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Patrick Kelly wrote:
>
> "Here we assume a "solar constant" of 1367 W m^-2 (Hartmann 1994),
> and because the incoming solar radiation is one-quarter of this,
> that is, 342 W m^-2., a planetary albedo of 31% is implied."
>
> Stripping this sentence to bare bones we are told that-- 'Because
> the incoming solar radiation is one-quarter of the "solar
> constant" a planetary albedo of 31% is implied', which is of
> course nonsense. So one has to ask what the authors might have had
> in mind when they wrote this.
>
>
> Hi again:
>
> I think that I have this sorted out, and I think part of it is that
> the original statement is worded poorly. I think what it should read is:
>
> Here we assume a "solar constant" of 1367 W m^-2 (Hartmann 1994),
> and because the incoming solar radiation is one-quarter of this,
> that is, 342 W m^-2. A planetary albedo of 31% is used.
>
>
> There key thing about the solar constant (1367 W per square metrr) is
> that it is for a plane that is perpendicular to the sunlight
>
> For the first point, the total power available to the Earth would be
> 1367 W per square metre multiplied by the circular area of the Earth.
> which is pi x radius x radius; or 1367 x PI x r X r.
>
> Obviously not all of the Earth's surface is perpendicular to sunlight.
> The total area on which the amount just calculates impinges is the
> total surface area of the Earth facing the Sun, which = 4 X PI X
> radius x radius.
>
> Therefore on average, the each square metre of the Earth will receive:
>
>
> 1367 x PI x r x r
> ------------------------------------------------
> 4 x PI x r x r
>
> = 342 Watts per square metre.
>
> Now for the second point. The The Earth's albedo (or its
> reflectivity), as measured by satellites is 0.31, that is 31% of all
> incoming radiation is reflected away by clouds, ice, water, etc. In
> Figure 7, 31% of the 342 Watts per square metre would be 106 Watts per
> square metre. The left side of the figure shows 77 W/m^2 being
> reflected by the atmosphere, and 30 W/m^2 by the ground for a total of
> 107 W/m^2
Hi Patrick & All, Dec 17, 2006
Your interpretation of the sentence in question essentially echos
mine, except you of course meant that 4pi*r*r was the total surface area
of the earth, not the area of the half that was facing the sun.
And your take on the origin of the 31% albido figure nicely
illustrates my point that this paper must be read with more than usual
care. For the benefit of those who might appreciate a picture as
opposed to chapter & verse I explained their derivation of albido using
Fig. 7.
But reverting to chapter & verse---
1) Satellite observations at the time of writing (p.198 cols 1 & 2)
indicated albedo of near 30% or 33%.
2) Error in measurements of absorbed solar radiation (incoming) was
about 4 times as great as was error in outgoing longwave (p. 199, col 1).
3) They assumed net inflow of energy at the top of the atmosphere to be
zero (p 198, col 1); i.e. outgoing longwave assumed equal to incoming
absorbed shortwave (see also p. 203 col 1).
4) Consequently "...we use the ERBE outgoing longwave flux of 235 W m^-2
to define the absorbed solar flux." (p. 199 col 2).
5) Thus they calculated reflected as incident (342) minus absorbed
(235); yielding reflected of 107 W/m^2; not explicitly stated but it
follows from items 1 to 4.
It does help to know the altitude associated with 'top of the
atmosphere'; thanks.
No, I don't think you missed anything but bear in mind that there
are 10 pages of text & we focused attention on the mechanics of one
explanatory sentence.
There is more meat in these 10 pages than I will attempt to digest
but I wonder. Is it possible to obtain a realistic energy balance if the
model assumes static conditions ? Is a model that assumes no motion, no
temperature gradients, no ice or snow able to reflect reality in a
useful way ?
Yours truly, Dave Webster, Kentville
>
> i.e. no allowance is made for the greater surface area at the top
> of the atmosphere than at the surface of the earth.
>
>
> The difference thickness of the atmosphere is so small compared to the
> radius of the Earth that it can be ignored. If you take 20 km to be
> the thickness of the atmosphere, and the radius of the Earth to be
> 6378 km* then the surface are at the top of the atmosphere will be
> increased by a factor of (6398/6378)^2 or 0.6%
>
>
> Given that I don't really see any problem unless there is something
> that I missed.
>
> Pat
>
> * From the 2007 RASC Observer's Handbook, available from Atlantic News
> and The Book Room or on-line at http://store.rasc.ca. I would
> recommend this book even if I wasn't the new editor ;-)
>
>
>
> ==========================================================================
>
> Patrick Kelly
> Director of Computer Facilities
> ==========================================================================
>
> Faculty of Architecture and Planning
> Dalhousie University
> ==========================================================================
>
> PO Box 1000 Stn Central 5410 Spring Garden Road
> Halifax, Nova Scotia B3J 2X4 Halifax, Nova Scotia B3J 2X4
> Canada Canada
> ==========================================================================
>
> Phone:(902) 494-3294 FAX:(902) 423-6672 E-mail:patrick.kelly@dal.ca
> ==========================================================================
>
>
>
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