# NavList:

## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

**Re: More sunlight in Greenland than in the Amazon**

**From:**Frank Reed

**Date:**2020 Dec 26, 08:40 -0800

David Pike, you wrote:

"try looking at the angle the sun’s direction of travel crosses the horizon at sunrise or sunset today at say 20S and 65N"

Right. That's the key to this. If we're using the standard definition of sunrise and sunset, the Sun has to cross an extra 50' of altitude below the horizon. How long does that take? This is a small enough angle that we can use the instantaneous derivative, which happens to be quite simple:

dh/dT = 15ʹ × cos(Lat) × sin(Azm),

which means that the rate of change in altitude of the Sun is 15 minutes of arc per minute of time multiplied by the cosine of the latitude and the sine of the azimuth, and this is true at any altitude in the sky, too. And we can think about this geometrically in terms of the angle that the motion of the Sun makes with the horizon. At the equator on an equinox, the Sun rises vertically --it's motion is straight up from due East at the simple standard rate of the Sun. In the math here, cos(Lat) and sin(Azm) are both one for this scenario so the rate of change is just 15' per minute as expected. To cover 50' of altitude at 15' per minute takes 3m20s so the day will be longer than twelve hours by 6m40s. On other dates it's slightly longer, something like 7m14s on the solstices, so let's call it 7 minutes to first-order.

In other latitudes, the Sun rises at angle. The angle of its motion away from the vertical is approximately equal to the latitude, very nearly, when the Sun's declination is small. It may be helpful to re-write the rate of change in terms of the angle away from E/W, also known as the "amplitude". Then the rate of change of altitude in a minute of time is 15ʹ × cos(Lat) × cos(Amp). Consider a case where the observer is at a latitude of 45°N and the Sun is 10° from East as it rises. How long does it take to cover the extra 50' below the true horizon? Again, we can think about (and draw) the picture here. The Sun is rising on a slant. In fact, it is climbing at an angle of 45° relative to the vertical so it takes considerably longer to cross that 50' band below the horizon. The azimuth/amplitude makes little difference in this case (because cos10° is 0.985). The Sun will take about 41% longer (sqrt(2)-1) to cross the 50' band and if we do the math, the time is about 4m47s. And note that the rising angle is nearly the same on every date in that latitude. The length of daylight is longer than the "pure" true altitude length by 9.4 to 11.1 minutes --call it 10 minutes to first-order. The average day length is 12 hours 7 minutes at the equator, 12 hours 10 minutes at 45° latitude.

And finally we can consider the extreme case near (but just south of) the Arctic Circle. For some days out of the year, the Sun skims just below the horizon. But thanks to being elevated by 50' (SD+refraction), it's in the sky much longer than it should be. I can imagine some artwork for this, showing the Sun's daily motion as a circle skimming the horizon with its edge flattened and elevated by refraction, but I'll leave that for someone else to illustrate (if I can stop myself!).

Reminder: on top of these latitudinal effects, there's the elliptical orbit issue which we have already discussed. That would break the symmetry between N and S latitudes.

Frank Reed