Talk:Nuclear Breath

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Uhh, this does do something now, by the way. It seems to be a variable amount of hot damage based on how long you waited to use it (that is to say, the fewer turns of Taste the Inferno you have left, the more damage it does). Using it removes the effect, so you only get one shot. Sadly I forgot to save the usage text from yesterday. --Tcoop 21:24, 4 May 2011 (UTC)

50 turns remaining: 5,5,5,3,6,5,6,9,2,6,6,2,7,2,2,6,7,2,4,2,2,2,5,2,6,5,
 3,4,5,11,10,8,3,6,5,4,7,4,3,4,2,6,2,5,2,2,2,7,3,7,9,7,7,4,3,5,8,6
49 turns remaining: 4, 11
48 turns remaining: 9 
47 turns remaining: 9 
46 turns remaining: 5 
45 turns remaining: 11,10,3,2
40 turns remaining: 10, 9 
35 turns remaining: 9 
30 turns remaining: 13
10 turns remaining: 82, 79, 90,88, 90, 64
5 turns remaining: 351, 213
2 turns remaining: 1250, 1435,1538, 1425,1538
1 turn remaining: 5000, 5000

Appears to be wholly useless till the <10 turns range, still not sure on formula, hope the info helps tho. --Boozeogre 00:51, 9 May 2011 (UTC)

Removed pagebreaking length. --Raijinili 19:45, 9 May 2011 (UTC)

I added some additional data points to the table above. --Starwed 22:34, 9 May 2011 (UTC)

I suspect it is something like 5000/N1.75 -- 5000/N2 damage, when N turns are left. --Starwed 20:26, 10 May 2011 (UTC)

Ok, took a lot of data at N=50. It's noticeably skewed towards the lower end:

2    12
3    6 
4    6
5    10
6    8
7    7
8    2
9    2
10   1
11   1

This puzzled me at first, since there doesn't seem to be such an obvious skew towards the lower end at smaller N. But it could be explained if the number is generated by rolling the exponent X first, and then calculating the damage as floor(1/N^x). Look at this graph to see what I mean. There would still be some skew at low N, but it'll be less than what you see at high N. --Starwed 20:44, 24 July 2011 (CEST)