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Rocky's CD factory (mega explosion tube)

Started by RockyRaccoon, April 08, 2008, 06:55:02 PM

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RockyRaccoon

NNID: RockyRogue<br />3DS: 4699 8033 5044

Mega (Tre)

Quote from: Karoku (2100)© on November 12, 2008, 06:11:29 PM
Sorry, I had trouble explaining it.  Let me show you what I mean:



You'd have to reverse the escalators though.
What do you like for breakfast?

RockyRaccoon

July 06, 2009, 10:37:35 PM #47 Last Edit: July 06, 2009, 10:40:07 PM by RockyRaccoon
Quote from: Karoku (2100)© on November 13, 2008, 02:54:18 PM
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% .
NNID: RockyRogue<br />3DS: 4699 8033 5044

KarneraMythos

Quote from: RockyRaccoon on July 06, 2009, 10:37:35 PM
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.

Smashpower 10th Anniversary Anthem: https://www.youtube.com/watch?v=rmnZ0R8quWM
WE HAVE RETURNED!

RockyRaccoon

July 07, 2009, 05:02:33 AM #49 Last Edit: July 07, 2009, 05:04:50 AM by 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 .
NNID: RockyRogue<br />3DS: 4699 8033 5044

Biggoron

Quote from: RockyRaccoon on July 06, 2009, 10:37:35 PM
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. ;)

NinjaGoron

Quote from: RockyRaccoon on July 06, 2009, 10:37:35 PM
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

NNID: RockyRogue<br />3DS: 4699 8033 5044

Mega (Tre)

Quote from: RockyRaccoon on July 06, 2009, 10:37:35 PM
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.
What do you like for breakfast?

NinjaGoron

Quote from: Mega *just got owned* on July 08, 2009, 09:06:51 PM
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 (Tre)

gasp i made a better one and used it for battle!
What do you like for breakfast?

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