Crypt Posted April 16, 2009 Report Share Posted April 16, 2009 a page (which uses the logarithmic scale of STR) for your own usage. Do what you want with that... http://cryptmaster.free.fr/HERO/exp/ Code: http://cryptmaster.free.fr/HERO/exp/index.txt For instance STRx minus 11.61 = 20 % STRx Thus STR 188 is 20% of STR 200. STR -38 is 10% of STR-21 etc. 5% ====>-21.61 10% ====>-16.61 15% ====>-13.68 20% ====>-11.61 25% ====>-10 30% ====>-8.68 35% ====>-7.57 40% ====>-6.61 45% ====>-5.76 50% ====>-5 55% ====>-4.31 60% ====>-3.68 65% ====>-3.11 70% ====>-2.57 75% ====>-2.08 80% ====>-1.61 85% ====>-1.17 90% ====>-0.76 95% ====>-0.37 100% ====>0 A possible usage: for instance as you know the lesser the STR of a character the less tiring is using x% of his STR. It works with high STR but may seems unfair for low ones. For instance imagine you want to GM a gnomic campaign (eg 1/8th size and STR-5 on the average.) With the standard rules a gnome using 100% of his STR-5 to lift 12.5 kg won't lose END. The previous table may be used instead. For instance you may say that if the character has a STR below 15 he lose END this way: 25% to 45% STR (which is STR minus 10 to STR minus 6)= 1 END 50% to 70% STR (which is STR minus 5 to STR minus 3) = 2 END 75% to 100% STR (which is STR minus 2 to STR) = 3 END For instance a STR-8 gnome would use 25% to 45% STR when using STR-18 (2kg) to STR -14 (3.7kg) = 1 END 50% to 70% STR when using STR-13 (4kg) to STR -11 (5kg) = 2 END 75% to 100% STR when using STR-10 (6kg) to STR -8 (8kg) = 3 END this is just a 2c idea for tiny creatures EDIT: see also the encumbrance thread: http://www.herogames.com/forums/showthread.php?p=1802876#post1802876 Quote Link to comment Share on other sites More sharing options...
Sean Waters Posted April 16, 2009 Report Share Posted April 16, 2009 Re: Proportion of used STR a page (which uses the logarithmic scale of STR) for your own usage. Do what you want with that... http://cryptmaster.free.fr/HERO/exp/ ...................... Nice - thanks. Thinking about it it is daft that someone who can lift 25 kilotons with their 100 STR at a cost of 10 END has to pay 1 END to lift 100kg. They probably wouldn't even know if they lifted 100kg. Quote Link to comment Share on other sites More sharing options...
Crypt Posted April 16, 2009 Author Report Share Posted April 16, 2009 Re: Proportion of used STR Nice - thanks. Thinking about it it is daft that someone who can lift 25 kilotons with their 100 STR at a cost of 10 END has to pay 1 END to lift 100kg. They probably wouldn't even know if they lifted 100kg. i think this kind of problem will never be solved unless HERO become 100% logarithmic. But of course as a 100STR and 100CON character would have an amazing 40REC and 200END this is not a big problem. off topic PS: and IMVHO the Grail of RPG systems should use a decibel scale....... Quote Link to comment Share on other sites More sharing options...
Crypt Posted April 22, 2009 Author Report Share Posted April 22, 2009 Re: % str Here is a note about how to add or substract STR (or any stat on the same scale *) without making any conversion to his value. Addition: X+y where X is the biggest number. Substract y from X then see the following chart: 18+ ====> X 17 to 10 ====> X+1 9 to 7 ====> X+2 6 to 4 ====> X+3 3 to 2 ====> X+4 1 to 0 ====> X+5 For instance you want to add STR 15 and STR 13. 15-13 = 2 so the result is X+4 => STR 19 Another one: you want to add STR 100 and STR 20. 100 -20 = 80 so the result is X => STR 100. Bob (STR 8) helps Tom (STR 12) to lift a MASS 14 item. 12-8= 4 ===> X+3 = 12+3= STR 15 so they succeed. Substraction: X-y where X is the biggest number. Substract y from X then see the following chart: 21+ ====> X 20 to 13 ====> X-1 12 to 9 ====> X-2 8 to 7 ====> X-3 6 ====> X-4 5 ====> X-5 4 ====> X-6 3 ====> X-7 2 ====> X-10 1 ====> X-15 0 ====> -INF (= impossible in a logarithmic scale) For instance STR 15 - STR 13 15-13 = 2 so the result is X-10 ===> STR 5. * the scale is as following: value = (⁵√2)ʳ * 25 where r is the Stat. so Stat = the logarithm (in base ⁵√2) of v/25 Quote Link to comment Share on other sites More sharing options...
Alibear Posted April 22, 2009 Report Share Posted April 22, 2009 Re: Proportion of used STR Nice - thanks. Thinking about it it is daft that someone who can lift 25 kilotons with their 100 STR at a cost of 10 END has to pay 1 END to lift 100kg. They probably wouldn't even know if they lifted 100kg. Why not rule that you don't pay end until you exceed your casual strength? Quote Link to comment Share on other sites More sharing options...
Crypt Posted April 22, 2009 Author Report Share Posted April 22, 2009 Re: Proportion of used STR Why not rule that you don't pay end until you exceed your casual strength? IMHO there is no need to (for instance because of high REC.) This is more a question about the HERO philosophy which sometimes makes linear manipulations on logarithmic scales. A pure logarithmic system would not work this way. Quote Link to comment Share on other sites More sharing options...
Alibear Posted April 22, 2009 Report Share Posted April 22, 2009 Re: % str If 6e seperates Str from REC? I hope it does, I hate figured chars. Quote Link to comment Share on other sites More sharing options...
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