- 1,196
- 382
These 2 formulas here, the one for mid-air and ground explosions
-
This forum is strictly intended to be used by members of the VS Battles wiki. Please only register if you have an autoconfirmed account there, as otherwise your registration will be rejected. If you have already registered once, do not do so again, and contact Antvasima if you encounter any problems.
For instructions regarding the exact procedure to sign up to this forum, please click here. -
We need Patreon donations for this forum to have all of its running costs financially secured.
Community members who help us out will receive badges that give them several different benefits, including the removal of all advertisements in this forum, but donations from non-members are also extremely appreciated.
Please click here for further information, or here to directly visit our Patreon donations page. -
Please click here for information about a large petition to help children in need.
Where are the explosion formulas from?
- Thread starter Hypertornado099
- Start date
- Status
- 4,115
- 2,173
The first formula came from the Outskirts Battledome, which originally sourced it from this: https://nuclearweaponarchive.org/Nwfaq/Nfaq5.html
I think it's valid for nuclear explosions regardless because, well, duh! That's what it's made for.
The second one, I have no idea. I did a Google search on it and only came to the conclusion that some dumbo on the wiki made it up.
- 1,196
- 382
- Thread starter
- #3
I looked at the site and couldn't find the formula or anything mentioning it at all, just the number 0.28
- 32,747
- 37,900
The first one came from reverse-searching the code in this site.
The second formula originated from an EU explosives pdf. @DontTalkDT knows the link, but he keeps forgetting to add it as reference like I asked him to.
EDIT: Found the ground-based explosion formula.
Here and here
The second formula originated from an EU explosives pdf. @DontTalkDT knows the link, but he keeps forgetting to add it as reference like I asked him to.
EDIT: Found the ground-based explosion formula.
Here and here
- 1,196
- 382
- Thread starter
- #5
Thanks but it seems like this wasn't the blog for the current formula W = R^3*((27136*P+8649)^(1/2)/13568-93/13568)^2, so the rest of the math must be from somewhere else
- 32,747
- 37,900
That formula is from both those links. @DontTalkDT can elaborate more about it.
- 10,499
- 11,547
It's the result of taking the formula in the document and solving for W
- 1,196
- 382
- Thread starter
- #8
Since you found the link, should it be added as reference?
- 32,747
- 37,900
Already asked DontTalkDT before to do so.
- 10,499
- 11,547
Done.
- 1,196
- 382
- Thread starter
- #11
So the first formula ignores thermal radiation and fireball radius it seems? The site has those but the formula only mentions air blast radius
- 32,747
- 37,900
As our Explosion Yield Calculations page states,
- When you get the result of said explosion, you multiply the value by 0.5. This is because only 40 to 50 percent of the total energy of the explosion is actually from the blast.
- Keep in mind that if the explosion is an actual nuclear explosion, you can disregard the point above.
- 4,716
- 4,140
Question
It says above that you multiply the value of non-nuclear explosives by .5 because the blast is only 50% of the total energy and that nukes don't have to abide by this rule. However in the link provided it's explicitly referring to nukes themselves as only having 50% of the energy being from the blast.
Do normal explosives blasts only make up 25% of the total energy in conventional explosives or is their something I'm not getting?
- 32,747
- 37,900
I honestly don't know either, and the rule itself is exclusively for airburst explosions.
Probably @DontTalkDT or @Executor_N0 could answer that.
- 10,499
- 11,547
You want to calculate the original total energy released by the weapon in any form, including heat and radiation. For a magical explosion you don't know how much more energy the explosive had that just didn't go into the explosion, so you can only say that the total energy is to some unqantified amount greater than the explosion energy. For a nuke, you know that the amount is specifically 50% with the rest going into heat and radiation.
- 4,716
- 4,140
That's not really what I'm talking about
Currently, if I'm interpreting the explosion yield page correctly, we scale nukes to 100% of the found energy instead of 50% which is what's listed as the energy of the blast itself.
And for regular or fantasy explosions we just assume 50% of the energy is from the blast itself IE we treat it as being similar to a nuclear explosions values
- 10,499
- 11,547
I... don't really understand your problem?
Like, the explosion formula we use gives you 100% of the energy release of a nuke. Not just the energy of the explosion, but also the heat and radiation stuff included.
For fictional explosions, we only know that the explosion created is similar to that of a nuke. Not that the heat and radiation release is the same. So what we do is calculate the fictional explosion as if it were a nuke first, by that including the heat and radiation of a nuke which the fictional explosion may not include, and then take 50% of that value, to subtract the heat and radiation we do not assume the fictional explosion produces.
Like, the explosion formula we use gives you 100% of the energy release of a nuke. Not just the energy of the explosion, but also the heat and radiation stuff included.
For fictional explosions, we only know that the explosion created is similar to that of a nuke. Not that the heat and radiation release is the same. So what we do is calculate the fictional explosion as if it were a nuke first, by that including the heat and radiation of a nuke which the fictional explosion may not include, and then take 50% of that value, to subtract the heat and radiation we do not assume the fictional explosion produces.
- 866
- 564
If I want to calculate the energy output of a nuclear explosion at ground level, should I multiply the value of the ground explosion formula by 2?
- 10,499
- 11,547
No, as the ground explosion formula makes no such correction to begin with.
- 866
- 564
That is, they do not take into account other energies. To take them into account, shouldn't I multiply x2?
- 10,499
- 11,547
No, because we never subtracted other energies from it to begin with.
- 866
- 564
The Explosions page states that if an airburst explosion is non-nuclear, the result should be divided by 2. This implies that the formula is typically intended for nuclear explosions.
It also mentions that for groundburst explosions, there is no need to divide by 2, indicating that the formula is already applicable to non-nuclear explosions.
So, if I calculate a groundburst nuclear explosion, shouldn't I multiply the value by 2?
The formula supposedly does not account for radiation
- 10,499
- 11,547
The ground burst formula is based on chemical explosives. Chemical explosives are not 100% burst either. They quite possibly have heat components that are (in relative terms) larger than nukes and they have (non-nuclear, but heat based) radiation as well.
So again, no, you can't just multiply the formula by 2 as it never was designed to represent half of the nuke energy to begin with.
So again, no, you can't just multiply the formula by 2 as it never was designed to represent half of the nuke energy to begin with.
- 10,499
- 11,547
Don't think so. I will close it.
- Status
Similar threads
- Locked
