Insolation is short for ‘incoming solar radiation’. There are various precise things this might mean. Sometimes it means the daily average power of the solar radiation hitting the top of the Earth’s atmosphere at a given latitude and a given time of the year. This quantity is also known as ${\overline{Q}}^{\mathrm{day}}$. For details, see:
Scientists studying the effect of Milankovitch cycles on the Earth’s glacial cycles often want to know the quantity ${\overline{Q}}^{\mathrm{day}}$ at the summer solstice, typically at the latitude 65°. (Presumably the latitude here is a bit arbitrary, but the summer solstice makes some of the math easier). As you can see, this quantity changes in a complicated way with the passage of time:
The red line is the actual value of ${\overline{Q}}^{\mathrm{day}}$. The greenish-brown line is what ${\overline{Q}}^{\mathrm{day}}$ would be if the eccentricity of the Earth’s orbit were zero. This graph was taken from Wikicommons. It as prepared by someone with the monicker ‘Incredio’. Should you believe him? Or should you check his calculation yourself?
To check it, you’ll need this formula for ${\overline{Q}}^{\mathrm{day}}$:
This is derived in the above Wikipedia article using some spherical trigonometry. Here:
${R}_{0}$ is the mean distance from the Earth to the Sun.
${R}_{E}$ is the current distance from the Earth to the Sun (since the Earth’s orbit is an ellipse, this changes during the course of a year).
${S}_{0}$ is the solar constant, about $1367W/{m}^{2}$.
$\varphi $ is the latitude.
$\delta $ is the declination.
$h$ is the hour angle. $\pm {h}_{o}$ is the hour angle when the sun comes up or down, obtained by solving $\mathrm{cos}({h}_{o})=-\mathrm{tan}(\varphi )\mathrm{tan}(\delta )$ when $-1\le \mathrm{tan}(\varphi )\mathrm{tan}(\delta )\le 1$. When $\mathrm{tan}(\varphi )\mathrm{tan}(\delta )>1$, the sun never sets and we say ${h}_{o}=\pi $. For example, this may happen above the Arctic circle in summer. When $\mathrm{tan}(\varphi )\mathrm{tan}(\delta )<-1$, the sun never rises and we can skip to the end of the calculation, since we instantly know ${\overline{Q}}^{\mathrm{day}}=0$. This may happen above the Arctic circle in winter. Warning: The minus sign in the definition of the hour angle for sunrise/set may be due to a different definition of $\delta $ than as given above.
Let $\theta $ be the polar angle describing the Earth’s position in its yearly orbit around the sun. For convenience, set $\theta =0$ at the vernal equinox. Then the declination is given by
where $\epsilon $ is the obliquity, i.e. the angle of title of the Earth’s axis relative to its orbital plane, where obliquity zero means its axis is at right angles to this plane.
The conventional longitude of perihelion $\varpi $ is defined relative to the vernal equinox, so for an elliptical orbit:
where $e$ is the eccentricity of the Earth’s orbit. In other words:
At the summer solstice, things simplify:
and
The longitude of perihelion $\varpi $, the obliquity $\epsilon $ and the eccentricity $e$ change slowly due to the Milankovitch cycles. They are somewhat complicated functions of time, but luckily this website will compute them for you:
So, with some work, you could check Incredio’s graph for yourself!
For what happens after the Sun’s radiation hits the top of the atmosphere, see Solar radiation.