The Value Of Material Things V

Three examples of Babylonian pottery, from an exhibit in the Oriental Institute Museum, University of Chicago, Chicago, Illinois, USA. This work is old enough so that it is in the public domain. Photography was permitted in the museum without restriction. Photograph taken by Daderot and used under the Creative Commons CC0 1.0 Universal Public Domain Dedication License, via Wikipedia.
Welcome
…to the latest installment of the “Value Of Material Things” series. Today’s article will focus on Pottery & Ceramics, and Peridot, with room for a sideline along the way if anything crops up.
Although Jade and Ivory are also laid out and ready to write, I lost a lot of time over the last week-and-a-half, so I’ve decided to hold off on them until Part VI.
This series is being driven by game prep; I simply collate my notes for that prep, organize them a bit to get a post with (hopefully) less work involved, and then expand on that beginning. Beginning below, I’m going to index the different goods and commodities that have been detailed.
AN INDEX
General Subjects:
- Depreciation & Appreciation – Part III -> General System (Part I)
- Generic Valuation System – Part I (The Asset Valuation Worksheet)
- Demand, Under-supply & Oversupply – Part II
Specific Subjects:
- Diamonds Value – Part I (The Asset Valuation Worksheet)
- Diamonds, Size, Cut – Part II
- Earthenware & Ceramics – Part V (this post)
- Earthenware Engine – Part V (this post)
- Ebony – Part III
- Emeralds Value – Part I (The Asset Valuation Worksheet)
- Emeralds Size, Cut – Part II
- Frames – Xmas post
- Gemstones (Cut) – Part II
- Gold Bricks – Part I (The Asset Valuation Worksheet)
- Gold Pieces – Part IIa correction to the content in Part II
- Metal Cups and Plates – Part V (this post) -> Earthenware Engine (ditto)
- Moonstones – Part IV
- Olivine – Part V (this post)
- Peridot – Part V (this post)
- Rock Weight – Part I (The Asset Valuation Worksheet)
- Rubies, Value – Part I (The Asset Valuation Worksheet)
- Rubies, Size, Cut – Part II
- Sandstone – Part V (this post)
- Sapphires, Value – Part I (The Asset Valuation Worksheet)
- Sapphires, Size, Cut – Part II
- Semi-Precious stones – Part I (The Asset Valuation Worksheet)
- Silk – Part IV -> Textiles Engine (Part III)
- Spices – Part I (The Asset Valuation Worksheet)
- Tapestries, Tapestry Engine – Part III
- Textile products, Textile Engine – Part III
- Tesserae and other Mosaics – Part IV -> Tapestry Engine (Part III)
Still Pending: Jade, Ivory, Historical Artifacts I (Ancient) & II (Recent)
Pottery & Ceramics
Right off the top, if there are any gems or jewels involved, value them separately.
The main thrust here will be toward clay-based pots, but at the end of the section I’ll talk a bit about Metal pots.
There are four numbers that are central to the value and other traits of pottery, ceramics, and other such. These are the external surface area, the external volume, and the internal volume (subdivided into below waterline and above). In some cases, the internal surface area will also be needed.
The easiest way to show you how to do this is to actually do one.
- Protruding Lip: There are always 3 elements to analyze – the bottom, the top, and the chord that connects them. In this case, the visible ‘space’ in the chord confirms the impression formed by the square-framed section: the average is going to be a lot closer to the lower value than the higher one. How much so? Estimate the height of the square-framed section, which extends up to the point where the edge becomes more horizontal than vertical – it’s about 80% of the height of the whole section, so we’re talking an 80:20 ratio, or 4:1:
Average radius = (A + 4 × B) / (4 + 1 = 5).
- Bowl Shape: Again, it’s the chord that is the most revealing, showing that for almost the entire section, the width is greater that the minimum. So, starting at the wide end, estimate how far down it is that the curve becomes more horizontal than vertical. problem: in this case, it doesn’t. So estimate how far before the curve begins to bend significantly. When I do that, I get 60%, so a 60:40 split is correct, or 3:2:
Average radius = (2 × A + 3 × B) / 5
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- Compound Curve: This shape has a virtually vertical section, and if accuracy was needed, it should be split into three smaller sections. But if a rougher answer will suffice, I use two characteristics: the relative position of the vertical section (is it closer to the longer radius, the shorter radius, or in the middle, as in this case?); and the top and bottom positions of the section (does it force a sharper curve on one side or the other)? In this case, you could argue that the bottom (shorter radius) part of the chord is showing a hair more substance than the top is showing empty air, but for my money they are pretty close to identical. So I would not bias the result at all:
Average radius – (A + B) / 2
- Compound Curve #2: This is a far more subtle and tricky shape to assess, because most of it is straight edge at an angle. Looking at the chord, the part at the bottom seems twice the size of the much flatter curve at the top, so there should be a bias toward the longer radius. Lacking the usual diagnostics, I use the point where the chord crosses the edge of the pot section, and find it to be very close to the half-way mark. Unless this section (and the whole pot) is absolutely huge – 20″ or more – the error from using an unbiased number will be so small as to be trivial. So again:
Average radius – (A + B) / 2
- Assume that the cross-section is circular.
- Assume that the change in thickness is even and uniform.
- Average (a) and (b). Convert to a radius.
- Find the average cross-sectional area.
- Count the length of the handle around the outside. I’ve used 1/2″ increments marking each one off in red, but normally I would just point and count. You usually only need to be approximate, but if precision is required, repeat for the inside and average the two.
- Average (a) and (b):
a + b / 2 = (1 + 3/8) / 2 = 11/8 / 2 = 11/16 = 0.6875 in.
- Convert to a radius:
0.6875 / 2 = 0.34375 in.
- Find the average cross-sectional area.
A = π × r^2 = π × 0.1181640625 = 0.371 in^2.
- Get the length – I’ve already done this, it’s 7″.
- Salt (Sodium Chloride) – density 2.16 g/cm3
- Lead (II) Oxide – density 9.53 g/cm3
- Barium Carbonate – density 4.29 g/cm3
- Strontium Carbonate – density 3.5 g/cm3
- Chromium (III) Oxide – density 5.22 g/cm3
- Uranium Oxide – density 10.97 g/cm3
- Vanadium Oxide – density
- Sodium Oxide – density 2.27 g/cm3
- Potassium Oxide – density 2.35 g/cm3
- Calcium Oxide – density 3.34 g/cm3
- Tin Oxide – added to other glazes to make them more opaque – density 6.95 g/cm3
- Zirconium Oxide – added to other glazes to make them more opaque – density 5.68 g/cm3
- Iron Oxide – colorant, added to give color – density 5.24 g/cm3
- Copper Carbonate – colorant, added to give color – density 4 g/cm3
- Cobalt Carbonate – colorant, added to give color – density 4.13 g/cm3
- Silica Glass, also known as fused quartz if derived from melting quartz crystals, is used where high service temperature, very high thermal shock resistance, high chemical durability, very low electrical conductivity, and good ultraviolet transparency are desired.
- Soda Lime Silicate Glass, as suspected, is what most of us think of when we hear the word glass – used in windows, bottles, etc.
- Sodium Borosilicate Glass is very resistant to thermal shock, and is used to make reagent bottles, flasks, lighting, electronics, cookware, and so on.
- Alkali silicate is used for precision optics and lenses.
- Aluminosilicate glass is resistant to acids, is scratch resistant, and is commonly used to make mobile phone screens and the like.
- If the earthenware is glazed – even if it’s a solid single color – n =1. Otherwise, use n=0.
- If the earthenware is glazed in a fancy or complex decorative pattern, add 0.5 to 1.5 to n.
- If the earthenware has a complicated relief sculpture or carving incorporated, add 0.5 to 1.5 to n.
- if the earthenware has embedded contents – gems, leaves, flowers, whatever – add 0.5 to n.
- If there is an especially complicated process involved, add 0.5 to n.
The Pot
This is the basic pot that we’re going to be working with. There are also a couple of handles but they won’t be important for a while, so I left them out. As you can see, it subdivides into four sections of different dimensions.
1. RadiI
Step 1 is to convert those (measured) diameters into radii.. Going from top to bottom:
12.4 / 2 = 6.2
11.3 / 2 = 5..65
14 / 2 = 7
5.76 / 2 = 2.88
2a. Exterior Volume, Top Section
To get the volume, we turn this into a cylinder by averaging the top and bottom radius. Note that this only works if the sides are basically straight lines, or close to it. If it’s a curve, I bias the result to get something close enough – see the “sidebar” below.
Sidebar: Pot Curves & Biased Averages
The diagram to the left illustrates four very common situations.
But our basic pot has none of that going on, because I have deliberately simplified it.
2b. Top Section
Average radius = 6.2 + 5.65 / 2 = 11.85 / 2 = 5.925 inches.
Cylindrical Area = π × 5.925 ^2 = π × 35.105625 = 110.29 in^2.
Volume = area × height = 110.29 × 4 = 441.16 in^3.
2c. Second Section
Average radius = 5.65 + 7 / 2 = 12.65 / 2 = 6.325 inches.
Cylindrical Area = π × 6.325 ^2 = π × 40 = 125.66 in^2.
Volume = area × height = 125.66 × 3.89 = 488.82 in^3.
2d. Third Section
Average radius = 7 + 5.76 / 2 = 12.76 / 2 = 6.38 inches.
Cylindrical Area = π × 6.38 ^2 = π × 40.7044 = 127.877 in^2.
Volume = area × height = 127.877 × 14 = 1790.28 in^3.
2e. Fourth Section (the base)
Average radius = 5.76 + 5.76 / 2 = 5.76 inches, obviously.
Cylindrical Area = π × 5.76 ^2 = π × 2.4 = 7.54 in^2.
Volume = area × height = 7.54 × 1 = 7.54 in^3.
2f. Total:
Add these four results up and we get 2727.8 in^3.
3. Internal Volume
To determine the internal volume, we need to know the thickness of the pot. I generally work on the principle of “an eighth for six” – i.e. for every 8″ of height, the thickness has to increase an eighth of inch. But because this is just a guideline, and potters would prefer to err on the side of safety, I’m quite happy to round it.
3a. Thickness
Our basic pot is a total of 22.89 inches tall; that says 3.815 eighths, or 0.477 inches thick. Close enough to 1/2 an inch.
The top will be considerably thinner than the base, for stability – probably 1/2 our average value, or 1/4″ in this case. The bottom will be 1.5 or 2 times our average, i.e. 0.75-1 inch. If the pot is reasonably solid and “chunky”, I’d go for the higher value; if it’s more delicate and better-crafted, I’d go for the lower value; and the rest of the time, I’ll pick something in between.
This time, I’ll do that and call it 0.8 inches.
3b. Base
I’m going to say that the base is solid clay, so it has no interior volume to contribute.
3c. Waterline
For convenience, I’m also going to assume that the waterline, i.e. the height to which liquid rises when the pot is “full”, is 4″ from the rim, i.e. the wasp-waist.
That shrinks the total internal height to 14 + 3.89 = 17.89″. Half-way along that is where our average will occur – 8.945 inches. That’s part-way down the long side of the pot.
3d. Differential
It’s also necessary to consider the difference from the average of the two extremes, because sometimes the average isn’t actually the average!
0.5 – 0.25 = 0.25″
0.8 – 0.5 = 0.3″
Aha! Just as I told you – the midpoint of the thickness is not where we though it was; it’s 0.25 : 0.3 or 5/11ths of the total length from the thin end.
5 / 11 × 17.89 = 8.13″ from the top. And, at that point, the thickness will be ( 0.8 + 0.25 ) / 2 = 1.05 / 2 = 0.525″ thick.
3e. Thickness at Waist
All this matters because we need to determine the thickness at the ‘bend’ in the pot, 3.89 inches down.
Thickness at waist = [ 3.89 / 8.945 × (0.525-0.25) ] + 0.25
= [ 0.435 × (0.275) ] + 0.25
= [ 0.119625 ] + 0.25
= 0.37″.
3f. Internal Structure
What we end up with, then, is what is shown in the above diagram, in which:
a = 0.25″
b = 0.25″
c = 0.37″
d = 0.8″
3g. Top Section (air)
6.2 – a = 6.2 – 0.25 = 5.95
5.65 – b = 5.65 – 0.25 = 5.4.
Average radius = 5.95 + 5.4 / 2 = 11.35 / 2 = 5.675 inches.
Cylindrical Area = π × 5.675 ^2 = π × 32.205625 = 101.177 in^2.
Volume = area × height = 101.177 × 4 = 404.71 in^3.
3h. Second Section (contents)
5.65 – b = 5.4.
7 – c = 7 – 0.37 = 6.63
Average radius = 5.4 + 6.63 / 2 = 12.03 / 2 = 6.015 inches.
Cylindrical Area = π × 6.015 ^2 = π × 36.180225 = 113.66353 in^2.
Volume = area × height = 113.66353 × 3.89 = 442.15 in^3.
3i. Third Section (contents)
7 – c = 6.63
2.88 – d = 2.88 – 0.8 = 2.08
Average radius = 6.63 + 2.08 / 2 = 8.71 / 2 = 4.355 inches.
Cylindrical Area = π × 4.355 ^2 = π × 18.792225 = 59.0375 in^2.
Volume = area × height = 59.0375 × 14 = 826.525 in^3.
4. Capacity
Capacity Total = 442.15 + 826.525 = 1268.675 in^3.
Conversions:
1000 in^3
= 554.113 fluid ounces
= 17.316 US quarts
= 4.329 gallons
= 16387.1 cm^3
= 16.38 liters
= 0.01638 m^3
So, 1268.675 in^3 = 20790 cm^3 = 20.79 liters = 21.97 US quarts = 5.49 US Gallons.
5. Total interior Volume
Total interior volume = 1268.675 + 404.71 = 1673.385 in^3.
5a. Difference, External – Internal Volumes
External Volume Total = 2727.8 in^3
Internal Volume Total = 1673.385 in^3
Difference = 1054.415 in^3
5b. Handles
If I add in an estimated volume for the handles, that difference is all pot, enabling a determination of the (empty) weight.
There are two common types of handle shape, and innumerable more decorative choices. I’ll get to those in a moment. First, the common types, the C-shape and the Question Mark:
Sidebar: The handles process
The illustration above not only depicts the basics of both, it illustrates how to deal with them.
The C-shape is simplest of them all. It starts with a thick end (a) and ends with a thinner end (b). For smaller pots, (a) might be thin and (b) thick, but the principle remains the same:
The second is also a C-section, but there is now a bulge at the top of the c; b is now the thickest point. That means that this has to be handled in two lengths – (a) to (b) and (b) to (c). The method remains the same.
The third figure shows a question-mark handle. This is an order of magnitude more complicated, as the profile shows. (b) is still the thickest point; (c) is now the thinnest, and there is a section that is uniformly that thick. At the bottom of the question-mark, the handle starts to thicken again.
The first step in processing such a handle is to establish the length at which the ‘recurve’ starts (c), because it’s at that point that what was ‘inside’ becomes ‘outside’. Aside from that, the process is the same, just with more lengths.
5b cont.
In the case of our example pot, I went with the simple C-shape. The (a) diameter is 1″ thick, the (b) is 3/8″ thick, and the total length is 7″.
Multiply by the number of handles – in this case, 2 – and you’re done.
2.597 × 2 = 5.194 in^3.
6 Weight
Clay has a density of 0.08686 lb / in^3 after it is fired, making it’s shape as permanent as it gets.
So, total the volume of the handles and the difference between exterior and interior volumes, and multiply by that density to get the weight of the pot:
Difference = 1054.415 in^3
Handles (2) = 5.194 in^3
Total = 1059.609
Weight (empty) = 1059.609 × 0.08686 = 92 lb = 41.73 kg
Weight (full, water) = approx 0.0163871 m^3 × 1000 kg / m^3 + 41.73
= 16.39 + 41.73 = 58.12 kg (128.13 lb).
Pots with loaded weights of more than ~100kg that are NOT made of metal might as well have no handles because the handles will break under such loads or be too thick to be useful under most circumstances.
All about Glazing
Ceramic Glazes can give a glossy finish to pottery and other earthenware. Aside from being decorative, it makes the pot stronger and impermeable to water.
Most glazes change color when applied to earthenware that it then fired for a second time. To prevent sticking, a small portion (frequently on the base) is left unglazed, though special refractory “spurs” may be used in modern times instead.
Colors in glazes usually come from various kinds of wood ash or from metal oxides. They can be applied by spraying, dipping, trailing, or the brushing on of an aqueous suspension of the glaze. The water content then dries off, and the earthenware is ready for its second firing.
More exotic techniques can be used – I have seen curtain lace used to give a textured look to a glaze, and a mulberry leaf used for the same purpose.
Small leaves and objects can be embedded in the clay and affixed with a glue and then covered in the glaze; if it is sufficiently transparent, the object will show through, but be tinted by the glaze to match the rest of the surface.
Overglaze decoration is applied on top of a fired layer of glaze, and generally uses colors in “enamel”, essentially glass, which require a second firing at a relatively low temperature to fuse them with the glaze. Because it is only fired at a relatively low temperature, a wider range of pigments could be used in historic periods. If an Overglaze is intended, the first firing is referred to as the “Ghost Firing”.
There are also numerous more advanced techniques.
Metallic Oxides
Metals are heavy. Their oxides tend to be less so, but even a thin coating can still add significantly to the weight of an object. Some of the oxides commonly used are:
Typically, such compounds would only make up 2-4% of the glaze, though sometimes more is required to achieve the color density that is desired.

This describes the methodology. Image by ceramicscape, licensed as Attribution-NonCommercial-NoDerivs via flickr.
Glaze Colors
I tried to find a glazing color reference chart, but the only ones on offer are not free of copyright restrictions. Recipes and blends are a closely-guarded secrets within the industry. But the below will give you some idea, and clue you in a little more as to the process.

This shows the methodology being put into practice. The tile on its own to the left is a reference, either unglazed or glazed in a very transparent material. Image by ceramicscape, licensed as Attribution-NonCommercial-NoDerivs via flickr.
If you really want to view a color chart, Here’s one on Pinterest, and another by Ceramic Glazes of Australia can be found on this site. But since neither of them is specifying the compounds and blends, they are of value only in displaying a sample of the range of color options.
7 Surface Area
To estimate how much glaze adds to the weight, we need to know the surface area of the pot and the handles. There are also a couple of crucial decisions to be made along the way. Once we know the surface area, we simply multiply by the depth of the glaze to get the total volume; deduct the % of the glaze that’s a metallic compound, multiply that by the density of glass to get weight, then do the same for the metallic compounds added to the glaze.
Simple, right?
The key formula is
C = 2 × π × r for the circumference of a circle,
then multiply by the length to get the surface area.
7.1 First Critical Question: Inside, Outside, or both?
Outside is for decoration and durability. Insider is for waterproofing and durability. As a general rule, inside won’t have much in the way of fancy colors (except maybe at the very top) and will have a relatively thin glaze (0.5 – 1 mm), while outside glazes may be multiple layers and quite thick – but I’ll get to that in a moment. First, the broader question: inside, outside, or both?
7.2 Second Critical Question: Full or partial?
There’s more nuance to this question than might initially appear.
Interior: I’ve already suggested that below the first couple of inches, the glaze might be treated differently. Why bother coloring deeper than that if no-one’s ever going to see it?
Exterior: This where artistic style enters the picture in a big way. Don’t worry, you don’t have to be a designer / artist yourself; you just need to think about what someone who is might do.
Maybe only the top half of the pot is colored – cream-colored, maybe. Maybe there’s a pattern in the second band on top of that cream color – charlie-brown zig-zags.
Or maybe the artist is using the natural brown color of the pot to represent the trunks of palm trees, adding green-glazed leaves at the top and silhouetting the trunks in a dark brownish green.
Ultimately, you don’t care; you just need the answers to two questions for each band: % coverage.
7.3 Third Critical Question: Depth.
Thin glazes are 0.1 – 0.5 mm in thickness. Thick glazes are going to be 2-2.5 mm thickness, maybe up to 4mm in exceptional cases.
What we’re actually doing with these critical questions is defining the glaze parameters that are to be applied in the actual calculations.
So, fish or cut bait time: Our example pot has heavy glaze (2mm) on the top section interior, and on the top 2 exterior sections. It has a thin glaze (0.2 mm) over the rest of the pot, both interior and exterior. The handles have an extra-thick glaze (3mm).
7.4a Exterior, Top section
Average Radius = (6.2 + 5.65) / 2 = 11.85 / 2 = 5.925 in.
Circumference = 2 × π × 5.925 = 37.228 in.
Height = 4 in.
Surface Area = 37.228 × 4 = 148.912 in^2.
Depth = 2 mm = 0.07874 in.
Volume of Glaze = 0.07874 × 148.912 = 11.725 in^3.
7.4b Exterior, Second section
Average Radius = (5.65 + 7) / 2 = 12.65 / 2 = 6.325 in.
Circumference = 2 × π × 6.325 = 39.741 in.
Height = 3.9 in.
Surface Area = 39.741 × 3.9 = 154.99 in^2.
Depth = 2 mm = 0.07874 in.
Volume of Glaze = 0.07874 × 154.99 = 12.204 in^3.
7.4c Exterior, Long Section
Average Radius = (7 + 2.88) / 2 = 9.88 / 2 = 4.94 in.
Circumference = 2 × π × 4.94 = 31.039 in.
Height = 14 in.
Surface Area = 31.039 × 14 = 434.546 in^2.
Depth = 0.2 mm = 0.007874 in.
Volume of Glaze = 0.007874 × 434.546 = 3.472 in^3.
7.4d Exterior, Base
Radius = 2.88 in.
Circumference = 2 × π × 2.88 = 18.096 in.
Height = 1 in.
Surface Area = 18.096 × 1 = 18.096 in^2.
Depth = 0.2 mm = 0.007874 in.
Volume of Glaze = 0.007874 × 18.096 = 0.142 in^3.
7.4e Interior, Top section
Average Radius = 5.675 in (from 3g).
Circumference = 2 × π × 5.675 = 35.657 in.
Height = 4 in.
Surface Area = 35.657 × 4 = 142.628 in^2.
Depth = 2 mm = 0.07874 in.
Volume of Glaze = 0.07874 × 142.628 = 11.231 in^3.
7.4f Interior, Second section
Average Radius = 6.015 in (from 3h).
Circumference = 2 × π × 6.015 = 37.793 in.
Height = 3.9 in.
Surface Area = 37.793 × 3.9 = 147.393 in^2.
Depth = 0.2 mm = 0.007874 in.
Volume of Glaze = 0.007874 × 147.393 = 1.161 in^3.
7.45 Interior, Long section
Average Radius = 4.355 in (from 3i).
Circumference = 2 × π × 4.355 = 27.363 in.
Height = 14 in.
Surface Area = 27.363 × 14 = 383.082 in^2.
Depth = 0.2 mm = 0.007874 in.
Volume of Glaze = 0.007874 × 383.082 = 3.016 in^3.
7.4h Interior, Base
This is a little different since it is the base – we need the area of the flat disk. Also, because glaze is applied as a liquid, there tends to be more of it settle at the bottom of the interior, so it may be double the thickness expected.
Average Radius = 2.08 in (from 3g).
Surface Area = π × 2.08^2 = 13.592 in^2.
Depth = 0.4 mm = 0.015748 in.
Volume of Glaze = 0.015748 × 13.592 = 0.214 in^3.
7.4i Handles (each)
Average Radius = 0.34375 in (from 5b).
Circumference = 2 × π × 0.34375 = 2.16 in.
Length = 7 in.
Surface Area = 2.16 × 7 = 15.12 in^2.
Depth = 3 mm = 0.11811 in.
Volume of Glaze = 0.11811 × 15.12 = 1.786 in^3.
Well, that’s not quite correct – there will be no glaze where the handles contact the pot, and the pot should also have no glaze where the surface is interrupted by the handles. But the resulting error is so small that it’s swamped by other factors, like unevenness in the glaze depth. So forget it.
7.4j Total Surface Area
This is just a matter of adding up the values you’ve just determined. You may be wondering why you need it? I’ll get to that in a minute.
148.912 + 154.99 + 434.546 + 18.096 + 142.628 + 147.383 + 383.082 + 13.592
+ (15.12 × 2 handles) = 1473.469 in^2.
7.4k Total Glaze
Same deal.
11.725 + 12.204 + 3.472 + 0.142 + 11.231 + 1.161 + 3.016 + 0.214
+ (1.786 × 2 handles) = 46.737 in^3.
That’s a significant amount of what is basically glass.
Except that this isn’t especially useful – we really need to subtotal these values by glaze thickness. And it would be useful to convert them to cubic cm at the same time –
1 in^3 = 16.3871 cm^3
2mm: 11.725 + 12.204 + 11.231 = 35.16 in^3 = 576.17 cm^3
0.2mm: 3.472 + 0.142 + 1.161 + 3.016 = 7.791 in^ = 127.67 cm^3
0.4mm: 0.214 in^3 = 3.51 cm^3
3 mm; 2 × 1.786 = 3.572 in^3 = 58.53 cm^3
7.5 Weight Of Glaze
To get the weight of the glaze, we have to multiply by the density – which is assumed to equal the density of glass plus the coloring agents.
Except that it’s not that simple.
Sidebar: Glass Ain’t Glass
According to ResearchGate, there are 5 different types of glass, all with different densities. They are Silica Glass (2.20 g/cm3), Soda lime silicate glass (2.49 g/cm3), Sodium borosilicate glass (2.23 g/cm3), Alkali silicate (3.02 g/cm3) and Aluminosilicate glass (2.64 g/cm3). Wikipedia helpfully advises that ‘ordinary window glass’ has a density of 2.5 g/cm3, meaning it’s probably Soda Lime Silicate glass.
So, to make sense of this, I need to look up each of these and see what they are used for.
So, none of those sound quite what we’re looking for.
Density of Ceramic – about 6 g/cm3
From that value alone, the correct answer is “none of the above”.
7.5 cont.
6 g per cubic cm is a very interesting number because it’s heavier than almost all the oxides listed earlier.
But, again, it’s not that simple: The additives (essentially) dissolve into the glass, altering it’s chemical structure – that’s how they achieve colors in the glaze. Some undergo chemical reactions along the way. The bottom line is that the metallic additions add to the density that’s already there.
As a general rule of thumb, and fir the purposes of the example in particular, I’m going to simplify that list of compounds to a generic one with the density of 4 g/cm3. But I’m also going to assume that there’s a color intensifier of some kind – so that’s a second substance, also of 4 g/cm3, for a total of 8 g/cm3.
I’m further going to assume 6% is the average concentration – while some of it may be 7.5% or more, most of the time, it’s going to be less than that. But these assumptions can and should be varied when that suits the item being created more – in other words, use the generic rule unless you feel an exception is warranted..
7.5a Doped Density
The thick parts of the density are therefore going to be
6 g /cm3 + 6% × 8 g/cm3 = 6 + 0.48 = 6.48 g / cm^3.
7.5b Applying Density
The 2mm and 3mm thickness will be at the higher density, the rest will be at the lower.
2mm + 3mm volumes = 576.17 + 58.53 = 634.7 cm^3;
x density 6.48 g / cm^3 = 4112.856 g = 4.113 kg = 9.0676 lb.
0.2mm + 0.4mm volumes = 127.67 + 3.51 = 131.18 cm^3;
x density 6 g / cm^3 = 787.08 g = 0.787 kg = 1.735 lb.
Total weight of glaze = 4.9 kg = 10.8026 lb.
7.6 Total Weight
In (6), we determined that the clay of the empty pot weighed 92 lb or 41.73 kg. We can now add the weight of the glaze to get a total of 102.8026 lb (46.63 kg).
It still holds 16.39 kg (36.134 lb) of liquid when full, for a total of 63.02 kg (138.9366 lb).
According to ThingsOnAScale.com, 60kg is about the same as 20 standard house bricks, so throw in 1/3 of a small bag of cement – or an extra brick – and you have some idea of the weight of the pot when it’s full – of liquid of water density. Earth is heavier, I suspect.
8 Value
It’s really hard to get a solid understanding of how size and multiple firings combine to determine the price of a piece of earthenware. Ultimately, I had to blend my own high-school experience with pottery and hints from multiple sources to devise a system for approximating the cost in terms of the equivalent number of hours of labor by a skilled artisan, and that’s what I am presenting in this section.
8.1 Surface Area × n
Once you have a total for n, multiply it by the surface area in square inches, as determined in 7.4j above:
To our example:
– Glazed, check. n=1. Decorative pattern – fairly minimally, +0.5. Complicated sculpture, no. Embedded contents, no. Complicated process, no. so n=1.5.
1.5 × Surface Area = 1.5 × 1473.469 in^2 = 2210.2035.
8.2 Add the external volume in in^3
2210.2035 + 2727.8 = 4938.0035.
8.3 Divide by 600.
4938.0035 / 600 = 8.23.
8.4 Take the cube root
8.23 ^ (1/3) = 2.019.
This is the number of hours of skilled labor involved in the creation and decoration of the object. It assumes that some costs, like firing, will be amortized (spread over) many objects – dozens or even hundreds – and so this will be a relatively trivial contributing factor.
8.5 Skilled Labor
Now for the trickiest part of this process –
(A) what is the going rate for skilled labor at the time of creation of the earthenware? I’m not talking about doctors or archaeologists or rocket scientists, nor the owners of large corporations – but a plumber or electrician should give a meaningful comparison.
(B) What is the minimum skill necessary to qualify for the description “skilled labor” as used in (A), in your game system? and
(C) What is the approximate skill of the person who created the specific item in question (plus 1/5 the skill of their supervisor, if they are an apprentice).
Note that (B) and (C) are pretty much game-system agnostic, because it’s the ratio that we are going to need. But there may need to be tweaks if they have a non-zero base level – see the example below.
Once A, B, and C are determined, Base Cost = Hours × A × C / B
I’m going to pluck a date out of the air at random 1985. In 1985, Glaziers earned about US $13.80. Plumbers earned US$15.50. So there’s a roughly 11% error. That said, glaziers are not exactly the same as potters. So either might be correct. I’ll use the $15.50 just to be consistent within the process as I’ve described it.
B – From a base of -150, in my superhero game system, 11% represents the minimum skill at which one can earn a basic living from that skill, and 21% is an expert (out of 150). Splitting the difference gives 16% as a reasonable threshold.
C – Another arbitrary number – let’s pick 18%.
Hours × A × C / B = 2.019 × 15.50 × (18+150) / (16+150)
= 31.4545 × (168) / (166)
= $31.67.
8.6 Profit Margin, Costs, Taxes, etc
Lastly, we need to adjust that for all the other cost components that go into the final price of an object.
Profit Margin: +50%
Administrative costs +25%
Material costs +10%
Studio Space +10%
Kiln +10%
Kiln operating costs +15%
Licensing: +5%
General Overheads +10%
Federal Taxes /0.65
State Taxes /0.85
Sales Taxes +10%
Totals: 50+25+10+10+10+15+5+10 = +135%
$31.67 × 2.35 = $74.4245
$74.4245 / 0.65 / 0.85 × 1.1 = $148.18.
So there it is – the value of this specific pot, made in this specific time and place.
But wait – we’re not done yet!
Metal Cups & Plates
The same process, with a few tweaks, can also be used to calculate the price of metal cups and plates.
- Metal costs more as a material through most of history than earthenware. There’s a reason why “China” is sometimes used as an alternative name for the latter, and why we’re still using it more than 6000 years after unlocking the secrets.
- But Metal is a lot more robust and resilient, and so lasts a lot longer, so people are happy to pay a bit extra.
- There are three basic phases to the production of metalware:
- Pattern, in which a single master copy is produced;
- Pre-production, in which that master copy is transformed into multiple molds;
- Replication, in which those molds, in batches, are used to make multiple copies of the master, in metal, at the same time.
On top of that, there may be a fourth phase, decoration.
- Being able to produce the item in batches of 4/6/8/10/12/15/16/20/25 amortizes the cost of the first two steps over the whole production run, and that’s assuming that the molds can’t be reused to make still another batch. If you are going to end up with 50 copies of the end product, you can afford to spend 20x as long in those early phases and still make out like bandit.
- Metal is generally so thin that the internal volume is essentially the same as the external volume. There can be exceptions.
- The softer the metal, the thicker it will have to be for strength, as a rule of thumb. So Copper will be thicker than Pewter, which will be thicker than Bronze, which will be thicker than Brass, which will be thicker than Iron.
- This rule of thumb doesn’t generally apply to precious metals, where the expense makes thickness undesirable, so delicacy rather than strength is the usual goal.
- As a result, for as long as man has employed metalworking, people have been looking for ways to cheat. Plating, which puts a thin layer of precious metal over the top of a base material, for example.
- Decoration usually consists of an enamel which is then fired to make it glass-like, the mounting of gems, and/or sculpted features. In modern times, you can add powder-coating and electroplating to that list. But they all get worked out the same way as ceramic glaze on earthenware, except for the gems, which should be valued separately.
I can’t think of too much more that you would need to know in order to apply the earthenware valuation techniques to metal objects.
Bonus Extra: Sandstone
I had need to work out how much Sandstone costs in order to value a block of the material.
As an everyday construction material (non-fibrous), it will gain value with age at a fairly glacial rate, I think, but – particularly if it comes from a historically-important or famous building – it would nevertheless appreciate in value. And that was the circumstance I faced – a partial block of sandstone allegedly from the Lighthouse at Alexandria.
There are just enough breadcrumbs of plausibility that the story could be true, so I needed to work out values either way. And a third set for an in-between state in which the story acquired increased plausibility (but still, no proof).
Illustrating The Stone Block
Illustrating the block was fun. Because it had exactly the shape that I wanted, I started with a block of granite, shown in a quarry. After cutting away everything that wasn’t part of the stone block, and rotating it to sit “flat”, I removed all the dark inclusions that make granite what it is, which left only the white parts – with the all-important shading and shadows that defined the shape.
Then I found a flat sheet of sandstone, rotated copies of that face in three dimensions to create the visible faces of the block, cut them to fit the shape defined by the once-was-granite, and assembled them against a background of marble tiles and wood-grain.
Finally, I took the once-granite, made it the top layer, and duplicated it – the second-from the top layer set to addition, and the top layer to multiplication. Next, for each face, I tweaked the opacity of the layers and also the brightness and contrast of the underlying sandstone until each of the faces looked natural, given the lighting.
A few additional tweaks and enhancements to the shadows and shape completed the block itself.
An excerpt from the resulting image, reduced in size to fit on the left, and actual size on the right. At the top is part of the original base image.
The last step was to add a shadow so that it looked solidly grounded (that’s the job of a shadow in artwork and illustration). It took only about twice as long to do as this recounting of the process, and looks unbelievably real. Call it 40 mins from start to finish.
1 Dimensions
Most importantly, in the course of achieving this, I had to decide on the size of the block in order to get the texture. The measurements I came up with were 12.5 × 10.5 × 8.5 inches (note that I didn’t size the tiles to match because I didn’t think of using them as a scale until afterwards).
2 Volume
That gives an approximate volume of 1115.625 in^3. But, to be useful information in value terms, I needed it to be in one of two scales: cubic cm, or cubic ft.
The latter is achieved by dividing by 12^3 = 1728, and gives a volume of 0.6456 ft^3.
3 Density
Those scales weren’t chosen capriciously; those are the scales given for the weight of rock (Bonus Content, the Asset Valuation Worksheet 2.0).
Sandstone is listed as 143.6 lb / ft3.
4 Weight
So,
Wt = Size × Density = 0.6456 × 143.6 = 92.70816 lb.
Again, this is not the most useful unit, because construction materials are typically sold by the ton, at least in the US.
1 short ton = 2000 lbs.
Again, converting to the more useful units gives,
Wt = 92.70816 / 2000 = 0.04635408 tons.
5 The Scales Of Money
Adjusting one era’s valuations to those of another is always a challenge. In a lot of cases, you can use Inflation between years A and B as a guide, and I have a bookmarked site that I use for that very purpose – https://www.in2013dollars.com/.
But there’s a complication – labor pay scales and the time required to complete a task; automation and mechanical tools have had a big impact.
The value of goods needs to be broken into two parts of this is likely to be a relevant factor, and each part adjusted separately. So let’s do that:
Basic Cost = Reference Value [year A] – Labor Component
Labor Component adj for labor efficiency, adj for labor pay-scale
Value [year B] = Basic Cost + Adj Labor Component, then adj for inflation
Take your time and understand the logic and the process, because there are all sorts of other values that can be derived if needed – for example, how much would “X” cost if it were still being extracted by “old” methods but at modern pay-scales?
Labor Efficiency
How many more or less man-hours are required to produce a certain amount of “X”?
Ideally, you want to represent this as a ratio so that you can simply multiply by the Labor Component.
The more closely the technology of A matches that of B, the more you can ignore the labor efficiency element, setting the ratio to 1.
Labor Pay-scale
This can actually require more detective work than most people would expect.
It’s relatively easy to find out what the modern pay-scale is for a particular job – it can be a lot harder finding out what it was in a particular past date, assuming that the profession even existed.
Quite often you’ll need to hand-wave an approximate value out of whole cloth and just live with it.
That all means that you should select the reference year (A) price after considering the target year (B) and the social and technological impact of the interval (B to A).
There will be any number of times when you can’t track down the required information anywhere. It’s for this reason that some writers build up collections of Almanacs from every year that they can find.
Short-cutting the process
Of course, if you can find actual prices for “X” on the target date, all of this complication goes away. This is so beneficial that it’s almost always my method of first resort.
But, often, it’s not that easy, and you will need to make adjustments.
6 Back to the Sandstone
So, my target dates were 1938, 1860, and 1852.
I couldn’t find prices for any of these, so I knew I would have to get a bit messier in my approach.
I did track down a price from 1950 in USD: $2613.67 / short ton.
While there would have been some advances between 1950 and 1938, I felt that they would be close enough in technology that I could ignore the labor component and simply apply inflation, so that became my reference year.
The older target dates would be a different story, but once I had one, they would be close enough that I could again step directly from that to the other..
6a Value 1950 -> Value 1938
$100 in 1950 was about $58.51 in 1938 (it’s more commonly expressed the other way around, but this is the more useful version for my purposes).
So $2613.67 [USD 1950] × 0.5851 gives me the 1938 value = $1529.26 [USD 1938] / short ton.
But, because this is the date in which the current campaign is set, I also worked out the value in a whole bunch of other volume units – just in case. I’ll get to those in short order.
6b Value 1938 -> Value 1860: Labor Efficiency
In 1850, they would have had jackhammers and powered saws and the like, and those would also have been around in the late 1930s; and none of it would have existed in 1860.
Or would it? They had steam power, back then. The first steam engines date all the way back to 1712, and James Watt had achieved success with his much more refined version in 1776.
There is a popular myth in some places that steam was less powerful than internal combustion. That wasn’t true at the time; they had something close to Parity. But the internal combustion engine doesn’t have to continually stop for water and coal, and doesn’t need one or more people feeding the fuel to the engine; that was the big advantage over steam, and it proved decisive.
It’s fair to expect that there would have been improvements in materials and engineering, and that these would have at least doubled the efficiency of mechanical tools over the 70-odd years in between, and probably quadrupled it in terms of the man-hours required to extract a ton of rock. When you have a gang like that, you usually need a supervisor; and you need a couple of people fueling and monitoring the steam engine. 4+3=7, so as a rough rule of thumb, I would suggest a factor of 7 in terms of labor efficiency.
Hold up, not so fast – there would normally be two people working with those pieces of heavy equipment, one operating the machine and the other loading the product. So that should be 8 to 2 – or 4 to 1.
6c Value 1938 -> Value 1860: Labor Costs
The base pay rate in 1938 relative to 1860 is x2.5 for unskilled labor, x5 for skilled labor. I defined those rules of thumb back when I was starting the current project. Is mining going to be higher or lower than those, and which one to use?
There’s a lot of nonsense that occasionally gets bandied about concerning how “unskilled” blue-collar workers are relative to white-collar occupations. How “uneducated” they might be is a different story, but I have a high regard for how “skilled” carpenters, plumbers, electricians, and yes, miners (and farmers) actually are. There are exceptions, but that was definitely a factor in setting those general pay scales – some very highly-skilled occupations like jewelers and senior engineers might get 10 or even 20 times the unskilled pay rate, but the doubling at the base level seems closer to accurate to me.
So:
1938 – x2.5 pay scale, but only 1/4 as many people needed.
1860 – 2/5 pay scale, but 4 times as many people employed.
6d Value 1938 -> Value 1860: Labor Component
Sometime after 1950, technology improved to the point that the labor costs of extraction become a relatively negligible component. Let’s set that point to 1980. Back in 1860, over a century earlier, the labor costs would have been the dominant component. Somewhere in between, then, the situation would have been one of approximate equity between the two.
1980-1860 = 120; 1/2 of 120 is 60; 1860+60 = 1920.
1938-1920 = 18 years. 18 years is 30% of 60 years.
Rule of thumb result: 1.3:1, non-labor to labor.
$1529.26 [USD 1938] / short ton in 1938 therefore divides into a labor component of 1529.26 / 2.3 = $664.90, and a non-labor component of $1529.26 – 664.90 = $864.36.
6e Value 1938 -> Value 1860: Adjust Labor Component
Apply the conversion rate determined earlier:
$664.90 [USD 1938] × 2/5 × 4 = $1063.84 [USD 1860, 1938 dollars]
Add the non-labor component back in: $1063.84 + $864.36 = $1928.20.
6f Value 1938 -> Value 1860: Inflation
Another value that I took careful note of is the effect of inflation from 1860 to 1938 – x1.7595.
$1928.20 [USD 1860 in 1938 dollars] / 1.7595 = $1095.88 [USD 1860].
So that is the price of Sandstone per short ton in 1860, as best as I can estimate it.
6g Value 1860 -> Value 1852
There won’t be much change in the labor component over such a short range. Simply apply inflation. Of course, that means I need to know what the inflation was, from 1852 to 1860.
Plugging the numbers into the website I linked to earlier, I get:
$100 in 1860 is equivalent in purchasing power to about $92.77 in 1852
That’s a ratio of 0.9277 – so multiply the 1860 price by that to get the 1852 price:
$1095.88 [USD 1860] = $1095.88 × 0.9277 = $1016.65 [USD 1852] per short ton.
One-stop conversion
You can carry this one step further and work out the overall conversion factor to go from one date to another:
The procedure is simple: Divide a base value (I’ve used $1000) by the reference price and multiply it by the final price for each target date.
$1000 [USD 1950]
= $585.10 [USD 1938]
= $419.29 [USD 1860]
= $388.974 [USD 1852].
The problem with doing so is that you can forget that these values are specifically relevant to sandstone only. Other industries will have different labor components, and hence be impacted differently. At best, this is an approximation.
6h Sandstone Prices
What’s generally going to be more useful is a bunch of prices by different weights.
USD 1950 $2613.67 / sht ton
= USD 1938 $1529.26 / us ton = USD 1860 $1095.88 / us ton = USD 1852 $1016.65 / us ton
= USD 1938 $829.31 / m^3 = USD 1860 $594.29 / m^3 = USD 1852 $551.32 / m^3
= USD 1938 $23.4834 / ft^3 = USD 1860 $16.8284 / ft^3 = USD 1852 $15.6117 / ft^3
= USD 1938 $1.686 / kg = USD 1860 $1.2082 / kg = USD 1852 $1.12085 / kg
= USD 1938 $0.76463 / lb = USD 1860 $0.54794 / lb = USD 1852 $0.508325 / lb
Sidebar: A brief notation about nomenclature
You will have noticed that I am very careful to specify exact “units” when it comes to dollar valuations. I don’t know of a faster expressway to confusion than simply labeling everything $, on the presumption that the text will provide the context needed to understand which $ you’re talking about.
It gets even more confusing if you are also converting from one form of $ to another – from US currency to Australian dollars, for example. That’s actually a really bad example, to be honest, because in 1966, we shifted from Australian Pounds (and shillings and pence) to a decimal dollar. Just to add one more layer of confusion.
Peridot
I’ve written before about the collapse in the semi-precious stones market – essentially, a couple of speculators (independently, but at the same time) took it upon themselves to flood the market with Tiger-Eyes, then selling at about $6 a carat ($11,200 / lb). Predictably, the market collapsed (down to $0.25 / lb) – and is only now starting to recover, after adjusting for inflation. That process is not yet complete, which is to say that in the commodities directly affected, the inflation-corrected value is not yet back to what it was before this incident.
ALL the semi-precious stones were affected to some degree. Those which suffered the least didn’t fall in value as far, but their price recovery has been just as slow.
Which brings me to Olivine and Peridot.
Olivine
Olivine is what gets found in the wild. When they are cut into gemstones, they are known as Peridot. Olivine/Peridot are found all over the place – Pakistan, China, Africa, Europe, and the US to name just a few locations. The color ranges from a milky green to yellow to yellow-green to an emerald-green.

The above image is a composite of several photos from Wikimedia Commons: Peridot2.jpg by Azuncha, Licensed under the Creative Commons Attribution-Share Alike 3.0 Unported License; Peridot_-_Mineral_Cabinet_(Arppeanum)_-_DSC05499.jpg, a photograph by Daderot, taken of an exhibit in the Arppeanum, Helsinki, and dedicated to the public domain under the Creative Commons CC0 1.0 Universal Public Domain Dedication License; Chryzolit_(perydot,_oliwin)_-_Kohestan,_Pakistan..jpg, a photograph by Elade53, who has released it into the public domain; Forsterite-Olivine-es21a.jpg, photographed by Rob Lavinsky, iRocks.com and licensed under the Creative Commons Attribution-Share Alike 3.0 Unported License; Forsterite-Olivine-tmu14d.jpg, photography by Rob Lavinsky, iRocks.com and licensed under the Creative Commons Attribution-Share Alike 3.0 Unported License; and two excerpts from Olivine-Hawaii.jpg, photographed by Alain COUETTE and licensed under the Creative Commons Attribution-Share Alike 3.0 UnportedLicense.
As required by the licenses, this compound image is also licensed under that license and can be used or adapted provided the attributions remain intact, and any resulting image is also licensed in this way. Between them, they give a fair representation of the color range of Olivine. Note that relative sizes have been adjusted to be roughly identical (except in the case of the two excerpts).
Peridot
In general, Peridot are yellow-green in color when cut. Below, I have compiled examples of 4 popular cuts for the gemstones, and one final example in which a relief has been carved onto the blank face of the gemstone.

The image above is a composite of four images from Wikimedia Commons. The Emerald-cut image, Gemperidot.jpg, was photographed by Humanfeather (aka Michelle Jo) and released into the public domain; the Oval-cut and Teardrop-cut gemstones are from Peridot-China.jpg, photography by DonGuennie (G-Empire The World Of Gems) (German site) and licensed under the Creative Commons Attribution-Share Alike 4.0 InternationalLicense; the diamond-cut image PB030020_peridote.jpg was photographed by Ronald Werner and also licensed under the Creative Commons Attribution-Share Alike 4.0 International License; and the Peridot_ring_stone_MET_DP141725.jpg image, with it’s Roman carving (which dates from 100BC to 200 AD) was donated to Wikimedia Commons by the Metropolitan Museum Of Art (New York City, USA) as part of a project by the Museum, Licensed to the public domain under the Creative Commons CC0 1.0 Universal Public Domain Dedication. It is 2.2 cm (7/8 in) in length, and was donated to the Museum by John Taylor Johnston in 1881. It should be noted that the relative sizes of these gemstones are not going to be the same as depicted – in fact, I have deliberately attempted to resize them to approximately the same dimensions.
Because of the terms of the Attribution-Share Alike 4.0 International, this compound image is also bound by that license and can be used or adapted provided the attributions remain intact, and any resulting image is also licensed in this way.
One more Historical anecdote
Peridot has often been mistaken for emeralds, even to the point where a royal scepter – I forget which country it was – that was supposedly decorated with priceless emeralds was discovered, more than a century after the fact, to in be actually encrusted with Peridot (the scepter was not retired, but it has been used rather less frequently, since).
Valuing Olivine
The principle is simple: Volume to Weight to Cut to Base Value, factor in Quality to get Final Value, then convert to whatever units you require.
In practice, it’s a little more complicated. I covered a lot of the relevant ground in Part II of this series.
1 Image
To start with, you need an image of the collection (there’s never just one) to be valued. You can either gather together images of various pieces of Olivine and compile them into a set (which is what I did) or you can grab a sheet of paper and draw a lot of rough ovals and triangles and shapes, something like this (you don’t have to use a green pen or pencil, but it adds to the presentability of the image):
Next, grab a highlighter (preferably green) and without overthinking it, crudely sketch in a 3rd dimension to indicate depth, like so:
A lot of Olivine are rounded, so where the depth looks like it would be suitable, add a rounded end, like this:
You don’t have to then color in the gems, but it does make it more visually clear:
Finally, number them. Oh, and some sort of a scale will be handy.
You’ll notice that I’ve made NO effort to be neat and precise. Quick and dirty is the way to go.
2 List
Next, you make a list (leaving a few lines of space at the top of the page), with one line for each numbered example. Next to it, you write one of several possible values (leave lots of room for further info): Sq, Co, Eg, Cy, Py..
- Sq = square or rectangular.
- Co = cone-shaped
- Eg = Egg shaped. A lot of them should fit. This includes spherical ones.
- Cy = Cylinder.
- Py = pyramid-shaped. Or wedge shaped.
At the top of the page, in that black space, list the formulas for the volume of that shape:
- Sq = a b c.
- Co = π a b^2 / 6
- Eg = 4 π a b c / 24
- Cy = π a b c / 4
- Py = a B C / 4
You may note that some of these aren’t the same as you were taught in maths class. ‘a’ is always the longest dimension; b is the across, and c is the depth. In rounded shapes, these are diameters, not radii, so they have to be halved to get the radius. That halving is built into these formulas.
The other note is the capitals for Pyramid – they signify the length and height of the triangle at the base. Which is another way of saying the usual b and c, generally.
3 Long Axis
With a soft ruler of some kind (as used by dressmakers) the next step is to fill out the values for a, b, and c for each of the examples, using the scale given:
1 cy 3.5
2 cy 3.4
3 eg 3.8
…
…
…
19 sq 2.4
20 cy 3
4 Across Axis
After that, you measure the b dimension and add that to your list:
1 cy 3.5 1.2
2 cy 3.4 1.5
3 eg 3.8 1.4
…
…
…
19 sq 2.4 2.5
20 cy 3 1.2
5 Depth
Next, you estimate the c dimension (based on the b dimension that you just measured to get your 5th entry:
1 cy 3.5 1.2 0.4
2 cy 3.4 1.5 0.25
3 eg 3.8 1.4 0.6
…
…
…
19 sq 2.4 2.5 0.8
20 cy 3 1.2 0.6
6 Volume
Using the formulas at the top of your page, now work out the volumes and write them down. I recommend doing them by groups – all the “Sq” entries, then any Co, and so on.
1 cy 3.5 1.2 0.4 1.32
2 cy 3.4 1.5 0.25 1.00
3 eg 3.8 1.4 0.6 1.671
…
…
…
19 sq 2.4 2.5 0.8 4.8
20 cy 3 1.2 0.6 1.6965
7 Weight
Next, you need to work out a conversion to the units of Peridot density: Density 3.275 g/cm3.
Because the scale was in cm, our volumes are already in cubic cm, so no conversion is necessary. But if you’ve used a quarter-inch as your scale, it may not look all that different, but the numbers will.
Multiply the volume by the adjustment and then by the density to get a weight in grams, then multiply that by 5 to get carats, and write it down:
1 cy 3.5 1.2 0.4 1.32 21.615
2 cy 3.4 1.5 0.25 1.00 16.375
3 eg 3.8 1.4 0.6 1.671 27.35
…
…
…
19 sq 2.4 2.5 0.8 4.8 78.6
20 cy 3 1.2 0.6 1.6965 27.78
8 Quality Rating
This is easier with a photo than with a sketch, but it can be done.
Rate the quality of the gemstone from 0.5 to 5, where higher is better and write it down.
With a photo, look for dark inclusions – spots of non-gem. If there are none, it’s high-quality; if there are lots, it’s not.
With a sketch, look for pale areas, where your quick-and-dirty coloring job was a little too rough.
Or just roll randomly.
- d10 /2 works…
- …but I would prefer to use (d6+d5-1)/2 because the average will have a higher probability.
- I would also contemplate (d3+d4+d5-2)/2, which gives a reasonable bell curve centered on the average.
9 Weight Adjustment
If you want to get technical, you should now adjust the weight (but not the number of carats) to account for the inclusions. The formula is:
Added weight = (0.5 / quality) × volume × 162.5 × 453.6 / 28317
= (0.5 / quality) × volume × 2.6
But this is not really necessary.
10 Cut Size
Next, we convert the uncut gem size to the size of the cut gemstone.
There are two ways of doing this: by calculation, or by guesstimate.
Cut Size = dens × ({1 / [(6-quality) × volume / 10] – 0.2} × 0.6)+0.2
I expect most GMs to use the guesstimate approach.
0.5 = 20%
1 = 27%
1.5 = 33.3%
2 = 40%
2.5 = 47%
3 = 53.3%
3.5 = 60%
4 = 67%
4.5 = 73.3%
5 = 80%
‘
If you’ve assigned quality values that aren’t an even 0.5 , use the following:
+ 0.1 = +4/3 of a %
11 Value, [USD 2023]
$ Price 1 ct = 50 + (quality × 6)
Apply the size^2 rule
So, a quality 2 gemstone of estimated (cut) size 3.2 carats would have a value of $634.88 [USD 2024].
12 Currency Unit Conversion
Of course, for my purposes, I needed 1860 and 1938 values:
1860 = 2023 x1.1 / 500
1938 = 1860 × 46.6123
13 Cutting Costs
But that’s how much they will be worth when cut. You really need to take into account the cost of that cutting.
In 1938, a skilled gem-cutter would earn $15 / hour, and a gem would take 1 + (6-quality) / 3 hours to cut. I haven’t looked up what the modern-day pay-scale would be, but USD $80 an hour would not surprise me, and neither would USD$240 an hour.
14 Uncut
But that’s still all theoretical. All sorts of things can go wrong with the cutting – it’s an art form as much as a science. A really skilled gemsmith might save half the losses, resulting in a larger cut stone; an unskilled craftsman might double them.
Because of these unknowns, uncut gems are (basically) worth about 1/10th the estimated cut value. So the uncut Olivine that could become a Peridot worth $634.88 [USD 2023] would have an uncut value of $63.45.
Other types of semi-precious stone
Using the information provided in earlier parts of the series, this technique can be applied to most other forms of semi-precious stone.
There are a few exceptions – Moonstone is an obvious one.
Peridot
Valuing Peridot is much simpler – most of the time. You no longer need to estimate the cut size, you already know it – but you may need to calculate it in the usual way (measuring, calculating volume, getting weight, converting to carats). And you will need to rate the quality.
Once you have that size and quality assessment, just plug the answers into the calculations shown above.
That’s a wrap
In the next one of these: Ivory and Jade! After that, I’m not sure what there is left to cover in any part VI, but you can be sure that if anything presents itself, I’ll be right on it.
Actually – now that I think about it – high-tech and magic goodies still aren’t covered… And then there’s rare books… and maybe furniture, both fancy and crude (but old)…
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February 28th, 2024 at 4:40 am
I absolutely love this but I’m curious how can we compute this into dnd values? Like gold or platinum piece value’s?
February 28th, 2024 at 5:11 am
Nothing simpler, Anthony. The Value Of Material Things IIa showed how to convert US dollars to D&D Gold Pieces. So it’s a simple currency conversion. Of course, the working is shown in full so that if your world has different assumptions built into it, you can calculate your own answer to match.
February 28th, 2024 at 8:36 am
I will read all your articles. Love stuff like this! Would love to see something on furniture, tapestries, clothing, anything at all! I play kobold presses world of Midgard I believe which is based on 13th to 14th century. Maybe Renaissance era? Thank you!
February 29th, 2024 at 2:14 am
Welcome aboard! I’ve done Tapestries and cloth/textiles (the two are obviously related). Furniture is still to come.