Area Effect Aspect Volume Computations

The OpenD6 Fantasy rules define an Area of Effect aspect that applies to magical spells and religious invocations. This aspect has a relatively simple set of rules:

Two-dimensional circle (a few centimeters thick): +1 per half-meter radius.

Three-dimensional sphere (for explosions and 3D illusions): +5 per meter radius and +1 bonus to hit one target (bonus is applied to the same target).

One alternate shape: +1 to area effect modifier.

Several alternate shapes (specific one chosen at time of casting): +3 to area effect modifier.

Fluid shape (shape may change any time during spell’s duration): +6 to area effect modifier.

We can break the rule text into two kinds of statements:

  1. Shape definitions (2D circle vs. 3D sphere).
  2. Modifiers (alternate shapes and fluid shape).

However, from spells in the OpenD6 Fantasy Locations, it appears ths sphere definition harbors a subtle error. This was uncovered by wrestling four four spells that have “fluid shape” modifiers. This involves a short journey into some relatively simple math to uncover a small piece of treasure.

Further, the (unpublished) OpenD6 Magic Guide has a number of shapes, and the rules for these seem to be broken. This will be a longer journey, deep into a wilderness that involves more serious algebra. This will yield a delightful, but, hardly useful piece of treasure. A kind of trophy with little practical value.

OpenD6 Fantasy Locations spells

Our journey begins when we uncover four spells that are examples of the Fluid Shape variant of the Area of Effect aspect in the OpenD6 Fantasy Locations rules:

  • The Basic Shelter spell, has an area effect difficulty of 6. The difficulty corresponds to the “Fluid Shape” modifier, but the spell fails to explicitly state a shape. The spell does describe a volume of \(1.25 \times 2 \times 2 = 5\) cubic meters. The OpenD6 Fantasy rules provide no definition for a rectilinear (“cuboid”) shape.
  • The Improved Hut spell, with an area effect difficulty of 11. This could be the “Fluid Shape” modifier with an implied shape with a difficulty of 5. The spell describes a volume of \(4 \times 4 \times 2 = 32\) cubic meters.
  • The Small Long-Lasting Tower and Keep-in-the Air spells have an area effect difficulty of 26. This also seems to use the “Fluid Shape” with an implied shape with a difficulty of 20. The spell describes a volume not to exceed 268 cubic meters.

The problem here is that “Fluid Shape” appears to be a modifier to a shape definition. There’s no irrefutable evidence of this interpretation, since there are no examples of fluid shapes in the OpenD6 Fantasy rules. It is a pretty clear intent, however.

With the assumption that Fluid Shape as a modifier (worth 6), the conclusion is a 5 cubic meter sphere has a difficulty of zero. The volume of a sphere is \(V = \frac{4}{3} \pi r^3\). The radius for volume of 5 cubic meters means \(r = \sqrt[3]{\frac{3V}{4\pi}} \approx 1.06\).

Superficially, this does not follow the rules. The rules say “+5 per meter radius”; this statement means a 1m radius sphere must a cost of 5.

Either the cost is wrong for some of these spells. Or the “+5 per meter radius” statement is wrong. There are two distinct consequences:

  • Assuming the proper rule was supposed to be +5 per meter after the first; a 1m sphere has a cost of zero., then the Small Long-Lasting Tower and Keep-in-the Air spells should have a difficulty of 21, not 26.
  • Assuming the proper rule is +5 per meter, then the Basic Shelter spell difficulty needs to be 11 (not 6) and the Improved Shelter spell difficulty needs to be 16 (not 11).

When we look at other spells, i.e. Fantasy Apportation spell Carrying Wind, we see “Area Effect (+15): 3-meter sphere”. This is conclusive evidence the rule – as written – was correct. The two spells (Basic Shelter and Improved Shelter) have the wrong difficulties for their areas of effect.

[The rules have numerous “off-by-one” errors, many due to rounding. This isn’t unusual.]

This leads to the following table of sphere sizes and difficulty values:

from IPython.display import display, Markdown
from contextlib import redirect_stdout
from io import StringIO

from sympy import pi

buffer = StringIO()
with redirect_stdout(buffer):
    print("| r (meters) | volume (cubic meters) | difficulty |")
    print("|---|--------|------------|")
    for r in range(1, 7):
        print(f"| {r}| {(4/3)*pi*r**3:.2f} | +{r*5} |")

md = Markdown(buffer.getvalue())
display(md)
r (meters) volume (cubic meters) difficulty
1 4.19 +5
2 33.51 +10
3 113.10 +15
4 268.08 +20
5 523.60 +25
6 904.78 +30

We’ve started this journey by uncovering a tiny editing error in an obscure corner of the rules. Two spells with incorrect computations for difficulty.

This is only the start of a much longer journey into a wilderness of TTRPG rules. We’ll turn to the never-officially-published OpenD6 Magic Guide.

Magic Guide Additional Shapes

The (unpublished) OpenD6 Magic Guide has a number of other shapes that are part of the Area of Effect aspect:

  • 2D shapes. We’ll skip these, they rules seem simple enough.
    • Circle. \(A = \pi r ^ 2\).
    • Wall. \(A = h w\).
    • Blast. \(A = \pi r ^ 2\). The radius varies with the distance from the caster. This seems to matter when sorting out the targets of a spell during play.
  • 3D shapes.
    • Hemisphere. \(V = \frac{2}{3} \pi r^3\).
    • Cone. \(V = \frac{1}{3} h \pi r^2\).
    • Cuboid. \(V = h w d\).
    • Cylinder. \(V = h \pi r^2\).
    • Pyramid. (Incorrectly stated in the rules) \(V = \frac{1}{3} h w_b l_b\).

The rules provide some “rule of thumb” computations for difficulty. The idea is to avoid the complex-looking math required to work out a difficulty based on volume. The rules of thumb seem vaguely wrong.

Further, when we look specifically at cuboid and compare the difficulty results with the four Fantasy Location spells, we have a serious discrepancy. The cuboid rules for volume of 268 cubic meters seem to assign a difficuilty of 268!

That’s not sensible.

We can try to use the Value-Measure table, but the difficulty winds up 12 instead of 26.

Since these rules were unpublished, it seems sensible to assume that this detail suffered from a lack of play-testing and editorial oversight. It’s simply wrong.

What can we do? We’ll need to wander through the forest of shapes to see how we can work with other shapes as simply as spheres. We’ll start with a deep dive into spheres.

Core Difficulty and Volume

The rules overall seem clear enough on one point: for three-dimensional shapes, the difficulty grows with the volume. The volume of a sphere, \(V\), with radius \(r\), is stated explicitly, with one tiny omission.

The difficulty for a spherical Area of Effect is computed directly from the radius, \(r\): \(d = 5 \times r\).

We’ll take this as a golden truth. We can turn this relationship around: given a volume, we can work out the radius of a sphere with that volume. This approach lets us compute the difficulty for any given volume, \(V\).

In order to do the algebra correctly, we need help. We’re going to rely on software tools. Specifically, the sympy package. See https://docs.sympy.org/latest/index.html.

A little setup: we’ll import the package, initialize formula printing, and define a few symbols that will be used through the rest of the story.

from sympy import *
init_printing()
V, R, r, l, w, h, d = symbols("V R r l w h d")

Here’s the formula for volume of a Sphere.

We’ll start with spheres because the answers are known. This easily generalizes to any shape that has a volume. What matters is the value for \(V\).

sphere = Eq(V, Rational(4,3)*pi*r**3)
sphere
\begin{equation*} \displaystyle V = \frac{4 \pi r^{3}}{3} \end{equation*}

We can solve this for \(r\).

Eq(r, solve(sphere, r)[0])
\begin{equation*} \displaystyle r = \frac{\sqrt[3]{6} \sqrt[3]{V}}{2 \sqrt[3]{\pi}} \end{equation*}

In the sphere case, computing volume from radius and then solving for the radius is obviously pointless. In all other cases, however, we will start with some formula that gives us a volume. From that, volume, we can then deduce the equivalent sphere radius and the difficulty for the given shape.

The problem with solving for \(r\) is we get an ugly-looking formula. Really paralyzingly ugly. Kind of turn-to-stone if you look at it, Medusa-scale of ugly.

The good news, is it has a form of \(S_s \sqrt[3]{V}\), where the \(S_s\) is the shape-specific constant, appropriate for spheres. You don’t have to trust me on this. We’ll get out the crowbar and take these formulae apart to see how the work inside.

The goal is to have a different constant for each of the other shapes. We’re going to work out \(S_h\) for hemisphere, \(S_c\) for cone, \(S_u\) for cuboid, \(S_y\) for cylinder and \(S_p\) for a pyramid.

We can, with some patience, manually extract the constant factor ($:raw-latex:frac{sqrt[3]{6}}{2 sqrt[3]{pi}} $) from the above equation. This is an unpleasant, error-prone manual operation. Here’s what it looks like.

N((6**(1/3))/(2*pi**(1/3)))*V**Rational(1,3)
\begin{equation*} \displaystyle 0.6203504908994 \sqrt[3]{V} \end{equation*}

We can confirm that the algebra was done error-free, by trying to simplify it. The messy-looking code blob represents the \(R_s\) constant value from the original equation.

nsimplify((6**(1/3))/(2*pi**(1/3)), [pi])
\begin{equation*} \displaystyle \frac{\sqrt[3]{6}}{2 \sqrt[3]{\pi}} \end{equation*}

Doing algebra manually to separate constants from variables is unpleasantly complicated. We can do better. We have tools.

The formula crowbar

Specifically, the sympy toolbox has a crowbar we can use to pry the formula into two distinct parts: a part with variables, and a part with the constants.

How does this work?

Let’s visualize the formula by depicting the computation operations. Each operation is shown in an oval which points to the arguments for that operation. The final literal values (and the variable, \(V\)) are in ovals that don’t point anywhere.

What we see is the overall form of the equation is a multiplication. It has four arguments, the first three of which are constants, and the fourth includes a volume variable, \(V\).

import graphviz

graphviz.Source(dotprint(solve(sphere, r)[0]))

We can see the four arguments, with errors flying out of the overall Mul operator node. A cube root operation, \(\sqrt[3]x\), is represented internally using a Pow node, meaning \(x^\frac{1}{3}\), which seems kind of awkward at first. However, the pretty-printed output looks like an ordinary cube-root, allowing us to disregard the nuances of the internal structure.

solve(sphere, r)[0].args
\begin{equation*} \displaystyle \left( \frac{1}{2}, \ \sqrt[3]{6}, \ \frac{1}{\sqrt[3]{\pi}}, \ \sqrt[3]{V}\right) \end{equation*}

We can compute the product of the first three factors, the constant part. From these we can compute a simple-looking weighting factor that converts the shape’s volume to the radius of a sphere.

N(solve(sphere, r)[0].args[0] * solve(sphere, r)[0].args[1] * solve(sphere, r)[0].args[2])
\begin{equation*} \displaystyle 0.6203504908994 \end{equation*}

Pragmatically, we can’t guarantee that all of the shapes will have the same node structure. To overcome the variability, we’ll define a function to parse the formula, separating the constant factors from the part of the expression with variables in it.

This only works for volume computations that are a simple multiply. This is true for the five shapes we’re working with.

It avoids the need to do algebra manually.

def constants(expr):
    """Only looks at the top-level Mul() node for constants and non-constant parts."""
    assert isinstance(expr, Mul), "Not a simple product"
    vars = [arg for arg in expr.args if not arg.is_constant()]
    cons = [arg for arg in expr.args if arg.is_constant()]
    if len(vars) == 1:
        return Mul(*cons), vars[0]
    else:
        return Mul(*cons), Mul(*vars)

Here’s how we use this function to decompose a formula into the constants and the variables. Note that when we multiple the constant product by the variable product, the equation is unchanged. This is proof we haven’t damaged anything.

cons, vars = constants(solve(sphere, r)[0])
cons * vars
\begin{equation*} \displaystyle \frac{\sqrt[3]{6} \sqrt[3]{V}}{2 \sqrt[3]{\pi}} \end{equation*} 
cons * vars == solve(sphere, r)[0]

True

The ugly formula gives us an exact value for converting a Volume to a radius. It doesn’t seem useful because it involves a bunch of irrational numbers.

We can compute a useful approximation for the contant factor, $:raw-latex:frac{sqrt[3]{6}}{2 sqrt[3]{pi}} $, which isn’t so ugly.

N(cons) * vars
\begin{equation*} \displaystyle 0.6203504908994 \sqrt[3]{V} \end{equation*}

This approach lets us use volume equations for any shape. We can solve for the radius of a sphere with the same volume. We can extract the constants and have a tidy \(R = S_x\sqrt[3]V\) kind of equation that yields a radius that can be plugged into the difficulty computation, \(d = 5R\).

There will be one exception, the hemisphere. It’s simpler than the others.

Hemisphere

A Hemisphere has a smaller volume for a given radius, \(r\). As with a sphere, the volume is characterized by a single number, \(r\).

hemisphere = Eq(Rational(2,3)*pi*r**3, Rational(4,3)*pi*R**3)
hemisphere
\begin{equation*} \displaystyle \frac{2 \pi r^{3}}{3} = \frac{4 \pi R^{3}}{3} \end{equation*}

We can solve for \(R\) to find the effective radius of a sphere with the same volume.

Eq(R, solve(hemisphere, R)[0])
\begin{equation*} \displaystyle R = \frac{2^{\frac{2}{3}} r}{2} \end{equation*}

We can extract the constants and compute an approximate value that’s easier to use.

cons, vars = constants(solve(hemisphere, R)[0])
N(cons) * vars
\begin{equation*} \displaystyle 0.7937005259841 r \end{equation*}

Here’s a depiction of the formula, just to show that there are only two constant arguments, and single variable argument to the top-level Mul node.

graphviz.Source(dotprint(solve(hemisphere, R)[0]))

Cone

A cone’s volume is described by a height, \(h\), and a radius, \(r\). We’ll define an equation using the cone’s parameters and the radius, \(R\), of the equivalent sphere.

cone = Eq(Rational(1,3)*h*pi*r**2, Rational(4,3)*pi*R**3)
cone
\begin{equation*} \displaystyle \frac{\pi h r^{2}}{3} = \frac{4 \pi R^{3}}{3} \end{equation*} 
Eq(R, solve(cone, R)[0])
\begin{equation*} \displaystyle R = \frac{\sqrt[3]{2} \sqrt[3]{h r^{2}}}{2} \end{equation*}

We can decompose this into a constant value and a volume-related expression. This gives us a computation for \(R\), from which we get difficulty, \(5R\).

cons, vars = constants(solve(cone, R)[0])
N(cons) * vars
\begin{equation*} \displaystyle 0.629960524947437 \sqrt[3]{h r^{2}} \end{equation*}

Here’s a depiction of the formula showing the first two arguments, which are the constants, and the third argument, which has the variables in it.

graphviz.Source(dotprint(solve(cone, R)[0]))

Cuboid

A Cuboid’s volume is decribed by the height, \(h\), width, \(w\), and depth, \(d\). We can compute the cuboid volume, and then solve for the equilavent sphere’s radius, \(R\).

cuboid = Eq(h*w*d, Rational(4,3)*pi*R**3)
cuboid
\begin{equation*} \displaystyle d h w = \frac{4 \pi R^{3}}{3} \end{equation*} 
Eq(R, solve(cuboid, R)[0])
\begin{equation*} \displaystyle R = \frac{\sqrt[3]{6} \sqrt[3]{d h w}}{2 \sqrt[3]{\pi}} \end{equation*} 
cons, vars = constants(solve(cuboid, R)[0])
N(cons) * vars
\begin{equation*} \displaystyle 0.6203504908994 \sqrt[3]{d h w} \end{equation*} 
graphviz.Source(dotprint(solve(cuboid, R)[0]))

Cylinder

A Cylinder’s volume is described by the a height, \(h\), and a radius, \(r\).

cylinder = Eq(h*pi*r**2, Rational(4,3)*pi*R**3)
cylinder
\begin{equation*} \displaystyle \pi h r^{2} = \frac{4 \pi R^{3}}{3} \end{equation*} 
Eq(R, solve(cylinder, R)[0])
\begin{equation*} \displaystyle R = \frac{\sqrt[3]{6} \sqrt[3]{h r^{2}}}{2} \end{equation*} 
cons, vars = constants(solve(cylinder, R)[0])
N(cons) * vars
\begin{equation*} \displaystyle 0.90856029641607 \sqrt[3]{h r^{2}} \end{equation*} 
graphviz.Source(dotprint(solve(cylinder, R)[0]))

Pyramid

A Pyramid has a height, \(h\), and a base width and length, \(w\), \(l\). This is the general case of a pyramid with a base that’s not square. As we’ll see, a square base is not a signficant simplification.

pyramid = Eq(Rational(1,3)*h*w*l, Rational(4,3)*pi*R**3)
pyramid
\begin{equation*} \displaystyle \frac{h l w}{3} = \frac{4 \pi R^{3}}{3} \end{equation*} 
Eq(R, solve(pyramid, R)[0])
\begin{equation*} \displaystyle R = \frac{\sqrt[3]{2} \sqrt[3]{h l w}}{2 \sqrt[3]{\pi}} \end{equation*} 
cons, vars = constants(solve(pyramid, R)[0])
N(cons) * vars
\begin{equation*} \displaystyle 0.43012700691405 \sqrt[3]{h l w} \end{equation*}

For a square-based pyramid, \(l=w\). Not a very intersting complication.

graphviz.Source(dotprint(solve(pyramid, R)[0]))

Additional Shapes and Difficulty Computations

We’ve got all of the necessary computations reduced to a \(R = S_s\sqrt[3]V\) kind of formula. We can use this to compute difficulty, as \(5R\).

Let’s summarize this in a tidy table we can use when designing spells with complicated areas of effect.

from IPython.display import display, Markdown
from contextlib import redirect_stdout
from io import StringIO

details = {
    'Hemisphere': ("radius $r$", hemisphere),
    'Cone': ("height $h$, base radius $r$", cone),
    'Cuboid': ("height $h$, width $w$, depth $d$", cuboid),
    'Cylinder': ("eight $h$, radius $r$", cylinder),
    'Pyramid': ("height $h$, base width $w$, base length $l$", pyramid),
}

buffer = StringIO()
with redirect_stdout(buffer):
    print("| Shape  | Measures                                    | Equivalent Sphere R     |")
    print("|--------|---------------------------------------------|-------------------------|")
    print("| Sphere | radius $r$                                  | $R = r$                 |")
    for shape, (arg, eq) in details.items():
        cons, vars = constants(solve(eq, R)[0])
        print(f"| {shape} | {arg} | $R = {latex(cons.evalf(2) * vars)}$ |")
    print()
    print(r"Difficulty is $5 \times R$.")

md = Markdown(buffer.getvalue())
display(md)
S hape Measures Equivalent Sphere R
Sp here radius \(r\) \(R = r\)
He misp here radius \(r\) \(R = 0.79 r\)
Cone height \(h\), base radius \(r\) \(R = 0.6 3 \sqrt[3]{h r^{2}}\)
Cu boid height \(h\), width \(w\), depth \(d\) \(R = 0 .62 \sqrt[3]{d h w}\)
Cyli nder eight \(h\), radius \(r\) \(R = 0.9 1 \sqrt[3]{h r^{2}}\)
Pyr amid height \(h\), base width \(w\), base length \(l\) \(R = 0 .43 \sqrt[3]{h l w}\)

Difficulty is \(5 \times R\).