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The Science Behind Teleportation (Now With Images)


Faust

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I think I found something you might find interesting. So when you teleport in zombies, it's like an hourglass effect, with each individual grain of sand passing through being your atom, each going through until it reconsitiutes itself as a cylinder at the bottom of the hourglass. In that, the space grows more and more compact before passing through the absolute zero point at the center of the glass. This is what we call the aether. The reason Sam and Maxis got trapped in aether was because richtofen screwed with the controls as they went in, breaking the frequancy and garbling them all together like on The Fly (I was the first one here to connect that, right? meh, whatever. lol.)

Now as for the aether, who here took alegbra? (See lots of hands) Who remembers what a parabola was? (Sees two or three.) Well, when graphing an equation, if an object is squared, you form a parabola when graphing. On both side of you y-axis you will see that the two numbers are exactly the same on the y even though the two x are opposite. Eventually the two converge and you chart will hit absolute zero, then begin to reconsitute itself on the other side of the y-axis.

parabola-for-math.gif

Let's say that your graph is you atom as they pass through a teleporter. You want to make a copy of yourself, in other words, you squared... See where I'm going with this. Now, your atoms as they pass though will gradually loose mass before hitting that absolute zero. Now, there is no absolute zero mass in our reality. Instead, you will poke a tiny hole in space time as it warps around you to form a zero point where you no long exist. Think equationary explosive decompression. Black holes. You will be sent to an absolute negative space where the rules of physics don't apply. This is aether. Nothing exists but yourself and the other beings trapped there.

IMG_0010.jpg

Because when you were sent in energised, the universe will still want that energy you left with and will attempt to pull you back through into our world instantly. You would not even be in aether a second, as you are pulled though so fast and time would have no meaning at that zero point without space-time anyway! Additionally, as space and time are both woven into eachother, you could be thrown out of time aswell. The only way you could escape is if some time in the future someone would poke an exit hole in time.

wormhole_graphic.jpg

Now here comes the trippy bit. Parabolas can be altered. If you already have a set y-axis, like say if at the time of testing the teleporters they tried aiming them with a frequancy, your axis you be moved. Your zero would not hit zero, but eather a Y-axis or an X-axis, but that remaining energy would fry you once you left space-time. That is why the first subjects in the teleporter got turned into goo. That is also what the teleporters could not be directed with the Gerche and the red teleporters. It looked like the Die Glockes could be aimed, but in fact it was because the mainframe was the only exit point connected to them, and space-time just immediatly pulled them through the nearest point.

Also, if the amount of atoms is multiplied, like if you were to add people, the more energy you would have coming out. Meanwhile, if you have only fractions to work with, like say with fractions of atoms created from nuclear half-lives, there would not be much energy coming out. As the zombies ARE created via radiation to 115, that could explain why zombies could not teleport in the Gerche and MDT, but as all the zombies in Five were fresh, they could with the red teleporters. Meanwhile, an entity that could alter it's energy (hellhounds) could free itself from aether and pull themselves through holes in space time. Say the plasma in lightning strikes or static energy you get in clouds. Seeing as fog is a type of cloud, that would explain why the fog comes in before a hellhound round.

IMG_0019.jpg

But the thought of an absolute zero bothered me... A living purgatory and if you were to say that the afterlife was a connected... That could mean that aether could also be a portal to hell.

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Guest Faust

I'll improve on this over the month. I know it's not perfect seeing as I could use my desktop for this. Please leave comments, one comment is equal to about three brains for me! ^_^

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Guest liltaz23

That was a very informative post, but I have a question. What makes the zombies in Five "fresh"? Like, freshly made?

Sorry if thats a stupid question, but I didn't fully understand.

And I'd give you brains for your excellent post, but I need to spread them around, first.

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Guest Faust

I'm thinking that they were freshly made because they imediatly began to attack once they zombiefied instead of the other outbreaks where they just sat around for a bit.

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Guest liltaz23

I'm thinking that they were freeshly made because they imediatly began to attack once they zombiefied instead of the other outbreaks where they just sat around for a bit.

Ahh, Ok, gotcha. Thanks for explaining. That makes sense.

And I don't think the Five teleporters use 115. I know its a joke map, (probably, at least) but I doubt Kennedy, Mcnamara, Castro, and Nixon are all immune to Uup. They'd be turning almost instantly when exposed to it, if I have my facts straight.

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I'm thinking that they were freeshly made because they imediatly began to attack once they zombiefied instead of the other outbreaks where they just sat around for a bit.

How do we know it's right when they were Zombified?

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Guest Scorpion

For the JFk crew, they could have found a medicine that could have made them immune. Remember how Richtofen is on their wall in the DEFCON? Perhaps Rich found the cure or prevention of infection and extracted it from his own crew.

Good informative post btw! Parabolas do have two identical points and etc. If only we had a visual of what went on in Der Riese and elsewhere. Perhaps that's what they'll do for the summer of 2012 and release it as a separate game (I can only hope)?

Your last sentence made this post even more interesting as seeing that there is some evidence of the underworld being tied in with the Nazi experimentation.

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Guest Faust

Thanks all. I'm going to have to go with liltaz on this one. They may not have thought everything through. That, or the reds work in a totally different way.

The above idea makes a lot of sence, seeeing as if Mac gets down he shouts "Give me the shot dammit!"

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Guest cjdog23

is this the one the mod moved? if *this* is tl;dr than obviously he has never read one of Alpha Snakes in depth posts before pics are added ;)

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Guest Faust

I know right? Oh well, I guess I'll work on this topic's looks later, I don't feel like graphing anything right now.

I feel like a dork for remembering all this, seeing as I always said that I'd never have to use this again and that math is for []s. :lol:

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is this the one the mod moved? if *this* is tl;dr than obviously he has never read one of Alpha Snakes in depth posts before pics are added ;)

A quick question... What does tldr stand for?

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Guest Monopoly Mac

Thanks all. I'm going to have to go with liltaz on this one. They may not have thought everything through. That, or the reds work in a totally different way.

The above idea makes a lot of sence, seeeing as if Mac gets down he shouts "Give me the shot dammit!"

Wait... What if the syrette we have been using all this time actually contains a cure of somesort?

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Guest cjdog23

although... if you really want to get into this... there are imaginary numbers to take into account ;)

take y=4 on the parabola, easy to solve, right? y=mx^2+b, so 4=x^2+0 (y-intercept being 0). this means x= ?4. ?4 is both 2 and -2, right? well... thats not quite it. when taking into account *all* possible numbers (not just rationals/irrationals) the ?4 technically equals 2, -2, 2i, and -2i. the imaginary numbers being in the III and IV quadrants of the graph (the 'negative space' parts)

so... what does this mean? frankly, i have no clue haha

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although... if you really want to get into this... there are imaginary numbers to take into account ;)

take y=4 on the parabola, easy to solve, right? y=mx^2+b, so 4=x^2+0 (y-intercept being 0). this means x= ?4. ?4 is both 2 and -2, right? well... thats not quite it. when taking into account *all* possible numbers (not just rationals/irrationals) the ?4 technically equals 2, -2, 2i, and -2i. the imaginary numbers being in the III and IV quadrants of the graph (the 'negative space' parts)

so... what does this mean? frankly, i have no clue haha

I love this lol, just got done with this in Algebra 2

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Guest cjdog23

although... if you really want to get into this... there are imaginary numbers to take into account ;)

take y=4 on the parabola, easy to solve, right? y=mx^2+b, so 4=x^2+0 (y-intercept being 0). this means x= ?4. ?4 is both 2 and -2, right? well... thats not quite it. when taking into account *all* possible numbers (not just rationals/irrationals) the ?4 technically equals 2, -2, 2i, and -2i. the imaginary numbers being in the III and IV quadrants of the graph (the 'negative space' parts)

so... what does this mean? frankly, i have no clue haha

I love this lol, just got done with this in Algebra 2

2 years ahead AP math FTW! haha, i took it this year as a freshman haha (btw, am i the only one on summer vacation or what!?)

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although... if you really want to get into this... there are imaginary numbers to take into account ;)

take y=4 on the parabola, easy to solve, right? y=mx^2+b, so 4=x^2+0 (y-intercept being 0). this means x= ?4. ?4 is both 2 and -2, right? well... thats not quite it. when taking into account *all* possible numbers (not just rationals/irrationals) the ?4 technically equals 2, -2, 2i, and -2i. the imaginary numbers being in the III and IV quadrants of the graph (the 'negative space' parts)

so... what does this mean? frankly, i have no clue haha

I love this lol, just got done with this in Algebra 2

2 years ahead AP math FTW! haha, i took it this year as a freshman haha (btw, am i the only one on summer vacation or what!?)

I am! Got out about 2 weeks ago, parabolas was our last topic

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Guest Anti Earth

I just sat my Specialist Maths, unit 3, Mid year exam today!

:D

(And I can assure you that the mathematical basis of this is....

entirely retarded :|

No offense...)

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Guest Faust

although... if you really want to get into this... there are imaginary numbers to take into account ;)

take y=4 on the parabola, easy to solve, right? y=mx^2+b, so 4=x^2+0 (y-intercept being 0). this means x= ?4. ?4 is both 2 and -2, right? well... thats not quite it. when taking into account *all* possible numbers (not just rationals/irrationals) the ?4 technically equals 2, -2, 2i, and -2i. the imaginary numbers being in the III and IV quadrants of the graph (the 'negative space' parts)

so... what does this mean? frankly, i have no clue haha

Erg... I'll work on it. :?

Thank you for asylum clearance.

Wait cj, doesn't that go into quadratic formula? x=-b+/- the square root of b*-4ac devided by 2a?

(Here's the last people who had a shred of getting this leave the room.)

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Guest cjdog23

still standard parabola ;) imaginary numbers cant physically be graphed, however, they're more or less of an abstract concept.

altho it is an interesting concept on how the transportation of matter might be achieved, if x in a parabola can equal ? (all real numbers)? what i mean is, at what point on the x axis would the transporation be complete and all information transfered (i.e. the top and bottom of the hour glass).

furthermore, the sand at the bottom of an hour glass is not in the same order as it was in the beginning, transferring it to the parabola theory, wouldnt the characters be in any random order of their atoms when they materialize on the other side?

the only way i can think to stop this is to make a circuit that records the order of each characters individual atoms real-time at the time of transportation, but the computational power to calculate out this amount of data is nearly incomprehendable in theory, isn't it?

idk, maybe im thinking to much into it, hey, i only got a B in math anyway haha

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although... if you really want to get into this... there are imaginary numbers to take into account ;)

take y=4 on the parabola, easy to solve, right? y=mx^2+b, so 4=x^2+0 (y-intercept being 0). this means x= ?4. ?4 is both 2 and -2, right? well... thats not quite it. when taking into account *all* possible numbers (not just rationals/irrationals) the ?4 technically equals 2, -2, 2i, and -2i. the imaginary numbers being in the III and IV quadrants of the graph (the 'negative space' parts)

so... what does this mean? frankly, i have no clue haha

Erg... I'll work on it. :?

Thank you for asylum clearance.

Wait cj, doesn't that go into quadratic formula? x=-b+/- the square root of b*-4ac devided by 2a?

(Here's the last people who had a shred of getting this leave the room.)

I'm still here! (barely)

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