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{3(TEC + 150)/5500 + EFF / 440}
Moderators: Duke, trewqh, korexus
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{3(TEC + 150)/5500 + EFF / 440}
/me happily joins the side-trackSaladin wrote:I was wondering what other languages they standard teach in the UK and the US besides English?
French, German, and Spanish can be found at almost all high schools and quite a few elementary/middle schools. My 7yo daughter, for example, took her first year of Spanish this past school year.Saladin wrote:This applies to just universities? As i was refering to basic highschools. Is that the same there?
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trewqh's version:
ARM * { (2 * LEV - 1) + (2 * DEF - 1) / 5}
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korexus's version: - Vortan don't you dare pick me up on the apostrophe, I know what I'm doing!
ARMt * { (2 * LEVa - 1) + (2 * DEF - 1) / 5}
It's not a required class ... but it was available through the school as an option (had to pay a fee and attend right before the normal school day begins), and she wanted to learn it.Saladin wrote:Cool, a second language at only 7. They don't start quite that young over here.
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[ARM * { (2 * LEV - 1) + (2 * DEF - 1) / 5}] * min[int_val{(TEC + 150)/50},7]
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[ARMt * { (2 * LEVa - 1) + (2 * DEF - 1) / 5}] * (TEC + 150)/50
Thank god i'm not Vortan.korexus wrote:Code: Select all
korexus's version: - Vortan don't you dare pick me up on the apostrophe, I know what I'm doing!
Wouldn't this always result in "7"? If TEC=0, then the first part of the equation will be 50 ... which is greater than 7. Perhaps you meant some other denominator other than 3?korexus wrote:min[int_val{(TEC + 150)/3},7]
Sorry, yes. The denominator should be 50. I was trying to pull out the number of attacks. I'll edit accordingly...Brykovian wrote:Wouldn't this always result in "7"? If TEC=0, then the first part of the equation will be 50 ... which is greater than 7. Perhaps you meant some other denominator other than 3?korexus wrote:min[int_val{(TEC + 150)/3},7]
-Bryk
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(H/40) * {30 * { (2 * LEVa - 1) + (2 * DEF - 1) / 5}] * (TEC + 150)/50
+ 30 * (S)
+ 90 * {A * (2 * LEV - 1) + (2 * DEF - 1)/5 } /50
+ (2/5) * ARM * (2 * LEV - 1) * (TEC + 150) / 50}
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POP + 2 * WOK + 4 * [ARM * { (2 * LEV - 1) + (2 * DEF - 1) / 5}] * min[int_val{(TEC + 150)/50},7] + (missiles)
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POP + 2 * WOK + 4 * [ARM * { (2 * LEV - 1) + (2 * DEF - 1) / 5}] * (TEC + 150)/50 + (missiles)
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{ 5H / (200 - 2H) } * { I + 15 + (DEF / 10) +
30 * { (2 * LEVa - 1) + (2 * DEF - 1) / 5}] * (TEC + 150)/50 +
90 * {A * (2 * LEV - 1) + (2 * DEF - 1)/5 } /50
(2/5) * ARM * (2 * LEV - 1) * (TEC + 150) / 50}
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(1/5) * {(2 * M) + [{(2/5) * ARM * (2 * LEV - 1) * (TEC + 150) / 50} / 10] + 15 + I}
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POP + 2 * WOK + (4 * [ARM * { (2 * LEV - 1) + (2 * DEF - 1) / 5}] * min[int_val{(TEC + 150)/50},7] + M + S) * EFF modifier
where
S = (1/5) * {(2 * M) + [{(2/5) * ARM * (2 * LEV - 1) * (TEC + 150) / 50} / 10] + 15 + I}
and
M = { 5H / (200 - 2H) } * { I + 15 + (DEF / 10) +
30 * { (2 * LEVa - 1) + (2 * DEF - 1) / 5}] * (TEC + 150)/50 +
90 * {A * (2 * LEV - 1) + (2 * DEF - 1)/5 } /50
(2/5) * ARM * (2 * LEV - 1) * (TEC + 150) / 50}
for I and H our information and Hitrate constants.
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(POP + 2 * WOK + (4 * [ARM * { (2 * LEV - 1) + (2 * DEF - 1) / 5}] * TEC + 150)/50 + M + S) * EFF modifier
where
S = (1/5) * {(2 * M) + [{(2/5) * ARM * (2 * LEV - 1) * (TEC + 150) / 50} / 10] + 15 + I}
and
M = { 5H / (200 - 2H) } * { I + 15 + (DEF / 10) +
30 * { (2 * LEVa - 1) + (2 * DEF - 1) / 5}] * (TEC + 150)/50 +
90 * {A * (2 * LEV - 1) + (2 * DEF - 1)/5 } /50
(2/5) * ARM * (2 * LEV - 1) * (TEC + 150) / 50}
for I and H our information and Hitrate constants.
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POP + 2 * WOK + (4 * [ARM * { (2 * LEV - 1) + (2 * DEF - 1) / 5}] * min[int_val{(TEC + 150)/50},7] + M + S) * EFF / 100
where
S = (1/5) * {(2 * M) + [{(2/5) * ARM * (2 * LEV - 1) * (TEC + 150) / 50} / 10] + 15 + I}
and
M = { 5H / (200 - 2H) } * { I + 15 + (DEF / 10) +
30 * { (2 * LEVa - 1) + (2 * DEF - 1) / 5}] * (TEC + 150)/50 +
90 * {A * (2 * LEV - 1) + (2 * DEF - 1)/5 } /50
(2/5) * ARM * (2 * LEV - 1) * (TEC + 150) / 50}
for I and H our information and Hitrate constants.
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Code:
(POP + 2 * WOK + (4 * [ARM * { (2 * LEV - 1) + (2 * DEF - 1) / 5}] * TEC + 150)/50 + M + S) * ({3(TEC + 150)/2750 + EFF / 220}) + (1/5) * [TEC * { (5M + 2S) / 150 + 26 * {ARM * (2 * LEV - 1) * (TEC + 150) } / 175000 + 9 * { ARM * (2 * DEF - 1) * (TEC + 150) } / 175000]
where
S = (1/5) * {(2 * M) + [{(2/5) * ARM * (2 * LEV - 1) * (TEC + 150) / 50} / 10] + 15 + I}
and
M = { 5H / (200 - 2H) } * { I + 15 + (DEF / 10) +
30 * { (2 * LEVa - 1) + (2 * DEF - 1) / 5}] * (TEC + 150)/50 +
90 * {A * (2 * LEV - 1) + (2 * DEF - 1)/5 } /50
(2/5) * ARM * (2 * LEV - 1) * (TEC + 150) / 50}
for I and H our information and Hitrate constants.