Guest essay by Robert A Cook, PE
Sea ice concentration, north and south poles as observed by satellite. Image from University of Illinois Cryosphere Today
In particular, for the twenty-second of each month, we will calculate and present for discussion:
- that day’s solar radiation level at top of atmosphere (TOA),
- that day’s declination angle (the tilt of the earth’s axis towards or away from the solar plane),
- that day’s average Antarctic and Arctic sea ice area and extents,
- an estimate of the latitude of the edge of that day’s Antarctic and Arctic areas,
- at the edge of the sea ice for that day, estimate the total reflected and absorbed solar radiation into open water and sea ice for a clear day. (This requires an estimate of the sea ice albedo for that day, the solar elevation of the sun for each hour of that day, an estimate of the open ocean water albedo each hour at each solar elevation angle, and an estimate of the atmosphere’s clarity that day, and the air mass attenuating the sun’s energy each hour of that day at that latitude. )
- an estimate of the average additional heat losses each hour on that day from the open ocean and from the sea ice.
Summary
The Antarctic sea ice continues to be far above average for this time of year: rising from +23% Feb 1 to 33.4% on Feb 28. This DOES matter, because the excess Antarctic sea ice this time of year reflects significant amounts of sunlight, and this loss continues to cool the planet. A lot.”
“The Arctic sea ice remains slightly below average for this time of year at -7%.
It doesn’t matter. There is almost no sunlight hitting the Arctic sea ice at this time of year. However, losing this Arctic sea ice cools the planet now, which often leads to additional Arctic sea ice area later in the year, which can reflect more sunlight then, then – again – cooling the planet.
I appreciate Anthony’s patience in delivering this report several days after Feb 22. As an excuse, I could claim that I needed the Cryosphere to process its data for the 22nd, or to claim that I was waiting breathlessly for the Antarctic sea ice minimum to finally arrive ( Minimum looks like it happened 28 Feb, based on Cryosphere increases reported 2-3 March), but we should all be humble as we observe the planet. Its schedule does not recognize our months and days and hours.
Antarctica first?
As usual, Antarctic sea ice goes first for several reasons.
First, it is almost always ignored by the CAGW press agents because the Antarctic sea ice reflects badly on several of their predictions about the effects of CO2 in particular and global warming in general. As observers of the global warming debate, you need to know what is happening all over, not just what the press agents want you to know, and what they don’t want you to know.
We will continue to show through the next few months just how much more important the Antarctic sea ice area actually is to the world’s heat balance: The much-hyped Arctic amplification is a very real effect. But it does NOT only occur in the limited area of the Arctic (where sea ice has been receding for several decades) but around the unlimited seas and ever-increasing sea ice surrounding the Antarctica. Down south, where the sun is always higher in the sky and the solar energy reflected back into space much greater, sea ice area really does matter.
Up north? Not so much 9 months of the year.
22 February 2015, Day-of-Year (DOY) = 053
Antarctic Sea Ice Area (SIA)
The Antarctic sea ice continued to melt through February as sea ice area decreased towards its usual its summer minimum. The Antarctic sea ice anomaly remained positive all month (more sea ice than “normal” for every day in February. The Antarctic sea ice anomaly itself decreased during the month, even though the percent of excess sea ice increased. At 0.618 Mkm^2 on 22 Feb, this “excess” sea ice is now represents a reflecting surface about half the size of Hudson’s Bay, at a latitude slightly further north than Hudson’s Bay.
The Antarctic sea ice has been more than 2 standard deviations above normal for almost every day of the past 2-1/2 years now, and February 2015 only continues that trend towards more sea ice.
SIA 1979-2008, DOY 53, = 1.874 Mkm^2, Average area this date
SIA 2015, DOY 53, = 2.492 Mkm^2, Actual area this date
SIA Anomaly, 2015, DOY 53 = 0.618 Mkm^2, Anomaly this date
Percent increase of Antarctic SIA = 33.0% more Antarctic sea ice than normal for this date
Today’s total Antarctic Ice = 14.0 + 1.5 + 2.492 = 18.0 Mkm^2.
The edge of the Antarctic sea ice is at latitude -68.3 south, slightly closer to the South Pole than the Antarctic Circle at -66.5 south latitude.
(Antarctica’s ice now covers a total area of 18.0 Mkm^2 = 14.0 mkm^2 of continental land ice + 1.5 Mkm^2 of permanent shelf ice plus 2.5 Mkm^2 of total sea ice.) Today’s Antarctic sea ice area represents Antarctica’s annual minimum area.
General Observations: The Antarctic sea ice completed its annual retreat towards the minimum sea ice area in 27-28 February, DOY = 57-58. This year’s minimum was no single sharp “point” but rather a slow flattening of the sea ice area over the last 13 days. You can never predict everything about the sea ice, but it is certainly expected to continue growing from now (2 March) through September’s maximum of 16+ million sq kilometers.
Below, the 1979-2010 avearge Antarctcic sea ice measurements are in green, this year’s actual measurements are in red. The Antarctic sea ice area anomalies are below in blue.
The remaining sea ice tends to be very close to the Antarctic land mass. The large open area (polynaya) in the Ross Sea region in January expanded somewhat, but the “ice island” offshore remained intact. This open area between the edge of the sea ice and the Antarctic continent mass is somewhat unusual, but the open water is expected to re-freeze shortly as air temperatures continue to decline. Most of the time in most years, the Antarctic sea ice lies right up close to the coastline, with the sea ice touching the coast (grounded on the beaches) called “fast ice”. (It is held fast by the land.)
Antarctic Sunlight, DOY = 53.
Solar radiation at Top of Atmosphere (TOA) = 1390 watt/m^2 this date (whole earth exposure) based on a yearly average TSI = 1362 watts/m^2. As it always does, solar radiation at TOA will continue to decrease from its yearly maximum of 1407 watts/m^2 on January 5 to its yearly minimum of 1315 watts/m^2 July 5. As far as the total planet heat balance goes, this means each day-of-year later means the sea ice at each pole will be able to reflect less and less between now and July 5.
Declination Angle on Feb 22 was = -0.183 radians/-10.48 degrees, Tau (the Day Angle) = 0.90
We are still in the Antarctic summer, but February represents late summer – compare it to early August up north. (Australian and South African readers do not need a summer-winter conversion table.)
At the edge of the Antarctic sea ice, at -68.0 latitude, sunrise occurred before 05:00 AM on Feb 22, sunset was 14 hours later after 19:00 PM.
At noon, at -68.0 latitude, air mass = 1.867; direct sunlight on a perpendicular surface = 813 watts/m^2 (Direct radiation on Feb 22 is down from January 22 due to increased air mass (greater attenuation), lower TOA radiation, and a slightly higher latitude of the sea ice edge. All as expected, since Feb 22 is later in the solar year, is right near the point of the annual minimum point for Antarctic sea ice, and has fewer hours of sunlight.)
At noon today, peak radiation on the sea surface = 434 watts/m^2 at a 32.3 solar elevation angle
At noon today under clear skies, the Antarctic Sea Ice albedo = 0.750: of the 434 watts hitting every sq meter of “excess” sea ice, 109 watts are absorbed, and 326 watts are reflected into space.
At noon today under clear skies at 32.3 SEA, the Open ocean albedo = 0.069: of the 434 watts hitting open ocean at the sea ice edge, 404 would be absorbed, and only 30 watts reflected.
Today, this day of year, from each and every “excess” meter of Antarctic sea ice, you can see that an “excess” of 294 watts/m^2 are reflected back into space (326 watts/m^2 – 30 watts/m^, clear day, at noon).
Well, “sunlight” occurs for 14 of the 24 hours down south at latitude -68.0 today, so it’s better to total the 14 hours that the sun is above the horizon. (We’ll compare this value later to what little sunlight is available up north.)
| DIR_Rad Horiz. | Hour | DIR Ocean Albedo | Dir Ocean Absorbed | Dir Ocean Reflected | Dir Ice Absorbed | Dir Ice Reflected |
| 0 | 0.00 | 0.000 | 0 | 0 | 0 | 0 |
| 0 | 1.00 | 0.000 | 0 | 0 | 0 | 0 |
| 0 | 3.00 | 0.000 | 0 | 0 | 0 | 0 |
| 3 | 5.00 | 0.682 | 1 | 2 | 1 | 2 |
| 44 | 6.00 | 0.352 | 29 | 15 | 11 | 33 |
| 124 | 7.00 | 0.205 | 98 | 25 | 31 | 93 |
| 216 | 8.00 | 0.137 | 187 | 30 | 54 | 162 |
| 304 | 9.00 | 0.101 | 273 | 31 | 76 | 228 |
| 374 | 10.00 | 0.082 | 343 | 31 | 93 | 280 |
| 419 | 11.00 | 0.072 | 389 | 30 | 105 | 314 |
| 434 | 12.00 | 0.069 | 404 | 30 | 109 | 326 |
| 418 | 13.00 | 0.072 | 388 | 30 | 105 | 314 |
| 373 | 14.00 | 0.082 | 342 | 31 | 93 | 280 |
| 302 | 15.00 | 0.102 | 271 | 31 | 76 | 227 |
| 214 | 16.00 | 0.138 | 185 | 30 | 54 | 161 |
| 121 | 17.00 | 0.208 | 96 | 25 | 30 | 91 |
| 42 | 18.00 | 0.359 | 27 | 15 | 11 | 32 |
| 3 | 19.00 | 0.698 | 1 | 2 | 1 | 2 |
| 0 | 21.00 | 0.000 | 0 | 0 | 0 | 0 |
| 0 | 23.00 | 0.000 | 0 | 0 | 0 | 0 |
| 3391 | 3391 | 3034 | 357 | 848 | 2543 | |
| Delta: | 2186 | 2186 |
So, over 24 hours, 2186 more watts/m^2 were reflected from each sq meter of “excess” Antarctic sea ice at -68.0 latitude on Feb 22 2015.
Arctic Sea Ice Area (SIA)
22 February 2015, Day-of-Year (DOY) = 53
The Arctic sea ice continues to slowly expand towards its spring maximum in late March. As expected, even as every individual day grows longer after the winter solstice on Dec 22, the Arctic continues to lose heat into space. This heat loss is seen as an increase every day in the Arctic sea ice area.
The sun rises earlier each morning, the sun sets a little later each afternoon +> Again, both as must happen as we approach the spring equinox March 22 when both north and south poles get an equal 12 hours of sunlight, and 12 hours of darkness.
Today’s Arctic sea ice anomaly remains negative at -0.979 Mkm^2. This continues its decade long negative value, and this value continues the steady negative sea ice anomaly started in early 2013 and continued through all of 2014. However, today’s anomaly is significantly smaller than both 2007 and 2012’s record low sea ice anomaly, and it represents an increase in Arctic sea ice area since 2005. Today’s Arctic sea ice anomaly remains within 2 standard deviations of the 1979-2008 mean, and that continues a trend begun in 2013 and continued through most the days since.
Today’s Arctic sea ice anomaly is obviously negative, and represents an area of “lost sea ice” roughly 81% the size of Hudson’s Bay’s 1.2 Mkm^2.
From Cryosphere (the Arctic Climate Research at the University of Illinois) for Feb 22, 2015:
SIA 1979-2008 Average, = 14.005 Mkm^2, (Average area this date)
SIA 2015, DOY 53 Actual Area = 13.027 Mkm^2, (Actual area this date)
SIA Anomaly, 2015, DOY 53 = -0.979 Mkm^2, (Anomaly this date)
Percent of Arctic SIA = only 7.0 % less Arctic sea ice than normal for this date
Total Arctic Sea Ice Area = 13.027 Mkm^2
The edge of the Arctic sea ice lies approximately at latitude 71.6 north, well north of the Arctic Circle at latitude 66.5. (This assumes a circular Arctic sea ice cap, centered at the north pole. The actual Arctic sea ice is only roughly circular, and its geometric center lies closer to the Canadian coast than to the Russian coast.)
Arctic Sunlight, DOY = 53.
Solar radiation at Top of Atmosphere (TOA) = 1390 watt/m^2, (same as Antarctica)
Declination Angle on Feb 22 was = -0.183 radians/-10.48 degrees, Tau (the Day Angle) = 0.90
At the edge of the Arctic sea ice, at latitude 72.0 north, the sun pokes its head above the horizon a little before 09:00 AM, and sets 6 hours later a little after 15:00 PM.
At noon today, 22 Feb, air mass = 6.780, solar elevation angle = 8.1 degrees
At noon today, 22 Feb, peak radiation on the sea surface = 28 watts/m^2 at only 8.1 degrees solar elevation angle.
At noon today, 22 Feb, the average Arctic sea ice albedo = 0.830, (Arctic sea ice albedo is still at its winter maximum) but almost no energy is available: 5 watts would be absorbed, 23 would be reflected.
At noon today, 22 Feb, at 8.1 degrees SEA, the open ocean albedo = 0.419 (Pegau & Paulsen, 2006), so 16 watts/m^2 would be absorbed, 12 watts/m^2 would be reflected.
The Arctic sea ice anomaly is negative (meaning sea ice has been lost from its 1979-2010 average), so open water dominates the reflection exchange: At noon on Feb 22, each sq meter of open ocean absorbed 16 – 5 watts/m^2 => the Arctic Ocean absorbed an additional 11 watts/m^2.
Over the 24 hour day, this was a total of 34 watts/m^2. (The math, if anybody is interested: if open ocean, the water absorbed 4+12+16+12+4 watts = 50 watts/m^2. If sea ice were present, the ice would have absorbed 2+4+5+4+2 = 16 watts/m^2. The difference = 50 – 16 = 34 watts/m^2.)
Today, this day of year, for every “lost” square meter of sea ice, the open Arctic ocean loses more energy from 24 hours of increased losses (increased long wave radiation from the open ocean water, from increased convection and conduction up to the sea surface, and from increased evaporation) than it gains from a few hours of increased absorption in the open Arctic Ocean. In all cases, at this latitude at all hours of the day, more energy is lost from the open Arctic Ocean water than from ice-covered Arctic waters.
Today, this day of year, less Arctic sea ice = more heat loss from the Arctic ocean.
Net Planetary Sea Ice Heat Balance (at noon, this day of year).
Arctic sea ice area anomaly x net solar energy absorbed/m^2 – Antarctic sea ice anomaly x net solar energy reflected /m^2
Over a 24 hour day on Feb 22 2015, the net effect of today’s sea ice was
.979 Mkm^2 x 34.0 watts/m^2 – 0.618 Mkm^2 x 2186 watts/m^2 = -1317.6 MWatts reflected back into space, thus cooling the planet.
Earlier Reports:
January 2015 http://wattsupwiththat.com/2015/01/24/state-of-the-sea-ice-january-2015/
Why regularly discuss sea ice area?
Well, with the “pause” now extending 18 years – 3 months, and with every other CAGW prediction regularly failing as CO2 steadily increases, Arctic sea ice loss is just about the only defense left of the CAGW’s basic predictions. It is regularly hyped and used, so you need to know the details of why they think it is important, and the limits to that assumed importance. (Certainly, the CAGW proponents will not tell you of any limitations or constraints Arctic sea ice poses to their theory of Arctic amplification!) Antarctic sea ice, on the other hand, is failing every assumed CAGW result, and is just uniformly ignored. On the other hand, because it disproves the basic CAGW predictions, you need to know the details of Antarctic sea ice, the problems it poses, and the threats it poses.
Why the twenty-second of each month?
It is a convenient and exciting (well, interesting at least) day for almost all of the changes in all areas we need to look at through the year: solar radiation levels, the earth’s declination, the Antarctic and Arctic sea ice minimums and maximum areas.
The summer and winter solstices (longest day and longest night of the year occur on or about the 22 Dec and 22 June each, the fall and spring equinox fall 90 days later on 21-22 March and 21-22 September each year.
The Antarctic sea ice maximum occurs during the two weeks after 22 Sept each year, the Arctic sea ice minimum occurs a few days earlier: now it is averaging 15-20 Sept. The Antarctic sea ice minimum occurs around 22 Feb each year, the Arctic sea ice maximum occurs a the weeks after 22 March.
Solar radiation is not quite as convenient scheduled, but it is at least completely predictable: maximum solar TOA occurs halfway between 22 Dec and Jan 22 each year, solar TOA minimum occurs 5 July, halfway between 22 June and 22 July.
References and Boilerplate.
Challenge any item or equation you disagree with or wish to expand upon. I will in general treat any specific equation sourced from common geometry (such as a conversion of area into latitude, or the solar elevation angle calculated for a day-of-year and hour-of-day and latitude as a specific ‘thing”. It is not a model, nor an approximation. At sea level, the sun really is exactly that high in the sky on that hour of that day of the year at that latitude on earth. If you disagree with an equation, cite your source and justify the difference.
Cryosphere (Arctic Climate Research at the University of Illinois)
Note: There are several other reliable international labs and institutions reporting daily sea ice areas and extents. Cryosphere at the University of Illinois is unique in reporting both Arctic and Antarctic sea ice areas. All SIA and SIE numbers from each different lab differ from each other day by day, so for consistency across both poles, I will only use Cryosphere’s values for area. (The Cryosphere data and graphs on WUWT Sea Ice pages is released one to two days after processing completes. Their on-line data files use January 1 as DOY = YYYY.0000, so check very carefully the all numbers you when download any of their files. Compare their areas and decimal dates with care during leap years and across seasons.)
Everywhere possible, I will quote the experimental data for actual measurements taken in the Antarctic and Arctic itself (sea ice albedo, air temperatures, water temperatures, winds and wind direction, sea ice area, cloudiness and direct/indirect radiation levels); or from the measurements of real seas and real winds and real waves in the open ocean (ocean albedo). Recognize the original experiment results ARE the data! Equally, each real world measurement has its own limits and its own assumptions.
Again, each source will be discussed in detail over time. Each experimental source will be cited as each detailed equation is discussed – and there will be disagreements between measured results from different sources writing in different journals at different times. Where the source article does present a specific equation or approximation of his or her own work, that equation or constant will usually be used “as-written” for that time frame or those conditions. (For example, in 2001, Dr Judith Curry reported Arctic sea ice albedo as 0.83 That value will be used for all arctic sea ice between January and early May. Her data (confirmed by Dr Perovich, 2002) showed a significant decrease in Arctic sea ice albedo between May and early September, and so her reported values will be curve-fit, and used for all Arctic sea ice albedo’s between those dates. Dr’s Curry, Perovich, and Warren (and others) have few recorded values for albedo during winter, so their 0.83 approximations need to be assumed constant in both hemispheres. Antarctic sea ice albedo reported per Warren, 2005.
There are very substantial differences in sea ice albedo and open ocean albedo, and between atmospheric absorption and diffuse radiation on clear days and cloudy days. For now, we will only evaluate clear days.
Solar radiation will usually be calculated much as the measured source data was obtained: in terms of direct radiation only under clear skies with typical Arctic conditions at typical arctic latitudes. Diffuse radiation and cloud cover and relative humidity levels are very important, but we need to get through many other things first.
Land area and sea ice area will generally use millions of square kilometers as units (abbreviated as Mkm^2) . Angles are in degrees and as Solar Elevation Angles (SEA) not Solar Zenith Angles (SZA), unless otherwise noted.
Your additions and questions about any value are encouraged of course, but always cite exactly what item or quote you question and why you feel it needs to be corrected.
I will not take credit for the basic research results discussed here – all of the hardest field work has already been done years before by many people and many teams from many nations and many institutions, nor of the basic equations and fundamentals used each time. Others deserve that credit, and they will be credited as each detail is discussed. I do acknowledge integrating their work together, and am responsible for the results discussed each time in this series.
Arctic Ice Albedo
1. Curry, J.A. and Schramm, J.L.; Applications of SHEBA/FIRE data to evaluation of snow/ice albedo parameterizations; Journal of Geophysical Research, Vol 106, D14, July 2001.
2. Korff, H.C., Gailiun, J.J. and Vonder Haar, T.H; Radiation Measurements Over a Snowfield at an Elevated Site; Department of Atmospheric Science, Colorado State University, January, 1974.
3. Warren, S.G.; Optical properties of Snow; Review of Geophysics and Space Physics, Vol 20, No. 1; Feb 1982
4. Brandt , R.E., Warren, S.G., Worby, A.P., Grenfell, T.C.; Surface Albedo of the Antarctic Sea Ice Zone, September 2005, Link: http://journals.ametsoc.org/doi/pdf/10.1175/JCLI3489.1
5. Perovich, D.K., Grenfell, T.C., Light, B., Hobb, P.V.; Seasonal evolution of the albedo of multiyear Arctic sea ice; Journal of Geophysical Research, Vol 107, C10, 2002.
[My thanks to reader Matt in January for recommending two references above: Brandt 2005 for Antarctic sea ice albedo measurements, and a link to Perovich 2002 with additional photos and errors not available in Curry 2001.]
North and South Poles: Important Climate Differences
Stemming ice loss, giant atmospheric rivers add mass to Antarctica’s ice sheet
Al Gore, wrong again – Polar ice continues to thrive
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Looking forward to six months down the track when below average ice in the Arctic means warming and above average ice in the Antarctic also means warming (the opposite to now)
It will also be interesting to see the total albedo calculations at the end of 12 months to see if sea ice actually makes any difference to the energy absorbed.
If more sea ice in summer reduces absorbtion but reduces heat loss in the winter, it could be they cancel each other out.
They are not equal. Robert can and has explained better then I. However the basics are that the SH sea ice is always on average at lower latitudes, and so reflecting more insolation then the NH land bound sea ice. In addition the SH sea ice is reflecting stronger insolation during the SH summer, as compared to the NH summer, as the earth is closer to the sun in December, and further away in July. The SH ice is cooling more then warming, (Reflecting more energy from the sun then preventing heat from the oceans escaping)
The NH ice is warming more then cooling. The sun is weaker for two reasons, it is both further away and its energy is impacting the ice at a greater angle.
Yes, but in July and August, the extra sea ice in the SH will mean less heat is lost from the Southern Oceans than normally would be if the ice were at average levels, thus leaving less open water to conduct, convect and radiate heat away.
So now we have sunlight at max strength and SH extra sea ice slowing warming, in Winter we have sunlight at minimum and SHesi slowing cooling.
So I restate: it will be interesting to see what the affect will be for the whole 12 months. Maybe one Jan/Feb effects will be bigger than opposite July/ Aug effects, maybe the reverse, maybe cancel each other out. Time and results will tell. That’s science.
This is the negative feedback to a naturally evolved homeostatic system regulation. The Southern Ocean SST anomaly is has been running negative, and it’s sea ice anomaly is positive. The Arctic SST and air temp anomalies are positive, the Arctic sea ice anomalies are running negative. The Earth is opening and closing its radiator shutters. This means for the current period the system has elevated ventilating of NH heat anomalies and a slowing heat loss in a cooling Antarctic Southern Ocean.
The remarkable thing about our climate for the last 150 years, since exiting the LIA, is how stable the global temps have been. To my mind, this negative feedback regulation of our polar radiators is why the effective climate sensitivity to an enhanced GHE has been essentially down in the noise, confounding the alarmists.
Ultimately, the long term trends in Global climate are dictated by the solar activity variations and TOA insolation changes due to Milankovitch cycles as the polar regions regulate the heat loss at the high latitudes.
Eyes should now be on what the sun is doing as cycle 24 winds down in one of the lowest magnetic cycles seen in at least 100 years.
Say RAC, “””””…..•that day’s declination angle (the tilt of the earth’s axis towards or away from the solar plane),……”””””
What means “the solar plane.” ??
I can think of a plane perpendicular to the earth orbital plane, that passes through the sun and the earth, and the earth rotation axis would have some projected tilt angle ” IN ” that plane, but not ” towards or away ” from it.
Then there is the earth orbital plane that passes through the sun and the earth; but then the Earth rotational axis tilt angle relative to that plane is constant, at least on a daily or monthly basis, but it would have a variable projected tilt angle onto that previously thought of plane.
So I’m not tuned into your geometry Robert; if you could clarify. I’m sure I’m not the only one who doesn’t see the picture.
But now I’ll get back to reading your dissertation in detail.
ThanX for the work you put into this.
George
george e. smith
Yes. Both questions are right. But that short answer doesn’t answer your question, does it? 8<)
Declination Angle. Over short periods of time (decades – NOT the Milankovich Cycle of many centuries!), the earth's axis is tilted 23.45 degrees, and that angle does not change.
Solar Plane. By convention, the solar plane is that perfectly flat surface that the earth's orbit around the sun lies on. The declination angle is a measure of the angle between the earth's axis and that plane – but relative to the sun's position in that plane, not to some "perfect point in the far heavens.
22 Dec, DOY = 356: the north pole points "away" from the sun's north pole, and the decl angle is at its yearly minimum. (A negative value, decl = -23.45).
05 Jan, DOY = 5: The earth is closest to the sun, and is traveling the fastest in its orbit. TOA radiation is at its maximum at 1408 watts/m^2.
22 Mar, DOY = 81: The spring equinox. The earth's tilt is "zero" with respect to the sun's position. On this day, the sun rises due east, and sets due west. We have 12 hours of darkness, and 12 hours of sunlight. Now remember, the earth's axis is still tilted those same 23.45 degrees as in December, January and February – aimed right at that far distance polar star – but with respect to the sun on the solar plane, its decl angle = 0.0
22 June, DOY = 173: The summer solstice (northern hemisphere.) The earth's north pole is now tilted towards the sun position on the solar plane (summer up north), the south pole is tilted away from the sun (winter down south), but the earth itself is nearing maximum distance from the sun in its elliptical orbit. Decl angle = +23.45
05 July, DOY = 186: The earth is at its furthest point from the sun, but is traveling slowest in its elliptical path. Decl angle = +22.88 degrees, obviously down a little bit from its 22 Jun maximum. TOA radiation is down to 1316 watts/m^2, but will begin rising again as we continue towards Jan 5.
22 Sept, DOY = 265: Autumn Equinox. The earth's tilt towards the north Star is still 22.45 degrees, but its declination angle with respect to the solar plane is back to 0.0 again. The sun rises in the east once again, and sets due west once again; and we have our second 12 hour day of the year.
Formulas.
There are several approximations – and I emphasize each is an approximation – of the declinatin angle, but the most accurate is
=0.006918-0.399912*COS(TAU)+0.070257*SIN(TAU)-0.006758*COS(2*(TAU))+0.000907*SIN(2*(TAU))-0.002697*COS(3*(TAU))+0.00148*SIN(3*(TAU))
Above is in Excel format using radians, assuming TAU (the day angle, also in radians) is defined for the day-of-year, and then for each hour of the day.
TAU_Day =2*3.1415*(DOY-1)/365
TAU=2*3.1415*(DOY+D9/24-1)/365
when cell D9 = the hour of the day from 00 to 23
Why so complex?
Because the earth’s orbit is an ellipse, its speed changes as it goes around the sun. Thus, the declination angle changes as the day-of-year changes. Further, when the declination angle is changing the fastest (as declination plot crosses its zero axis at the two equinox) the sea ice is at its most important two points of the year: Near March 22, Maximum Arctic sea ice, Minimum Antarctic sea ice; then – in September, maximum Antarctic sea ice and minimum Arctic sea ice.
I use the hourly declination angle every day-of-year in the spreadsheet for calculating solar exposure: From declination angle and day-of-year, you only need to specify a latitude to get: sunrise, sunset, elevation of the sun at every hour of the day, air mass, atmospheric attenuation, and the "tilt" of the sea surface from the sun.
It's complex no doubt – there are more simple formulas approximating declination angle – but once entered one time, the accurate declination angle is accurately available for every hour of every day at every latitude.
Further thoughts to my post on the polar region sea ice as shutters on a radiator (above)
The polar ice levels are the regulators acting on a time constant of decades to years.
In any human engineered regulator, the set point can be usually be adjusted. For example:
– In a nuclear reactor, the control rods move in or out adjusting the available neutrons to set a higher or lower power output by controlling the rate of fission.
– In a simple gas pressure regulator, a turn screw mechanism changes the force on a diaphragm valve to set an outlet pressure differential.
Some will argue the higher pCO2 will drive a higher set point. I would say yes, if the polar sea ice didn’t respond as it does. But, the underlying physical reason is that the physical properties of seawater don’t change is that it is the seawater that is the control fluid. That is the latent heat properties, the enthalpies of fusion don’t change year to year. Sea water will always freeze (at 1 bar) at around -1.5º C, (the effect of the salt ions are also critical is this process). And thus sealing off the water underneath from further evaporative cooling. And when it melts in the summer sun, it melts at a the same temp (slightly higher than it froze), regardless of pCO2 changes. Year after year, the seawater re-freezes at the same temp as it did the previous years, as the summer sun incidence angle wanes and the nighttime cooling cycles lengthen.
So it is the unique but constant physical properties of sea water freezing and the freshwater ice melting points that control the sea ice regulator set points at the poles, not pCO2.
The “solar plane” would be synonymous with “the ecliptic”, using astronomy terms.
[Correct, that is the better term. .mod]
wickedwenchfan
I don’t see how increased insolation’s effect could offset the decrease in energy input into the system and if it did I don’t see how you could have maintain the increased ice levels year after year.
The thing that concerns me is if we continue to have these levels of “low” ice in the NH and at the same time the “high” level of ice in the SH reflecting more of the input energy what offsets this loss of energy to the planet? does this signal a long term global temp down turn which many here are predicting? I live in the NE US and personally prefer to have the mild winters over the ones we had in the 70’s, I personally have been wishing for a little more Global warming another degree or 2 would be could for my personal outlook during the winter.
What is the mechanism causing the NH and SH ice to transition over time, this is what interests me and is it a stable process? it seems pretty evenly balanced and almost like it was designed to shed energy or accumulate energy when need into or out of the system.
Bob Boder, asking wickedwenchfan
Good questions.
Simple answer? We don’t know.
Oh, by the way, you are not allowed (per common consensus “science” to wonder about Who was the Intelligent Designer who first balanced all these things out. It’s just random chance and simple physics after all. Just another random cloud of perfectly balanced ions randomly gathered together in space in one sphere so gravity could begin gathering them together. All 10^58 ions and nuclei and electrons just perfectly balanced randomly drifting in space only 5 billion years ago.
/sarchasm
Robert A Cook,
I want to thank you for this equation, I’ve got running code, and it’s running away (I think ~10 more days of run time). I went ahead and calculated an accumulated watts for a 24 hour day based on the hourly values, it is already really slow, using a smaller timestep would just take too long.
I’ve download the solar constant data, so each day is calculated based on that. I’d like to add aerosols, if anyone know of a good daily data source please let me know.
For dates without real data, I think I’m going to use 100, and then the result will be a percentage of the solar, instead of a fixed unit. This is similar to the base field that BEST calculates.
Good article as well!
Well RAC, Mother Gaia knew all that which you just posted, except she uses Lotus 123.
And she doesn’t tell us any of the details anyhow, so I will have to pay attention, and read your expose, carefully. Thanx for that and for the Excel form too.
Now I just have to figure out how Kevin Trenberth gets your 1408 to 1316 Wm^-2 down to 342.
I just assume that it averages 1362 Wm^-2 , for 24 hrs per day. (somewhere).
Interesting that your max / min nummers do average to 1362, which is the latest NASANOAA figure that I have seen, and which I adhere to.
When you look at your geometry, it is quite clear that the sun shines 24 hours per day, every day of the year; just not all in the same place.
Thanks again.
G
Bob Boder asked. “What is the mechanism causing the NH and SH ice to transition over time, this is what interests me and is it a stable process? it seems pretty evenly balanced and almost like it was designed to shed energy or accumulate energy when need into or out of the system.”
If you watch ( http://www.netweather.tv/index.cgi?action=stratosphere;sess= ) for a season you will notice warm/cold air breaking of the vortex and going down past the equator. It seems the poles transfer heat via the stratosphere (the Doub effect). And hence you cant take one pole only as the two are linked. This transfer is not constant and we have a area of warm air over the artic at present which has stopped a strong vortex forming and pulling in more cold air, so giving us a week jet stream and our present NH weather.
This is my understanding of what happens.
Dose anyone have ideas about what is disrupting the vortex`s?
I see a really conspiracy here, there are an awful lot of contributors WUWT that are named Robert/Bob. What up with that!
I trust you mean “the opposite situation means warming”, because I’m sure you know that the current situation also means warming, as do all possible, and several impossible, situations.
Excellent article. You bring up many salient points with regard to why the Antarctic is more important globally than the Arctic. I feel sometimes that we do not point out the fallacies in the CAGW arguments that stress Arctic ice anomalies. You do that very well here. Thank you.
For future reports since this one was so long, could you provide an executive summary at the beginning that imparts the information you are conveying in the report? For example, and as you state about 3/4 of the way through the article, “Over a 24 hour day on Feb 22 2015, the net effect of today’s sea ice was
.979 Mkm^2 x 34.0 watts/m^2 – 0.618 Mkm^2 x 2186 watts/m^2 = -1317.6 MWatts reflected back into space, thus cooling the planet.”
This line should be near the top of the article imo, with the proofs under it. There should also be some tie-in of that number back to real life, such as what effect it would have on global temps if it continues. Unfortunately, most people will never take the time to learn what a watt is, let alone understand all the number you typed, That is certainly NOT a criticism of you, but an acknowledgement by me that few lay people care to take the time to understand climate science. By having an executive summary, it allows some of us to cut and paste the summation onto other sites where less investigative minds only want the cleft notes. If we are to challenge the meme of the CAGW followers, we need simple to grasp talking points.
Just my pennies.
I should add that I know you had a summary at the beginning, but it was in percentages, and doesn’t relate well to real life. I think that is where we have missed the boat, and where CAGW zealots have done a good job. “The world has warmed by 0.5C over 30 years. We’ll now see more tornadoes, less ice, more heat waves, etc, etc, etc …”. Confirmation bias. What will -1317.6 MWatts mean over time if it continues? We need stories as well, but it is to be hoped that we don’t go over the top like our misguided friends have.
“So, over 24 hours, 2186 more watts/m^2 were reflected from each sq meter of “excess” Antarctic sea ice…”
====================================
Keep talking like that and pretty soon we will be talking of how many Hiroshima bombs of energy are not being absorbed into the oceans.
Seriously I appreciate the detail of your work showing the net affect of energy gain and or loss associated with disparate sea ice levels. I wish the summaries to each hemisphere could be put in a spread sheet showing the gain / loss / net gain or loss / and the combined net gain or loss of both hemispheres. (Similar to your daily sheet, but perhaps a simplified monthly summary combing both hemispheres into a monthly average, and then showing a global net gain or loss.) Thanks for all your work.
Meteorologists doing daily weather talk are probably the best target for this so I’d say a daily release would fit the needs of that core constituency rather nicely. I polar energy balance number that can be dropped into the weather report as a data feed on the web site and a small mention in the broadcast would be really helpful.
Layman query; wouldn’t exposed sea water around the Arctic radiate more energy up through the ‘window’ than ice?
That depends on date/time/solar declination, etc… items in list at beginning of RAC’s post.
Richard111
March 5, 2015 at 3:54 am
The “extra losses” are important – particularly in the “crossover weeks” of early April and late August. The rest of the year, the solar energy absorbed and reflected when the sun is up is much greater than even these losses over the entire 24-hour day and night. They matter – but are a smaller factor.
Clouds above all. But 2 meter air temperature, relative humidity or dew point temperature, wind speed, water temperature, and Tsky (absolute temperature receiving the LW radiation from the ocean or sea ice)
affect the amount of each “extra losses”. And, most of the time, they vary hour-by-hour. Makes it tricky to calculate. Most papers ignore them, or use “average” monthly values that add up to “near-zero” values.
Yes, those are “extra losses” compared to the primary “gain” (of solar energy heating the darker, newly-exposed ocean waters) that everybody ignores. And, in truth, compared to the POTENTIAL solar energy gains into the open ocean, these extra losses are smaller. NOT zero, but smaller. However, those “potential gains” mentioned above are all too often approximated using nominal values of ocean albedo for an iceberg melting under the direct summer sun high above the horizon on the equator through a equatorial (1.0 air mass) or a temperate latitude’s1.5 air mass!
“24 hours of sun” doesn’t mean much when 10 of those hours the sun is only 8 to 12 degrees above the horizon.
At those elevation angles, the ocean reflects 20 – 35% of the sun’s energy – NOT a high-SEA albedo of 0.066.
At those SEA’s, the sunlight is getting attenuated by 4.0, 8.0, 10.0, 16.0 (up to as much as 31) air masses.
At those times of the year, the Arctic sea ice albedo is DOWN to a measured low in July at 0.42 – 0.38.
Many hours of the day even in mid-summer, there is actually very little “extra” energy absorbed into the Arctic Ocean.
But we will calculate every variable each month every day-of-year at every degree different SEA as the year unfolds.
With the excess SH ice preventing so much insolation from entering the oceans and the NH ice decrease allowing ocean heat to escape to the atmosphere, it is hard to imagine the oceans gaining the missing heat the CAGW proponents claim. Do you know if the space derived TOA measurements capture the polar latitude’s atmospheric energy gains and losses well?
Good question. I do not have an absolute answer right now.
We may find one as the discussion proceeds month-by-month.
You seem to have a problem with the units you are using. The Watt is a unit of power, and if you aggregate it over a period (such as 24 hours), you end up with a quantity of energy (or work, or heat), which should be expressed as either watt-hours, or converted to joules. You have expressed it as megawatts.
You say
“”So, over 24 hours, 2186 more watts/m^2 were reflected from each sq meter of “excess” Antarctic sea ice at -68.0 latitude on Feb 22 2015.””
but this is greater than the total top-of-atmosphere radiation of 1362 watts/sq.m
True. But your mixing instantaneous top-of-atmosphere radiation (watts/sec) with the total radiation reflected from the sea ice – down at the bottom-of-atmosphere – over an entire 24 hour period.
On Feb 22, Radiation received at top-of-atmosphere (TOA) = 1390 watts/m^2 per sec.
Over a 24 hour day, that is actually 1390 x 24 x 60 x 60 = 120 watts/m^2 x 10^6.
But, the 2186 watts/m^2 is the simple sum of 24 individual measurements, each taken one hour apart.
Thus, rather than integrating the solar radiation received, refracted, absorbed in the atmosphere, reflected from the ice or from the open water, absorbed, etc. over 24 hours, I am merely taking 24 hourly “results” and simply adding them up.
Technically, that doesn’t even make the sum = watt/sec-hour. Which doesn’t really exist. Then again, you don’t get an accurate daily total by adding each hour’s value to every other hour either. Supposedly, you could multiply each hourly result by 3600 (the number of seconds/hour), but that implies radiation received, reflected, absorbed and lost in transmission is constant over the entire 3600 seconds of each hour. Which isn’t right either.
Nevertheless, over a day, repeated each day of the year, you CAN compare what happens over the summer and winter periods at both poles. Which is our goal in this exercise.
As the reader above mentioned, you have to compare like-to-like. Regardless of what units are used.
Over a 24 hour day, that is actually 1390 x 24 x 60 x 60 = 120 watts/m^2 x 10^6.
Again your units are wrong, as calculated that should be 120×10^6 J/m^2 or 120 MJ/m^2.
If you’re writing an article about the energy received by the earth, you must surely understand that the term “watts per second” makes no sense at all. One watt is one joule per second, so you can count the number of joules in 24 hours, or you can work out the average number of watts received over a 24-hour period, but you can not add up the number of watts received each hour and add them up and express the answer in watts. It is meaningless to do so.
A 100-watt bulb in your house consumes 100 watts day in and day out if you leave it on. At the end of 24 hours it has consumed 2400 watt-hours, or 8640000 joules. (i.e 100 joules per second)
““But, the 2186 watts/m^2 is the simple sum of 24 individual measurements, each taken one hour apart.”
That explanation still makes no sense. That would be like saying “I was driving 50 km/hr for an hour, and 60 km/hr for the next hour, and 70 km for the final hour. 180 km/hr is the simple sum of the 3 individual measurements.” Adding speeds or adding power values like this is nonsensical.
“Over a 24 hour day, that is actually 1390 x 24 x 60 x 60 = 120 watts/m^2 x 10^6.”
NO. Over 24 hours that is 120 x 10^6 JOULES/m^2, not WATTS/m^2.
“Thus, rather than integrating the solar radiation received, refracted, absorbed in the atmosphere, reflected from the ice or from the open water, absorbed, etc. over 24 hours, I am merely taking 24 hourly “results” and simply adding them up.”
Actually you ARE integrating — or at least doing a simple rectangle rule approximation to the integration. You are (even though you don’t seem to realize it), integrating and getting a result that should be written in units of energy per square meter (W*hr/m^2), not power per square meter (W/m^2)
Actually you ARE integrating — or at least doing a simple rectangle rule approximation to the integration. You are (even though you don’t seem to realize it), integrating and getting a result that should be written in units of energy per square meter (W*hr/m^2), not power per square meter (W/m^2)
true.
But a “Rectangle” approximation using hourly values is a very, poor approximation of the true intregal. Worse, it uses “corner” values for something that is not even a true sine wave. For rectangles, you’d need to use the 1/2 hour values across the hourly max/min. A 12:00 noon value (local solar time) only only valid for one sec, then it is “too high” for the next 3599 seconds until 13:00 PM.
A far better approximation would be trapezoids for each hour – approximating the rising (morning) values between the start and end points, and the falling (afternoon) values between their start and end points.
RACookPE1978 commented
Dang………
Now, what ya gotta help me with is them there “unit” thingies ….
So, iffen you start with 1390 watts/m^2 up at the top of atmosphere at the beginning of each hour, and youse figger out some sort of approximated and variated value getting reflected off of the sea ice at the beginning of each hour after the sun’s ray reflect and reverberate and invertebrated and pale-anthro-propagated a few micro-seconds later at the bottom of the atmosphere, what units do you use at the end of a 24-hour day for the total all of the seconds that went by that day?
Good points. We will use Watt-hour values next month.
tjfolkerts – exactly.
I thought of using the speed analogy as soon as I’d submitted my earlier comment. The whole article shows a lack of understanding over the difference between power and energy, which confuses whatever point it was trying to make
“A far better approximation would be trapezoids for each hour – approximating the rising (morning) values between the start and end points, and the falling (afternoon) values between their start and end points.”
That certainly is a bit better in a general, abstract sense for numerical integration. However, in this case with endpoints that drop to zero, the result is exactly the same.
From 4:00-5:00, the trapezoid has an area of (0+3)/2 W/m^2 * 1 hour = 1.5 W*hr/m^2. From 5:00-6:00 is (3+44)/2 W/m^2 * 1 hr = 23.5 W*hr/m^2. Continue through each hour and you get the following:
1.5
23.5
84
170
260
339
396.5
426.5
426
395.5
337.5
258
167.5
81.5
22.5
1.5
Add them up and you get exactly the same 3391 W*hr/m^2 as the simple rectangles (which is easy to prove mathematically as well).
The only real way to improve the numerical integration would be to divide the time into finer intervals (eg 30 min or 10 min).
tjfolkerts commented…
Yeah!
30 minute intervals would ~double the 10 day run time to calculate Total daily watt – hours accumulated for just 1978 on, the period I have for solar constant data for the lat of every station I have surface data for.
tjfolkerts
Well, yeah – Right after I added up the whole bunch manually … and came to the same conclusion. 8<) (Can I complain to the mod's that you copied my homework – before I handed it in?
Caveats. That works only if equal time intervals are available.
And only if the morning (first data value!) at 00:00 is = 0.0 and the last data value (23:00) also equals 0.0 (Or if first_value (day 1) = last_value (day 1) = first_value (day 2). Otherwise, add up the time intervals through the day, then first_value/2 + last_value/2 will work pretty well as long each day-to-day change is relatively small.
So, you're concluding watt-hours is the right unit for the day-to-day total?
tjfolkerts
Actually, the infamous Wikipedia does have a nice trapezoid animation showing the method, and its approximations.
http://en.wikipedia.org/wiki/Trapezoidal_rule#mediaviewer/File:Trapezium.gif
“what units do you use at the end of a 24-hour day for the total all of the seconds that went by that day?”
I multiple watt seconds x 3600 for watt hours.
When i do annual power it will probably be kwh.
The appropriate unit would be Joules, on a daily basis MJ would be best as shown above.
Phil. commented
Then why is the Solar Constant in Watt/M^2?
Then why is the Solar Constant in Watt/M^2?
That’s the incident instantaneous power, the author was referring to the daily integrated value,
W=J/sec so W*sec= J*sec/sec = J
When the wind blows hard, as it often does in polar regions, the heat lost from open water will increase significantly compared to that lost from ice. It is implicit in the post but bears repeating.
“So, over 24 hours, 2186 more watts/m^2 were reflected from each sq meter of “excess” Antarctic sea ice at -68.0 latitude on Feb 22 2015….
Over the 24 hour day, this was a total of 34 watts/m^2. (The math, if anybody is interested: if open ocean, the water absorbed 4+12+16+12+4 watts = 50 watts/m^2. If sea ice were present, the ice would have absorbed 2+4+5+4+2 = 16 watts/m^2. The difference = 50 – 16 = 34 watts/m^2.)”
You can’t ADD W/m^2. You can add J/m^2, or you can average W/m^2.
He should have written, “So, over 24 hours, 2186 more watt-hours/m^2 were reflected. . .”
Robert, Have you considered the impact of clouds and/or the atmosphere?
1) Even on a “clear” day, a significant fraction of the light will be absorbed by the atmosphere before ever reaching the surface. If Trenberth’s diagram is to be believed, the average absorbed by the atmosphere is ~ 78/341 = 23%. At low angles of incidence (ie always near the poles) the absorbed fraction would go up at the sunlight travels longer distances through the atmosphere. This means that your numbers for reflected solar energy should all be much lower (ie reduced by at leaet 23%). This reduces the net warming/cooling effect.
2) How cloudy are these areas? On cloudy days, much less energy yet would reach the surface, and the difference in albedo would be less important. I have no idea about the general weather patterns, but if some areas or some times of the year have heavy cloud cover, that would throw your numbers off quite a bit.
Good point.
Also worth considering that because air is less dense than water or land it takes less energy to raise its temperature. As the atmosphere absorbs much of the suns energy before it hits the planets surface it pulls into question the whole notion of the surface warming the atmosphere and stage 1 of the standard greenhouse effect diagram.
Thought experiment: Graphite rocks are more absorbant of sunlight than water. If 50cm of water is covering a shallow lake with a graphite rock bed, does the sun warm the water or does the graphite warm the water?
Near the poles the atmosphere is much less thick. The troposphere is an average of 56,000 ft thick at the equator and only an average of 30,000 ft thick at the poles. This makes the poles able to lose heat much easier and at the same time even though low angles of incidence do increase the distance light travels through the troposphere the decrease in atmosphere thickness along with the much drier atmosphere at the poles will attenuate the resulting absorption of light. What would the final numbers be? I have no idea..
“Near the poles the atmosphere is much less thick. “
Sort of. The surface pressure is (roughly) the same, so the total mass (or total # of moles) above each square meter is (roughly) same. The atmosphere rises to a smaller height, but the density of the cold air is greater, basically offsetting each other.
“the decrease in atmosphere thickness along with the much drier atmosphere at the poles will attenuate the resulting absorption of light. “
The point about water is important. The reduced absolute humidity will me less absorption by water (both for incoming sunlight and outgoing thermal IR. Reduced levels of dust could also be important. But (as pointed out in the 1st paragraph), a thinner overall atmosphere is NOT important here.
Actually, it looks like the atmospheric absorption is built in to the calculations, including effects due to angle above the horizon, The reduction I was concerned about is included.
So, my point #1 is moot. Kudos to Richard on including this factor.
tjfolkerts
March 5, 2015 at 4:52 am
Clouds are critical five ways. Maybe more.
Yes, right now I am ignoring them – but at least I know I am ignoring them. And trying desperately in each clause to make sure the readers know I am ignoring them!
Clear skies, arctic conditions (both poles!) of low temperature, little water vapor, little dust and soot and little pollen (Antarctic in particular) = very low atmospheric attenuation. Measured values ~ 0.85
Clear skies temperate latitudes ~ 0.75
Temperate latitudes, but pollen or dirt or pollution? As low as 0.60 or below in local areas.
Equatorial (high humidity) but no pollution? About 0.75
Fine, those were clear skies. Clouds start by reflecting between 65% to 75% of the potential solar radiation directly from the cloud top. But how much each day? How much on different latitudes on different days? In different seasons?
Regardless, assume you make some assumptions about the “number of cloudy days” each season. Then these clouds attenuate the left-over (non-reflected) solar energy (either strongly or weakly) the as they pass through the cloud layers – and there may be several layers from stratosphere down to the bottom of the dark gray cumulus cloud raining on the solar meter than hour.
Now, you have anywhere between 15% of the original potential solar energy at ground level to 5%. So, you can see that “clouds matter” but the amount of energy left over to calculate after the clouds are factored in is much less than the original “clear sky” case.
But the albedo of both sea ice and open ocean change with clouds!
Sea ice. More clouds occur over open ocean than over sea ice.
Albedo of sea ice is lower under clear skies (by about 8% – 10%) than under cloudy skies in both hemispheres.
But – the albedo of open ocean is a function of solar elevation angle ONLY under clear skies. Under clouds (diffuse radiation) the open ocean albedo remains constant at 0.066 – that old hoary classic Wikipedia value all of us have memorized by comparing it to clean snow-covered Arctic sea ice in January (0.83). Under clear skies, measured open ocean albedo goes from 0.45 (at the horizon) to 0.066 (at solar elevation angles over 33 degrees). So, SEA matters in clear skies, but not cloudy skies.
However, under cloudy skies, winds are typically higher (convection losses increase) but the LW radiation losses (from open ocean or from the top of sea ice) go down. Average air temperature, and hourly air temperature – changing through the year from -25 deg C to +1.5 degree C changes the film coefficients for conduction and convection losses.
Higher humidity reduces LW radiation losses.
Evaporation losses reduce as humidity increases – which always happens under the higher winds typical underneath storm clouds! – but the higher winds increase other factors in those evaporation loss rates as well.
High winds (typical of clouds) reduce the film coefficient between sea surface and the air, so that increases convection losses from the ice or from the top of the sea surface.
High winds reduce the open ocean albedo, but under clouds, you have almost only diffuse radiation available anyway, so there is little real effect on ocean albedo.
Emissivity (gray body LW radiation) is about the same for both sea ice and open ocean water.
So true!
Warmists of course replace humbleness with hubris. The link below takes you to Googles ngram viewer. Type in hubris as the search word and you will see usage for the word hubris tracks surprisingly well with CO2 and Warming temperature claims. Using hubris as my guide the only valid conclusion is that CO2 is a direct cause of hubris as evidenced in alarmist claims.
Ngram Viewer
If the above link does not work the link is http://books.google.com/ngrams/
As interesting as this article is, it ignores the opposite effect of sea ice – namely the insulation effect reducing heat loss from the water below.
This wouldn’t matter if the article was just discussing how the albedo effect works over different seasons and latitudes, but rather, it seems to be trying to draw conclusions about the effect on climate change. I don’t know which effect is larger, or whether they net off, but it does leave the question hanging.
See my comment above. Agree with you.
Actually he doesn’t totally ignore the loss of Arctic sea ice.
Today, this day of year, for every “lost” square meter of sea ice, the open Arctic ocean loses more energy from 24 hours of increased losses (increased long wave radiation from the open ocean water, from increased convection and conduction up to the sea surface, and from increased evaporation) than it gains from a few hours of increased absorption in the open Arctic Ocean. In all cases, at this latitude at all hours of the day, more energy is lost from the open Arctic Ocean water than from ice-covered Arctic waters.
A nitpick:
For angles near the horizon, you can’t ignore atmospheric refraction.
A few years ago, some Eskimos observed, among other things, that the stars seemed to be coming up in different places. The environmental activists, who were using the Eskimos’ testimony, tried to censor that particular bit. They said it would hurt their credibility. Finally someone pointed out that the Eskimos were right. Changes in atmospheric refraction, caused by temperature and humidity changes, would affect the position of the stars on the horizon.
Any equation is just a model of reality. (That could be my pet peeve #1.)
There are several versions of “air mass” and “declination angle” available. Some correct for refraction, some do not. Some Air_Mass formulas correct for the earth’s curvature, some do not. Some correct for the colder temperatures (greater density) near the poles, some do not. Worldwide, Kasten & Young has become widely accepted with its integrated values for both. It is NOT clear whether Kasten & Young, in any of their versions, corrected for the shorter “height” of atmosphere near the poles, but the greater density becomes more important anyway. The slight difference in radius between pole and equator is not apparently corrected against in any formula I have found so far. Neither is the different radius for the earth’s oblate spheroid difference between between north and south pole.
Bottom line? Kasten & Young 1989 is accepted, I used it.
The one used here (Kasten and Young, 1989) is also used by the NOAA for their air mass values, and IS valid for low solar elevation angles and DOES correct for refraction near the horizon. The often-quoted “latitude cosine” function is wrong, and is only an approximation valid between south 45 and north 45 latitudes.
Expressed as an Excel equation for solar elevation angles, Kasten & Young, 1989 is:
Air_Mass = (1/(COS(3.14159/2-H9)+0.50572*(6.07995+I9)^-1.6364))
where H9 is the SEA in radians, and cell I9 is the SEA in degrees.
More frequently, you will see this written for Solar Zenith Angle (SZA) as :
Air_Mass_SZA_Deg = (1/(COS(SZA_Deg)+0.50572*(96.07995-SZA_Deg)^-1.6364))
SEA (or SZA) are calculated from latitude, declination angle, and hour-angle for each specific day-of-year and that unique hour-angle.
SEA_Radians =ASIN( (SIN(F12)*SIN(LAT)) +(COS(F12)*COS(LAT)*COS(G12)) )
for cell F12 = That declination angle (radians) on that day-of-year for that hour-of-day
cell G12 = the hour-angle in question (radians)
and LAT = the spreadsheet variable for latitude in question, (also in radians).
if you have declared everything in variables, then this becomes:
SEA_Radians =ASIN( (SIN(DECL_Rad)*SIN(LAT_Rad)) + (COS(DECL_Rad)*COS(LAT)*COS(HRA_Rad)) )
For other approximations, see also Young, 1994.
Young 1994, for cell d17 = SEA (radians)
AirMass (Appx) =(1.003198 * (sin(d17) + 0.101632)/( (sin(d17))^2 + (0.090560 * sin (d17)) + 0.003198)
Bear in mind that this is a nitpick. 🙂 If we’re talking about watt-hours per day, the contribution of the sun at the horizon can be ignored.
You can’t predict exactly where the stars will appear at the horizon unless you take the local conditions at the time of observation into account. Any Eskimo knows that.
More nit-pick (of mine) from my military days: “after 15:00 PM”, the PM annotation is not needed when using 24 hr time.
Joel O’Bryan
Yes, but most readers are not familiar enough with military time to convert back and forth. Things are confusing enough trying to “force” a 24 hour day at both poles at the same day-of-year that I elected to “duplicate” an PM (and sometimes an AM as well) for hourly numbers.
Wind direction and speed does more to determine Sea Ice Coverage than does temperature. Sea Ice Volume, much more difficult to detect, tells the temperature story, but this of course includes both air temperature and much more important water temperature.
I live on Lake Michigan, 36th floor, high enough to be able to see the Lake ice up and open up based on the WIND!!! Today I see maybe 10-15% ice coverage. Last week it was 100% in this area, and the week before that it averaged maybe 60%. What does this tell us? Cold, windy winter in the Midwest…
The thing that must be kept in mind is the wind may reduce ice coverage, but prior to 22 March, but the effect is still cooling (as the nights are longer than the days). As the wind now blows over a large fetch of open water, and with air temps still far below freezing, the overturning effects will keep ice from reforming but it markedly accelerates heat loss to depth (convection, in the form of downwind lake effect snow). Once the wind dies down, and the air temps remain solidly below freezing at night, the ice will reform rapidly.
+1
Was it just me or are there some issues in the Arctic section? The heading says Jan 22 and DOY=22. Is this last month’s report in this part or did the heading fail to be updated? It looks like all the data is for Feb 22 and DOY=53, so I am guessing the section titles for the Arctic portion need to be updated. Nice report otherwise. There are a lot of Watts going missing in the current configuration. So much for “energy budgets”!
[DOY values are corrected, thank you for the “blogosphere’s” quick peer-review. .mod]
While I generally agree with your statement “33.0% more Antarctic sea ice than normal for this date” using your chart, this other chart of “Daily Global Sea Ice Anomaly” http://nsidc.org/data/seaice_index/images/daily_images/S_stddev_timeseries.png shows much more sea ice area so the percentages are less. I’m also confused about your claim Antarctic sea ice “rising from +23% Feb 1 to 33.4% Feb 28. ” I’m not seeing that in either chart – on the nsidc chart it looks like the opposite happened comparing to the 1981-2010 average?
They call it “global warming” so I always argue in terms of global sea ice and that recovered dramatically over two years ago and remains at mid 1980’s levels http://arctic.atmos.uiuc.edu/cryosphere/IMAGES/global.daily.ice.area.withtrend.jpg
The striking difference to me is how long it is currently remaining at the average level. It started down in 2001 and became less stable with a slow decline with several brief recoveries from ~2006 to and including one at the end of 2011.
But then “something” must have happened because at the end of 2012 it recovered and appears to be stable again. Does anyone know why that happened?
If as they say CO2 somehow “controls” earth’s temperature and CO2 is now at a record high concentration, one so high that they said it would cause catastrophic warming – how was it possible for global sea ice to recover back to the same as what it was 30 years ago and stay there for over two years?
Mike M
You are reading a different laboratory’s interpretation of the daily satellite sea ice numbers. There are differences between JAXA, DMI, NSIDC, Cryosphere, and the others each day, but – in general – over longer periods of time, they track fairly well. But whenever you compare number-to-number, you do have to use the same lab. That’s why I chose Cryosphere for all Sea Ice reports – they are one of the few that reports both north and south poles.
But, more important, you have to look at either “Area” (ocean waters 100% covered with sea ice) or “Extents” (ocean waters covered by at least 15% of sea ice – just about the maximum point where ships can travel at slow speed by steering between ice floes and the larger, deeper, more dangerous icebergs). Again, for consistency between north and south, because the Cryosphere reports “sea ice area” each day, I’ll report and use “sea ice area”.
Area, of course, is smaller that “Extents” – so percentages comparing the two are going to be different.
Note: The Cryosphere does not issue or plot “standard deviations” in its reports, so those have to be either assumed the same as another lab’s values, or left out entirely.
See the following for Antarctica’s longer percentage trends
And, for the Arctic.
Ooops – wrong title for the first chart link, S/B “Antarctic Sea Ice Extent”
The below quoted statement doesn’t make sense to me. The South Pole is at 90.0 South Latitude.
“The edge of the Antarctic sea ice is at latitude -68.3 south, slightly closer to the South Pole than the Antarctic Circle at -66.5 south latitude.”
It is farther south than the Antarctic Circle, but it is closer to the Antarctic Circle than to the South Pole.
Tim
it’s a bit tricky addressing both poles – when so many millions of people (in the northern hemisphere) are so conditioned to “north” “up” “pole” “further away” = “colder air, less light, more ice, more freezing, etc … For these many millions, “further south” ALWAYS means “closer to the equator” and “hotter, drier, more deserts, less ice, more and longer summer weather, more hurricanes, etc. July is always “hot” and December 22 is always “in winter” . Mentally, remember people have been conditioned by 40 years of propaganda to believe “Global Warming” means Central England will become as dry and barren as the Sahara Desert. Upstate New York will become a disease-ridden malaria-infested central American swamp. (As if the insect-ridden malarial swamp waters they cleared to dig the Eire Canal and the Russian city of St Petersburg can be forgotten in history’s muck.)
Across all 12 months of the year, the Arctic Sea ice is concentrated north of the Arctic Circle’s 66.5 North latitude. At its minimum point in mid-September, the Arctic sea ice edge lies between 80 north and 78 north latitude, with only a few hundred sq kilometers further south between Canada’s islands. At its late March maximum of 14.0 Mkm^2, the Arctic sea ice just about fully covers the Arctic Ocean’s 14.0 Mkm^2 sq area, plus all of Hudson Bay, and areas of the Denmark Strait, about half of the Bering Sea, and partial areas in the waters between Greenland and Canada. Europe’s far smaller Baltic and Norwegian waters add just a little bit.
The important point to remember is that virtually all of the Arctic sea ice is always north of the Arctic Circle every day of the year.
For 10 months of the year, the Antarctic sea ice area is a band AROUND the antarctic continent whose icy areas are ALL between the Antarctic Circle and the equator. Only in January and February does the irregular Antarctic sea ice cross the Antarctic Circle, coming as you correctly point out, between the Antarctic Circle at 66.5 south and the South Pole. But those are the summer months in Antarctica, and so the edge of the Antarctic sea ice is hit with more than 16 hours per day of sunlight in November, almost 20 hours of sunlight in December, 18 hours per day in January, and 14 hours per day in February.
And, those are the days of the year when the sun’s rays shine brightest, with TOA radiation at its yearly peak.
I see your point. But try reading RAC’s sentence this way: 68.3S is slightly closer than 66.5S to the S Pole.
MJ
March 5, 2015 at 12:21 pm
True, it really should be: “68.3S is closer to 66.5S than to the South Pole; but 68.3S is much closer to the Equator than 72N is to the Equator.”
“The much-hyped Arctic amplification is a very real effect.”
It’s like saying that the climate made the AMO warm up, when in reality the AMO made the climate warm up. If the ice extent is reduced with a negative North Atlantic Oscillation, then it’s an amplified negative feedback. Increased forcing of the climate increases positive NAO, e.g. when the ice extent increased through the late 1970’s, and the 2013/14 increase.
Might be a good place to mention that the great lakes are approaching record ice levels also. GK
http://iceweb1.cis.ec.gc.ca/Prod20/page2.xhtml?grp=Guest&mn=&lang=en
Great Lakes ice levels should be topping out in the next week, at least based purely on temperatures.
After this last blast of frigid air eases up, next week will see temperatures get above freezing. The Southern Great Lakes will have some max readings approach 50.
Next week is the 2nd week in March, so we are long over due for melting.
I will guess that the peak in ice coverage for the Great Lakes would happen, based on the climatological average during the month of February, probably around the 2nd/3rd week.
Obviously, peaking very late(in early March) correlates closely with cold Winters, especially late cold. Having 2 years in a row with ice coverage greater than 80% has not happened since the end of the previous natural global cooling cycle in the 1970’s,
There is a strong regional weather element involved here. The core of the coldest global temperature anomalies the past 2 years has been over the Great Lakes and areas surrounding them.
By no coincidence, the weather has been completely the opposite along the West Coast during much of that period with the Ridiculously Resilient Ridge, often built up well into the Northeast Pacific and at times last Winter, to Siberia teleconnecting well, with a downstream, deep upper level trough, at times a cut off upper level low, which when amplified has periods of being a “Polar Vortex”.
Even last July, when we here in Indiana, (one state south of the Great Lakes) had our coolest July since records have been kept, this pattern occurred. However, it’s been most pronounced during the past 2 Winters.
These past 2 Winters have featured a weather pattern very similar to the Winters of 76/77 and 77/78 for much of the US. Upper level ridge and drought in the West, extreme cold, Polar Vortexes at times in the East. The timing and similarities make for a strong connection to the natural cycle than also results in a -PDO,
The PDO in the last year has spike positive(+PDO).
Either we are prematurely switching back to a +PDO regime(10-15 years early) with the possibility of a pattern change which will favor milder Winters again or the +PDO spike higher will be short lived, similar to the late 1950’s, in the middle of the previous -PDO regime, when the +PDO only lasted briefly and was followed by another 15 years back to a -PDO, along with more modest global cooling(until that ended in the late 1970’s)
http://research.jisao.washington.edu/pdo/PDO.latest
http://research.jisao.washington.edu/pdo/
What the PDO does during the next 2 years is huge in telling us about the natural cycle.
Thanks, Mike Maguire, for the PDO info.. When I wrote
here: http://wattsupwiththat.com/2015/03/05/noaa-claims-elusive-el-nino-arrives-the-question-is-where/#comment-1875829
I thought of you and here you are
(so I posted the PDO part of your comment here: http://wattsupwiththat.com/2015/03/05/noaa-claims-elusive-el-nino-arrives-the-question-is-where/#comment-1876013)
Thanks for the great info.,
Janice
The temperature of the southern polar circle.
http://www.cpc.ncep.noaa.gov/products/stratosphere/strat-trop/gif_files/time_pres_TEMP_MEAN_JFM_SH_2015.gif
“SIA 1979-2008, DOY 22, = 1.874 Mkm^2, Average area this date
SIA 2015, DOY 22, = 2.492 Mkm^2, Actual area this date
SIA Anomaly, 2015, DOY 22 = 0.618 Mkm^2, Anomaly this date”
Did you mean DOY 53 (Feb. 22)?
[Corrected, thank you. .mod]
Well done, R.A. Cook! Your attention to detail is wonderful. That you have written a scientifically meaningful report is demonstrated in all the great discussion it has elicited on this thread.
Thanks for all the hard work to keep WUWT a first class science site!
(politics is also important…. but it is refreshing for all the wonderful scientists and engineers here to have threads like this one appear).
How can this be first class science when the author clearly doesn’t understand that the Watt is a unit of power not energy (and therefore can’t be added over time), and apparently also struggles with the 24 hour clock?
You don’t read sarcasm well, do you? 8<)
“Today, this day of year, from each and every “excess” meter of Antarctic sea ice, you can see that an “excess” of 294 watts/m^2 are reflected back into space (326 watts/m^2 – 30 watts/m^, clear day, at noon).”
Watch out, carbon pollution MIGHT increase DLWIR by 0.2 watts/m^2 per decade!
Even back in the late sixties the papers were in cloud cuckoo land.
In 1969 the New York Times predicted an ice-free Arctic by 1989, and a new ice age – during the same week.
https://stevengoddard.wordpress.com/2015/03/05/settled-science-from-1969/