SourceOp TF2 Tourny (for server regulars)

[FrostDragonfly]

hi, im [Cow-Lie]FrosT of the 24/7 goldrush server

and i thought it'd be cool if we could organize a thing where the regulars from each sourceop server play against each other

i brought it up amongst some of the regulars of the goldrush server and there is definitely some interest.

if u guys think its a good idea, maybe we should spread the word and get it rollin.

[Half-Duplex]

Hrm, I'd be down...

...friendly little competition is always fun. But what about those who play on multiple SourceOP servers? Who will they identify themselves with?

[FrostDragonfly]

yea u bring up a good point.

but im sure we wont be THAT strict, just pick a group u wanna play with.

[Half-Duplex]

so in other words...the way it already is now?

There needs to be some sort of team tree..or something...

...but if we even decide to go that far, might as well join a league and compete there...right?

Its a good idea, and I'd be down if you could get it all together. But I see fail in the future.

I could be wrong. Sorry to be a negative-nancy.

[Drunken F00l]

I like the idea and if there was enough interest I would help with it.

[FrostDragonfly]

first things first though, we need a quick and easy way to spread the word.

i suggest using a MoTD to get the word out, or even have an op go around on each server asking around

maybe have the motd telling them to join the sourceop group and post a reply of nick + server u play on to start getting rosters up

after that we can set a time/server for these to take place.

[Adder]

I think we all know what server would win.

[Hell-met]

I think we all know what server would win.

I concur.

#1 is populated with team-stacking faggots, hence we cannot be beaten.

[Zeitgeist]

I think we all know what server would win.

I concur.

#1 is populated with team-stacking faggots, hence we cannot be beaten.

+1


also, i am interested in fighting for the team-stacking faggots! this sounds awesome!

[avy]

Do old regulars count? Or ringers? d0=