SME
Geophysics · Problem type
Radar depth from two-way time
Use this for “Ground with a relative permittivity of 9 returns a radar echo 40 ns after the pulse. How deep is the target?”
Diagram
v
=
0.3
÷
√
9
=
0.1
m/ns.
d
=
vt
2
=
2
m:
the
pulse
goes
down
and
back,
so
half
the
time
is
the
trip
down.
This is an example. Type your own values to start a new problem.
d (m)
=
v (m/ns)
×
t (ns)
÷
2
εᵣ
Relative permittivity
example
Formulas
v
=
0.3
÷
√εᵣ
0.1
=
0.3
÷
√
9
d
=
v
×
t
2
2
=
0.1
×
40
2
Assumptions
•
Radar travels at the speed of light in air, c = 0.3 m/ns; in the ground it is slower by √εᵣ.
•
The time is two-way: down to the target and back up to the antenna.
•
The ground above the target has one permittivity all the way down.
Step-by-step
WE KNOW
εᵣ = 9, t = 40 ns
FIND
radar speed (v), depth (d)
Step 1 · Find radar speed (v)
Use v = 0.3 ÷ √εᵣ
Light’s 0.3 m/ns in air, slowed by the square root of the permittivity.
v
=
0.3
÷
√
9
v
=
0.3
3
v
=
0.1
m/ns
Step 2 · Find depth (d)
Use d = v × t ÷ 2
Speed times the two-way time, halved for the trip down.
d
=
0.1
×
40
2
d
=
4
2
d
=
2
m
Check
Put the numbers back in. Both sides should match.
0.1
=
0.3
÷
√
9
✓
2
=
0.1
×
40
2
✓
Topic check
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