@Shos: Yes. Yes.
@Shos again: Your first step should be isolating yourself completely from society. If people know you are still alive, you are doing it wrong. The last few important hypothesises proven were all proven by hermits that have dedicated years of their lives to nothing else but that single hypothesis.

Is it possible for a snake to disappear by eating itself? Me and my buddies had a fierce arguement over this at the lunch table.

How long will it take to reach my goal of 200 avatars on Neo?

Why do you have about about -1030957 times more karma than you have posts as of now?

@Yaya: No. It is able to kill itself, but the conservatíon law of mass says that it can't disappear in nothingness.
It would kill itself very quickly anyway, so even if it had some sort of interdimensional hole in his mouth, he still wouldn't be able to completely eat himself.

@soccerboy: Depends on your current amount of avatars, your motivation and which avatars you already have. In the old Interguild when I had about 180 avatars, I had a little competition with Shos who could reach 200 first. I got those avatars within two weeks, but you should remember that I already had many retired and random event avatars, and also several which aren't difficult but still take a long time to get, such as the NQII avatars.
I suggest avatar.starried.net to easily find out which avatars you still need and how to get them.

@soccerboy again: I asked Accel to reduce my karma to the minimum amount possible. It was mostly done as a joke, but also to let myself not worry about karma. I don't care if I gain or lose karma, because my karma is a joke anyway.

but, for the record, jell, you have 61 karma.(2^31+your current karma-1=your karma).

note: while at -2^31+1 if you get a -1 it doesn't matter, so actually, you may have less than 61.
note2: i wonder what happens if oyu have -2^31+1 and i -1 you, and then i cancel that -1. would that add 1? ;D

If I'm at -(2^31)+1 and you -1 me, I would either loop back up to 2^31-1 or I nothing would happen at all. If you delete that rate, I would probably go back down to -(2^31)+1 again.

here's a question:
if i have two identical balls which have a distance of 1 meter between them, how long will it take them to hit  a massless, tiny block between them?

Fg= G*m*M/d^2
Frot=I*r
Fres=0.5*ρ*Cd*v^2*A
Friction can be neglected if it is assumed the angular speed of the balls is the same as their lineair speed divided by their radius, which would be the case if the stationary friction coefficient is high enough.
This is assuming the sphere don't contain an electric or magnetic charge and any molucular level forces, such as the Van der Waals force, are neglected.

G=6.67300*10^-11
m=M
I=2/5*m*r^2
ρ=1.2
Cd=0.47
A=Ï€*r^2

Ftot=Fg-(Frot+Fres)=G*m*M/d^2 - (I*r + 0.5*ρ*Cd*v^2*A) = G*m^2/d^2 - (2/5*m*r^3 + 0.5*ρ*Cd*v^2*π*r^2)=6.67300*10^-11*m^2/(1-2x)^2 - (2/5*m*r^2 + 0.89*v^2*r^2)
a=d^2x/dt^2=Ftot/m=6.67300*10^-11*m/(1-2x)^2 - 2/5*r^2 - 0.89*(dx/dt)^2*r^2/m
Unfortunately this can't be solved.

this is not what i meant. i meant, supposing there are no other forces at all in the universe, and they're in vacuum. each ball has the mass of 1 kg, radius of 0(they're dots).

If there are no forces in the universe, nothing will ever happen. They won't ever accelerate and just stay in their own spot for ever and ever.
In fact, if there are no forces in the universes, the quarks in the atoms won't get hold together anymore by the strong nuclear force and the balls will simply dissolve.

no, they will not dissolve; they will stay as a balll, because no force allows them to dissolve.

anyways, the balls are neutrons if you'd like. no electrical charge. the only force is the gravitational force between them.

If they were neutrons, there would still be the strong and weak nuclear forces. Besides that, they will break down into a proton, elektron and anti-neutrino long before they reach eachother. And once that happens there is also the Coulomb force to account for.
And the balls will dissolve thanks to Heisenberg. It is impossible for a particle to stand still, especially light particles like quarks. They will gain temporary energy from the vacuum. According to the chaos theory their positions will become disordered and they will move in random directions, mostly away from eachother.

sure. now let's pretend we have two round rigid bodies, when the only force that exists ON THEM is the gravitation between them.

you're avoiding my question, jell. if you keep thsi way, i'll be forced to answer...

How many finger's am I holding up?

Probably none by the time he reads your message.

true... I might though...

How many fingers am I not holding up? Hint: The answer may or may not be 69.

@shos: so we have two spheres with a radius of 0 meter, a weight of 1 kilogram, who are 1 meter apart from eachother. I should neglect the strong nuclear force, the weak nuclear force, the elektromagnatic force and any associated forces such as resistance, Einstein's special relativity, Heisenberg's uncertainty principle, the particle-wave duality, rotational intertia and anything else, besides gravitational force, for which I should not use general relativity but Newton's laws.

Fg=G*m*M/d^2=6.7*10^-11 / (1-2x)^2
a= d^2x/dy^2 = Fg/m = 6.7*10^-11 / (1-x)^2
I suck at differential equations, so I'm just going to take the Euler method to solve this with dt=0.1. I do have a feeling the equation will be an exponential equation, because it is positive feedback.
Anyway, the Euler method comes with t=95950 s.

@soccerboy: One. In your nose. Admit it.

@Isa: I know for sure that you aren't holding up e^π+(1/2*sqrt(2)*1/φ)^(e/2)*i fingers.

well yeah, that's probably almost correct. see, the differential equation should be d^2/dx^2(f)=G/f^2; where f is the distance function, and x is the time. to solve that, we multiply the equation by d/dx(f) and integrate, and then we recieve
0.5d/dx(f)^2=-G/f,
which is another differential equation we have no idea how to solve. voila! computer be here!

Wolfram Alpha - doesn't know how to solve.
MATLAB - doesn't know how to solve.
MapleSoft 12 - give you this answer:

f(x) = (1/2)*_C1*((exp(RootOf(-2*_Z*G*_C1^3*exp(_Z)-csgn(1/_C1)*(exp(_Z))^2*_C1^2+csgn(1/_C1)*G^2*_C1^4-2*csgn(1/_C1)*x*exp(_Z)-2*csgn(1/_C1)*_C2*exp(_Z))))^2+2*G*_C1*exp(RootOf(-2*_Z*G*_C1^3*exp(_Z)-csgn(1/_C1)*(exp(_Z))^2*_C1^2+csgn(1/_C1)*G^2*_C1^4-2*csgn(1/_C1)*x*exp(_Z)-2*csgn(1/_C1)*_C2*exp(_Z)))+G^2*_C1^2)/exp(RootOf(-2*_Z*G*_C1^3*exp(_Z)-csgn(1/_C1)*(exp(_Z))^2*_C1^2+csgn(1/_C1)*G^2*_C1^4-2*csgn(1/_C1)*x*exp(_Z)-2*csgn(1/_C1)*_C2*exp(_Z))), f(x)=(1/2)*_C1*((exp(RootOf(-2*_Z*G*_C1^3*exp(_Z)-csgn(1/_C1)*(exp(_Z))^2*_C1^2+csgn(1/_C1)*G^2*_C1^4+2*csgn(1/_C1)*x*exp(_Z)+2*csgn(1/_C1)*_C2*exp(_Z))))^2+2*G*_C1*exp(RootOf(-2*_Z*G*_C1^3*exp(_Z)-csgn(1/_C1)*(exp(_Z))^2*_C1^2+csgn(1/_C1)*G^2*_C1^4+2*csgn(1/_C1)*x*exp(_Z)+2*csgn(1/_C1)*_C2*exp(_Z)))+G^2*_C1^2)/exp(RootOf(-2*_Z*G*_C1^3*exp(_Z)-csgn(1/_C1)*(exp(_Z))^2*_C1^2+csgn(1/_C1)*G^2*_C1^4+2*csgn(1/_C1)*x*exp(_Z)+2*csgn(1/_C1)*_C2*exp(_Z)))


...yeah, we're screwed.

The problem is that it isn't a linear differential equation and you can't substitute anything, as far as I know. So I don't know if it is even possible to find an exact solution for it. Which is strange, because it is a pretty basic equation.

Are you psychic?

@soccerboy: One. In your nose. Admit it.   Actually no, I was playing a waltz on the clarinet

Is this a question?

What is the surface area of a pentakaidecagon with every other 3 sides being 2 in. and those 3 sides being 5.25 in. and the height being 27.2851094157?