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Samus Aran [Discussion]

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Re: Samus Aran [Discussion]

Beams, bombs, and sandstone
Samus can destroy sandstone with her beams and bombs. Sandstone has a hardness of 6.5 to 7 on Mohs scale of hardness. Sandstone also has a density of 2.3 to 2.4 kg/cm^3. While the power beam and bombs seem weak, they're capable of destroying sandstone withour any problem.
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Re: Samus Aran [Discussion]

Omega Pirate's height and claws
In this image, we see Samus standing next to the Omega Pirate. (This is an official concept art from Retro Studios, which can be unlocked in the game.) Samus is 23 px., so 190 cm. the equivalent. (I want to determine how sharp these claws are.) The Omega Pirate is 160 px. 160 px. / 23 px. is 6.9565217391304348. 190 cm. * 6.9565217391304348 is 1,321.739130434782612. (13.21739130434782612 m.; 43.36 ft.) The tip of the claws are 1 px. 1 px. / 160 px. is 0.00625. * 1,321.739130434782612 cm. is 8.260869565217391325 cm., which is a little thicker than my thumb.
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Re: Samus Aran [Discussion]

Samus and water pressure
In a previous calculation concerning the temperature at Samus can withstand without the Varia Suit, I considered how deep groundwater can go, which I ended up with 9,144 m. (30,000 ft.) because Samus travels further below Maridia and Maridia is just above Norfair. Fluid pressure requires the density of liquid (water in this case), the acceleration of gravity, and the height of the water relative to the person below it.

P = ρgh
P = (999.97 kg/m^3)(9.81 m/s^2)(9,144 m.)
P = 8.97 * 10^7 kg/m/s^2, or 89.7 MPa, which is equal to 885.3 atmospheres (approx. pressure at the bottom of the Pacific Ocean (approx. depth 5.5 km.)) (6 * 10^7 Pa)

Edit: All right, so Samus never actually goes underwater that's 30,000 ft, as not all of Maridia is submerged in water, so this calculation for Samus cannot be used. However, I'm not sure how deep Samus can travel in Maridia. For now, I want to start with the lowest part I can think of where the entrance to Maridia begins. In this image, I took the distance from the bottom all the way to the blue door above. This will have to be an approximation, since I cannot accurately determine the distance. What I mean is there will be slightly off, but it should be negligible. Samus in Super Metroid is 86 pixels. The distance from the ground to the door, using Samus as a ruler, gives a distance of 1,083 px.

1,083 px. / 86 px. = 12.5930232558139535

1.9 m. * 12.5930232558139535 = 23.92674418604651165 m.

P = (999.97 kg/m^3)(9.81 m/s^2)(23.92674418604651165 m.)
P = 234,700 kg/m/s^2 (Pa), approx. (0.4 to 1.5) * approximate impact pressure of a fist punch (150,000 to 550,000 Pa), according to Wolfram | Alpha.

Of course, just above the door Samus passes through there is more water, so the water pressure will be much higher.
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Re: Samus Aran [Discussion]

This will be my last post here. After this, I'll be taking my information to a blog I'm working on.

How much sheet of ice can Samus break with the Speed Booster?
Samus is 190 cm., but because her knees are slightly bent, I'm going to go with 182.88 cm. (6 ft.) This would be equivalent to 112 px. The length of the ice sheet is 415 px. This is only taking the certain portion of the wall, which will take a perfect square shape. The height is more than 300 px., but I'm going to just work with 300 px. for now because the other part will take time for me to work on at the moment. I'll cover that in my blog.

Length
415 px. / 112 px. = 3.7053571428571429
182.88 cm. * 3.7053571428571429 = 677.635714285714293552 cm.

Height
300 px. / 112 px. = 2.6785714285714286
182.88 cm. * 2.6785714285714286 = 489.857142857142862368 cm.

Now, I don't know how thick this ice sheet is, but I'd presume it's thicker than the ice barriers at the beginning of Phendrana Drifts. Let's say 3 inches thick, or 7.62 cm. So the wall is 2,529,418.57512244903582822722416326561818565632 cm^3. Ice has a density of 0.9167 g/cm^3 at 0°C. The location Samus is in is likely colder than that, but let's work with that. This would mean that Samus is smashing through 2,318 kg. of ice (2.319 metric tons) without slowing down. Keep in mind it would be more, actually, considering I did not address the entire upper and lower parts of these ice sheets.
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Re: Samus Aran [Discussion]

I am currently putting all things Metroid here. I'll probably add some variety for other games.
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Re: Samus Aran [Discussion]

Vorash's mass
The Max Factory Samus Aran model from Metroid: Other M has Samus scaled at 1/8, which would put her height at 5'9", not 6'3". Considering the Metroid II: Return of Samus manual gives us the specifications of Samus with an image of her in her suit, this must mean Samus' height and mass are regarding the suit itself, not Samus outside of it. By the way, the 1/8 model is Samus outside of her suit. But I'll go with 6'3" anyway. From here, Vorash's length is 29.18 meters. According to Wikipedia, "Gottfried and colleagues introduced a method to determine the mass of the great white after studying the length–mass relationship data of 175 specimens at various growth stages and extrapolated it to estimate C. megalodon's mass. According to their model, a 15.9 metres (52 ft) long megalodon would have a mass of about 48 metric tons (53 short tons), a 17 metres (56 ft) long megalodon would have a mass of about 59 metric tons (65 short tons), and a 20.3 metres (67 ft) long megalodon would have a mass of 103 metric tons (114 short tons)."

It's obvious that Vorash is longer, but he becomes thinner where the tail begins. The length of a C. megalodon (maximum) is 20 meters, while the C. megalodon (conservative) is less than that. (Image) Assuming Vorash's tail makes up the extra 9 meters, I could use the information and work with 103 metric tons for Vorash. However, it doesn't stop there. Because Vorash sinks in the lava, that means his body is denser than it. 3 g/cm^3 would be higher than the density of lava, so if I decided to find out the mass of Vorash, I would need to take 103,000,000 grams (103 metric tons) and divide it by 1 g/cm^3 (density of water), which would result in a volume of 103,000,000 cm^3. Multiply this by 3 g/cm^3 and I end up with 309,000,000 grams, or 309 metric tons.
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Re: Samus Aran [Discussion]

How much pressure can Samus withstand under water?
In Super Metroid, the first time Samus enters Maridia, there is a red door on the right of her, and on the left is a plant immediately next to the blue door Samus has gone through. On a normal scale, this plant is 32 pixels tall and Samus is 86 pixels tall. Dividing 32 px. over 86 px. and then multiplying this by Samus' height of 1.9 meters gives the height of the plant 0.7069767441860466 meters. From the Super Metroid map for Maridia, the height of the plant is 18 px. tall. In order to find out how much percentage this has reduced, I must take 32 px. - 18 px., which gives me 14 px., and then I must divide this by 32 px., which gives me 43.75%.

Next, I need to multiply Samus' heigh of 86 px. by 43.75%, which gives us 37.625 px. I'll have to round this to 38 px. as there is no way for me to work with 37.625 px. In order to find out how much pressure there is under water, this requires using the equation, p = ρgh, where ρ is density, g is gravity, and h is the height or depth of the water. Since I want to go with how much water is above Samus, I'll take the total amount of pixels 976 px. - 38 px., which would give us 938 px. Taking 938 px. over 38 px. would give me a whopping 2,468.42%! This multiplied by Samus' height of 1.9 m. would result in 46.9 m. Before I can find the pressure, I'll need to consider Zebes' gravitational acceleration, which is not 9.81 m/s^2, but 9.36 m/s^2. This is based on the mass and diameter of Zebes.

p = (999.97 kg/m^3)(9.36 m/s^2)(46.9 m.)
p = 439,000 kg/m/s^2 (439 kPa)

I'm sure Samus could dive deeper, as Herbert Nitsch dove 253.2 meters under water.
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Re: Samus Aran [Discussion]

Vorash's mass recalculated
The Queen Metroid’s length is 11 m. , but from Val’s game blog, the queen’s length is 24.55 m. , so there’s some discrepancy. Vorash’s length is 29.18 m. Taking this length and dividing it by 24.55 m. means there was an increase of 4.63 m. Using this value and dividing it by 24.55 m. gives us a percentage increase of 189%. Multiplying this by the Queen Metroid’s actual length of 11 m. gives us 2.07 m., which means Vorash’s actual length should be 13.07 m. I said before that the length of a whale shark is 12.2 m. . So I suppose a whale shark would have been a better choice after all. The question is, what is the mass of a whale shark?

The average mass of a whale shark is 18.69 metric tons.  This is equal to 18,688,005.6 g. In order to find the volume of an object, you need to divide it by its density. Water has a density of 1 g/cm^3, which means the volume is 18,688,005.6 cm^3. For andesitic magma, I used the low-end of 2.45 g/cm^3, which would be 45,785,613.72 g., or 45.79 metric tons. This is significantly lower than my last calculation, but even then, Samus is still physically strong. The mass could be greater, considering Vorash’s fins and the fact that he pulls in the opposite direction of Samus before she throws it into the air. Of course, if there was any increase in Samus’ physical strength, 50 metric tons would be the safest assumption.

More here.
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Re: Samus Aran [Discussion]

How much energy can the plasma beam produce in Metroid Prime 3: Corruption?
While it is unknown as to what kind of metal is in Metroid Prime 3: Corruption, working solely off of color, we can get an idea of what metal steambots are made of. Of course, steambots are likely made up of an alloy, since “metals” is used. If we use color, then copper would be a likely metal. Two common metals alloyed with copper is zinc and tin. The former makes brass and the latter makes bronze. It’s possible that steambots are alloyed with the latter, since their weaker variant are tinbots, which from the name alone likely are made of tin. While I could simplify steambots by working only with copper, I’m going to focus this calculation on tinbots instead, since I am personally finding it difficult to find the heat of fusion and heat of vaporization for bronze.

Tin has a heat of fusion of 59.2 J/g, a heat of vaporization of 2,491.8 J/g, and a specific heat capacity of 0.228 J/g°C. It has a melting point of 231.93°C and a boiling point of 2,602°C. With these in mind we can find out how much energy is required to vaporize tinbots. I will assume that tinbots are generating a heat of 37°C, which is the body temperature of a human. Also, considering the size of these tinbots, I will assume they have a volume a little over a human skeleton. If a human skeleton makes up 15% of the body mass, and if the mass of an average adult male is 70 kg., then that’s 10.5 kg. I will assume that this mass is equal to that of the volume, which means if tin has a density of 5,769 kg/m^3, then a tinbot has a mass of 60,574.5 kg (60,574,500 g.). That’s heavy.

In order to add heat into a solid object to melt it, it requires energy. So the first phase would require Q = m * c * ΔT, where m is mass, c is specific heat capacity, and ΔT is change in temperature. This will require the melting point and initial temperature. Using this equation gives us an energy of 2,692,175,500.98 J. This is already ridiculous, as it is already close to the required energy to atomize humans, but let’s keep going. Heat of fusion is Q = m * Hf, which would give us 3,586,010,400 J. This alone already passes the energy required to atomize a human. Observing this happening to space pirates should not surprise me, since it happens, but to see these numbers is insane!

Like before, I will be using Q = m * c * ΔT, since the transition from heating up a solid and converting it to a liquid is now at the point of causing this liquid to become a gas. Instead of using 37°C, I’ll be using the melting point and boiling point for my initial and final change. This results in 32,733,003,589.02 J. Next is heat of vaporization, which is Q = m * Hv, which is 150,939,539,100 J. By this point I would use Q = m * c * ΔT once more, but I don’t think there is a point, so what I will do now is add all of this energy together, which is 189,950,728,590 J, which is 45.4 tons of TNT if anyone was interested in the energy output. While this can be ascribed to a partially or fully charged plasma beam, it doesn’t take long to melt these tinbots with ordinary shots.

--------------------

I recalculated and ended up with 1.6 megajoules. This is only for the tinbot, however. The energy required to vaporize steambots and atomize space pirates would need to be higher.
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Re: Samus Aran [Discussion]

Energy required to vaporize a space pirate
It’s a fact that space pirates are taller than Samus, so the question is just how much mass does a typical space pirate have? To find out, I decided I would use Hamwi’s method.  Devine’s formula made no sense, as a typical space pirate would be a little over a tonne. For males, the Hamwi formula is 106 lbs. (48.08 kg.) and then an additional 6 lbs. per inch (2.72 kg. per 2.54 cm.). This means that if a typical space pirate stands at 7’6″ (2.286 m.), then 30 in. is an additional height over 5′. This means that a typical space pirate should weigh 286 lbs. (129.727 kg.). That looks like a more reasonable weight.

Based on scan images, we can see space pirates have bones,  so what we’re working with here is similar to that of finding out just how much energy is required to atomize a human being. Space pirates appear to be unaffected by high temperatures, as we can observe flying pirates in Metroid Prime standing in Magmoor Caverns next to lava. Samus’ powered armor cannot withstand this temperature, so we should at least expect more energy than 2.99 gigajoules to atomize a space pirate. I will be using the calculation required to find out how to vaporize a human, but apply it to a typical space pirate for the sake of simplicity. After all, organisms are carbon-based and require water to live.

The average density of a human is 985 kg/m^3.  A typical space pirate has a mass of 129.727 kg., so 129.727 kg. / 985 kg/m^3 is 0.131702538071066 m^3. This is approximately twice the volume of a human. The composition of a human is 65% oxygen and 9.5% hydrogen.  Since water contains two hydrogen atoms and one oxygen atom, adding these together gives us a total of 84%. This means the human body is made up of that much water. I will apply this to space pirates as well. 129.727 kg. multiplied by 84% means that a typical space pirate’s body is made up of 108.97068 kg. of water. Carbon makes up 18.5%, so 129.727 kg. multiplied by 18.5% is 23.999495 kg.

The heat of vaporization of water is 2.257 * 10^6 J/kg. This multiplied by 108.97068 kg. of water gives us a total energy of 245,946,824.76 J. The heat of vaporization of carbon is 2.9624 * 10^7 J/kg. This multiplied by 23.999495 kg. gives us 710,961,039.88 J. Adding these values together gives us the total energy to vaporize a typical space pirate, which would be 956,907,864.64 J. Now that we have this information, we can find out how much energy is required to atomize a space pirate. I expect it to be greater than 2.99 GJ, simply because of the size difference.

We can now use the volume of a typical space pirate, which is 0.131702538071066 m^3. This will be multiplied by 84%, which will give us 0.11063013197969544 m^3. This will be multiplied by the atomization value of water, which is 5.138416 * 10^10 J/m^3. This is 5,684,636,402.47 J. Next is carbon. The volume of a typical space pirate will be multiplied by 18.5%, which will give us 0.02436496954314721 m^3. The atomization value of carbon is 1.565 * 10^11 J/m^3. This multiplied by the volume gives us 3,813,117,733.5 J. Adding these two values of atomization gives us a total of 9,497,754,135.97 J. This is equal to 2.27 tons of TNT.

With these all in mind, this should give us an idea of the amount of energy that also comes from the plasma beam in Metroid Prime and Metroid Prime 3: Corruption, as well as the light beam in Metroid Prime 2: Echoes. Based on the color temperature of the light beam, it may very well produce more energy than the plasma beams, since the hotter an object is, the more energy it’ll produce.

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