Fairy Tail- Wendy vs. Sherria

Fairy Tail episode 169 Wendy vs Sherria

Wendy Marvel the sky dragon slayer and Sherria Blendy the sky god slayer can as part of their magic eat air to regain their strength.  During their match in episode 169 Wendy vs Sherria, the girls both choose to eat air to regain their strength.  As they are enjoying their mid match snack the announcer asks them not to eat it all as everyone else in the stadium needs it to breathe.  Just how much air could the girls eat and not cause undue problems for the spectators?

Wendy Marvel

Wendy Marvel having a snack

Sherria Blendy

Sherria Blendy having a snack

 

Considerations

Yes, I know that as the girls eat air the empty space created by their gorging will be filled in by the surrounding air.  Thus, we’ll assume that they are eating it fast enough to cause a problem for spectators and that replacement air is not flowing into the stadium to fill in the void.

 

Volume of Domus Flau

The Domus Flau is similar in appearance to the Roman Colosseum and this will provide the numbers needed to determine the internal volume of the stadium.  If you compare the two side by side you will see that while the Colosseum is oval in shape the Domus Flau is circular.  As such, I will be using the length of the Colosseum for the diameter of the Domus Flau.

Domus Flau

Domus Flau

Colosseum

Colosseum

Numbers

Total Diameter- 189 meters

Fighter area diameter– 87.5 meters

Total height– 50 meters

Area wall height– 4.6 meters

 

I wish it were as simple as taking the volume of a cylinder for the volume of the colosseum, but that would be forgetting about the seating area for the spectators, which is solid and should not be included in the volume calculation.  I will, however, start by imagining the Domus Flau sitting inside a cylinder 189m in diameter and 50m tall.  I can then subtract out the areas occupied by the physical structure of the Domus Flau.  The remaining volume would then be the air held inside of it.

At least that’s how I thought I would do it originally, but then I discovered an easier method.  This method involves finding the volume of a cylinder and the volume of a frustum.  The cylinder is the fighting area of the stadium up to a height of 15m where the stadium seating starts.  A frustum is a cone with the top cut off and it will be the remaining internal volume of the stadium.  Yes, I know this will not be 100% accurate, but I think it will be good enough for our purposes.

Frustum

Frustum

 

Volume of the fighting area– cylinder

V=πr2h

V = 3.14 x (43.75m)2 x 4.6m

V = 27,660.74 m3

 

Volume of the frustum- the rest of the stadium

V = (πh/3)(R2 + Rr + r2)

V = ((3.14 x 45.4) /3)(94.52 + (94.5 x 43.75) + 43.752)

V = 712,128.277104m3

 

Total internal volume of the stadium = 27660.74m3 + 712128.277104m3

Total internal volume of the stadium = 739,789m3

 

Pressure

Pressure is the force applied to the surface of an object and gasses like air can and do apply pressure to the objects around them including humans.  Imagine if you will a column starting at the top of your head all the way up to outer space.  It is this column of air weighing down on you that creates the air pressure that you feel and affects you.  At sea level the pressure of this column of air is 1 atmosphere.

air pressure

air pressure

 

As you move up higher in altitude by going up a mountain this column of air will get smaller and as such the air pressure exerted on your body will decrease.  This really isn’t a problem until you get to around 1,500 meters.  It is at this point that the atmospheric pressure will have decreased to a point where it begins to impact your ability to breathe.  The percentage of oxygen in the atmosphere hasn’t changed, but the pressure has decreased which impairs the ability of our lungs to function.  This is why individuals with respiratory issues will have trouble after getting off the plane in places like Denver, which are a mile above sea level.

The pressure of a confined space can be calculated using the ideal gas law or PV = nRT.  This means that it is possible to determine just how much air the girls can eat before impacting all of the spectators.

P = 1 atmosphere

V = 739,789m3 converted to 739,789,000L

n = amount of air in moles which was converted into grams

R = Gas constant = 0.08206 L·atm/(mol·K)

T = 22 degrees Celsius converted to 295 degrees Kelvin

 

At this point all we really need to do is solve for n and we’re done, right?  Sadly, no it’s not going to be that easy.  Air is actually a mixture of gases, with the five major components being oxygen (20.95%), nitrogen (78.09%), argon (0.93%) and carbon dioxide (0.4%). (The fifth component, water vapor was left off due to how it is highly variable, which would make doing this particular thought experiment much more difficult.)  This means that we need to is figure out how much each gas contributes to the overall pressure and use that number to determine the amount of each individual substance in air.  I know this sounds complicated but it’s actually not that bad.  The partial pressure of each gas, or the amount it contributes to the total pressure, is equal to the percentage of the substance in the mixture times the total pressure… oh heck just look at the equations.

Total pressure = the sum of all the partial pressures

Partial pressure = percentage of the substance in the mixture x the total pressure

 

Pressures at sea level

1 atmosphere = (1 x 0.2095) + (1 x 0.7809) + (1 x 0.0093) + (1 x 0.004)

Partial pressure of oxygen = 0.2095 atmosphere

Partial pressure of nitrogen = 0.7809 atmosphere

Partial pressure of argon = 0.0093 atmosphere

Partial pressure of carbon dioxide = 0.004 atmosphere

 

Amount of air before eating

Using the partial pressures, it is possible to determine the amount of each substance present at sea level or before the girls had their little snack.

n = (PV) / (RT)

Mass of oxygen = 204,874,720 grams

Mass of nitrogen = 668,202,024 grams

Mass of carbon dioxide = 5,378,560 grams

Mass of argon = 11,368,360 grams

FYI- the formula gives you the number of moles which I converted into grams.  This is done by multiplying the number of moles by the molar mass.

 

Amount of air at 1,500 meters

Let’s assume that the girls stop eating when the air pressure equals that of the air at 1,500m, or 0.95 atmospheres, because they don’t want to hurt anyone.  This means we have to recalculate the partial pressures using the formula from before but changing the 1 to 0.95.

0.95 = (0.95 x 0.2095) + (0.95 x 0.7809) + (0.95 x 0.0093) + (0.95 x 0.004)

Partial pressure of oxygen = 0.199 atmosphere

Partial pressure of nitrogen = 0.741855 atmosphere

Partial pressure of argon = 0.008835 atmosphere

Partial pressure of carbon dioxide = 0.0038 atmosphere

 

Mass of oxygen = 194,606,528 grams

Mass of nitrogen = 634,791,948 grams

Mass of carbon dioxide = 5,109,632 grams

Mass of argon = 10,799,920 grams

 

Now it is just a matter of subtracting the mass of air at sea level from the amount at 1,500 meters.

Total mass of air at sea level = 889,823,664 grams

Total mass of air at 1,500 meters = 845,308,028 grams

Mass of air eaten = 44,515,636 grams

Or

44,516 kg of air

 

If the mass of air doesn’t do it for you it is possible to determine the volume of air they ate by using the following formula.

V = w / p

V = volume

w = weight or mass which in this case 44,516 kg

p = density (The density of air at sea level where they were eating is 1.19 kg/m3)

V = 44,516 kg / 1.19 (kg/m3)

V = 37,408 m3 or 37,408,000 liters

 

The standard SCUBA tank used by recreational divers holds 2,265 liters of air so the girls ate the equivalent of 16,516 scuba tanks.  I typically get about 60 minutes out of a shallow water dive which is anywhere from 0-15 meters.  This means I could stay underwater for 661 days.

 

FYI- I realize now that I could have saved a lot of time by just using the V = w / p and the density of the air at sea level and 1,500 meters, but a little extra science never hurt anyone.