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#1 Edited by TheOneWhoPullsTheStrings (2746 posts) - - Show Bio

Here's a better calculation for the moon feat. First we need to find out how big the picture on the moon is. If we set the background to green, and then the moon and then simply the red, we can run it through a script, like on python.

The picture:

No Caption Provided

The results are:

Green: 26617 pixels

Red: 8983 pixels

Since the pic is 256x256 we can deduce the number of pixels of the moon by a simple subtraction:

Moon: 256*256 - 26,617 = 38,919 pixels

Figure the percentage now...

Percentage of paint: 8,983/38,919 = 23.08%

Now we need to use that percentage to get he overall surface area we are talking about. The total surface area of the moon is: 37.9 million square kilometers.

37.9*0.2308[=to23.08%]= 8.74732 million square km.

Since we want to factor out the fact that this is only one half of the moon, with the other half being on the other side, we can divide that by half.

8.74732/2 = 4.37366 million square km.

Move the decimals over to get the whole number back...

Total area that Hancock managed to paint: 4,373,660 square kilometers, give or take some small rounding related imperfections.

Let's assume he just managed to do this in one go, and just flew in a straight line, so we don't need to do any extra work with multiple passes. That is just adding that foot in a straight path of about a foot KILOMETER LONG brush (pretty big brush...), we should add this number to the distance to the moon.

Distance to the moon: 384,472 km.

Distance he moved during painting the entire red surface area: 4,373,660 km.

Adding this together, we get = 4,758,132 km that he needed to go to paint the moon.

Now he needs to come back to earth in time. So add another 384,472 km...

And we arrive at 5,142,604 km needed to go for him to leave to the moon, paint the entire huge area he needs to, then come back and be finished.

Assuming he had to take no extra trips and painted in nice huge KILOMETER long brush (assume he could), as a very low estimate (and if you play with a realistic number, wow, does this skyrocket in speed, and even combat/turning speed for this), and no flying way back to see his progress and make corrections, no mess-ups, etc, and the most perfect flight path with no wasted KM possible...

Let's take the time of the scene now...

We see the previous scene was 'one month later', then this. We have a full length normal conversation between the characters Mary and Ray in the park, oblivious to the moon. No one in the background is remotely noticing, and no one knows about the moon so far that we know of nearby. Ok, this is a HUGE development, how long do you think it would take the average person to become caught up in the drama that would have happened? In PUBLIC no less. I would say the conservative estimate, is what? 4 hours? Heck, I will go ultra conservative, and let's say no one noticed no one painting or starting to paint large portions of the moon for 12 hours (lol, these people hiding in caves).

Now let's plug these numbers in. I will assume to help make it even more conservative that he just finished right at that moment and called Ray, so his speed doesn't even get faster here...

We have 12 hours, 5,142,604km, and we have speed = 266,289 miles per hour (instead of doing this on pen/paper, I just used calc at https://www.calculatorsoup.com/calculators/math/speed-distance-time-calculator.php)

Guess how much machs that is? About mach 350.

On an estimate that he painted with a paint brush as long as a kilometer perfectly the first time with absolutely no delays. As the most possible, conservative estimate you can possibly give hancock here on all variables, from time to how big of an area he could paint at once without slowing down.

Does anyone want to see realistic estimates, that place him well within Mach 30k+? Because just about every possible thing was lowballed to ridiculous levels here on these numbers. There is no way he is painting a km at a time without slowing down, and having no artistic delays. Just subtle adjustments will increase the distance traveled by orders of thousands. And thus the speed will go up even faster, especially under more sane time frames.

And the India->Delaware, 7532.5 miles via air straight journey - even if accepted scene transition to 3 seconds is about mach 12k, or 9,039,000mph.

So yeah, in DCEU, if you lowball with all conservative metrics, supes is an order of magnitude faster. But by simply being reasonable with the numbers, Hancock gets even faster than that. Both with their highest end speed feats. I tend to not try and be conservative, and realistically - Hancock is an order of magnitude faster than supes in travel speed.

I think this thread pretty much proves one thing: It is quite comparable, and both sides have been going on for so long about it because they tend to give different scales of how quick everything really was in Hancock or not. If you give him very conservative estimates, he is slower. Give him high estimates, some much faster than superman, give him moderate, and still outclasses him; but the scale is very big here, and that is so debatable to some still I think. So again, it proves that it is idk, probably, but not conclusively? And then I will still stand by my statement. With the peak speed feats of both on table - Hancock should be faster than DCEU superman.