We will construct our chess set on a suitably grand scale - every thing in metres to make life easy. We will also round off every thing to the nearest unit and have things utterly simple for a change - especially when I have chosen simple geometric shapes for the chess pieces. You will also have to do a bit of imagining - not often that I let you fantasise!

The two tables below define the basis for our chess set. Note that our board will be made up of 8 x 8 = 64 squares with sides of 4m each. Thus our board will be 32m square equating to 1,024 square m (about one-tenth of a hectare - or just over a quarter of an acre for you real estate chaps).

Table 1. Our simple chess set.

Piece	Number	Shape	Dimensions (m)	Individual surface area (except base - m2)	Total surface area of all pieces (m2)	Total surface area of bases (m2)
Pawns	16	Cylinder	1 Ω x 3 tall	10	163	13
Knights	4	Rectangular prism	1 x 2 x 4 tall	26	104	8
Bishops	4	Right cone	2 Ω x 4 tall	4	16	12
Rooks (Castles)	4	Square prism	2 x 2 x 4 tall	26	144	16
Queens	2	Right cone	3 Ω x 6 tall	14	28	14
Kings	2	Cylinder	3 Ω x 6 tall	64	128	14
Totals	32 pieces				583	77

Notes: Ω = diameter. All areas rounded to the nearest metre.

Table 2. Basis for calculations

Shape	Surface area ignoring base
Cylinder	π x R2 + 2 x π x R
Rectangular prism	(2 sides x W1 x H) + (2 sides x W2 x H) + (W1 x W2)
Right cone	(π x R2 x H)/3
Square prism	(4 sides x W x H) + (W + W)

Where R = radius, W = width, H = height and π = pi (or 22/7 for some).

What's the old fool on about this time, I hear you cry?

You will note that in Table 1 we have added up the surface areas of all of our pieces (excluding their bases) and the surface areas of all of the bases. Lets put our pieces on the board - whether at random or in two lots of two rows on either side of the board doesn't matter. Neither does the colour of piece or square matter. Who is going to play with a chess set where the pawns weigh a bit over six tonnes each! Picture 1 shows part of a karstic chess set.

Because the pieces are going to be standing on the board we need to subtract their total basal area from the board area: 1,024-77 = 947m2.

And then we add the surface area of all our chess pieces 947+ 583 = 1,530 m2.

Hands up all those who noted that the surface area of our board and pieces has increased by about 50%! And if we get another full set of 32 pieces and have a piece in every square the surface area has doubled.

I will get to the point shortly ...

Let's look at another arithmetic exercise. We will take a limestone cube one metre a side and cut it in some tricky fashion without losing any material into litre-sized lumps (10cm x 10cm x 10cm). And then we will take each of these and cut them into centimetre-sized cubes. And then do the same to produce millimetre-sized cubes!

What happens to the surface area?

1 metre cube =	6m2
0.1 metre cube x 1,000 =	60m2
0.01 metre cube x 1,000,000 =	600m2
0.001 metre cube x 109 =	6,000m2 - over half a hectare from a one metre cube!

[A bit like the chap who elected to be paid for inventing the game of chess by having grains of rice placed on each square of a chess board as 1 grain, 2 grains, 4 grains, 8 grains, 16 grains and so on. Amounts to 264 = 6.2 x 1027 grains (for really useful trivia at 10 to 35mg per rice grain this equals 6.2 x 1021 to 2.2 x 1023 tonnes!)].

Let's assume fairly regular metre-scale jointing in a large block of limestone; that the block is ten metres each side, five metres high and has vertical joints one metre apart and that there are no other joints.

Effectively we will have 10 x 10 blocks each five metres tall. Ignoring their bases, each of the joint-bounded blocks will have a surface area of 4 sides each 5 x 1m + one m2 for the top - 21 m2 all up for each block. Thus the total surface area is 2,100 m2. If there were no joints the surface area would be 4 sides each 5 x 10m and + 10 x 10m for the top. This equals 300m2 - one seventh of the jointed total.

We are going to assume that all the joints are open and that water can drain down each surface equally dissolving limestone as it goes. Referring back to annual erosion rate determinations actually estimated (jargon for "measured") for bare rock surfaces at Yarrangobilly we find the figure of 0.0027mm/year in Spate et al (1985) and a better estimate of 0.011mm/year reported in Smith et al (1995). The latter was taken over a longer timeframe (11.25 versus 4.46 years) so errors should have been reduced.

So how much limestone was lost from our jointed block as compared to the unjointed one using the two different erosion rates over several time periods

Short-term erosion estimate (0.0027mm/year) and limestone density of 2.66kg/litre

	Surface area (m2)	1 year	100 years	1000 years
Jointed block	2,100	0.006m3 (6 litres)	0.600m3 (600 litres)	6m3
		16kg	1.6 tonne	16 tonne
Unjointed block	300	~0.001m3 (800 ml)	0.08m3 (80 litres)	0.8m3
		~2.1kg	~213kg	~2.1 tonne

Long-term erosion estimate (0.011 mm/year) and limestone density of 2.66 kg/litre

	Surface area (m2)	1 year	100 years	1000 years
Jointed block	2,100	0.023m3 (23 litres)	2.3m3	23m3
		61kg	6.1 tonne	61 tonne
Unjointed block	300	~0.003m3 (>3l)	0.33m3 (330 litres)	3.3m3
		~8.8kg	~880kg	~8.8 tonne

Yes, I know that litre is not a SI unit for the measurement of solid volumes! I could have used cubic decimetres!

Astute people will have realised that the surface area will decrease with time so we would have to use calculus (which I can't) to work out how long our 10 x 10 x 5 m block would last. Assuming the surface area (and the climate - remember earlier ANDYSEZs on the dangers of manipulating rates over long periods of time!) don't change our block would vanish in about 22,000 years at the higher erosion rate and about 83,000 years at the lower. See what a difference jointing and thus an increase surface area makes!

Three pieces of our epikarst chess set! The Pinnacles, Nambung National Park, Western Australia

What's the point of all of this? The epikarst is the soil mantle and its interaction with the limestone beneath which frequently weathers into pinnacles increasing its surface area. A number of authors (e.g. Smith and Atkinson 1976) have pointed out that two-thirds of all limestone solution takes part in the cutaneous zone (~10% at the surface and in the soil) and in the subcutaneous zone (~60% in weathered limestone and fissures - in joints and as pinnacles) which together make up the epikarst. Most of the rest takes place in joints in the rock mass and very little (<1%) in caves.

Think of all the particles and spaces in the soil. The enormous surface area here allows for retention and transmittal of water, for the biologic process that produce the all-important high levels of carbon dioxide to enhance our limestone solution.

We will continue this discussion - especially on limestone soils in the next ANDYSEZ.

References

Smith DI and Atkinson TC, 1976, The Erosion of Limestones, Chapter 5 in Ford TD and Cullingford CHD (eds) The Science of Speleology, Academic Press, London

Smith DI, Greenaway MA, Moses C and Spate AP, 1995, Limestone weathering in Eastern Australia. Part 1: Erosion Rates, Earth Surface Processes and Landforms, 20:45-463

Spate AP, Jennings JN, Smith DI and Greenaway MA, 1985, The Micro-Erosion Meter, use and limitations, Earth Surface Processes and Landforms, 10:427-440