guess the number of shells to win!

[Joe Botting]

It's all very well talking about reconstructing these ecosystems in artistic form, but that does tend to gloss over some of the details a tad... How are we to convert from a rock with scattered bits of dead thing to what was living there to start with? I guess that's part of what this entire board is about, really, but we've got to start somewhere.

Let's take a sandstone with shells in. Easy enough. There are, let's say, two or three shells on the upper surface of a 10-cm-square slab, and you can see a couple more in cross-section on the sides. Anyone want to guess how many there were actually living there at the time? (I'll give you a hint - you can ask extra questions if you want! ;-)

Joe

[Froggie]

Okay, I'll bite (since this topic seems a bit quiet).

1. Thickness of slab?
2. Dimensions of shell fragments?

[Joe Botting]

Hi there Froggie, and welcome to our little empire... :-)

Thanks for taking pity on a poor lonely thread, which is exceedingly grateful. To answer your questions:

1. Thickness... ooh, let's say 3 cm. It's easy to carry, then.

2. Well, the complete ones are 1 cm diameter, and approximately circular. Well done for spotting the fragments - being mostly less that a millimeter, though, they're only visible under your hand-lens! Sadly, there aren't all that many fragments that you can see... perhaps a dozen bits up 1 mm², and you can ignore anything smaller...

Joe

[Froggie]

Well, you know me, always ready to tackle a problem quite unrelated to what I'm supposed to be doing.

I should have asked you what exactly "a couple on the sides" means, but for the sake of argument lets go with what you've given me. We'll get back to the "couple on the sides" later.

I'll assume that there are no visible shells on the bottom face, and that the shells are flat and 1mm thick (just to make my life easier).

Now, the volume of a 1mm surface layer of the top and bottom faces is 2x100x100 = 20 000mm³. We have 3 shells in that, giving a number density of 0.00015 shells/mm³.

Total volume of the block is 300 000 mm³, so there should be 45 complete shells within it.

As a check, 0.00015 shells/mm³ should give us approximately 1.6 full shells visible in the 1mm surface layer of the sides (volume 11088mm³) ... which sounds rather close to the "couple more" in the description, particularly if we take into account that we see cross-sections not full shells on the sides. Cool.

Now, the fragments. These I'll assume are 1mm³ cubes for ease of calculation. We can see a dozen or so, so again looking at the 1mm thick layer of the surface (this time top, bottom, and sides) we have a density of 12 in 31088mm³ or 0.000386 fragments/mm³. Total volume of the block is still 300 000mm³, so ... 1158 fragments in the whole block.

Assuming that the shells are 1cm across and 1mm thick, their volume will be 78.5 mm³. Therefore 1158 1mm³ fragments must have originated from 148 or so shells.

Thus we come to the conclusion that there must have been 148+45 or about 200 complete shells in this volume to give the density of fragments and shells we see on the surface. This is probably a lower limit -- for example, if there are larger fragments or if we see more shells on the sides and bottom then there will be correspondingly more shells in the block.

[hallucygenia]

200! Blimey! That seems like a lot, but I can't argue with the maths.

Of course, it's a bit more complicated than that. How much time does the thickness of the block represent? The sand that became the block could have been washing around on the sea floor for a good while - which would explain all the shell fragments. Let's further that a brachiopod lives for five years (I have no idea how long a brachiopod lives, but this probably isn't unreasonable). Let's also assume that any dead brachiopod becomes fragmented pretty quickly. So complete brachiopods represent those living at the time of burial. There are 45 living brachiopods, and 148 fragmentary shells. That's about 3 times as many dead ones as living ones, and living ones are a maximum of 5 years old, so the rock as a whole represents at least 20 years worth of deposition. In a shallow-water environment, with a lot of sediment being deposited, that may not be too unreasonable.

There are a lot of holes that could be picked in the above argument, not least that I've assumed that only living brachiopods can be preserved as whole shells - this is undermined by the fact that there are shells in the middle of the block, and as brachiopods (at least the ones in this example) don't burrow, this could be a problem. There's also the problem of shell fragments becoming too small to see, and the problem of shells being transported in from elsewhere.

The figure of 45 living brachiopods is probably too high for a fairly obvious reason - living space. Brachiopods take up, say, 1 square centimetre each. The slab has a surface area of 100 square centimetres. That means that almost half the surface of the slab would be taken up by brachiopods. This is possible, but seems rather crowded. Thus, at least some brachiopods must stay as whole shells after death.

In the above, I've been assuming that the only thing destroying brachiopods after death is mechanical damage, i.e. being broken down into sand-sized fragments.

To come back to the original question that started this thread, the number of brachiopods that were living on the area of the sea floor that became the sandstone slab is more than three and fewer than 45 at any one time.. The number living during the total time that the rock was deposited is at least 200. (Time for rock to be deposited I define as the time from the deposition of the top of the underlying bed to the time of deposition of the bottom of the overlying bed.)

I'm beginning to waffle a bit, so I'll stop now. If I think of anything else that makes sense I'll post it later on.

Lucy

[Froggie]

Ah yes, time. I'd forgotten about that one. Your points about living space, washing down of fragments and layers of deposition are very good. Its clearly been too long since I last did some geology, I'm not thinking in the right way.

I think between us we've given a lot of food for thought to anyone reading this - hopefully there'll be some more comments and discussion.

Anyway, does Joe have an answer to this puzzle or was he just fishing?

[Joe Botting]

Good work, guys!

There's no actual answer to the problem, but I was hoping to get to something sensible, which actually seems to have happened, almost. I'm impressed! :-)

A few more things to consider, though:
1) A brachiopod has two shells! Therefore, divide all answers by two, and it's already looking more reasonble.
2) Mechanical breakage is a minor part of the production of fragments - most of it is erosion by microbes, algae, fungi and so on, that weaken them until they fall apart. This requires a significant length of time washing around on the surface...
3) You need to consider the distribution of the shells vertically - if they're only on a couple of planes, and the rest is barren, then it's probably a result of winnowing, and perhaps even concentration of shells into one layer. That, of course, completely changes things... and if there are both fragments and complete shells, you know that it wasn't absolutely continuous deposition.

In this case, let's assume that the shells are indeed fairly evenly scattered through the bed, that none of them are articulated, and that 20 years is a good estimate. (In shallow water, it could all have been deposited in an hour or so, but the shells would still represent a longer time interval.)

So, taking all this into account, let's have a final answer for the average density of individuals at any one time, and let the criticisms begin! After that we'll try something more difficult... ;-)

Joe

[hallucygenia]

Final answer?

We're allowed to assume that the deposition of the bed took 20 years. I will further assume that any whole brachiopod shell lived at that place on the sea floor at some time during that 20 years, and any shell fragments can be ignored. There are 45 complete shells, which corresponds to 22 and a half brachiopods. That's about one a year living on that piece of the sea floor, on average.

Criticism, please!

[Joe Botting]

Yup, so I guess if they each lived about five years (which may or may not be reasonable!), then we should have five individuals at any one time in a 10 cm² area. In other words, it's surprisingly close to what we see in the fossils. I wasn't actually expecting that!
Of course, it could have been rather different if there was any winnowing, but I reckon that's a good start.

Right then. Sponges. The sediment is full of fragments of their siliceous spicules, and every few slabs you find a more-or-less complete specimen. These things are much bigger - up to 30 cm tall and 5 cm diameter, with spicules up to 10 cm long. Most of the spicules are hollow, indicating various degrees of dissolution - as you would expect of opal, which is unstable in warm, shallow water...
What are we going to do with these, then?

Joe

[hallucygenia]

What to do with sponges? Well, there's that game where you throw lots of long thin sticks on the floor and then have to pick them up without disturbing any of the others...

Seriously, don't know. Will have to give it some thought.