like the thread title states.
Forum > FAQ's, Player Guides and Newbie Help > What is the highest possible effective level?
84 is the highest you'll see.
85 is technically the max, but that's unattainable.
85 is technically the max, but that's unattainable.
Originally posted by bigbean
why would it be technically the max but unattainable?
Game coder is a tease
why would it be technically the max but unattainable?
Game coder is a tease
Originally posted by bigbean
why would it be technically the max but unattainable?
Probably because 85 is the 100th percentile (which is impossible to be). You could in the 99.99th percentile but you would never reach 85.
why would it be technically the max but unattainable?
Probably because 85 is the 100th percentile (which is impossible to be). You could in the 99.99th percentile but you would never reach 85.
Edited by InRomoWeTrust on Oct 18, 2012 17:55:22
BagO'Chips
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\textstyle\frac{n}{0}=\infty
\frac{6}{3}=2 \frac{6}{0}=\,? ?\times 0=6. \frac{0}{0}=\,? ?\times 0=0.
\lim_{b \to 0^+} {a \over b} = +\infty \lim_{b \to 0^-} {a \over b} = -\infty \lim_{(a,b) \to (0,0)} {a \over b} \lim_{x \to 0} {f(x) \over g(x)} \mathbb{R}\cup\{\infty\}
There's a lot more, but you get my drift, right?
\frac{6}{3}=2 \frac{6}{0}=\,? ?\times 0=6. \frac{0}{0}=\,? ?\times 0=0.
\lim_{b \to 0^+} {a \over b} = +\infty \lim_{b \to 0^-} {a \over b} = -\infty \lim_{(a,b) \to (0,0)} {a \over b} \lim_{x \to 0} {f(x) \over g(x)} \mathbb{R}\cup\{\infty\}
There's a lot more, but you get my drift, right?
Originally posted by Jay89
I wouldnt say Unattainable, but simply "Unattained at this point in time"
wat
I wouldnt say Unattainable, but simply "Unattained at this point in time"
wat
Jay89
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I just dont understand how 85 is the highest there is, but your saying its "Impossible" to reach. All im simply implying is that nobody has found a way to reach 85, unless it truly has been made to be unreachable, in which case why waste the time to make it 85?
BagO'Chips
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Originally posted by Jay89
I just dont understand how 85 is the highest there is, but your saying its "Impossible" to reach. All im simply implying is that nobody has found a way to reach 85, unless it truly has been made to be unreachable, in which case why waste the time to make it 85?
Originally posted by BagO'Chips
\textstyle\frac{n}{0}=\infty
\frac{6}{3}=2 \frac{6}{0}=\,? ?\times 0=6. \frac{0}{0}=\,? ?\times 0=0.
\lim_{b \to 0^+} {a \over b} = +\infty \lim_{b \to 0^-} {a \over b} = -\infty \lim_{(a,b) \to (0,0)} {a \over b} \lim_{x \to 0} {f(x) \over g(x)} \mathbb{R}\cup\{\infty\}
1/0=\infty0/0\textstyle\frac{1}{x}2x = 2\textstyle\frac{a}{b}\frac{\lim\limits_{x \to 0} 1 }{\lim\limits_{x \to 0} x}
I just dont understand how 85 is the highest there is, but your saying its "Impossible" to reach. All im simply implying is that nobody has found a way to reach 85, unless it truly has been made to be unreachable, in which case why waste the time to make it 85?
Originally posted by BagO'Chips
\textstyle\frac{n}{0}=\infty
\frac{6}{3}=2 \frac{6}{0}=\,? ?\times 0=6. \frac{0}{0}=\,? ?\times 0=0.
\lim_{b \to 0^+} {a \over b} = +\infty \lim_{b \to 0^-} {a \over b} = -\infty \lim_{(a,b) \to (0,0)} {a \over b} \lim_{x \to 0} {f(x) \over g(x)} \mathbb{R}\cup\{\infty\}
1/0=\infty0/0\textstyle\frac{1}{x}2x = 2\textstyle\frac{a}{b}\frac{\lim\limits_{x \to 0} 1 }{\lim\limits_{x \to 0} x}
Originally posted by Jay89
I just dont understand how 85 is the highest there is, but your saying its "Impossible" to reach. All im simply implying is that nobody has found a way to reach 85, unless it truly has been made to be unreachable, in which case why waste the time to make it 85?
Well, an admin modified dot reached 85, but it was at some insane SP value. Nothing possible to actually achieve, regardless of how you build the dot.
Here's how I believe it hit 85. Being that 85 would be the 100th percentile, the only way to reach 85 would be to be in the 99.999999th percentile, so far out with the repeating 9s that the code rounds you to the 100th percentile (85).
It's just all how it was coded. 85 isn't realistically reachable.
I just dont understand how 85 is the highest there is, but your saying its "Impossible" to reach. All im simply implying is that nobody has found a way to reach 85, unless it truly has been made to be unreachable, in which case why waste the time to make it 85?
Well, an admin modified dot reached 85, but it was at some insane SP value. Nothing possible to actually achieve, regardless of how you build the dot.
Here's how I believe it hit 85. Being that 85 would be the 100th percentile, the only way to reach 85 would be to be in the 99.999999th percentile, so far out with the repeating 9s that the code rounds you to the 100th percentile (85).
It's just all how it was coded. 85 isn't realistically reachable.
Edited by InRomoWeTrust on Oct 18, 2012 18:32:48
jdbolick
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Yeah, no one thought 85 was even possible until that ridiculous dot with 9999 everything was created.
Jay89
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hmmm. Seems stupid that somebodys 1400 Sp value QB has the same Eff level as Qb's, with 1700+. Hell even 1900 is an achievable # for a QB (Without any aeq of course)
fogie55
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Originally posted by InRomoWeTrust
84 is the highest you'll see.
85 is technically the max, but that's unattainable.
with or without X Factor?
84 is the highest you'll see.
85 is technically the max, but that's unattainable.
with or without X Factor?
Originally posted by Jay89
hmmm. Seems stupid that somebodys 1400 Sp value QB has the same Eff level as Qb's, with 1700+. Hell even 1900 is an achievable # for a QB (Without any aeq of course)
Most of that disparity is caused by SP cost of attributes over 100, you can have 1 more point in Vision, Strength, and Throwing and be 90 SP ahead in value and essentially be the same dot.
hmmm. Seems stupid that somebodys 1400 Sp value QB has the same Eff level as Qb's, with 1700+. Hell even 1900 is an achievable # for a QB (Without any aeq of course)
Most of that disparity is caused by SP cost of attributes over 100, you can have 1 more point in Vision, Strength, and Throwing and be 90 SP ahead in value and essentially be the same dot.
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