Well hell yeah, he's totally right! Fantastic observation! Let's look at this more closely:

First of all, we know that Gygax initially presented the "rebound" rule in OD&D Vol-3, p. 9, as a response to "some referees" who otherwise let fireballs and lightning bolts blast out stone areas equal to their full area (which is admittedly crazy). Then it got more detail in the AD&D PHB p. 73 spell description, in a bracketed note: "The area which is covered by the fireball is a total volume of roughly 33,000 cubic feet (or yards)." Now already this is super-sketchy, because the "or yards" qualifier doesn't makes any sense -- the area can't ever be in yards due to the all-caps rule on p. 39: "IT IS IMPERATIVE THAT OUTDOOR SCALE BE USED FOR RANGE ONLY, NEVER FOR AREA OF EFFECT (which is kept at 1" = 10')". Recall that this is a new rule, first appearing in the 1E PHB; is it so new that it post-dates the AD&D text to fireball itself?

In any event, the volume given is basically correct: granted the "Area of Effect: 2" radius sphere" (interpreted here as 20 feet radius), we do have a spherical volume pretty close to the given "roughly 33,000 cubic feet", more specifically:

Okay. So let's consider the fairly simple and normal case where a fireball is directed so that it explodes at ground-zero height. Then obviously the resulting "compressed" area is actually a half-sphere with this same volume V above the ground. Therefore, we need to find the new radius r such that the half-sphere area has the same volume. Solve (with appropriate reductions at each step):

So, as John expected (and me as well, once he'd pointed this out), a standard fireball ground burst under the official rule does indeed get some extra area -- although only an extra 5 feet radius, less than I might have intuited. That is one extra square if you're using a 1"=5 feet battlemap; or maybe you're happy to ignore the difference, you philistine, you. If so, stop here, because the rest of the post is more complicated and makes even less difference.

Still with me? Okay, then -- You see, distressingly, that's not what John actually asked. What he asked was a more prickly problem, namely: "... the situation where a human caster is casting it at a man sized creature (I would say the most common use? e.g. a group of orcs) [so] the centre of the fireball will be roughly five feet off the ground..." Oooookay. Well, 5 feet off the ground isn't a simple half-sphere any more. It's calculusin' time! Consider the following:

On the left is a cross-section of the fireball area that we are about to rotate around to find the full volume. To make the notation easier, I've turned it 90 degrees on its side and placed the center of the attempted sphere at the origin (the "ground" is now vertically 5 feet to the left, where the shape gets flattened out; and the ball otherwise extends some unknown radius r from the origin in every direction, where we can guess that this r is something between 20 and 25 feet). On the right, you can see what one slice looks like as we start to revolve it around the x-axis: a circle of some other radius (note that every part of that upper arc is exactly r feet from the origin: that's what having a sphere means, after all). And what is the radius of that new-circular slice? Well, it's the height y, where, per the Pythagorean theorem, x

^{2}+y

^{2}=r

^{2}, and so y

^{2}= r

^{2}-x

^{2}. And thus if y is the radius of this sub-circle, then its area must be A= πy

^{2}= π(r

^{2}-x

^{2}); and if its width is some infinitesimal dx, then its part of the volume can be written as the product π(r

^{2}-x

^{2})dx. Now we just need the integral of this between -5 and r, as shown in either illustration above.

This isn't novel: What's happening here is just the exact same formulation for the volume of a regular sphere that you'll find in any calculus textbook, just replacing the usual lower bound "-r" with the number "-5" for our purposes. Let's compute an expression for this integral (excuse handwritten notes below):

And it's this expression for volume that still has to maintain the value of 33,510 ("roughly 33,000") under our fireball-compression rule. So set it equal to that number and try to solve for r:

Now, if the highest exponent was 2 (a quadratic), then I could just run it through the ol' quadratic formula and get a solution; but it's not, it's got an exponent of 3, so I have to use the cubic formula. It's like the quadratic formula but it's more complicated. Probably easiest if I just let some software do it, like by going to the site www.cubicsolver.com, and punching in the coefficients (2, 15, 0, -32,125). Then we see that the 3 possible values of r include two imaginary numbers as well as the value r = 22.96... ≈ 23 feet.

So to answer John's actual, insightful, very profound question -- Yes, if fireballs are defined to be spheres, and you launch one 5 feet from the ground, then the extreme edge actually swells to 23 feet from the center point. Better watch your face, or you'll burn it right off!

(Want to see the cubic formula? Go here. What comes next? How about a quartic formula. After that? A proof that no higher-degree solving formulas can exist.)

Now, just to come up from the depths at the end -- Stuff like this is actually why I'm no longer a fan of the "compress to space" rule for fireball. In some ways it's kind of fun or even arguably a balancing factor for the otherwise all-powerful fireball -- but c'mon, the complications due to this rule are just crazy. The very fact that the PHB has to mention a figure of "roughly 33,000 cubic feet" is kind of ridiculous for a tabletop game. And the completely overlooked logical implication being that it's not really a 20-foot radius, but rather a 25-foot radius (or thereabouts) in normal usage on the ground. Blast it!

I hate to say it (after calculus was done) but I think you're doing this backwards.

ReplyDeleteThe spell assumes a target at ground level, so the area of effect is a 2" radius, assuming the extra volume from the hemisphere in the ground.

The real question is, how much smaller is a fireball that detonates into a true sphere shape.

Also: and I don't mean to be annoying here, but 33,000 cubic feet was given in the rulebook, because the sphere will expand to fill 33 10'x10'x10'

ReplyDeletecubeswhich is easy for a tabletop game to simulate.See, most ceilings in dungeons are 8'-12' in height, and since your radius is a radius, basically the DM floods 330' feet of corridor with fireball, 160'-170' from ground zero, or am I missing something?

"the sphere will expand to fill 33 10'x10'x10' cubes which is easy for a tabletop game to simulate."

ReplyDeleteThat's how I've been adjudicating fireballs since at least the 90's. A 5th level magic-user with a ring of fire resistance and few scruples can easily clear out big chunks of some cramped dungeon maps.

To -C and the illustrious Mr. Rients:

ReplyDeleteThings aren't as simple as you paint them.

1)-C's first comment says the spell assumes a target on the ground. That doesn't follow from the spell description, but perhaps -C relies on common sense here. At any rate, the spell

doesgive the area of effect as a 2" (20') radius sphere. If Gygax meant for the area actually to have been a 20' radiushemisphere, as -C seems to suggest, he would presumably have said that.2) -C's second comment assumes that the fireball is set off in a corridor; if so, it will expand as described (and the caster will probably fry some portion, if not all, of his party). But what if it's set off in a room? It will start expanding as a sphere (i.e. equally in all directions from the center point) until it hits walls or ceiling, at which point it will expand laterally along said walls or ceiling. Accurately modeling this is complex and tedious. The recourse of most DMs seems to be what Jeff Rients suggests here: fill up 33 squares on your map with fireball. It's practical, but not an accurate representation of the fireball's behavior in many cases. Granted, one might not care about absolute accuracy, but if that's so, why wade through a post so devoted to such accuracy that it subjects readers to math that is probably above most of their heads?

Fireballs are magic and therefore do not behave according to calculus but according to the rule of appropriateness-at-the-time and this DM judges that to mean that they are 20' radius indoors and 20 yard radius outdoors, regardless of targeting.

ReplyDeleteSo there.

OK, Nagora: an easily workable adjudication, which is probably all that's important in most cases of actual game play. But, as Delta shows very well in his post, this adjudication doesn't adhere to the parameters given in the PHB. As I read it, that's the point of Delta's post: accurately modeling the Fireball spell as described in the 1e PHB is not worth it.

ReplyDelete-C said: "I hate to say it (after calculus was done) but I think you're doing this backwards. The spell assumes a target at ground level, so the area of effect is a 2" radius, assuming the extra volume from the hemisphere in the ground."

ReplyDeleteGood to think about, but that's contradicted twice by the 1E text. (a) The area-of-effect really is written as "2-inch radius sphere" (not hemisphere). (b) The 33,000 cubic feet volume is also only correct for a 20-foot radius sphere (it would be half that for a hemisphere).

Although maybe you think the 1E text is itself incorrect, which would be a different story.

I also do agree with everything else JHB wrote above, including the fact that room heights are always a lot different from corridor heights in classic materials (so that the 33-map-squares method doesn't match the written rule).

Next refinement on this stuff would be to look at the actual physics involved. Here's an interesting company -- http://www.blastanalysis.com/

ReplyDelete"We have been examining a surface burst explosion, in which the explosive charge is assumed to be hemispherical and placed on flat, hard, smooth ground surface, so that the blast wave will also be hemispherical. In the Free Air configuration, the blast wave is assumed to be spherical. (AirBlast has separate databases for the surface burst and free air configurations, both based on experimental measurements of blast wave properties; it is

notassumed that a surface burst explosion is equivalent to a free air explosion of twice the size)."http://www.blastanalysis.com/abtour9.htm

(Which is to say: the "keep same volume" rule doesn't line up with actual physics.)

"The effects of an urban environment on the physical properties of a blast wave may not be immediately obvious. For example, an explosion at a T-junction produces the maximum blast effects along the side arms and not along the main stem."

http://www.blastanalysis.com/WordDocuments/Blast%20Effects%20Dist%20MABS17.pdf

Hypothesis -- If we really dug into the physics of all this, a more realistic rule* might be to increase damage in the constricted area (reflected shockwaves bouncing around multiple times**), rather than increasing area of the blast.

ReplyDelete* Not that there's anything wrong with declining that.

** Like

lightning bolts. In fact, jointly it was called the "rebound" rule in OD&D Vol-3.This is the nerdiest post and discussion ever. Therefore I love it.

ReplyDeleteBy the way, I've been adjudicating fireballs like Jeff for the past 17 years or so and no one ever complained. Even when the backblast killed the whole party.

Delta, Im afraid the flaw in your model is that you are maintaining that from the moment the expanding fireball makes contact with the ground the truncated sphere will continue to expand *preserving proportion* as a truncated sphere.

ReplyDeleteIt wouldn't. The reflection of the blast would contribute most at ground level and a diminishing amount as we rise.

You would always see a deformed hemisphere with the radius of any horizontal slice greatest at ground level and decreasing monotonically to the apex.

Kent -- I do agree with that, and it's not something that I initially took into consideration, and I thank you for pointing it out here.

ReplyDeleteSo if we start thinking about the actual

physicsof an explosion like that (instead of simply keep-the-volume math), then I think we can agree on two things -- (a) the area would indeed be larger than indicated at ground level to some degree (I could imagine it being less than what I computed here, for a given psi-level), and (b) the very idea that volume-is-constant is itself an inaccurate rule (as per the blastanalysis.com link above).Do you have better data or source for what exactly a blast of that level would look like? (I'm guessing that it's in the range of 1s or 10s of kg of TNT...)

(a) Yes

ReplyDelete(b) Sure. I don't have any more information than you have. I assume blast energy is lost into the interfering material and some materials would be more reflective than others.

For game purposes one might approximate the blast as hemispherical. A blast over metal ground might lose say no volume; over solid stone ground, mts, cliffs, caves might lose 1/6 volume and over earth 1/3 of the volume might be lost. Yielding new radii larger by 1.26, 1.17 and 1.1 respectively. So even on a battlefield the radius could be 10% larger. Of course I just made up the absorption fractions. Though it does have me wondering if a fireball detonating in woods would have a *smaller* hemispherical radius than the original sphere on account of absorption. Once fixed for a game the tactical effect of terrain might be interesting for a wizard general.

As someone who is reading Chainmail and has an interest in skirmishes in D&D up to the two hundred per side I think this stuff is worth considering if a simple set of modifications can result.

Kent: That's all good and I think I agree with your physics intuitions. In practice for myself I don't plan on applying any of those modifiers since they do make pretty small adjustments. (Even moreso if you zoom out, say, 1:4 for a mass battle.)

ReplyDelete