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In lugnet.org.us.nelug, Eric Joslin writes:
> > The old formula for throwing range, ((5xPower)/Mass)+Skill" is both too
> > complicated and generates ranges that are much too long. Instead, the new
> > formula will be updated to (Skill x Power / Mass)".
>
> Ok, already I have a fairly universal question about this formula.
>
> The standard trooper has 1d6 skill, and 1 power. We'll assume he's trying to
> throw something with a mass of 1, as well.
>
> So, we look at the formula, and we get 1d6 x 1 / 1 = 1d6". I assume this
> means that the trooper decides to throw a grenade, and you roll to see how
> far he can throw it. You roll a 4, he can throw it up to 4". You roll a 1,
> and he fumbles and drops it at his feet, and has a few seconds to regret not
> having married Betty when he had a chance.
That's why in real life you try to throw grenades through, around, or over
obstacles that you can then hide behind. And also I think that bad things
should happen to any unit who rolls ones.
> But, now let's assume the crazy bastard is trying to throw something with a
> mass of 2! You look at the formula, and you get: 1d6 x 1 / 2 = 1d6 / 2.
> What does this mean in BrikWars terms?
>
> Does it mean that you roll the D6 and divide by two? That seems reasonable.
> So, you roll a 3, divide by two and get 1.5, and he throws it 1.5 inches.
That's about the size of it. You could also try taking a jigsaw to your die
and cutting it in half so that you could roll any .5d6's that crop up.
However your no-jigsaw solution is probably the more elegant one.
> What if he has a skill of, say, 1d10+4, and is throwing a mass of 2? Do you
> roll a d10, divide that by 2, and add 2?
Sounds good so far.
> What if you have a skill of 1d6, and a power of 3, and you're throwing
> something with a mass of 2? Do you roll 3d6, add up the total, and divide it?
You could divide 3d6 by two ((SkillxPower)/Mass), or you could multiply 1d6
by 1.5 (Skillx(Power/Mass)), whichever you prefer. The second method is
more likely to hit the maximum or minimum numbers than the first method, but
the two methods are otherwise equivalent.
> I know these questions seem anal, I just like to know these kinds of things.
Regardless of what order in which you do the adding and multiplying and so
forth, any method you might choose will be algebraically identical. The
shapes of the probability curves vary slightly, but not enough so that I'd
want to force players to conform to one method as opposed to another. So,
just do whatever seems natural and everything should go smoothly.
- Mike Rayhawk.
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