Sea Ice 101 – Insolated, Isolated, Insulated Icebergs in Space, A Mid-Term Exam

(Or, How Climate Science Is Taught Today)

Guest essay by Robert A. Cook

The following begins a series of topics on Arctic and Antarctic Sea Ice. And, since we (behind the keyboard) will be talking to you (the readers, and perhaps even the learners) as if we were openly addressing you in a classroom or hall, we will obviously begin class with the mid-term exam.

Don’t worry, the answers are already provided, but we are going to start the mid-term exam by questioning both the answers to the mid-term exam, but also the questions on the mid-term as they are already written.

More appropriately, we will question the ASSUMPTIONS and APPROXIMATIONS needed to make the “answers” work as written. Finally, we will assign the Sea Ice 101 Course Final Exam.

The class may now open its exam; any notes, articles and journals desired; the web, any available textbook or reference, and their minds. This particular mid-term exam is from the Institute for Structure and Nuclear Astrophysics (ISNAP), Department of Physics, University of Notre Dame. “The Physics of Climate” examples used were taught in spring semester, 2011, by Dr Michael Wiescher. Since Dr Wiescher is currently teaching a 2015 version of this class just down the hall, students are reminded to be civil, be quiet, and remain in their seats unless called upon. Guests may arrive in the classroom, they too are to be treated with all of the respect they earn.

Syllabus (2011) = http://isnap.nd.edu/Lectures/phys20054/class_syllabus.pdf

Lecture Notes (2015 class) = http://isnap.nd.edu/Lectures/phys20054/15Lecture%201%20Physics%20of%20Climate%20overview.pdf

20054 Mid Term Exam Solutions (2011) =

http://isnap.nd.edu/Lectures/phys20054/midterm_exam_solution.pdf

Lecture Notes, Earth Energy Budget and Balance (2011) =

http://isnap.nd.edu/Lectures/phys20054/Lecture_5_Energy_balance-1.pdf

Dr Wiescher’s lecture notes (Earth Energy Budget and Balance, hereafter EEBB 2011, are a fairly good summary and background of the earth’s heat budget. Incomplete, but a good start to some parts of the problem.

The mid-term based on those notes? A different story. In particular, let’s look at problem 4 on that exam, and the “solution set” for that problem. (I found this problem solution set in 2012, while looking for Arctic albedo calculations and assumptions.

The required solution to Problem 4 are reproduced below in their entirety.)

image

Let’s look at each error (or questionable assumption) in each sentence and phrase above.

1. “Consider the emission temperature of the polar ocean …”

image

1.A. Yes, “Emission temperature” is covered on page 15 of EEBB 2011. But the actual equations on pg15 do not include any of the formulas expected to be used when the student answers the question on the exam. Certainly misleading and confusing.

So, as we will see in a few minutes, even IF the student used this formula to answer Problem 4, that answer would be graded as “WRONG”!

1.B. But do the expected answers to Problem 4 actually calculate any “emission temperature”?

Regardless of the problem statement, the four required “answers” to Problem 4 are actually: Energy reflected summer, energy reflected winter, energy absorbed summer, and energy absorbed winter.

But look at pg 14 of the class notes (EEBB) for “energy absorbed” … (Not “emission temperature”)

image

We will see in a few minutes that even if the student used pg14 to calculate his or her answer to problem 4, that answer would still be graded as “WRONG” …. This is NOT the required answer either!

Finally, regardless of whether formulas from pg 14 or from pg 15 were used, an “emissions temperature” is still NOT a “heat flux”.

1.C. Next, regardless of what was “written” as the correct answer to Problem 4, or what was “expected to be written” as the answer to Problem 4, the ACTUAL information “IMPORTANT WHEN STUDYING CLIMATE CHANGE” when you answer Problem 4 is:

“4.A. What is the CHANGE in “Total Energy absorbed” AND “Total Energy lost” when sea ice is melted in the Arctic? Does your answer change between summer and winter conditions?”

Then, because the Antarctic sea ice behaves completely differently, Problem 4B is needed:

“4.B. What is the CHANGE in “Total Energy absorbed” AND “Total Energy lost” when sea ice is melted in the Antarctic? Does your answer change between summer and winter conditions? ”

2. “… for an incoming solar flux of F=1370W/m^2.”

2.A. First, the SORCE solar research team established a “new” correct solar irradiation level for Total Solar Irradiation (TSI) at the sun’s surface at 1362 Watt/m^2 in 2008. Here, the class is expected to use the wrong value for solar radiation levels.

The IPCC claims the entire “forcing” if atmospheric CO2 doubles is only 3.7 watts/m^2. Thus, by starting with a TSI radiation level “too high” by more than double the IPCC’s own 3.7 watts/m^2, every fundamental heat balance calculation after that statement is invalid and misleading. (Yes, 1370 watts/m^2 is the reference TSI used in the EEBB class notes, but that TSI value was already 3 years out-of-date when written.)

2.B. However, even IF the assigned 1370 watts/m^2 WERE the correct TSI radiation level, the actual radiation at top-of-atmosphere at the earth varies day-to-day over the year as the earth rotates around the sun in an elliptical orbit.

To calculate actual surface radiation levels at the earth’s surface, actual DAILY solar radiation levels at the top-of-atmosphere must be used.

Those ever-changing daily solar energy levels – which vary by 90 watt/m^2 from January through July – are NOT even expected to applied in the “approved answer” to Problem 4. But Problem 4 explicitly DOES require the student to compare heat balances at two different times of the year, so ANY season-to-season (really, day-by-day) change MUST BE correctly applied to the solar radiation at top-of-atmosphere. The earth does receive 1370 watts/m^2 at TOA twice a year: The first is March 23 (the spring equinox, near maximum Arctic sea ice extent each year) and 17 October, about a month after the fall equinox. But, the Arctic Ocean sea ice is in near total darkness (22 of 24 hours of the day) by that October 17.

If he or she chooses March 23, the sun is up in the sky above the Arctic, but the sun is NOT very high in the sky for very many hours each day, AND the Arctic sea ice is at its maximum extent for the year. Thus, ANY choice of a “summer conditions = smooth water” = “maximum solar energy absorbed” assumption for March 23 MUST be considered dead wrong.

So again, once again, IF the student answers Problem 4 accurately and completely, he or she will be graded “WRONG ANSWER” because neither March 23 nor October 17 is valid for the photographed conditions in Problem 4!

2.C. Atmosphere attenuation is ignored in the required answer to Problem 4.

In-atmosphere attenuation will reduce actual solar radiation levels at ground by MORE THAN 42% at noon in mid-September on the Arctic Ocean sea ice. (By 96.5% if noon, October 17 were chosen!) But atmospheric attenuation is ignored completely in the required answer to Problem 4. (Both air mass (length of passage) and atmospheric absorption (atmospheric clarity) need to be included when you calculate attenuation, and some assumptions have to be made for both, but the student MUST calculate something for attenuation loss before he can assume the solar energy hits the ice at sea level. )

So again, once again, IF the student answers Problem 4 accurately and completely, he or she will be graded “WRONG ANSWER.”

2.D. But the ocean surface is horizontal, right? And, any sea ice floating on a flat ocean surface is also horizontal.

So, to calculate “received radiation on a horizontal surface near the pole”, the student MUST correct for the altitude of the surface, the latitude of the surface, and the earth’s axial tilt on the day the surface is struck by sunlight. Now, these are not “original” discoveries to science – the formulas have been in use even before Magellan’s crews circled the globe, but they have to be done. (And be done properly. One problem: Even the 20054 class notes cover this topic very poorly, so the 20054 class notes need to be addressed as well.)

Yet again, IF the student spends the extra time to answer Problem 4 accurately and completely, he or she will be graded “WRONG ANSWER.”

3. “Choose the albedo and emissivity parameters for summer and winter conditions….”

3.A. Albedo changes over the year for Arctic sea ice are shown on pages 3,4,5, and 8 of EEBB 2011. In fact, pages 4 AND 8 are actually taken from Perovich 2002 (see also Curry 2001) during actual SHEBA albedo measurements on the actual arctic icecap. Clearly, the “right answer” for Arctic sea ice albedo over the summer arctic ocean is available to the class. (We do not know if it was introduced accurately, or whether the in-class discussion was accurate and detailed, but it was available to the class. Has been discussed in class. )

But the required “answer” for Problem 4 ignores BOTH the EEBB sea ice albedo plot, AND the sea ice photographs used in class.

Instead, the “answer” to Problem 4 uses a single “average” albedo for “ice” (obviously assuming “winter = ice”) of 0.80. Worse, as shown in few minutes, even this expected “winter albedo” of “sea ice” in Problem 4 is itself wrong!

3.B. Still worse than those two problems, getting the correct “answer” to Problem 4 requires assigning an albedo of 0.06 for “summer conditions”, again requiring the student to equate “summer (in the arctic) = only open ocean water (in the Arctic).”

Now remember your geography. All of the Arctic Ocean is above the Arctic Circle. So, from the middle of October through the end of February (the “winter conditions” for problem 4) there is NO sunlight at all ANYWHERE across the Arctic Ocean sea ice. It is completely dark all day that far north. Therefore, the photograph in Problem 4 showing “winter conditions” was actually taken sometime during the arctic summer, and – yes! – it’s cloudy, wind-swept snow and block ice does show what 80% of the arctic “summer” days are like on the icecap !

So, yet again, if a student answers Problem 4 accurately and to the best of his or her ability, he or she gets graded with a “WRONG ANSWER”.

3.C. Worse, 0.80 is not the correct value for winter sea ice, mid-summer sea ice, snow-covered sea ice, Arctic sea ice nor Antarctic sea ice under any conditions or seasons.

Again, look at the EEBB plot of measured Arctic sea ice albedo on page 4: The summer “ice” albedo varies by date: beginning as high as 0.85 in May, going as low as 0.38 to 0.40 in mid-July 30 through August 12, returning to 0.75 in late August, then rising back towards 0.85.

Using this plot from their class notes, even a high school student “should” be able to pick a date, then chose the correct “sea ice” albedo for that date and that latitude, right? These are college students, paying $45,000.00 (or more) per semester to learn the “Physics” or “Climate Science”. If I pay a beginning engineer $65,000.00 yearly salary to calculate solutions in the real world, a student paying 90,000.00 per year should be required to read a single graph in a single class.

Therefore, the correct summer and winter sea ice albedo should be required to answer Problem 4, right? Well, the “summer conditions” albedo is not correct.

3.D. Well, 0.07 is not the correct albedo for water either.

The albedo of the open ocean varies from 0.06 up to 0.45 (and higher) by:

solar elevation angle (or solar zenith angle);

ratio of direct and diffuse solar radiation present;

and by the wind speed and area of water present (which controls wave height).

Specifically BECAUSE the Arctic sunlight levels change so much from winter to spring to summer to fall and back to winter in both elevation angle AND duration (how high the sun any given day and how long the sun is visible each day), the student MUST be able to calculate solar elevation angles for ANY day of the year at ANY time of day.

In Problem 4, the need is particularly important: The albedo of open ocean water changes from 0.06 (at noon, in mid-June or July) to 0.35 – 0.40 (each morning, evening and night in the summer; all day in the spring and fall). Once again, IF the student answers Problem 4 accurately and completely, he or she will be graded “WRONG ANSWER.”

4. “…. as shown in the pictures …”

4.A. We will repeat a little bit from above. Let’s look at the pictures in more detail.

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The photograph for “summer conditions” shows absolutely calm water( no waves, no winds), no clouds (direct radiation only), and clear skies (no aerosols or particulates implied.) The near-absence of shadows even on the highest icebergs imply at least a 36 – 45 degree solar incidence angle. These conditions mandate the student assign a different albedo for BOTH “summer ice” and “summer ocean” conditions!

4.B. The presence of several high icebergs in the background of the “summer conditions” means that photograph was taken much further south than the “arctic Ocean.” (The clumpy, broken large chunks of ice floating in the foreground of the same image show the same evidence, but not as strikingly as the three big icebergs.) This is because large stable icebergs do not form in the central arctic. Large icebergs actually break off from mountain glaciers found down both coasts of Greenland, then drift south with the ocean currents in the Davis Strait and Denmark Strait.

ALL sea ice in the Arctic itself (where the “arctic sea ice” loss is measured!) forms as low 1 and 2 meter thick ice. Some 25-35% of first and second year sea ice is 2 meters thick, this sea ice remains through the summer and forms flat-bottomed 3, 4 and 5 year sea ice. This multi-year sea ice is often trapped against the northern Canadian coasts and islands by winds and currents. At times, the flat “sea ice” is pressed together into thin, tall ridges, but when these ridges are freed and allowed to float freely, they immediately fall sideways back to “flat sea ice”. When the Arctic sea ice flows south, these pieces stay flat. Thus, there are no icebergs in the Arctic Ocean – they simply don’t form up there. Arctic sea ice is flat 2-3 meter floes, with less than 30 cm of ice above water.

The presence of icebergs, calm seas (and the ocean is notoriously rough east of Greenland, in the Bering Sea, and north of Iceland) and a high solar elevation angle indicate that this photograph was taken in mid-summer about latitude 50-60 north – probably near the Labrador sea.

And the Labrador Sea, the St Lawrence River estuary, the Davis Straits are never ice-covered in summer, and some areas of each just ” touched” by sea ice in much of winter.

So, no student should compare “summertime” heat absorbed in an open ocean under clear skies near Labrador with an Arctic Ocean sitting in the dark under ice 20 degrees further north! Once again, IF the student answers Problem 4 accurately and completely, he or she will be graded “WRONG ANSWER.” (It would be like comparing the sun tan a student gets in Miami Feb 10 with one in Indiana the same day.)

4.C. The “winter conditions” photograph is even more mysterious. The Arctic sea ice between mid-October and mid-March is far north of the Arctic Circle, and thus has NO sunlight exposure at all at ANY time of day. The photo shows high winds (blowing ice in ½ to 1 meter swirls), a light gray even diffuse light (as if there were no heavy storm clouds at all, despite the evidence of high winds!), and a rough mid-ocean type of low pressure ice ridges.

4.D. The shadows in this picture are lightly shaded coming from the low right at perhaps 15-20 degrees. This implies either a early April “noon” sunlight at the time of maximum Arctic sea ice area, or a limited area further south such as Hudson Bay (near latitude 60 north) sometime between mid-March and mid-May. Hudson Bay completely melts by Mid-July, so it is not likely to be a late May or June timeframe.

4.E. The ice covering in the “winter conditions” means there is no evaporation losses, no convection losses from the open water, and increased conduction temperature differences between the ocean water under the ice and the atmosphere. Further, the presence of high-level clouds means significantly MORE diffuse solar radiation – which changes both the albedo of the sea ice and that of the open water the sea ice supposedly replaces. But higher wind speed increases convection losses over the “summer” calm ocean waters, but would increase evaporation losses if water were present. (Conflicting, isn’t it?)

Worse, the presence of clouds is MORE likely over open ocean waters (as in the summer conditions photo) and less likely over ice-covered seas (the winter conditions photo). Again, the photos contradict hundreds of days of actual on-site observations of both Arctic and Antarctic winter and summer conditions! (See the lecture notes on EEBB pg 4 (from Perovich, 2002) for photographs of actual winter (pre-summer melt) and summer sea ice conditions. Clearly, the class has been shown photo’s of what real sea ice looks like under real summer conditions. Changing fundamental information in the middle of a test problem is hardly reasonable nor prudent if you want new students to understand a complex subject.)

Cloud albedo also affects solar radiation available at ground level, and the effect of cloud albedo differences is covered in the class notes, EEBB pg 5. But cloud albedos are ignored here in the Problem 4 solution set, where clouds matter a great deal in all parts of the problem solution!

Once again, IF the student answers Problem 4 accurately and completely, he or she will be graded “WRONG ANSWER.”

4.F. The actual daily AVERAGE air temperatures vary from month-to-month in both the Arctic and Antarctic. Hourly temperature changes – even at the same daily “average” temperatures! – are much greater under clear skies with no clouds and low winds than under cloudy skies with high winds. Dry bulb temperature, wet bulb temperature, wind speed, relative humidity and air pressure all change relative humidity calculations. Relative humidity and wind speed changes greatly affect the film temperature and heat losses from both ice-covered surfaces AND from the open ocean in both winter and summer conditions.

But NO convection, conduction, evaporation, nor radiation heat losses or gains are addressed in the problem 4 solution set.

Once again, IF the student answers Problem 4 accurately and completely, he or she will be graded “WRONG ANSWER.”

4.G. Long wave LW radiation losses again differ significantly in both picture conditions – again in opposite directions. Relative humidity and surface skin temperature and sky temperatures (no clouds, high thin clouds, or low dark clouds) ALL change how much long wave radiation is lost to space each hour.

Again, these real-world differences and their effect on the radiation losses are NOT covered in the required “answer set” to this problem. IF the desired answer was “emission temperature” under both conditions, or temperatures of any kind under either conditions, then all heat exchange MUST be addressed and correctly calculated for full credit to the problem.

More fairly, such calculations are a bit long, so a full calculation should either be given more time, or a more limited question asked that can be answered in the time provided.

Once again, IF the student answers Problem 4 accurately and completely, he or she will be graded “WRONG ANSWER.”

5. Finally, having probably spent entirely too much time looking at just the problems within Problem 4 in the problem statement of Problem 4, let’s look at the “equation solutions” of problem 4. First, look at how the student is expected to calculate the reflected energy in the summer and in winter.

Summer albedo ~ 0.07 Fref = 0.07 * 1370 ~ 96 W/m^2

Winter albedo ~ 0.80 Fref = 0.80 * 1370 ~ 1096 W/m^2

My judgment? If you answered both parts as required by the answer sheet, you should flunk. But, IF the student answers Problem 4 accurately and completely, he or she will be graded “WRONG ANSWER.”

5.A. Both albedos are incorrect.

First, 0.07 is not the correct albedo for water. 0.06 is closer, but even that value is ONLY valid for wind-swept sea water under diffuse radiation, NOT the very calm summer arctic ocean water in direct sunlight pictured

The albedo for “pure water” at various solar elevation angles (not even solar zenith angles) is “somewhat” and very poorly shown in EEBB pg 6 below. (Various albedo changes for several different substances at difference solar zenith angles are also shown on class handout EEBB pg 7, but, admittedly, these plots are for solid land varieties.)

However, even this attempt at displaying Fresnel’s graphs is misleading: The plot shown is valid. But ONLY for Fresnel’s approximations of two perpendicular polarized light waves showing the theoretical reflections from a perfectly calm pure water surface at room temperature in the lab. The real-world albedo for the actual “open ocean” has been measured at sea, in the arctic leads, and from open water towers since Payne published his first papers in 1972. The values for open ocean albedo are referenced by Perovich (and Curry and many other sources) in the documents already available to the student. There is no excuse for allowing anybody anywhere to even imply Fresnel equations are valid for ocean waters.

The actual albedo of calm ocean water in mid-September (80 degree north, at noon, when the arctic sea ice is at its minimum and the sun is at its highest point) is 0.316.

So, the expected albedo for “summer conditions” in problem 4 is wrong. Is the expected albedo for “winter conditions” in Problem 4 any better?

Well, no.

0.80 is NOT accepted for ANY period of the year in either hemisphere for sea ice under ANY conditions.

0.83 is the measured albedo for both Arctic and Antarctic winter sea ice covered with fresh snow – but only until late spring in either hemisphere. Arctic sea ice albedo drops down to 0.40 in mid-summer. Bare sea ice (if snow is blown off by high winds) is about 0.60; but bare sea ice rarely occurs at any time of year.)

Antarctic sea ice albedo also goes down during the melt season, but not as far. Its albedo drops to 0.75 in mid-summer (Dec-Jan-Feb). Again, look at the plots in the class notes, pg 4. [See Brandt and Warren, 2005; Curry, 2001, Perovich, 2002 for sea ice albedos; and Burt, 1954; Budyko, 1956; Grishchenko, 1959; Davies, 1969; Payne, 1972; Perovich, 1984; Briegleb, 1986; Frank Hansen, 1993; Bacellar, 2002; Rutlege & Schuster, 2004; Jin, 2004; Pegau & Paulsen, 2008 (and others) for open ocean albedos. ]

5.B. The TOA solar radiation level in Problem 4 is wrong. (Since TOA = 1370 watt/m^2, the solar radiation levels are for TOA levels in mid March or mid-October ONLY.)

5.C. No atmospheric attenuation nor air mass calculation is made. Thus, the student is expected to assume that the sea ice is being irradiated up in space, right?

5.D. No latitude correction is made. Thus, the student must assume that the ocean water is tilted perpendicular to the sunlight, right?

Oddly, Problem 7 on the same mid-term exam does require the student to calculate atmospheric attenuation – although the method required in Problem 7 is not valid for other latitudes and air clarity values – but clearly the students are expected to know how to make the calculation correctly. And, Problem 3 on the mid-term exam requires the student to address albedo differences on different days of the year. (Perhaps they are just not required to know WHEN to make to make the correct calculation the correct way using the correct values.)

6. Fabsorbed.

Frankly, this calculation puzzles me. Clearly, the intent is to apply Kirchhoff’s Law to the absorbed energy going into the sea ice as shown in the class notes, EEBB, pg 2.

But the assumptions required when you derive Kirchhoff’s law mean it is ONLY valid for a perfect blackbody perfectly radiated in a perfect black hole in a perfect space.

The only input energy allowed in is from solar radiation at a perfect temperature spectrum and uniform absorptivity, perfectly emitted from a radiating black body onto a perfect “grey body” of a perfectly uniform emissivity and perfectly uniform heat capacity. (Thus, the sea ice must be in space, directly exposed to the sun’s “average” energy.)

The area receiving solar (SW or IR) radiation must be the same size and orientation as that radiating thermal (LW) radiation. (No heat loss is permitted from the back wall of the sea ice. No radiant losses in any other direction are permitted when Kirchhoff’s Law is derived as shown.)

Both receiving and radiating areas must be exactly perpendicular to the original radiating blackbody. (No correction is made for angle of the sea ice to the radiant energy coming in. radiant energy outbound is assumed perpendicular to the gray body. )

The “viewing angle” between both black bodies is perfect (No occluded or hidden surfaces. This goes with the perpendicular assumption above, and is hard to accomplish in real-world spacecraft radiating energy from their power sources.)

The perfect gray body radiating energy must be at perfect thermal equilibrium with the space around it. (In this particular derivation, the surrounding space is itself a perfect black body at 0.0 K degrees. All parts of the gray body are assumed at the same temperature.)

No other heat loss can be allowed when you use Kirchhoff’s Law. (Hence, my claim that the iceberg is “insulated” and “isolated” on the front, back side and all walls. No evaporation, no convection, no conduction losses permitted. No heat loss to the atmosphere, to the salt water underneath, no sublimation, no phase changes allowed in melt ponds above the surface, no freezing of sea water below the surface if air temperature lowers. Completely isolated sea water.)

No light energy is allowed to go through the sea ice. (Opacity is 0.0. This condition, at least, is somewhat reasonable.)

But the required equation itself, after the student is shown Kirchhoff’s Law?

Energy absorbed = (1-albedo) * (radiation available at TOA) * (1-emmisivity)

No. Absolutely not.

But a good equation for opaque, insolated, isolated, insulated icebergs in space however.

7. Final exam, Sea Ice 101.

There will be a one-question Final Exam when the Sea Ice 101 Class ends later this year. Students should keep their notes ready for this question.

Calculate the actual energy absorbed and energy released from the open water and from sea ice on ONE date and ONE latitude at any selected single hour. Chose both Arctic or Antarctic latitude, and any day-of-year – BUT! – sea ice must actually be exposed to solar radiation on that hour of that day! (You may not select 85 north on Dec 22. You may not select 85 south on 22 Dec. You may NOT select 0.0 latitude on any day-of-year.)

Further, since all Final Exam answers will be available to all members, you cannot select any latitude or day-of-year already chosen by somebody else. (First chosen, first choice.)

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86 thoughts on “Sea Ice 101 – Insolated, Isolated, Insulated Icebergs in Space, A Mid-Term Exam

    • Oh, You are absolutely right. But, then again, I wasn’t trying to do a full explanation of a complex and intertwined subject here. Nor even a full explanation of each section of the full topic.
      Rather, I wanted to “get started”. Declining Arctic sea ice is the SINGLE remaining “weapon” that the CAGW community has left in its arsenal. It is the first and last “evidence” that is called up for “climate alarmism” and for hysterical forecasts of future doom and heat-filled daze of boiling arctic nights.
      But, when you actually calculate the heat gained, and the heat reflected from Arctic and ANtarctic sea ice … When you actually try to answer Problem 4 AS YOU SHOULD, then you find that … Well, there is no problem with any future loss of Arctic sea ice. Instead, you find that the ever-growing Antarctic sea ice IS a great threat for cooling, and one that is many times greater than the hysteria about the loss of arctic sea ice.
      Yes, later, l intend to cover each section in much greater detail. For example:
      1. How much solar radiation falls on the top-of-atmosphere, and how much makes it through the atmosphere for different latitudes and different climates and clouds. The result? We’ll be able to predict a reasonable value for solar radiation falling on the sea surface anywhere on earth for any hour-of-day on any day-of-year. Why? Because it matters.
      2. Where the sea ice is, and when the sea ice is present (or absent) each day-of-year on an average year, and over the past twenty years as the Antarctic sea ice has grown, and the Arctic sea ice has grown smaller. Why? Because the answers from the first topic affect the second, and because the second topic (actual sea ice area and dates) affect the angle of the sun each day over the sea ice.
      3. So, why are area and dates and latitudes that important? Because the angle of the sun controls the albedo of the open ocean, and the day-of-year controls the albedo of the Arctic and Antarctic sea ice.
      So why does albedo matter? Because albedo controls how much solar energy is absorbed that second of that hour on that day at that latitude.
      4. But albedo (solar heat gain into the Arctic Ocean or Southern Ocean) is only a part of the other losses: evaporation, conduction, convection, and thermal LW radiation. Each of those is affected by the presence, or absence, of sea ice above the ocean. And, like everything else, those calculation depend on things that vary with latitude, day-of-year, and hour-of-day.
      You cannot answer Problem 4 without addressing all of the little points properly and accurately.

      • If I am understanding that correctly, what you are essentially saying is that the data provided is insufficient to answer the question.
        Please correct me if I am wrong.

      • This is why clouds are so important and why we have insufficient data on clouds to even begin to model.
        The time of year when a cloud forms, the time of the day it forms, how quickly it forms, the time it dissipates, how quickly it dissipates, the area of the cloud that is formed/forming, the volume of the cloud that forms, the height at which the cloud forms, the composition of the cloud that forms, the latitude/longitude over which the cloud forms and the daily albedo of the shaddow area (and albedo can change on a daily basis, even in oceans with plancton blooms), the absorption characteristics of the surface that has been shaddowed by clouds, all such factors have to be known.
        There is an assumption that cloudiness can be averaged and it does not change from year to year, but that is an assumption and one that seems to be questionable given the chaotic nature of clouds. No two years have the same cloudiness.
        When you are dealing with a system that is more than 4 billion years old, it would not be at all surprising to see a lengthy span (20 to 50 or even 150 years) where changes in cloudiness could result in a sustained warming or cooling trend for 20 or 50 or so years.
        The most likely explanation for the 20th century warming is changes in cloudiness (and possibly a reduction in aerosol polution in the 1960s/80s which is now no longer happening due to the advent of China’s inductrialisation). Whether there are forcings/factors that have brought about that change is a different matter

  1. This whole exercise reminds me when I was at university doing an advanced quantum course. For obvious reasons I will leave out identities. Anyway, being a bit of an individual student (I didn’t rely on my Profs to teach me Quantum, I taught myself), I completely solved a particularly difficult exam of ten questions, something that shouldn’t had happened, as I was told later. Anyway, after returning the paper with my own individually thought out solutions, I learned later I got a BIG FAT ZERO. I took this to higher ups in the University, and it wasn’t very well appreciated, for you see, I believed I answered all questions not only correctly, but better than what the Prof. did himself, and I challenged him, and that didn’t go down well with all the egos in the department. Long story short, is that this Prof. got very upset that I didn’t solve the questions as he would have liked me to, I quote you what he said (its nearly 30 years ago but I can’t forget it), “you didn’t do them my way”, I replied, “I don’t read minds, I just do physics. And I do it better than you.” As you can imagine now, at the time I didn’t, it was war…. I lost.
    Modesty aside, I did do physics, and still do physics, better than him. But it wasn’t enough, students must do as they are indoctrinated. Nothing has changed, at universities around the world. It doesn’t matter how good you are, or how well you understand the science, if it doesn’t follow the gospels of high priests of science, you are out of the church.
    To end my story, that was the start of the end of my career in physics, at least any chance of an academic career. When you test the competency of the false gods, especially in physics, you pay a heavy price.
    So it is with this above question in climate science. You can show them how it should be done, it doesn’t matter. Academia is not a place to test science and theories, it is now a place where students are tested on their ability to conform to practices that academics seek to instill in their students, that is, students that will make academics look good. So what does a student need to be able to do today in science…
    – Students must never question textbooks,
    – Students must follow an academic’s step-by-step process to explaining and solving problems. If you can read the academics mind, so much the better,
    – Students must never show independent brilliance, or to better their Prof;
    – Students must do all the work, write all the papers, and let the Prof, get all the praise,
    – Students must protect and bolster accepted theories, no matter how flawed they are, or worse, show an alternative theory.
    So, you can write, hundreds of pages of proofs to show why a question is lame or just bad. You can show how all the possible answers are wrong. It doesn’t matter. For YOU ARE NOT AN ACADEMIC, AND ANYTHING YOU WRITE OR CONJURE UP IS JUST WRONG, FOR THE ACADEMIC IS RIGHT BY EX CATHEDRA. For you see, if you shine light on a particular academics incompetence, you are not just arguing against an individual, you taking on the entire department and the whole university. How will it look, if say Harvard, MIT or Cal Tech came out and stated that [they] have a dud physics professor? Even though if they did (and in reality they do!), these universities would lose incredible face. So they are always right, just like Michael Mann is always right, and the university will always defend him. Even though it is proven that Mann, is a no good lying weaseling cheater.
    Its not about the science, or the math, or the data. Its about the status and position of office. Science today is driven by title not facts. I learned this the hard way. When are you guys are going to understand, it ain’t about science. It never was, and never will be. Its about prestige, title and money. [Competency] and good hard science has been slowly being killed off since the 80’s. And its going to get worse.
    Facts are not going to change anything, you have to start kicking the lousy bums out of office, and that means TENURE MUST END.

    • Dorian
      Amen to all of that! It is generally known there are silos all over a university campus but what is not so well known is there are science sects in each one. Students end up liking sects while they are there, most of them anyway.
      Re the article above:
      It is a bit shrill and contains bloopers like this, “Now remember your geography. All of the Arctic Ocean is above the Arctic Circle. So, from the middle of October through the end of February (the “winter conditions” for problem 4) there is NO sunlight at all ANYWHERE across the Arctic Ocean sea ice. ”
      I just flew over the Arctic ocean just near the pole and there is plenty of sunlight north of the Arctic Circle in the first week of February. Observation beats theory every time.
      🙂

      • Is it?
        Values are for Feb 3, day-of-year = 34.
        85 north? Dark all 24 hours.
        80 north? Dark all 24 hours.
        75 north? Dark all 24 hours.
        72 north? The sun is 1.2 degrees above the horizon at noon. Dark the remaining hours.
        70 north? Two hours of sunshine 11:00 to 13:00, the sun is 3.2 degrees above the horizon at noon.
        68 north? About the same.
        But, you see, there is less than 20% of the Arctic sea ice below 70 north latitude (Bearing Sea and Hudson Bay are the two large exceptions.) So, does it matter is there is two hours of sunlight in an area where there is no Arctic sea ice at all?
        Oh. By the way, remember when I pointed out sea ice is at sea level in the Arctic? Flying ABOVE the arctic clouds at 35,000 feet – 40,000 feet requires a different calculation for “hours of sunlight” than being on the ice below the clouds on that same day at that same time of day at that same latitude.
        So, February 3 this year? The arctic sea ice (excepting small bits of the Bering Sea and Hudson Bay near noon) was in the dark. But I would like to know what latitude you were over top of, and which day you were flying.

        • garymount

          Sunlight in the arctic, January 25

          Yes. And you will notice that the furthest edge of the sunlight at local solar noon is barely kissing that 70 north latitude on 25 January – Not even 1 degree above the horizon yet. The actual edge of the Arctic sea ice is NOT the “arctic” that you have been trained so lovingly to “know.”

      • Crispin, instead of ‘silos’ you should have used the more politically correct term ‘cylinders of excellence.’

      • Some sunlight illuminating the atmosphere occurs even when the sun [solar disc] is below the horizon. Hence, civil twilight is defined as -6 degr below the horizon. It’s the evening sky after sunset but before nighttime darkness. Since the solar disc is hidden there is no insolation on the surface, no radiant heating. But there might be a bit of light to see by.

        • Crashex
          Some sunlight illuminating the atmosphere occurs even when the sun [solar disc] is below the horizon. Hence, civil twilight is defined as -6 degr below the horizon. It’s the evening sky after sunset but before nighttime darkness. Since the solar disc is hidden there is no insolation on the surface, no radiant heating. But there might be a bit of light to see by.

          True. Very true.
          But, we are not talking about “visibility” but about “solar heating.” Further, and you ARE making an important point that I DO want to keep talking about the subtle details, we started by limiting the discussion to “a clear day, looking at direct radiation.” Not diffuse radiation.
          But it is NOT a trivial difference, and your point is important. But, please, keep the difference clear.
          Diffuse radiation (all-sky radiation) has very, very little “power” to heat things up with absorbed SW (IR) radiation. But, it does exist. Global radiation measurements (depending on the type of pyrometer used and the specific type of of radiation shield mounted) include both diffuse + direct radiation, diffuse radiation only, or direct radiation only. We “hope” that anybody using the term “global” radiation in a paper actually and really and truly measured “global radiation” as a sum of both direct and diffuse radiation …. But I often find that is not the case. Blah.
          Oh well. Life happens.
          The albedo of water is very, very different for diffuse radiation than for direct radiation: Under almost ALL circumstances, I have found that diffuse radiation albedo of the open ocean is 0.06 under ALL solar elevation angles. This is the typical “Wikipedia value” for the albedo of water you always see.
          The albedo of water (open ocean) for solar elevation angles OVER 40 degrees is also about 0.06. Again, notice that this IS correct “per Wikipedia” for the ocean between say latitude 48 north and 48 south. Not bad, actually. (For you extremists, a solar elevation angle of “over 40” is equal to a “solar zenith angle” of “under 50” … “I” personally, don’t like solar zenith angles, so “I” won’t use them. Live with it.)
          So, let us confuse the issue a bit. And, I’m cheating: These words can be used for the the topic on Albedo of the Open Ocean.
          Under real world skies and real world atmospheres, there is ALWAYS a little bit of diffuse radiation getting back-scattered and front-scattered even in the clearest of all skies. Thus, even under a perfectly clear sky at noon on the equator on March 22, there will be about 10% of the direct radiation present and shining down like from the sun’s spotlight as diffuse radiation shining from “everywhere” in a gentle glow from all angles.
          Now, as the solar elevation angle decreases, and as the atmosphere gets more and more “scattering” as air mass increases, the percent of diffuse-to-direct increases substantially. Near the horizon, the Russians in their long-running Arctic measurements on ice islands up north, measured the diffuse-to-direct measurements as almost 0.90 to 1.00.
          Thus, as the sun sets in the Arctic – a period that can take 1-2 hours! – the percent of total sunlight goes down, right?
          But the percent of diffuse sunlight in that total constantly increases.
          And you – properly! – pointed that out.
          BUT! “I” – anticipating your question – will point out that
          (1) I wanted to limit myself to direct radiation effects right now – but “you” wanted more detail.
          (2) I need to point out the reflection of clouds in that diffuse radiation value actually reaching the surface = (75% – 85% of the available radiation is reflected from the top and upper surfaces of the clouds before it can create the diffuse radiation that is visible)
          (3) Visible light is not “heating” solar energy on the sea surface.
          Finally, as you will see, the actual solar radiation on the earth’s surface is 25 – 50 watts/m^2 on the day’s of interest. Diffuse radiation adds perhaps 5-10 watts/m^2 to that value. Noticeable most days of the year, but not significant in the total scheme of things.
          8<)

      • My note was simply addressing these two prior comments.
        Crispin–“I just flew over the Arctic ocean just near the pole and there is plenty of sunlight north of the Arctic Circle in the first week of February. Observation beats theory every time.”
        Crispin is talking about illumination to see, while the post is focused on insolation on the surface from direct solar radiation; two different things.
        RACookPE–“Values are for Feb 3, day-of-year = 34.
        85 north? Dark all 24 hours.
        80 north? Dark all 24 hours.
        75 north? Dark all 24 hours.
        72 north? The sun is 1.2 degrees above the horizon at noon. Dark the remaining hours.”
        I take exception to the term Dark as used here. You should say something like–no sunrise, no direct insolation”. Further, you seem to be talking past Crispin’s point that there is suficient ambient illumination for visual observations. Civil twilight is 0 to -6 degr. and nautical twilight is from -6 to -10 degr. At ~75N on February 3 there is 6 hours of twilight between 0 and -6 degr. and plus a few hours between -6 and -12. Calling it “Dark for all 24 hours” is incorrect because “Dark” references the more complete absence of ambient illumination than the conditions present at twilight.
        To some extent, you are both right and both wrong at the same time because you are talking past each other about two slightly different things.
        You noted well that I was referencing “direct” solar radiation and bookkeeping the diffuse radiation, even if small, is important for accuracy.

    • And that is EXACTLY how academics works in 2015. And that only scratches the surface of the absurdity. Dorian’s post is worthy of its own thread!

    • This is a double-edged sword. Without tenure there wouldn’t be a single “skeptic” left in most universities. I do agree that tenure often protects incompetents.

    • Thanks for your story Drian. it brought back a vivid, to this day, memory of being expelled! from a high school for the ( Mathematics so not so advanced as you) same reason, I went through the same process but like you lost I also have to add another comment, word got around to other schools and teachers and I actually (in the long run thankfully) quit school and have lived happily ever after. I hope you have as well.

    • Dorian
      Things weren’t any better back in the late 60’s when I challenge An economic professor about sale tax being regressive and income tax being fair. Having worked for my father’s plumbing business doing estimating, bookkeeping, and payroll as well as digging ditches, I answer the question that all taxes are regressive. The end user of goods and services pay all taxes related to goods and/or services. Sale tax rate is known and collected at POS. All other taxes are priced into goods and services and hidden from view but still paid for by the end user.

    • Very interesting Dorian. I was an instructor in the AirForce. I was arguing with a fellow instructor about a minor technical matter. His reasoning that he was correct and that I, a newbee, was wrong, was that he had been teaching that technical point that way for years. After showing him the correct info in the technical manual he was a changed man. I found several times that some things were being taught wrong. Most of the errors were made by attempts to simplify things. I told my students that if an instructor or teachers says something that doesn’t make sense to them that it may be because it really doesn’t make sense. I was apalled by the lack of understanding by some of my fellow instructors.

  2. This is what Gail Combs had to say on this subject explained so much better.
    Now, at the equinoxes, when both Arctic and Antarctic are both hit by the same solar intensity, the Antarctic Sea Ice receives between 2x (Feb-March) to 5x (September-October) the energy that the Arctic sea ice receives. Thus, to reflect equal energy into space, the “gain” of even 1.0 Mkm^2 of southern sea ice extents needs to be balanced by a loss 2 to 5 LARGER in the Arctic.
    Instead, we see near even sea ice changes. So, since at today’s levels of sea ice extents, “more Arctic sea ice loss” up north means a net loss of energy from the planet; and “more sea ice extents gain” down south means a net loss of energy from the planet, we are facing a future big problem. If today’s trends continue.
    And, as we can always predict, if today’s rates of Antarctic sea ice extents gain continue, the sea route around Cape Horn [Drake Passage] will be blocked to shipping within 8-10 years!
    In another comment Mr. Cook gets more specific
    http://wattsupwiththat.com/2014/02/18/how-much-sunlight-actually-enters-the-system/#comment-1571201
    ….Albedo of Arctic sea ice changes only based on day-of-year. Albedo starts high at 0.82, stays steady at 0.82 until May, decreases through the summer to a low of 0.46, then rises again to 0.82 until about September, then remains at 0.82 until the end of December. This is from Dr Curry’s measured data.
    1. Albedo of sea ice does NOT change with latitude.
    2. Albedo of open ocean changes with every HOUR of every day as the solar elevation angle changes each minute. Specifically, open ocean albedo does NOT change explicitly with latitude, but latitude affects the overall SEA change over day-of-year AND latitude and hour-of-day (HRA), These changes are based on the earth’s declination and geometry and is strictly and specifically defined. But, Hour-of-day and day-of-year CANNOT be separated from latitude.
    3. Opposite the above, the yearly maximum solar radiation occurs in early January at 1410 watts.m^2. The minimum solar top-of-atmosphere radiation occurs July 3, when the Arctic sea ice is decreasing strongly day-by-day, BUT while Arctic sea ice is between min and max. Roughly, the edge of Arctic sea ice is between 74 and 76 north.
    At the point of maximum solar radiation at TOA, the ANTARCTIC sea ice is is a wide “ring” slowly varying from 59.2 south (last October under 1370 watts/m^2) to about 64 south latitude (in January under 1410 watts/m^2) to a minimum sea ice extent at 3 Mkm^2 (in March at 70 south latitude back down to 1360 watts/m^2). So, when the TOA solar radiation is at its maximum, ARCTIC sea ice is dark. When the top-of-atmosphere radiation is at its max, Antarctic sea ice is not at its minimum.
    Net effect: As a whole, Antarctic sea ice is MUCH, MUCH closer to the equator every day of the year.
    Overall, increased heat losses from open ocean in the Arctic (when Arctic sea ice is at a minimum in late August-September) are much greater than increased heat absorbed into that open water. More sea ice loss in the Arctic => More heat loss from the planet and a net cooler planet.
    The opposite happens in the Antarctic: More sea ice around Antarctica means more heat reflected from the planet and a net cooler planet.
    It is not really necessary to “combine” or group the other two parts of the Antarctic
    Up north, the Arctic Ocean STARTS at 70 north latitude, and this IS the southern limit of the Arctic Ocean. Essentially ALL “Arctic sea ice” then cycles between 70 north
    latitude (at MAXIMUM extents at 14.0 Mkm^2) and 80 north (if 4.0 Mkm^2). In the future, this minimum could go even closer to the pole: if there were 1.0 Mkm^2, all the arctic sea ice is a little beanie cap from the pole to 85 north latitude.
    The Antarctic sea ice is INCREASING at all times of the year.
    The Antarctic sea ice cycles between a minimum of of 4.0 Mkm^2 at latitude 70 south, to a maximum of of 19.5 Mkm^2 at latitude 59.2 south.
    The Arctic sea ice only varies between 72 north and 82 north.
    On EVERY day of the year, Antarctic sea is exposed to 2 to 5 times the radiation that Arctic sea ice receives, and is therefore Antarctic sea ice is 2 to 5 times MORE important to the earth’s heat balance than the Arctic sea ice. (But the tropics are even more important.)
    That is the first of several posts by Mr. Cook and others where they have a ‘discussion’ with Willis E. Several good papers on the subject are cited.
    []

      • Curious George
        Link, please. I question the 2x to 5x claim strongly. I even won’t take your word for it.

        I don’t care if you take my word for it or not. I DO want to be able to be sure YOU can make that calculation yourself. Further, if you disagree with any other references or calculations, speak up!
        Complete the rest of the class.
        What? Are you not capable of doing the original work yourself?

      • No. To the best of my knowledge, no other person has ever calculated that value: The relative importance of Antarctic and Arctic sea ice in the earth’s energy budget compared over the full year has never been published.
        Are you afraid of original research?
        Are you afraid of “thinking” on your own and “demand” that each sentence be written by somebody else first?

      • Oh, sorry. I just realized that you were the author of this post.
        While I agree with you that there is an abysmal “science” taught in the name of Ecology and Climate Change, countering them with false claims does not help. I still take an exception to your claim that on June 21st “Antarctic sea is exposed to 2 to 5 times the radiation that Arctic sea ice receives”. Numbers, please?

    • Thank you for the summary of those earlier posts.
      We will get the actual source of each of those calculations and the background references and papers and source documents for every value quoted above in the next few classes. But they are too many, too complex for one post all at once.

  3. Robert, thank you for a surprisingly entertaining piece. I have very little science training yet I was able to follow rather well. I have a feeling that no engineering student would be taught in such an appalling way.
    Those two iceberg photos at almost all that you need to make the case. Photos don’t lie, but captions can. 🙂

  4. It’s similar to my own kids asking me what the answer is for a question in their junior and senior high text books. i always tell them I can give you my answer, but the answer that the teacher is looking for is in the book – go read the book. As for teaching arctic climate, seems we don’t know very much but the experts will never admit that.

  5. Sorry for being impatient.
    But what actually is the point of all these exams and albedo?
    Sorry also for asking without reading the papers related to the subject.
    Are the albedo physics renovated, or just another new try on the wrong old approach?
    I do not doubt whatsoever all the good and knowledge about its effect on whatever, but as I most interested in climate perse, and the assumed role and effect of albedo, I am asking these questions.
    If I am not wrong the last time they tried such physics they ended up with a huge error about the predictions on the arctic warming and the famous Runway Global Warming.
    When there never even a single sign of a RGW, there was definitely a much more than ever predicted or calculated warming in the Arctic, in such an error that the albedo physics seems laughable when climate considered.
    I don’t know how much more wrong than that they could be found with their physics before starting to consider that whatever albedo effect is, has no much effect, for not saying not whatsoever, in climate, let alone the earth energy budget and balance.
    The albedo impulses on climate are too small and end up cancelling each other out before could even be registered in climate, same as any other background noise.
    The last “RGW experiment” gave this as a result, as far as I can tell, so what is the new catch, apart from still not being able to distinguish between the earth energy balance budget and that of the atmosphere?
    cheers

    • An appropriate question. Thank you.
      Look back above, at my answer to Salvatore’s first challenge right at the top of the replies.
      And, yes, Gail and I have covered parts and pieces of these topics before. But they need to be done in detail. As you’ll see reading that answer above.

  6. I was hoping that it was going to be a “multiple choice” exam. I would have done better on that. Thanks for the Gail Combs reference posting to explain it better for those such as me…

  7. Ten years from now, I can look back and say, “see, I knew there was no way that the Drake passage would become frozen over and blocked by ice.”

    • You may not select 85 south on 22 Dec.
      Should this be June?

      Ah. No, but it was a trick question. I was taking advantage of some subtle geography here that you may not be aware of yet. See, the Antarctic continent spans a rough circle around the south pole whose “average” coast line is right at 70 south. There is no Antarctic sea ice at 85 south in May, June, July, December, or March. It’s all land ice at 85 south all year around; so, NO, you don’t get to measure energy reflected from land ice if the sea ice melts in December!
      Now. Always remember to flip-flop your mind here: Summer temperatures around Antarctica are highest in December-January-February.
      On the other hand, the lowest-ever satellite-era Arctic sea ice area was 3.0 million km^2 in September 2012. That DOES correspond to a “cap” around the north pole down to 85 north latitude. So, 85 north latitude could be a valid latitude in September. But not any other time of year!

      • I must be missing something here. Looking up ‘arctic circle,’ I find:
        the imaginary circle around the earth, parallel to the equator, at latitude 66° 32′ S; it marks the southernmost point at which the sun appears above the level of the horizon at the winter solstice
        At 85 S on the solstice the sun would never come up. But the prof just said:
        sea ice must actually be exposed to solar radiation on that hour of that day!
        Hmmm…. Looks like a trick question all right. : > )

  8. Minor quibble. Those don’t look like icebergs to me. I thought they were simply chunks of ice protruding maybe 20-30 cm above the water at most. It is hard to judge the scale in a picture like this.

    • I agree, the photo is not explicit.
      But, yes, those are “icebergs” whose top is 5-15 meters above sea level. At a 1:10 ratio of “above-water” to “below-water”, the remainder of the icebergs below water (invisible) can only be due to glacier ice breaking off from Greenland.
      In the “winter conditions” photo, the camera is at eye level 1.5 to 1.7 meters above top-of-ice, focusing dead level at the top of the clumps. This establishes their maximum relative height of 1.5 to 1.7 meters. 80% of Arctic Sea ice is 1.5 to 2.0 meters thick, and it must be presumed that that this is ‘average” Arctic sea ice (not multi-year ice.
      The icebergs in the back of the “summer conditions” are viewed from a helicopter at 150 – 200 feet elevation, and show the icebergs in the background. It is unclear, but a green ship (4 deck high @ 2 meter per deck) may be near the iceberg in back left corner. (If so, it is unusual to be that close to a iceberg, even in pack ice.)

  9. Could be a minor point but the “Students” may have already realized they cannot win.
    What to do (What do they do … What can they do)?
    [trimmed] the “Prof” and staff. Without a winnable resolution the students are right in “[trimmed] the Prof., Staff and Admin. Staff and perhaps add a few Deans and Provosts and Vice Chancellors and Chancellor to the mix including the President of the University and the hallowed “Descendent From God” the Board of Regents.”
    Children of the Prof., Deans, Provosts, Admin. Staffs and the Chancellor and President and Board of Regents are fair game [trimmed].
    Automobiles burning, Apartments burning, Houses burning [rest trimmed. .mod]
    FU

  10. I’m impressed by the Sherlock Holmes-like detective work on the phoney Arctic photos. Amazing that a tenured univ lecturer could forget that the poles get dark.
    I’m not great at math but here are my summary conclusions:
    1. Albedo.
    Needs sunlight. In terms of hours per year everywhere on earth gets the same. But due to both incident angles and long attenuating path through atmosphere, less sunlight at poles.
    Therefore albedo of ice or anything else is weakest and at a minimum at the poles, increasing toward the equator. Ice and snow being whitish are important players in albedo. But since there is no albedo in the dark, for ice and snow to exert significant albedo cooling effect they must exist at low latitudes away from the poles.
    2. Heat loss from water.
    This happens day and night independent of sunlight.
    Where there is open water, heat loss depends on how cold the overlying air is. Air gets colder toward the poles. Thus, opposite to albedo, heat loss from open water increases toward the poles.
    At and near the poles sea ice cover warms the system by reducing open water heat loss. Albedo less relevant.
    At low latitudes sea ice cover warms the system via albedo in stronger sunlight – heat loss from open water to warmer air less relevant.
    Thus Gail Combs’ point – both open water at the smaller Arctic and growing sea ice at lower latitudes at the larger Antarctic – both are cooling the planet. Not warming.
    A critical latitude exists at both poles, above which sea ice warms, below which sea ice cools. At the Arctic it’s above, at the Antarctic sea ice is below this level.
    Correct me if this is wrong or over simplistic.

    • Yuppers. About right in all regards.
      Now, it is actually worse than you think.
      Imagine an open water at 75 north near the north pole.
      If the surface is water, it is radiating from a surface at -2 or -4 deg C, right. 270 K say.
      If ice, the surface is radiating from a surface at near the air temperature -25 C or 248 K, right?
      Both sea ice and water have the same emissivity, so final radiation heat loss between the two to the same environment is only proportional to T^4, right?
      Thus, what loses LW radiant heat more readily? Open sea water in the Arctic or sea ice? Open water, clearly.
      What has more evaporation losses? Open water, right?
      What has greater convection losses and conduction losses? Open water, right?

  11. Correction:
    At low latitudes sea ice cover warms cools the system via albedo in stronger sunlight – heat loss from open water to warmer air less relevant.

    • Phlogiston
      At low latitudes sea ice cover warms cools the system via albedo in stronger sunlight – heat loss from open water to warmer air less relevant.

      Ok. So, what do you define as “low latitudes” and where does today’s sea ice reside at these “low latitudes”?
      I’m not disagreeing with your premise – because I have no reason to agree nor disagree right now, but – let’s be realistic, what day-of-year, what latitude, what hour-of-day is your statement true?

    • At low latitudes sea ice cover cools the system via albedo in stronger sunlight – heat loss from open water to warmer air less relevant.

      Now, expanding on the above, what is the latitude of the expanding Antarctic sea ice edge? (Hint: 58-67 degrees south)
      What is the latitude of the contracting Arctic sea ice edge? (Hint: 72 to 80 degrees north)

  12. “”The IPCC claims the entire “forcing” if atmospheric CO2 doubles is only 3.7 watts/m^2.””
    I still cannot find a valid scientific explanation for how atmospheric CO2 delivers this ‘forcing’.
    Low IR band photons CANNOT warm any substance already warm enough to emit those same IR band photons.

  13. Did Dr. Wiescher write the text book for the class?
    ‘So, as we will see in a few minutes, even IF the student used this formula to answer Problem 4, that answer would be graded as “WRONG”!’
    If so then he could sell a supplement text book to explain why the answers are wrong.

  14. Robert, I am confused and I would like to understand (might not be alone).
    Could you begin Lesson #2 with a synopsis of Lesson #1 learning – what were wrong assumptions, approaches, why and such? Provide some solid ground for moving ahead, perhaps. Imagine that I am a representative person who needs to understand why this is important and how to get there.
    As example – I’ve never taken a “Climate Science” course – not even sure such a thing is viable – do they actually make calculations? thought it was all bs models, predictions, claims for money, alarms. . .
    I’m not able to ask a good question yet – I don’t know enough, but I would like to.
    Start with this – why does sea ice matter? You say it is their last leg, but if I were to write a letter in newspaper, would anyone understand? care? Please feel free to recommend background readings for me. I know I am missing something that might be important.
    Thanks.

  15. This is an unfortunate post. Most of Dr Balls physical objections are correct but this is vitiated by his ignorance of conditions in the Arctic. Three examples:
    “At times, the flat “sea ice” is pressed together into thin, tall ridges, but when these ridges are freed and allowed to float freely, they immediately fall sideways back to “flat sea ice”.”
    Completely false. Compression ridges are caused by pressure, but they mostly float on invisible “keels” and will to a large extent remain long after the ice-pressure has ceased. The Arctic sea-ice is anything but flat when seen up close.
    “The presence of several high icebergs in the background of the “summer conditions” means that photograph was taken much further south than the “arctic Ocean.””
    Also false, conditions like these are easily found in the coastal areas of East and West Greenland, Baffin’s Land and around Svalbard where there are glaciers nearby. As a matter of fact I think the unusual clear, calm and cloudless conditions are probably a lot easier to find in e. g. East Greenland than in the Labrador Sea.
    “The Arctic sea ice between mid-October and mid-March is far north of the Arctic Circle, and thus has NO sunlight exposure at all at ANY time of day.”
    Once again false. This is only true at the North Pole, where there is indeed dark during most of this period. Further south the dark period is much shorter. Remember that on March 23 the day is 12 hours long everywhere, from the North Pole to the South Pole. And this is just a few days after the Arctic sea-ice maximum. Anyone wanting to verify this for himself can take a look at a webcam in Longyearbyen in Svalbard (76 degrees N) around local mid-day, for example: http://longyearbyen.livecam360.net/
    Even at this time of year (Feb 10) there are several hours daylight there. And this late in the season Isfjorden is covered by sea-ice.

    • Once again false. This is only true at the North Pole, where there is indeed dark during most of this period. Further south the dark period is much shorter. Remember that on March 23 the day is 12 hours long everywhere, from the North Pole to the South Pole. And this is just a few days after the Arctic sea-ice maximum. Anyone wanting to verify this for himself can take a look at a webcam in Longyearbyen in Svalbard (76 degrees N) around local mid-day, for example: http://longyearbyen.livecam360.net/
      Even at this time of year (Feb 10) there are several hours daylight there.

      We discussed the important differences to varying degrees of detail between “dark”, visible light (enough light to see with), diffuse light (Before and after “official” sunset), diffuse solar radiation levels (as heat energy), and direct solar radiation levels (as heat). And, the details between those distinctions and those different points matter. Sometimes.
      No, peak Arctic sea ice area is reached in the weeks AFTER March 22. The gradual peak in arctic sea ice area is now usually March 28-April 10, several very important days later. (Minimum Antarctic sea ice is reached in late February.)
      Solar radiation exposure (hours of direct and indirect (diffuse) sunlight available each day) changes very, very rapidly around the equinoxes in March and September. Be specific! A valid claim made one day is wrong even three days later. You’re crossing not 1 or 3 days, but over 5 weeks here.
      Is there sea ice near Svalbard at 76 north latitude in March 23? Of course! That is much further north and much later in the year than the 60 degrees (“dark” on Feb 3) actually used as an example.

    • Completely false. Compression ridges are caused by pressure, but they mostly float on invisible “keels” and will to a large extent remain long after the ice-pressure has ceased. The Arctic sea-ice is anything but flat when seen up close.

      These high, narrow “keels” are what I did describe as “falling back flat” when the Arctic sea ice melts.
      That glaciers are actually present on Svalbard Island at latitude 76 does not change the absence of massive icebergs across the 14 million sq kilometers of Arctic sea ice across the rest of the sea ice area.
      The presence of mountains on Svalbard is important to Svalbard: But its land area is 61,022 km^2, and glacial ice covers 36,502 km^2 of that 61,000 sq km^2. So, does a photo of a sea ice in front of Svalbard glacier (36,000 km^2 / 14,000,000 km^2) change the result when you calculate how much sunlight is reflected from the rest of the Arctic?

    • This is an unfortunate post. Most of Dr Balls physical objections are correct but this is vitiated by his ignorance of conditions in the Arctic.

      Dr Ball is not commenting here, nor is he writing here.
      I am responsible for the words written. Now, if Dr Ball DOES wants to take blame for something I wrote down incorrectly, or something that was not clearly written, then HE needs to get over here and take the blame himself. “I” won’t do it for him. 8<)

      • I was just savouring the irony of these “shortcomings” being imputed to Tim Ball, since he was flying Search and Rescue in the Arctic for years and might have a notion or two.

  16. Bubba Cow February 10, 2015 at 3:58 am
    …why does sea ice matter? You say it is their last leg, but if I were to write a letter in newspaper, would anyone understand? care?
    No idea, but the rapid rise in the summer see ice extent to the east of the Arctic Ocean should be of interest or even mild concern.
    Here is September 2007 (source nsidc)
    http://nsidc.org/sites/nsidc.org/files/images/asina/20070926_extent.png
    magenta line is 30 year average
    and here (right hand side image) September 2014 (source noaa)
    http://www.arctic.noaa.gov/reportcard/images-essays/fig4.1-perovich.gif

  17. Here, this class, is a perfect opportunity to write the text as a IPython interactive ‘book’, an ‘I-book’ (a la p-book and e-book). I-books do the math before your eyes on data pulled in real time. Arrgh! It is hard for an old dog to learn new tricks.

    • Doug Huffman

      Here, this class, is a perfect opportunity to write the text as a IPython interactive ‘book’, an ‘I-book’ I-books do the math before your eyes on data pulled in real time.
      Arrgh! It is hard for an old dog to learn new tricks.

      Please! Show us how.
      Examples will be very welcome!

  18. Sorry not patient enough to go through all this, especially if its wrong!! Perhaps the prof is terrible but I must say, teaching 101 stuff has to be somewhat airy and simplified. I found lots of shortcomings with my profs, of course, but at my age, I realize that in broaching a subject, you really can only give a flavor of the types of problems the science deals with. I had a prof who taught a 101 course (over 50 years ago) who in the 200 course told us to forget half of what we he’d said in 101 – it was wrong!! Then we delved into more details. It works. Oh I had wrangles in graduate school although I found there that there was a bit more equality. When I finally got out and off to work, I found I had a new learning phase. The stuff I had been taught wasn’t up to date and I had to learn what was in the gap.
    In one job, in a report I prepared and was proud of, I had so many details and analyses of the project we were doing (still in university mode) that my boss came back to me and told me that all the minutae were unnecessary – we know what the chain rule is, etc. etc. He said a good carpenter taps the nail a few times to get it started and then drives it home with 2 or 3 whacks! I respectfully pass this advice onto you.

    • Understood. Your point is correct.
      But…. Using 101 is deliberate: We are just hitting the basics of heat transfer down to the ice surface, into it, and then back off of it. But, if we start with basics, then everybody can follow along, adding details as they see fit. And, just as happened with the short detour on direct and diffuse radiation levels – and how they affect heat gains and how they affect “visibility” changes – the detours are informative as well.
      Yes. As I looked up all of these formulas and references, I did find a few of my original values were wrong. For example, what Gail quoted above was what I first used – but now I recognize that May, June, and July do “add” more direct solar energy to the Arctic than the Antarctic. But the other 9 months of the year? Antarctic is more important, by an even larger ratio.
      So, yes, going through the tedious details is important. Particularly since no others have ever done this comparison.
      When walking across melting icepack (or running in the dark near cliffs), it is always necessary to mimic Theodore Roosevelt in the jungle:
      “Walk softly, but probe first with a big stick.”

    • From experience, I know that a good number of nails come out of the pouch upside down or bent and then quite a few are downstairs nails when it’s the roof you’re framing! There’s no problem that can’t be fixed by more hammer blows or a bigger hammer.

    • I don’t know that I agree with this entirely. Maybe the 101 material ought to be simplified and airy, but it ought to be approximately correct too. Students may only recall a few percent of what the prof presents, but wouldn’t it be better that this few percent provides a solid foundation, rather than something that has to be unlearned at a later time?

  19. At low latitudes sea ice cover cools the system via albedo in stronger sunlight – heat loss from open water to warmer air less relevant.
    Above from Phiogiston which is key and I agree with. Let me add Antarctic Sea Ice does just that because it is at much lower latitudes then Arctic Sea Ice. Therefore should be a bigger climatic player.

  20. This mid-term exam from ND is very strange. What pedagogical value is Problem 2, for example? If 50F air were pushed up slope from the Mississippi river valley in Southern Illinois all the way to Boulder, without interaction with the ground and without precipitation then it would arrive at Boulder very nearly 268K, and Boulder would have an annual temperature around 23F. OK, no quarrel with that. The prof. asked students to compare the computed temperature to the recorded one, but never gives the recorded one. Is the point just to demonstrate that students can do the calculation? Doesn’t the context suggest strange stuff that rarely, if ever happens?
    Well, one thing I learned is this. When I was in college course numbering was three digits: Physics 121 for example. Now we have expanded generally to four digits to make room for new offerings. Physics 121 has become physics 1210. Notre Dame must be planning a huge expansion of course offerings. Physics 20054? Really.

  21. RACookPE1978 February 9, 2015 at 4:58 pm

    Despite a lot of nit picking and noise in this thread, your point here is well made. One can use the most advanced models and computers available and the end result means very little if one critical item in the calculation path is quite poorly known.
    A couple of weeks ago someone posted a link to work by Duncan Steel who had expounded at length on insolation and albedo–which was dismissed unfortunately by people who hadn’t read the work at ll or hadn’t read it carefully. Steel mused that albedo may not be determinable to any great precision. Often people use a value like 0.3 for earth average albedo to make a point (much like the ND Prof in this example did) but the actual earth average value at any time may fluctuate from this value, and the fluctuations could have time scales of a day to hundreds of years or longer. This fluctuation in absorption then confounds the use of forward modeling (climate models) to make predictions regarding CO2 forcing.

  22. Thank you for a terrifying, if not unexpected, look at what passes for science and math coupled with “climate science” in one of today’s university courses. Just the instructor’s failure to understand the basics of the formulas he’s presented and their real relationships with the world they describe speaks volumes to the uphill fight any student would have trying to learn from this course. One could only guess at the poor grade a good student, properly applying the math and doing accurate work would receive. I love posts that bring the real world back into the argument. Thanks.

  23. nice piece but will need a “re read” here not everything did got well soaked in hahahahahaha
    just a hint for formatting: Not everrybody is Native English speaking, so maybe a short simplified summary through analogy may be a help. However it is just a detail. Of course i understand from this article that it is even this expanded only “scratching the surface”.
    however with the comments maybe one nitpicking point on the article itself: the term “dark”.
    I understood the term dark most likely as you wanted it to be understood: no visible sun nor a partial visible sun. Of course that doesn’t mean there is a radiation, but the diffuse radiation of twilight is “to low to be an importance or make a difference in the points you make” Perhaps maybe a short note about that as introduction would be at it’s place. of course in the way you made the article it “should” be considdered.
    but to me i was able to get the points. it is true that academics and real science are two things that don’t go hand in hand….Science in the real way is to dare to do trial but also go in error… and daring to be honest that you tried made a hypothesis and were wrong. that’s something that science doesn’t do anymore. As academics are even worse then scientists…. the point is very sadly true: often wrong things are taught and the critical investigating mind is often sent home with a ban from colleges…

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