SME
Physical Geology
Geologic structures (folds, faults, strike and dip)
Use this for “A normal fault dips 60° and has 10 m of dip slip. Find the throw and the heave.”
Diagram
sin δ = t/s = 8.66/10 ≈ 0.866
cos δ = h/s = 5/10 ≈ 0.5
tan δ = t/h = 8.66/5 ≈ 1.7321
This is an example. Type your own values to start a new problem.
t (m)
=
s (m)
×
sin(
δ
°)
h (m)
=
s (m)
×
cos(
δ
°)
Formulas
t
=
s
×
sin(
δ
°)
8.6603 = 10 × sin(60°)
h
=
s
×
cos(
δ
°)
5 = 10 × cos(60°)
tan(
δ
°)
=
t
h
tan(60°)
=
8.6603
5
Assumptions
•
Pure dip slip: the blocks slide straight up or down the fault plane, with no strike-slip part.
•
Throw is the vertical offset of a layer across the fault; heave is the horizontal gap between its broken ends.
•
A normal fault’s hanging wall moves down (the crust stretches); a reverse fault’s moves up (it shortens).
Step-by-step
WE KNOW
δ = 60°, s = 10 m
FIND
throw (t), heave (h)
Step 1 · Find throw (t)
Use t = s × sin(δ°)
The throw is the side opposite the dip, with the slip along the fault as the hypotenuse.
t
=
10
×
sin(60°)
t
=
10
×
0.866
t
=
8.6603
m
Step 2 · Find heave (h)
Use h = s × cos(δ°)
The
heave
is
the
side
next
to
the
dip,
so
it
is
the
slip
times
cos
δ
.
h
=
10
×
cos(60°)
h
=
10
×
0.5
h
=
5
m
Check
Put the numbers back in. Both sides should match.
8.6603 = 10 × sin(60°) ✓
5 = 10 × cos(60°) ✓
tan(60°)
=
8.6603
5
✓
Topic check
See what you know: 5 questions on this topic
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