Why Albedo is the Wrong Measure of Reflectivity for Modeling Climate

Guest essay by Clyde Spencer

The total reflectivity of Earth is important in energy budget calculations and in Global Circulation Models used to predict climate. If the value used is too low, it will contribute to models predicting more warming than is correct. I’ll demonstrate below why measured albedo is in fact too low as an estimate of total reflectivity.

One frequently sees reference to the nominal 30% albedo of Earth with respect to the energy budget and alleged anthropogenic global warming. Although, the CRC Handbook of Chemistry and Physics lists a value of 36.7% in recent editions. Albedo is a measure of the apparent reflectivity, of visible light, of celestial bodies such as the moon, Mars, and asteroids. This works well to estimate the type of surface cover and particle size for objects with only bare rocks and regolith. To properly measure albedo over an observed hemisphere, the sun, measured body, and the observer (Earthlings!) should be in approximate alignment. Earth’s albedo is estimated by measuring the albedo of the moon (12%) during a Full Moon, and then measuring the Earthshine near the time of a New Moon. This measurement mostly captures diffuse-reflectance of vegetation, soil, regolith, and especially clouds. However, it includes some minor specular reflections from smooth objects (such as rocks, but especially water waves) whose surfaces are oriented such that incoming rays reflect towards the moon.

However, the illuminated portion of the Earth will always be slightly different, varying with time and location of the measurements, and this probably contributes to the observed variation in measured albedo over time. Because the moon is not in the plane of the ecliptic, any measurements of Earthshine over time are going to be from different viewing angles, and of different regions of Earth’s surface, along with changes resulting from weather and seasons. That is, the albedo, as estimated from Earthshine measurements, can be expected to vary considerably for geometric reasons. These measurements provide an apparent reflectivity integrated over an entire hemisphere of the Earth.

Now, I say “apparent” reflectivity because not all light arriving at Earth is reflected back towards the sun. Albedo is, at best, a lower-bound on the amount of light reflected from Earth. It would only be appropriate to use if Earth were completely covered by clouds, like Venus (albedo = 65%), or completely covered by regolith, like Mars (albedo = 15%), with no water, clouds, or vegetation.

Additionally, albedo is an instantaneous measurement. To the extent that vegetation and clouds change with the seasons, a further, more complicated correction has to be made for those effects to derive an annual average.

For purposes of illustration (or perhaps I should say illumination) consider what the situation would be if Earth were a Hollywood Waterworld (ala Kevin Kostner). Assume that there is only a sediment-free ocean, no clouds, and gentle winds, minimizing whitecaps. Water strongly reflects specularly (as any fisherman or sailer knows only too well), in contrast to the strong diffuse-reflectance of most other things on Earth.

Specular-reflectance is quantitatively calculated with Fresnel’s Reflectance Equation ,using the index of refraction and the extinction coefficient (sometimes called absorption coefficient), which constitute the Complex Refractive Index (CRI). The index of refraction that most readers are probably familiar with is defined simply as the ratio of the speed of light in a vacuum to the speed of light in the material being measured. The extinction coefficient for water varies with wavelength – being larger for red and infrared than the other end of the spectrum – but we can consider it to be negligible for the following illustration. Thus, the CRI for water has an imaginary component that can be considered to be zero. This somewhat simplifies calculations. The CRI varies with temperature, salinity, and the wavelength of the incident light. I’ve used an approximation (n = 1.34) for typical seawater at 550nm wavelength, which is the peak emission of sun light.

Remember that the angle of reflection equals the angle of incidence, where incidence is measured from the surface normal. Assume that we are looking at our Waterworld from a geostationary orbit and that this Waterworld is a perfect sphere instead of the oblate spheroid (Geoid) that the real Earth is. Light is coming from the sun in rays that are parallel at this distance. A ray hitting normal to the surface of the water has an angle of incidence of 0°. All other rays intercepting the surface of the illuminated hemisphere have larger angles of incidence, out to a maximum of 90° at the limbs, or terminator. Ninety degrees is grazing the surface; there is 100% reflectance, and the character of the light is spectroscopically the same as the source (the sun).

The graph below shows what the reflectance is for all angles of incidence for seawater. Note that above about 40° the reflectance starts to rise noticeably, by 60° the reflectance is rising sharply, and above 79° it is above the nominal albedo (30%) commonly cited in Global Warming energy budget discussions.

albedo-seawater

Figure 1. Total Reflectivity of seawater

From the position of the geostationary satellite, looking down, one would see a spot of light reflecting back from the bundle of rays approximately normal to the surface of the water. That would be about 2% reflectance. Everything else would look black, save for some random white caps reflecting some flashes of light. However, that doesn’t mean that all the light is being absorbed! The orbital position would just be wrong for observing the light reflected out into space, away from the sun. The rays hitting at greater distances from the normal ray will experience a larger angle of incidence because of the curvature of the Waterworld. Thus, they will experience a greater percentage of reflectance. The importance of this is that as the reflectance increases with increasing angle of incidence, the absorption, and thus warming, decreases proportionately.

Fresnel’s equation is not integrated easily in its native form; at least I’m not up to it! A 6th-degree polynomial gives a reasonably good fit, which would allow me to somewhat mechanistically integrate it and obtain the area under the curve. However, some readers have expressed objections to high-order polynomials for fitting curves. I think it has something to do with a phobia about wiggling elephant trunks. Therefore, I took the easy way out and just weighed the total area of the graph on paper, and then cut out the area under the curve and weighed it. The ratio of the two weights gives the proportion of the two areas.

The area under the reflectance curve for seawater (Fig. 1, above) is about 9%. Because the sun moves with a constant angular velocity, the value is the time-averaged reflectance of a point on the surface of Waterworld for a 6-hour period. That means that for any particular spot on the surface of the ocean, starting at the terminator at sunrise, on the Equator, it would have an initial instantaneous reflectance of 100%. As the planet rotates under the sun, the percentage of light reflected from the point will decrease until it reaches the minimum of ≈2% at local solar noon. The process will then be reversed until the point again reaches maximum reflectance at the position of the sunset terminator. The total light reflected from that point, over a nominal 12-hour sunlit period, will be about 18% of the incident light. This is more than twice the value listed for the diffuse-reflectance of water and is in the range of values given for vegetation.

Furthermore, it should be noted that a point on the terminator at sunrise, at 60° latitude (N&S), would experience a decline in reflectivity until local noon, just as at the equator, but it will never get below six percent! The minimum reflectivity increases with increasing latitude, leading to larger average reflectances.

However, that doesn’t adequately describe the situation because the surface areas with the greatest reflectivity have larger areas than the lowest reflectivity areas. Because the Earth is rotating, it is important where the water is with respect to land and clouds. While an average reflectivity value may be useful for an initial first-order approximation, the variation with time/location has to be taken into account for accurate modeling.

I also did a discreet summation of the frustums of a hemisphere (Af = 2πR DX). Multiplying the normalized (to a unit area for the hemisphere) frustum areas by the average reflectivity for the angle of incidence, for each of the frustums, gives the area-weighted reflectivity for each frustum. Summing them gives an area-weighted average reflectivity of about 18%. This is the instantaneous area-averaged reflectance over a hemisphere. This is almost an order of magnitude larger than the sunlight reflected from a small spot on the surface of the ocean directly below the local noon sun during an equinox. It is far greater than the apparent albedo (≪2%) of our hypothetical Waterworld!

Even on the real Earth, these specular-reflection effects are significant because about 71% of the surface is covered by water, which NASA claims has a reflectance of about six percent. That means the actual total reflectivity of Earth must be higher than the estimate of diffuse reflectivity obtained from Earthshine albedo measurements (See “Terrestrial Albedo”).

Now, things get more complex as we add land and clouds because they behave differently than water. Bare land and vegetation generate a combination of specular and diffuse-reflection that is best described with a Bidirectional Reflectance Distribution Function (BRDF). It is a mathematical description of how light is distributed with the two effects of diffuse and specular-reflection, with varying angles of incidence. There is always a strong forward lobe of reflectance (anisotropy) for oblique illumination, even for snow. That is because snowflakes (and plant leaves) are smooth, planar, and tend to be sub-parallel to the surface of the Earth. Thus, measurements of albedo from directly above will result in values that are lower than the true total reflectance that would be obtained by integrating the BRDF over an entire hemisphere.

There has been work done with modeling CERES satellite measurements; however, judging from the following illustrations, they don’t have it right. The right-hand illustration of Fig. 2 shows a hemisphere with a large amount of land. The oceans are shown as darker than vegetated land (8% –18% albedo). Indeed, a value of 6% for open ocean seems to be totally inappropriate. Therefore, this illustration appears to be primarily the diffuse-reflectance water-albedo, and not total reflectivity. Because the value is so low, it apparently does not take into account how the specular-reflectance changes with time and position. That is, it is an average of the small incidence-angle (nadir) specular-reflectance (≈2%) of direct sunlight, as obtained from satellites; whitecaps; bottom reflectance from shallow water; diffuse-reflectance of suspended particles near the water surface; and scattered skylight coming from all angles.

olr-terra-ceres2

Figure 2. Terra/CERES views the world in outgoing longwave radiation (left) and reflected solar radiation (right). Image Credit: NASA

Vegetation behaves differently from inorganic reflectors. First, leaves tend to be smooth, with a waxy coating that favors specular reflection. Next, while plants look green because red and blue light are absorbed (It is also highly reflective in the near-infrared.), not all incident absorbed light contributes to warming. The plant chlorophyll converts incident light to carbohydrates and it does not result in warming. The estimates of efficiency with which this conversion takes place vary with the plant and the growing conditions. However, it is generally thought to be in the low single-digit percentage range. Thus, the effective reflectance of visible light by vegetation should be adjusted upward in warming calculations to account for the lack of warming. Phytoplankton and algae in the oceans similarly capture sunlight and convert it to biomass instead of warming the water to the extent that a first-order estimate from reflectivity alone would suggest.

Common dry sand can have a diffuse-reflectance as high as 45%! Regolith, sand, and soils typically have a dominantly diffuse-reflectance. Soil exposed by agriculture can vary from very bright yellow or whitish, calcium-rich desert soils, to dark organic-rich soils; when the soils are wet they get darker and have a stronger specular component. And, of course, little if any soil is exposed during the growing season. Any ‘average’ value assigned to bare soils will probably be wrong more often than right. Estimates for soil reflectivity should be derived from soil maps to better obtain area-weighted values for different regions of the world, and take into account the growing season.

The other extreme is diffuse-reflectance from clouds, which approaches true Lambertian Reflectance. Lambertian Reflectance is a condition where light is reflected equally in all directions and the clouds appear uniformly white from all directions, except under the clouds and in shadow areas between them. Clouds can vary widely in their albedo, but a commonly accepted average value is around 50%.

What clouds contribute to the energy balance is the difference between their reflectance and the reflectance of the surficial materials they are covering. Clouds also re-direct light, making calculations more challenging. However, clouds have their greatest impact on reflectivity and albedo when they are within about 50 degrees of the surface-normal pointing towards the sun.

The Polar Regions are notoriously cloudy. That is one reason the Vikings invented the use of the sunstone to help them navigate where a compass and stars were unusable. Much of the year, it doesn’t really matter whether there is open water or ice because clouds interfere with sunlight reaching the surface during the Arctic Summer, and there is no sunlight during the Arctic Winter! When sunlight does reach the surface, the 100% reflectivity at the Earth’s limbs helps explain, in part, why the poles are so cold. (This also occurs around the perimeter of the oceans at the planet’s terminators, not just at the poles.) Snow on top of ice actually scatters light in all directions, including downward. Thus, there is actually more absorption at a glancing angle with the presence of snow than there would be with calm, open water! There are tradeoffs that complicate the situation and I don’t think they are being taken into account by most climate modelers. The simplistic explanation of decreasing Arctic ice being responsible for reinforcing warming is probably a stretch by those unfamiliar with Fresnel’s equations.

Clouds are the best evidence that the naive “Science is settled” claim is false. Clouds are highly variable in their position, extent, and albedo, often moving rapidly under the influence of winds. There is reason to believe that a decrease in cloudiness has had more impact on the retreat of glaciers than has the supposed average increase in air temperature (≪1°C/century at mid-latitudes). Melting of ice is not directly proportional to the increase in temperature if the temperature never gets above the melting point of ice. At best, one would expect that glacier retreat would follow the lapse rate if air temperature were the controlling factor. What one finds is that the retreat of glaciers is more pronounced on mountain slopes with south-facing aspects. That suggests that there has been a decline in clouds that formerly protected the snow and ice from direct sunlight.

More frustrating yet, the phase-change energy exchanges that take place in clouds cannot be modeled at a scale necessary to capture details, for inclusion into Global Circulation Models. There simply aren’t any computers powerful enough to handle the calculations, and simplifying assumptions must be made. There is an old aphorism that “The Devil is in the details.” That applies quite aptly to the problem of energy exchange in clouds. Because clouds change so rapidly, and have an important impact on both albedo and total reflectivity, it is probably inappropriate to talk about the Earth’s albedo as though it were some constant. Clouds are changing all the time!

In summary, albedo is commonly defined as the “whiteness of a surface.” Water is commonly characterized as having low reflectivity because albedo is used. Water can be illuminated obliquely and have a reflectivity higher than snow, but it will appear black to an observer not in the plane of reflection, and observing at the angle of reflection. The albedo commonly used by climatologists is on the low-end of a range of measured values. Even the high-value albedos, are too low to be used as an estimate of total terrestrial reflectivity because albedo excludes almost all specular reflections. Albedo is an under-estimate of the reflectivity of Earth surficial materials; it is only appropriate for clouds, and to a lesser extent, for bare sands. At the very least, specular-reflection has to be taken into account, as well as diffuse-reflectance, for the oceans. The most accurate representation would be through measured BRDF for all angles of solar incidence for the major surficial materials found on Earth.

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193 thoughts on “Why Albedo is the Wrong Measure of Reflectivity for Modeling Climate

  1. This article makes a good bookend relative Monckton’s error in flux density articles.
    Incident radiation, reflectance, absorption, emission, & conversion. It is the sun interacting with earth.

  2. True that albedo changes seasonally. But polar orbit satellites can account for this by integrating measurements over years. And they can measure total reflected light in their ever changing field of view by simply measuring upwelling shortwave radiation (the Sun’s spectrum) at TOA. And integrate that into a ‘whole earth’ over time. And they can compute the albedo fraction by dividing by total incoming in the solar spectrum. So although the post gives many reasons albedo may be underestimated, in practice it is a well observed variable in the satellite era.

    • Ristvan. I do agree with your comments. We can analyze albedo in many ways but it is not needed, because it can be measured pretty accurately by satellites.

    • Rud,
      I think that you missed the main point. Classic determinations of albedo miss specular reflections, which alone are greater than what is typically attributed to water. For satellites to measure that specular reflection, they have to have off-nadir views of over 50 degrees AND have to be looking TOWARDS THE SUN or they won’t see it! It is more likely that an equatorial orbit will have the correct viewing direction than a polar orbit would.

      • At a minimum, you would need both polar and equitorial orbiting satellites. They would also need special sensors to not only look straight down as most satellites do, but all angles back towards the edge of the earth.
        There are no current satellites that do this. The best the current satellites could do is get a slightly better guestimate.

      • Clyde, I carefully read your arguments. I understand the physics of Fresnel lenses. I understand specular reflection, and angles of incidence. I still think you are mostly wrong on undercalculated albedo, and the satellite observations mostly right. For example, low angle specular reflection is largely’captured’ in the constantly changing satellite viewing window aperture. Come on, its just a lens with a backing photoreceptor CCD. Like cellphones, except fancier and more expensive. So it is not mostly missing from observational data.
        IMO, arguing albedo is understated so AGW is overstated is another one of those fringe skeptic arguments that should be banished because just feeds ammo to warmunists claiming we are all science illiterate nutters. Same as my peeves about feedbacks in a very recent guest post–which brought out the usual discreditable skeptics in droves. Which was the point of the post.
        I don’t care about your feelings, beliefs, or pet theories. I care only about truth, and that we win the warmunist war against untruth. Now, if truth says CAGW is real, I will switch sides. If you can show observationally that albedo is mismeasured by present satellite means, likewise I will switch sides.
        As JMK said, “When presented with new facts I change my mind. What do you do?”
        That is highly unlikely. Climate wars are like Syria. A mess. I am personally trying to tidy up the mess before fighting the base war.

      • For example, low angle specular reflection is largely’captured’ in the constantly changing satellite viewing window aperture. Come on, its just a lens

        Really? Just a lens? Do you mean all those different camera and telescope lens are all the same and just a big marketing scam?
        ;)

      • For example, low angle specular reflection is largely captured’ in the constantly changing satellite viewing window aperture. Come on, its just a lens with a backing photoreceptor CCD.

        I don’ t think this is correct. Many satellites like the ones of the ERBE project use wide field of view instruments but they also have a shutter to protect the optics as the view approaches the limb at dawn and dusk. So this low angle signal is probably being systematically excluded.

        Also any values which are more than about 2 S.D. off the average tend to get removed by automated QA controls. So even if it got picked before the shutter came down it would likely get dropped from the processed data.

        This may be one cause of the 5 W/m^2 net imbalance they were have on CERES.

      • Greg,
        Thank you for the information. This is the kind of insightful information that I was hoping to elicit from those more familiar with the various missions than I am.

    • ristvan,
      Keep in mind that the CERES satellites (Terra, Aqua, Aura) are in solar-synchronous orbits. Which are designed to see any given point below it’s maximum latitude at the same time of day. Therefore, while they have nearly full coverage over the earth, they remove the variable of time of day from the monitoring.
      Terra passes south of the equator at 10:30 am local time.
      Aqua passes north of the equator about 1:35 pm local time with Aura following 7 minutes behind it.
      http://www.zarya.info/Diaries/Launches/GeoSS/ss-Age.php

      • CERES also flew on TRMM for many years, which covered all local times as the orbit precessed. The CERES people have done great work over many years (starting with ERBE) estimating the Earth’s radiative energy budget as accurately as possible. I would not lightly dismiss their work.

      • …from what I can tell, the CERES retrievals take into account specular reflection. Earthshine from the moon is a very error-prone way to measure the Earth’s reflectivity. People who have tried it got quite different answers as CERES gave on different satellites in different sun-synchronous orbits, which agree pretty well.

      • Roy,
        As I implied in my article, I would expect the Earthshine to vary considerably both because of the moon not being in the plane of the ecliptic, and because seasonal and weather changes will actually change the diffuse reflectance.

        If the CERES researchers actually took specular reflection into account, then there is a disconnect between what they are showing as their results and what I’m presenting.

        Actually, I suspect that most satellite sensors would be’ blinded’ if they looked directly into the sun or the reflections of the sun when near 100%. After all, they were designed to look at a ‘15% grey card’, not the sun.

      • Roy,
        I’m expecting, and looking forward to, criticism. But, I think that what it all comes down to is how the CERES people explain the oceans, as shown in Fig. 2, having a reflectance less than the land when my calculations argue that it should be at least approximately equal to land, if not higher. I’m not suggesting that their work should be dismissed lightly. But, as Ricky was so fond of saying, “Lucy, I think you have some splain’ to do.”

      • Roy Spencer,

        The CERES people have done great work over many years (starting with ERBE) estimating the Earth’s radiative energy budget as accurately as possible. I would not lightly dismiss their work.

        Yes, they are serious about what they do but as you know satellite retrieval is a complex game and all sorts of unanticipated effects can jump out.

        You are aware of the aliasing problem in ERBE that resulted in them dropping the monthly averaged time series after Trenberth pointed out the alias problem.

        Here is what the daily ERBE data looks like :

        Most of the massive alias of the the diunal cycle was due to some pretty naive assumptions about tropical weather being the same all day ( well we all make mistakes ).

        Now taking 36 or 72d orbital averages removes a lot of the alias problems but there is more alias than signal there and something more complex than just the two aliased copies of the 12h lit portions. There is a 3-2-3-2 pattern in the alias itself and I have not worked out what that is about ( and neither have they ).

        So respect to the ERBE/CERES guys, they have been MOST helpful to me in getting the best out of the data, but there are remaining issues.

        I think there may be something in this specular reflections at low incidence but I’d be hard pressed to see it being a large as Clyde is suggesting.

      • While I know it’s just an image, it does look to be fully light traced, the PS3 Earth screensaver, it’s obvious there are huge reflections off into space.

      • Greg,
        Just for the record, by convention, angles of incidence and reflection are measured from the surface normal. That is, we are talking about high reflectance being associated with HIGH angles, not low angles.

      • Stephen and Clyde: Doesn’t this sun synchronous orbit mean that the tropics are always being sampled with the sun high overhead (low angle to normal) and high latitudes are always being sampled with the sun near the horizon? If so, satellites collect data from a fraction of “global” perspectives.

      • Frank,
        I suppose the crux of the problem is whether the fleet of satellites that are in orbit are the optimal data collection system to properly capture all the outgoing radiation. To measure the reflectance at high angles of incidence (near grazing) requires the satellite to be positioned on the dark side of the planet and looking towards the terminator. That is a very demanding dynamic range for intensity. When there is an eclipse, people are warned not to look directly at the sun. Yet, that is what we would have to demand of a satellite if we are to capture the high reflectance values.

      • @Frank,
        My main point was that the CERES equipped satellites (TERRA, AQUA, AURA) are in relatively low sun-Sync orbits. This means that they only see any one spot on the earth in the tropics only once per day and that spot is only at the same time every day. TERRA sees the equator at 10:30 am local time on every pass. AQUA at 1:35 pm and AURA 8 min later. That means, NO CERES instrument records the tropics at 9:00 am, 12:00 pm 3:00 pm, 4:00 pm, 5:00 pm. etc. Given how quickly afternoon clouds build up, especially above sun-lit warm water, I conclude the CERES data is sampling way below the Nyquist diurnal frequency of changes to albedo.

        (From: Evidence that Clouds Actively Regulate the Temperature, WUWT Oct 6, 2013
        Sun-Synchronous orbit parameters (daylight side. same goes for night passes)
        http://en.wikipedia.org/wiki/Terra_(satellite)
        Perigee (h) = 705. km
        Apogee = 725. km
        Inclination = 98.1991 °
        Period = t = 98.8 min
        Earth Radius = r = 6350. km
        Distance to Horizon = r/(r+h)*Sqrt(2rh + h^2)
        Distance to Horizon at Perigee = 2767. km
        Distance around Equator = c = 39898. km
        speed of rotation of Earth at equator = Ve = 27.7 km/min
        Distance between passes = Ve*t = 2737 km
        Width of 1×1 grid cell at equator = 111 km
        Grid Cells between passes = 24.7 cells
        Grid Cells between 45 deg oblique on each pass = 12.7

        So every day, only 1/2 of the grid cells near the equator is within the 45 degree oblique view of TERRA (CERES) satellite. These fortunate cells are sampled only once that day. You have to be up near 60N and 60S before you get full coverage of the cells within 45 deg of vertical.

        This was confirmed in the last comment in the thread with a quote from NCAR:

        To minimize temporal sampling errors, the CERES team uses geostationary satellite imagers calibrated against MODIS and CERES to capture changes in clouds and radiation between CERES observation times. (NCAR, Guidance tab

        That gives the “CERES dataset” the hourly coverage in grid cells outside the 10:30am and 1:30 pm CERES passes. But calling it CERES data isn’t exactly “truth in lending” since 85% of it is GOES imager data “calibrated” to look like CERES.

        The problem is compounded in that the calibration is derived from 10:30 am and 1:30 pm data with probably substantial and critical help from the TRMM. Personally, I’d love to see the raw calibration data and look at its variance.

      • @Clyde,
        To measure the reflectance at high angles of incidence (near grazing) requires the satellite to be positioned on the dark side of the planet and looking towards the terminator. That is a very demanding dynamic range for intensity. When there is an eclipse, people are warned not to look directly at the sun. Yet, that is what we would have to demand of a satellite if we are to capture the high reflectance values.

        It took a while to fully sink in, but you make an important point. Just how close to the sun (in angular distance) does any albedo measuring device record data? From that answer, we can figure out how much of the earth’s surface have un-measured albedo and make estimates of the potential error between the assumed level and the range of possible values.

        When someone views the video of cloud movement from a GOES satellite, you see almost a continuous sequence. There is always a jump, a period of no pictures, every day. It must be the satellite shutting down near local midnight when the sun is too close to the field of view.

      • As the author here apparently has not waded through all of the literature documenting the measurements of Earth albedo, I could not go through all of the discussion on the CERES measurement of albedo. But I feel some clarifying comments are necessary. First, I agree, moon shine is a crude way to measure the Earth’s albedo and it is baffling why anyone would consider it in this day and time. Roy Spencer’s comments should be taken to heart. The Earth radiation budget community has been well aware of the issues involved in measuring albedo for well over 40 years and improvements have been ongoing throughout that time.

        To date, CERES provides the most accurate measurement of the Earth’s albedo. In addition to the broadband measurements by CERES on the Terra and Aqua satellites, we use 5 geostationary satellites to fill in the missing times. A considerable amount of effort has been expended to first normalize all of the GEO imager channels to their well-calibrated MODIS counterparts. The GEO data have been analyzed, as the MODIS also were, to retrieve cloud properties to determine the content of the scene corresponding to each pixel. For the CERES broadband measurements, that scene information comprises vegetation type, snow coverage, water surface amount, and cloud phase, altitude, optical depth, and particle size. Empirical models of reflectance anisotropy (BRDFs) have been derived from multi angle measurements as functions of all of those parameters in order compute the albedo from the given measurement. As the measurements are limited to 1030 and 1330 LT +/-1h or so, the albedo at each missing hour is estimated from the narrowband GEO imager measurements after those narrowband measurements have been converted to broadband albedos using empirical functions based on matched GEO imager data and CERES broadband fluxes measured from Terra and Aqua, and TRMM to cover the relationships for other times of day. To further improve those measurements, the GEO fluxes are normalized morning and afternoon to the CERES broadband measurements to ensure consistency. The integration of the fluxes over the daylight hours provides the measurement of albedo, accounting for all angle dependencies and scenes throughout each day. That is the state of the art and the errors are small.

      • P. Minnis,

        You said, “The integration of the fluxes over the daylight hours provides the measurement of albedo, accounting for ALL angle dependencies and scenes throughout each day. That is the state of the art and the errors are small.”

        There is an old joke: “We’re from the government and we’re here to help you.” You have added a new twist — “We’re from the government; trust us. We have everything under control.”

        What I have read consistently, from those who are apparently intimately involved in the research, is that numerous satellite measurements have been used to create BRDF models to estimate total reflectance, primarily the diffuse reflectance of snow, vegetation, granular surficial materials, and clouds. However, apparently those observations have been binned and the bin for the highest zenith angles run from 75 to 90 degrees; water varies from 21% to 100% reflectance in that range of angles. I have concerns that the sensors cannot be used near the grazing angle out of concern of either permanently blinding them or at least going beyond the range of calibration and/or saturating.

        In summary, I have received criticisms that my concerns are “fringe,” “peanuts,” and “We are doing everything correctly and the errors are small.” However, I remain unconvinced that the situation is as well characterized as you would have me believe. NO ONE except Donald K has attempted to quantitatively demonstrate that my concerns or calculations are invalid.

    • True, the satellite era has opened the chance for measuring total reflectance. Most ‘polar’ orbits are canted many degrees, we may still have holes in coverage at the poles. Does anyone know if there is a ongoing collection of incoming vs outgoing SWR/LWR (with enough time/location data to create a global reflectance map) or better, a global reflectance map product available?

      • commieBob,
        But they have to be the RIGHT measurements! I’m suggesting that so far a significant contribution to total reflectivity has been overlooked. And, I’m waiting for someone to put a stake in the heart of the vampire that I claim has been sucking energy from Earth.

      • Clyde Spencer says: September 12, 2016 at 4:12 pm

        … I’m suggesting that so far a significant contribution to total reflectivity has been overlooked …

        Satellites use scanning radiometers to detect visible light and IR. As the satellite travels along its path, the radiometers scan from side to side across the path. The scan is close to horizon to horizon.

        The result is that the satellites capture almost all the light leaving the planet at all angles. Eventually data is gathered for all angles of illumination and for all angles of view for all locations on the planet’s surface.

        Here’s a good paper that describes how the data is collected and albedo is calculated.

        Here’s a paper that has definitions for some of the terms used in the other paper above.

      • commieBob,
        You said, ” As the satellite travels along its path, the radiometers scan from side to side across the path. The scan is close to horizon to horizon. The result is that the satellites capture almost all the light leaving the planet at all angles. Eventually data is gathered for all angles of illumination and for all angles of view for all locations on the planet’s surface.”

        No! Any light directed away from the sensor will not be recorded. And, the maximum reflectivity will not be recorded unless the whiskbroom path is aligned with the sun and ground target. That can happen near the equiniox for a polar-orbiting satellites, but cannot be achieved unless the telescope is steerable and the path is kept pointed towards the sun. That is, there has to be a great circle connecting the whiskbroom path and the surface normal pointing at the sun.

      • commieBob,
        There is nothing in the article whose link you provided that I haven’t seen before. I have no argument with anything presented. Indeed, it argues for the use of BRDF for diffuse reflectors, which I also promoted. However, nowhere is water mentioned!

        In the link with “definitions,” note that the zenith angle for the experiment never gets larger than 60 degrees, where water has a reflectance of 6%. Things only get interesting with water at larger angles.

      • Clyde Spencer says: September 13, 2016 at 2:27 pm

        No! Any light directed away from the sensor will not be recorded.

        That’s trivially true.

        Eventually, we get samples for every location on the planet for all angles of incidence and all view angles. In other words, all the incoming and outgoing visible light is accounted for.

        … the maximum reflectivity will not be recorded unless the whiskbroom path is aligned with the sun and ground target.

        The maximum reflection isn’t always in the obvious place. link Check out Figure 3 in the link. The maximum albedo for sea water appears to be at 15°.

        I will say this one last time. All that is necessary is to account for all the incoming and outgoing visible light. The satellites do a good job of that over time.

      • commieBob,

        Thank you for the link to to surface reflectance model article. It is a good summary of the physics and mathematics of the situation.

        Note that in section 1.3, which gives what is called “typical values,” the ~2% value for green light is for a near nadir-viewing sensor. The value is lower than predicted from Fresnel’s equation because the sun is at 35 deg and the sensor is at 10 deg. That is, the criteria for measuring specular reflection is not met and it appears that the “whitecap” component and suspended particles component are the major contributors. This is not a defect, it is a result of the purpose being to account for aerosols with typical imaging satellites. It is important to realize that the article states explicitly that the largest viewing angle AATSR encounters is about 55 degrees.

        Note also in section 2.3 that an uncertainty of 50% is given for the whitecap component.

        Section 3 is what is applicable directly to my original comments. What they are properly calling glint is equivalent to what I called the specular reflection. Note especially that Fig. 2 essentially agrees with my Fig. 1, except that it only goes out to about 32 degrees. Things don’t get really interesting until past 60 degrees!

        You asked me to look at Fig. 3 in the link. I did. I’m not sure that you read the caption. The figure relates to how wind speed impacts glint.as measured to the normal to the surface, and is not measuring the reflected beam for specular reflectance. That is, whitecaps will spread out the reflection and re-diredt some of the incident light vertically (and elsewhere). Please note that in my article I presented the caveat that in the theoretical Waterworld, winds were minimal and therefore whitecaps negligible, despite Kevin Kostner having a sail boat.

        You said, “I will say this one last time. All that is necessary is to account for all the incoming and outgoing visible light. The satellites do a good job of that over time.” It is good that you are not going to say this again because I see nothing in this link that supports your contention that all possible viewing angles are sampled for estimation of specular reflection.

      • @commieBob,
        Eventually, we get samples for every location on the planet for all angles of incidence and all view angles.
        Is that really true? As Clyde has pointed out elsewhere, to cover “all angles” we must ask the sensors to look into the sunset and sunrise. I bet we don’t do that. The CERES sats never look in this grazing angle near the tropics because of the time of day of their passes. CERES sats experience the sunrise and sunsets only in the high latitudes of their orbits, but how many degrees from the suns direction do they actually record data?

    • Rud,

      OK, you have given me your opinion, and I do respect it. You are obviously a bright, well-educated individual. However, all you have provided is opinion. You said, “I still think you are mostly wrong …, and the satellite observations mostly right. OK, that is your opinion, but I’d like something more substantive. Like, explain why the NASA figure shows water to be much less reflective than land cover.

      You go on to say, ” low angle specular reflection is largely’captured’ in the constantly changing satellite viewing window aperture…. So it is not mostly missing from observational data.” What you are missing is that high reflectance is a result of very high angles of incidence (NOT low), which direct the reflected light AWAY from the recording sensor. So, it IS missing from the observational data!

      Like you, I care only about truth, not opinions. I have provided numerous links to definitions and accepted literature values, and a logical argument as to why my thesis is correct. Rather than dismiss me because you think that it is “fringe,” I’d appreciate a counter argument that is more robust than “satellites are just like cell phone cameras.”

      • Clyde it would seem to me that a satellite would capture the values of reflected light and specular reflection as the satellite orbited away from the point of incidence. For example, if the Satellite is directly above, ie., between the sun and the earth, it would be capturing the brightness of the 0 degree angle, but as it orbits to the 45 degree angle and on it will capture the brightness of specular reflection, provided the sensor is shielded from any direct light coming from the sun. As it rotates past 90 degrees, “the dark side of earth” reflection drops to zero.

        Given the spherical shape of the earth, it should be possible to just account for different land, water conditions and come to a pretty close rough estimate as to what albedo is at that time. Granted, I’ll give to you that “clouds” are a major obstacle here, … but then clouds are a major obstacle for Climate Science anyway.

      • Dr. D,
        The key to this problem is that the satellite has to be able to direct its view to where it is viewing the surface at the angle of reflection (=incidence), and has to be in the plane of the normal to the surface and the incident ray — and then not be blinded by the bright light from water. That fundamentally excludes nadir-viewing sensors — most imaging satellites. That is, the telescope has to be steerable in all directions so that whatever its orbit, it has to be able to look in the general direction of the sun. It will generally be easier for equatorial-orbit satellites to achieve this geometry, but then only around the time of the Equinox. At the time of the solstice, steerable telescopes in a polar orbit could position themselves such to pickup reflections at approximately 90 minute intervals. The bigger question, which I don’t have detailed information on, is the dynamic brightness range of the sensors. They are typically intended for capturing relatively dark scenes (0-30%). Whether any of them could handle full solar brightness I rather doubt. It is often a concern of operators to be careful NOT to look directly at the sun to avoid damaging the CCD array. Even so, I suspect that there are some geometries that will be difficult to achieve because they will require looking at the limbs or just beyond, while the satellite is on the dark side.

      • Clyde, it might be easier and more reliable to observe albedo from more remote positions in space. Since we know the amount of insolation that planet receives from the sun, one could observe from a more remote platform (the Moon for instance) the planetary brightness should be proportionate to illumination and distance. More remote positions would also help by effectively averaging the visible surface (ocean-land-clouds-atmosphere). Earth-orbiting satellites are probably not the best solution for measuring the albedo.

      • Duster,
        More to the point, albedo is the wrong measure for the intended application. It is missing the specular reflections that bounce off the water AWAY from the sun or your proposed alternative platforms.

      • The satellite measurements show that land is more reflective than water because, well, it is. Albedo is the integration of reflectance (outgoing radiance in a given direction divided by incoming radiance) over the entire upward hemisphere. Radiance is given in W/m2/sr. The specular component covers a small portion of the reflectance “dome”. Its weight in the integration over the dome is small. As noted by one of the replies, the reflectance increases with increasing solar zenith angle (SZA), but the energy available for reflecting is diminishes as the cosine(SZA). The albedo is the combination of the atmosphere and surface reflectance and increases with SZA over land also, but to a lesser degree relatively. However, the land is generally much more reflective than ocean at most wavelengths, especially those in the near infrared. Hence, despite the large specular reflectance, it is peanuts when considering the radiation exiting in all directions and the differences in inherent reflectivity differences among vegetation, minerals, rocks, and water.

      • P. Minnis,

        You said, “The satellite measurements show that land is more reflective than water because, well, it is.” That is a simplistic claim that, when taken in its generalization, is wrong. Land can never reach 100% reflectance. A damp, freshly-plowed Mollisol will be darker than seawater. Forests of evergreens with a high shadow content can be quite dark. In work that I have done, Sonoran desert creosote fields commonly classified as asphalt. I think that your generalization is too broad brush to stand.

        You then said, “The specular component covers a small portion of the reflectance ‘dome’.” In the case of the oceans, if 100% is being reflected from a “reflectance dome,” there isn’t anything left over to worry about!

        You then say, “but the energy available for reflecting is diminishes as the cosine(SZA).” I have no argument with that except it is a non-sequitur. I have never made any statements about the magnitude of outgoing radiance. I have only pointed out that there is energy escaping that doesn’t seem to be explicitly accounted for.

        I’m afraid I can’t agree with your characterization of “peanuts” when the value of reflectance of water can be as high as 100% When it is weighted for surface area, it might only be a small percentage, but the energy imbalance is only a few percent!

        Speaking of “peanuts,” the near-IR component of sunlight is off on the shoulder of the solar spectrum.

    • Ristvan— Take a look at the newest US Navy ships, the ones that are near invisible to Radar. Why do you think they have those angles? Hint, they reflect the radar beam away from the ship with the radar trying to find it. Look at the Stealth aircraft and how they are designed which is the model for these ships. If th signal is reflected into nothingness it is the same as “invisible.” Now apply what you have learned to your incorrect comments.

  3. I started digging into this topic after watching the Earth Sun sim screen saver on my PS3, and noticed the ocean sized reflection of solar back out to space.

  4. And integrate that into a ‘whole earth’ over time.

    But individual measurement uncertainty is pretty large, multiple that by thousands of measurements, an hour apart(or whatever the orbit period is), The orbits don’t overlap squarely at the poles, it’s a mess and I wouldn’t trust anyone saying they can measure it to higher uncertainty than what we can measure the uncertainty for Co2’s forcing.

    It’s all based on peoples opinions and biases written into code, and then extrapolated out to makebelieve land. It’s BS.

  5. Product shows the Outgoing Longwave Radiation (OLR) from the Advanced Very High Resolution Radiometer (AVHRR) and High Resolution Infrared Sounder (HIRS). The AVHRR and HIRS OLR products are divided into day and night (ascending and descending) products.
    Monthly Mean

    Product shows the average solar radiation absorbed (W/m2) in the earth-atmosphere system. It is derived from AVHRR Channels 1 and 2. The mean is displayed on a one degree equal area map on a seasonal basis. This product is also referred to as Shortwave Absorbed Radiation (SWAR). Absorbed solar radiation is the difference between the incoming solar radiation at the top of the atmosphere and the outgoing reflected flux at the top of the atmosphere.
    Monthly Mean

    • ren,
      I think that this is a paradigm change for some of you. We cannot measure the absorption directly from orbit. So, we measure the reflectance (R) and assume that absorption is 1-R. If the AVHRR sensor is obtaining only nadir views, then AGAIN, the specular reflectance is being missed!

      • I agree with you Clyde. Using LW to estimate the absorption of SW at any given point in time ignores that most absorption is below the surface of the oceans, and does not account for a delay in the release of LW due to ocean circulation. All one sees with LW measurements is the immediate release of absorbed SW as long wave at the skin and on land, … or an addition of LW to the immediate release due to upwelling warm water in the oceans.

      • Dr. D,
        You said, “…Using LW to estimate the absorption of SW at any given point in time ignores …” I won’t disagree with what you say. However, the crux of the problem is that it is ASSUMED that the proportion of incident radiation that isn’t reflected back, and measured as albedo, is absorbed. My argument is that some of the light that is assigned to absorption can be reflected by specular reflection outside the view of the camera recording albedo. Even a fisheye lens isn’t going to capture the reflected light if it is not aimed in the direction of the camera! It is often the unexamined assumptions that get one into trouble.

    • Henryp,
      Classically, albedo has been defined as the ‘whiteness’ of an object in visible light. I used green light for my graph as a convenience of selecting a single index of refraction. Green also happens to be the peak output of the sun, and the dominant color of vegetation. Also, most dark-colored rocks are either some shade of red or green due to the ferrous and ferric ions, respectively.

      • A closer reading of the definition indicates, to me at least, that it is diffuse reflectance only. Absolutely NOT specular.

      • For purposes of Earth radiation budget. Albedo is defined as the amount of all outgoing radiation at all wavelengths divided by the total incoming solar irradiance (all wavelengths). When computing the albedo the incoming is integrated separately over the day and the outgoing is integrated over the day. Those quantities are rated to obtain the albedo.

      • P. Minnis,
        I think it is an unfortunate choice of terminology to use albedo, which has a historical use in astronomy, to define total reflectivity. Also, the definition conflicts with another common definition, which is the apparent whiteness of an object. Water can have 100% reflectance and yet appear black if viewed from any but the geometrically correct position to intercept the specular reflection.

    • Alex,
      You get a gold star for today. It would appear that you are the first to really grasp that there is a difference between diffuse and specular reflectance. It seems that many here are having a hard time wrapping their head around the idea that nadir-viewing satellites (and that is most imaging satellites) are not going to capture the specular reflectance that is directed out into space away from either the moon or Gorersat. Some satellites pick up water bright spots, but it is my recollection that the sensor is usually saturated and not useful for any kind of quantitative measurement.

      • I’ve been putting together some information for a possible post on reflectance, albedo and specular reflection. I was going to have a water planet too. So I guess I cheated.You are quite right that most people just don’t get it. My calculation for diffuse sky was 17.7 % specular reflection. There is about 100watts/M^2 diffuse light at midday on the equator at the equinox. Thats a lot of radiation- 17 Watts/ M^2 goes spearing off to space with no further interaction with surface.

      • Alex,
        I did my calculations for the reflectance of the hemisphere of a Waterworld twice. First, I used 90 frustums and got a value of 18.3%; I then did it for 900 frustums (thank god for spreadsheets!) and got 17.6%. I rounded up to 18% for the article. As I remarked to Donald K., any calculation of reflected intensity or flux derived from reflectance has to account for the intensity of light reaching the surface, which decrease with the angle of incidence because of more absorption with a longer slant range, and the increasing size of the footprint with increasing angle of incidence.

        Would you please define “diffuse sky” for me?

      • Diffuse irradiance refers to sunlight that is scattered by the atmosphere. I was trying to separate sky from diffuse light from clouds.

  6. Great post thanks Clyde Spencer. I will bookmark and re-read. Can you comment on the fact that incoming radiation at a low angle of incidence has more atmosphere to travel through and hence more chance of being absorbed by water vapour and cloud?

    I love your first earth image. Just look at the LW pouring out of the mid latitudes. Here is your radiator folks. I suggest that cloud cover analysis in these regions over several years and ENSO cycles would be very revealing

    • Michael,
      Your are right that long slant-ranges will increase the path length of incoming radiation, and ALSO the outgoing. I haven’t seen any studies related to this and I’m not sufficiently acquainted with the techniques used for determining atmospheric absorption for other than nadir views. Off hand, I’d speculate that this insight should cause some atmospheric physicists to re-think their work. Incidentally, long-slant ranges come into play in UV absorption in Antarctica because the sun NEVER gets directly above the so-called “ozone hole” and there are usually anomalously high concentrations of ozone sitting outside the circumpolar vortex.

  7. I have often wondered about how several of Clyde’s examples have been accounted for in this magical Albedo number they cam up with. One only has to look out the window on an airplane trip and you can observe many of these phenomena and even examples of how the differences in angle affect them.
    50 years ago, when I was making algorithms for computer simulations/calculations, I was taught to make physical observations, take the observed data, determine the relationship and fundamental function. Using that function to develop an algorithm to model these changes. Then, using that algorithm, see if it simulates the physical observations/measurements I had observed. But that was back when computers and computer science was in the domaine of the mathematicians. Are they teaching a new way to make computer models today?

    • Are they teaching a new way to make computer models today?

      Yeah, write whatever they’ll pay for, even better you get a bonus if they can’t tell if you got the right answer or not for another 50 years.

    • userbrain,
      I’m old enough to harken back to those days of which you speak. I did some early modeling (in the 1990s) on the problem of surface UV, based on TOMS ozone data.

  8. Distilling Earthshine from Moonshine seems a rather crude measure for estimating terrestrial albedo. I think it would be simpler to just integrate image frames from the DSCOVR satellite (“GoreSat”) and compare those values to the expected values of a perfect reflector. (DSCOVR is locked into the Lagrange L1 point 1.6 million km away between the Earth and Sun.)

    http://epic.gsfc.nasa.gov/

    • Johanus,
      Like Rud, you miss the point. Viewed from that position, Earth will only reveal the diffuse reflectance and little if any of the specular reflection that is going off into space in the direction AWAY from the sun.

      • You really need 24×7 constant 360 full coverage, say at least 6 satellites (in geo orbit) or more, and you’d really want both poles… Um, maybe add it to gps, how many satellites do they use?

      • True, but the terrestrial processes which impede diffuse radiance also tend to impede specular reflections too. So it seems that the total reflection could be accurately estimated from a model based on measured reflections, assumed to be approximately and constantly Lambertian, with cosine angle always close to zero.

        In any case, albedo estimation was one of the principal objectives described in the DSCOVR proposal (“Triana Report”) , including estimating and comparing albedo estimates at various wavelengths (UV to IR -> 0.2 um to 100 um) using the onboard 4-channel NISTAR radiometer, in conjunction the EPIC polychromatic imaging camera:
        http://www-pm.larc.nasa.gov/site/doc-library/references/DSCOVR/NAS.Triana.report.12.99.pdf

      • Johanus,
        I wouldn’t put much stock in anything associated with the name “Al Gore.” His original proposal for Triana was for it to be used in a classroom environment. I did a ‘back of envelope’ calculation and came to the conclusion that during a 50-minute class, even a relative fast-moving event like a storm would only move a couple of pixels. That would be about as exciting for the students as watching grass grow.

        Again, I think that you are missing the point. Yes, the atmosphere should have symmetry and impact outgoing radiation similarly whether it is back-scattered or forward-scattered. But, my calculations strongly suggest that the forward scattered radiation simply has been overlooked. Thus, the albedo being used is short a few bricks of a full load.

      • “… fast-moving event like a storm would only move a couple of pixels. …”

        Storms are meso-scale or synoptic scale events, which are best viewed by geostationary spacecraft.

        DSCOVR was intended for viewing atmospheric phenomena in at a planetary scale. Its multispectral radiometer and imager allow it to recognize and measure ozone, SO2, aerosols, clouds/water vapor and vegetation on a daily scale over the entire surface of the earth facing the Sun.

        Some data is made available to the public, e.g. a dozen or so images per day, but not real time. (The latest imagery now is from 11-Sept. Transmission from space to Earth is bandwidth limited, so the 2048×2048 images recorded by EPIC are averaged down to 1024×1024 for transmission.

        It is also instrumented to measure solar wind (3d magnetic field, electron/proton flux etc) and will be a replacement for the aging ACE satellite.

        I’m no fan of Al Gore, but have to admit that his idea of launching a satellite to make Apollo-17-style “Blue Marble” images has now evolved into a really versatile and useful space research platform.

      • Johanus,
        OK, maybe I was being a little harsh and judgmental about Gore, and the reincarnation of Triana is probably providing us with useful information. But, does it address the point of this discussion?

      • Johanus,
        The original Apollo 17 Blue Marble photograph.
        Beautiful.

        NASA Photo ID: AS17-148-22727 File Name: 10075945.jpg
        Film Type: 70mm Date Taken: 12/07/72
        Title: View of the Earth seen by the Apollo 17 crew traveling toward the moon
        Description:
        View of the Earth as seen by the Apollo 17 crew traveling toward the moon.
        This translunar coast photograph extends from the Meditierranean (sic) Sea area
        to the Antarctica south polar ice cap. This is the first time the Apollo
        trajectory made it possible to photograph the south polar ice cap. Note the
        heavy cloud cover in the southern hamisphere (sic). Almost the entire coastline
        of Africa is clearly visible. The Arabian Peninsula can be seen at the
        northeastern edge of Africa. The large island off the coast of Africa is
        the Malagasy Republic. The Asian mainland is on the horizon toward the
        northeast.

        http://science.ksc.nasa.gov/mirrors/images/images/pao/AS17/10075945.htm

      • Johanus,
        Thank you for the link. I don’t have time to read it tonight. I’ll try to remember to read it when I get back from Denver next Sunday.

      • Johanus,

        My apologies for my tardy response. I just got back from a trip to Denver on professional business.

        Thank you for the link. I just read it. I’m not sure you have. I don’t see any way that the DSCOVR can measure sunlight reflected AWAY from the sun, only light reflected back towards the sun. That is the essence of the problem. The authors suggest that this approach can be used for determining albedo of exoplanets. That may be true if the exoplanets are lacking water, but is isn’t strictly applicable to Earth.

      • Actually the paper seems to be an extended abstract, so lacking details on how albedo was extracted from the DSCOVR data. Some of the references may provide more info on this. Thx.

  9. Great article Clyde,

    Re, (my bold): “Thus, there is actually more absorption at a glancing angle with the presence of snow than there would be with calm, open water!

    The exaggerations around “ice albedo feedback” in various global warming claims have long irritated me, and:

    If wave action is parallel to the line of sight, strong specular reflection is seen with the sun at higher elevations so that in the arctic it will be evident at lower latitudes. If the waves align across the line of sight there will still be specular reflection but away from the line of site.

    Some believers argue that that your observation does not apply because there is no such thing as smooth water in the Arctic.

    • bobgj,
      As you suggested, even WITH waves, the orientation of the waves is important. I have, in the past, read studies that predicted waves and whitecaps based on wind speed and fetch. That is a refinement I didn’t want to get into here because the submission was already 2,800 words. However, from my experience from working in remote sensing, the first-order effect of other than a smooth water surface is to spread out the apparent locus of reflectance. Thus, in the real world, my simple Waterworld breaks down. The question is, “Is water always so disturbed as to invalidate my claim that ignoring specular reflectance leads to an underestimate of total reflectance?” I suspect not!

    • John_C,
      What I provided in my article is basically the current, state-of-the-art reflectance map of a hemisphere averaged over time. The units are in w/m2 rather than reflectivity, but I assume you can make the conversion;

    • bobgi,
      Incidentally, I should have remarked that surface ‘roughness’ becomes less important at very large (grazing) angles. The surface reflectance of an object is most sensitive to changes in the angle of incidence where the slope of the line is changing most rapidly, somewhere around 70 to 80 degrees.

  10. ALbedo is one of the most direct modulator of insolation and it clearly needs a more thorough modelling.

    • Robert or Ottawa,
      I basically told Monckton of Benchley the same thing. He assumed that 0.3 was essentially correct and of sufficient precision to be useful. I don’t think so.

  11. The non-greenhouse effect component of climate change, known as Ecological Changes — land & water use & cover changes with the time, termed as urban-heat-island & rural-cold-island effect — with the space & time do affect the albedo and other radiative processes.

    Dr. S. Jeevananda Reddy

    • Reddy,

      I would agree that land use changes are important, and probably more important than CO2, but I’m not prepared to defend that anymore than what I did in my first essay several months ago.

      A recent comment by Nick Stokes (I believe) suggested that land use changes have increased reflectivity. It was a rather broad-brush claim without citations. That is why I remarked in my essay that the reflectivity of soils varies widely. To make his claim stick, one would need maps of before and after surface conditions, and in the case of agriculture, have information on the rates of growth and the leaf-area index for the various stages of growth. I think that is largely beyond the state-of-the-art for existing land use maps.

      • Clyden Spencer — in internet there temporal variations of different cities like Washington DC. You can try and see wheather there is any change in reflectivity over this region as a first step. I look at practical issues. This is exactly what I did in building evaporation and evapotranspitration models and soil water balance/crop weather models.

        Theoretical excercize lead nowhere. This is exactly what the warmist groups are trying and destroying the science of climate change. We must look at local issues first, though this regional issues. Beyond this, there is no useful science. Even with the little data collected by boat — invasion — search on winds, we have excellent wind patterns.

        Dr. S. Jeevananda Reddy

      • Dr. Reddy,.
        I’m afraid I’ll have to respectfully disagree about the value of theorizing. Einstein was famous for his thought experiments that preceded actual empirical experiments by others. One of the problems we have with science today is people not actually carefully thinking through their experiments before they start gathering data. They then find that they either didn’t collect what they needed, or did so without sufficient control and metadata. I’m getting the feeling that those building the climate models didn’t carefully consider the physics of the problem in its entirety before they started gathering data and writing out formulas. I’m attempting to demonstrate that is the case.

  12. There is also the question of topography (large-scale roughness) In mid-low latitudes the angle of incidence is often near-perpendicular on one side of a range and reflecting into more landscape on the other side of the range. Work that lot out chaps :-)

    • And topography varies in slope and orientation. Mix this with varying seasons and vegetation then I suggest it cannot be modeled. Remote measurement of both incoming and outgoing at ground level over given areas must be the only way to quantify, surely? Now that is hard

    • Michael,
      Topography can and does complicate things. However, the dominant reflectance from rocks, regolith, and soil is diffuse. About the only time you are probably going to have specular reflection from rocks is when they are glacially polished. And, unless they are sub-horizontal, tilting away from the surface normal, the reflections probably won’t go off into space.

  13. Clyde, you have touched on one of the many many shortcomings of climate science (ahem).

    The inability to make accurate measurements of the fundamental optical quantities involved in the climate models will always make these models useless for any practical engineering work to solve the alleged “climate crisis”.

    As someone who has actually made NIST traceable absolute radiometric measurements I can assure everyone that it a royal pain in the rear.

    Just to measure an absolute radiometric quantity in the visible takes about a million dollars worth of laboratory equipment; a calibration standard (a very high priced light bulb measured by NIST), precision current standards, a spectroradiometer, a room with all the walls painted black (we call ours the “bat cave”).

    Oh and also very careful distance measurements, to get to 1% absolute radiometric knowledge the position of everything has to be known to thousandths of an inch. It’s that 1/r squared law after all.

    And even then you can only do about 1% absolute knowledge of radiometric quantities in the visible wavelength spectra. Near-IR, M-IR and Far-IR are even harder. And that is here on the ground in a laboratory. Check out some of the NIST (US Govt bureau of measurement standards) websites about making absolute radiometric measurements,

    Measuring reluctance has the added difficulty of all the angles involved, You have to consider the angular dispersion of the illuminating light (the Sun) and the specular/dispersive nature of the illuminated object,

    Trying the measure the Albedo of the Earth is hopelessly difficult, Anyone that claims knowledge of this number to better than 5 percent is fooling themselves.

    And using it as an input to a computer model “projecting” the temperature in one hundred years is insane.

    Cheers, KevinK

    • Kevin,
      Monckton’s analysis of sensitivity used an albedo with one significant figure and mixed it with other measured values with up to 6 significant figures. It is key to getting good results. I’m suggesting that using diffuse reflectance alone may be an underestimate of total reflectivity of more than 8% [(18-6)*0.71]

  14. Opps, I meant to write “reflectance” instead of “reluctance”, must be the EE part of my training overcoming my Optical designer side.

  15. For several years I drove to work beside the Pacific. The sun shining off the water varied considerably, day by day, hour by hour. Even a small amount of chop spread the bright image of the reflected sun out over a wide area, well before the sun reached the horizon.

    • jorge,
      And my point being is that while waves can spread out the ‘reflection’ of the sun, you have to be at an oblique viewing angle and in line with the sun to generally see it. An orbiting satellite looking down (nadir) will start to see some light as the surface roughens, but any measurements will only be a fraction of the total that is reflected. And, those conditions you observed are not typical of the geometry used for albedo measurements.

  16. ‘ What one finds is that the retreat of glaciers is more pronounced on mountain slopes with south-facing aspects. That suggests that there has been a decline in clouds that formerly protected the snow and ice from direct sunlight.’

    I’ve been pointing out for years that increased air temperatures and hence GHGs can’t produce this effect, which is observed in many places; Cascades volcanoes, Himalayas, Greenland. It can only be caused by increased solar radiation at the surface and/or albedo changes. The latter I believe is more significant than most think, due to embedded dark material in the ice accumulating at the surface during periods (over years) of melt. The same thing has occurred with Arctic sea ice resulting in the disproportionately rapid melt of older sea ice in recent years.

  17. Regarding: “The area under the reflectance curve for seawater (Fig. 1, above) is about 9%. Because the sun moves with a constant angular velocity, the value is the time-averaged reflectance of a point on the surface of Waterworld for a 6-hour period. That means that for any particular spot on the surface of the ocean, starting at the terminator at sunrise, on the Equator, it would have an initial instantaneous reflectance of 100%. As the planet rotates under the sun, the percentage of light reflected from the point will decrease until it reaches the minimum of ≈2% at local solar noon. The process will then be reversed until the point again reaches maximum reflectance at the position of the sunset terminator. The total light reflected from that point, over a nominal 12-hour sunlit period, will be about 18% of the incident light.”:

    How does 9% reflected become 18% reflected in a 12 hour whole daylight period that consists of two 6-hour periods each having 9% of the incident sunlight being reflected?

    Meanwhile, not even this tells the whole story because percentage reflected should be weighted by amount of sunlight hitting each square meter of surface, which is proportional to the cosine of the angle from zenith (neglecting effects of the atmosphere). I estimate this would reduce the reflected percentage from 9% to about 4-4.5%.

    But also not all of the world is at the equator – and I think the overall correction would be to weight that curve by the product of the sine and the cosine of the sun’s angle from zenith – I did not really work out the math, but I think this will give a correct (or nearly so) answer for the percentage of sunlight reflected by a Waterworld, assuming the sea all over Waterworld is flat and mirror-smooth, and the atmosphere does not complicate optical calculations. This would make reflection less than 9% – I estimate 6-7%. But waves (both whitecap ones and smooth ones) and the atmosphere (even without clouds) and reflection from stuff under the water surface would increase this figure.

    • Donald,
      First off, I have to congratulate you. I would consider this the first substantive criticism of my work.

      Let me see if I can re-phrase the scenario. Assume a point on the surface at the terminator at sunrise. Because the sun moves across the sky at a constant angular velocity, we can map time to the angles. The graph will start out at 100%, decline to 2% at local noon, and then start to climb again to 100% at sunset. The area under that graph of reflectance over time, will be 18%. That is, any particular point will reflect 18% of the light that impinges on it during the time it is illuminated by the sun.

      I haven’t worked out all the math either. But, in the next paragraph I demonstrate that near the poles the minimum reflectance at local noon will be much higher, so that the average for the whole hemisphere would appear to have to be higher than what a point on the equator experiences.

      If we were calculating the flux at the surface, then we would have to account for the change in the footprint of the bundle of rays. However, the definition of reflectance is the the ratio of light leaving the surface to the light impinging on the surface. So, less light per unit area reaches the limbs, but the reflectance calculation should be right. You are mixing apples and oranges.

      • You said the area under the curve in the Figure 1 graph is 9% of the area of the graph. Try splicing your graph to a mirror image of itself and determining the percentage of the graph’s area that is under the curve. The figure is 9%, the same as for each half, despite the maximum occurring twice and the minimum occurring once.

      • Donald,
        Congratulations! I think that you have me here. You did what no one else with their “opinions” was able to do. Now, can you find a flaw in my claim that the instantaneous reflectance for the hemisphere of the Waterworld is 18%? Also, do you have any insights on why the average reflectivity of a point on the surface of the water is 9% when the hemispherical reflectance is apparently 18%?

      • Sometime in the next day or two I will do a brute force integration (numerically) of reflectance of sunlight by Waterworld. In my model, Waterworld will be an 8,000 mile wide sunward-facing hemisphere with refractive index of 1.34 in a square sunbeam 8,000 miles by 8,000 miles, divided into 64 million little square sunbeams 1 mile wide. Ones whose centers pass more than 4,000 miles from the center of Waterworld (about 13.74 million of them) will be counted as missing Waterworld. Percentage reflected by the other roughly 50.26 million will be calculated on basis of the angle of incidence of the center of that 1 mile wide beam on Waterworld, and an oversimplification assumption will be made that all of each beam does what the center of that beam does – either hits or misses Waterworld. Percentages reflected of each of them will be added up and divided by the number of beams counted as hitting Waterworld. I am expecting a lot less than 18%, I think about 7%. I will repeat for the 8,000 mile wide square sunbeam being divided into 6.4 billion square sunbeams 1/10 mile wide, to see if this changes anything significantly – I expect not. I will post my results, and I will post a link to my code which will then be on the web.

    • Donald,
      When you report the results of your integration, would you please also provide the total area you obtained for your hemisphere. I figure you should get an area of about 1.01 x 10^8 mi^2.

      • OK, I’ll add code that counts the area of my Waterworld hemisphere that intercepts each of my 1 square mile sunbeam cells, each as 1 square mile divided by the cosine of the angle of incidence, and then tally up all of these. I expect the counted area of intercepted sunbeam cells as close to 6.4 E7 square miles times pi/4 (5.0265 E7 square miles), and the amount of hemisphere area counted as doing the intercepting being close to the actual area of the hemisphere which is 1.0053 E8 square miles.

        My figures will not be exactly these because I will be counting 1 square mile sunbeam cells as entirely missing if their centers miss Waterworld and entirely hitting if their centers hit Waterworld, and each 1 square mile sunbeam cell counted as hitting being assigned an intercepting area of Waterworld of 1 square mile divided by the cosine of the angle of incidence between that sunbeam cell’s center and the surface of a curved hemispheric Waterworld.

        A small percentage of my 1 square mile sunbeam cells will have large errors, but the large errors will affect a smaller percentage of the intercepted sunbeam cells if I make them smaller. I plan to run my model with square sunbeam cell width and height of 1,000 miles, 100 miles, 10 miles, 1 mile, and I said I would do this again for .1 mile. (I am now not confident that a .1 mile cell size run will run in a timely manner.) While I am developing my model, I will use a 1,000 mile cell width/height, because that means each quadrant of my modeled sunward-facing hemisphere of Waterworld will intercept (or not) only 16 sunbeam cells, and I can hand-check the math at that point, and I can verify that the results are appropriately mirror-image-repeated in all 4 quadrants.

        Also, now I see a possibility that my code will have line count small enough that I can post it here instead of putting it on the web.

      • Results:

        For sunbeam cell size of 1,000×1,000 miles:
        Reflected percentage = 5.694%
        Intercepted sunbeam area = 5.200E7 square miles
        Reflecting hemisphere area = 9.134E7 square miles

        For sunbeam cell size of 100×100 miles:
        Reflected percentage = 6.752%
        Intercepted sunbeam area = 5.024E7 square miles
        Reflecting hemisphere area = 9.625E7 square miles

        For sunbeam cell size of 10×10 miles:
        Reflected percentage = 6.750%
        Intercepted sunbeam area = 5.027E7 square miles
        Reflecting hemisphere area = 1.000E8 square miles

        For sunbeam cell size of 1×1 mile:
        Reflected percentage = 6.751%
        Intercepted sunbeam area = 5.027E7 square miles
        Reflecting hemisphere area = 1.005E8 square miles

        Code is at:
        http://donklipstein.com/wtrwrld.bas

        While I was developing the code, I used cell size of 1,000×1,000 miles and I checked incidence angle and reflected percentage of all 52 sunbeam cells that hit Waterworld. The incidence angle of the outermost 16 sunbeam cells hitting Waterworld were either 62.11 or 72.17 degrees, and reflected percentages agreed with the Figure 1 graph at 7.099 or 16.465% respectively. The four sunbeam cells that were closest to the center of Waterworld had incidence angle of 10.18 degrees and reflected percentage of 2.112%.

      • If I remember my Feynman correctly, a single air/glass interface reflects 4%

        Having looked at the sun’s reflection off a lake, it is very bright, but 1/10 of the Sun’s brightness is still going to be bright, just maybe not so much compared to full strength.

      • Donald,

        What you have calculated appears to be the reflectance from a flat disk of the same diameter as the Earth. This is analogous to what is called a Bond albedo. A disk has a diameter of exactly 1/2 of the area of a hemisphere. What I presented was the instantaneous reflectance of the entire hemisphere, weighted by area, with bands of known area multiplied by the calculated reflectance for the center of the bands.

        I’m not going to try to wade through your code to understand the details of what you did, in part because I’ve already pointed out that your description indicates you haven’t taken the right approach. Assuming that you have assigned the correct angle of incidence to each sunbeam cell, you provid no claim about angles being present above 72.17 degrees.

      • I now have results for a sunbeam cell size of .1x.1 mile:
        Reflected percentage = 6.751%
        Intercepted sunbeam area = 5.027E7 square miles
        Reflecting hemisphere area = 1.005E8 square miles

        To the number of significant digits that I had my code reporting, these figures match those for a sunbeam cell size 1×1 mile.

      • Donald,
        Let’s assume, just for the sake of argument, that your value is correct. The value of approximately 6% used by NASA includes both specular reflection (glint) and diffuse reflectance and is approximately equal to your specular calculation. Which one is wrong?

      • Regarding “What I presented was the instantaneous reflectance of the entire hemisphere, weighted by area, with bands of known area multiplied by the calculated reflectance for the center of the bands”. The bands also have to be weighted by the amount of sunlight incident upon each square meter of each band (which is proportional to the cosine of the angle of incidence), as well as by their area.

        If you want, I could divide an 8,000 mile diameter sunbeam into 10 rings with outside radius being 400 miles greater than the inside radius, and calculate the area of each ring and percentage reflected for each ring, and total these up.

        ring_mean_radius area angle_incid reflec_ coeff area*coeff
        miles mi^2 degrees mi^2
        200 5.026E6 2.866 .02111 1.061E5
        600 1.508E7 8.627 .02112 3.185E5
        1000 2.513E7 14.47 .02116 5.317E5
        1400 3.518E7 20.49 .02132 7.500E5
        1800 4.524E7 26.74 .02176 9.844E5
        2200 5.529E7 33.37 .02288 1.2650E6
        2600 6.534E7 40.54 .02562 1.6740E6
        3000 7.540E7 48.59 .03267 2.4633E6
        3400 8.545E7 58.21 .05410 4.6228E6
        3800 9.550E7 71.80 .15924 1.5207E7

        total 5.0265E8 2.7927E7

        2.7927E7 / 5.0265E8 is .05556. This is erroneously low because the average reflectivity for each ring, especially the outermost one, is greater than the reflectivity at the ring’s mean radius. The same thing happened in my integration effort when I used a large cell size. Using a larger number of narrower rings will increase this figure towards .06751.

        I am aware that the area of a band of Waterworld getting each ring of sunlight is the ring’s area divided by the cosine of the angle of incidence. However, the solar flux incidence on that band, in terms of watts per square meter of water surface, is 1360 W/m^2 times the cosine of the angle of incidence. So these two issues cancel each other out.

        As for your mention of me not claiming any angles of incidence above 72.17 degrees in my integration effort: This is true when I choose a cell size of 1,000 by 1,000 miles, not true when I choose smaller cell sizes.

      • Donald,
        Do what you please, but I don’t think that you are taking the right approach. I demonstrated that the lower-bound for a particle on the surface of the ocean should be (caveats in place) >9% at the equator for the sunlit period of the day. As one moves away from the equator, the minimum reflectance value incrreases, leading to a larger average. I also demonstrated that the instantaneous reflectance for the Waterworld should be about 18%. So, how can your value of 6.8% be right?

      • Looks like my column headers did not format well.

        1st column is ring mean radius in miles
        2nd column is ring area in square miles
        3rd column is angle of incidence at the ring’s mean radius in degrees
        4th column is the reflection coefficient, unitless

        5th column is the product of the second column and the 4th column, and this has the unit of square miles

      • Regarding the NASA figure for specular and diffuse reflection combined being about 6%, and my figure of 6.75% being for specular reflection alone:

        I expect the diffuse reflection in the real world to increase my figure a little, not by a lot, since nearly all light penetrating into the ocean is absorbed. I expect waves to increase my figure by a little, not by a lot, since most of the light hitting the ocean hits at angles of incidence where reflection coefficient does not change much if the angle of incidence changes by 10 degrees. I expect whitecaps to increase my figure only slightly, due to whitecaps and seafoam covering a very small percentage of the ocean. Most of the waves far from land are gentle swells and ripples, especially outside the cloudy areas of storms.

        But there is a factor that can reduce the actual figure – sunlight hitting at a low angle of incidence has to get through more atmosphere and has increased probability of running into cumuliform clouds. So for a global average, incident solar flux drops more than in proportion to the cosine of the angle of incidence as angle of incidence increases. It seems to me that the NASA figure is 6% of the sunlight that gets past the atmosphere and clouds. I wonder if that 6% figure is after the loss from the atmosphere scattering and the clouds reflecting some of the sea-reflected sunlight back to the surface for another chance to be absorbed.

      • Donald, have you seen this document?
        http://journals.ametsoc.org/doi/pdf/10.1175/1520-0493%281979%29107%3C0775:TAOWAA%3E2.0.CO%3B2

        Analyzing the Arctic, my rough estimates is that open water cools more than warms, if it’s clear a lot.
        Polar water, the polar region, would have both leading and trailing reflections, as well as a reflections across the northern edge.
        I know above somewhere around 75-80 degrees, the albedo is nearly the same as ice and snow.

      • Donald,
        Remember that the definition of reflectance is the ratio (unitless) of reflected light to incident light. The things that you are mentioning impact the intensity of light reaching the surface, and probably add to the amount scattered in the atmosphere. However, they should have no impact on the reflection coefficient.

      • Clyde: Your 9% for the sunlit part of the equator is average reflectance during the sunlit part of the day, without weighting for insolation varying during the sunlit part of the day. Insolation in terms of watts per square meter of surface is proportional to the cosine of the angle of incidence. Average insolation-weighted reflectance of the sunlit part of Waterworld’s equator is about 4.6%.

      • Donald L. Klipstein (replying to Clyde:)
        Your 9% for the sunlit part of the equator is average reflectance during the sunlit part of the day, without weighting for insolation varying during the sunlit part of the day. Insolation in terms of watts per square meter of surface is proportional to the cosine of the angle of incidence. Average insolation-weighted reflectance of the sunlit part of Waterworld’s equator is about 4.6%.

        Hmmmn. Define your terms there.
        1. Where is the sensor? If the sensor is above the clouds (and therefore above the water and land) then you’re going to get a different result than the “tested” or measured open ocean albedoes at ALL incident angles: The albedo measurements at sea are made by suspending a small stabilized (flat horizontal) sensor to measure the available total (or direct) solar energy, and the reflected solar energy from the ocean’s surface at that particular incident angle and wind speed/wave height.

        2. Obviously, the sensor is below the clouds, and is measuring the diffuse albedo as well. Even in the arctic, more clouds means a higher measured sea ice albedo because the available downward light energy is increased. Some open ocean measurement methods shade the receiver to minimize diffuse radiation, some accept it, some measure both direct and diffuse reflected energy and “shade” the direct incoming and reflected direct energy to try to measure only diffuse, etc.

        3. Regardless, if the sensor is at the satellite height and level, the measured albedo is either land, desert, jungle, forest, or farmland/steppe; or is open ocean, or is cloud only, or is m mix of all four.

      • Donald.
        You still aren’t getting it. Reflectance is a ratio of output to input. it is when you convert to radiance or flux that you take into account the variation in insolation. The variation in insolation affects everything else as well.

      • Clyde Spencer (In reply to Donald L. Klipstein.)

        Donald.
        You still aren’t getting it. Reflectance is a ratio of output to input. it is when you convert to radiance or flux that you take into account the variation in insolation. The variation in insolation affects everything else as well.

        Sorry, I’m not sure what you are getting at here.

        Assume 100 watts/m^2 are measured DOWN at the sea surface at 19:00 hours on summer evening, before the sun has set, with a solar elevation angle of 12.0 degrees. Clear skies, 0.0 meter/sec wind speed.
        What do you expect the measured UP radiation to be into a sensor at the same position?
        OK. Now that everybody is at the same latitude and assumptions, what do you believe the ocean albedo to be, and why?
        50 watts/m^2 are diffuse, 50 watts/m^2 are judged to be direct radiation. Radiation at TOA is 1340 watts/m^2.

      • RAC,

        According to Fresnel’s equation for reflectance, an angle of incidence of sunlight at 78.0 degrees on a placid seawater surface should reflect 28.7% of the incident (green) light. Assuming that the sensor is in the principal plane, and looking in the direction of the sun at the same downward angle, it should measure <14.4 watts/m^2 of direct, collimated radiation, and some fraction of the diffuse, uncollimated radiation. The exact amount will vary with the fiield of view of the sensor. If it is a narrow FOV, it will only measure that fraction of diffuse radiation that has essentially the same angle of incidence as the direct radiation (Although, I'm not sure how you will tell them apart.). If it has a large FOV, it will integrate radiation form other directions. A sensor at any other position will only record the diffuse radiation, which will be some fraction of the 50 watts. At this low angle, there will be a further correction needed to compensate for attenuation by the atmosphere of both reflected direct and diffuse radiation. I'm assuming your diffuse radiation is scattered by the atmosphere in all directions. Should I further assume that Rayleigh scattering has shifted the spectral character?

        Does this adequately answer your question?

      • Clyde:
        Regarding “Donald. You still aren’t getting it. Reflectance is a ratio of output to input. it is when you convert to radiance or flux that you take into account the variation in insolation. The variation in insolation affects everything else as well.”

        Maybe I committed a grammar error. But the sunlit half of a narrow equatorial strip on Waterworld will reflect about 4.6% of the sunlight despite the average reflectivity being about 9%. I will show my work below:

        Consider a narrow strip on Waterworld’s equator, whose AM half of its daylight half is divided into 18 pieces, each covering 5 degrees of longitude. Make the width of this strip such that the area of each piece is 1/1360 square meter, and assume equinox, so that a 5 degree piece of this strip would get 1 watt of incident sunlight if it was noon everywhere on that piece.

        First column: Time, AM at the center of each 5 degree strip segment
        2nd column: Angle of incidence in the center of the segment, degrees
        3rd column: Reflectivity at the center of the 5 degree strip segment
        4th column: Incident sunlight on each 5 degree segment, watts
        5th column: Reflected sunlight from each 5 degree segment, watts

        06:10 87.5 .7629 .0436 .0333
        06:30 82.5 .4515 .1305 .0589
        06:50 77.5 .2732 .2164 .0591
        07:10 72.5 .1697 .3007 .0510
        07:30 67.5 .1090 .3827 .0417
        07:50 62.5 .0731 .4617 .0337
        08:10 57.5 .0517 .5373 .0278
        08:30 52.5 .0389 .6088 .0237
        08:50 47.5 .0313 .6756 .0212
        09:10 42.5 .0268 .7373 .0198
        09:30 37.5 .0242 .7934 .0192
        09:50 32.5 .0227 .8434 .0191
        10:10 27.5 .0218 .8870 .0194
        10:30 22.5 .0214 .9239 .0198
        10:50 17.5 .0212 .9537 .0202
        11:10 12.5 .0211 .9763 .0206
        11:30 07.5 .0211 .9914 .0209
        11:50 02.5 .0211 .9990 .0211

        Sum of the reflected watts in the 5th column: .5305
        Sum of the incident watts in the 4th column: 11.4627

        Dividing the 5th column sum by the 4th column sum yields .0463 reflected watt per incident watt. This is for the equatorial strip, not Waterworld as a whole where I expect a higher figure (around .0675).

        This is despite the average of the reflectivities in the 3rd column being .1202. (I wonder why you got .09 using your paper cutting/weighing method and I got .12 with my method, even though I get reflectivity for every angle reported agreeing with your Figure 1 curve. I used water refractive index of 1.34 and air refractive index of 1.)

        I repeated this for dividing the AM half of the sunlit half of the equatorial strip into 180 .5-degree pieces each with area of 1/1360 square meter, without printing each line, and:
        Sum of reflected watts is 5.2713
        Sum of incident watts is 114.592
        Reflected/incident = .0460 for the equatorial strip

        Despite the average of the reflectivities of each piece being .1214.

      • Donald,

        I originally did my weighing with an analytical balance. It appears that I misread what was stamped on one of the weights. When I re-weighed tonight with an electronic balance, I got 12.2%, which is essentially what you got for the lower-bound. That answers one question.

        Keep in mind that the whole purpose of this exercise is to account for ALL of the energy leaving Earth. By reducing the incident flux to account for the footprint of the light bundle increasing, you are ignoring some of that energy. It doesn’t magically disappear. There are potentially two ways to deal with this. One, it to NOT correct for the increase in the footprint. The other, which is more rigorous and requires more computation, is to do it your way, but to integrate the reflected intensity over the entire area that the beam has spread out to. That is, you started with a vertical bundle of 1 m^2; you reported a flux that was reduced by the cosine of the angle, but that misses the energy outside that one-square meter area that is incident.

        In any event, it now looks like the total specular reflection from the Waterworld is somewhere between 12 and 18%, which makes it look like NASA/NOAA still have some explaining to do as to why their combined specular and diffuse reflectance is 6%..

      • Clyde:
        Regarding “The other, which is more rigorous and requires more computation, is to do it your way, but to integrate the reflected intensity over the entire area that the beam has spread out to. That is, you started with a vertical bundle of 1 m^2; you reported a flux that was reduced by the cosine of the angle, but that misses the energy outside that one-square meter area that is incident.

        In any event, it now looks like the total specular reflection from the Waterworld is somewhere between 12 and 18%”:

        I’ll repeat my equatorial strip calculations for bands defined by the segments of the equatorial strip, concentric with a line passing through the center of Waterworld and the center of its sunlit hemisphere, all of which combined account for the whole sunlit surface of Waterworld.

        First column: Time, AM at the center of each 5 degree strip segment
        2nd column: Angle of incidence in the center of the segment, degrees
        3rd column: Reflectivity at the center of the 5 degree strip segment
        4th column: Band area, square meters
        5th column: Incident sunlight on each band, watts
        (band area * 1360 * cosine of angle of incidence)
        6th column: Reflected sunlight from each band, watts

        06:10 87.5 .7629 2.270E13 1.347E15 1.027E15
        06:30 82.5 .4515 2.253E13 3.999E15 1.806E15
        06:50 77.5 .2732 2.218E13 6.530E15 1.784E15
        07:10 72.5 .1697 2.167E13 8.862E15 1.504E15
        07:30 67.5 .1090 2.099E13 1.093E16 1.191E15
        07:50 62.5 .0731 2.016E13 1.266E16 9.284E14
        08:10 57.5 .0517 1.916E13 1.400E16 7.239E14
        08:30 52.5 .0389 1.803E13 1.492E16 5.812E14
        08:50 47.5 .0313 1.675E13 1.539E16 4.823E13
        09:10 42.5 .0268 1.535E13 1.539E16 4.129E14
        09:30 37.5 .0242 1.383E13 1.492E16 3.610E14
        09:50 32.5 .0227 1.221E13 1.400E16 3.176E14
        10:10 27.5 .0218 1.049E13 1.266E16 2.765E14
        10:30 22.5 .0214 8.696E12 1.093E16 2.340E14
        10:50 17.5 .0212 6.833E12 8.683E15 1.880E14
        11:10 12.5 .0211 4.918E12 6.530E15 1.380E14
        11:30 07.5 .0211 2.966E12 3.999E15 8.444E13
        11:50 02.5 .0211 9.911E11 1.347E15 2.843E13

        Sum of column 5 incident watts is 1.773E17
        Sum of column 6 reflected watts is 1.207E16
        Column 6 sum / column 5 sum = .0681

        Average of area-weighted reflectivities of each of the bands is .1727

        I checked that the sum of the column 3 areas (divided by 2.59E6 square meters per square mile) is 1.0056E8 square miles. I checked that the sum of the column 5 incident watts is 1360 times the area in square meters of an 8,000 mile diameter circle.

        Repeating for bands being .5 degree wide instead of 5 degrees wide, not reporting results for each of the 180 bands:

        Total incident watts is 1.771E17
        Total reflected watts is 1.195E16
        Reflected watts / incident watts = .0675

        Average of the area-weighted reflectivities of each of the band is .1747

        Sum of band areas converted to square miles is 1.0053E8

      • Donald,
        I’m beginning to feel like we are flogging a dead horse here. One can get results similar to what you report for your 0.0675 average using somewhat arbitrary numbers as follows:

        reflectivity intensity-in intensity-out intensity ratios
        1.00 0.01 0.01 1.000
        0.90 0.12 0.11 0.900
        0.80 0.23 0.18 0.800
        0.70 0.34 0.24 0.700
        0.60 0.45 0.27 0.600
        0.50 0.56 0.28 0.500
        0.40 0.67 0.27 0.400
        0.30 0.78 0.23 0.300
        0.20 0.89 0.18 0.200
        0.10 1.00 0.10 0.100
        Totals
        5.50 5.05 1.87 5.50
        Averages
        0.550 0.505 0.187 0.370
        Column 3 is the product of reflectivity and intensity-in.

        The average reflectivity is 0.550 (column 1), while the ratio of the average intensity-out to average intensity-in is 0.370. That is, one can expect that the ratio of the average intensitties will be less than the average of the reflectivities that were used to create intensity-out.

        However, we aren’t really interested here in the AVERAGE of the intensity ratios. Rather, we are interested in the TOTAL intensity-out integrated over the total area reflecting.

        I don’t think that any further discussion on this will be constructive.

      • Clyde:
        “However, we aren’t really interested here in the AVERAGE of the intensity ratios. Rather, we are interested in the TOTAL intensity-out integrated over the total area reflecting.”

        I reported total out, total in and total area. Total out / total in is what is .0675, and the average of the intensity ratios is what is .1747. Can you find a specific error in my most recent calculations (using concentric bands) that show total_out / total_in is .0675-.0681?

      • Clyde: You wrote:”
        Totals
        5.50 5.05 1.87 5.50
        Averages
        0.550 0.505 0.187 0.370
        Column 3 is the product of reflectivity and intensity-in.

        The average reflectivity is 0.550 (column 1), while the ratio of the average intensity-out to average intensity-in is 0.370. That is, one can expect that the ratio of the average intensitties will be less than the average of the reflectivities that were used to create intensity-out.

        However, we aren’t really interested here in the AVERAGE of the intensity ratios. Rather, we are interested in the TOTAL intensity-out integrated over the total area reflecting.”

        In your example, total_out / total_in is your column 3 total divided by your column 2 total, which is .370, which is the same as your column 3 average divided by your column 2 average.

      • Regarding:
        “The average reflectivity is 0.550 (column 1), while the ratio of the average intensity-out to average intensity-in is 0.370. That is, one can expect that the ratio of the average intensitties will be less than the average of the reflectivities that were used to create intensity-out.

        However, we aren’t really interested here in the AVERAGE of the intensity ratios.”

        The “ratio of the average intensitties” and the “AVERAGE of the intensity ratios” are two different things. The former is your column 3 average divided by your column 2 average, and the latter is your column 1 average since column 1 (reflectivities) is the individual intensity ratios.

  18. Regarding: “I also did a discreet summation of the frustums of a hemisphere (Af = 2πR DX). Multiplying the normalized (to a unit area for the hemisphere) frustum areas by the average reflectivity for the angle of incidence, for each of the frustums, gives the area-weighted reflectivity for each frustum. Summing them gives an area-weighted average reflectivity of about 18%. This is the instantaneous area-averaged reflectance over a hemisphere. This is almost an order of magnitude larger than the sunlight reflected from a small spot on the surface of the ocean directly below the local noon sun during an equinox.”: This sounds to me like it does not take into account weighting for amount of sunlight per square meter, neglecting effects of the atmosphere, being proportional to the cosine of the angle of the sun from the zenith. I think taking that into account reduces reflectivity by at least a half.

    • Donald,
      See my previous remark about the difference between reflectance and the flux reflected. However, it strikes me that your criticism might also be applied to diffuse reflectance because the intensity of the light striking the limbs will be smaller than what strikes the surface directly under the sun.

      • At this point, I see the lower light intensity at the limbs as my preferred explanation for why I expect high specular reflection at the limbs to fail to get more than (my estimate) ~7% of the solar flux incident upon the surface of an ideal Waterworld being reflected.

      • Donald,
        You said, “I see the lower light intensity at the limbs as my preferred explanation for why I expect high specular reflection at the limbs to fail to get more than (my estimate) ~7%” I don’t think you grasp the definition of “reflection.” It is a unitless ratio of incident to reflected light. Even if the incoming flux is lower at the limbs, that doesn’t effect the reflection coefficient. The proper way to handle things is to determine the reflection coefficient, calculate the incident light flux, taking into account the larger footprint, and then multiply the input flux by the incident flux to determine the outgoing flux.

        I’ll respond to your “brute force” experiment tomorrow.

      • Clyde: You just said to multiply two fluxes by each other? Isn’t one supposed to do this instead?
        Determine the reflection coefficient, calculate the incident light flux, taking into account the larger footprint, and then multiply the reflection coefficient by the incident flux to determine the outgoing flux. Also – the incident flux (per square meter of surface) is proportional to the cosine of the angle of incidence.

        For example, if a 1 mile wide square sunbeam hits Waterworld at the point on one of the 60th parallels where it is high noon, it illuminates two square miles at half the intensity it would if it hit the point of the equator where it is high noon. (Assuming equinox.) So in my integration, I calculated percentage reflected of 1 square mile of face-on sunlight using the appropriate reflection coefficient for the angle of incidence.

      • Donald,
        I’m afraid that was jet lag typing. I intended to say, “The proper way to handle things is to determine the reflection coefficient, calculate the incident light flux, taking into account the larger footprint, and then multiply the REFLECTION COEFFICIENT by the incident flux to determine the outgoing flux.”

  19. Regarding “Vegetation behaves differently from inorganic reflectors. First, leaves tend to be smooth, with a waxy coating that favors specular reflection”: I was outdoors a lot when I was a kid, and I noticed that leaves are mostly closer to being diffuse reflectors than specular ones.

    • Donald,
      Go to, say a golf course, set your camera to get a good exposure looking towards the sun. Lock the settings, turn around and take a picture of the grass looking away from the sun. See what happens.

      • I’m not at a golf course now, but I did just look out my second floor window at the edge of my house, looking at a tree and my lawn in almost the same direction the sun’s rays were going. I saw diffuse reflection.

      • Donald,
        You missed the point, perhaps because you want me to be wrong. So, I’ll spell it out for you. The BRDF of grass is very diffuse, but it has a strong forward reflectance lobe that is composed of specular reflectance. Your eyes generally make a poor photometer, which is why I suggested you use a camera. Try using a pair of Polaroid sun glasses outdoors and you will observe that a lot of plants have a strongly polarized reflectance, indicative of specular reflectance.

      • Specular retroreflection by a surface (assuming its refractive index is the same as that of water) is 2%. If you are in the sunward bright lobe of reflection by grass in a lawn, what you are seeing is the opposition effect, or that’s what it’s called when this phenomenon happens with Saturn’s rings – the leaves of grass are hiding their own shadows.

      • Donald,
        The ‘Hot Spot’ observed in remote sensing imagery, where the shadows are hidden, occur when there is alignment with the sun, sensor, and ground. I’m saying that even looking in the direction that allows you to potentially see shadows, it will be brighter than when the shadows are hidden because of the specular reflections.

      • What direction of looking is where I will see brighter reflectance than where I see the shadows being most-hidden by the shadow-casting objects? Are you claiming that specular retroreflection creates a bright ring instead of a bright spot? Does remote sensing imagery indicate this happens? How do you propose this occurs, in a way that rules out effects of the viewer’s shadow including its penumbra? How do you reconcile this with where a camera should look on a lawn for the brightest recording of grass? At this point, I am maintaining that the main factor here, at least on lawns, is the opposition effect.

      • Donald,
        The remote sensing ‘hot spots’ that I have seen are spots and not rings, but they are not uniform. Also, as I recollect, the polarization is not uniform or even radial. The simplest explanation is the the spot is bright because there are no shadow pixels mixing with the illuminated pixels. However, I think that it isn’t that simple.

      • I took a look at that grass model. The spike seems to peak at a reflection coefficient around .1. A lot of that graph surface is around .045-.05. I “eyeball estimate” the average reflection coefficient to be about .06.

        This is at a wavelength of 600 nm which is visible orange light, where I expect the reflectivity of green grass to be less than for surface-reaching solar radiation as a whole. For that, I expect a little higher than .06.

  20. Regarding “The plant chlorophyll converts incident light to carbohydrates and it does not result in warming”: Only the carbohydrates that don’t get metabolized or burnt by animals, fungi, bacteria, other life forms, or fires have their chemical energy failing to have their sunlight energy used in their creation becoming heat.

    • Donald,
      OK, eventually the latent heat of photosynthesis may be released, if it isn’t buried for eons. However, what we are considering here is a time scale that is relatively short. That is, if light is absorbed and immediately converted to heat, it contributes to sensible heating on a short time scale. I’m saying that the predicted atmospheric heating for this year, the next, or even decades from now will be too high if we don’t account for photosynthetic activity delaying the release of heat energy.

      • If the amount of biomass worldwide is not growing from one year to the next, the amount of energy stored in the biomass is not growing from one year to the next. So the amount of energy being released each year from consumption and decay of biomass is the same as that stored by production of biomass each year – for that matter even more if the world is losing biomass from one year to the next, which is the case due to deforestation.

      • Donald,
        OK, this is in part semantics. As is often the case, Earth acts like a time delay. The energy from sunlight eventually finds its way into the system. However, in a precise description of the energy cycle, the energy used by photosynthesis should be brought in at a later date as energy of oxidation, and not be ascribed to immediate heating from molecular agitation. The reason for this fine distinction is that the time delay can be tens of millions of years if the organic material is buried in the oceans.

      • The sum of biomass and carbon that is sequestered (including in the lithosphere and peat bogs) is not increasing from year to year, in terms of heat of combustion of these carbon products. Overall, we are burning these faster than they are being produced. Therefore, over 100% of a year’s conversion of solar energy to biomass chemical energy is chemical energy from biomass (or former biomass) converted to heat energy each year. This is true in a large majority of all years when the annual global carbon budget was reasonably well known, meaning after the Mauna Loa CO2 record started being in good condition – IIRC around 1959.

      • Donald,
        You said, “The sum of biomass and carbon that is sequestered (including in the lithosphere and peat bogs) is not increasing from year to year.” On that you are wrong. There are tremendous quantities of organic materials that find their way into the ocean, as in the Mississippi delta, that get buried. Not all the phytoplankton that die become oxidized as they drift down the water column; they subsequently get buried by inorganic materials also drifting down. Almost all of the Paleozoic limestones in the Midwest are so rich in organics that hammering on them can release aromatics that are sickening to the smell. Sequestering has been going on since the creation of life.

        We are certainly burning coal faster than it is being produced, but then the environment was quite different, when the major coal beds were laid down, than it is today.

      • We are burning carbon faster than it is being sequestered in all ways combined, and one strong piece of evidence is the fact that CO2 is increasing. Each year on average, about half the CO2 from a year’s worth of fossil fuel combustion is removed by the world’s net natural sinking (and not all of that CO2 is sunk as sequestered biomass) and the other roughly half accumulates in the atmosphere.

  21. Regarding “When sunlight does reach the surface, the 100% reflectivity at the Earth’s limbs helps explain, in part, why the poles are so cold”: What about the poles getting less sunlight year-round to reflect? Even though they get a lot during their summers…

    So what is the reflectivity of the ice at the poles? It’s not 100%. What about reflectivity of smooth open water to sunlight from the sun when the sun is 10 degrees above the horizon? The figure 1 graph indicates about 34% – much less than 100%. What about when the sun is 15-plus degrees above the horizon 24/7 – which is the case for the majority of the sunlit 6 months per year at each pole? I don’t see the situation being less favorable for absorbable sunlight 20 or 30 degrees from the poles, where there is sometimes significant open water.

    • Donald,
      I didn’t say the reflectivity was the only reason for it being cold at the poles. Now is when you are supposed to remind me about the larger footprint of light.

      Well, to put my remarks in context, one often sees the remark made that the “dark water” in the Arctic will absorb more energy. What I’m pointing out is that the reflectance can be as high as 100% with the right geometry. And even during the Arctic Summer, the sun doesn’t pop up to 15 degrees above the horizon. Its elevation depends on where you are. And even where it gets fairly high, it doesn’t stay at that elevation but moves up and down during the day. But even you acknowledge that 30% might be a reasonable average reflectance in contrast to 2% over the tropics. So, I’m suggesting that the role of heating water is probably overstated when you take into account the reflectivity and the decreased intensity of light at the poles.

      • About the sun not popping up to 15% above the horizon during the Arctic summer: At the North Pole on the summer solstice, it is 23.4 degrees above the horizon for 24 hours. At the Arctic Circle on the summer solstice, the sun’s elevation bottoms out on the horizon and tops out at 46.8 degrees above the horizon.

      • Donald,
        Good, you have done your homework. But, those values you quoted are for one day only, the solstice. It is all downhill from there!

        Incidentally, that “bottoms out” is when the sun’s light is grazing the surface and has a reflectance of 100%

  22. Regarding “There is reason to believe that a decrease in cloudiness has had more impact on the retreat of glaciers than has the supposed average increase in air temperature (≪1°C/century at mid-latitudes)”: Wasn’t a decrease of cloudiness as a result of warming something predicted by the IPCC-considered CMIP5 global climate models? Did those models get that right as well as the Arctic warming more than other parts of the world, while getting wrong the amount of post-2005 global and tropical middle/middle-upper troposphere warming (due to assuming no multidecadal natural cycles contributed to the rapid warming from the early-mid 1970s to around 2005)?

    • Donald,
      With respect to cloudiness, I’m going to do some hand waving here. My understanding is that warming is predicted to result in increased evaporation. The increase in Arctic warming is supposed to be a result primarily of less radiative cooling. As to whether more water vapor will result in more or less clouds, I don’t know. However, from what I have read, the IPCC predictions of precipitation are even worse than the temperature predictions.

      • Clyde Spencer
        September 12, 2016 at 9:56 pm

        “Donald,
        With respect to cloudiness, I’m going to do some hand waving here. My understanding is that warming is predicted to result in increased evaporation. The increase in Arctic warming is supposed to be a result primarily of less radiative cooling. As to whether more water vapor will result in more or less clouds, I don’t know. However, from what I have read, the IPCC predictions of precipitation are even worse than the temperature predictions.”

        A warming as a result of less radiative cooling?

        A warming is equal to increasing cooling. Primarily. The decrease in excitation caused by less radiation is instantaneous. The same moment that we observe a lowered flux density, there is less power in the system. Less power leads only to one thing, and is only caused by one thing, less energy flowing through the system. If there is a decreasing output, it can only be caused by a decreasing input. Unless there is an energysource adding to the system.

        The same way that less cooling at the poles have no possibility be a result of heat, the TOA spectrum showing how co2 effectively decrease radiation by absorption is a result of warming.

        Albedo is obsolete. We only need to focus on what is absorbed, and that is equal to what the surface absorbs from direct insolation. Everything in the earth-atmosphere system is a result of solar irradiation, nothing that happens is a cause. Clouds espescially, as albedo they contain energy, They are absorbed energy that is caused by irradiation. They represent the energy that exceeds the capacity of water to absorb at the moment of irradiation. Molecules exceed the level of excitement to stay in liquid phase and dissolves it`s structure in the same process that melts metal and carry energy away from the heat.

        What happens to reflections coming from the bottom of the system is irrelevant besides if we measure it from space. All radiation is inside the system, the outer boundary is TOA. The surface absorbs an equal intensity to what it radiates to space. There is only one source of heat, the sun. That means that the surface gets all it´s energy from the sun, directly. A cloud is absorbed irradiation, what it reflects is not more than the energy that caused it. The surfaces of water will never reflect more than they absorb. The poles will never warm enough to raise the mean temperature of earth, because they cool down.

        Everything in the system is instantaneous as it is open without an outer wall, cooling from twice the surface that is heated. And on top of that it radiates from a icecold gasvolume globally. There is not a single watt of surplus or accumulation above the surface, unless the radiant energy density inside the system gets an increased input from the sun.

        See for yourself, calculate radiant energy density in J/m^3 for the atmosphere and the solid mass at the surface. It´s only geometry and everything inside it is caused by the sun.

        1 cubic metre of solid earth mass is excited continously to a temperature equal to 770J . Two cubic metres radiate through a surface of 1m^2 according to it´s temperature which is equal to 385J/m^3. 1027J is the energydensity in solar radiation present in 1m^3 air at the surface in shape of a sphere. Every mean temperature in the system can be calculated by geometry. It think albedo can be ignored.

      • Simplemind,
        You said, “We only need to focus on what is absorbed, and that is equal to what the surface absorbs from direct insolation.” But, the problem is how do we determine what is absorbed? What is typically done is to measure the reflectance and assume that reflectance and absorption account for all the incoming energy and therefore absorption = 1 – reflectance. Therein is the problem. If you don’t account for ALL of the reflectance, you get the absorption wrong.

  23. A fine article rich in detail and common sense. Two comments resonated with me in particular:

    1. Albedo is an instantaneous measurement. To the extent that vegetation and clouds change with the seasons, a further, more complicated correction has to be made for those effects to derive an annual average.

    2. Clouds are the best evidence that the naive “Science is settled” claim is false. Clouds are highly variable in their position, extent, and albedo, often moving rapidly under the influence of winds.

    There is no better illustration of the importance of clouds and the influence that they have on surface temperature than to consider that global temperature is at its annual maximum in July when the Earth is furthest from the sun and irradiance 6% weaker than in January. That’s when global cloud cover is at its annual minimum due to the energy returned to the atmosphere by the continents of the northern hemisphere, so much greater as a proportion of the surface of the hemisphere than is the case in the southern hemisphere.

    It is observed that the surface warms as geopotential height increases at 500 hPa. GPH is a measure of the temperature (and the density) of an air column below the point of height assessment. Anomalous increases in height are due to an increase in the ozone content of the air. Ozone heats the air by absorbing long wave radiation from the Earth and the efficiency of the mechanism increases with atmospheric density.
    GPH is observed to increase with surface pressure and the temperature of the air at 200 hPa and all points above. So, the temperature at the surface is intimately connected with the temperature of the upper air that varies with its ozone content. Why? Because as the air warms it holds less water in the condensed floating form we call cloud and more as invisible, non reflective gas.

    This mode of climate change, all natural and all driven by changes in the external environment in which the Earth and its atmosphere resides is described, for those few curious enough to look, at https://reality348.wordpress.com/

  24. ” What one finds is that the retreat of glaciers is more pronounced on mountain slopes with south-facing aspects. That suggests that there has been a decline in clouds that formerly protected the snow and ice from direct sunlight”

    Oops – that Northern Hemisphere mind set again – the same blinkers that directs so much attention to the Arctic and the Atlantic Sea

    • Good point (and I live in the SH). Although I would point out the absence of east west running mountains and hence north and south facing mountain slopes in the SH. So there are few potential examples. Except perhaps on the sub-arctic Southern Ocean islands and they are little studied because of remoteness. Kerguelen’s north facing glaciers are showing retreat, and one can infer the south facing aren’t (or not as much) and glaciers that aren’t retreating are of less interest to the AGWers and therefore not reported.

      https://glacierchange.wordpress.com/tag/kerguelen-glacier-retreat/

    • Michael,
      Guilty as charged. However, in my defense, there is more land in the Northern Hemisphere and there is nothing comparable to the Himalayas or even Glacier National Park in the Southern Hemisphere.

  25. Clyde, I think you need to get familiar with the actual obital geometry of the satellites concerned and look at viewing angle of the instruments. There is too much hand waving about what is probably being seen and under what conditions.

    I’m fairly familiar with ERBE but not CERES.
    CERES did have a large ( impossible ) net energy gain of about 5 W/m^2 . This could just be instrument calibration not living up to spec. or it could be a failure to capture some reflected insolation which is falsifying the results.

    I think they have done several revisions and reduced the problem but I’m always very sceptical of post hoc “corrections” which usually end up being parameter tweaks in some model involved in the extraction process.

    There were several adjustments to the ERBE SW data which sounded a little speculative. They seemed credible but still speculative. There was one storey about the dome material degrading under UV light which was not validated by materials tests but seemed to be “consistent with ” the drift. If I was in charge of the science I would want and identical dome tested under high UV in the same geometry to prove there was a degradation. Then I’d ask who chose a UV sensitive material for a dome on an ultra high precision instrument that would be exposed to strong UV twice per orbit.

    Last time I looked the “revision 3” data which used this correction was still not officially archived, so maybe someone else is equally a little sceptical.

    The degree of accuracy and precision which is required for this kind of energy balance if extremely difficult to achieve, so it’s an amazing feat anyway but there may be some mileage in the idea that some reflected light is being missed.

    Anyway, to get traction I think you need to get into specifics rather than guessing about the instruments.

    • Greg,

      Despite being retired, I don’t have the time to devote to studying this issue in the detail you suggest. I have several other projects and responsibilities that eat up my time and have higher priorities in my life. I’m currently doing research on the optical constants of opaque minerals and compiling a database of values. There is little that is actually known about the specifics of individual mineral species because the calculations are not trivial and most of the original theoretical work was done before computers.

      I saw what I thought was a probable issue when I realized that diffuse reflectance was being used for calculating absorption, and consequently heating and climate sensitivity. I wanted to bring the potential problem to the attention of those like yourself who might be willing and able to run with it. Even though “The Science is settled” I think that there are a lot of questions left to be answered. In any event, an albedo of one-significant figure, which might be missing a significant fraction of energy leaving the surface of Earth, does not leave me feeling confident that anybody really has the problem under control.

      At the very least, I hope that by sticking my neck out I will have encouraged some people to examine the assumptions behind the claims about how climate sensitivity is calculated. Unfortunately, it appears that the trust in government employees is so high that many are willing to dismiss the possibility that anything has been overlooked. There may be some “not invented here” issues too. It seems evident to me that any calculations that depend solely on diffuse reflections are missing part of the ‘big picture.’ And, if the reflectivity of sea water as calculated by NASA and NOAA is less than the theorectical specular reflectance alone, then I think something is being missed. The reflectivity should be the sum of diffuse and specular!

    • Greg, I think it is unfair to say “you need to get familiar with the actual orbital geometry of the satellites concerned and look at viewing angle of the instruments.”

      Clyde has brought up some interesting questions and observations. If experts can easily answer those questions, experts can and provide links to greater detail and schematics and everyone is better off. If the questions are hard to answer, or cannot be answered, well that’s good to know, too.

      It is not, however, the job of the unpaid to “get familiar” with why the data has a large ( impossible ) net energy gain of about 5 W/m^2 . No, sir. It is the job of those who get paid by the project managers to be upfront with the “bust” and not hide it behind post-hoc data adjustments to make the data fit expectations. It is quite possible what most needs adjusting are the assumptions. Which brings us back to why Clyde wrote the article.

      • Stephen,
        Thank you for the support. If I observe that “The king has no clothes,” I’m not sure that it is my responsibility to prove my statement.

  26. Clyde: Really enjoyed and appreciated the work that went into t his interesting article.

    Have you thought about this issue from the perspective of climate models rather than the perspective of satellites? For every grid cell on the planet, the model needs to correctly deal SWR often arriving far from normal, or the model will do a lousy job, Someone must have looked into how well different models handle this problem and therefore used raw satellite data to decide whose code worked best.

  27. Frank,
    Thank you. Whether or not satellite observations can correctly characterize reflectance is open to question. However, the whole point of the exercise was to challenge some of the assumptions that go into climate models. I think that to do it right would require a dynamic simulation taking into account how the reflectance changes with changing angles of illumination, and not rely on averages. Fallow fields will be different from fields ready for harvest. Wet soils will be darker than light soils. Snow will dramatically change mid-latitude reflectivities. Turbid water after storms will significantly change the diffuse reflectance while not impacting specular reflectance. It is an immensely complex problem.

    • Clyde: Very interesting article, and a decent, thought-provoking discussion free of the usual political rants. It does, as you say, highlight an inherent problem of energy balance estimations. If I can characterize the whole energy-balance issue: it is trying to measure the very small difference between two large quantities. Tiny errors resulting from not capturing all the outgoing energy, will lead to massive errors in estimating the imbalance. Hence I have distrusted all energy-balance calculations from the outset. You have given some solid logic to support what was formerly an intuitive stance. Thanks.

      • Smarter than a rock,
        Thank you for your endorsement. The check is in the mail. Seriously, there is more to this issue than most of the ‘experts’ acknowledge and I think that it deserves more than a cavalier dismissal with “I don’t believe it.” One (usually) doesn’t know what they don’t know. It is then easy to have confidence in declarations that “We have everything under control.”

  28. Clyde,

    A very enjoyable article, thank you.

    One aspect of reflectance I have been wondering about lately has to do with something I serendipitously “discovered” a few decades ago (while pondering decking surface slipperiness in the then prospective small wooden boat I was about to build), which was tiny glass beads used to make highway line paint highly reflective. (worked beautifully, by the way ; )

    My thought is that water vapor droplets/particles (not large drops) are dominated by water surface tension forces, and so maintain a nearly spherical shape in air, and thus, I figure, ought to act as directional reflectors to sunlight, much as those tiny glass beads do to auto headlight. Which is to say I imagine there is a diffuse water vapor “albedo” effect that is sun-facing atmosphere wide . . I’ve never seen any discussion of this effect, and wonder if you have . .

    • JohnKnight,
      I can’t speak with any great expertise on your question, but I believe what you are referring to are generally called “retroreflectors.” Because clouds are so bright, I think that they are effectively acting as retroreflectors. When a light ray enters a water droplet it can potentially pass straight through, but is probably more likely to suffer one or more total internal reflections before exiting. The total internal reflection happens with essentially 100% efficiency, so the light exiting will be nearly as bright as the ray entering. (There is some loss as a result of the very small, but non-negligible extinction coefficient, i.e. absorption.)

  29. Clyde,

    I’ve been doing some surface energy-budget stuff. Do you know of any CMIP5 (or newer) GCMs that don’t account for illumination angle when calculating reflectance?

    Why didn’t you talk about multi-angle imaging instruments that are used to measure BRDF?

    • MieScatter,

      I didn’t talk about multi-angle imaging instruments because this wasn’t intended to be a summary of the NASA orbiting satellite programs.

      The point of this is that there is a good reason to believe that the models that are used to predict albedo aren’t doing it right. Namely, if the total of diffuse and specular reflection of water is estimated to be 6%, and Fresnel’s equation predicts that the water average should be something like 3X that value, then there is a problem that needs to be explained.

      It is beginning to look to me like the values being used by NASA for water represent the nadir-to-near-nadir specular reflectance (2%) plus about 4% for diffuse reflectance from turbidity. Therefore, the correct value might be as high as 22% (4% + 18%).

      • Clyde Spencer (replying to Miescatter)
        The point of this is that there is a good reason to believe that the models that are used to predict albedo aren’t doing it right. Namely, if the total of diffuse and specular reflection of water is estimated to be 6%, and Fresnel’s equation predicts that the water average should be something like 3X that value, then there is a problem that needs to be explained.

        It is beginning to look to me like the values being used by NASA for water represent the nadir-to-near-nadir specular reflectance (2%) plus about 4% for diffuse reflectance from turbidity. Therefore, the correct value might be as high as 22% (4% + 18%).

        No.

        There are two very different albedoes for open ocean water.
        One for diffuse radiation.
        The open ocean albedo for reflecting diffuse radiation remains near-constant constant across essentially all solar incident angles (solar elevation angles above the horizon) at the “classic” open water albedo = 0.067.

        The open ocean albedo for reflecting direct radiation is also 0.67, but ONLY for solar elevation angles above 33 degrees above the horizon.
        Below 33 degrees SEA, the direct radiation albedo increases sharply as solar elevation angle decreases, going above 0.45 at SEA less than 10 degrees. (Few measurements are published for SEA < 10 degrees.) See Pegau and Paulsen for wind speed corrections to the basic formula for open ocean direct radiation albedo with respect to solar elevation angle. (Also measured as Solar Zenith Angle in many papers.)

        However, the solar radiation received at sea level to a flat surface is greatly attenuated by the atmosphere at low Solar Elevation Angles, and a good part of that attenuation is scattering. So, all solar elevation angles, there is some diffuse radiation and some direct radiation. The percent of each in the total varies strongly SEA itself, and with atmosphere contamination (day of year (season) and latitude control the amount of dust, amount of pollen, number of clouds and type of clouds and altitude of clouds, and time of pollen.) Roughly, at very low SEA angles, the ratio of diffuse to direct varies from 0.76 to 0.90 At high SEA angles and clear sky, the diffuse radiation may be as low as 0.05 the direct radiation.

        All of these factors affect the ratio and final albedo you began describing above.

      • RAC,

        You tell me that there are two albedos. P. Minnis tells me that there is only one. I think that the following links do a better job of explaining the situation:
        http://www2.hawaii.edu/~jmaurer/albedo/
        http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19880018293.pdf

        You remark that few measurements have been published for angles of incidence greater than 80 degrees. That is exactly the crux of my concerns! There is an old saying that “The Devil is in the details.” I’m concerned that the paucity of data for very high angles of incidence is a ‘detail’ that is just being ignored.

  30. “I’ve been doing some surface energy-budget stuff. Do you know of any CMIP5 (or newer) GCMs that don’t account for illumination angle when calculating reflectance? ”

    Does that imply that you know that some/most recent GCMs DO take angle into account.

    • Hi Greg,

      I’ve never met a scientist working on surface reflection who doesn’t know about this and I’d be shocked to find one. Here’s a 1988 guide to albedo calculations as an example.
      http://ntrs.nasa.gov/search.jsp?R=19880018293

      Papers have been saying for at least a decade that GCM ocean reflectance accounts for incidence angle. I tested it by looking at the seasonal cycle of ocean albedo as a function of latitude in CMIP5. I think all CMIP5 models except the CMCC ones account for this, but afaik I’ve not used the CMCC models in my analysis.

      I haven’t checked the source code so I’d be happy to be corrected. But it’s also possible that Clyde didn’t realise that the models include this, which would make most of the blog redundant.

      • MieScatter,

        I gave a cursory review of the NASA article on calculating ‘directional albedo’ that you provided. I see that the procedure uses binning. What’s more, the binning for observations runs from 75 to 90 degrees for the highest zenith angles. The reflectance of water varies from 21% to 100% in that range. Obviously, any calculations performed with that kind of imprecision is going to leave a lot of wiggle room for a claim that the models are using angular observations. From the perspective of my claims, things don’t get interesting until 75 degrees is reached.

        If you are seeing a seasonal variation in ‘albedo’ with latitude, then you are almost surely looking at diffuse reflectance from inorganic and organic turbidity. This contribution is additive and doesn’t really speak to my concerns about why the theoretical values appear to be higher than the values used in the models.

  31. Energy density uses the same calculation as radiation, but has units of Joule/m^3. It says that it is a synonym for magnetic field behaviour, pressure and enhalpy at the same time as it can express temperature in Kelvin and fluxdensity in W/m^2.

    I cannot not see albedo in there. Where is it hiding?

    “This article is about energy per unit volume. For energy per unit mass or energy density of foods, see specific energy.
    Energy density
    SI unit J/m3
    In SI base units kg·m-1s-2
    Derivations from
    other quantities
    U = E/V

    Energy density is the amount of energy stored in a given system or region of space per unit volume or mass, though the latter is more accurately termed specific energy. Often only the useful or extractable energy is measured, which is to say that chemically inaccessible energy such as rest mass energy is ignored.[1] In cosmological and other general relativistic contexts, however, the energy densities considered are those that correspond to the elements of the stress–energy tensor and therefore do include mass energy as well as energy densities associated with the pressures described in the next paragraph.

    Energy per unit volume has the same physical units as pressure, and in many circumstances is a synonym: for example, the energy density of a magnetic field may be expressed as (and behaves as) a physical pressure, and the energy required to compress a compressed gas a little more may be determined by multiplying the difference between the gas pressure and the external pressure by the change in volume. In short, pressure is a measure of the enthalpy per unit volume of a system. A pressure gradient has a potential to perform work on the surroundings by converting enthalpy until equilibrium is reached.

    https://en.wikipedia.org/wiki/Energy_density

  32. Energy density calculated from irradiation from the sun at TOA in units of J/m^3 in shape of a sphere gives the right mean surface excitance and solid mass temperature, at the same time it tells us how the magnetic field behaves, what the pressure is and the relative enthalpy inside a system, which shows us the direction of the flow of energy towards immediate higher entropy.

    Why do we include co2 or albedo when we get f****d up numbers and have to use an icecold atmosphere of -18C as a heat source? Why should I not use the definition of energy density?

    • Uh, because “energy density” has nothing to do with reflection coefficient, which is a nondimensional number between zero and one, where zero is non-reflecting (“black”) and one is all-reflecting (“white”). See https://en.wikipedia.org/wiki/Albedo for a reasonable discussion.

      I have no idea what you are driving at in your first paragraph, and I rather suspect neither do you.

  33. Regarding “One frequently sees reference to the nominal 30% albedo of Earth with respect to the energy budget and alleged anthropogenic global warming. Although, the CRC Handbook of Chemistry and Physics lists a value of 36.7% in recent editions. Albedo is a measure of the apparent reflectivity, of visible light, of celestial bodies such as the moon, Mars, and asteroids”:
    This seems to suggest that the discrepancy between the 30% and 36.7% figures is from the 36.7% figure being “a measure of the apparent reflectivity, of visible light”, and the 30% figure being of total solar radiation so as to be suitable for energy budget calculations.

  34. As for some specific spectral ranges of solar radiation where Earth’s albedo seems to be lower than with visible light: One is ultraviolet of wavelengths shorter than 315 nm, about 4% of the sun’s output, which is mostly absorbed in the ozone layer. Another is infrared of wavelengths longer than 1.5 micrometers, about 12% of solar radiation, which is readily absorbed by liquid water and ice. This means clouds, ice and snow mostly absorb these wavelengths and diffuse reflection of them by radiation penetrating into water is nil. In addition, a few percent of solar radiation is at wavelengths in infrared shorter than 1.5 micrometers but highly absorbed by water vapor, and these wavelengths generally don’t reach the surface.
    Another factor is that most of the sun’s invisible radiation is infrared, which is scattered by the atmosphere less than visible light is.

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