---
title: "The Stratigraphic Scale"
book: "University Earth and Space Sciences — Year 1"
subject: earth
language: en
chapter: 24
exercises: 12
source: https://one-course.com/books/earth/3/en/chapter/24-the-stratigraphic-scale
license: CC-BY-NC-SA-4.0
credit: "One Earth and Space Sciences Book, One Course (one-course.com)"
---

# Chapter 24 — The Stratigraphic Scale

On the beach of Itzurun, at Zumaia on the Basque coast, the cliffs are made of thin beds of limestone and marl, tilted and repeated for hundreds of metres. At one contact, where a pale limestone bed ends and red marls begin, a bronze disc stamped with four letters is set into the rock (figure below). It is a golden spike. It fixes, for the whole planet, the instant at which one division of geological time — the Selandian [Stage](#def-b1-stratigraphic-scale-chronostratigraphy) — begins. A rock in Patagonia or in Siberia is Selandian if it was laid down after the instant recorded at that contact and before the one recorded at the next spike up the scale. This chapter explains how geologists came to need such a marker: three different ways of dividing a pile of rocks, the principles that order them, the lines that join sections far apart, and the international chart that gathers the result.

![The golden spike of the Selandian Stage in the Zumaia section: a bronze marker stamped GSSP, set at the top of the uppermost limestone bed of one formation, at the base of the red marls of the next. (Josugoni, CC BY-SA 4.0, via Wikimedia Commons.)](https://one-course.com/images/onecourse/chapters/earth-3/b1-stratigraphic-scale/img-a36a57c5b66e.jpg)

*The golden spike of the Selandian [Stage](#def-b1-stratigraphic-scale-chronostratigraphy) in the Zumaia section: a bronze marker stamped GSSP, set at the top of the uppermost limestone bed of one [formation](#def-b1-stratigraphic-scale-lithostratigraphy), at the base of the red marls of the next. (Josugoni, CC BY-SA 4.0, via Wikimedia Commons.)*

**You already know.**

In a pile of layers that has not been turned over, each layer is younger than the one below it; a structure that cuts across rocks is younger than they are; fossils change from older layers to younger ones in the same order everywhere; the Earth’s history is divided into eras and geological periods, each with an age in years for its start, gathered in an international chart (Book 1, grade 7). A [borehole log](https://one-course.com/books/earth/3/en/chapter/5-cross-sections#def-b1-cross-sections-log) records the units met against depth, and correlation matches the same units from log to log ([Definition 5.9](https://one-course.com/books/earth/3/en/chapter/5-cross-sections#def-b1-cross-sections-log)); an [unconformity](https://one-course.com/books/earth/3/en/chapter/5-cross-sections#def-b1-cross-sections-unconformity) is a surface of erosion or non-deposition ([Definition 5.12](https://one-course.com/books/earth/3/en/chapter/5-cross-sections#def-b1-cross-sections-unconformity)). A [sedimentary facies](https://one-course.com/books/earth/3/en/chapter/21-clastic-sediments-and-their-environments#def-b1-clastic-sediments-facies) is a body of rock whose characters record its environment of deposition ([Definition 21.12](https://one-course.com/books/earth/3/en/chapter/21-clastic-sediments-and-their-environments#def-b1-clastic-sediments-facies)). The biology volumes describe how species appear and become extinct, and the few mass extinctions in which a large share of all species disappeared within a short time.

## 24.1 Three ways to divide the rocks

A geologist logging a cliff can divide the same pile of beds in three ways: by what the rocks are made of, by the fossils they contain, and by the time during which they were laid down. The three divisions answer different questions, and their boundaries need not coincide. The rules for naming and defining each kind of unit are set by international agreement.

**Definition 24.1 (Formation, member, group).**

A *lithostratigraphic unit* is a body of rock defined and recognised by its lithology and its position among other bodies, whatever its age. The primary unit is the *formation*: a body of rock distinct enough from its neighbours to be mapped at the scale of the region, into which the whole column should be divided. A *member* is a named subdivision of a formation, distinguished by its lithology; a *group* is a succession of two or more formations with significant lithological characters in common. A formal name joins a local geographical name to the rank or to the dominant rock: the Kell Sandstone, the Mardon Shale Formation.

A [formation](#def-b1-stratigraphic-scale-lithostratigraphy) may be less than a metre thick or several thousand metres. It is extended away from its type section only as far as its lithology can be recognised, so that its boundaries follow the rock, not time.

**Definition 24.2 (Biozone, index fossil, first appearance datum).**

A *biozone* is a body of strata defined or characterised by the fossils it contains. A *range zone* holds the known occurrences of one taxon; an *interval zone* is the body of strata between two specified fossil horizons, for example the lowest occurrences of two taxa. The *first appearance datum* (FAD) of a taxon is the horizon of its lowest occurrence, the *last appearance datum* that of its highest. An *index fossil* is a taxon used to recognise a biozone: the best are widespread, abundant, easy to identify and short-lived.

A fossil horizon has no thickness; a [biozone](#def-b1-stratigraphic-scale-biozone) may be a few metres or a few hundred thick depending on how fast the sediment accumulated while the taxa lived. Because species evolve and never reappear once extinct, the assemblage of any one age is unique. Some of the best [index fossils](#def-b1-stratigraphic-scale-biozone) are microfossils, such as foraminifera and conodonts: widespread, abundant, and small enough to be recovered from the cuttings of a well.

**Definition 24.3 (Chronostratigraphic and geochronologic units).**

A *chronostratigraphic unit* is the body of all the rocks formed during a given interval of geological time, and only during it; its boundaries are surfaces of equal age. The interval of time itself is a *geochronologic unit*. The ranks, from the smallest, are the *stage* (its time: an age), the *series* (an epoch), the *system* (a period), the erathem (an era) and the eonothem (an eon); each unit is made of several consecutive units of the rank below.

The two words describe the same division seen as rock or as time ([Table 24.1](#tab-b1-stratigraphic-scale-ranks)): rocks of the Jurassic [System](#def-b1-stratigraphic-scale-chronostratigraphy) were laid down during the Jurassic Period, and the Lower Jurassic [Series](#def-b1-stratigraphic-scale-chronostratigraphy) during the Early Jurassic Epoch. The [stage](#def-b1-stratigraphic-scale-chronostratigraphy) is the smallest unit recognised all over the world.

| [chronostratigraphic unit](#def-b1-stratigraphic-scale-chronostratigraphy) (rock) | [geochronologic unit](#def-b1-stratigraphic-scale-chronostratigraphy) (time) | example |
| --- | --- | --- |
| eonothem | eon | Phanerozoic |
| erathem | era | Cenozoic |
| [system](#def-b1-stratigraphic-scale-chronostratigraphy) | period | Paleogene |
| [series](#def-b1-stratigraphic-scale-chronostratigraphy) | epoch | Paleocene |
| [stage](#def-b1-stratigraphic-scale-chronostratigraphy) | age | Danian ($66.00\text{ to }61.66\,\mathrm{Ma}$) |

***Table 24.1.** The ranks of the chronostratigraphic and [geochronologic units](#def-b1-stratigraphic-scale-chronostratigraphy), with the units that contain the Danian, the first [stage](#def-b1-stratigraphic-scale-chronostratigraphy) of the Cenozoic.*

## 24.2 The principles and correlation

Before any rock could be dated in years, the order of events was read from the rocks themselves. The basic principles of stratigraphy were set out in 1669 by Nicolas Steno; between 1785 and 1800 Hutton and William Smith added the use of the layers as intervals of time and of their fossils to recognise them.

**Definition 24.4 (Principles of stratigraphy).**

1. The *principle of superposition* : in a sequence of sedimentary layers that has not been overturned, each layer is younger than the one on which it rests.
2. The *principle of original horizontality* : sediments and lava flows are laid down horizontally, or nearly so — slightly sloping at the edges of a basin.
3. The *principle of lateral continuity* : within a basin, a layer extends continuously in every direction until it thins out at the basin’s edge or against an obstacle; a layer cut by a valley continues on the other side.
4. The *principle of cross-cutting relationships* : a body or structure that cuts across rocks — a dyke, a fault, an erosion surface — is younger than the rocks it cuts.
5. The *principle of faunal succession* : fossil taxa succeed one another in the same order in every region, so that rocks holding the same assemblage are of the same age.

The principles are not equally strong. Superposition gives an order that every other method must respect. Original horizontality is what makes a tilted bed a record of later deformation. Lateral continuity is what allows a geologist to draw the same layer on both sides of a valley, and, within a limited distance, a bedding plane traced on the ground is the best line of equal age there is. Faunal succession is the one that crosses oceans; it is the subject of the next paragraphs.

### Correlation by biozones

[Correlation](https://one-course.com/books/earth/3/en/chapter/5-cross-sections#def-b1-cross-sections-log) by fossils compares the order of first and last appearances in two sections (figure below). The two are not equally reliable. The highest fossil of a taxon found in a section depends on how many were preserved: where they are rare, the last one found may lie well below the level at which the taxon really disappeared, and a fossil eroded and redeposited can turn up above it. A first appearance can be displaced too, but a widespread taxon that spread quickly is reached at much the same level across a basin.

![A range chart and a correlation by biozones (invented taxa a to d). The thick bars are the ranges of the taxa in each section. Three interval zones are bounded by the first appearances of b, c and d, which are joined from section to section (dashed). The last appearance of a falls in the upper part of zone c in section 1 and near its base in section 2: it is a poorer guide. The zones are thinner in section 2, where the sediment accumulated more slowly.](https://one-course.com/images/onecourse/chapters/earth-3/b1-stratigraphic-scale/fig-87afd3416244.svg)

*A range chart and a correlation by [biozones](#def-b1-stratigraphic-scale-biozone) (invented taxa *a* to *d*). The thick bars are the ranges of the taxa in each section. Three [interval zones](#def-b1-stratigraphic-scale-biozone) are bounded by the first appearances of *b*, *c* and *d*, which are joined from section to section (dashed). The last appearance of *a* falls in the upper part of zone *c* in section 1 and near its base in section 2: it is a poorer guide. The zones are thinner in section 2, where the sediment accumulated more slowly.*

**Method 24.5 (Correlating sections by biozones).**

1. Plot, for each section, the range of each taxon against height: from its lowest to its highest occurrence.
2. Choose the fossil horizons that define the zones — preferably first appearances of widespread taxa that are independent of the facies.
3. Draw the zone boundaries in each section at these horizons.
4. Join the same horizons from section to section; check that the lines do not cross, as superposition requires.
5. Read, between two lines, the thickness of each zone in each section; where a zone is missing, look for an [unconformity](https://one-course.com/books/earth/3/en/chapter/5-cross-sections#def-b1-cross-sections-unconformity) or a change of facies before concluding that the taxon is absent.

A [biozone](#def-b1-stratigraphic-scale-biozone) is only as good as the independence of its fossils from the environment. A bivalve that lived only on sandy shores is found only in sandstones; following it from section to section follows the sandstone, not time. Biostratigraphic correlation is therefore not necessarily time correlation: it may follow a facies whose age changes from place to place.

### Boundaries that cross time

Different sediments are laid down at the same time in different places: sand on the shore, mud offshore, lime farther out on a shelf starved of sand. A line of equal age therefore passes from one lithology to another. Conversely, a lithological boundary traced across a basin need not be a line of equal age at all.

**Definition 24.6 (Diachronous boundary).**

A boundary is *diachronous* when its age changes from place to place along it. The boundaries of [formations](#def-b1-stratigraphic-scale-lithostratigraphy) and of facies are commonly diachronous; the boundaries of [chronostratigraphic units](#def-b1-stratigraphic-scale-chronostratigraphy) are, by definition, not.

The classic case is a transgression: [relative sea level](https://one-course.com/books/earth/3/en/chapter/23-diagenesis-and-sedimentary-basins#def-b1-diagenesis-basins-eustasy) rises faster than sediment can fill the room made, and the shoreline moves inland. The beach sand is laid down at the shoreline at every instant, so a sheet of sandstone spreads inland with it; the offshore mud follows on top. The top of the sandstone is then the trace of the moving shoreline: young inland, old seaward (figure below). The order of facies seen in one vertical section — sandstone under shale under limestone — repeats their order across the shelf at one instant; the Year 3 volume makes this the basis of its facies models.

![A sandstone sheet laid down during a transgression (schematic, vertical scale exaggerated). The dashed lines t_1 to t_6 join points of equal age, t_1 the oldest. Where one meets the basement, it starts at the shoreline of its time; seaward it passes from sand to mud and, in the younger ones, to lime. The top of the sandstone lies between t_1 and t_2 in section 2 but between t_3 and t_4 in section 1: the formation boundary is diachronous, younger towards the land.](https://one-course.com/images/onecourse/chapters/earth-3/b1-stratigraphic-scale/fig-eb99f7682f19.svg)

*A sandstone sheet laid down during a transgression (schematic, vertical scale exaggerated). The dashed lines $t_1$ to $t_6$ join points of equal age, $t_1$ the oldest. Where one meets the basement, it starts at the shoreline of its time; seaward it passes from sand to mud and, in the younger ones, to lime. The top of the sandstone lies between $t_1$ and $t_2$ in section 2 but between $t_3$ and $t_4$ in section 1: the [formation](#def-b1-stratigraphic-scale-lithostratigraphy) boundary is [diachronous](#def-b1-stratigraphic-scale-diachronous), younger towards the land.*

How fast does the boundary’s age change along it? On a coastal plain that rises gently inland, the answer is a matter of geometry.

**Proposition 24.7 (Speed of a transgression).**

A coastal plain rises inland with the uniform slope $s$, and the [relative sea level](https://one-course.com/books/earth/3/en/chapter/23-diagenesis-and-sedimentary-basins#def-b1-diagenesis-basins-eustasy) rises at the constant rate $R$, with no sediment added at the shore. The shoreline then moves inland at the speed

$$
u = \frac{R}{s},
$$

and a boundary that follows the shoreline is younger, at a point a distance $d$ farther inland, by $\Delta t = d/u = d\,s/R$.

**Proof.** Take $x$ as the distance inland from a fixed point and $z$ as the height; the plain is the line $z = s x$, and the sea surface at time $t$ is $z = z_0 + R t$. The shoreline is where they meet: $s\,x(t) = z_0 + R t$, so $x(t) = (z_0 + R t)/s$ and $\mathrm dx/\mathrm dt = R/s$. The shoreline reaches a point a distance $d$ farther inland after the time $d/u = d\,s/R$. ∎

The ratio is large because coastal plains are nearly flat. A rise of $10\,\mathrm{m}$ per million years over a plain sloping $1$ in $2000$ moves the shore $20\,\mathrm{km}$ inland per million years; a sandstone top that looks like one surface in a cliff is then a million years younger $20\,\mathrm{km}$ inland. Sediment brought to the shore slows the transgression or reverses it.

### Lines of equal age

What correlates time, then, if lithology does not and fossils only approximately? The best lines are made by events that leave a trace everywhere at once. A volcanic ash falls over a wide area in what is, geologically, an instant, and settles on sand, mud and lime alike: an ash bed is a line of equal age that crosses facies boundaries, and it can often be dated in years by the methods of [Chapter 25](https://one-course.com/books/earth/3/en/chapter/25-radiometric-dating-methods#ch-b1-radiometric-methods). Reversals of the Earth’s magnetic field, taken up in the Year 3 volume, are recorded in rocks of many kinds and can be recognised worldwide once their order is known; so are sharp shifts in the isotopes of carbon, such as the one that marks the base of the Ypresian ([Table 24.2](#tab-b1-stratigraphic-scale-gssps)). Between two dated levels, the age of any bed is estimated by assuming that the sediment accumulated at a constant rate.

**Proposition 24.8 (Age between two dated levels).**

Two beds at the heights $z_1 < z_2$ of a continuous section have the ages $t_1 > t_2$. If the sediment accumulated at a constant rate between them, a level at the height $z$ between them has the age

$$
t = t_1 - f\,(t_1 - t_2), \qquad f = \frac{z - z_1}{z_2 - z_1},
$$

and the rate of accumulation is $r = (z_2 - z_1)/(t_1 - t_2)$. If the two ages carry the independent standard uncertainties $\sigma_1$ and $\sigma_2$, the interpolated age carries the uncertainty

$$
\sigma_t = \sqrt{(1 - f)^2 \sigma_1^2 + f^2 \sigma_2^2}.
$$

**Proof.** At a constant rate $r$, the height grows linearly with time: $z - z_1 = r\,(t_1
- t)$, since ages decrease upwards. At $z_2$, $z_2 - z_1 = r\,(t_1 - t_2)$, which gives $r$; dividing the two relations gives $t_1 - t = f\,(t_1 - t_2)$. The age is the linear combination $t = (1 - f)\,t_1 + f\,t_2$ of two independent quantities; the variance of $a X + b Y$ for independent $X$ and $Y$ is $a^2
\operatorname{Var} X + b^2 \operatorname{Var} Y$, which gives $\sigma_t$. ∎

The uncertainty is smallest between the two dated beds and is never larger than the larger of $\sigma_1$ and $\sigma_2$ there. The real limit is the assumption: sediment accumulates in pulses, may stop at hidden breaks, and compacts differently in sand and in mud ([Proposition 23.4](https://one-course.com/books/earth/3/en/chapter/23-diagenesis-and-sedimentary-basins#prop-b1-diagenesis-basins-solid)), so that the interpolated age is no better than the regularity of the section.

## 24.3 Boundary stratotypes

A [chronostratigraphic unit](#def-b1-stratigraphic-scale-chronostratigraphy) is defined in the rock before it is dated. The older practice described each [stage](#def-b1-stratigraphic-scale-chronostratigraphy) by a type section of the whole unit; but the top of one type section and the base of the next, chosen in different places, could leave a gap of unassigned time between them or overlap. The modern practice defines only the lower boundary of each unit at one point in one section; the upper boundary is the lower boundary of the next unit. No interval of time can then fall between two units or belong to both.

**Definition 24.9 (Boundary stratotype, unit stratotype, GSSP).**

A *stratotype* is the designated section of rock that serves as the standard of reference for a unit or a boundary. A *unit stratotype* is the type section of a whole unit; a *boundary stratotype* is the section that contains the specific point defining a boundary between two units. A *global boundary stratotype section and point* (GSSP) is the boundary stratotype of a unit of the international chronostratigraphic scale: a single point in a single section, chosen by international agreement, which defines the base of a [stage](#def-b1-stratigraphic-scale-chronostratigraphy) and of every larger unit that begins with it.

The GSSP is a point in rock, not an age. Its age in years is whatever the best dating of that point gives, and it may be refined without moving the point. The section is chosen to make the point findable elsewhere: deposition continuous across the boundary (never at an [unconformity](https://one-course.com/books/earth/3/en/chapter/5-cross-sections#def-b1-cross-sections-unconformity), where the missing time would make the boundary ambiguous), preferably marine, rich in well-preserved fossils of widespread taxa, without a change of facies at the level, little deformed or altered, accessible and protected. Above all, the point is placed at a level that coincides with one chosen signal — usually the first appearance of a taxon — inside an interval rich in other signals (ash beds, magnetic reversals, isotopic shifts) by which the level can be recognised far away. A permanent marker, the golden spike, is placed at the point.

**In the field — How a boundary is chosen.**

A boundary is studied by a working group of the subcommission in charge of the [system](#def-b1-stratigraphic-scale-chronostratigraphy), which compares candidate sections, logs them bed by bed, samples their fossils, ash beds and isotopes, and writes a proposal. A proposal needs a supermajority — more than $60\,\%$ — of the voting members of the subcommission, about twenty specialists; it then passes to the voting members of the International Commission on Stratigraphy, the executive officers and the chairs of its seventeen subcommissions, and finally to the executive committee of the International Union of Geological Sciences, whose approval ratifies it. The results are published, the marker is set in the rock, and no change to a newly established boundary may be proposed for ten years.

| base of | age ($\mathrm{Ma}$) | section | marker of the point | ratified |
| --- | --- | --- | --- | --- |
| Fortunian (Cambrian) | 538.8 | Fortune Head, Newfoundland, Canada | first appearance of the trace fossil *Trichophycus pedum* | 1992 |
| Lochkovian (Devonian) | 419.62 | Klonk, near Prague, Czech Republic | first appearance of the graptolite *Monograptus uniformis* | 1972 |
| Induan (Triassic) | 251.902 | Meishan, Zhejiang, China | first appearance of the conodont *Hindeodus parvus*, base of Bed 27c | 2001 |
| Danian (Paleogene) | 66.00 | Oued Djerfane, El Kef, Tunisia | base of a $50\,\mathrm{cm}$ dark boundary clay; iridium anomaly | 1991 |
| Selandian | 61.66 | Zumaia, Spain | base of red marls; second radiation of a nannofossil group | 2008 |
| Ypresian (Eocene) | 56.00 | Dababiya, near Luxor, Egypt | base of a carbon-isotope excursion | 2003 |
| Chibanian | 0.774 | Chiba, Japan | base of a tephra, just below a magnetic reversal | 2020 |
| Meghalayan | 0.0042 | Mawmluh Cave, Meghalaya, India | a level $7.45\,\mathrm{mm}$ deep in a speleothem | 2018 |

***Table 24.2.** Eight GSSPs. The markers are first appearances of taxa, a geochemical anomaly, an isotope excursion, a magnetic reversal and a climatic event recorded in a cave deposit.*

[Table 24.2](#tab-b1-stratigraphic-scale-gssps) shows how varied the chosen signals are. The Paleozoic and Mesozoic boundaries of the table rest on first appearances of fossils, a graptolite and a conodont; at the base of the Cambrian the fossil is a trace fossil, the mark left in the sediment by an animal’s activity rather than the animal itself. In the Cenozoic examples the signals are chemical and physical: the iridium of the Danian, the carbon-isotope excursion of the Ypresian, the magnetic reversal that lies $1.1\,\mathrm{m}$ above the Chibanian point. The youngest boundary of the table is not in marine rock at all: the base of the Meghalayan is a level inside a speleothem, a deposit grown in a cave, recording a climatic event of 4200 years ago.

**Figures as of 2026.**

In October 2026 the table of the International Commission on Stratigraphy listed 80 ratified GSSPs, and two more proposed. The first was ratified in 1972, at Klonk; the most recent in 2024, when two boundaries of the Silurian, until then defined in Wales, were moved to new sections in Spain and in the Czech Republic. Of the 101 [stages](#def-b1-stratigraphic-scale-chronostratigraphy) of the Phanerozoic on the chart, 21 still had no ratified GSSP; their bases are provisional.

Below the Phanerozoic, the scale is mostly divided differently. Except for its youngest [system](#def-b1-stratigraphic-scale-chronostratigraphy), the Ediacaran, the Precambrian is cut at round ages chosen by agreement — the divisions are defined in years, not by points in rock.

## 24.4 Mass extinctions as markers

The biology volumes describe the mass extinctions as moments when a large share of species disappeared within a geologically short time. For stratigraphy, such an event is a gift: a horizon found on every continent and in every ocean, across which the fossil assemblages change completely. Two of them bound the three eras of the Phanerozoic.

### The end of the Cretaceous

In 1980, L. W. Alvarez, W. Alvarez, F. Asaro and H. V. Michel measured iridium — one of the platinum metals, depleted in the Earth’s [crust](https://one-course.com/books/earth/3/en/chapter/9-the-layered-earth#def-b1-layered-earth-layers) compared with their abundance in the cosmos — in deep-sea limestones across the boundary between the Cretaceous and the Tertiary — today the boundary between the Cretaceous and the Paleogene, the K–Pg boundary. They found it enriched about 30 times above the background in Italy, 160 times in Denmark and 20 times in New Zealand, exactly at the level of the extinctions. They proposed that an [asteroid](https://one-course.com/books/earth/3/en/chapter/3-comparative-planetology#def-b1-comparative-planetology-small-bodies) $10\,\mathrm{km}$ across, give or take four, had struck the Earth and thrown about 60 times its own mass of pulverised rock into the atmosphere; the dust would have spread worldwide and darkened the sky. The clay at the boundary, they noted, differs chemically from the clay of the limestones above and below.

Later work tied the boundary to one impact, Chicxulub. Argon–argon dating places that impact and the boundary with its extinctions within $32\,000$ years of each other, and a review of the evidence concluded that the impact triggered the extinction, which also fell within the time of a great outpouring of basalt in India. The GSSP of the base of the Danian — the base of the Paleogene and of the whole Cenozoic — is placed at the base of the dark boundary clay at El Kef in Tunisia, at the iridium anomaly. The same boundary can be seen on land, in Colorado (figure below).

![The boundary between the Cretaceous and the Paleogene, marked by the dashed line, at Trinidad Lake State Park, Colorado: sandstone above, dark beds below and along the boundary. (Nationalparks, CC BY-SA 2.5, via Wikimedia Commons; the dashed line is in the original.)](https://one-course.com/images/onecourse/chapters/earth-3/b1-stratigraphic-scale/img-003926ef1426.jpg)

*The boundary between the Cretaceous and the Paleogene, marked by the dashed line, at Trinidad Lake State Park, Colorado: sandstone above, dark beds below and along the boundary. (Nationalparks, CC BY-SA 2.5, via Wikimedia Commons; the dashed line is in the original.)*

### The end of the Permian

The extinction at the end of the Permian was the most severe loss of marine and land life of the last half-billion years. At Meishan in China, the section that holds the GSSP of the base of the Triassic, five ash beds have been dated by the uranium–lead method on zircon. They place the extinction between $(251.941 \pm 0.037)\,\mathrm{Ma}$ and $(251.880 \pm 0.031)\,\mathrm{Ma}$: an interval of $60 \pm 48$ thousand years. A sharp negative shift of $3\,\text{‰}$ in the carbon isotopes of the carbonates precedes it, and the carbon cycle remained unstable for about $500$ thousand years.

The base of the Triassic itself, $(251.902 \pm 0.024)\,\mathrm{Ma}$, is not placed at the extinction but at the first appearance of the conodont *Hindeodus parvus*, a level that falls within the interval of the extinction dated at Meishan. An extinction is a set of last appearances, which preservation blurs; one first appearance of a well-known species makes a sharper line. The causes of the two extinctions — an impact, the great basalt provinces — are the subject of the Year 2 volume.

## 24.5 The international chart

The International Chronostratigraphic Chart gathers all the units, from the eonothems to the [stages](#def-b1-stratigraphic-scale-chronostratigraphy), in one table: their names, the base of each in millions of years with its uncertainty, a mark for each base defined by a ratified GSSP, and a standard colour for each unit, that of the Commission for the Geological Map of the World. The Phanerozoic part, reduced to its eras and periods, is drawn below from the chart’s own values.

![The Phanerozoic on the international chart: its three eras and twelve periods, to scale, in the chart’s colours, with the age of the base of each period (uncertainties omitted). The triangles mark the two bases that are mass extinction horizons. The Quaternary, 2.58\, Ma long, is the thin band at the top.](https://one-course.com/images/onecourse/chapters/earth-3/b1-stratigraphic-scale/fig-85c0b504bce1.svg)

*The Phanerozoic on the international chart: its three eras and twelve periods, to scale, in the chart’s colours, with the age of the base of each period (uncertainties omitted). The triangles mark the two bases that are mass extinction horizons. The Quaternary, $2.58\,\mathrm{Ma}$ long, is the thin band at the top.*

The periods are very unequal (figure above). The Cretaceous lasted $77.1\,\mathrm{Myr}$, the Neogene $20.46\,\mathrm{Myr}$ and the Quaternary only $2.58\,\mathrm{Myr}$; the mean of the twelve is $44.9\,\mathrm{Myr}$. The inequality is inherited: the [systems](#def-b1-stratigraphic-scale-chronostratigraphy) were named in early classifications, before any rock was dated in years — some names give a position (Paleogene, Neogene, Quaternary), some a rock (Carboniferous, Cretaceous), some a people or a place (Ordovician, Silurian, Devonian, Permian) — and the chart keeps the names while defining their bases by GSSPs. The ages in years come from the isotopic dating of volcanic rocks close to the boundaries, such as the ash beds of Meishan ([Chapter 25](https://one-course.com/books/earth/3/en/chapter/25-radiometric-dating-methods#ch-b1-radiometric-methods)), interpolated as in [Proposition 24.8](#prop-b1-stratigraphic-scale-interpolation); a new version of the chart may shift an age while the GSSP stays where it is.

**Figures as of 2024.**

The Anthropocene, a name proposed in 2001 for a new interval marked by human activity, was studied from 2009 by a working group of the Subcommission on Quaternary Stratigraphy. In late 2023 the group proposed a [series](#def-b1-stratigraphic-scale-chronostratigraphy) beginning about 1950, with its GSSP in the sediment of Crawford Lake, Ontario. The subcommission’s vote, closed on 4 March 2024, rejected it: 4 votes for, 12 against, 3 abstentions, 3 members not voting. The voting members of the International Commission on Stratigraphy recognised the vote. The Anthropocene is not a unit of the chart; it remains an informal term, widely used.

## 24.6 Exercises

**Exercise 24.1 ★.**

Say whether each name is that of a lithostratigraphic, a biostratigraphic, a chronostratigraphic or a [geochronologic unit](#def-b1-stratigraphic-scale-chronostratigraphy), and give its rank where it has one: (a) the Jurassic [System](#def-b1-stratigraphic-scale-chronostratigraphy); (b) the Mardon Shale [Formation](#def-b1-stratigraphic-scale-lithostratigraphy); (c) the Early Cretaceous; (d) the *Hindeodus parvus* Zone; (e) the Danian [Stage](#def-b1-stratigraphic-scale-chronostratigraphy); (f) the Paleocene Epoch; (g) a sandstone [member](#def-b1-stratigraphic-scale-lithostratigraphy) of the Mardon Shale.

**Solution of Exercise 24.1.**

(a) Chronostratigraphic, a [system](#def-b1-stratigraphic-scale-chronostratigraphy). (b) Lithostratigraphic, a [formation](#def-b1-stratigraphic-scale-lithostratigraphy). (c) Geochronologic, an epoch (its rock is the Lower Cretaceous [Series](#def-b1-stratigraphic-scale-chronostratigraphy)). (d) Biostratigraphic, a [biozone](#def-b1-stratigraphic-scale-biozone). (e) Chronostratigraphic, a [stage](#def-b1-stratigraphic-scale-chronostratigraphy). (f) Geochronologic, an epoch. (g) Lithostratigraphic, a [member](#def-b1-stratigraphic-scale-lithostratigraphy).

**Exercise 24.2 ★.**

With the bases of the eras in the figure of the chart (those of the Cambrian, the Triassic and the Paleogene), compute the duration of the Paleozoic, the Mesozoic and the Cenozoic, and the share of the Phanerozoic that each represents. What share of the time since the base of the Hadean, $4567\,\mathrm{Ma}$, does the whole Phanerozoic represent?

**Solution of Exercise 24.2.**

Paleozoic $538.8 - 251.902 = 286.9\,\mathrm{Myr}$, $53.2\,\%$ of the Phanerozoic; Mesozoic $251.902 - 66.00 = 185.9\,\mathrm{Myr}$, $34.5\,\%$; Cenozoic $66.0\,\mathrm{Myr}$, $12.2\,\%$. The Phanerozoic is $538.8/4567 = 11.8\,\%$ of the time since the base of the Hadean.

**Exercise 24.3 ★.**

A cliff shows, from the bottom: beds A, B, C and D, tilted at $40{}^{\circ}$; a dyke E that cuts A to D; an erosion surface that cuts the top of D and the dyke; horizontal beds F and G above it; a fault that offsets A to F but not G. Put the events in order, and name the principle that justifies each step.

**Solution of Exercise 24.3.**

A, B, C, D were laid down in that order (superposition), horizontally (original horizontality), so their tilt is a later event. The dyke E cuts A to D, so it is younger than D (cross-cutting); whether it came before or after the tilting cannot be told from the description. The erosion surface cuts D and E, so it is younger than both (cross-cutting). F and G were then laid down on it (superposition). The fault offsets A to F but not G: it moved after F and before G (cross-cutting). Order: A, B, C, D; tilting and the intrusion of E (order undecided); erosion; F; faulting; G.

**Exercise 24.4 ★.**

Two ash beds $30\,\mathrm{m}$ apart in a shale are dated $85.40\,\mathrm{Ma}$ (the lower) and $84.60\,\mathrm{Ma}$ (the upper). The first appearance of an [index fossil](#def-b1-stratigraphic-scale-biozone) lies $12\,\mathrm{m}$ above the lower ash. Compute the rate of accumulation and the age of the first appearance, assuming the rate constant.

**Solution of Exercise 24.4.**

$r = 30/(85.40 - 84.60) = 37.5\,\mathrm{m}$ per million years. The first appearance lies $12/37.5 = 0.32\,\mathrm{Myr}$ after the lower ash: $85.40 - 0.32 =
85.08\,\mathrm{Ma}$.

**Exercise 24.5 ★★.**

In a section, four taxa have the ranges (heights in metres): A 0–34, B 12–50, C 26–62, D 41–70. (a) Draw the range chart. (b) Give the extent and the thickness of the concurrent-range zone of B and C, and of the [interval zone](#def-b1-stratigraphic-scale-biozone) between the first appearances of B and D. (c) In a second section, B is found from 3 to 15 m and C from 9 to 22 m, and nothing else is known. Which zone boundaries of (b) can be recognised there, and which cannot? Which horizons can be correlated between the two sections with the most confidence?

**Solution of Exercise 24.5.**

(a) Four vertical bars against height. (b) B and C occur together from $26\text{ to }50\,\mathrm{m}$: a concurrent-range zone $24\,\mathrm{m}$ thick. The [interval zone](#def-b1-stratigraphic-scale-biozone) from the first appearance of B to that of D runs from $12\text{ to }41\,\mathrm{m}$, $29\,\mathrm{m}$. (c) In the second section the first appearances of B ($3\,\mathrm{m}$) and of C ($9\,\mathrm{m}$) can be placed, so the base of the [interval zone](#def-b1-stratigraphic-scale-biozone) of (b) and the base of the concurrent-range zone are recognised; its top (the first appearance of D) and the overlap’s top as seen in section 1 cannot be checked, D being absent. The first appearances of B and C ($12\,\mathrm{m} \leftrightarrow
3\,\mathrm{m}$, $26\,\mathrm{m} \leftrightarrow 9\,\mathrm{m}$) are the safest lines; the last appearances (B at $50\,\mathrm{m}$ and $15\,\mathrm{m}$) depend more on preservation.

**Exercise 24.6 ★★.**

During a transgression the shoreline moved inland at $20\,\mathrm{km}$ per million years. (a) By how much does the age of the top of the shoreline sandstone differ between two wells $60\,\mathrm{km}$ apart along the direction of the transgression? (b) An ash falls one million years after the shoreline passed the seaward well, where the offshore mud accumulates at $30\,\mathrm{m}$ per million years. How far above the top of the sandstone is the ash bed found there? (c) Where was the shoreline when the ash fell? Is the ash found above or below the sandstone in the landward well, and why may it not be found there at all? Explain with a sketch.

**Solution of Exercise 24.6.**

(a) $60/20 = 3\,\mathrm{Myr}$: the top of the sandstone is three million years younger in the landward well. (b) One million years of mud at $30\,\mathrm{m}$ per million years: the ash lies $30\,\mathrm{m}$ above the top of the sandstone. (c) When the ash fell, the shoreline had moved $20\,\mathrm{km}$ inland of the seaward well, so it was still $40\,\mathrm{km}$ seaward of the landward well, which stood on dry land. The ash fell on the land surface there, below the sandstone that the sea laid down two million years later — if it was not washed away first, as a loose layer on land may well be. The same ash bed is above the [formation](#def-b1-stratigraphic-scale-lithostratigraphy) boundary in one well and below it in the other: the boundary crosses time.

**Exercise 24.7 ★★.**

(a) Compute the relative uncertainty of the ages of the bases of the Lochkovian ($(419.62 \pm 1.36)\,\mathrm{Ma}$) and of the Induan ($(251.902 \pm 0.024)\,\mathrm{Ma}$), in per cent. Suggest why they differ so much. (b) From the two ages that bracket the end-Permian extinction at Meishan, $(251.941 \pm 0.037)\,\mathrm{Ma}$ and $(251.880 \pm 0.031)\,\mathrm{Ma}$, recompute the duration of the extinction and its uncertainty, assuming the two errors independent. Compare with the published $60 \pm 48$ thousand years.

**Solution of Exercise 24.7.**

(a) Lochkovian $1.36/419.62 = 0.32\,\%$; Induan $0.024/251.902 =
0.0095\,\%$, about 34 times smaller. The base of the Induan lies among ash beds dated precisely in the GSSP section itself; a large uncertainty suggests a boundary whose age must be carried from dated levels farther away. (b) $251.941 - 251.880 = 0.061\,\mathrm{Myr} = 61$ thousand years, with the uncertainty $\sqrt{37^2 + 31^2} = 48$ thousand years: the published $60 \pm 48$ thousand years.

**Exercise 24.8 ★★.**

Take the [asteroid](https://one-course.com/books/earth/3/en/chapter/3-comparative-planetology#def-b1-comparative-planetology-small-bodies) of 1980 as a sphere $10\,\mathrm{km}$ across, of density $3000\,\mathrm{kg}\,\mathrm{m}^{-3}$ (an assumed value), throwing up 60 times its mass as dust. (a) Compute the mass of the [asteroid](https://one-course.com/books/earth/3/en/chapter/3-comparative-planetology#def-b1-comparative-planetology-small-bodies) and of the dust. (b) Spread the dust evenly over the Earth (diameter $12\,756\,\mathrm{km}$): what mass falls on each square metre? (c) Laid down at a density of $2500\,\mathrm{kg}\,\mathrm{m}^{-3}$, how thick is the layer? (d) Repeat (c) for diameters of $6\,\mathrm{km}$ and $14\,\mathrm{km}$, the range of the estimate. Compare with the thickness of the boundary clay at the Danian GSSP.

**Solution of Exercise 24.8.**

(a) $m = 3000 \times \frac43\pi\,(5000)^3 = 1.57 \times 10^{15}\,\mathrm{kg}$; the dust, $60m = 9.42 \times 10^{16}\,\mathrm{kg}$. (b) The Earth’s surface is $4\pi(6.378\times10^6)^2
= 5.11 \times 10^{14}\,\mathrm{m}^{2}$: $184\,\mathrm{kg}$ per square metre. (c) $184/2500 =
0.074\,\mathrm{m}$, about $7\,\mathrm{cm}$. (d) The thickness goes as the cube of the diameter: $1.6\,\mathrm{cm}$ for $6\,\mathrm{km}$, $20\,\mathrm{cm}$ for $14\,\mathrm{km}$. The order of magnitude, centimetres to decimetres, is that of the boundary clay, $50\,\mathrm{cm}$ thick at El Kef; the comparison can only be rough, since the dust cannot have fallen evenly and a clay also gathers the ordinary sediment of its time.

**Exercise 24.9 ★★.**

(a) In a condensed section where sediment accumulated at $2\,\mathrm{m}$ per million years, how thick is the interval of $60$ thousand years of the end-Permian extinction? How thick would it be for the shortest and the longest durations allowed by its uncertainty? (b) In a deep-sea section accumulating at $10\,\mathrm{m}$ per million years, how thick is the interval of $32\,000$ years within which the impact and the K–Pg boundary are synchronous? (c) What do these numbers say about sampling a section to study an extinction?

**Solution of Exercise 24.9.**

(a) $60 \times 10^{3} \times 2 \times 10^{-6} = 0.12\,\mathrm{m}$, $12\,\mathrm{cm}$; for 12 and 108 thousand years, $2.4\,\mathrm{cm}$ and $21.6\,\mathrm{cm}$. (b) $32 \times 10^{3} \times
10 \times 10^{-6} = 0.32\,\mathrm{m}$. (c) An extinction occupies centimetres to decimetres of rock: samples must be taken centimetre by centimetre, and a single sample $10\,\mathrm{cm}$ thick in the condensed section averages fifty thousand years — most of the event.

**Exercise 24.10 ★★★.**

A coastal plain slopes $1$ in $2000$ and the [relative sea level](https://one-course.com/books/earth/3/en/chapter/23-diagenesis-and-sedimentary-basins#def-b1-diagenesis-basins-eustasy) rises at $10\,\mathrm{m}$ per million years. (a) With [Proposition 24.7](#prop-b1-stratigraphic-scale-transgression), how fast does the shoreline move inland, and by how much does the age of the top of the shoreline sandstone differ between two wells $30\,\mathrm{km}$ apart? (b) Sediment is now delivered to the shore, which builds the coastal plain upward by $a$ metres per million years at the shoreline. Show that the shoreline moves inland at $(R - a)/s$ as long as $a < R$. What happens when $a > R$? (c) Explain why a sandstone top whose age changes by a million years can look, in a single cliff, exactly like a surface of equal age.

**Solution of Exercise 24.10.**

(a) $u = R/s = 10 \times 2000 = 20\,000\,\mathrm{m}$ per million years, $20\,\mathrm{km}$ per million years; over $30\,\mathrm{km}$, $\Delta t = 1.5\,\mathrm{Myr}$. (b) The plain is now the surface $z = s x + a t$. The shoreline satisfies $s x + a t = z_0 + R t$, so $x(t) = (z_0 + (R - a)t)/s$ and $\mathrm dx/\mathrm dt = (R - a)/s$, positive (inland) while $a < R$. When $a > R$ the speed is negative: the shoreline moves seaward, a regression, and the shoreline sandstone is then younger seaward. (c) A cliff $200\,\mathrm{m}$ long spans, at $20\,\mathrm{km}$ per million years, $0.2/20 = 0.01\,\mathrm{Myr}$ — ten thousand years of diachronism, far below what fossils or most dates can resolve; the rock on either side of the contact is the same everywhere along it.

**Exercise 24.11 ★★★.**

Two ash beds lie at $10\,\mathrm{m}$ and $50\,\mathrm{m}$ in a section and are dated $(100.50 \pm 0.05)\,\mathrm{Ma}$ and $(99.70 \pm 0.03)\,\mathrm{Ma}$. (a) Interpolate the age of a level at $25\,\mathrm{m}$ and its uncertainty with [Proposition 24.8](#prop-b1-stratigraphic-scale-interpolation). (b) Where between the two ashes is the uncertainty smallest? Find the value of $f$ that minimises $\sigma_t$, and the minimum. (c) Show that $\sigma_t$ never exceeds the larger of $\sigma_1$ and $\sigma_2$ between the two beds.

**Solution of Exercise 24.11.**

(a) $f = (25 - 10)/(50 - 10) = 0.375$ and $t = 100.50 - 0.375 \times 0.80 =
100.20\,\mathrm{Ma}$, with

$$
\sigma_t = \sqrt{(0.625 \times 0.05)^2 + (0.375 \times 0.03)^2} = 0.033\,\mathrm{Myr}.
$$

(b) $g(f) = (1 - f)^2\sigma_1^2 + f^2\sigma_2^2$ has $g'(f) = -2(1 - f)\sigma_1^2 + 2f\sigma_2^2$, zero at $f^* =
\sigma_1^2/(\sigma_1^2 + \sigma_2^2) = 0.0025/0.0034 = 0.735$, nearer the better dated bed; there $\sigma_t = \sigma_1\sigma_2/\sqrt{\sigma_1^2 + \sigma_2^2} =
0.026\,\mathrm{Myr}$, smaller than either. (c) $g$ is a parabola opening upward, so on $[0, 1]$ it is largest at an end: $g(0) = \sigma_1^2$, $g(1) = \sigma_2^2$. Hence $\sigma_t \le \max(\sigma_1, \sigma_2)$ between the beds.

**Exercise 24.12 ★★★.**

A well passes through a uniform shale. An ash dated $30.0\,\mathrm{Ma}$ lies at $1000\,\mathrm{m}$ depth and an ash dated $29.0\,\mathrm{Ma}$ at $200\,\mathrm{m}$; a [stage](#def-b1-stratigraphic-scale-chronostratigraphy) boundary lies at $600\,\mathrm{m}$. (a) Interpolate its age from the depths. (b) The grains, not the pores, arrived at a constant rate. Using the shale of [Table 23.1](https://one-course.com/books/earth/3/en/chapter/23-diagenesis-and-sedimentary-basins#tab-b1-diagenesis-basins-athy) and [Proposition 23.4](https://one-course.com/books/earth/3/en/chapter/23-diagenesis-and-sedimentary-basins#prop-b1-diagenesis-basins-solid), compute the solid thickness between each ash and the boundary, and interpolate the age again in proportion to it. (c) By how many thousand years do the two estimates differ, and which is the better?

**Solution of Exercise 24.12.**

(a) The boundary is halfway in depth: $f = 0.5$, $29.50\,\mathrm{Ma}$. (b) With $\varphi_0\lambda = 0.63 \times 1960 = 1234.8\,\mathrm{m}$, the solid thickness from $600\text{ to }1000\,\mathrm{m}$ is $400 - 1234.8\,(e^{-600/1960} - e^{-1000/1960}) =
232.2\,\mathrm{m}$, and from $200\text{ to }600\,\mathrm{m}$, $194.2\,\mathrm{m}$; in all $426.3\,\mathrm{m}$. The fraction of the grains laid down between the lower ash and the boundary is $232.2/426.3 = 0.545$, and the age is $30.0 - 0.545 =
29.46\,\mathrm{Ma}$. (c) The second estimate is $45$ thousand years younger. It is the better one: the deeper half has lost more of its water, so it holds more grains, and more time, per metre than the shallower half (mean porosity $0.42$ against $0.51$).

## 24.7 Problem: Correlating three sections across a basin

**Problem 24.1.**

Weekend problem — how fossils, ash beds and a diachronous sandstone combine to date a stage boundary

Three sections are logged along a line across an invented basin: A in the west, B $25\,\mathrm{km}$ east of A, and C $20\,\mathrm{km}$ east of B. In each, the Kell Sandstone is overlain by the Mardon Shale. Heights are in metres above the base of each section:

|  | A | B | C |
| --- | --- | --- | --- |
| total thickness logged | 80.0 | 66.0 | 32.0 |
| top of the Kell Sandstone | 45.0 | 20.0 | 6.0 |
| ash bed 1 (dated $(152.40 \pm 0.04)\,\mathrm{Ma}$) | 5.0 | 8.0 | 3.0 |
| ash bed 2 (dated $(151.20 \pm 0.04)\,\mathrm{Ma}$) | 71.0 | 56.0 | 27.0 |
| first appearance of *Vardenoceras medium* | 36.6 | 31.2 | 14.8 |
| first appearance of *Vardenoceras ultimum* | 65.8 | 52.0 | 24.8 |
| last appearance of *Vardenoceras primum* | 49.0 | 40.0 | 19.0 |

*V. primum* is found from the base of every section. The bivalve *Ostreola* is found only in the Kell Sandstone, in all three sections. Three [interval zones](#def-b1-stratigraphic-scale-biozone) are defined: zone 1 up to the first appearance of *V. medium*, zone 2 up to that of *V. ultimum*, zone 3 above. A [stage](#def-b1-stratigraphic-scale-chronostratigraphy) boundary is defined, at its GSSP elsewhere, by the first appearance of *V. medium*.

**Part I — Biozones.**

1. Draw the three sections side by side at a common vertical scale, with the Kell Sandstone, the ash beds and the zone boundaries.
2. Give the thickness of zone 2 in each section, and the thickness of the part of zone 3 logged.
3. Why can the base of zone 1 not be placed in any of the three sections?
4. Why is *Ostreola* a poor [index fossil](#def-b1-stratigraphic-scale-biozone) , however common it is?
5. In which zone does the last appearance of *V. primum* fall in each section? Could it serve as a fourth zone boundary?
6. Correlate the sections by the first appearances of *V. medium* and *V. ultimum* . What does the ratio of the thicknesses of zone 2 suggest?

**Part II — The Kell Sandstone.**

7. In which zone does the top of the Kell Sandstone lie in each section?
8. Is the top of the Kell Sandstone a surface of equal age? Which way did the shoreline move, and what kind of event is this?
9. Ash bed 1 lies in the Kell Sandstone in all three sections, ash bed 2 in the Mardon Shale. Why are the ash beds better time lines than the [formation](#def-b1-stratigraphic-scale-lithostratigraphy) boundary?
10. Which of the principles of [Definition 24.4](#def-b1-stratigraphic-scale-principles) justify following a single ash bed from A to C?

**Part III — Rates.**

11. Compute the rate of accumulation between the two ash beds in each section, in metres per million years.
12. Suggest why the rate falls from A to C.
13. How many years does a sample $10\,\mathrm{cm}$ thick represent in each section? Which section gives the finest resolution in time?
14. Assuming constant rates, compute the age of the top of the Kell Sandstone in each section.
15. Compute the speed of the shoreline between C and B and between B and A, in kilometres per million years. Did the transgression speed up or slow down?
16. Compute the age of the first appearance of *V. ultimum* in each section.
17. Do the three ages of question 16 agree within the uncertainty of the ash dates? What does this say about the first appearance?

**Part IV — The age of the boundary.**

18. Interpolate, in each section, the age of the first appearance of *V. medium* .
19. Compute the uncertainty of each age from the uncertainties of the ash dates with [Proposition 24.8](#prop-b1-stratigraphic-scale-interpolation) .
20. Compute the mean of the three ages and their spread. Is the spread larger or smaller than the uncertainty of question 19?
21. In section A the first appearance of *V. medium* lies in the Kell Sandstone, in B and C in the Mardon Shale. How could a change of lithology between the ash beds bias the interpolation in A?
22. Which of the three sections would make the best [boundary stratotype](#def-b1-stratigraphic-scale-gssp) , and why? Which would be the worst?
23. How long did zone 2 last?
24. What is the interpolated age of the boundary?

**Solution of Problem 24.1.**

1. Three columns at, say, $1\,\mathrm{cm}$ for $10\,\mathrm{m}$ , joined by the ash beds and by the zone boundaries; the Kell Sandstone wedge thins from $45\,\mathrm{m}$ in A to $6\,\mathrm{m}$ in C.
2. Zone 2: A $65.8 - 36.6 = 29.2\,\mathrm{m}$ , B $20.8\,\mathrm{m}$ , C $10.0\,\mathrm{m}$ . Zone 3 logged: A $14.2\,\mathrm{m}$ , B $14.0\,\mathrm{m}$ , C $7.2\,\mathrm{m}$ .
3. Zone 1 begins at the first appearance of *V. primum* , which is found from the base of every section: its first appearance lies below them.
4. It lives only in the sandstone: it follows the facies, and the facies is of different ages from place to place. In A it occurs in zones 1 and 2, in B and C only in zone 1.
5. In zone 2 in all three (A $49.0\,\mathrm{m}$ , B $40.0\,\mathrm{m}$ , C $19.0\,\mathrm{m}$ ). It is consistent from section to section here, and so could subdivide zone 2, but a last appearance depends on preservation; it should be checked in more sections first.
6. The lines join $36.6\,\mathrm{m}$ , $31.2\,\mathrm{m}$ and $14.8\,\mathrm{m}$ , then $65.8\,\mathrm{m}$ , $52.0\,\mathrm{m}$ and $24.8\,\mathrm{m}$ . Zone 2 thins in the ratio $29.2 : 20.8 : 10.0$ , about $2.9 : 2.1 : 1$ : if the zone lasted the same time everywhere, the sediment accumulated about three times faster in A than in C.
7. A: zone 2 ( $45.0 > 36.6$ ); B: zone 1 ( $20.0 < 31.2$ ); C: zone 1 ( $6.0 < 14.8$ ).
8. No: it is in zone 2 in A, zone 1 in B and C, so it is younger in the west. The marine shale spread westward over the shoreline sand: the shoreline moved west, inland, in a transgression.
9. An ash falls over the whole basin in a geological instant and settles on any sediment; the top of the sandstone follows the moving shoreline and is [diachronous](#def-b1-stratigraphic-scale-diachronous) .
10. Lateral continuity (the bed continues across the basin), and superposition (ash 1 below ash 2 in every section, which checks the correlation).
11. A $(71.0 - 5.0)/1.20 = 55.0\,\mathrm{m}$ per million years; B $48.0/1.20 = 40.0\,$ ; C $24.0/1.20 = 20.0\,$ .
12. A is nearest the land, where the sediment was supplied; less of it reached the deeper water to the east.
13. $0.1/55 = 1.8$ thousand years in A, $2.5$ in B, $5.0$ in C: A resolves time most finely.
14. A $152.40 - 40.0/55.0 = 151.67\,\mathrm{Ma}$ ; B $152.40 - 12.0/40.0 =  152.10\,\mathrm{Ma}$ ; C $152.40 - 3.0/20.0 = 152.25\,\mathrm{Ma}$ .
15. C to B: $20/(152.25 - 152.10) = 133\,\mathrm{km}$ per million years; B to A: $25/(152.10 - 151.67) = 58.5\,\mathrm{km}$ per million years. It slowed down.
16. A $152.40 - 60.8/55.0 = 151.29\,\mathrm{Ma}$ ; B $152.40 - 44.0/40.0 =  151.30\,\mathrm{Ma}$ ; C $152.40 - 21.8/20.0 = 151.31\,\mathrm{Ma}$ .
17. They span $16$ thousand years, less than the uncertainty of each (about $28$ thousand years, question 19): the first appearance is synchronous across the basin within the resolution of the dates, which is what makes it a good index.
18. A $152.40 - 31.6/55.0 = 151.826\,\mathrm{Ma}$ ; B $152.40 - 23.2/40.0 =  151.820\,\mathrm{Ma}$ ; C $152.40 - 11.8/20.0 = 151.810\,\mathrm{Ma}$ .
19. $f = 0.479$ , $0.483$ , $0.492$ ; $\sigma_t = \sqrt{(1 - f)^2 + f^2} \times  0.04 = 0.028\,\mathrm{Myr}$ in each, $28$ thousand years.
20. Mean $151.819\,\mathrm{Ma}$ ; spread $151.826 - 151.810 = 16$ thousand years, smaller than the uncertainty of each age: the three agree.
21. The interpolation assumes one rate from ash 1 to ash 2. In A that interval holds shoreline sandstone up to $45\,\mathrm{m}$ and shale above; sand and mud need not have accumulated at the same rate, and they compact differently ( [Chapter 23](https://one-course.com/books/earth/3/en/chapter/23-diagenesis-and-sedimentary-basins#ch-b1-diagenesis-basins) ). If the sand accumulated faster, the sandstone part holds less time than its thickness suggests, and the first appearance is older than computed — the age from A is indeed the oldest of the three.
22. B: the first appearance lies in marine shale, $11\,\mathrm{m}$ above the top of the sandstone, with no change of facies near the level, and the rate gives $2.5$ thousand years per $10\,\mathrm{cm}$ . A resolves time better but places the level in the shoreline sandstone, $8.4\,\mathrm{m}$ below a change of facies; C is condensed, $5$ thousand years per $10\,\mathrm{cm}$ . C is the worst for resolution, A for facies.
23. From the mean ages of the two first appearances, $151.819 - 151.302 =  0.52\,\mathrm{Myr}$ : about half a million years.
24. The boundary is $(151.82 \pm 0.03)\,\mathrm{Ma}$ .