Combine daylight duration and solar angle

Duration and angle contribute together

Noon altitude describes one instant. Daylight duration describes how long the Sun’s center is above the horizon. Daily solar energy depends on the full curve of projected input, combining both. A lower noon Sun can still contribute substantial daily energy when illumination lasts many hours.

Start at latitude +45° and compare the two solstices. Record noon altitude, daylight hours and daily mean irradiance separately. Then compare the June case with the equator and North Pole. Avoid ranking daily input from noon altitude alone; inspect the integral represented by the full-day summary.

At June solstice, +45° has a noon Sun at 68.44° and about 15.43 hours of daylight, producing about 43.14 MJ/m² daily. The North Pole has a lower Sun at 23.44° but 24 hours of illumination, producing about 46.78 MJ/m². This concrete counterexample shows why noon altitude alone cannot rank the daily totals. It does not predict that the pole is warmer.

The model assumes a 24-hour solar day and holds the selected seasonal geometry fixed during that day. Its input is direct radiation on a horizontal plane at the top of the atmosphere. Cloud cover and atmospheric transmission do not explain any differences because they are not included.

Define daylight before counting it

This investigation counts only times when the Sun’s center is strictly above an ideal horizon. It ignores the finite solar disk, atmospheric refraction and terrain. Published sunrise and sunset conventions can differ: NOAA’s calculation reference uses a zenith angle of 90.833° to approximately include refraction and solar-disk size. NOAA sunrise/sunset conventions.

For ordinary rise-and-set cases, define A = sin φ sin δ and B = cos φ cos δ. The horizon condition is A + B cos H = 0. The positive sunset hour angle H₀ therefore satisfies cos H₀ = −A/B. When H₀ is in radians, daylight duration is 24H₀/π hours.

This formula needs separate polar cases rather than inserting values outside the inverse-cosine range or dividing by zero. Continuous above-horizon illumination gives 24 hours; continuous below-horizon conditions give zero. The model reports those cases explicitly.

Inspect the exact-pole equinox limit

At an exact pole and equinox, the Sun’s center lies on the ideal horizon throughout the day. It is neither strictly above nor below. The label is horizon all day, with zero hours strictly above and zero direct horizontal flux.

Do not describe that limiting case as an ordinary polar night. The geometric position is different even though the strict above-horizon count and direct horizontal input are zero in both cases. The zero-tilt comparison has this horizon-all-day behavior at the exact poles at every orbital angle.

At North Pole in June, the Sun center remains above the horizon with constant altitude +23.44° in the fixed-day model. In December it remains below at −23.44°. Compare the three labels before interpreting the plotted number of daylight hours.

Convert a daily mean into daily energy

The daily mean averages over the full 24 hours, including zero direct input at night. Daily energy in MJ/m² is mean irradiance in W/m² multiplied by 0.0864, because 86,400 seconds divided by one million converts joules to megajoules.

At an equinox on the equator, the ideal day lasts 12 hours and noon irradiance is 1361 W/m². The daily mean is 1361/π, about 433.2 W/m², and daily energy is about 37.43 MJ/m². Simply multiplying noon irradiance by half a day would overestimate the total because morning and afternoon sunlight arrives obliquely.

On paper, sketch a daylight hump and its corresponding constant daily-mean rectangle over 24 hours. Their areas represent the same daily energy. The rectangle is an equivalent average, not a claim that sunlight stays constant overnight.

Transfer to a reported daily mean

A hypothetical geometric case has daily mean irradiance 250 W/m². Convert it to daily MJ/m². Does that value alone tell you its noon altitude or hours of daylight?

Check the day-length transfer

The daily energy is 250 × 0.0864 = 21.6 MJ/m². The mean alone does not separately determine noon altitude or daylight duration; different time profiles can have the same integrated energy. The complete geometry or daily curve is needed to distinguish them.