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

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before.

W/m^2=20

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

Reply to
JosephKK
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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 ( snipped-for-privacy@misty.com)

Reply to
Don Klipstein

part:

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before.

W/m^2=20

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Within the polar circles it gets stranger than that. Have to do some = good=20 spherical geometry/trigonometry to get it. Think Barrow, Alaska. ~71 N.

Reply to
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 ( snipped-for-privacy@misty.com)

Reply to
Don Klipstein

baser=20

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Your previous posts not liking Al's film being presented in little=20 kids classrooms is validated. As for the other videos you certainly=20 be specific if you wish. Personally i find Vaclav Klause to be very=20 much on point on questioning "climate change" / "weather variability"=20 and what to do about it.

Reply to
JosephKK

=20

before.

W/m^2=20

W/m^2.

W/m^2.

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day. =20

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Rough cuts are not that impressive here, back-of-the-envelop should be=20 stated as such, because it will surely be seen as such.

Reply to
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 ( snipped-for-privacy@misty.com)

Reply to
Don Klipstein

wrote:=20

before.

435 W/m^2=20

W/m^2.

172.8 W/m^2.

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~71 N.

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Have to give you chops on that.

Reply to
JosephKK

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