Here we go,

If you've ever debated Post-Crisis Superman, you've probably at least heard the name Mageddon crop up once or twice. But what is Mageddon?

Basically, it's a weapon created by the Old Gods (the precursors to the New Gods of New Genesis and Apokolips) which played a role in their eventual destruction. After that, it was trapped in a gravity well beyond the event horizon of the visible universe for some 15 billion years, until it somehow escaped, destroyed the planet of superpowered beings assigned to guard it, and headed for Earth.

Why do all of the dangerous cosmic threats always make a direct beeline for Earth? Well Sinestro Corps War actually explained that, thus depriving me of any jokes I could make about it here. But I digress.

Via a lot of plot devices and DEM (including everyone on Earth temporarily being given superpowers by an alien superbeing), Mageddon was defeated. Superman has two notable feats involving Mageddon, the first being when he was captured by it and forced to turn its wheels to help move its body. The second was when he absorbed the power in the "anti-sun" warhead designed to wipe out the galaxy. As the latter involves funky FTL physics, I'm going to attempt to quantify the former.

First of all, how big is Mageddon? In a word, Big. This is the clearest picture I could find of its whole body:

Yes, that's the sun or anti-sun in the foreground.

Now if I were to simply pixel scale it from the sun, I would be underestimating its real size, since it's further away from the viewer than the sun. So in order to get a slightly more accurate value, I'll try something I don't think has been attempted in any other calcs. (I hope I can explain it clearly, since it's rather complicated).

First I'll scale the sun and then use the angular size calculator to find the distance from the viewer:

Height of the page = 1526 px = 135 degrees.

Diameter of the sun = 1,392,684 km = 187px = 16.54325033 degrees

Distance to the sun = 4,789,900 km

Redoing with new angsize rules.

Panel width = 990 px

Panel height = 1526 px

2*atan(tan(70/2)*(990/1526) = 48.86108984882531

(187/1526) * 48.86108984882531 = 5.98756474460321118908 degrees.

Redoing yet again.

2*atan(187/(1526/tan(70/2))) = 9.80853240447804 degrees.

Distance to the sun = 8,115,400 km

Now as we were measuring the diameter of the sun from one visible edge to another, and the farthest edges would be on the plane bisecting the sun perpendicular to our angle of vision (crappy diagram to illustrate what I'm talking about):

This means the distance we got above is the distance from the viewer to the closest edge of the sun + the radius of the sun. Now since Mageddon is behind the sun, it is at least a further solar radius away, so it minimum distance would be 8,811,742 km (again, the real distance would be much greater than this, since Mageddon is obviously not physically touching the sun and there is likely a great deal of space between the sun and its body, but it's closer than the previous number).

Now I'll use the angular size calculator again, but this time I'll input the diameter of the sun and the distance I got above, then solve for the angle. I got 9.0367 degrees, which corresponds to 172.284989263888339368388306 px. This means the sun would appear this size if it was one further solar radius away. This is a difference of 108.5410869507457368286906617486%. So now I can pixel scale Mageddon from the sun, and multiply the numbers by this factor to get his size if he were one further solar radius away, which would be closer to his true size.

Since the edges of Mageddon's body extend past the page, I used the ellipse select and brush tools in GIMP to extend the outlines of its body. I think these are pretty accurate, considering they resemble the contours of an octopus, which Mageddon's shape seems to be based on.

Now to actually scale the thing.

Length = 1315px = 9,793,473.048 km. Applying the multiplier, we get 10,629,942.09652752877132682988964198651733 km.

Height = 1484px = 11,052,101.9 km. With the multiplier, it's 11,996,071.5331640216442127203722395938234 km.

We can't see the depth, but as it's a rough ellipsoid I believe the standard is to use the shortest known axis as the third axis.

Using the ellipsoid volume calculator, I get a volume of 710,013,062,229,702,200,000 km^3.

Now for the tentacles. I couldn't find a pic showing their total length, so I'm going to have to lowball it and scale only the parts shown on-panel. There are 6 visible tentacles in the picture, 4 in the foreground and 2 in the background. They should all be the same size, roughly (just appearing different due to perspective). Scaling one of them gives it a width of 216.8893727 px, which equals (after applying the multiplier) 1,753,248.26848862672371842037765084468754 km. The visible length of the tentacle (394.5072876 px), using the same method, is 3,189,041.53892112151524991989311794619585 km.

Volume of a cylinder = pi * r^2 * h. So the volume of the visible part of one tentacle = 7,699,045,653,143,722,653.287106675773443006019493 1798365898948 km^3. Multiplying this by 6, we get 46,194,273,918,862,335,919.72264005464065803611695 90790195393688 km^3. Adding this to the volume of the head, we get a total volume for Mageddon of 756,207,336,148,564,535,919.7226400546406580361169 590790195393688 km^3.

Now for density. It was never stated what Mageddon was actually made of, but it must have been some pretty hardcore shit, considering it can not only maintain its non-spherical shape despite its size, but it casually shrugs off planet busters with no visible damage and it essentially pulled itself out of a black hole (not to mention its chains could restrain even Superman). Given these facts, I think the density of osmium would be a very conservative estimate. Now obviously Mageddon is not completely solid, but as we are only measuring the part Superman pulled, which would be solid, that's immaterial (no pun intended).

Gallery

I'll estimate that 60% of Mageddon is empty space, so Superman would be pulling 40% of its volume: 302,482,934,459,425,814,367.8890560218562632144467 8363160781574752 km^3. Using the density of osmium, that's 6,833,089,489,438,429,146,570,613,775,533,732,986. 0143528422380205577364768 kg.

Now we just need to find the speed at which he pulled it. There was no indication that he was turning the wheels at any kind of insane relativistic or FTL super speed, but he was doing it fast enough to move the entire thing around in a reasonable amount of time. So as a low end I'll use the speed required to move Mageddon one of its own body lengths (measured by its shorter axis) in one hour, which is equal to 2952.7616934798691031463416360116629215 km/s.

Now KE = 0.5 * m * v^2. So the total energy involved in this feat is 29,788,175,849,874,416,436,740,588,643,341,052,300 ,501,390.153934195361739869296143779199863126295 joules, or 7,119,544,897.197518268819452352614974259202053096 0645782 yottatons

It would put Clark far beyond planet level

@dorukesin1: I didnt get past the original art you used to determine size. This calculation will be very hard for 2 or 3 reasons.

1) Look how large Orion is in the picture you posted compared to the Sun. If your math is right, Orion is thousands of miles tall, granted he is closer to the viewer, but my point being I am fairly sure based on that we can assume this is not drawn to scale. This means Megaddon could be much larger or smaller than your calc assumes IMO.

2) We can assume Mageddon has gears inside that spin and cause it to move. These gears tend to have power ratios. So it's hard to determine if he was moving the whole machine or just providing enough force to turn a gear that translated that force into enough motion to move Mageddon' millwheels. There's a reason we use machines, and that is to get more work out of the same amount of force. If Clark picked Mageddon up and moved it that would be one thing. But it seems he was simply a cog "turning the endless millwheels". He seems to have been simply powerful enough to power the machinery there. Not moving the entire thing.

@avatar_of_green: Orion isn't too big "12 light-years west" Orion and the others examining some destruction that Mageddon had done. Orion isn't too big

You just copy-pasted this from the narutoforums blog.

You Endlessmike or not?

Good heavens...

Why should we assume that Superman wasn't turning it at FTL speeds? "It doesn't seem like it" isn't good enough justification if you accept that he's millions and billions of times faster than light.

@jogga: I didn't say "hey look at my superman calculation" ,look at the title "Interesting Calculation of Superman's Mageddon Feat"

No credit to EM?

Great plagiarism.

1. That's not the actual sun, that's the Anti-Sun on top of Mageddon's head
2. As others have already mentioned, pixel scaling isn't accurate, look how big the other characters are...,
3. We have no idea what the the endless millwheels do, for all we know they could be used to turn on some lights, so Superman turning them doesn't mean anything IMO

You certainly didn't give EM any credit tho.

That's actully the sun, the anti-sun is different in size and color iirc

It was angasizing and pixel sailing, iirc

The wheels were for pulling the megaton, since it needed the buff.

But how much did superman contribute to pulling is the point? Machines increase the work done by similar amounts of force. That's the whole point of the Mageddon machine. It destroys all on its own. But, how do we know how much Supes contributed to the movement and how much force he had to put in to make it move? "Superman is pulling the full weight" is definitely not accurate.

Also, definitely can see planets in the foreground which should be closer or as close as Orion and he still looks much too large, so I'm gonna stick with my original pixel-scaling isn't accurate in all instances argument.

I've never had too much joy in arguing in Vs. Battles and calcs. I'll just give you the link and you can argue there, better than here with me. http://www.narutoforums.com/blog.php?b=18097

@jogga: Those guys are trolls. No thanks. I am here for a reason.

@avatar_of_green: To endlessly debate One-Punch Man?

@dorukesin1: I've never had to give a plagiarism warning for forums before. This is yours. Create your own content and don't represent others work as your own.