The size of the planet Cybertron in the Generation One cartoon is somewhat of a mystery, due largely to the fact that it is rarely shown simultaneously with anything of a known size to make a comparison. If we go simply by the oft-used establishing shot artwork of the planet, which features spires and fissures which are at least 10% as deep/tall as the planet's radius, we can conclude that the planet appears very, very small, unless these features are intended to be thousands of kilometers in size. However, if we dismiss the towers and gaps as simply artistic license (after all, similar artwork was used sometimes in the comic, as well, but comic Cybertron was explicitly stated to be "Saturn-sized"), there is an opportunity to make a different sort of assessment based on the three-part episode, "The Ultimate Doom".
With the G1 S1 DVD set in my posession, I finally have a copy of "The Ultimate Doom" with which to make observations of Cybertron and the Earth and attempt to make some sort of estimate of its size in the cartoon universe. Throughout the miniseries, there are only two shots which I consider to be useful:
Below are screencaps of these two shots. Click on the them for full-sized diagrams, complete with superimposed measurements:
With Shot 1 we can make measurements of the apparent radii of both planets. It is unclear whether the planets are meant to be at the same "depth" into the screen -- it's possible that one or the other is much further away, like in this famous photo of the Earth and Moon taken by the Galileo probe (the Moon is significantly closer than the Earth in this picture, making it appear both larger and closer to Earth than it really is):
The only hint to depth in Shot 1 is the column of atmosphere between the two planets. It doesn't appear to me that the artists were trying to portray foreshortening here, since the column's width is uniform, but that's not exactly a solid argument. For simplicity, I'll assume -- for now -- they are at an equal depth.
Based on measurements I've made of this shot, Cybertron's radius is roughly 1/25 of Earth's. The smallest planet in our solar system, Pluto, is ~1/6 of Earth's radius. (Our moon is a little over 1/4.) These measurements indicate that Cybertron has a truly miniscule radius of ~250 km (diameter ~500 km). That makes it about as wide as the state of New Mexico. There are plenty of objects in our solar system of that size or less, of course, but few or none that most people will have ever heard of.
In other words, if Cybertron and Earth are at the same depth, then Cybertron is very small. Too small, I think, to realistically believe that it's the home to millions of giant robots who have been waging war for many eons. On the other hand, we know that at least some of Cybertron's lower levels are accesible, which would increase the habitable area. Also, in the cartoon mythos the whole planet was constructed (or modified, perhaps) by the Quintessons. This is more believable if the planet is small.
We can tell from the way humans operate on Cybertron that the surface gravity there is roughly the same as Earth's, maybe with a tolerance of 10% or so. An Earth-like surface gravity for a spherical body with a radius of 250 km implies an average density of 140 g/cm^3. Water, you may remember, has a density of 1 g/cm^3. The Earth's average density is ~5-6 g/cm^3. Lead is a mere 11, and gold only 19. Unless Cybertron has powerful artificial gravity generators in operation planetwide, it is unrealistically dense by more than an order of magnitude when under the assumption of equal depth in Shot 1. (Needing to leave gravity generators on continuously would help explain the energy drought, though.)
My last analysis of Shot 1 under the equal-depth assumption regards the orbital distance. The center-to-center distance between Earth and Cybertron works out to 8040 km, or 1.256 times Earth's radius. Earth's moon, for refernece, orbits at a center-to-center distance of 60 times the Earth's radius. Since Cybertron did not arrive in this position by natural means, this is not really an argument against equal-depth, but I do think it's worth noting that equal-depth places Cybertron almost 50 times closer to us than the Moon.
We can alleviate the density problem, the too-small problem, and the really-close problem all by assuming the Cybertron is further "into" the screen than Earth in Shot 1. The further-in Cybertron is situated, the larger it must be to appear at the size shown. Sadly, there is no way based only on Shot 1 to determine how much further it is.
This is where Shot 2 could, conceivably, be of help. If we interpret Shot 2 as being a depiction of sunrise/sunset as seen from near the surface of the Earth, we can compare the apparent sizes of Cybertron and the Sun and learn more about Cyb's size. The Sun, as seen from Earth, is roughly half a degree of angle in diameter. Based on the relative sizes of Cybertron and the Sun in this shot, Cybertron has an angular diameter 2.87 times larger, or 1.435 degrees. I'm going to do some math, now. If you don't like math, you can decide to trust me and my answer:
The projected center-to-center distance between Earth and Cybertron is established by Shot 1 as 8040 km. The true c2c distance in km is
c2c = Sqrt[ 8040^2 + z^2 ]
where z is the "depth" of Cybertron into the screen beyond Earth. Since Cyb's angular diameter is small, we can use a small-angle approximation (size=distance*angle) to find the linear diameter. That means we need to know the distance from our sunrise-observer to Cybertron. We can find that with a right triangle:
d Obs ________________ Cyb | . | . R_Earth | . c2c |. Center of Earth
So the distance d is found from
c2c^2 = d^2 + R_Earth^2 d = Sqrt[ c2c^2 - R_Earth^2 ]
Combine this with the small-angle formula:
D_Cyb = d * (1.435 degrees) = Sqrt[ c2c^2 - R_Earth^2 ] * (1.435) = Sqrt[ 8040^2 + z^2 - R_Earth^2 ] * (1.435) = Sqrt[ 8040^2 - 6400^2 + z^2 ] * (1.435) R_Cyb = Sqrt[ 8040^2 - 6400^2 + z^2 ] * (1.435) / 2
where I've stuck in the Earth's radius, 6400 km. The radius of Cybertron is now written as a function of only one variable: the depth, z, of Cybertron as depicted in Shot 1. What happens if we stick in z=0? Do we get something even remotely close to the 250 km radius that we got directly from Shot 1? No.
For Shot 2, if we assume a larger c2c distance than the minimum of 8040 km (determined from Shot 1) this implies a *larger* Cybertron such that it can maintain its apparent size despite being far away. The smallest Cybertron we can get is for z=0, which gives a radius of 3490 km. So if we assume z=0, Shot 1 gives a radius of 250 km, and Shot 2 gives a radius of 3490 km. Increasing z results in a larger radius as determined by both shots, and at roughly the same rate, so the smaller Shot 1 radius isn't going to catch up to the larger Shot 2 radius and give us a definite value. The obvious conclusion is that The Ultimate Doom is not internally consistant.
Although 3490 km is a nicer radius for a terrestrial planet than 250 km is (it's about the same as Mars'), that figure should not be trusted because it can only be arrived at by assuming that Shot 1 and Shot 2 are consistant, which is clearly an untrue assumption. The size cannot be determined from Shot 2 alone because we need to know how far Cybertron is and there is no basis for that judgement in Shot 2.
In conclusion, the size of Cybertron cannot be determined from The Ultimate Doom with any precision beyond saying "it's probably smaller than Earth" because the miniseries itself is highly inconsistant.
Aren't you glad you read the whole page just to learn nothing?