SME
Physical Geology · Problem type
Distance to a quake from the S − P lag
Use this for “The S wave arrives 25 s after the P wave. With vP = 6 km/s and vS = 3.5 km/s, how far away is the focus?”
Diagram
P arrives after 210 ÷ 6 = 35 s; S after 210 ÷ 3.5 = 60 s.
The farther the station, the longer the lag.
This is an example. Type your own values to start a new problem.
d (km)
=
Δt (s)
×
vP (km/s)
×
vS (km/s)
÷
(
vP (km/s)
−
vS (km/s)
)
k
Km per second of lag
worked out (not typed) · km/s
Formulas
k
=
vP
×
vS
vP − vS
8.4
=
6
×
3.5
6 − 3.5
d
=
k
×
Δt
210 = 8.4 × 25
Assumptions
•
Both waves leave the focus together and travel at steady speeds, true near the surface.
•
Each second of lag stands for k = vP vS ÷ (vP − vS) km; with 6 and 3.5 km/s that is 8.4 km.
•
Distant stations need travel-time curves, since deeper rock is faster.
Step-by-step
WE KNOW
Δt = 25 s, vP = 6 km/s, vS = 3.5 km/s
FIND
km per second of lag (k), distance to the focus (d)
Step 1 · Find km per second of lag (k)
Use k = vP × vS ÷ (vP − vS)
Each
km
adds
1
vS
−
1
vP
seconds
of
lag;
flipped,
that
is
vP
vS
vP − vS
km
per
second.
k
=
6
×
3.5
6 − 3.5
k
=
6
×
3.5
2.5
k
=
21
2.5
k
=
8.4
km/s
Step 2 · Find distance to the focus (d)
Use d = k × Δt
Each
second
of
lag
is
k
km,
so
multiply
by
the
lag.
d
=
8.4
×
25
d
=
210
km
Check
Put the numbers back in. Both sides should match.
8.4
=
6
×
3.5
6 − 3.5
✓
210 = 8.4 × 25 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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