Rocky's CD factory (mega explosion tube)

[RockyRaccoon]

good .

[Mega]

Sorry, I had trouble explaining it.  Let me show you what I mean:

You'd have to reverse the escalators though.

[RockyRaccoon]

Hehe, I was thinking the same over lunch.  Just because the stage is expanded in anyway to "increase" chances, is only equivalent to splitting a dollar into two 50 cent coins.  The fraction of the result will equal the same.

that's not true . the number of items rendered is proprotional to the concentration of sandbags x the concentration of attacks .

I = c[X][A]
I = items , c = constant , X = sandbags , A = attacks , [] = concentration

if you split the sandbags into 2 groups , you halve their concentration . if you split the players into 2 groups , you halve the concentration of attacks .

I' = c[X]'[A]' = c([X]/2)([A]/2) = 1/4 c[X][A] = I/4

by doubling the size of the stage , you cut the yield down to 25% .

[Karoku (2100)©]

that's not true . the number of items rendered is proprotional to the concentration of sandbags x the concentration of attacks .

I = c[X][A]
I = items , c = constant , X = sandbags , A = attacks , [] = concentration

if you split the sandbags into 2 groups , you halve their concentration . if you split the players into 2 groups , you halve the concentration of attacks .

I' = c[X]'[A]' = c([X]/2)([A]/2) = 1/4 c[X][A] = I/4

by doubling the size of the stage , you cut the yield down to 25% .

What about the amount of players?  You would hereby expand the rate in that sense? 

I said that previous statement because of the speed of the conveyor belts, slowly bringing the sand bags as they fall down into each one.  Does movement from belt-to-belt stall their vanishing time, you know, until they disappear one-by-one?  If the laws of physics do slow down the sandbags, then the size of the factory itself doesn't matter after all.

[RockyRaccoon]

if you add more players , you increase the concentration of attacks if and only if the players stand in the same place . if they stand in different places , the concentration of attacks remains the same while the larger stage size decreases concentration of sangbags , since the total number of sandbags are constant .

so if you double the stage size and add a 2nd attacking player , the overall yield will be

a) unchanged if the two stand together
b) reduced by 50% if they stand apart

since the best you can possibly achieve is an unchanged yield , there is nothing to gain by duplicating the stage once over .

this is also an approximation that is more generous toward your position . taking a more accurate calculation , you will see that the number of sandbags in the air or on their way back down will increase as the sandbags are seperated , resulting in an even lower concentration of attacks .

[Biggoron]

that's not true . the number of items rendered is proprotional to the concentration of sandbags x the concentration of attacks .

I = c[X][A]
I = items , c = constant , X = sandbags , A = attacks , [] = concentration

if you split the sandbags into 2 groups , you halve their concentration . if you split the players into 2 groups , you halve the concentration of attacks .

I' = c[X]'[A]' = c([X]/2)([A]/2) = 1/4 c[X][A] = I/4

by doubling the size of the stage , you cut the yield down to 25% .

You'll never see Brawl Central or another Brawl site give you information about the game like this.

[<Ninja Goron]

that's not true . the number of items rendered is proprotional to the concentration of sandbags x the concentration of attacks .

I = c[X][A]
I = items , c = constant , X = sandbags , A = attacks , [] = concentration

if you split the sandbags into 2 groups , you halve their concentration . if you split the players into 2 groups , you halve the concentration of attacks .

I' = c[X]'[A]' = c([X]/2)([A]/2) = 1/4 c[X][A] = I/4

by doubling the size of the stage , you cut the yield down to 25% .

...what the crap?

please say this took like 5 or 6 minutes to figure out and wasn't like a week long project or something

[RockyRaccoon]

it's common sense .

[Mega]

that's not true . the number of items rendered is proprotional to the concentration of sandbags x the concentration of attacks .

I = c[X][A]
I = items , c = constant , X = sandbags , A = attacks , [] = concentration

if you split the sandbags into 2 groups , you halve their concentration . if you split the players into 2 groups , you halve the concentration of attacks .

I' = c[X]'[A]' = c([X]/2)([A]/2) = 1/4 c[X][A] = I/4

by doubling the size of the stage , you cut the yield down to 25% .

lol, only Rocky would try to confuse a majority of the American population by putting a simple problem into a simple formula.

[<Ninja Goron]

lol, only Rocky would try to confuse a majority of the American population by putting a simple problem into a simple formula.

hey its summer. i dont want to learn.

[Mega]

gasp i made a better one and used it for battle!