# NavList:

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

**Re: Sun's average declination**

**From:**Frank Reed

**Date:**2015 Mar 15, 13:42 -0700

It would have been interesting if we could come up with a way to see this from the analemma, Greg, but I don't think there's any "clever" way to do that.

Don Seltzer's comment provides a really good way to get a quick estimate of the actual value of the average declination. We know that a graph of the Sun's declination through the year starting from the Spring equinox looks a lot like a graph of the sine function starting from zero. It's not exact though. The shape isn't quite right, and the positive part of the graph is not balanced by the negative part. In fact the positive part, when the Sun's declination is north, is about 7.5 days longer than the negative part. That's another way of saying that northern hemisphere Spring and Summer (as normally defined, from equinox to equinox) are together a week longer than northern hemisphere Fall and Winter (or equivalentally 7.5 days longer than southern hemisphere Spring and Summer). So if we take a graph of the sine function and stretch out the positive arch a bit, we can see that the average declination will be north. To get the average value, we need the area under the positive portion and the area "under" the negative portion. The positive portion reaches up to a maximum value of 23.45° and has a width of 186 days. The negative porition has a minimum of -23.45° and a width of 179 days. For a rough approximation to the area, we can treat that sine arch as a triangle. Then the area is (1/2)·width·height. The actual area under the arch is somewhat greater --it's bigger than a triangle-- so let's call it (2/3)·width·height. The area under the positive Spring+Summer arch is then (2/3)·23.45°·186d while the negative arch is (2/3)·23.45°·179d. Subtract those and divide by 365 to get the average declination: (2/3)·23.45°·(7d/365d) = 0.3°. Robin Stuart and Herbert Prinz worked it more exactly and found 0.4°. That's 0.4° North, of course. So the average declination of the Sun is about 24' north of the equator. It takes just about a day for the Sun to reach that declination after the Spring equinox. The Sun is moving north at "1 knot" at the equinox. So next week, that's a little trivia to contemplate. The equatorial value of the Sun's declination, right at the instant of the equinox, is nice and symmetrical, but the real "average" value in the Sun's declination is reached 24 hours later.

As for the weather, * this keeps the glaciers away!* It's part of the Milankovitch cycles. Because the Sun is at aphelion in early July, it lingers longer over the northern hemisphere. That's why the Sun's annual average declination is north of the equator. Without that convenient timing, and a few other related astronomical cycles, the snow would last and last and some of us would be looking at snow in our backyards all year. Snow and glaciers.

That's interesting, Herbert, that the average declination is slowly falling with the passing centuries. I would not have guessed that. I think Antoine has spotted the key factor here: this is due to the overall reduction in the Earth's axial tilt. Back in December there was a media story running around claiming that the solstice night would be the "longest night in history" (for northern hemisphereans). Apparently the creator of this story, deliberately attempting to spawn a popular "viral" story, based this theory on the fact that the absolute lengths of days (full rotations of the Earth) are slowly increasing. That's only true in the long term so even his basic model was flawed. But additionally, the Earth's axial tilt is decreasing which means that days and nights are actually becoming, very gradually, more equal in length. That's the dominant factor affecting long-term changes in the length of nights, by my calculations at least.

Frank Reed

Conanicut Island USA