Introduction: what is C-Moon?
I recently watched a certain anime series about a bizarre adventure, when a thought occurred to me, as I assume it does to most viewers upon seeing it: “wait a minute, would this be structurally sound enough to withstand such force”. Ok hold on, what am I even referencing? Of course the subject of this post is the ending of part 6 of jojo’s bizarre adventure, where (spoilers) the villain Pucci gains the ability to turn gravity down towards where his feet point within a three kilometre radius using his stand (an ability system in this anime) called C-moon. But, for those who don’t know what nonsense i’m spouting, the major takeaway is that at a certain point in this series gravity for that three kilometre gets “reversed” or points to what we would call up, which made me think if the rock of the upper crust withstand this force or would Cape Canaveral (which is where this takes place) be torn off of the earth and launched away?
First things first though, Pucci clearly doesn’t believe in the force of gravity, because there isn’t really a way for this to happen like it does in the show. The most feasible possibility is that an extremely large mass appears and moves around at some distance of the characters, which exerts twice the force as earth does in that area. However, this of course wouldn’t cause the precise boundary where gravity goes from how it is normally to upside down and it would be a gradient (and the earth would most likely be pulled towards that object as a whole). But for this post I’ll be looking at what the consequences would be, because there may be more to learn about our world if we do so.
So, naturally, I set out to answer this question, not leaving one stone in the ocean of information about mechanical stresses unturned.
Dividing up the tension (circle explanation)
This brings us to a very important question though, how would Cape Canaveral actually break loose from its earthy prison? To understand this, we must take a look at how much force sedimentary rocks in the earth’s upper crust can withstand; but it isn’t so easy to figure this out. Let’s start things off easy with what stresses, or ways for it to break, the rocks would be under and, because the area is sphere, we can safely divide them up into two categories, shear stress and tensile stress. To show how this can be concluded I made a little handy dandy figure right under this, which also includes how you can visualise the different stresses.
As you can see at the bottom of the shape, right “below” where Pucci is, the force is perpendicular to the surface area of that part of the earth’s crust, which generates a stress like it’s being pulled apart, like the stress in a rope in tug of war (not like the stress in PTSD, mind you), which we call tensile stress. On the other hand you have the stress which acts on the rocks at the surface of the ground, you can see this stress as if it were scissors; the force pushes parallel to the surface, the outside region stays put while the inside part gets pulled upwards.
Because of the perfect symmetry of a sphere both these stresses occur at exactly the same frequency even if it’s only half a sphere, which makes sense because if it weren’t symmetrical, and you would double it, it still wouldn’t be symmetrical, but a sphere is what it means to be perfectly symmetrical. At any given point you would be able to calculate how much percentage of the stress goes to which type using the sine and cosine, but for this research the only part about that which is important is that these percentages always add up to hundred percent (With precisely in the middle being fifty fifty for example).
Modelling the breakpoint
So how do we then decide when these stresses, or pressures, are enough to shatter the rock and for Cape Canaveral to make a hasty escape from the earth’s surface? Well, here I’ll be working with two models, both on the extremes of what could theoretically happen, so real life is somewhere in between. Why use models that aren’t the truth, the whole truth and nothing but the truth? It’s because the problem for when larger structures like this actually break hasn’t really been solved yet and most scientists aren’t interested about whether Pucci could break apart earth’s crust or not (a real shame, I know). So the best we can do, with my knowledge and capabilities of learning material physics, is to make a guesstimation of sorts based on different simplified models of what could happen. Furthermore, it is important to note that if one of the types of stress exceeds the value at which it mechanically fails it breaks, so only the weakest link is important in that case as you would assume.
Model number one is if we assume all parts are perfectly inelastic and interact with each other. This means that the entire structure can only break if everything breaks at once, so as long as there is a single point where it can neither break due to shear stress or tensile stress, the entire structure would stay intact because the part with sufficient stress can’t be separated because of a lack of wiggle.
A more formal mathematical expression of this would be that the structure breaks if:
σ(T) + σ(S) ≤ σ(average), Where σ is the stress (so not the grindset), TC is Tensile Strength and SF is Shear Strength; where strength refers to the maximum at which mechanical failure occurs (it breaks).
This would mean that at every possible point the one or both of the stresses exceed the threshold and it breaks, which causes the entire thing to break. It has to be lower or equal because if it was a cumulative 110% of the applied pressure you would always have the following situation: Take for example 65% for TF and 45% for SF. There would be a small point where it would be divided as 60% TF and 40% SF meaning that neither would break an the entire structure holds.
Model number two, the somewhat more GROUNDED one (I couldn’t withstand the pun), says the opposite, it assumes that the ground is elastic enough so that every part can break individually so it acts as if everything was separate and didn’t interact. However, this also means that as soon as one tiny part can break, the entire structure also breaks apart. ‘How come?’, you might wonder. Well if the entirety’s average pressure is enough to break, say, the bottom only tensile stress part, that small mass it was holding down now has to be taken over by another part, increasing the average pressure. This would make the 99% tensile stress part break, and then the 98% etc, until the entire thing starts breaking. Even if the shear stress failure threshold was ludicrously high the SHEAR (darn it i did it again) amount of stress on that one small point — remember even if only one percent is tensile if that part breaks that entire section does because it’s the weakest link — would easily tear it to shreds. So if we were to put this into a similar mathematical expression we would get that it breaks apart if: σ(T) ≤ σ(average) ∧ σ(S) ≤ σ(average).
This might seem as a small change but it makes all the difference, because now both have to be above the average stress, meaning that the rocks would have to be around twice as sturdy to remain structurally sound.
This leads us right into how we go about applying these models.
Will it break?
So, figuring out how we know at what stress value (in MPa) rocks split due to shear stress and tensile stress was a rabid hole I was not prepared for and which would be too long to explain here, which is why instead you can expect a separate blog post about how that whole shebang works so I can explain it in more detail. But for this post the conclusions are important so they’re listed here rapidfire style alongside other necessary data, which we obtained from Determination of Mechanical Properties of Altered Dacite by Laboratory Methods, A Review of the Tensile Strength of Rock: Concepts and Testing and wikipedia.
The average density of the earth’s crust’s upper layers is 2,66 • 10^3 kg/m3, and the average rock composition in the earth’s crust’s upper layer is akin to that of dacite.
Dacite has a Tensile Strength of 3,2092 MPa (from experiments and 𝜏 = -UCS/mi)
and a Shear Strength of ~23 MPa. Which in short normally comes from Mohr-Coulomb failure criterion (just a formula for a line) and experiments, with a confining pressure (σ3) of zero and compressive stress (σ1) of 100, and then drawing a Mohr’s circle and then pinpointing the point of overlap with the criterion, although this time the real relationship was used between the normal stresses and shear stress because of our points proximity to the origin (where the line doesn’t correlate anymore). However, in the future i’ll explain it in more detail but you can always do some research yourself into these terms.
Now we have two ways to look at what happens: model 1 and model 2, however, the result for both is the same. Firstly in model 1 we find that σ(T) + σ(S) = 3,2092 + 23 ≈ 26 MPa (significant figures) which brings us to that σ(T) + σ(S ) ≈ σ(average) (26,1 MPa). The values are quite close so we can’t say definitely what happens, but because this model is the best case scenario it’s pretty obvious that it wouldn’t actually hold up. Model number 2 gives the same result, but more conclusively, which is to be expected because in this model the breakpoint is lower. Both the tensile and shear strength are lower than 26,1 MPa so σ(T) ≤ σ(average) ∧ σ(S) ≤ σ(average) applies, which was our breaking condition, so this too would end catastrophically.
In light of this we can safely conclude that Cape Canaveral would under no circumstance hold, but at what radius of C-moon would it?
What is the minimum radius for it to break?
To calculate what the minimum radius is for it to break loose, we actually only have to make a small change, we keep the radius as undefined and calculate from the answer backwards. We can do this because we know what conditions must be fulfilled and the corresponding mathematical expression with values. Expectedly, the radius differs if you use number 1, where everything acts as a whole and there is no wiggle room, versus when you use number 2, where everything is segmented.
Model number 1 we already know because that is the result of the main calculation, the one accurate to the show, so we know that to be equal to 3 km because of the max values and the pressure of the half sphere were within a rounding error, although most importantly the error between how it would actually take place versus the theory.
For model 2 we do more of the same this time with σ(T) ≤ σ(average) ∧ σ(S) ≤ σ(average), where in this case we only have to look at the weakest link, in this case the tensile strength. This gives us that the same 0,0086982 • r MPa as before now has to be equal to 3,2092 MPa (σ(T)). Calculating this in the same way gives us a minimum radius for model 2 of r = 3,2092/0,086982 = 369,94.., which means that it’s only a measly radius of 369 metres.
Final remarks
Once again thank you for reading, there is still much to explore on this topic but this blog is long enough as it is so I had to cut it short. Material physics ends up being one of the most difficult fields to try to grasp, in my opinion at least, but it is definitely worthwhile.
Join us next episode on phyzards.com where we’ll see the conclusion on how the rock strengths are actually measured and or calculated.
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