SME
Remote Sensing · Problem type
Radiance and reflectance from DN
Use this for “A pixel reads 100 with gain 0.8 and offset −1.5. ESUN is 1,550, the Sun is 35° from overhead and Earth is 1 AU away. What are its radiance and reflectance?”
Table
Pixel value (DN)
Reflectance (ρ)
This is an example. Type your own values to start a new problem.
L (W/(m²·sr·μm))
=
G (W/(m²·sr·μm))
×
DN
+
(
B (W/(m²·sr·μm))
)
ρ
=
π
×
L (W/(m²·sr·μm))
×
(
d (AU)
)
2
ESUN (W/(m²·μm))
×
cos(
°)
G and B are the gain and offset that turn DN into radiance L; d is the Earth–Sun distance in AU, and θₛ the Sun’s angle from overhead.
Formulas
L
=
G
×
DN
+
B
78.5 = 0.8 × 100 + (−1.5)
ρ
=
π
×
L
×
d
2
ESUN × cos(θₛ°)
0.1942
=
π
×
78.5
×
1
2
1,550 × cos(35°)
Assumptions
•
The gain and offset come from the image’s metadata; they turn stored numbers into radiance.
•
Reflectance ρ = πLd² ÷ (ESUN cos θₛ), with π = 3.14159…: the share of the band’s sunlight sent back, at the top of the atmosphere.
•
ESUN is the band’s mean sunlight at 1 AU; d² corrects it for Earth’s distance from the Sun that day.
•
Haze is not removed, so this is top-of-atmosphere, not surface, reflectance.
Step-by-step
WE KNOW
DN = 100, G = 0.8 W/(m²·sr·μm), B = −1.5 W/(m²·sr·μm), d = 1 AU, ESUN = 1,550 W/(m²·μm), θₛ = 35°
FIND
radiance (L), reflectance (ρ)
Step 1 · Find radiance (L)
Use L = G × DN + B
Each step of DN adds the gain in radiance, starting from the offset.
L
=
0.8
×
100
+
(−1.5)
L
=
80
+
(−1.5)
L
=
78.5
W/(
m
2
·sr·μm)
Step 2 · Find reflectance (ρ)
Use ρ = π × L × d² ÷ (ESUN × cos(θₛ°))
Radiance back over sunlight in: π spreads radiance over the half sky, cos θₛ slants the sunlight.
ρ
=
π
×
78.5
×
1
2
1,550 × cos(35°)
π
×
78.5
×
1
2
=
246.62
1,550 × cos(35°) = 1,269.7
ρ
=
0.1942
Check
Put the numbers back in. Both sides should match.
78.5 = 0.8 × 100 + (−1.5) ✓
0.1942
=
π
×
78.5
×
1
2
1,550 × cos(35°)
✓
Topic check
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