How to get the damaging leaked University of East Anglia CRU filesabout AGW

[Don Klipstein]

If you cannot do geometry let alone physics any better than that you
deserve all the disregard you can possibly get.

This came from the lower color coded map at:

http://en.wikipedia.org/wiki/File:Insolation.png

The lower one is after atmosphere and clouds. The upper one is before.

Both maps appear to me to be reasonable.

Insolation at the equator on an equinox averaged over a day is 435 W/m^2
above the atmosphere.
(1366 W/m^2 divided by pi)

Insolation at a pole on summer solstice above atmosphere is 543 W/m^2.
(1366 W/m^2 times sine of 23.439 degrees)

Insolation at a pole year-round average above atmosphere is 172.8 W/m^2.
(543 W/m^2 divided by pi)

- Don Klipstein ([email protected])

 

[Don Klipstein]

~~~~~~~~~~~~~
[1] http://www.usatoday.com/money/economy/housing/2009-10-23-foreclosures-interactive_N.htm
[2] http://www.washingtonpost.com/wp-dyn/content/article/2009/12/13/AR2009121302442.html
The Coming Debt Panic. 12/14/09 "It's time to stop worrying about the
deficit -- and start panicking about the debt."

[3] estimate of the Actuary, http://www.cms.hhs.gov/ActuarialStudies/
(http://www.cms.hhs.gov/ActuarialStudies/Downloads/
S_PPACA_2009-12-10.pdf)
Estimated effect on insurance by 2019, numbers in millions of
Americans

(view table in fixed font)
Coverage current law Reid bill difference
-------- ----------- --------- ----------
Medicare 60.5 60.5 0

Medicaid
& CHIP 63.5 83.6 20.1 million

Employer-
sponsored
insurance 165.9 160.7 -5.2

Individual
coverage 25.7 45.9 20.2

Uninsured 56.9 23.6 -33.3
----- ----- -----
TOTALS 372.5 374.3 1.8

NET ADDITIONAL INSURED -------------------^^^

20.1 -5.2 +20.2 is 35.1, not 1.8. 1.8 million is the discrepancy
between a gain in 35.1 million insureds and a reduction of 33.3
million uninsureds.

- Don Klipstein ([email protected])

 

[Don Klipstein]

When the solar rays hit the ice or water surface, part of the power is
reflected directly back to space, while part is refracted into the ice
or water and finally absorbed.

If I understand correctly, the reflected part is determined by the
reflection coefficient according to the Fresnel equations.
http://en.wikipedia.org/wiki/Fresnel_equations

The reflective constant depends of the refractive index and the angle,
at which the rays hit the surface.

The refractive index for water is 1.333, while for ice, 1.31 seems to
be listed. Thus, when comparing the reflective coefficient and hence
the transmission coefficient into ice/water for a flat ice surface and
a calm water surface, the amount of power refracted into ice/water is
practically the same.

Thus, from the power balance point of view, what difference does it
make, if the Arctic ice melts and there are open waters a few months
each year ?

Ice is generally more reflective than water, since it tends to be more
white in color. I is usually not transparent but white to light gray, as
photos of Arctic and Antarctic sea ice show. It often even has snow on
it.

While fresh clean white snow will reflect most of the power back to
space, in the arid polar desert, how often are there fresh snow ?

It snows there, and actually somewhat often - just usually not a lot.

How long will the fresh snow remain white and clean, when the
contamination from industry etc. will increase the absorbtion,
speeding up the melting of the snow and ice ?

- Don Klipstein ([email protected])

 

[JosephKK]

This came from the lower color coded map at:

http://en.wikipedia.org/wiki/File:Insolation.png

The lower one is after atmosphere and clouds. The upper one is before.

Both maps appear to me to be reasonable.

Insolation at the equator on an equinox averaged over a day is 435 W/m^2
above the atmosphere.
(1366 W/m^2 divided by pi)

Insolation at a pole on summer solstice above atmosphere is 543 W/m^2.
(1366 W/m^2 times sine of 23.439 degrees)

Insolation at a pole year-round average above atmosphere is 172.8 W/m^2.
(543 W/m^2 divided by pi)

- Don Klipstein ([email protected])

Please why you use pi as the divisor. I might find it right after you explain it.

 

[Don Klipstein]

Please why you use pi as the divisor. I might find it right after you
explain it.

For one day at the equator at an equinox, the sun is up half the day.
During that half of the day, the sun moves through the sky at constant
angular velocity.

Excluding atmospheric effects, insolation is equal to that with sun at
zenith, multiplied by sine of angle above horizon. A graph of insolation
as a function of time will be a halfwave rectified sinewave. Average
value of that over a half cycle (12 hours) is peak times 2/pi. Average of
that for a full cycle (24 hours) is half that, or peak divided by pi.

In the case of year-round insolation at a pole excluding atmospheric
effects, I just now realize this is not exactly correct but an
oversimplification. For the half of the year that the sun is up, its
angle above the horizon as a function of time graphs as a halfwave
rectified sine wave. Average angle of the sun above the horizon is peak
angle times 2/pi.
I oversimplified insolation to be proportional to angle, since sine of
an angle is close to proportional to angle as the angle varies through a
range of 0 to 23.44 degrees. So as an oversimplified approximation,
average insolation over the 6 months with sun is peak insolation times
2/pi. (Excluding effects of atmosphere.) I just realized several
minutes ago that the exact figure is slightly greater. If I calculated
right with a brute force method, it's about .93% greater.

- Don Klipstein ([email protected])

 

[JosephKK]

For one day at the equator at an equinox, the sun is up half the day.
During that half of the day, the sun moves through the sky at constant
angular velocity.

Excluding atmospheric effects, insolation is equal to that with sun at
zenith, multiplied by sine of angle above horizon. A graph of insolation
as a function of time will be a halfwave rectified sinewave. Average
value of that over a half cycle (12 hours) is peak times 2/pi. Average of
that for a full cycle (24 hours) is half that, or peak divided by pi.

In the case of year-round insolation at a pole excluding atmospheric
effects, I just now realize this is not exactly correct but an
oversimplification. For the half of the year that the sun is up, its
angle above the horizon as a function of time graphs as a halfwave
rectified sine wave. Average angle of the sun above the horizon is peak
angle times 2/pi.
I oversimplified insolation to be proportional to angle, since sine of
an angle is close to proportional to angle as the angle varies through a
range of 0 to 23.44 degrees. So as an oversimplified approximation,
average insolation over the 6 months with sun is peak insolation times
2/pi. (Excluding effects of atmosphere.) I just realized several
minutes ago that the exact figure is slightly greater. If I calculated
right with a brute force method, it's about .93% greater.

- Don Klipstein ([email protected])

Within the polar circles it gets stranger than that. Have to do some good
spherical geometry/trigonometry to get it. Think Barrow, Alaska. ~71 N.

 

[Don Klipstein]

Within the polar circles it gets stranger than that. Have to do some good
spherical geometry/trigonometry to get it. Think Barrow, Alaska. ~71 N.

Yes, true. At all latitudes between equator and poles, as well as
equator on a day other than an equinox, I would brute-force it with motion
in spherical coordinates. I would have to calculate angle of sun from
horizon for every minute of every day of the year, take sines of these,
change all negative ones to zero, and average them.

It's just that at the poles yearround and at the equator on an equinox,
there are easier special cases to support my contention of how much
insolation there is at the poles and how much there is at the equator.

- Don Klipstein ([email protected])

 

[JosephKK]

Please retract that claim.

BTW your memory is failing. I last stated my position on this on s.e.d
in a reply to you about 18 months ago - Google groups is handy:

http://groups.google.co.uk/group/sci.electronics.design/msg/19d16752dad82548?hl=en&dmode=source


I am on record from the release of Al Gores film saying that I did *not*
think it appropriate or helpful to show it in UK schools and that I
think he is a hypocrite. My position is clear enough. eg.

http://groups.google.co.uk/group/sc...&lnk=gst&q=Gore+Martin+Brown#4e0c812701c65810

Regards,
Martin Brown

Your previous posts not liking Al's film being presented in little
kids classrooms is validated. As for the other videos you certainly
be specific if you wish. Personally i find Vaclav Klause to be very
much on point on questioning "climate change" / "weather variability"
and what to do about it.

 

[JosephKK]

Yes, true. At all latitudes between equator and poles, as well as
equator on a day other than an equinox, I would brute-force it with motion
in spherical coordinates. I would have to calculate angle of sun from
horizon for every minute of every day of the year, take sines of these,
change all negative ones to zero, and average them.

It's just that at the poles yearround and at the equator on an equinox,
there are easier special cases to support my contention of how much
insolation there is at the poles and how much there is at the equator.

- Don Klipstein ([email protected])

Rough cuts are not that impressive here, back-of-the-envelop should be
stated as such, because it will surely be seen as such.

 

[Don Klipstein]

Rough cuts are not that impressive here, back-of-the-envelop should be
stated as such, because it will surely be seen as such.

I did only mention two special cases. One is exact as far as I know.
I thought the other one was at the time I stated it, but I admitted my
error.

- Don Klipstein ([email protected])

 

[JosephKK]

I did only mention two special cases. One is exact as far as I know.
I thought the other one was at the time I stated it, but I admitted my
error.

- Don Klipstein ([email protected])

Have to give you chops on that.