Re: OUDF - Railguns??

Sorry for not replying to this before, but I've been kinda busy:

Matthew Smith wrote:

>In order to allow me more accuracy in creating a replacement, do you
>think that you could tell me the formulae that you used to work out
the >damage values above?

Basic combinatorics. Since I don't know how much maths you know (beyond
the "really no good" you wrote :-/ ) I'll describe it fairly
thoroughly. The damage values were:

Range:	R6:	B4 (vs level-0/1/2 screens)
0-12	2.91	3.2/2.53/1.87
12-24	2.33	2.4/1.9/1.4
24-36	1.75	1.6/1.27/0.93
36-48	1.17	0.8/0.63/0.47
48-60	0.58	0/0/0

The average damage of an R6 hit is the average result of 1d6, which is
(1+2+3+4+5+6)/6 = 3.5. If you multiply the average damage per hit with
the probability that the shot hits in the first place you get the
average damage at that range. Hitting on 2+ means that you hit on a 2,
3, 4, 5 or 6, ie. 5 times out of six, so the hit probability is 5/6;
similarly hitting on 3+ gives you a 4/6 hit probability, etc. This
gives you the R6 column above.

The average damage of one beam die against an unscreened target
*excluding the re-rolls* is (0+0+0+1+1+2)/6 = 2/3. The re-rolls (which
are not affected by screens) add on average
2/3*(1/6+(1/6)^2+(1/6)^3+...+(1/6)^infinity) = 2/3*1/5 = 2/15, so the
total average damage of a single beam die is 2/3+2/15 = 4/5 = 0.8.
Multiplying 0.8 by the number of beam dice the weapon fires in each
range band gives you the "0 screens" column for the B4.

Only the *initial* die is affected by screens (the re-rolls are not),
so against a level-1 screen one beam die inflicts on average
(0+0+0+0+1+2)/6+2/15 = 1/2 + 2/15 = 19/30 = 0.6333.... and against
level-2 screens it inflicts on average (0+0+0+0+1+1)/6 + 2/15 =
0.46666.... Multiplying these values with the number of dice fired in
each range bands gives you the "1" and "2" columns for the B4 above.

Regards,

Oerjan Ohlson
oerjan.ohlson@telia.com

"Life is like a sewer.
  What you get out of it, depends on what you put into it."
- Hen3ry