The Physics of Cybertron FAQ Version 1.0 11-17-97 by Robert E. Powers (repowers@shell.faradic.net) There's been some long-standing confusion about Cybertron -- where it is, how big it is, its atmosphere. Here, using the scanty facts we have from the show and comics, are some answers. Each answer will treat both the comic and the cartoon as the seperate universes which they are. The purpose of all this is not to provide definative answers, but to see just how much psuedo-science can be wrung out of the few and conflicting facts we're given, and lay to rest some myths... 1) How big is Cybertron? Cartoon: Best guess, considerably smaller than the moon. The only comparison we have comes from "The Ultimate Doom", where Cybertron is brought to Earth -- and shown to be much smaller than it. The moon is about 1/4 the diameter of the Earth. Cybertron is around the same size, maybe a bit smaller, so I'd guess 3000 km diameter. Comic: According to issue #1, the same as Saturn -- many times the size of the Earth, around 120,000 km in diameter. Later views seem to contradict this. Many images show Cybertron to be of small scale, the curve of its horizon visible to those on its surface. Unicron was about the same size as Cybertron, and a letters colomn answer instructed us to think of Unicron as "planetoid size". Unicron dwarfed the Transformers, but they weren't all that much smaller than him. Also, both Cybertron and Unicron were supposedly formed from "asteroids" (issue #60), which would limit the size to about 500 km. 2) How far is Cybertron from Earth? Cartoon: No clear answer. Before the events of "The Ultimate Doom", Cybertron was presumably stationary, though it was never shown to be orbiting a star. After that it was pushed into motion throught the cosmos. In "The Rebirth" it was again brought to Earth; no telling what happened after that. Comic: Cybertron originally orbited Alpha Centauri, four light years away. It was shaken loose from orbit (in the UK comics, this was the result of a prototype planetary drive built by Megatron), and 1000 years later passed through our system. 4,000,000 years have passed since then. Simple math: 4,000,000 yrs * (4 ly/ 1000 yrs) = 16,000 light years away For perspective, galaxies range in size from around 3000 up to 750,000 light years across; ours is thought to be about 25,000 light years across. So Cybertron is on the other side of the galaxy by now! Of course, that's assuming it wasn't caught by a star's gravity and slowed down or put on a different course, or simply drawn into orbit around the center of the galaxy like stars are. (Our galaxy rotates around its axis; all stars in it are in orbit around the galactic core.) OTOH, Cybertron supposedly passed close to Nebulos around 1987, and issue 42 mentioned that Nebulos is a "several dozen light years" from Earth. What, the comics in a self-contradiction?? Never!... 3) How dense is Cybertron? How high is its gravity? Hoo-boy, that's a tough one... but it's the main reason I wanted to do this FAQ. Let's use what little we 'know' about Cybertron, and a bunch of assumptions, to calculate the density of Cybertron -- and put some limits on its size. We have to start with two assumptions: Cybertron has an Earth-like atmosphere, and a gravitational field similar to Earth. In both the comics and the cartoons, these are safe assumptions, since we see humans running around on the planet with no protective breathing gear, and (seemingly) without getting flattened by excessive G's. See: issues #74-78 in the comics; "Divide and Conquer", "The Ultimate Doom", "Desertion of the Dinobots", and numerous 3rd season eps in the cartoon. With these assumptions, we can conclude that Cybertron has about the same mass as Earth, in order to hold onto that atmosphere. One could counter that the atmosphere is continually resupplied by some artificial source, allowing for a lower-gravity Cybertron. Given the results below, you might want to do that for the cartoon. But that's one assumption too many for me, and I'm not going to bother with it. Consulting our handy Physics text, we find that Earth's mass = 6 x 10^24 kg = Cybertronian mass To get a volume out of this, many more assumptions must be made. First, we've seen in both cartoon and comic that Cybertron is not solid; it is riddled with tunnels, passages, and chambers. (issues 60 & 61; "Key to Vector Sigma"; "Dweller in the Depths") Second, it is more or less made of metals of some kind or another, so where it has structure it's probably pretty dense. But this dense structure is laced with areas that are hollow. Consider some common metal densities... Aluminum = 2.7 g/cm^3 Iron = 7.9 g/cm^3 Lead = 11.35 g/cm^3 Titanium = 4.54 g/cm^3 Also consider the densities of materials at pressures like those found in the interior of the Earth: density of Earth's mantle = 3400 to 5500 kg/m3 = 3.4 to 5.5 g/cm3 density of Earth's inner core = 13000 kg/m3 = 13 g/cm3 average density of Earth = 5.5 g/cm3 We'll assume we want to shoot for about 15 g/cm3 = .015 kg/cm3 for average density of Cybertronian material, since it's largely metal like the Earth's core. First, let's look at what happens if you assume a solid planet at this density: volume of Cybertron = (mass/density) = (mass/(mass/volume)) = (6x10^24 kg/(.015 kg/cm3)) = 4 x 10^26 cm3 volume = 4/3 * pi * r^3 r = cube root (.75/pi * vol.) r = 460,000,000 cm = 4,600 km, about .75 times the size of Earth (r = 6,378 km) But, in the cartoon, you've got a Cybertron that's much smaller than Earth, with r = 1500 km. volume = 4/3*pi*1500^3 = 1.41 * 10^10 km^3 = 6x10^24 kg / density of Cybertron density = 4.24 * 10^14 kg/km3 = 424 g/cm3!!! 400 times more dense than common Earth metals and interior densities! And to account for missing interior space, the materials have to be even denser... so a sub-moon sized Cybertron with Earth-like gravity isn't realistic, unless it's made of some ultra-dense materials. Common sense backs up the calculations: how are you going to cram Earths' mass into such a (relatively) tiny volume without coming up with ultra-dense materials? Now let's consider the question of hollow space inside of Cybertron. Hollow space effectively reduces the total density of the planet, as follows: ratio of solid volume to total volume * metals density = actual density The more hollow space you have, the smaller the ratio on the left becomes, and the smaller your actual total density becomes. Suppose Cybertron really IS Saturn-sized, as comics issue #1 says, and Earth-massed, as the evidence says... how much hollow space is there? r=120,000 km v=4/3 pi r3 = 7.2 * 10^15 km3 6*10^24 kg / 7.2*10^15 km3 = 830,000,000 kg/km3 = 833 g/m3 = .000833 g/cm3 .000833 g/cm3 / 10 g/cm3 = .0000833 as ratio of solid-to-total volume... a nearly empty sphere with only .008% of its volume filled with solid matter! So the Saturn-sized Cybertron just isn't realistic -- such a thin shell couldn't support its own weight in all probability. Again, common sense backs this up, indicating that to spread Earth's mass over a volume hundreds of times its own results in a very 'dilute' mass, not very dense at all. My guess for a 'realistic' Cybertron is .5 to .75 solid-volume-to-total-volume, that is, about 25% to 50% of the interior volume is hollow. Using this and the same density as before gives: (.25 -- .5) * 15 g/cm3 = (7.5 -- 3.75) g/cm3 = density volume = mass/density = 6*10^27 g / (7.5 -- 3.75) g/cm3 = 8*10^26 -- 1.6*10^27 r = cube root (3/4 vol. / pi) = 5.76*10^8 cm -- 7.26*10^8 cm = 5.76*10^3 km -- 7.26 * 10^3 km radius = 5,760 -- 7,260 km diameter = 11,520 -- 14,520 km ...roughly in the neighborhood of our own mudball planet (which weighs in at 12,600 km) -- compare that to the cartoon's 3000 km, or the comic's 120,000 km. Also, remember that this is assuming an *average* density of materials that is actually a bit higher than the average of Earth's core, where its average density is highest. The mantle and crust of Earth have much lower densities, about half to one-forth that of the core. It might be more realisic to lower the assumed Cybertron density a bit, say by 1/3; this would give a larger Cybertron. All things considered, the physics of the situation seem to dictate that even a reasonably hollow Cybertron should only be a few times the size of Earth at the very most, unless it's made of ultradense metals, the only thing that would allow a much smaller volume; and it should not be significantly smaller than Earth. So, in the end, the most realistic answer to "How big is Cybertron" would seem to be "Earth-sized or a bit bigger." Please, feel free to check my math, correct my formulas and data, and challenge my assumptions...