SME
Remote Sensing
Electromagnetic radiation and sensors
Use this for “A sensor 705 km up has an IFOV of 42.5 μrad and a 15° field of view. How big is one ground pixel, and how wide is the swath?”
Diagram
Pixel = H × IFOV = 705 km × 42.5 μrad = 705,000 m × 4.25 × 10⁻⁵ = 30 m.
Swath
=
2H
tan(
FOV
2
)
=
2
×
705
km
×
tan
7.5°
=
185.6
km.
H, the fan and the swath are drawn to scale; the pixel is enlarged in the box.
This is an example. Type your own values to start a new problem.
p (m)
=
H (km)
×
IFOV (μrad)
FOV
Field of view
example · °
S
Swath width
calculated · km
Formulas
p
=
1,000
×
H
×
IFOV
×
10
-6
29.9625
=
1,000
×
705
×
42.5 ×
10
-6
S
=
2
×
H
×
tan(FOV°
÷
2)
185.63 = 2 × 705 × tan(15° ÷ 2)
Assumptions
•
The sensor looks straight down over flat ground; pixels toward the swath’s edges are larger than the one below.
•
The IFOV is a tiny angle, so the pixel is the altitude times the angle in radians (1 μrad is 10⁻⁶ rad).
•
The field of view fans out evenly on both sides of the point straight below the satellite.
Step-by-step
WE KNOW
H = 705 km, IFOV = 42.5 μrad, FOV = 15°
FIND
ground pixel (p), swath width (S)
Step 1 · Find ground pixel (p)
Use p = 1,000 × H × IFOV × 10⁻⁶
A
small
angle
in
radians
times
the
distance
is
the
length
it
spans
on
the
ground:
×
1,000
puts
H
in
metres,
×
10
-6
the
IFOV
in
radians.
p
=
1,000
×
705
×
42.5 ×
10
-6
p
=
705,000
×
(42.5
×
10
-6
)
p
=
29.9625
m
Step 2 · Find swath width (S)
Use S = 2 × H × tan(FOV° ÷ 2)
Each
half
of
the
fan
reaches
H
×
tan(
FOV
2
)
to
the
side
of
the
point
below,
and
the
swath
is
both
halves.
S
=
2
×
705
×
tan(15°
÷
2)
tan(7.5°) = 0.1317
S
=
185.63
km
Check
Put the numbers back in. Both sides should match.
29.9625
=
1,000
×
705
×
42.5 ×
10
-6
✓
185.63 = 2 × 705 × tan(15° ÷ 2) ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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