SME
Physical Geography · Problem type
Daily sunlight at the top of the atmosphere
Use this for “How much sunlight, averaged over 24 hours, reaches the top of the atmosphere at 40° N on the June solstice?”
Diagram
The Sun is overhead at δ = 23.4°; the place is at φ = 40°.
Noon sun angle = 90° − |φ − δ| = 90° − 16.6° = 73.4°.
cos
H
=
−tan
φ
tan
δ,
so
H
=
111°:
61.9%
of
the
parallel
is
lit,
day
=
2H
15
=
14.8
h.
This is an example. Type your own values to start a new problem.
φ
Latitude
example · °
δ
Declination
example · °
H
Sunrise hour angle
calculated · °
Q
Daily mean sunlight
calculated · W/m²
Formulas
cos(
H
°)
=
−tan(
φ
°)
×
tan(
δ
°)
cos(111.3342°) = −tan(40°) × tan(23.44°)
Q
=
(
1,361
π
)
×
(
H
×
π
180
×
sin(
φ
°)
×
sin(
δ
°)
+
cos(
φ
°)
×
cos(
δ
°)
×
sin(
H
°))
498.8599
=
(
1,361
π
)
×
(111.3342
×
π
180
×
sin(40°)
×
sin(23.44°)
+
cos(40°)
×
cos(23.44°)
×
sin(111.3342°))
Assumptions
•
The solar constant S₀ = 1,361 W/m², with Earth at its mean distance from the Sun.
•
Q averages the sunlight over all 24 hours, night included, before the air absorbs or reflects any.
•
H is the sunrise hour angle: 180° under the midnight sun, 0° in polar night (Q = 0).
Step-by-step
WE KNOW
φ = 40°, δ = 23.44°
FIND
sunrise hour angle (H), daily mean sunlight (Q)
Step 1 · Find sunrise hour angle (H)
Use cos(H°) = −tan(φ°) × tan(δ°)
The sunrise hour angle: how far Earth turns from sunrise to noon, where the parallel meets the circle of illumination.
−1 < tan(40°) × tan(23.44°) < 1: the Sun rises and sets
H
=
arccos(−tan(
φ
°)
×
tan(
δ
°))
H
=
arccos(−tan(40°)
×
tan(23.44°))
tan(40°) × tan(23.44°) = 0.3638
H
=
111.3342°
Step 2 · Find daily mean sunlight (Q)
Use Q = (1,361 ÷ π) × (H × π ÷ 180 × sin(φ°) × sin(δ°) + cos(φ°) × cos(δ°) × sin(H°))
Add
up
the
sunlight
on
a
level
surface
from
sunrise
to
sunset
(
H
in
radians
in
the
first
term),
then
spread
it
over
24
hours.
Q
=
(
1,361
π
)
×
(111.3342
×
π
180
×
sin(40°)
×
sin(23.44°)
+
cos(40°)
×
cos(23.44°)
×
sin(111.3342°))
111.3342
×
π
180
=
1.943
rad
1.943 × sin(40°) × sin(23.44°) = 0.4969
cos(40°) × cos(23.44°) × sin(111.3342°) = 0.6547
Q
=
498.8599
W/
m
2
Check
Put the numbers back in. Both sides should match.
cos(111.3342°) = −tan(40°) × tan(23.44°) ✓
498.8599
=
(
1,361
π
)
×
(111.3342
×
π
180
×
sin(40°)
×
sin(23.44°)
+
cos(40°)
×
cos(23.44°)
×
sin(111.3342°))
✓
Topic check
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