# Chaitanya's Random Pages

## June 23, 2012

### The shortest day of the year

Filed under: climate and weather,mathematics — ckrao @ 1:14 pm

June 21 was the shortest day of this year here in Melbourne, Australia and I saw on timeanddate.com that the length of the day was 9h 32m 31s. I was curious to see how close this was to the rough estimate that one can derive for the length of day on the winter solstice based on the calculation I give below. We make the following simplifying assumptions.

• The earth is spherical
• On the winter solstice for the southern hemisphere the sun is directly over the tropic of cancer (i.e. directly overhead there at some time during the day)

We use the figure below to visualise the situation at this time. Assume the sun is a long way to the right of screen. It shows the sun directly overhead at the tropic of cancer (C) and Melbourne at solar noon. The left half of the image experiences night. Imagine the earth rotating about the axis joining the north and south poles (NS). Melbourne will then rotate around this axis and in this image it will always be somewhere on the line passing through D and M. When it is at the point D it corresponds to dawn or dusk. Hence the length of its daylight will the time in which it is in the red region.

Note that the angle $\theta$ is equal to the earth’s angle of tilt, which is known to be approximately 23.5 degrees. Below is another look at the red region from a different angle of perspective. In this figure below the line $DQD'$ corresponds to the single point $D$ in the figure above. Let $\alpha = \angle MPD$. Then the fraction of the day in daylight will be $\alpha/\pi$ where $\alpha$ is measured in radians.

Using the notation below, the daylight time will be $\alpha/\pi \times 24$ hours. We use trigonometry to find $\alpha$. Firstly in triangle $OPM$, $\displaystyle r = R \cos \phi \quad \quad (1),$

where $\phi$ represents the latitude of Melbourne (38 degrees).

Secondly $OP = R \sin \phi$. Hence $\displaystyle PQ = OP \tan \theta = R \sin \phi \tan \theta. \quad \quad (2)$

Finally, in triangle $PDQ$, $\cos \alpha = PQ/r$ so combining this with (1) and (2) gives $\displaystyle \cos \alpha = \frac{R \sin \phi \tan \theta}{ R \cos \phi} = \frac{\sin \phi \tan \theta}{\cos \phi}.$

Hence $\alpha = \arccos\left(\tan \phi \tan \theta\right)$ and for Melbourne the predicted length of daylight is $\displaystyle \alpha/\pi \times 24 = \frac{24 \arccos\left(\tan 37.783^{\circ} \tan 23.438^{\circ}\right)}{\pi} \approx$ 9h 22m 54s.

This is about 10 minutes (1.8%) shorter than the actual time, the difference being due to effects such as the sun not being a point object, diffraction of light through the atmosphere and the approximations listed above. However the above calculation is a handy guide. A more precise calculation is shown here.

Incidentally, the shortest day here this year was curious temperature-wise as the graph below shows (data from the Australian Bureau of Meteorology). The temperature hovered between 9.5°C and 9.9°C (at the half-hour sampling instances) between 6:30am and 7:30pm, with higher temperatures on either side! The maximum (after 9am) of 10.5°C was only reached at around 10pm.