SME
Historical Geology · Problem type
Deposition rate between dated beds
Use this for “An ash bed is 252 Ma and a lava flow 50 m below it is 254.5 Ma. How fast did the beds pile up, and how old is a fossil bed 30 m up?”
Diagram
The shale is older than 252 and younger than 254.5 million years.
This is an example. Type your own values to start a new problem.
t₁
Age of the top bed
example · Ma
t₂
Age of the base bed
example · Ma
H
Thickness between them
example
Δt
Time between the dated beds
worked out (not typed) · Myr
r
Deposition rate
calculated · mm/yr
h
Height of the layer above the base
example
t
Age of the layer
calculated · Ma
Formulas
Δt
=
t₂
−
t₁
2.5 = 254.5 − 252
r
=
H
1,000 × Δt
0.02
=
50
1,000 × 2.5
t
=
t₂
−
h
1,000 ×
r
253
=
254.5
−
30
1,000 × 0.02
Assumptions
•
The beds piled up at a steady rate with no gaps; an unconformity between the dated beds breaks this.
•
Compaction is ignored: the thickness today is taken as the thickness laid down.
•
One meter in a million years is 0.001 mm a year, so the rules carry a factor of 1,000.
Step-by-step
WE KNOW
t₁ = 252 Ma, t₂ = 254.5 Ma, H = 50 m, h = 30 m
FIND
time between the dated beds (Δt), deposition rate (r), age of the layer (t)
Step 1 · Find time between the dated beds (Δt)
Use Δt = t₂ − t₁
The base bed is older, so the time between the beds is its age less the top bed’s.
Δt = 254.5 − 252
Δt = 2.5 Myr
Step 2 · Find deposition rate (r)
Use r = H ÷ (1,000 × Δt)
The thickness over the time it took; the 1,000 turns meters per million years into millimeters a year.
r
=
50
1,000 × 2.5
r
=
50
2,500
r
=
0.02
mm/yr
Step 3 · Find age of the layer (t)
Use t = t₂ − h ÷ (1,000 × r)
The layer is younger than the base by the time its height above the base took to pile up.
t
=
254.5
−
30
1,000 × 0.02
t
=
254.5
−
30
20
t
=
254.5
−
1.5
t
=
253
Ma
Check
Put the numbers back in. Both sides should match.
2.5 = 254.5 − 252 ✓
0.02
=
50
1,000 × 2.5
✓
253
=
254.5
−
30
1,000 × 0.02
✓
Topic check
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