About these notes:¶
This document uses the interactive IPython notebook format (now also called Jupyter). The notes can be accessed in several different ways:
- The interactive notebooks are hosted on
githubat https://github.com/brian-rose/ClimateModeling_courseware - The latest versions can be viewed as static web pages rendered on nbviewer
- A complete snapshot of the notes as of May 2015 (end of spring semester) are available on Brian's website.
Many of these notes make use of the climlab package, available at https://github.com/brian-rose/climlab
These notes closely follow section 2.7 of Dennis L. Hartmann, "Global Physical Climatology", Academic Press 1994.
The amount of solar radiation incident on the top of the atmosphere (what we call the "insolation") depends on
- latitude
- season
- time of day
This insolation is the primary driver of the climate system. Here we will examine the geometric factors that determine insolation, focussing primarily on the daily average values.
Solar zenith angle¶
We define the solar zenith angle $\theta_s$ as the angle between the local normal to Earth's surface and a line between a point on Earth's surface and the sun.
from IPython.display import Image
Image('../images/Hartmann_Fig2.5.png')
From the above figure (reproduced from Hartmann's book), the ratio of the shadow area to the surface area is equal to the cosine of the solar zenith angle.
Instantaneous solar flux¶
We can write the solar flux per unit surface area as
$$ Q = S_0 \left( \frac{\overline{d}}{d} \right)^2 \cos \theta_s $$where $\overline{d}$ is the mean distance for which the flux density $S_0$ (i.e. the solar constant) is measured, and $d$ is the actual distance from the sun.
Question:
- what factors determine $\left( \frac{\overline{d}}{d} \right)^2$ ?
- under what circumstances would this ratio always equal 1?
Calculating the zenith angle¶
Just like the flux itself, the solar zenith angle depends latitude, season, and time of day.
Declination angle¶
The seasonal dependence can be expressed in terms of the declination angle of the sun: the latitude of the point on the surface of Earth directly under the sun at noon (denoted by $\delta$).
$\delta$ currenly varies between +23.45º at northern summer solstice (June 21) to -23.45º at northern winter solstice (Dec. 21).
Hour angle¶
The hour angle $h$ is defined as the longitude of the subsolar point relative to its position at noon.
Formula for zenith angle¶
With these definitions and some spherical geometry (see Appendix A of Hartmann's book), we can express the solar zenith angle for any latitude $\phi$, season, and time of day as
$$ \cos \theta_s = \sin \phi \sin \delta + \cos\phi \cos\delta \cos h $$Sunrise and sunset¶
If $\cos\theta_s < 0$ then the sun is below the horizon and the insolation is zero (i.e. it's night time!)
Sunrise and sunset occur when the solar zenith angle is 90º and thus $\cos\theta_s=0$. The above formula then gives
$$ \cos h_0 = - \tan\phi \tan\delta $$where $h_0$ is the hour angle at sunrise and sunset.
Polar night¶
Near the poles special conditions prevail. Latitudes poleward of 90º-$\delta$ are constantly illuminated in summer, when $\phi$ and $\delta$ are of the same sign. Right at the pole there is 6 months of perpetual daylight in which the sun moves around the compass at a constant angle $\delta$ above the horizon.
In the winter, $\phi$ and $\delta$ are of opposite sign, and latitudes poleward of 90º-$|\delta|$ are in perpetual darkness. At the poles, six months of daylight alternate with six months of daylight.
At the equator day and night are both 12 hours long throughout the year.
Daily average insolation¶
Substituting the expression for solar zenith angle into the insolation formula gives the instantaneous insolation as a function of latitude, season, and time of day:
$$ Q = S_0 \left( \frac{\overline{d}}{d} \right)^2 \Big( \sin \phi \sin \delta + \cos\phi \cos\delta \cos h \Big) $$which is valid only during daylight hours, $|h| < h_0$, and $Q=0$ otherwise (night).
To get the daily average insolation, we integrate this expression between sunrise and sunset and divide by 24 hours (or $2\pi$ radians since we express the time of day in terms of hour angle):
$$ \overline{Q}^{day} = \frac{1}{2\pi} \int_{-h_0}^{h_0} Q ~dh$$$$ = \frac{S_0}{2\pi} \left( \frac{\overline{d}}{d} \right)^2 \int_{-h_0}^{h_0} \Big( \sin \phi \sin \delta + \cos\phi \cos\delta \cos h \Big) ~ dh $$which is easily integrated to get our formula for daily average insolation:
$$ \overline{Q}^{day} = \frac{S_0}{\pi} \left( \frac{\overline{d}}{d} \right)^2 \Big( h_0 \sin\phi \sin\delta + \cos\phi \cos\delta \sin h_0 \Big)$$where the hour angle at sunrise/sunset $h_0$ must be in radians.
The daily average zenith angle¶
It turns out that, due to optical properties of the Earth's surface (particularly bodies of water), the surface albedo depends on the solar zenith angle. It is therefore useful to consider the average solar zenith angle during daylight hours as a function of latidude and season.
The appropriate daily average here is weighted with respect to the insolation, rather than weighted by time. The formula is
$$ \overline{\cos\theta_s}^{day} = \frac{\int_{-h_0}^{h_0} Q \cos\theta_s~dh}{\int_{-h_0}^{h_0} Q ~dh} $$Image('../images/Hartmann_Fig2.8.png')
The average zenith angle is much higher at the poles than in the tropics. This contributes to the very high surface albedos observed at high latitudes.
Here are some examples calculating daily average insolation at different locations and times.
These all use a function called daily_insolation in the module climlab.solar.insolation.py to do the calculation. The code implements the above formulas to calculates daily average insolation anywhere on Earth at any time of year.
The code takes account of orbital parameters to calculate current Sun-Earth distance.
To look at past orbital variations and their effects on insolation, we use the module climlab.solar.orbital.py which accesses tables of values for the past 5 million years. We can easily lookup parameters for any point in the past and pass these to daily_insolation.
%matplotlib inline
import numpy as np
import matplotlib.pyplot as plt
import netCDF4 as nc
from climlab import constants as const
from climlab.solar.insolation import daily_insolation
from climlab.solar.orbital import OrbitalTable
First, get a little help on using the daily_insolation function:
help(daily_insolation)
Here are a few simple examples.
First, compute the daily average insolation at 45ºN on January 1:
daily_insolation(45,1)
Same location, July 1:
daily_insolation(45,181)
We could give an array of values. Let's calculate and plot insolation at all latitudes on the spring equinox = March 21 = Day 80
lat = np.linspace(-90., 90., 30.)
Q = daily_insolation(lat, 80)
plt.plot(lat,Q)
plt.xlim(-90,90)
plt.xticks([-90,-60,-30,-0,30,60,90])
plt.xlabel('Latitude')
plt.grid()
plt.title('Daily average insolation on March 21')
In-class exercises¶
Try to answer the following questions before reading the rest of these notes.
- What is the daily insolation today here at Albany (latitude 42.65ºN)?
- What is the annual mean insolation at the latitude of Albany?
- At what latitude and a what time of year does the maximum insolation occur?
- What latitude is experiencing either polar sunrise or polar sunset today?
Global, seasonal distribution of insolation (present-day orbital parameters)¶
Calculate an array of insolation over the year and all latitudes (for present-day orbital parameters). We'll use a dense grid in order to make a nice contour plot
lat = np.linspace( -90., 90., 500. )
days = np.linspace(0, const.days_per_year, 365. )
Q = daily_insolation( lat, days )
And make a contour plot of Q as function of latitude and time of year.
ax = plt.figure( figsize=(10,8) ).add_subplot(111)
CS = ax.contour( days, lat, Q , levels = np.arange(0., 600., 50.) )
ax.clabel(CS, CS.levels, inline=True, fmt='%r', fontsize=10)
ax.set_xlabel('Days since January 1', fontsize=16 )
ax.set_ylabel('Latitude', fontsize=16 )
ax.set_title('Daily average insolation', fontsize=24 )
ax.contourf ( days, lat, Q, levels=[0., 0.] )
plt.show()
Time and space averages¶
Take the area-weighted global, annual average of Q...
Qaverage = np.average(np.mean(Q, axis=1), weights=np.cos(np.deg2rad(lat)))
print 'The annual, global average insolation is %.2f W/m2.' %Qaverage
Also plot the zonally averaged insolation at a few different times of the year:
summer_solstice = 170
winter_solstice = 353
ax = plt.figure( figsize=(10,8) ).add_subplot(111)
ax.plot( lat, Q[:,(summer_solstice, winter_solstice)] );
ax.plot( lat, np.mean(Q, axis=1), linewidth=2 )
ax.set_xbound(-90, 90)
ax.set_xticks( range(-90,100,30) )
ax.set_xlabel('Latitude', fontsize=16 );
ax.set_ylabel('Insolation (W m$^{-2}$)', fontsize=16 );
ax.grid()
plt.show()
Credits¶
The author of this notebook is Brian E. J. Rose, University at Albany.
It was developed in support of ATM 623: Climate Modeling, a graduate-level course in the Department of Atmospheric and Envionmental Sciences, offered in Spring 2015.
%install_ext http://raw.github.com/jrjohansson/version_information/master/version_information.py
%load_ext version_information
%version_information numpy, climlab