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• ## Problems with the current model

Now if you are to go to the Explosion Yield Calculations page you'll find instructions on how to find the yield of an explosion going off of the air blast radius (near-total fatalities). But see there is a problem with that, that requires you know the radius of the air blast radius. When it comes to nuclear explosion you'll find that it's almost impossible to measure said radius, often because it not shown in it's entirety on screen but even worse is the fact that even if is shown you have no way of telling what is a "near-total fatalities" and what is a "widespread destruction" blast radius.

## Solutions

I'm sure many of you have noticed that calculators such as the Nuclear Weapon Effects CalculatorÂ (NWEC for short) and NUKEMAP Â produce far more accurate results, the downside is that you can only enter in yield and get the radius of the explosion but not the other way around.

Well fortunately enough I went through the studies they cite, mainlyÂ The Effects of Nuclear WeaponsÂ a 1977 research done by the United States Departments of Defense and Energy. I basically found the formulas that the calculators use to determine the radiusÂ of the fireball via the yield. This is important because the fireball is the part of the explosion that is most often visible and the most consistent too. Unlike the air blast the fireball's radius is not influenced by by height of detonation all that matters is if it's in contact with the ground i.e. whether it's an airburst or a surface burst.

### Formulas used

Looking at the NWEC we see 3 different values for a fireball radius:

• R(minimum) = 90*(Y^0.4)
• R(airburst) = 110*(Y^0.4)
• R(ground-contact airburst) = 145*(Y^0.4)

Where R is in feet (will be accounted for) and Y is yield in Kilotons.

Radius of an airbust fireball (sphere) is equal to R(minimum) + R(airburst)

Radius of a surface fireball (half a sphere) is equal to R(airburst) + R(ground-contact airburst)

## Formulas for Yield via fireball radius

These formulas are more so consistent with the NWEC rather than NUKEMAP, any differences in the airburst likely stem from the way the sites round numbers more than anything else, but when it comes to surface NUKEMAP will always give you a bigger radius for the same yield, possibly because it can account for the the split second in the begging of the explosion when the fireball and air blast interact.

Note: The times 3.28084 is to convert meters into feet as R is in meters and Y is in Kilotons of TNT.

### Airburst

Y = ((R*3.28084)^2.5)/(200^2.5)

### Surface

Y = ((R*3.28084)^2.5)/(255^2.5)

• As far as I'm aware the fireball of a nuclear weapon makes much sense with nuclear weapons not so much with other explosion types.

Overpressure is less reliable on the specific mechanic as far as I am aware.

I aslo think that one actually sees the pressure wave of the explosion more frequently then the fireball. At least for large scale explosions.

• Yes this is strictly nuclear type explosions. Overpressure depends on detonation height, we can't accurately tell what the overpressure is and we can't determine where the 5 psi range ends and where the where the 1 psi range ends so fireballs are far more reliable.

• DontTalkDT wrote: I also think that one actually sees the pressure wave of the explosion more frequently then the fireball. At least for large scale explosions.

Personally my brain has like, always defaulted to going for the fireball

Though yeah, you should be measuring the pressure wave

• We should add a section for radiation applicantion, such as a criteria that would make the energy fall under the category of radiation for when the feat is discussed wether to divide or not

• Okey-dokey.Â

• So does any of our instruction pages need to be adjusted here?

• Can I get people to look at this thread for adding for nuke based calculations?

• I replied.

• Antvasima wrote: So does any of our instruction pages need to be adjusted here?

• I'd think so. Again using "near total fatalities" is unreliable as that's the 5 psi range but the air blast extends farther and we can't accurately determine it's borders.

The fireball is a way better solution as an explosion with a set yield will produce a fireball with a set radius with the only influencing factor being if it's in contact with ground or not.

Also I don't see the formula we use for surface blasts listed on instruction pages and what we define as an explosion of nuclear origin is unclear.

• Calc group members opinions would be appreciated.

• Well this calculator says fireball radius scales with Y^0.4, that's what I have been using all along for fireball explosion calcs.

• Spinosaurus75DinosaurFan wrote:
Well this calculator says fireball radius scales with Y^0.4, that's what I have been using all along for fireball explosion calcs.

Yes I know, but aside from the fact that Y^0.4 only works for airbursts the results that it produces are inaccurate. The result you getÂ for the radius of a set yield explosion is always higher than the one displayed by the actual calculators.

• Maybe we should write guidelines about which formula should be used for which situation?

• That seems like a good idea. Would you be willing to help DontTalkDT?

• So is this going anywhere or nah?

• It depends on if our calc group members think that it would be a good idea, and are willing to revise the relevant instruction page accordingly.

• So what formula should we use for fireballs now?

• Spinosaurus75DinosaurFan wrote:
So what formula should we use for fireballs now?

Well, the ones I found? I mean those are for nuclear explosions only and I don't know how we determine what's a nuclear explosion and what's not.

If you are referring to just a fireball that isn't necessarily caused orÂ  accompanied by an explosion I believe I've seen some way for calcing that floating around, I can search for it if you want.

• Somebody should preferably politely remind DontTalkDT to comment here again.

• 100 Megaton Tsar Bomba wrote:
Yes this is strictly nuclear type explosions. Overpressure depends on detonation height, we can't accurately tell what the overpressure is and we can't determine where the 5 psi range ends and where the where the 1 psi range ends so fireballs are far more reliable.

IIRC the formula we usually use assumes optimal detonation height, making the result always an acceptable low end.

The end we use is 20 psi overpressure, if I'm not wrong, and usually we assume that that's the range of the visible explosion, as basically everything in that range is destroyed. For a shockwave that only shatters the glasses of buildings or something we would not use that formula.

Fireballs are for general explosions hardly reliable. If you have an explosion caused by a real nuclear weapon, sure, use any parameter you can determine, but in all other cases I doubt that the fireball formulas for nuclear weapons are a good approximation. Consider that in chemical explosives fire is caused due to the burning of the chemicals and for explosions caused by impact the fireball, if present, is created by stuff like friction and pressure.

• @DontTalkDT

Thank you for the input. So do you think that any instruction page needs to be updated from this, or can they be left as they currently are?

• Yeah 20 psi is a decent assumption as that's demolitions and heavy damage to reinforced buildings and shattering windows is like 0.45 psi. Lowballing it is also fine.

But when I say fireball I mean something like this.Â Those sort of things often get shown on screen and meassuring them can yield far better results if they count as nuclear (not that I know what is needed in order for an explosion to qualify as nuclear in regard to this wiki).

We already go off of nuke calculator for air blast radius it seems whu not do the same with fireball?

And yes I know it doesn't work for non nuclear explosion but there isn't much of a fireballÂ or differing air blast radiuses for different overpressures in regards to those.

• bump

• Antvasima wrote: @DontTalkDT

Thank you for the input. So do you think that any instruction page needs to be updated from this, or can they be left as they currently are?

Can somebody ask DontTalkDT to comment here again via his message wall? Just make sure to describe what you want in the thread title. He said that he tends to ignore "Hello" messages.

• Ok, lemme list some problems I have with the instructions page first:

1. We tell people on the Explosion Yield Calculations page to devide by 2 for any non-nuclear explosion because only half the energy is from the blast but we never say that we are referring to an air blast and that the rest of the energy is from various types of radiation emited (mostly thermal).

2. Nothing is said about what would and wouldn't be considered nuclear. Most explosions in fiction are not caused by atomic bombs and they do not emit ionizing radiation but many of them are likely to emit thermal which is pretty much the remaining 50% of the energy of an explosion.

3. This formula Radius in meters^3*((27136*1.37895 + 8649)^(1/2)/13568 - 93/13568)^2 gets used quite often and there is no mention of it on the page. Also I've seen a few instances where a fireball radius was meassured and plugged into the equation which will obviously produce inaccurate results but no one seemed to notice a problem.

Anyways I'll go bother DT with this now.

• Okay.

• Yeah the 3rd point really irks me a lot that it isn't on the page.

• You can also ask some other calc group members, such as Kepekley23, Damage3245, Ugarik, Executor N0, and AlexSoloVaAlFuturo, to comment here.

• I agree with the suggestions. And yes, I never understood why this calculation method isn't mentioned in any official page (I think that the page should explain various explosion methods and not only nuclear explosion, but I think that this should be done after this deal with our nuclear explosion standards are finished).

• Thank you for helping out.

• 100 Megaton Tsar Bomba wrote:

But when I say fireball I mean something like this. Those sort of things often get shown on screen and meassuring them can yield far better results if they count as nuclear (not that I know what is needed in order for an explosion to qualify as nuclear in regard to this wiki).

We already go off of nuke calculator for air blast radius it seems whu not do the same with fireball?

And yes I know it doesn't work for non nuclear explosion but there isn't much of a fireball or differing air blast radiuses for different overpressures in regards to those.

A fireball of a nuclear weapon is fundamentally nothing different than that of meteors or chemical explosives. It's gas heated to the point of becoming a bright plasma, which is exactly what a flame is.

However its behaviours are possibly very different due to how it's produced. The fireball of a nuclear weapon is to my knowledge produced by the air absorbing a certain spectrum of the weapons x-ray and growing heated as a result.

Hence the requirement of it being an actual nuclear explosion caused by a nuclear weapon. Something like being caused by radiating energy would for example not suffice here, as differences in spectrum can change things a lot.

Shockwave estimates should work pretty well for non-nuclear stuff as one can actually approximate results of nuclear explosions based on the shockwave without knowing details of nuclear weapon functionality.

100 Megaton Tsar Bomba wrote: 1. We tell people on the Explosion Yield Calculations page to devide by 2 for any non-nuclear explosion because only half the energy is from the blast but we never say that we are referring to an air blast and that the rest of the energy is from various types of radiation emited (mostly thermal).

I'm not sure why it's relevant what the rest of the energy is emitted as, as the point is that it just can't be assumed to be emitted (in the same ratio as for a nuclear weapon). But sure we can mention that and that explosion means air blast, which is basically assumed to be the visible explosion.

2. Nothing is said about what would and wouldn't be considered nuclear. Most explosions in fiction are not caused by atomic bombs and they do not emit ionizing radiation but many of them are likely to emit thermal which is pretty much the remaining 50% of the energy of an explosion.

Everything not nuclear weapon based isn't considered nuclear. Yes, many other explosions also emit thermal energy, but there is no reason the thermal energy / shockwave ratio would be the same as the ratio emitted by nuclear weapons. Hence the thermal radiation can't be considered for the calculation (unless you can quantify the amount completely separate, but that is an entirely different topic).

3. This formula Radius in meters^3*((27136*1.37895 + 8649)^(1/2)/13568 - 93/13568)^2 gets used quite often and there is no mention of it on the page. Also I've seen a few instances where a fireball radius was meassured and plugged into the equation which will obviously produce inaccurate results but no one seemed to notice a problem.

The formula was never formally accepted, I think, which is why it was never added. We might as well add it at this point, though.

As for the fireball: As said, that part of the explosion can safely be assumed to have 20 psi overpressure and hence gives a valid low end. If the fireball is measured that is most likely because the shockwave outside of it can't be reasoned to be at least 20 psi in intensity.

• @DontTalkDT

Thank you for helping out. So what do you think should be changed in our instruction pages?

• The main differences between a nuclear and a non-nuclear explosion other than it's origin are the yield of the explosion, the way in which it emits it's energy and the temperatures reached. Event the most powerful non-nuclear explosions were only in the kilotons and usually don't get above a few thousand degrees while the gore of a nuclear fireball reaches millions.

Nuclear fission does indeed release a sizable amount of it's energy in the form of radiation that heats up the air but toÂ my knowedge most of heat is released due to the exothermic nature of fission of heavy and unstable radioactive elements but it doesn't have to be this way. Fundamentaly a nuclear explosion is just an explosion caused by a nuclear reaction which can be either fission or fusion.

A pure fusion explosion would still be considered nuclear and would release nearly all it's energy in the form of thermal radiation and a blast. And I understand the whole ratio problem but ultimately if the explosion is considered nuclear the ratio is not that important as it varies depending on even something like detonation height and obviously some thing like a thermonuclear bomb would release more thermal energy (higher percent) than a regular fission bomb. Ultimately the combined values of allthe different types of energies equal 100% which is what the formulas calculate, the toatl yield so there is no need to divide by anything. Again this isn't about regular explosions only nuclear ones.

The explosion also doesn't need to be caused by a litteral atomic bomb in order to be nuclear, many explosions would fit the description of a nuclear explosion even without being caused by an isotop of sorts. That is the presence of a large, bright fireball that causes severe burns, lasts several second, leadingÂ to a powerful shockwave that demolishes everything in it's path.

• 100 Megaton Tsar Bomba wrote:

The explosion also doesn't need to be caused by a litteral atomic bomb in order to be nuclear, many explosions would fit the description of a nuclear explosion even without being caused by an isotop of sorts. That is the presence of a large, bright fireball that causes severe burns, lasts several second, leadingÂ to a powerful shockwave that demolishes everything in it's path.

No, because these explosions might seem similar but are caused entirely differently, which means a different spread of the energies we don't directly consider in the formula.

You can reach that same phenomenon with a big pile of TNT, but that won't have the same energy radiated as a nuclear bomb of equal shockwave.

For magical explosions that is even more so. Their efficency in regards to shockwave generation (or also fireball generation) can be much higher than that of any real life weapon, to the point that they have 0% energy spend outside of the measured effect.

• Antvasima wrote:
@DontTalkDT

Thank you for helping out. So what do you think should be changed in our instruction pages?

Mostly just the inclusion of the alternative formula.

The rest is some changes in wording and additions to explanations.

• @DontTalkDT

Okay. Thank you. Feel free to do so.

• Yes I understand, a nuclear explosion (an A-bomb one) is caused by a chain nuclear fission of unstable, heavy, radioactive isotops. When the the nuclei get split they release a large amount kinetic energy, radiation, and heat. The reason why they are orders of magnitude more powerful than regular explosions is largely due to the fact that part of their mass gets converted into energy which doesn't happen with regular explosion (at least not to such a large extent).

The general ratio of energy release is 50% blast, 35% thermal and 15% radiation (5% initial and 10% residual). This isn't always true, when it comes to H-bombs or bombs over 50 Kilotons in general the radiation plays a way smaller role. There are also ER weapons where the radiation is 50% (45% initial and 5% residual) of the total with the blast and thermal being 30% and 20% respectively. Your regular stack of TNT won't be releasing radiation and not much thermal either, it's mostly a blast, yes.

But in general energy distribution is a function of yield and can be altered depending on the condition of the detonation. Most often when someone wants to maximize the effects of either the blast, radiation or thermal radiation they change the height of detonation more than anything else.

I understand that magical explosions often release no radiation meaning all their energy is split between blast and heat. Going by the way we use only half of the calculated yield if the explosion in non-nuclear because it's mostly just a blast wave it seems that what would be appropriate to do with explosions that produce a large amount of thermal radiation via a massive fireball that lasts for a significantis amount of time is to use only 85% of the calculated yield because it's only a fireball and a blast wave.

Also are the formulas proposed being accepted or not?