Armadillo Run forum

Hello, this is where you can discuss your Armadillo run Addiction (AA).

I will go first.

Hi, I'm Drago.

[All: Hi, Drago]

I have been addicted for at least a year. My addiction has progressed such that it has become a part of my image. I have affixed the Armadillo Run Logo to anything I can get my hands on. My addiction rules my life. I find myself spending long hours with it.


<- My algebra textbook

I have come to terms with this addiction and I now embrace it!

Thank you.

Hi, I'm Sam

[All: Hi, Sam]

My real names Sam. I have lived under the alias of McGinge for a few months now... I like to play AR not only to satisfy my urge for creativity (:D), but to escape the hell that is "nagging parents"... I also have come to terms with my AR addiction... And i'm trying to cut down...

Thankyou all for listening...

Ei = 1/2mv^2 + Qpλln(1/Ri)/(2πεo)
Ef = Qpλln(1/Rf)/(2πεo)

Ei = Ef

I'm not addicted. I can stop anytime I want. I swear.

I have no stimulus to play at all, so I'm not addicted. I'll stop soon, because it takes too much time.

After retiring, I'll return when Orjan Flatseth will do!

Contest Veteran and Contest Legend are addicted.
Light Master and Champion of Light are not.

There is your logic, guys?

//Due to your logic, guys, I'm half-addicted ^_^

FF wrote:Contest Veteran and Contest Legend are addicted.
Light Master and Champion of Light are not.

There is your logic, guys?

//Due to your logic, guys, I'm half-addicted ^_^

Maybe:

Addiction = 1/Rank

So where would that put people who have never played the game? Armadillo Uber Killer OMFGer Light Force Winner King?

NeXFerret wrote:So where would that put people who have never played the game? Armadillo Uber Killer OMFGer Light Force Winner King?

no, the rank is the independent variable.

so it would be "Armadillo Uber Killer OMFGer" level of addiction

anyway I guess:

{x| 1/x, x<>0}

Ikerous wrote:Ei = 1/2mv^2 + Qpλln(1/Ri)/(2πεo)
Ef = Qpλln(1/Rf)/(2πεo)

Ei = Ef

Pliz, explain this formula. (just tell us what does every letter mean).

Why would you ask that? Once you get him started, he never stops. Once he get home he will have a blast explaining it, so be prepared.

Well, for one, we can ignore him if we want...


But go ahead, Ike! Explain away!

Ms_Ikerous wrote:Why would you ask that? Once you get him started, he never stops. Once he get home he will have a blast explaining it, so be prepared.

Shuddup!

FF wrote:Pliz, explain this formula. (just tell us what does every letter mean).

I was answering the batman question in drago's "algebra book."

It says that theres a proton (Of mass m and charge Qp) that's a certain distance (Ri) from a charged line (With charge density λ). The proton starts out moving towards the charged line with an initial velocity (v) and ends up stopping at some distance (Rf) from the line. εo is just a constant in physics.

It's just a basic energy problem. The proton starts out with a lot of kinetic energy and a small amount of potential energy and ends up with a lot of potential energy. (Ei = Ef)

I think that's how the question goes. It's been a while since i've seen that

Ikerous wrote:

Ms_Ikerous wrote:Why would you ask that? Once you get him started, he never stops. Once he get home he will have a blast explaining it, so be prepared.

Shuddup!

Lol

Ikerous wrote:

FF wrote:Pliz, explain this formula. (just tell us what does every letter mean).

I was answering the batman question in drago's "algebra book."

It says that theres a proton (Of mass m and charge Qp) that's a certain distance (Ri) from a charged line (With charge density λ). The proton starts out moving towards the charged line with an initial velocity (v) and ends up stopping at some distance (Rf) from the line. εo is just a constant in physics.

It's just a basic energy problem. The proton starts out with a lot of kinetic energy and a small amount of potential energy and ends up with a lot of potential energy. (Ei = Ef)

I think that's how the question goes. It's been a while since i've seen that

Thanks for that... =|

Ikerous wrote:It says that theres a proton (Of mass m and charge Qp) that's a certain distance (Ri) from a charged line (With charge density λ)........

Yes now I see. Checked. Correct .