- 1,259
- 418
These 2 formulas here, the one for mid-air and ground explosions
Follow along with the video below to see how to install our site as a web app on your home screen.
Note: This feature may not be available in some browsers.
The first formula came from the Outskirts Battledome, which originally sourced it from this: https://nuclearweaponarchive.org/Nwfaq/Nfaq5.htmlThese 2 formulas here, the one for mid-air and ground explosions
I looked at the site and couldn't find the formula or anything mentioning it at all, just the number 0.28The 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.
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
That formula is from both those links. @DontTalkDT can elaborate more about it.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
Since you found the link, should it be added as reference?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.
Already asked DontTalkDT before to do so.Since you found the link, should it be added as reference?
As our Explosion Yield Calculations page states,So the first formula ignores thermal radiation and fireball radius it seems?
QuestionAs our Explosion Yield Calculations page states,
Fireball stuff is another formula different from the airburst.
- 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.
I honestly don't know either, and the rule itself is exclusively for airburst explosions.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?
That's not really what I'm talking aboutYou 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.
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?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.
That is, they do not take into account other energies. To take them into account, shouldn't I multiply x2?No, as the ground explosion formula makes no such correction to begin with.
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.No, because we never subtracted other energies from it to begin with.