So without further adieu...

"To put some sort of focus on proceedings, I am going to attempt to answer the following question: How many quarter pounder cheeseburgers would Spider-man have to eat per day?

According to the Spider-Man 2 game statistics, then, Spider-man produces exactly 0.005 gallons of web fluid for each thread he spunks out for swinging on a building. Bear in mind that, in the game, like in the film on which it is based, Spidey generates web fluid organically inside his own body. I hate working with heathen imperial measurements so let's translate that to something a little bit more metric.

0.005 US liquid gallons = 18.927059 cubic cm

Now, let's assume that the average distance between Spider-man and a tall building when he's sailing happily through the air is 25 feet. So let's stretch those 19 cubic centimetres into a cuboid 25 feet long. First we convert 25 feet into centimetres.

25 feet = 762 cm

So now we divide the volume by the known length. Before I do that, though, I'd like to thank Windows calculator for their contribution to this project.

19 cubic cm / 762cm = 0.0249343832020997375328083989501312cm

We'll call that an even 0.025cm. This figure is the area in centimetres squared of a web-line's cross-section. That's roughly one quarter of a square millimetre. Since the square root of 25 is 5, then (if I'm doing this right) then the width of an average thread of web line is 0.05 millimetres. Go take a look at the nearest convenient ruler - a millimetre is really titchy, isn't it? Now divide that titchy space by 100 and multiply the result by 5. That's too titchy to think about. If Spider-man really does only use 19ml of web fluid in one line, the line would be so thin it could be used by amoebae as a gallows rope.

Now let's consider that this line has to support a full-grown human, which opens up a whole new can of worms. According to wikipedia, the average weight of a US teenager is 99 to 141 pounds. Spider-Man is *beep* built, o'course, so we'll say he leans towards the latter value and give him 140lbs. Once again I find myself swimming in the treacherous Imperial Sea, so let's get that converted into God's own metric.

140lbs = 63.502932kg

We'll call that 63 and one half. Now to calculate the pressure being applied to the webline in kg/square cm:

63.5 / 0.005 = 12700

And with one last bit of mathematical magic, we'll convert that into Pascals, the preferred unit of pressure:

12700 kg/square cm = 1245444550 pascals

So that means that Spider-Man's web fluid needs a tensile strength (as in, how much it can support without breaking) of AT LEAST 1245.44 megapascals (MPa). According to Wikipedia, spider silk has a tensile strength of 1150 MPa. But then, this isn't strictly speaking spider silk, this is spider silk scaled up to the kind of gooey spunk a human/spider hybrid would produce. So it wouldn't be completely out of the question for it to have a higher tensile strength than standard spider silk. Damn those smug wise-guys in the Spider-man 2 research department.

None of this, of course, answers the question of how many quarter pounders Spidey would need to eat every day. Since we have no idea of the exact nature of web fluid we can't really work out its density, so let's pluck a completely arbitrary number out of the air and say that the density is 1.5g/cubic centimetre. So 19 cubic centimetres (the average volume of a webline, remember) weighs 28.5g. Didn't I say this would be fun?

Now we have to figure out how much of the wretched stuff Spider-Man uses while out on the town. He maintains a double life, so we'll say he divides 12 waking hours exactly between his two personas and that he spends 6 hours of the day on patrol. Now for some completely made up figures for what he does during one hour of crimefighting:

Weblines used for transport - 50

Number of weblines needed to web up a thug - 20

Number of thugs webbed up - 15

Total number of weblines used in combat - 300

Total number of weblines used per hour - 350

Patrol = 6 hours, so number of weblines used per patrol: 2100

And the total mass of all those weblines? 2100 x 28.5 = 59850g, or 60kg, very nearly Spider-Man's entire body weight. And you can't just generate spunk fluid out of nothing, dear reader - all of that has to be produced from the nutrients in Spider-Man's sexy, sexy bod. So, how many quarter pounder cheeseburgers would Spider-Man need to eat before his patrol in order to prevent himself from becoming a poorly-dressed emaciated husk in ten seconds flat? Let's convert back to the hated imperial system to find out.

60kg = 132.277357lbs

132.277357 x 4 = 529.109428

A grand total of 529 complete quarter pounder cheeseburgers and maybe a bite or two of another. So, if a skinny youth ever pulls up to the drive-thru and orders five hundred cholesterol combos, take a moment to shake him by the hand and thank him for his service to justice. Or assume the London Philharmonic sent him out to pick up lunch, or something.

FUN FACT: In the comics, Peter Parker's webshooters were technological. Sam Raimi opted to have organic webshooters in the Spider-man films because he felt that an organism capable of generating seemingly limitless amounts of miracle adhesive from bodily nutrients and shooting it at high speed from glands in his wrists would be 'more realistic'.

ANOTHER FUN FACT: Sam Raimi went to university."

So there you have it, makes sence doesn't it. Mechanical shooters should be what the new movie does, and judging by the first Spidey pic released we will have it that way. Peter has always been a genius, and this movie needs to show that side of him, much more then Raimi's films did.

Posted By:

ThreeBigTacos- 1/14/2011, 10:37 AMXandera- 1/14/2011, 10:58 AMNice article nick...

GrandWrex- 1/14/2011, 11:09 AMnuck82- 1/14/2011, 11:55 AMcan you add me a cliff notes please? thank you sugar tits ; )

SOAL

MassExecutions- 1/14/2011, 12:35 PMselinakyle- 1/14/2011, 3:00 PMPaulley- 1/14/2011, 3:04 PM:P

superotherside- 1/15/2011, 9:26 AMPlease log in to post comments.

Don't have an account?Please Register.