### "Zenial Days" on Hawaii

I live in Kaunakakai, Hawaii, on the island of Molokai. The busiest intersection on the island is more or less at 21.0893 N, 157.0227 W. We'll use that for Kaunakakai.
As I understand it, **the longest day** of the year in our tropical town **is not the summer solstice**. We should have two longest days of the year, one shortly before the summer solstice, one shortly afterward. The sun should be directly overhead once as it travels north to the Tropic of Cancer, and once as it heads back south for its rendezvous with the equator and the equinox.

How can I find out exactly when those dates are? The sunrise and sunset times on our local tide calendar are too imprecise for me to figure it out that way. Is there a celestial almanac that will list the latitude of the sun for each date? Is there a formula a 7th grade English teacher can figure out?

### Reply

You indeed seem to have a clear understanding of the motions of the Sun.
Every day the Sun rises somewhere near east and sets somewhere near west, and at noon (in the continental US) it passes to the south and reaches its largest elevation angle above the horizon (largest for the day). In those states, that angle is always less then 90°, so the Sun never passes overhead--"at zenith." Also (because of the tilt of the Earth's axis), that angle depends on the date. In summer it grows larger as the date approaches midsummer day on 21 June, then it decreases again (see "From Stargazers to Starships").

Hawaii is close to the equator, so the Sun's noontime elevation there in midsummer actually exceeds 90°, making it pass **north** of the zenith. The noon elevations before and after that day get smaller, and you are asking, when are they 90° , so that the noontime Sun passes **exactly** overhead.

I would guess those dates are **May 26** and **July 18**. (These are "**zenial days**" and were apparently noted by Maya astronomers).

But you seem to want more than numbers--**you also want understanding**. I therefore ask you to **draw a picture** on a sheet of paper, a simplified **north-south cross-section** of the celestial sphere at Kaunakakai. Then go through the reasoning below, very slowly, making sure you understand every step. You might want to draw additional sketches, and also to consult suitable parts of "From Stargazers to Starships."

First draw a horizontal line (x-axis), which represents the north-south axis on the ground (north at right). Add to it (above it) a perpendicular line (y-axis) which points up to zenith, straight overhead. The origin is where **you** stand.

On the right, add a straight line northwards from the origin, making an angle 21.09° with the x-axis (you need no accuracy, this is just a sketch). It represents the direction to the celestial pole (or the pole star, very nearly) from Kaunakakai. And finally, a line from the origin, making a 90° angle with the preceding one--i.e. tilted southwards from the zenith, by an angle 21.09 degrees. It represents the direction to the celestial equator.

Over the year the Sun (in its apparent motion relative to the stars) traverses the **ecliptic**, a circle on the celestial sphere making an angle 23.5° with the celestial equator. At equinox it is on the intersection of equator and ecliptic, during the summer it is north of the equator, during winter south of it.

On any day of the year, it will have on that sphere a certain "distance" (actually, an angle) north or south of the equator. Suppose it is summer. Imagine a line on the celestial sphere, from the celestial pole to where the Sun is, and you continue it until it hits the equator. The size (in degrees) of that last section is **the Sun's "declination angle"** at that time, its elevation angle above the equator. On midsummer day it is 23.5°, at equinox, zero, and you can extend the definition to find its value on midwinter day as (–23.5°).

Now go back to your drawing, of the north-south sky at Kaunakakai. On midsummer day at **noon**, the direction to the Sun makes an angle of 23.5° with the direction to the celestial equator. You can add that direction to your sketch as a broken line, if you want, and because 23.5° is **larger** than the latitude of Kaunakakai, it is tilted **past** zenith, **northward** of the y-axis. Not the highest possible elevation! You want the Sun to be **at** zenith, which means the declination--the elevation (above the equator) of the point of the ecliptic where the Sun is located--has to be 21.09°, or about 21 degrees, 5.5 minutes.

The web has many sites which calculate or tabulate the position of the Sun, and the US Naval Observatory publishes a yearly **astronomical almanac**, but for your questions, you may find the required information at

That is a tabulation of many variables, some specific to a location at latitude 45 north--but all you need is the declination of the Sun, which is tabulated day by day. Look on what days it comes closest to 21 degrees, 5.5 minutes, and you have your answer.

Please let me know how clear the above explanation has been. I could probably do a better job in a short session with you in front of a blackboard or a sheet of paper, but I'm retired and no one will pay me to fly to Hawaii to do so. (Curses!).