Project sunlight onto a horizontal area

Spread one beam over different horizontal areas

A beam perpendicular to a surface concentrates its intercepted power on a smaller area than the same beam arriving obliquely at a horizontal surface. The incoming source need not change. Draw several parallel rays striking a horizontal line first steeply and then shallowly, and compare the footprint lengths.

Solar altitude α is the angle of the Sun’s center above the ideal horizon. For a positive altitude, the direct horizontal top-of-atmosphere irradiance is F = 1361 sin α, in W/m². The 1361 W/m² value describes an area normal to the rays; the sine projects it onto a horizontal area. Below the horizon, the direct term is zero rather than negative.

At α = 90°, the full normal irradiance reaches the horizontal plane in this geometry. At 30°, sin α = 0.5 and F = 680.5 W/m². At 0°, the rays are parallel to the horizontal plane and the direct horizontal term is zero. This is a geometric top-of-atmosphere calculation, not a measured ground-level radiation budget.

Calculate the noon angle

At local solar noon, the model uses α_noon = 90° − |φ − δ|, where φ is latitude and δ solar declination. The absolute value gives the angular separation regardless of sign. A negative noon altitude means even the highest daily Sun center remains below the horizon.

At latitude +45° during the June solstice, δ = +23.44°, giving noon altitude 68.44°. At an equinox it is 45°, and at December solstice it is 21.56°. Capture these cases and compare their projected noon fluxes using one common scale.

Do not divide altitude angles and expect the same ratio for flux. Sine is nonlinear: halving an altitude does not generally halve its sine. Draw a small right triangle or use a calculator to compare sin(60°) and sin(30°). The latter is about 57.7% of the former, not 50%.

Follow the angle across a day

The hourly geometry uses μ = sin φ sin δ + cos φ cos δ cos H. Here μ equals sin α, and H is the hour angle measured from local solar noon. Each solar hour corresponds to 15°, so H = (solar hour − 12) × 15°. Noon has H = 0; morning and afternoon have opposite signs.

The horizontal irradiance is 1361 × max(0, μ). The maximum operation excludes negative direct flux at night. NOAA’s solar-position equations use the corresponding relationship between latitude, declination, hour angle and zenith angle. NOAA solar-position equations.

Local solar time describes the Sun’s position, not a civil time-zone clock. The investigation does not convert longitude, time zone or daylight-saving conventions into observed sunrise times. Read its daily curves on their stated time basis.

Separate power from energy

A watt is one joule per second. Irradiance in W/m² is therefore a rate of energy delivery per area. Energy accumulated over a day also depends on how long each rate persists. A high noon value alone does not determine the daily total.

For a paper unit check, imagine a constant 500 W/m² lasting two hours. Multiplying by 7200 seconds gives 3,600,000 J/m², or 3.6 MJ/m². The actual modeled solar curve is not constant, so its daily total sums the changing input over the whole day rather than repeating the noon value.

Atmospheric absorption and scattering, clouds, surface reflection and heat storage are absent. No ground temperature follows directly from the plotted flux. A surface energy balance would require additional processes and quantities.

Transfer to an unfamiliar angle

At a positive solar altitude with sin α = 0.25, calculate the direct horizontal flux. If the same value persisted for one hypothetical hour, what energy per area would it deliver?

Check the solar-angle transfer

F = 1361 × 0.25 = 340.25 W/m². Over 3600 seconds that would deliver 1,224,900 J/m², or 1.2249 MJ/m². The one-hour constant-flux condition is a paper exercise, not the changing solar-altitude history generated by the model.