Official AA Thread

Discuss anything not related to Armadillo Run here.

Moderator: Moderators

Official AA Thread

Postby JustcallmeDrago » Tue Sep 11, 2007 1:09 pm

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.


Image<- My algebra textbook

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

Thank you.
User avatar
JustcallmeDrago
Contest Veteran
Contest Veteran
 
Posts: 904
Joined: Sun Mar 25, 2007 6:19 pm
Location: Earth

Postby McGinge » Tue Sep 11, 2007 5:16 pm

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...
In an ideal world, where anything and everything is in perfect form, is there a perfect form of imperfection?

Work THAT one out! :)
User avatar
McGinge
Contest Legend
Contest Legend
 
Posts: 936
Joined: Mon May 28, 2007 2:49 pm
Location: Sevenoaks, England

Postby Ikerous » Tue Sep 11, 2007 6:01 pm

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.
User avatar
Ikerous
Light Master
 
Posts: 977
Joined: Mon Dec 04, 2006 8:02 am
Location: California

Postby NickyNick » Tue Sep 11, 2007 7:10 pm

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!
AR & W:A forever
User avatar
NickyNick
Champion of Light
Champion of Light
 
Posts: 1330
Joined: Sat Dec 30, 2006 10:43 pm
Location: Ukraine - Kharkov

Postby FF » Tue Sep 11, 2007 7:36 pm

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

There is your logic, guys? :roll: :wink:

//Due to your logic, guys, I'm half-addicted ^_^
User avatar
FF
Light Master
 
Posts: 405
Joined: Thu Apr 05, 2007 4:33 pm
Location: Russia

Postby JustcallmeDrago » Tue Sep 11, 2007 8:40 pm

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

There is your logic, guys? :roll: :wink:

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


Maybe:

Addiction = 1/Rank
User avatar
JustcallmeDrago
Contest Veteran
Contest Veteran
 
Posts: 904
Joined: Sun Mar 25, 2007 6:19 pm
Location: Earth

Postby NeXFerret » Tue Sep 11, 2007 8:45 pm

So where would that put people who have never played the game? Armadillo Uber Killer OMFGer Light Force Winner King?
The bakeries begin to amass millions as a new product emerges; Armadillo Buns!
NeXFerret
 
Posts: 60
Joined: Fri Jan 19, 2007 8:02 pm

Postby JustcallmeDrago » Tue Sep 11, 2007 8:51 pm

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}
User avatar
JustcallmeDrago
Contest Veteran
Contest Veteran
 
Posts: 904
Joined: Sun Mar 25, 2007 6:19 pm
Location: Earth

Postby FF » Tue Sep 11, 2007 9:38 pm

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).
User avatar
FF
Light Master
 
Posts: 405
Joined: Thu Apr 05, 2007 4:33 pm
Location: Russia

Postby Ms_Ikerous » Tue Sep 11, 2007 9:41 pm

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.
You may be thinking...Who would marry Ikerous? I did!
User avatar
Ms_Ikerous
Contest Winner
Contest Winner
 
Posts: 55
Joined: Sun Aug 12, 2007 6:24 am
Location: Sitting Next To Ikerous

Postby JustcallmeDrago » Tue Sep 11, 2007 10:01 pm

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


But go ahead, Ike! Explain away! :P
User avatar
JustcallmeDrago
Contest Veteran
Contest Veteran
 
Posts: 904
Joined: Sun Mar 25, 2007 6:19 pm
Location: Earth

Postby Ikerous » Tue Sep 11, 2007 10:50 pm

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
User avatar
Ikerous
Light Master
 
Posts: 977
Joined: Mon Dec 04, 2006 8:02 am
Location: California

Postby JustcallmeDrago » Tue Sep 11, 2007 10:58 pm

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... =|
User avatar
JustcallmeDrago
Contest Veteran
Contest Veteran
 
Posts: 904
Joined: Sun Mar 25, 2007 6:19 pm
Location: Earth

Postby FF » Wed Sep 12, 2007 12:33 pm

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 ;).
User avatar
FF
Light Master
 
Posts: 405
Joined: Thu Apr 05, 2007 4:33 pm
Location: Russia


Return to Off Topic

Who is online

Users browsing this forum: No registered users and 7 guests

cron