How this instrument works
The sunset hour angle H₀ marks the moment, measured in degrees of Earth's rotation from solar noon, when the sun's centre crosses a flat, unobstructed horizon. It comes from pinning the sun's elevation at zero — horizon level — inside the general relation that ties elevation to latitude, declination, and hour angle, then isolating the angle itself: the latitude-declination sine product moves across the equals sign, the matching cosine product cancels from both sides, and what remains is cosH₀ = −tanφ·tanδ, exactly the arccosine this instrument evaluates.
That tangent product is also where the formula shows its limits honestly. Near the equator, or at modest latitudes through most of the year, it stays comfortably between −1 and 1, and acos returns an ordinary angle. Push latitude and declination high enough together — deep into an Arctic or Antarctic summer or winter — and the magnitude of tanφ·tanδ exceeds 1, so no angle satisfies the equation: the sun never sets, or never rises, on that day at that latitude.
Turning H₀ into a clock reading uses Earth's own rotation rate: 15° of hour angle per hour, so sunset = 12 + H₀ ⁄ 15°, adding the hours from solar noon onto noon itself. That result is solar time, not wristwatch time — it assumes the sun's geometric centre and a perfectly flat horizon, with no allowance for atmospheric refraction, an effect that lifts the setting sun's apparent disk by roughly 34 arcminutes and pushes the real, visible sunset a few minutes past what this geometric hour angle reports.
- Set Latitude for your site — degrees north of the equator count as positive, degrees south as negative.
- Set Solar declination for the date you care about; it reads 0° at the equinoxes and swings to about ±23.45° at the solstices.
- Read Sunset hour angle, rad — the angular distance from solar noon to sunset, expressed in radians.
- Read Sunset, solar hours after midnight for the clock reading, in solar time, at which the sun sets.
Worked example — 40°N latitude at the equinox
Pick a site at 40°N and check it on whichever equinox is nearest — the one date each spring and each autumn when the sun crosses the celestial equator and declination reads 0° everywhere on Earth at once. Enter Latitude = 40 and Solar declination = 0. The instrument works out H₀ = acos(−tan(40°)·tan(0°)) = acos(−0) = acos(0), and acos(0) is exactly π ⁄ 2 radians — 1.57079632679 rad, the value it reports for Sunset hour angle, rad.
With H₀ known, sunset = 12 + H₀ ⁄ 15° converts that angle into a clock reading: 12 + 1.57079632679 × 3.8197186342054885 = 18.0 solar hours, read out as 18.000000 for Sunset, solar hours after midnight — 6:00 PM in 24-hour solar time. That is exactly twelve hours after the 6:00 AM sunrise a companion instrument reports for the same latitude and declination, the defining symmetry of an equinox, when every latitude splits its day and night evenly.
Questions
Why does the formula use −tanφ·tanδ instead of sinφ and cosδ separately?
Because it is an algebraic simplification, not a separate law. The broader elevation formula ties how far the sun sits from the horizon to a latitude-declination sine product plus a matching cosine product times the cosine of hour angle. Pin that elevation at zero for sunset, move the sine product to the other side, and divide out the cosine product, and each sine-over-cosine pair collapses into a tangent — the reduced version of the same spherical relation, not a shortcut that trades away accuracy.
What does it mean if the calculator can't return a sunset hour angle?
It means the magnitude of tanφ·tanδ exceeds 1, which happens only at high latitude during the local summer or winter half of the year. Physically it signals the midnight sun or the polar night: the sun stays above or below the horizon for the entire solar day, so no angle satisfies cosH₀ = −tanφ·tanδ, and the formula has nothing real left to return.
Why might the actual sunset differ from this instrument's answer?
Because the formula treats the sun as a point on a flat horizon with no atmosphere. Refraction near the horizon lifts the sun's apparent position by about 34 arcminutes, and its visible disk adds roughly another 16 arcminutes of radius on top of that, so published almanacs use an effective horizon near 0.833° below geometric — delaying true sunset a few minutes past what this geometric hour angle predicts.
Why divide the hour angle by 15° to get a clock time?
Because Earth turns 360° in 24 hours, which works out to 15° every hour. An hour angle of H₀ degrees from solar noon therefore corresponds to H₀ ⁄ 15 hours of elapsed time, and adding that to solar noon (12:00) produces the solar-time clock reading this instrument reports as Sunset, solar hours after midnight.
Does this give the time on my wristwatch, or solar time?
Solar time — the reading treats solar noon as exactly 12:00, which rarely matches wristwatch noon. Clock time also depends on where you sit inside your time zone and on the equation of time, an offset that swings up to about sixteen minutes across the year, so turning this result into a watch reading needs that extra correction on top of the hour angle alone.
Who actually relies on a sunset hour-angle formula like this one?
Solar-energy engineers use it to bound the daily generation window of a photovoltaic array; agronomists build day-length models from it to predict when a photoperiod-sensitive crop will flower; and almanac compilers use the same relation, refined with a refraction correction, to print the sunset times that fill a daily newspaper's weather page.