You really should be looking at The Climate Laboratory book by Brian Rose, where all the same content (and more!) is kept up to date.
Here you are likely to find broken links and broken code.
This document uses the interactive Jupyter notebook format. The notes can be accessed in several different ways:
github at https://github.com/brian-rose/ClimateModeling_coursewareAlso here is a legacy version from 2015.
Many of these notes make use of the climlab package, available at https://github.com/brian-rose/climlab
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# Ensure compatibility with Python 2 and 3
from __future__ import print_function, division
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
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.
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.
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:
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).
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.
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.
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.
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
$$ \overline{\cos\theta_s}^{day} = \frac{\int_{-h_0}^{h_0} Q \cos\theta_s~dh}{\int_{-h_0}^{h_0} Q ~dh} $$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_insolationin the package
climlab.solar.insolationto 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.
We can look up past orbital variations to compute their effects on insolation using the package
climlab.solar.orbitalSee the next lecture!
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%matplotlib inline
import numpy as np
import matplotlib.pyplot as plt
from climlab import constants as const
from climlab.solar.insolation import daily_insolation
First, get a little help on using the daily_insolation function:
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help(daily_insolation)
Here are a few simple examples.
First, compute the daily average insolation at 45ºN on January 1:
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daily_insolation(45,1)
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Same location, July 1:
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daily_insolation(45,181)
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We could give an array of values. Let's calculate and plot insolation at all latitudes on the spring equinox = March 21 = Day 80
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lat = np.linspace(-90., 90., 30)
Q = daily_insolation(lat, 80)
fig, ax = plt.subplots()
ax.plot(lat,Q)
ax.set_xlim(-90,90); ax.set_xticks([-90,-60,-30,-0,30,60,90])
ax.set_xlabel('Latitude')
ax.set_ylabel('W/m2')
ax.grid()
ax.set_title('Daily average insolation on March 21')
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Try to answer the following questions before reading the rest of these notes.
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
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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.
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fig, ax = plt.subplots(figsize=(10,8))
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=[-1000., 0.], colors='k' )
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Take the area-weighted global, annual average of Q...
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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:
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summer_solstice = 170
winter_solstice = 353
fig, ax = plt.subplots(figsize=(10,8))
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()
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%load_ext version_information
%version_information numpy, matplotlib, climlab
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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
Development of these notes and the climlab software is partially supported by the National Science Foundation under award AGS-1455071 to Brian Rose. Any opinions, findings, conclusions or recommendations expressed here are mine and do not necessarily reflect the views of the National Science Foundation.
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