Ice Beam

Description

The ice beam is one of Samus’ numerous different weapons. Arguably one of her most important abilities, the ice beam allows Samus to freeze enemies in place and use them as solid platforms. Like any of her other beams, it can be combined with her charge beam to create a more devastating projectile. On top of this, in Super Metroid, it can be used to create a small shield around Samus that freezes anything it touches. This weapon in particular plays a crucial role in many of her adventures as it is often a necessary item to defeating the metroids themselves, as they harbor an extreme weakness to cold temperatures. The ice beam has featured in almost every game to date and often makes for some interesting platforming challenges.

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Process

To first analyze the numbers behind this incredible weapon, we need to know exactly what it is doing. My hypothesis is that the ice beam freezes the air surrounding the enemy, rather than the enemy itself. I find this to be the most likely case, as the enemies are completely unharmed when the ice around them thaws. Furthermore, it would take far less energy to freeze the air than the entire creature.

Unfortunately, this raises several questions (e.g. what part of the air is being frozen, what is the freezing point of that portion of the air, how much is being frozen, etc). To answer most of these questions, I had to do a lot of digging.

The first and most important question pertains to the composition of Zebes’ atmosphere. As can be seen in the Metroid manga detailing Samus’ origins, Samus is able to comfortably breathe when exposed to Zebes’ atmosphere. Therefore, it seems fairly safe to assume that the atmosphere of Zebes is very similar to that of earth. While this does make this process exponentially easier, earth’s atmosphere is still composed of many different gases. In order for the ice beam to freeze the air, it seemed most likely that it was freezing the water vapor in the air. However, in order to ensure this, I needed to figure out just how much of Zebes’ atmosphere was water vapor in comparison to earth, to do that I needed the density of Zebes’ atmosphere.

To find density, I first researched what variables I would need. In my research, I found that V = SqRoot((2mg)/(pAC)) where V = terminal velocity, m = mass of the falling object, g = gravity, p = density, A = projected area of the object, and C = the coefficient of drag. To find both acceleration to due gravity and terminal velocity, I used a similar process as I did for velocity of the speed booster.

I picked a couple locations with a large amount of verticality and individual tiles to more easily calculate displacement. In the case of terminal velocity, it was as simple as falling down the shaft I used frame by frame and calculating how many tiles Samus fell and within how many frames after reaching her terminal velocity. I used the same ratios from my previous calculations (60 fps, 1.9 meters = 48 pixels, 1 tile = 16 pixels) to convert those measurements into seconds and meters, and ended with a terminal velocity of approximately 10.86 m/s. For gravity, I instead jumped up, and observed how many frames it took her to reach her maximum jump height, as well as the height itself. Since Vf = 0 at this point I can use this data to solve for acceleration due to gravity which turned out to be approximately 15.08 m/s^2.

As for projected area, I took a look at her sprite when she is at her terminal velocity and determined that the area would be roughly that of an oval formed by the cross section of two lines. One from her elbow to her shoulders, and back to the other elbow, and the other from her back to her face. Based on these two diameters, I found the projected area under Samus to be approximately 0.6054 m^2.

Finally, I researched what exactly the drag coefficient was, which is a quantity used to quantify drag of an object. According to engineeringtoolbox.com, the drag coefficient of a person in upright position is approximately 1.0-1.3. I went with 1.3 given that the bulkier frame of Samus’ suit would likely have more force acting on it. With all my required values, I simply re-arranged the previous equation to solve for density, which I then calculated to be approximately 29.241 kg/m^3, just over 24 times the density of the earth (1.2 kg/m^3).

With this ratio in mind, I then set out to calculate how much of Zebes’ atmosphere was water vapor. To do that, I found out how much of earth’s atmosphere was water vapor, and adjusted that to fit the ratio. According to libretexts.org, the standard molar volume of any gas at STP is 22.4 L/mol which would be approximately 45 mol/m^3. About 4% of this is water vapor, meaning there are approximately 1.8 moles of water vapor per m^3. Again assuming Zebes’ atmosphere is near identical in composition to earth, and 24 times as dense, that would mean there are approximately 43.2 moles of water vapor per m^3 on Zebes. Based on that and the fact that it would have the highest freezing point, I felt that it was fair to assume that the water vapor was the portion of the air being frozen. Furthermore, since water vapor is being frozen, the substance that the enemies are encased in would in fact be ice

Now all that was left to do before my final calculation, was to check the freezing and boiling points of this vapor. To do this, I needed a reference pressure. Given that the ice beam can still easily freeze enemies in the hottest area in the game known as Norfair, I chose to find what the approximate pressure of Norfair would be. Norfair consists of a series of subterranean caverns, many of which are filled with a magma-like substance. Most of the area is too hot for Samus to access without the proper equipment, as such, I hypothesized that the temperature would be similar to that of the inside of an inactive volcano. While this information was difficult to find, I was eventually able to find some data on livescience.com, which details the inside of Mount Hood to be approximately 750 degrees Celsius. Ideal gas law states that PV = nRT, where P = pressure in atm, V = volume in liters, n = moles of a substance, R = the gas constant (0.082), and T = temperature in degrees Kelvin (in this case 1023). It can be assumed that molar volume and density of the air remain the same in Norfair, as Samus moves identically to how she does on the surface of the planet. Therefore, all the variables are accounted for. Based on the previously established ratio of Zebes’ atmosphere to earth’s (24:1) we can calculate that one liter of Zebes’ atmosphere is approximately equal to 1.07 moles ((1/22.4)*24). With all that in mind we can calculate the pressure of Norfair to be approximate 90 atmospheres or 9.12 Mega-pascals.

Based on this phase diagram of water, we can see that the freezing point of water is still 0 degrees Celsius at this pressure. the boiling point however, is now closer to about 275 degrees Celsius.

Finally, with all that digging out of the way, we can move onto the final calculations

Final Calculations

To calculate the energy output by the ice beam, the final piece of the puzzle we need is the mass of the ice around an enemy. As Samus can freeze many different types of enemies, all of which have varying different sizes, I chose the largest enemy I could find that I could calculate the volume of with decent ease.

The enemy I happened to choose, was the metroids themselves. The metroids are parasitic creatures that grow as they consume the energy of living beings. The average adult metroid (pictured right) is roughly spherical in shape and about the size of Samus herself. Given their particular shape, I felt it would be trivial to calculate their volume, and by extension the volume of ice that would surround them. Based on this sprite, the radius of the metroid would be 19 pixels (0.752 meters).

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Given that the sprite does not change size when frozen, the radius of the ice would need to be less than a pixel in length (about 4 cm). To ensure that the ice would be of a decent thickness and for the sake of generosity, let’s say the ice is about 2 cm thick. Taking the volume of a sphere with a radius of 0.752 meters and subtracting it from the volume of a sphere with a radius of 0.772 meters gets us a volume of approximately 0.15 m^3. As the density of ice (917 kg/m^3) would not change since solids and liquids are nearly impossible to compress, the mass of this volume of ice would be approximately 137.55 kg.

To calculate the final energy output, we will need to calculate the sum of the energy it would take to bring this mass of water vapor down to 0 degrees Celsius, as well as the energy it would take to change its state from vapor, to water, and then into ice.

The first calculation would be to find the energy (Q) required to bring the vapor to its boiling point (in Norfair approximately 275 degrees Celsius). for this calculation, we need the mass of the vapor, its specific heat capacity (how much energy it takes to change 1 kg of the substance by 1 degree) and the change in temperature (Q =mcΔT). The specific heat capacity of steam is approximately 1.996 kJ/kgK and the change in temperature would be approximately -475 degrees. Therefore the energy expended to bring the vapor back down to boiling point would be approximately 130,411.155 kJ.

However, there are still 3 more steps to this calculation. The next involves the energy needed to change the state from solid to liquid Q=mLv, where Lv = latent heat of vaporization of water (2258 kJ/kg), this gives us approximately 310,588 kJ.

The third calculation is the same as the first. However, this time the heat capacity used is that of water (4.187 kJ/kgK) and the change in temperature would be -275 degrees. Plugging these values in results in approximately 158,378.51 kJ.

The fourth calculation is the energy to change this water into ice using the latent heat of fusion of water (334 kJ/kg) Q=mLf. The resultant energy from this equation would be approximately 45,942 kJ.

Results

Taking the sum of all these values gives us a final energy output of approximately 645,320 kJ. To contextualize that, that would be more energy than the combined kinetic energy of 645 1 tonne cars moving at 161 km/h or the energy output by 154 kilograms of TNT. all that energy is contained in a single shot from her arm cannon, and that isn’t even taking into account her various other beam upgrades that can be stacked on top of each other. Overall, far more impressive than the speed booster (approximately 22.36 kJ) and extremely more complicated to research, it may be safe to say that this was and will likely remain the most complex topic covered in this blog. Interestingly enough, there are seemingly even more powerful abilities in her arsenal. However the ice beam sets the bar extremely high, so her arguably more powerful abilities may fall short when it comes to the math. Only time will tell if her other abilities live up to these immense expectations.