***Copied from the reddit post, here: http://www.reddit.com/r/teslore/comments/2phm72/on_nirn_lore_edition/***

Greets,

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Below, I have described my process for determining planet Nirn’s size, mass, and other characteristics relating to both of the aforementioned factors. To begin, I will describe my method for determining Nirn’s circumference.

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By creating a unit of measurement equal to the total width of Skyrim (which has been approximated to ~890.887 km), ***see this map: http://static.giantbomb.com/uploads/original/1/13836/742368-nirnmap.jpg***, I was able to compound that distance across the estimated vastness of Nirn at that latitude (45 degrees), ***see this map: http://th07.deviantart.net/fs70/PRE/f/2014/178/6/1/world_map_of_nirn_by_n_a_i_m_a-d7nx393.jpg***.

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C = 2π * r, where C is the circumference, and r is the radius of the circle. And C(45) = 890.887 km [est. Skyrim width] / ~8.6 cm [when zoomed in, and measured] = ~103.5915116 km/1 cm (on that map). And, counting the total amount of centimeters across latitude 45, yields 51.7 cm. Since this is only half the arc, we multiply by 2 to get 103.4 cm for the scaled down version of Nirn’s circumference at 45 degree latitude. And therefore, the circumference of Nirn at L=45 is ~10,711.3623 km, since 1 cm = ~103.5915116, and when multiplied by 103.4. And thusly, r(45) = C/2π = 10,711.3623 km/2π = 1,704.766257 km, or 1,701,472.02 m. ***This result is the radius from Skyrim at latitude 45 to Nirn’s vertical center, which we will denote as, a***.

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Since maps will always be distorted when planed from a spherical source (special thanks to ladynervar for pointing this out), we must determine the remaining distance of the radius, which can be achieved through trigonometric functions, namely the cosine function.

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Since we know the length of the chord, ***a*** to be 1,704.766257 km, then we can assume that this distance is also equatable to the distance from Nirn’s center, directly below the opposite end of the hypotenuse from the center (aka, Nirn’s radius). Thus, we can construct a square, which will be divided into a right triangle for the trigonometric analysis. Since we know both the angle, θ, to equal 45 degrees, and the length of the sides besides the radius, we can determine the radius to be equivalent to the following:

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a / cos (θ) = 2,410.903562 km = h = r, where a is the chord length, θ is our angle 45, and h is the hypotenuse, Nirn’s radius.

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So, with the radius now determined, we can move on to the other calculations, including: the circumference, volume, surface area, estimated mass, and density (gathered from the volume and mass measurements; naturally, this will come last).

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With radius r = 2,410.903562, then Nirn’s circumference = 2π*r = 15,148.15384 km; remember, it is was 10,711.3623 at latitude 45.

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and Volume = V = (4/3) * π * r^3 = 4.189 * 2,410.903562 km^3 = 5.869865163 e+10 km^3.

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And Surface Area = 4 * π * r^2 = 12.57* 2,410.903562 km^2 = 7.304147609 e+7 km^2.

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The in-game gravitational acceleration was calculated from the time it took an object to fall 1 m, which was ~0.46 seconds on average. Thus, the gravitational acceleration in Skyrim (and to the assumed Nirn properties) follows the gravitational equation, a = 2 * d / t^2. Where d is the distance traveled in t seconds. Therefore, we substitute the observed values in for d and t to get a = 2 * 1 m / 0.46 seconds^2 = 9.451795841 m/s^2. Earth’s gravitational acceleration is 9.807 m/s^2, so Nirn’s is slightly lower.

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***Nirn’s Mass:*** Using the following equation for planet mass, we can determine Nirn’s approximate mass, based on what we now know. So, M = (a * r)/G. Where M is Nirn’s mass, a is the gravitational acceleration of Nirn, r is the radius of the globe, and G is the gravitational accelerational constant of 6.674 e-11 N * M^2 / kg^2. The gravitational accelerational constant is applied to essentially all gravitational interactions at the macroatomic scale, and it is assumed to work the same way in the physics of the game. Therefore, the mass of Nirn is assumed from the following figures: M = (9.451795841 m/s^2 * 2,410.903562 km)/6.674 e-11 N * M^2 / kg^2 = 8.2333943751336 e+23 kilograms. The earth is ~6 e+24 kg.

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Using the equation for density, D = M/V, and after converting from kg/km^3 to kg/m^3, the density is as follows: D = 8.2333943751336 e+23 kg / 5.869865163 e+19 m^3 = ~14,026.54771 kg/m^3. The density of the Earth is ~5,520 kg/m^3.

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***For the TLDRers***

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Circumference of Nirn = 15,148.15384 km

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Radius of Nirn = 2,410.903562 km

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Volume = 5.869865163 e+10 km^3

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Surface Area = 7.304147609 e+7 km^2

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Gravitational Acceleration = 9.451795841 m/s^2

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***The mass of Nirn is 8.2333943751336 e+23 kilograms with a density of 14,026.54771 kg/m^3 , where Earth’s mass is 5.972 e+24 kg and has a density of 5,520 kg/m^3***

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Well, that’s about it I hope you all enjoyed this post!

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Farewell, and slay some stuff! ~bengle

***If any great modders are out there, please see if it’s possible to make something THIS big, using shittier graphics. I have no idea if this is possible, but it’d be really interesting to see how immense everything would be (land-scape wise) if it were magnified by the upscale…***