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What Is The Volume Of A Reactor Core (Using Volumes Of Revolution)?


Renegade343
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So while I was messing around in Mercury testing weaponry, I happened to play a Sabotage mission, and got my way to the core. When I was halfway into destroying all the tubes, I stopped and wondered: 

 

What is the volume of a Reactor Core?

 

So, getting my Oberon and Codex Scanner again, I went out for another Sabotage mission and started taking measurements needed to calculate the volume. From messing around with Pythagoras's Theorem, algebra, and a bit of GDC usage, I found out the radius to be 0.367m. 

 

Since the core is a sphere, we shall first make a 2-D formula for it (i.e.: When the sphere is a circle), which would be in the form of (x-z)^2 + (y-w)^2 = r^2. Thus, the circle formula is: 

 

x^2 + y^2 = 0.135 (since we would be placing this circle on the origin [0, 0])

 

Then, using the formula for the volume of revolution (since the circle formula will be rotated 2π around the x-axis): 

 

V = ∫π*(y^2) dx

V = π * ∫(0.135 - x^2) dx

V = π * (0.135x - 0.333x^3 + c)

 

Now, since the domain is -0.367 ≤ x ≤ 0.367, 

 

V = π * (0.135 * 0.367 - 0.333 * (0.367)^3) - π * (0.135 * -0.367 - 0.333 * (-0.367)^3)

V = 0.1039 + 0.1039

V = 0.208 m^3

 

Of course, to do a quick check, 

 

V = (4/3) * π * r^3

V = (4/3) * π * (0.367)^3

V = 0.207 m^3, which is very, very close. 

 

So the volume of a Reactor core is 0.208 m^3, if you have the sudden urge to know what it is while destroying it. 

 

And by the way, the name of the assault rifle 'Karak' is a palindrome. Just putting it out there. 

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