It’s been a long time since I’ve just had a ramble on about assorted semi-related matters but that’s what’s in store today.

I’ll move from topic to topic with what may at times seem gay abandon, but it will all come together in the end.

This is all stuff that’s derived from, or been inspired by, last week’s article.

We start with this:

How Large Is A (Cut) Gemstone?

Let’s say you have a gemstone of a known value. How large is it? I’ll work with a Ruby, but throw in what you need to work out other gemstones along the way.

You see, I had listed as a treasure in an adventure a ruby worth $25,000 USD 2023 dollars, but was concerned that it might not be large enough to suit the purpose, which was as the centerpoint of a turban.

I was picturing in my mind something about the size of a hen’s egg.

Weight (in Carats)

From last week’s post, there’s the rough estimation,

Value = value of an equivalent 1-carat gemstone × size squared.

The words ‘equivalent’ are important because they take every variable other than size out of the equation.

We start, then, with a standard gemstone of 1 carat size. If it’s particularly brilliant or whatever, we simply apply a value factor to take that into account.

Last week, I listed the base value (in 2023 US dollars) of a typical 1ct Ruby as $4625

I didn’t want this ruby to be all that exceptional in any other way, so I’ll worth with that figure.

Value = 25000 so

weight squared is 25000 / 4625 = 5.4054 so

weight in carats is the square root of this, or 2.325.

Weight in grams

A metric carat is 0.2 g, so

2.325 × 0.2g = 0.4645 g.

Volume

To get the volume, I need to know two things: the volume of the shape (when cut), and the density of ruby.

Density of major gemstones:

Diamond = 3.5-3.53 g / cm-3
Ruby = 4.02 g / cm-3
Sapphire = 3.98 g / cm-3
Emerald = 2.69 g / cm-3
Jade = 3.3 – 3.38 g / cm-3

I got this information from Rocks And Co dot com, where they list a vast number of precious and semi-precious varieties of stone, their hardness, their refractive index (relates to brilliance) and – most importantly – their relative density.. From Actinolite to Zultanite, they’ve got it covered.

So, if my weight is 0.4645 g, and ruby has a density of 4.02 g per cubic cm, then

Volume = weight / density = 0.4645 / 4.02 = 0.115547 cm3.

Is that a lot? It doesn’t sound like it, but we’ll see.

Volume of a cut

Some cuts are easy to work out.

A square cut is length × breadth × width, less 3-10% for the beveling.

A sphere has the volume 4/3 pi time radius cubed.

A pyramid shape has the volume 1/3 times height times base area.

A cylinder is pi times radius squared times length.

A conical shape is 1/3 pi times radius of the base squared times height.

A hemisphere is half the volume of a sphere.

A prism is base Area time height.

Most of these are primary-school maths.

‘Diamond’ Cut

Some shapes are more complicated – the typical “diamond cut” for example:

Here, you have two cones with their bases touching, and the top lopped off the upper one. To get this volume you need:

• the volume of the cone defined by r and h1,
  to which you have to add
• the volume of the cone defined by r and (h2+h3),
  from which you need to subtract
• the volume of the cone defined by r and h2
  – except that the r of this cone will be smaller, as you can see.
• As a rough rule of thumb, use (r-h2).

so that’s 1/3 pi times (h1 times r^2) + (h2 times r^2) – (h3 times (r-h2)^2).

It would be nice if we could assume symmetry, so that h1=h2+h3, but I don’t think we can. In fact, I think h1 is less than the other two combined, but that’s just an artists’ impression.

In the diagram, I deliberately made the top and bottom symmetrical and the angle of the facet ‘ring’ seems too flat to me – it should be more vertical to look right.

Technically. too, you shouldn’t use the “radius” (r) at all – as the bottom part of the diagram shows, you should use the average of r1 (what we’ve been calling r), measured along the long side to the corner of each facet, and r2, which measures to the flat surface in between – but as you can see, the error is so negligible that it’s not worth the effort.

Hearts

Hearts are another problem. There are three ways to do hearts: the complicated and messy way, the even more complicated and messy way, or the easy way.

Method 1
The complicated and messy way is to do a cone with it’s base the radius of the heart at the point of it’s greatest width, divide the volume by the width radius and multiply by the thickness radius to account for the flattening, then add two halves of identical spheroids at the top – radius of 1/2 r, two hemispheres.

This will; work, more or less – most heart cuts actually have a curved surface and not the straight lines of a cone, so your calculated value will be a little low, but it’s a workable ballpark number.

Method 2
This uses overlapping spheroids. There’s a ring around the conical shape that shouldn’t be there, but that’s a minor problem and could even represent a correction of the difference between a heart and a flattened cone.

It’s working out the overlap between the two hemispheres that’s the real headache. Of course, it can be done – but it’s a complicated process involving the distance between the two radii relative to the size of those radii, and frankly, not worth the effort.

Method 3
The third image might look more complicated, but it’s not. I started with the ellipse (actually, it’s closer to a circle, but that may not always be the case), selected half of it, and skewed it vertically.

If you remember the formula for the area of a regular trapezoid – which is a square that’s been skewed – it’s still the base times the height. Skewing a shape doesn’t add area, it just moves it around.

And the same should be true of a 3D, flattened, heart cut gem – you aren’t changing the volume, just moving it from the bottom of the heart to the top.

Which means you can use the formula for the volume of an elliptical spheroid (which I haven’t given yet), take the measurements of your radius from the widest part of the heart, and treat it simply as a distorted shape.

Spheroids

Which brings me back to what I needed for my Ruby.

An elliptical spheroid basically has three measurements that we care about: the thickness, the width, and the height. Half of each of these gives the radius along each of the three axes.

And then it’s a simple variation on the volume of a sphere:

Volume = 4/3 pi r1 r2 r3

…except that I used a, b, and c.

There will be some minor errors from the mitering of the edges to create the shape, but they are too small to worry about.

The size of a gemstone, elliptical cut – reality check

To start with, I decided to work out the size if the ruby was cut as a sphere.

V = 0.115547 = 4/3 pi r^3

Multiply both sides by 3/4 and you get

0.08666025 = pi r^3.

Now divide by pi:

r^3 = 0.08666025 / pi = 0.0275848

My calculator has x^3 and inverse-x^3 functions, which makes the result easy:

r = 0.302 cm = 3.02mm.

Diameter = 6.04 mm,

or about 3.8 sixteenths of an inch. Call it 1/4 of an inch, more or less, and you won’t be far wrong. About the size of a pea.

The size of a gemstone, elliptical cut – assumptions about shape

That reality check gave me some perspective. A ruby the size of what I was imagining would be about 100 times the size and 10,000 times the value – it would literally be worth millions.

On that basis, I could take a more realistic view of the results that I was about to get.

First, I needed to make some assumptions about the shape – specifically, I decided that (as a rule of thumb), the height would be 1 1/2 times the width. Or, if you prefer, the width would be 2/3 of the height. These are dimensions that I know look good, not to spindly or too fat; a definite oval shape.

With that done, I could simplify the formula, because A × B would be 1.5 B^2:

V = 4/3 × pi × 1.5 × C × B^2, or B^2 = V × 3/4 × 2/3 / pi C = 1/2 V / pi C
actual B × 1.5 = actual A

The size of a gemstone, elliptical cut – first calculation

I started with a C of 1.5mm, because that’s about as thin as I though a gemstone could possible get.

B^2 = 1/2 × 0.115547 / pi / 0.15 = 0.1226 cm^2

B = 0.35 cm = 3.5 mm

A = 1.5 × 3.5 = 5.25 mm

Gemstone is 3 mm thick, 7 mm wide, 10.5mm tall
Gemstone is approx 1/8th of an inch thick, a little over 1/4 of an inch thick, and 7/16ths of an inch tall.

That suddenly seemed disappointing, for the obvious reason that these were radii, not diameters. They were describing a gemstone 3mm thick, 4 mm wide, and 6 mm tall. (1/8th thick × 3/16 wide × 1/4 ” tall).

The size of a gemstone, elliptical cut – further calculations

To get a feel for how changes in thickness were affecting the size, I decided to try thicknesses of 2 and 2.25 mm.

C = 2 mm, so

B^2 = 1/2 × 0.115547 / pi / 0.2 = 0.09195 cm^2

B = 0.3 cm = 3 mm

A = 1.5 × 3 = 4.5 mm

Gemstone is 4 mm thick, 6 mm wide, 9 mm tall
Gemstone is approx 3/16, still about 1/4 of an inch wide, and 3 eighths of an inch tall.

C = 2.5 mm, so

B^2 = 1/2 × 0.115547 / pi / 0.25 = 0.073556 cm^2

B = 0.271 cm = 2.71 mm

A = 1.5 × 2.71 = 4.07 mm

Gemstone is 5 mm thick, 5.42 mm wide, 8.14 mm tall
Gemstone is still approx 3/16 of an inch thick, 3/16 wide, and 5/16 inches tall.

Finally, having realized that I was thinking of the C radius as thickness and not the C diameter, I tried a thickness of 0.8 mm:

C = 0.8 mm, so

B^2 = 1/2 × 0.115547 / pi / 0.08 = 0.22987 cm^2

B = 0.48 cm = 4.8 mm

A = 1.5 × 4.8 = 7.2 mm

Gemstone is 1.6 mm thick, 9.6 mm wide, 14.4 mm tall
Gemstone is 1/16 of an inch thick, 3/8 wide, and 9/16 inches tall.

Halving the thickness (more or less) certainly let the gemstone grow in its other dimensions.

At this point, I reflected that the smaller size might be more appropriate to the game circumstances anyway.

The size of a gemstone, elliptical cut – final calculation

Ultimately, and with that in mind, I decided that the proportions of the first calculation were actually closest to what I wanted, anyway – maybe a thickness of 2.5mm instead of 3 would be better, but it’s on the right track.

C = 1.25 mm, so

B^2 = 1/2 × 0.115547 / pi / 0.125 = 0.14712 cm^2

B = 0.3836 cm = 3.836 mm

A = 1.5 × 3.836 = 5.75 mm

Gemstone is 2.5 mm thick, 7.672 mm wide, 11.5 mm tall
Gemstone is 1/8 of an inch thick, 5/16 wide, and 7/16 inches tall.

If you have trouble working with the above, there are two traps that kept catching me out – convert mm to CM by dividing by ten in the B^2 line (so 1.25 becomes 0.125), and remember that these are giving radius values and not diameters, you have to double them to get the physical height, width, and thickness of the gemstone.

Special acknowledgment to Ginfab.com and their mm-to-fractions-of-an-inch converter, which helped get me out of a real tangle!

How much is a gp worth?

Of course, for this to be really useful, we need values not in 2023 US dollars, but in D&D / Pathfinder Gold Pieces!

Dimensions of common coins

There are two real-life gold-coin denominations that I can recall of the top of my head – florins and doubloons. So let’s start with them.

A florin is a 28.5mm diameter British and Australian coin. The weight was 11.3 grams. But it turns out that they were silver, not gold. Scratched.

Doubloons were 2.32″ (59mm) in diameter and
0.13″ (3.3mm thick). That’s about twice the size in both dimensions to what I always thought of as a GP, to be honest, but it’s a start.

From last week’s post: 1 cubic inch of gold is 313.54573 g if pure, and worth about 20,067 USD 2023. I’m going to start by assuming this isn’t pure gold – maybe it’s 15 ct, maybe even 18 ct.

volume = area × height = pi r squared × height

= pi × (2.32/2)^2 × 0.13 = 0.55 cubic inches.

0.55 cubic inches = 172.45 g = 2023 USD $11036.85 if pure.
= 82777.64 if 18-ct
= 6898.03 if 15-ct.

That seems a bit big to be a common currency as ubiquitous as the GP is. By a factor of at least 10 and probably 20 or 25 or more.

Applying a factor-of-twenty

Because it’s the value in the middle of my guesstimate, let’s do a factor of 20 scale reduction first.

To keep life (relatively) simple, let’s assume that radius and thickness are both reduced by the same amount.

That means that:

volume / 20 = old measurements × f^3

0.55 cubic inches / 20 = 0.0275 cubic inches.
= pi × (f × 2.32 / 2)^2 × 0.13 × f
0.0275 / pi = 1.16^2 × f^2 × 0.13 × f
f^3 = 0.55 / pi / 1.16^2 / 0.13 = 0.05004 (i.e. 1/20th)
f = 0.3685

So r = 1.62 × 0.3685 = 0.59697 ”
thickness = 0.13 × 0.3685 = 0.047905″.

About 1.2 inches across and less than 1/20th of an inch thick, eh? I think the thickness is now too small, and the radius still too large.

Tell you what, let’s try a dime:

What if a dime were made of gold?

There’s a purpose to this. For a start, a dime is roughly the size that I think of when I think GP. It For another, it’s a fairly ubiquitous coin size – not far away from the Australian 10-cent piece for example. And for a third, if I can say that a GP is the size of a dime, but the weight of × dimes, everyone will be able to relate to the numbers.

Weight (real dime) = 2.268g
Diameter = 17.91 mm = 1.791 cm
Thickness = 1.36 mm = 0.136 cm

Volume in cubic cm = pi × r^2 × ht
= 3.1416 × 1.791 ^2 × 0.136 = 1.3705 cm3.

Density of pure gold = 19.3 g / cm3

so weight = 26.45065 g
= 11.66 regular dimes.

Value, 24 ct = 26.45065 / 313.54573 × 20067 = $1692.85.
Value, 18 ct = $1269.636.
Value, 15 ct = $1058.03

But 15 ct gold would also reduce the weight – not all the way to 62.5%, because whatever impurities there are have to weigh something, but that’s a start. 26.45065 × 62.5% = 16.53165625 g = 7.29 regular dimes.

Okay, so that’s got a reasonable physical size sorted. The value still seems high, by a factor of about 10.

Next, we need to think about relative value.

What’s a GP worth?

We need to think about the sheer number of gp in circulation. 1,000 gp is a high price but it’s well within the bounds of ‘reason’ for an adventurer to be carting that around with him. Or 10,000. or 100,000.

100,000 gp, if each one weighs 26.45065g = 2,645,065g or 2,645.065 kg. For those who aren’t thinking metric, that’s 5831.37 lb.

Okay, yeah. 2.917 US tons. I can see that… not happening.

But, if they weigh something more than 16.53 g instead of 26.45, that’s a weight of 3644.6 lbs. That’s better, a step in the right direction.

But it’s still arguing in favor of an even lighter, smaller gp – maybe by a factor of 3 or so.

But that not only screws up the value of the ‘golden dime’ but the nice simple comparison that we derived from it.

The size thing with the dime works so perfectly that I’m loathe to mess with it. But that means that we’re left with adulterated (lighter) coins, which will reduce the value but not by as much as seems necessary, or abandoning the dime standard.

If the value is reduced to 1/3, so is the gold content, and so is the volume that it makes up within the coin. So to keep the dimensions, we need 2/3 of the coin to be made of something lighter. Like 0 grams, but that’s going a bit far.

How about reducing the gold content 75% from the 15-ct mark – that would be 4-ct gold.

The table I presented last time doesn’t go anywhere near that far. But the difference per carat is about 4.17%, so 4-ct gold would be:

Value, 24 ct = $1692.85.
Value, 4-ct = $282.37.

Wt, 24 ct = 26.45065g
Wt of gold in 4-ct = 4.412g = 1.945 regular dimes.

Density Of Adulteration

So, let’s do this: make up the weight to a number of standard dimes in total, derive the density required to keep the same physical coin size, and see if there’s something on the periodic table or list of common compounds that matches.

3 standard dimes = 6.804g, -4.412g gold = 2.392g other.
volume = 1.3705 cm3, – 4.412g gold / 19.3, = 1.1419 cm3.
density required = 2.392 / 1.1419 = 2.095 g cm-3.

4 standard dimes = 9.072g, -4.412g gold = 4.96g other.
volume = 1.1419 cm3 (this isn’t going to change).
density required = 4.96 / 1.1419 = 4.081 g cm-3.

5 standard dimes = 11.34g, -4.412g gold = 6.928g other.
volume = 1.1419 cm3
density required = 6.928 / 1.1419 = 4.88 g cm-3.

6 standard dimes = 13.608g, -4.412g gold = 9.196g other.
volume = 1.1419 cm3
density required = 9.196 / 1.1419 = 8.05 g cm-3.

7 standard dimes = 15.876g, -4.412g gold = 11.464g other.
volume = 1.1419 cm3
density required = 11.464 / 1.1419 = 10.02 g cm-3.

8 standard dimes = 18.144g, -4.412g gold = 13.732g other.
volume = 1.1419 cm3
density required = 13.732 / 1.1419 = 12.0256 g cm-3.

9 standard dimes = 20.412g, -4.412g gold = 16g other.
volume = 1.1419 cm3
density required = 16 / 1.1419 = 14.012 g cm-3.

10 standard dimes = 22.68g, -4.412g gold = 18.268g other.
volume = 1.1419 cm3
density required = 18.268 / 1.1419 = 16 g cm-3.

Okay, let’s get busy with my SI chemical data…
…
…
…

Results, in order of increasing density:

Zinc 7.14
Tin (white) 7.3
Iron 7.86
Cadmium Oxide 8.1
Bismuth Oxide 8.9
Nickel 8.90
Copper 9.0
Lead Oxide, Red 9.1
Silver 10.5
Mercury Oxide 11.1
Lead 11.4
Tungsten Oxide 12.1
Tantalum 16.6
Tungsten 19.3

8.1 to 8.05 required is a fair match. In fact, it’s the only SINGLE additive that comes close.

If I assume a mixture, all sorts of possibilities arise.

(x%) d 1 + (100-x%) d2 = target…

Let’s take that old standby, Copper + Lead:

10% copper + 90% lead = 0.9 + 10.26 = 11.16
20% copper + 80% lead = 1.8 + 9.12 = 10.92
30% copper + 70% lead = 2.7 + 7.98 = 10.68
40% copper + 60% lead = 3.6 + 6.84 = 10.44
50% copper + 50% lead = 4.5 + 5.7 = 10.2
60% copper + 40% lead = 5.4 + 4.56 = 9.96
70% copper + 30% lead = 6.3 + 3.42 = 9.72
80% copper + 20% lead = 7.2 + 2.28 = 9.48
90% copper + 10% lead = 8.1 + 1.14 = 9.24

7 standard coins requires 10.02.

10.02 = an × 9 / 100 + (100 – a) × 11.4 / 100
1002 = 9 a + 11.4 (100 – a) = 9 a + 1140 – 11.4 a
1002 – 1140 = 9 a – 11.4 a
1140-1002 = 11.4 a – 9 a = 2.4 a
138 = 2.4 a
a = 138 / 2.4 = 57.5

So 57.5 % copper + 42.5% lead will nail the target exactly.

The limits of speculation

so, let’s be clear about this – having demonstrated that a solution is possible, that doesn’t mean that I am suggesting people adopt it. I’m certainly not recommending it.

This is simply a demonstration of how to get your own answer. Personally, I would have loved an 8-dime solution – dividing by 7 is a lot harder to do in your head! And maybe using Iron (low density) and Tungsten or Tantalum (high density).

Let’s take stock

So, 7 standard dimes = the weight of a single (adulterated) 4-ct golden dime – as a working solution – and has a value in USD $2023 of $282.37.

The value is still too high, but we’ve pushed the purity option about as far as is possible. We need something else to get that down to somewhere around the $20-25 range, ideally, but I could live with $30 or even $50.

Maybe I should attempt to nail that target a bit more precisely as my next step.

A really good meal for 1 will cost about 1 gp (I call this the “Good Meal” standard). It will also cost $50-100 modern Australian dollars at the extreme, but $18-$30 is more reasonable, and that’s what I’ve been using as my yardstick. If I convert those numbers into US dollars at today’s exchange rate, I get:

$100 AUD = $63.76 USD
$50 AUD = $31.86 USD
$30.AUD = $19.13 USD
$18 AUD = $11.48 USD

But, maybe I can get more traction by going down a step?

There are (the last time I checked) 10 silver pieces in 1 gp, and a moderately-decent meal will cost 1-5 of those (I guess I’ll have to call this the regular fair standard)!

So that’s 0.1 gp – 0.5 gp.

The sort of meal I’m talking about is $10 – $20 AUD. That’s a Chinese takeaway, or McDonalds, or Kentucky Fried Chicken, or a Pizza – something along those lines.

0.1 gp = AUD $10-20 = USD $6.38-12.76;
1gp = USD $63.80 – $127.60 USD.

0.25 gp = AUD $10-20 = USD $6.38-12.76;
1gp = $25.52 – $51.04 USD.

0.5 gp = AUD $10-20 = USD $6.38-12.76;
1gp = $12.76 – $25.52 USD.

The highest value – $127.60 USD – is 2.2 times our current gp value.
The lowest value – $12.76 USD – is 22 times our current value.

That says that we need to cut the value of the gp maybe 5-10 fold.

So, what else affects the value of gold?

Supply vs demand.

What happens to a commodity when there is a significant oversupply? The price crashes. I’ve already commented that there seems to be a lot more gold around in a fantasy game, so it would not be unreasonable to expect that to have some devaluatory effect.

The trick is getting the oversupply right. In the previous post, I talked about Tiger-eyes:

Tiger-eyes (sister to Cats-eyes) were highly esteemed from 1880-1890 and once sold for $6 a carat or about $11,200 per pound. Two speculators independently flooded the market in 1890, causing the price to crash to just 25 cents a pound – a loss of 99.997% of their value.

99.997% is too extreme for what we want. I don’t know if any economists have modeled the impact on price of oversupply, and I’m not sure what to search for in order to find out. But that’s not the sort of problem that’s ever stopped me before…

Hmmm… Back again.

“Competition among producers to increase sales leads to downward pressure on prices. You can show excess supply on a graph as the horizontal distance between the demand and the supply curves at a price above the equilibrium price.” — Pricing in mass markets, Kwanhui Lim

…
…
…
… I may be interpreting this all wrong, but:

Demand Curve – the change in demand with rising price
Supply Curve – the change in supply with rising price? rising cost? not sure.
Equilibrium point – the point at which they meet.

….more searching

(10 minutes later) …okay, I think I;me getting the hang of this. The theory seems to be that there is an optimum price at which demand matches supply, all other things being equal. If you have an oversupply situation, competition between sellers to be the one that actually makes a sale drives the price down. Most of the example graphs I’ve been seeing have used straight lines for both, but a few have used curves, which makes more sense to me.

The supply curve runs from bottom left (low units, low expenditure) to top right.

The demand curve runs from top left to bottom left. The higher the price, the fewer units people will want to purchase because of the expense. The converse is that the more units are available, the lower the price that people are willing to pay.

.

Wikipedia’s example chart shows an increase in demand for whatever reason – it shifts the demand line up and to the right. The point at which the supply and demand lines cross gives the relevant price and quantity that will be sold. The increase in demand means that both more units will be sold and the price per unit will go up.

If you put the price up without increasing supply, it shifts the supply curve up and to the left, so the intersection point will be at a higher price, but with fewer units sold.

Logically, increasing supply without increasing demand produces a supply curve that is down (lower prices) and to the right (more units available to sell, the definition of an oversupply situation). The two shifts might not be the same, I’ve assumed that they follow the demand line, so the greater the oversupply, the lower the impact per unit on price – diminishing returns, in other words, .I’ve modified the original graph to show the result, as I understand it.

One point to note is that eventually, your supply curve will drop below a price value of zero If you increase the oversupply even further, eventually the intersection point with the demand curve will get down to the irreducible minimum as well.

Salting the tail

So, if gold is more abundant, in a game universe, the value of gold will drop. How much, isn’t all that important – in fact, we’ve already defined (roughly) how much we want it to be.

It’s possible that we could even beat a hasty retreat and revert to 24-carat pure gold coins, and do it ALL with oversupply. That would probably be more in keeping with the vast amounts of gold floating around the game universe, anyway.

The combination means that we can set the modern value-equivalent of the gold piece to whatever we want so long as it’s less than what it would be worth in 2023.

Some mathematical trickery

There are three ways to get a value in between two known limits. The first and most obvious one is to average the two. In this case, we have 2.2 and 22 – which gives an average of 12.1.

When we’re talking about values that are to be multiplied by something else, though, I’ve often found that an alternate method yields better results: Multiply the two together and take the square root.

2.2 × 22 = 48.4, square root is 6.957.

A variation is to take that number away from the maximum result.

22 – 6.957 = 15.043.

I always try out the different results to see what best fits.

Base value of a gp = $282.37.

Oversupply: option 1: 282.37 / 12.1 = $23.34.
Oversupply: option 2: 282.37 / 6.957 = $40.59.
Oversupply: option 3: 282.37 / 15.043 = $18.77

How do these compare to our target values?

$23.34:

The Very good meal standard:
$50 – $100 AUD = $31.86 – $63.76 USD – not even close
$18 – $30.AUD = $11.48 – $19.13 USD – a little above

The regular fair standard:
$63.80 – $127.60 USD.- completely out of the ballpark.
$25.52 – $51.04 USD. – nowhere near it.
$12.76 – $25.52 USD – very close.

$40.59:

The Very good meal standard:
$50 – $100 AUD = $31.86 – $63.76 USD – right in the middle
$18 – $30.AUD = $11.48 – $19.13 USD – too high

The regular fair standard:
$63.80 – $127.60 USD- not high enough
$25.52 – $51.04 USD – right on target
$12.76 – $25.52 USD – too high

$18.77:

The Very good meal standard:
$50 – $100 AUD = $31.86 – $63.76 USD – definitely not
$18 – $30.AUD = $11.48 – $19.13 USD – just scrapes into the range

The regular fair standard:
$63.80 – $127.60 USD – nowhere near high enough
$25.52 – $51.04 USD – not high enough
$12.76 – $25.52 USD – right in the middle

I make that 2 votes for $23.34, two for $40.59, and two votes for $18.77.

An appropriate result?

In a way, I guess that could be seen as an appropriate result. So many other aspects of the game world are left to the individual GM to determine, surely something so fundamental should be one of them?

So here’s the upshot: decide for yourself what the “exchange rate” between gp and USD, even if you use it for no other purpose than this.

Once you have that number, you can take a gem result – be it 1 gp, 10 gp, 50gp, 500gp, 5000gp, or 50,000 – and get the USD equivalent. Once you have that, and the information contained in this post and last week’s, and you can directly determine the weight and dimensions of the gem.

Special gems are easily handled simply by multiplying the base value of a 1ct gem with those special characteristics by an appropriate factor – times 1.5? Times 10? Times 100? – it’s up to you.

What’s more, if you do a little prep work and save the results, you can define a standard gem – in your game world – and up-scale or downscale it as needed.

Value / base 1-ct value will always give you the square of the number of carats. So if the dice say a 10,000 gp gem, you can simply divide by your base value, take the square root, and you have the number of carats.

An Afterthought

Carats are divided into points, each 1/100th of a carat. If the smallest gem worthy of the name is (say) 10 points, or 10% of a carat, and that’s worth 1 gp, you’re ready to go.

A 10,000 gp gem? Divide by 0.1 and take the square root. Instant gem size: 316.23 carats.

Weight and size are right behind.

I was intending to deal with a third item – the value of an Ebony statue – but I’m completely out of time. So that will have to wait for another occasion. And maybe I’ll talk about Persian Rugs at the same time, who knows? And Tapestries. Tapestries would be good.

There are a great many objects of value out there. The better that you can define them, the more real you can make them seem to your players – no matter how outrageous those definitions might be.

It’s inevitable – there will be a “The Value Of Material Things” number III!