.03 \times 9.8}{3} \times h^4$
$h^4 = \frac{25 \text{ billion} \times 3}{9.8 \times 15}$
$h = 150 \text{ meters tall}$
$total.mass = (2.467)^3 \times 150^3 = 50,625,000 \text{ kg}.$

Which is big enough for giant's feet to sink into the ground as he walks.

I don't think you have to worry about earthquakes from footsteps. If the giant was big enough for that to happen there would be other bigger problems to worry about.

• This is a superb answer, thank you! The fact that it's a ballpark doesn't matter: by showing that there effectively is no upper limit, you provided an answer that is precisely as exact as I needed it to be. As it so happens, "Loeper" seems like a pretty good name for a giant - hope you don't mind being an antagonist in my dnd world. – Pink Sweetener Jul 16 '18 at 18:15
• I'm not sure about this figure. It seems like if you took something heavier than the Earth you could drop it from a lot closer than half way to the moon and still completely wreck the place. – Daron Jul 16 '18 at 18:35
• Upon additional review, the "h to m" formula should look like m = 15h3. So the main formula is 25GJ = 9.8*15*h4, and the height is actually a fourth root of 170M, which results in a "dwarfish" 144 m height. – Alexander Jul 16 '18 at 18:56
• Coincidentally, the tallest Imperator Titans in the 40k universe are 140m tall. Probably a bit denser than fleshy giants, so even a smaller model would have this effect. A Bolo Mark XXXIII weighs 32,000,000 Kg; it has less impact in tracked operations, but a rough landing after counter-grav flight would do the trick. – Lofty Withers Jul 17 '18 at 7:08
• Don't forget ground pressure. Enough of that, and your giant won't be walking so much as wading... or sinking like a rock in jelly. – Monty Wild Jul 18 '18 at 9:08

A Bit Taller Than Mount Everest.

Alexander says we need each step to have a Tsar Bomb's worth of energy to be measurable. Wikipedia says that's $210$ petajoules or $210 \times 10^{15}$ joules. For the moment let's say the giant is $1$ mile tall and do some back-of-the napkin calculations to figure out the energy.

To simplify let's lift the giant $1$ mile off the ground and drop it. This will hugely overestimate the energy of a single step.

The energy is $E = mgh$ for $m$, $g$ and $h$ the mass, gravitational acceleration, and height respectively. The height is $1$ mile or $1600$ metres. The gravity is about $10$ $m/s^2$. The mass needs more work. . . .

If a human is $6$ feet tall the giant is about $880$ times taller. Then it should weigh $880^3$ times as much. If a human weighs about $70kg$ we're talking $47 \ 703\ 040\ 000 \ kg$ for the giant.

Multiply those together you get about $7.7 \times 10^{14}$ joules. Much less than the Tsar Bomb. Repeat the calculations you get the following figures:

$2$ miles $-$ $6 \times 10^{15}$ joules

$3$ miles $-$ $2 \times 10^{16} = 20 \times 10^{15}$ joules

$4$ miles $-$ $5 \times 10^{16} = 50 \times 10^{15}$ joules

$5$ miles $-$ $9 \times 10^{16} = 90 \times 10^{15}$ joules

$6$ miles $-$ $1.65 \times 10^{17} = 165 \times 10^{15}$ joules

$6.5$ miles $-$ $2.1 \times 10^{17} = 210 \times 10^{15}$ joules

So you can certainly get at least $6.5$ miles. Is that enough for you?

• Yes, stratosphere-height is definitely tall enough, thank you! – Pink Sweetener Jul 16 '18 at 19:01
• Most of the Tsar Bomba's energy was lost to the atmosphere, not the ground. Keep that in mind. An apples-to-apples comparison would have required a magical cover over the bomb that forced all the energy downward (much like a leg coming down on the earth). In other words, it's a poor comparison. – JBH Jul 16 '18 at 19:49
• Granted, however I've not assumed anything about how the energy is directed. – Daron Jul 16 '18 at 19:55
• @Daron I think JBH's point was that Alexander's comment wasn't quite correct. Tsar Bomba was indeed noticeable from the other side of Earth, but that was the case in spite of the fact that most of Tsar Bomba's energy didn't actually go into the ground. Thus, a substantially smaller amount of energy than the total Tsar Bomba energy release would still be detectable on the other side of Earth in the case OP asked about. – reirab Jul 17 '18 at 16:02
• @reirab effects noticeable on other side of Earth without instruments? I highly doubt that. – Alexander Jul 19 '18 at 0:11