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Unobtainable Viable Weapons

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NovaRain

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Post Fri Jun 29, 2018 2:25 am

Re: Unobtainable Viable Weapons

Muttie wrote:So simply multiplying with RoF may not be that accurate really. Nor seems reducing the RoF based on the hit chance. (?)
Still no idea how to figure out the total damage of a burst, to be honest.

There's a post by Atom on NMA explained how burst attack works in the engine: http://www.nma-fallout.com/threads/crit ... st-3645850
Dividing RoF by 3 and rounding down should be a more reasonable mean value for a burst attack, unless the attack is done in point-blank range. You need to use some hook script if you want to get the exact number of bullets that hit the target.
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Muttie

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Post Mon Jul 02, 2018 11:21 pm

Re: Unobtainable Viable Weapons

Thanks, for finding that old post. And thanks for the rule of thumb.

http://www.nma-fallout.com/threads/crit ... st-3645850
Atom wrote: All the bullets are divided into three groups (left, right, central, that will contain about 1/3, 1/3, and 1/6+n bullets respectively, where n are leftover bullets from the initial 1/6 volley that attacked the primary target).
These three groups will then travel along three "parallel" tracks (but hex adjacent to the attacker is common for all of them), trying to hit targets on the way. And skill does matter - determine_to_hit is called every time the valid target is found. Basically, when the group of bullets finds the target, the bullets try to hit it one by one until one of them fails. Then they move on to the next target.

The above was confirmed in-game. I used minigun throwing 1000 bullets at a time (I set ToHit to constant 95%), about 180 hit the primary target, the rest were hit by around 20 - which is the expected value, 1/(1-0.95). When I changed ToHit to constant 100%, the primary target was hit by 334, and the two at it's sides for 333.

I understand the example as following:
Of the 1000 bullets 1/6 will hit the target. That's 166,7 bullets. Of those 95% will hit, that's 158,3.
Then there are three groups of 1/3 (333), 1/3 (333) and 1/6 (166,7) bullets that will hit a target on their fly path until a bullet misses.
With 95%. (1/00,5=20). That's twenty hits before a miss.
So the targets to each side are hit by 20. And the primary target by 158,3+20=178,8 bullets.

And to sum it up:

Burst calculation:
1/6 bullets of the Round of Fire will hit the primary target and make hit rolls.
The rest of the bullets (RoF) are split into three groups: 1/3 of the bullets go to the right, 1/3 to the left and the remaining 1/6 stay central (probably going towards the primary target). Those three groups will check for hits until a bullet misses. When that happens the remaining bullets of that group will not make hit rolls, but travel past the target and try to hit targets behind it (same procedure).
So if I burst an opponent from point blank range 1/6 of the RoF will make a hit roll, while 5/6 of the bullets will also do so until one bullet misses, then the remaining bullets of that group (Note: those 5/6 bullets “may” still be split in three groups of 1/3, 1/6 and 1/3?) will travel past the point blank target to hit targets behind it (if any).

Is this correct? I think it is. The only thing I'm not clear about are the “leftovers”
Atom wrote:and 1/6+n bullets respectively, where n are leftover bullets from the initial 1/6 volley that attacked the primary target

Are those leftovers all bullets after the first, or any that missed, or fractions?

However, if this is correct and confirmed it actually could be added to: http://fallout.wikia.com/wiki/Burst_Shot
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