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Comments 129351 to 129400:

129351. Gord at 20:48 PM on 1 June 2009
        It's the sun
        Patrick - In your Post #461 you said... "Surface absorbs RSe from the sun and Rae from the atmosphere. Surface loses heat by Rea to the atmosphere, Res to space, and Cea (convection) to the atmosphere. The atmosphere absorbs RSa from the sun and Rea from the surface, and gains Cea from the surface, and loses energy by Rae to the surface and Ras to space. THESE EQUATIONS ARE BASED ON THE CONSERVATION OF ENERGY: Rate of energy gain by the surface = RSe + Rae - Res - Rea - Cea Rate of energy gain by the atmosphere = RSa + Rea + Cea - Ras In climatic equilibrium, the average of each rate of energy gain is zero. Where is energy being created or destroyed?" ---- Answer: + Rae = energy absorbed by the Earth's surface. Rae came from the Atmosphere that got it's energy from the Earth's radiation. The Earth's radiation cannot cause it's own radiation to increase. It's called Energy Creation....a violation of Conservation of Energy. Further, Rae IS GREATER THAN Rse and Rse is the energy absorbed by Earth's surface from the SUN ...THE ONLY ENERGY SOURCE!..another Energy Creation and violation of Conservation of Energy. ------------------- Your equations ARE BASED ON A VIOLATION OF CONSERVATION OF ENERGY. Like I have said...With Patrick, 1 + 1 = anything but 2. What great Comedy!
129352. paravantis at 20:09 PM on 1 June 2009
        It's Urban Heat Island effect
        ...I meant to write "Parker's" mediocre effort!
129353. paravantis at 20:08 PM on 1 June 2009
        It's Urban Heat Island effect
        As an addendum to my previous post, Parker mediocre effort (2006) conveniently omits Karl, Diaz and Kukla (1988) from its list of references.
129354. Dan Pangburn at 16:16 PM on 1 June 2009
        It's the sun
        Patrick 027 485 There can be no significant NET Climate Science type positive feedback. So there could be at least as much positive as there is negative from the temperature increase. I expect that additional negative Climate Science type feedback will be discovered that is associated with clouds. I modeled the K&E graph except I included thermalization as an unknown. (first model April 29, updated today) I used average cloud temperature 258 K, cloud emissivity .5, cloud coverage 50% (see Zero-dimensional models at http://en.wikipedia.org/wiki/Climate_model) and varied the fraction of cloud radiation that reaches the ground from 50% to 90%. (I assumed cloud radiation to be graybody type from the water/ice particles) Thermalization varied from 16% for 50% cloud-to-ground to 24% for 90% cloud-to-ground. Considering all of the GHGs that the cloud radiation needs to penetrate to reach the ground, the fraction that makes it is probably closer to 50% than 90%. That which is not thermalized is needed as back radiation to the surface. I was surprised to see such a low number for thermalization but can't find anything wrong. Using their values for 'absorbed by atmosphere', thermals and evapotranspiration and correctly combining numbers I get acceptable agreement with what they got.
129355. Patrick 027 at 14:53 PM on 1 June 2009
        It's the sun
        To whomever may wish to know: Regarding: ----------------------------- P = (e*ec*N)*BC*A(T^4 - Tc^4) Then let U = e*ec*N*BC*A The rest was correct: P = U * (T^4 - Tc^4) This can be rewritten as: P = U*T^4 - U*Tc^4 ----------------------------- This refers to the net radiant power transfer between two layers, that is emitted and absorbed by those layers. It does not include emissions and absorptions in other layers. It assumes the layers are thin enough for each to be nearly isothermal within itself vertically (horizontal temperature variations are generally far to spread out compared to optical thickness to matter much, though there could be occasional exceptions on a small scale). Relative to perfect blackbody layers with entirely transparent space in between, the amoung of radiant power emitted by one layer and absorbed in another is proportional to the emissivity e1 of the emitting layer and the absorptivity of the absorbing layer, a2, and the transmissivity of the intervening space, N. The radiant power emitted by the second layer and absorbed in the first is proportional to e2 * a1 * N. At local thermodynamic equilibrium, emmissivity = absorptivity (at each location and time, direction, and wavelength). It is generally a very safe assumption that each small unit volume (small enough to be nearly isothermal) is approximately in local thermodynamic equilibrium. (Fluorescence is an example of the type of radiation that can be emitted when a system is not in local thermodynamic equilibrium - the energy involved has not been thermalized.) Hence, a2 = e2 and a1 = e1. So, for the radiant power exchanged by two layers by emission and absorption, the radiant power in both directions, and thus, the net radiant power, are all proportional to e1*e2*N. In the above equations, e1 = e and e2 = ec, but more generally: Pjk = net P per unit area from layer j to layer k = ej*ek*Njk*BC* ( Tj^4 - Tk^4 ) ------ Of course, this formulation is for the most simple case where ej, ek, and Njk are not wavelength dependent, and also requires Njk to be the transmissivity for the total flux, integrated over direction (it also assumes the refractive index of each layer is close to 1). Still keeping the approximation of index of refraction = 1 (a very safe approximation for the atmosphere): For a single direction and wavelength, if the intervening space is made of layers l=j+1 to k-1 (assuming k is greater than j, otherwise ... you get the idea), and Nm is the transmissivity of layer m, then Njk = Nj+1 * Nj+2 * Nj+3 * ... * Nk-1 [where j+n,k+n are all subscripts], and in the absence of scattering or reflection, Nm = 1 - em. Identifying the direction by it's angle from verticle, q, and identifying the wavelength by L, and for each wavelength, measuring the layer thicknesses out so each layer has the same e and thus the same N, Ijk(L,q) = e'j*e'k*[(1 - e'j+1)*(1 - e'j+2)* ... *(1 - e'k-1)]*BC*[ Ibb(Tj) - Ibb(Tk) ] = e'^2 * (1-e')^(k-j-1) * BC * [ Ibb(Tj,L) - Ibb(Tk,L) ] Where Ijk(L,q) is the spectral intensity (monochromatic intensity per unit interval of the spectrum) of the radiation - the radiant power per unit solid angle per unit area normal to the direction of radiation, per unit interval of wavelength (it can also be formulated as per unit interval of frequency); Ibb(T,L) is the blackbody intensity at wavelength L at temperature T. **In this case, e' is the emissivity along a path at angle q through a layer, so it will be different for different q. What is e' if e is the emissivity along a vertical path through a single layer? Well, if N is the transmissity along a vertical path, and N' is the transmissivity in the direction q, then: N' = 1 - e' N = 1 - e and N' = N^sec(q), because sec(q) = 1/cos(q) = the path length through the layer in the direction q relative to a vertical path length, and the trasmissivity through a path is equal to the product of the transmissivities of subunits of the path. hence, e' = 1-N' = 1 - N^sec(q) = 1 - (1-e)^sec(q) --- Solid angle (symbol w used below) is measured in steradians (symbol sr), which is the projection (from the center of a sphere) of a solid angle onto a spherical surface area, divided by the square of the radius, and thus, like radians, steradians can be treated as unitless. The spectral power (monochromatic power per unit of spectrum) per unit horizontal area = Pjk(L) = (integral from q = 0 to q = pi/2)[ Ijk(L,q) * cos(q) * 2*pi*sin(q)*dq ]. The integrand has a factor of cos(q) because Ijk(L,q) is a value per unit area normal to the direction q; that unit area is a fraction cos(q) of its projection onto a horizontal surface. The factor 2*pi*sin(q)*dq is the the solid angle dwencompassed by the interval of directions dq (it is a ring centered around q=0, of angular width dq from the center and angular length 2*pi (full circle) around the center. cos(q) * 2*pi*sin(q) = pi * sin(2*q) *** pi/2 is a measure in radians, in case anyone was confused by that; pi ~=3.1415927 and pi/2 is a 90 deg angle. Integrating from q = 0 to q = pi/2 encompasses all the contributions from a hemisphere of solid angle. It is not necessary to integrate over the whole sphere of directions because the quantity being integrated is a net value that is the sum of radiant intensities in opposing directions. The same result would be obtained by integrating the contributions to power per unit horizontal area from the intensity from each direction, over the whole sphere of solid angle. PS: For isotropic radiation I(q) = I, the radiant flux per unit area F = (integral from q = 0 to q = pi/2)[ I * cos(q) * 2*pi*sin(q)*dq ]. = I * (integral from q = 0 to q = pi/2)[ pi*sin(2*q)*dq ] = I * pi * [ -cos(pi) + cos(0) ]/2 = pi * I The solid angle in the hemisphere is: (integral from q = 0 to q = pi/2)[ 2*pi*sin(q)*dq ] = 2*pi*[-cos(pi/2) + cos(0) ] = 2*pi There are 2*pi steradians in a hemisphere. --- There is a less clunky way to figure this stuff out - we can let the layer thickness as measured by optical thickness become infinitesimally thin, and use calculus: still at a given wavelength: In a unit volume, there is a density of emission cross section, the emission cross section per unit volume = ecsv. This is equal to the absorption cross section per unit volume if at local thermodynamic equilibrium. ecsv from one substance is equal to the product of the emission cross section of that substance per unit mass times the mass of that substance per unit volume. The exv of different substances and additional amounts of substances add linearly to give the total. The exv is the effective blackbody surface area facing a given direction - each molecule contributes to ecsv with a cross-section. It is possible for ecsv to be direction-dependent (analogous to the molecules having elongated optically-defined shapes with some non-random alignment) but it is generally the case in gases that exv is isotropic (analogous to each molecule being spherical or to random alignments of molecules, so that on average, each molecule looks like a spherical blackbody that presents the same cross-sectional area in each direction). (Of course, the image of a molecule having some well-defined shape with a clear edge is just a mathematical equivalent to what it contributes, on average, to bulk optical properties.) (It is possible to imagine some macroscopic analogue to molecular cross-sections - imagine how radiation propagates through a room filled with blackbody spheres - except, there will be some diffraction around those spheres, which corresponds to a contribution to the scattering cross section on the molecular or small particle scale. Scattering on the molecular scale has a macroscopic analogue of mirrored and/or clear glass spheres or other shapes. The total (extinction) cross section (emission + scattering = extinction) can be different in different directions but must be the same in a pair of opposing directions; however, it would be possible to construct a macroscopic analogue in which the scattering and emission cross sections vary between pairs of opposing directions - for example, spheres that are mirrored on one side and perfect blackbodies on the other - but I think this might not have a molecular analogue, at least for radiation with wavelengths longer than the molecular scale (??).) A unit volume is equal to a unit path length with a given area. Thus, the escv, the emission cross section per unit volume is equal to the emission cross section (an area) per unit area per unit path length, and is thus a fraction of area per unit path length. We can consider an infinitesimally thin layer, of thickness dx, so that the fraction of area per unit path length, escv*dx, is nearly zero. Since the fraction of area of the whole layer dx is nearly zero, the fraction of area of any sublayer must be nearly zero, so the fraction of any sublayer cannot significantly block the fraction of another sublayer (the molecules that contribute to the emissivity are spread out enough so as to be very unlikely to align on top of each other and block each other from view in the direction x). Thus, essentially the entire emission cross section within such a thin layer dx is visible in the direction x. Thus, the emissivity of that layer in the direction x is equal to the fraction escv*dx. And the emissivity of the layer 2dx will be approximately 2*escv*dx, but actually just slightly less than that; the emissivity approaches 1 and cannot increase farther. The transmissivity (in the absence of scattering, and with the absorption cross section = emission cross section) is 1 - escv*dx for the layer dx in the x direction. The transmissivity through a layer n*dx is approximately 1 - n*escv*dx if n is not large, but it is exactly (1-escv*dx)^n; The transmissivity decreases by a fraction escv*dx for each dx. Where the transmissivity is N, dN/N = -escv*dx N = 1 when x = 0. Integrating: ln(N) - ln(1) = -escv *(x-0) ln(N) - 0 = -escv *x N = exp[-escv *x] For a radiant intensity at x, I(x), going in the x direction, I(x) = I(0)*N(x) = I(0) * exp[-escv*x], which I believe is refered to as Beer's Law - this applies to the portion of the radiation that was present at position 0 that reaches x, and does not include radiation emitted along the path from 0 to x. ------------ For the spectral intensity Ixs emitted from a dx at position x that reaches position x = s (going in the positive x direction): Ixs = Ibb(T(x),L) * escv * dx * exp[-escv *(s-x)] Let Is is the intensity of radiation at s, in the x direction, emitted from the layer between x=0 and x=s: Is = (integral from x = 0 to x = s)[ Ibb(T(x),L) * escv * exp[-escv *(s-x)] * dx ] Which can be numerically integrated for any T(x) function, but if we assume T(x) is a constant T between 0 and s: Is = (integral from x = 0 to x = s)[ Ibb(T,L) * escv * exp[-escv *(s-x)] * dx ] Is = Ibb(T,L) * escv * exp[-escv *s] * (integral from x = 0 to x = s)[ exp[-escv *(-x)] * dx ] Is = Ibb(T,L) * escv * exp[-escv *s] * (1/escv) * [ exp[escv *s] - exp[escv *0] ] Is = Ibb(T,L) * exp[-escv *s] * [ exp[escv *s] - exp[escv *0] ] Is = Ibb(T,L) * [ exp[escv *(s-s)] - exp[escv *(0-s)] ] Is = Ibb(T,L) * [ exp[0] - exp[-escv *s] ] Is = Ibb(T,L) * [ 1 - exp[-escv *s] ] = Ibb(T,L) * [ 1 - Ns ] where Ns is the transmissivity of the layer from 0 to s in the x direction. Not surprisingly, this implies the emissivity of the layer, es, is 1 - Ns. Of course, the absorptivity of the layer is equal to the emissivity of the layer (local thermodynamic equilibrium) ------ So the net radiative transfer in the x direction, from emissivity to absorptivity, between two layers g and h, seperated by layer s (where g, h, and s are the thicknesses of the layers in the x direction), defined as positive in the direction from g to h, is: Igsh(L) = eg * ah * Ns * Ibb(Tg,L) - eh * ag * Ns * Ibb(Th,L) where ej, aj, and Nj are the emissivity, absorptivity, and transmissivity of the layer j. Since ej = aj (local thermodynamic equilibrium): Igsh(L) = eg*eh*Ns * [ Ibb(Tg,L) - Ibb(Th,L) ] eg = 1 - exp[-ecsv *g] eh = 1 - exp[-ecsv *h] Ns = exp[-ecsv *s] --- What if we want to know the contribution of this radiant transfer (which is per unit area normal to the x direction) to the radiant transfer PGSH(L) per unit horizontal area, where G, S, and H are the layer thicknesses measured in the vertical direction? Let x be an angle q from vertical. Consider the solid angle dw. Note that h*cos(q) = H, so h = H*sec(q), and g = G*sec(q), and s = S*sec(q) d(PGSH(L)) = Igsh(L) * cos(q) * dw = eg*eh*Ns * [ Ibb(Tg,L) - Ibb(Th,L) ] * cos(q) * dw = [ Ibb(Tg,L) - Ibb(Th,L) ] * (1 - exp[-ecsv *g]) * (1 - exp[-ecsv *h]) * exp[-ecsv *s] * cos(q) * dw = [ Ibb(Tg,L) - Ibb(Th,L) ] * (1 - exp[-ecsv *G*sec(q)]) * (1 - exp[-ecsv *H*sec(q)]) * exp[-ecsv *S*sec(q)] * cos(q) * dw Assuming layers G and H are isothermal over horizontal distance (generally a good approximation for the spatial scales involved in radiative energy transfers, except in some conditions), then the same value applies to all directions at an angle q from vertical, so dw can be replaced with the ring-shaped solid angle 2*pi*sin(q) * dq, and integration can be done over q from 0 to pi/2 to find PGSH(L). Of course, ecsv will vary over vertical distance (And anywhere the path runs into a cloud, or surface, etc.) Within the atmosphere, ecsv will tend to be proportional to air density at wavelengths dominated by well-mixed gases (in the absence of clouds), but at many wavelengths (in between absorption lines), decrease faster with height, while at some wavelengths (near line centers), decrease more slowly. Water vapor is concentrated downward, and there are clouds, etc. --- The mathmatics can be simplied by using optical depth or optical thickness-based coordinates. A unit 1 optical thickness is the distance x over which ecsv*x = 1, so that transmissivity is 1/exp(1). Various different vertical coordinates are used in atmospheric science, including geometric height z (as in x,y,z), pressure (x,y,p), potential temperature (x,y,theta), and sigma coordinates (x,y,sigma), where sigma is set to one value at the surface regardless of surface pressure, and one value at infinite height (p=0), and varies linearly with pressure in between. Any of those vertical coordinates can be mapped onto optical thickness coordinates and vice-versa based on the optical properties at each position. Different mappings will be necessary for different wavelengths, however, so integration over the spectrum to find total fluxes will require converting back to one of the other vertical coordinates. Pressure coordinates will tend to have nearly-constant heat capacity per unit vertical distance, which is convenient for converting an energy flux convergence (in the absence of horizontal net fluxes, -1 * the vertical derivative of the net upward energy flux (power) per unit area) to a rate of temperature increase. Redefining (to conserve on variables) G, H, and S to vertical layer thicknesses in optical thickness coordinates: eg = 1 - exp[-G*sec(q)] eh = 1 - exp[-H*sec(q)] Ns = exp[-S*sec(q)] d(PGSH(L)) = eg*eh*Ns * [ Ibb(Tg,L) - Ibb(Th,L) ] * cos(q) * dw = [ Ibb(Tg,L) - Ibb(Th,L) ] * (1 - exp[-G*sec(q)]) * (1 - exp[-H*sec(q)]) * exp[-S*sec(q)] * cos(q) * dw --- For dw = 2*pi*sin(q) * dq, d(PGSH(L)) = [ Ibb(Tg,L) - Ibb(Th,L) ] * (1 - exp[-G*sec(q)]) * (1 - exp[-H*sec(q)]) * exp[-S*sec(q)] * pi*sin(2*q) * dq Using optical thickness coordinates is convenient; the surface and space can be regarded as layers with infinite thicknesses, over which T is (for x,y, and time) constant. To figure out the total net radiant energy transfer emitted from one layer G and absorbed in all layers above Hj, one must integrate over the different Hj layers. If one wants to know the total net radiant energy absorbed from layer by all layers above and below, one must sum the net fluxes from the other layers. If one wants to know the net upward flux at a given position, one must integrate the net upward flux between each pair of layers across that position. Of course, one must also integrate over solid angle and over the spectrum (or the portion of the spectrum being considered). Fortunately, the qualitative effects are easy to visualize. One can see some fraction of perfect blackbody radiation radiation intensity from a layer according to that layer's optical thickness, the optical thickness of the space between you and that layer, and the angle; and putting yourself in the place of another layer at the viewing location, the fraction absorbed depends on your optical thickness in that direction - that, multiplied by the other factors, is how much 'you would see' of the other layer from that direction, and that other layer would see the same amount of 'you'. With radiation, 'what you see' (pretend you can see at any wavelength) is what you get. --- What about scattering? For the extinction cross section per unit volume xcsv and the scattering cross section per unit volume scsv, xcsv = ecsv + scsv. The transmissivity over distance s without scattering Ns = exp[-xcsv * s] There will, however, be a glow of diffuse scattered radiation, which will add to the radiant intensity in other directions; some scattered radiation still reaches some forward distance after scattering. ... it's complicated (but qualitatively easy to understand - scattering, and reflection in general, decrease transmission between layers going in any one direction but replace what they block from view with radiation from the same side of the scattering/reflection as the viewing position; for a mix of absorption and scattering - any blocking of radiation from behind will be replaced with a combination of radiation from the blocking layer and radiation from the same side of the blocking layer as the viewing position, but with varying intensities depending on the temperature of the emitting sources); to be continued if/when I get to it...
129356. Patrick 027 at 02:49 AM on 1 June 2009
        It's the sun
        Dan - "Maybe it doesn't make sense because you have not grasped that feedback as used in Control Theory is different from feedback as used in Climate Science. See 475." Well, maybe, ... but what do you think of the feedback in climate science now? (Point being that you can't say that feedback by one definition is not positive because feedback by another definition is not positive - When climatologists say that the warming from a doubling of CO2 would be a bit over 1 K without feedbacks, they mean without feedbacks except for the increased thermal radiation to space as a function of temperature; adding positive feedback to make the warming 3 K may be entirely compatable with a negative feedback in Control theory. ---------- "So what fraction of the energy radiated from the surface and absorbed by the GHGs is thermalized?" Essentially all of it.
129357. Dan Pangburn at 22:41 PM on 31 May 2009
        It's the sun
        Patrick 027 479 So what fraction of the energy radiated from the surface and absorbed by the GHGs is thermalized?
129358. Dan Pangburn at 22:27 PM on 31 May 2009
        It's the sun
        Patrick 027 478 "That doesn't quite make sense - what if there was positive feedback - then the speed could certainly change." Maybe it doesn't make sense because you have not grasped that feedback as used in Control Theory is different from feedback as used in Climate Science. See 475.
129359. Gord at 19:23 PM on 31 May 2009
        It's the sun
        Patrick - Re: 476 Original Equation (Stefan-Boltzmann Law) P = e*BC*A(T^4 - Tc^4) which when expanded and giving ec as the emissivity of Tc gives: P = e*BC*A*T^4 - ec*BC*A*Tc^4 If T = temp of the Earth and Tc = temp of the atmosphere, this is a clear subtraction of power produced by the Earth and the power produced by the atmosphere. Clearly, the emissivity of the atmosphere does not have any affect on the power produced by the Earth and vice-versa. ---------- Here is the "Patrick's Law" equation pulled out of his imagination with no logical development shown: What variable shall we use for transmissivity? How about N: P = (e*ec*N)*BC*A(T^4 - Tc^4) Patrick said..."When both the hot object and colder object have emissivities e and ec that could be less than 1, and there is a layer between with transmissivity T:" (Patrick later changed T to N.) ---------- Notice that "Patrick's Law" does not require a temperature for the layer between T and Tc! He just has N = transmissivity for this layer....THAT'S ALL! emissivity = 1 - Transmissivity - reflectivity http://www.optotherm.com/emiss-physics.htm Thus, the emissivity of this new layer of can be expressed as en = (1 - Tn - Rn) Substituting En into Patrick's Law gives: P = (e*ec*en)*BC*A(T^4 - Tc^4) and expanding gives: P = e*ec*en*BC*A*T^4 - e*ec*en*BC*A*Tc^4 Now the power produced by the Earth is changed by the new atmospheric layer emissivity and the higher atmospheric layer emissivity! A clear violation of the Stefan-Boltzmann Law....and ABSOLUTELY HILARIOUS! Only a total "incompentent" and/or someone with "delusions of grandeur" would consider "modifying" the Stefan-Boltzmann Law. What utter drivel.
129360. aplysiatoxin at 14:41 PM on 31 May 2009
        It hasn't warmed since 1998
        It has been cooling for ~8 yrs. Plotting a 11-y moving average conveniently allows you to ignore the last ~6 yr. Plotting a trailing 2-5 y average would catch the recent trend, which correlates with solar activity and most definitely does not correlate with the continuing increase in CO2. This website allows you to explore trends
        Response: There has been cooling over the past few years. However, the surface temperature record is a noisy signal - imposed upon the long term warming trend is much short term variability. Consequently, it's not uncommon for there to be short periods of cooling over the past 35 years of warming.
129361. Patrick 027 at 10:45 AM on 31 May 2009
        It's the sun
        Correction: ... Larger droplets may also be more likely to contain an effectIVE ice nucleus (?)...
129362. Patrick 027 at 10:37 AM on 31 May 2009
        It's the sun
        "If the K&E graphic was not misleading there would be no need for all the clarification/explanation/rationalization." Actually, I think that's a bit like saying that this is misleading: that rain and snow, and other precipitation, comes from clouds which form when water vapor condenses (and sometimes freezes); the water vapor comes from evaporation from wet surfaces. Well, are dew and frost considered 'precipitation'? Sometimes clouds and precipitation evaporate before reaching the surface. Etc. Some plants get moisture directly from fog that is not actually precipitating. Aside from all that, though, there are a few missing links in the water vapor -> cloud -> precipiation chain: How does condensation take place without a surface (a small droplet has a high internal pressure because the surface tension of the spherical surface is squeezing it; this raises the equilibrium vapor pressure so that it tends to evaporate; this makes it very hard to form a droplet from pure gas, because it has to start out very small)? How do cloud droplets and ice crystals gather into particles large enough to fall out at significant rates (it is hard to grow large enough from continual condensation on particles from the vapor phase in the short time periods that are observed)? (Answers: generally aerosols. Some aerosols are more or less effective at nucleating cloud droplets. (If hydrophilic, they provide a surface so that the initial droplet can have less internal pressure from surface tension; perhaps much more important, if soluble, they provide a solute that can decrease the equilibrium vapor pressure, and more so at high concentrations - such as occurs when the smallest amount of water vapor condenses on a large enough aerosol. Some particles(such as salt) are considered hygroscopic because they pull moisture out of the air even when below 100% relative humidity. Moist particles go through a haze particle phase in which increasing size increases the relative humidity (relative to a flat surface of pure water) necessary to keep it in equilibrium (because the size increase dilutes the solute); once enough particles per unit volume reach some size where the necessary supersaturation for equilibrium decreases with increasing droplet size (due to the decrease in internal pressure from the surface tenstion), those particles continue to grow as cloud droplets, bringing the relative humidity back to about 100 %, and causing the evaporation of remaining haze particles. See also "Kohler curve".) Some are more or less effective at nucleating ice crystals. When the temperature gets down toward -40 deg C, homogeneous nucleation of ice within liquid water becomes more likely; the rate of homogeneous nucleation is a per unit volume rate, so larger droplets will freeze faster at higher temperatures than smaller droplets will. Larger droplets may also be more likely to contain an effect ice nucleus (?). When droplets hit ice particles, ice crystals may be nucleated. After some ice is nucleated, the surface of a droplet may freeze first, leaving liquid water within it, which can break the ice particle into smaller pieces when it freezes and expands. Different size particles have different terminal velocities and fall at different rates, so they may run into each other and sometimes combine. (Turbulence might have a local centrifuge effect that would amplify the differential motion of particles, though I'm not sure this is significant). Ice surfaces have a lower saturation vapor pressure at a give temperature than liquid water surfaces, so in the presence of a few ice particles, many small liquid droplets tend to evaporate as water vapor diffuses and is deposited to form a few larger ice particles that can then fall at a higher speed.)
129363. Patrick 027 at 10:09 AM on 31 May 2009
        It's the sun
        "Patrick 027 468 Thermalized means that the energy is mostly conducted to adjacent non-GHG molecules that do not significantly re-radiate. "When a radiant flux is absorbed, it is assumed to be mostly thermalized" Where then does all that radiation back to the surface come from?" I explained that before when I explained what thermalization is. When a GHG molecule absorbs a photon, in most cases, it exchanges energy with other molecules before it would radiate a photon. This keeps the different gaseous substances in the same unit volume of air at about the same temperature in spite of different optical properties. But GHG molecules also radiate photons. They radiate photons, in each wavelength interval in each direction, at a temperature-dependent rate that statistically follows blackbody radiation as a function of temperature, multiplied by the emission cross section of each molecule. After a GHG molecule radiates a photon, it has less energy, but through molecular interactions, energy tends to be redistributed to keep all the gaseous substances near the same temperature in each unit volume. Hence, the air as a whole loses energy when the GHG molecules (and clouds, etc.) it contains emit photons. The net change in the enthalpy of the air (enthalpy is better to use with air because it expands as it gains heat, so that some of the energy actually does work, which increases the heat capacity (at constant pressure) relative to a constant volume value.) is caused by the absorbed radiant energy minus the emitted radiant energy, plus latent heat gain (if the latent heat present in water vapor is not counted toward the enthalpy of the air), and if within the first mm or so of the surface, conduction from the surface (and diffusion of water vapor from a wet surface, if water vapor latent heat is counted toward the enthalpy of the air). "If the K&E graphic was not misleading there would be no need for all the clarification/explanation/rationalization." Well, maybe it was intended for other scientists who would recognize what it means without much need to figure it out.
129364. Patrick 027 at 09:54 AM on 31 May 2009
        It's the sun
        Dan - "A familiar example of a feedback system is the cruise control for a car. When working properly, it controls on speed and the speed of the vehicle is held fairly constant. Now if the speed changes substantially one must conclude that the unit is not controlling on speed. That is, there is no feedback from speed." That doesn't quite make sense - what if there was positive feedback - then the speed could certainly change. ---- If the temperature goes up and down without forcing (which does happen - look at the all the wiggles about the longer-term trend of the last several decades), that doesn't mean that there are no positive feedbacks. (There are fluctuations in cloud cover patterns, humidity patterns, and evaporation and latent heating and temperature patterns associated with variability in circulation patterns in the atmospheres (which can also contribute to surface albedo fluctuations), so there are both thermal and mechanical (momentum) feedbacks through which circulation patterns cause changes in circulation patterns - but over time there tends to be a predictable statistical description of all that (which is the longer-term climate)... Heat can also be added to or removed from the surface environment by vertical transports in the ocean.) As I was explaining in the last comment to you, I think I saw where a major miscommunication may lie. In equilibrium (for the longer-term average over internal variability), the radiant energy out equals the radiant energy in. A solar TSI increase is a positive forcing, increasing the radiant energy input. An increase in CO2 is a positive forcing, decreasing the radiant energy output. In climatology, the equilibrium response without any feedbacks is generally described as the temperature increase required to change, directly through blackbody emission's dependence on temperature, the radiant energy output, so as to restore balance. However, I suspect that in Control theory, that is itself considered feeback. Remove that feedback, and what happens? The rate of net energy storage is constant - if the heat capacity doesn't change, temperature increases linearly over time to infinity. That is the climate response without any feedbacks, including the feedback of blackbody radiation as a function of temperature (let's call it the blackbody radiation feedback). The approx. 1 K increase due to a doubling of CO2 includes that feedback. And in the absence of forcings, this feedback is a negative feedback that must be stronger than the net positive feedback from all other feedbacks if the climate is to be stable. Consider the water vapor feedback again. Using the same values as before (and using a linear approximation to responses): in response to a positive radiant forcing, the temperature rises. Without any feedbacks including the blackbody radiation feedback, the temperature rises indefinitely. But with the blackbody radiation feedback, their is an equilibrium temperature increase of 1 K, and the difference between the equilibrium temperature and the temperature decays exponentially at a rate determined by the climate sensitivity and the heat capacity. If the water vapor feedback increases the equilibrium temperature change from 1 K to 2 K, then this implies that the water vapor response to a 2 K temperature change is sufficient to force a 1 K temperature increase - it would cause indefinite warming if the blackbody radiation feedback did not stop it. It might help to use some time-dependent formulas: Let T be the change in temperature from some reference value. Let R be the radiant forcing difference from some reference level. Let the blackbody radiant feedback be B, where B = b*T, and assume b is negative. Let F be the net of all other radiant feedbacks to the change in temperature, where F = f*T. Let the heat capacity (per unit area) be C. Time is t. Conservation of energy: dT/dt = (R+F+B)/C dT/dt = [R + (f+b)*T]/C dT/dt = R/C + T*(f+b)/C ------------------------------- For R = constant, Let T = Teq - A*exp(-Z*t) A*exp(-Z*t) = Teq - T then dT/dt = Z*[A*exp(-Z*t)] dT/dt = Z*(Teq - T) IF the climate is stable, Z is greater than 0. --- dT/dt = Z*Teq - Z*T and dT/dt = R/C + T*(f+b)/C --- therefore Z*Teq = R/C Z*T = -T*(f+b)/C --- Z = -(f+b)/C Z = R/(C*Teq) e-folding time of (Teq-T) = 1/Z = C*Teq/R = C * climate sensitivity -(f+b)/C = R/(C*Teq) -(f+b) = R/Teq climate sensitivity = Teq/R = -1/(f+b) Teq = -R/(f+b) ------------------------------- IN SUMMARY (where R, T, and Teq are defined as relative to reference values): Whatever has happened up to time t, if R is set to a constant value, then: If the climate is stable, there is a constant equilibrium T = Teq, of the same sign as R, that T tends to approach over time through the exponential decay of (Teq - T). This exponential decay has a time scale (a time constant, or e-folding time) of 1/Z = C * Teq/R, which is the product of the heat capacity (per unit area) and the climate sensitivity. Teq = -R/(f+b) 1/Z = C * Teq/R = -C/(f+b) ------ (Teq-T) = exp[-Z*t] = exp[ -t / (C * Teq/R) ] = exp[ t*(f+b)/C ] ------ The climate is stable if the climate sensitivity is either zero or positive and finite. The climate sensivity, defined as R/Teq, is equal to -1/(f+b), where B = b*T is the blackbody radiant feedback and b is negative, and F = f*T is the net effect of all other feedbacks. ------------------------------- An increase in R from an initial value to a new value shifts Teq. If (Teq-T) was zero before the change in R, it will not be immediately zero after the change; (Teq-T) will then exponentially decay to zero. Notice that what happens to (Teq-T) is the same if, instead of changing the external radiative forcing R, there is an unforced change in T. An unforced change in T from Teq will tend to exponentially decay over time if the climate is stable. A stable climate is compatible with a positive value of f, provided that f is not greater than the absolute value of b. When climatologists say the net feedback is positive, they are refering to f; when they state the climate sensitivity Teq/R without feedbacks, they mean with f = 0; they still include the effect of a negative b. The climate sensitivity with f = 0 is -1/b ---------- Revisiting an earlier example, considering water vapor as the only contributor to f: Let -1/b = 0.25 K / (W/m2) (which means that a 0.25 K increase in temperature increases the outgoing radiation by 1 W/m2 if f=0; b = -1/0.25 (W/m2)/K = -4 (W/m2)/K) Teq = 0 K when R = 0 W/m2 (these are both relative to a reference or baseline state). Then imposing a forcing R of 4 W/m2 increases the Teq by 1 K if f=0. If T = Teq before R was changed from 0 W/m2 to 4 W/m2, (Teq-T) will equal 1 K before T responds. As (Teq-T) decays to zero from 1 K, T rises from 0 to 1 K, at a rate depending on C and R/Teq for f=0. Before T has changed, C*T is increasing at a rate R = 4 W/m2. But for each 0.1 K increase in T, the radiative feedback -B increases 0.4 W/m2,But when T has increased 0.5 K, that rate has dropped to R+B = R+b*T = (4 - 2) W/m2 = 2 W/m2. For each halving of (Teq-T), the rate of energy gain is halved. Hence, if C is constant, (Teq-T) decays exponentially. The time for each halving of (Teq-T) is proportional to C and proportional to Teq/R. T (K), B (W/m2), R+B (W/m2), Teq-T (K), 0.0, -0.0, 4.0, 1.0 0.1, -0.4, 3.6, 0.9 0.2, -0.8, 3.2, 0.8 0.3, -1.2, 2.8, 0.7 0.4, -1.6, 2.4, 0.6 0.5, -2.0, 2.0, 0.5 0.6, -2.4, 1.6, 0.4 0.7, -2.8, 1.2, 0.3 0.8, -3.2, 0.8, 0.2 0.9, -3.6, 0.4, 0.1 1.0, -4.0, 0.0, 0.0 ----- Let f = 0.5 * (-b) = 2 (W/m2)/K. Thus, as T increases 0.1 K, there is a net radiative feedback (except for B) of F = 0.2 W/m2. F+B for each 0.1 K increase is half of B alone, F+B = -0.2 W/m2. Because f+b has been halved, Teq will double to 2 K for R = 4 W/m2. Because F and B are both 0 when T = 0 K, the initial rate of T increase is the same regardless of what f and b are; C*T increases initially at a rate R = 4 W/m2. Since halving the sum f+b doubles Teq/R, T has to change twice as much, so it makes sense that the climate response time will now be twice as long (PS this makes it hard to determine climate sensitivity based on the earliest state of a climate response to a change in R). T (K), B (W/m2), B+F (W/m2), R+B+F (W/m2), Teq-T (K), 0.0, -0.0, -0.0, 4.0, 2.0 0.1, -0.4, -0.2, 3.8, 1.9 0.2, -0.8, -0.4, 3.6, 1.8 0.3, -1.2, -0.6, 3.4, 1.7 0.4, -1.6, -0.8, 3.2, 1.6 0.5, -2.0, -1.0, 3.0, 1.5 0.6, -2.4, -1.2, 2.8, 1.4 0.7, -2.8, -1.4, 2.6, 1.3 0.8, -3.2, -1.6, 2.4, 1.2 0.9, -3.6, -1.8, 2.2, 1.1 1.0, -4.0, -2.0, 2.0, 1.0 1.1, -4.4, -2.2, 1.8, 0.9 1.2, -4.8, -2.4, 1.6, 0.8 1.3, -5.2, -2.6, 1.4, 0.7 1.4, -5.6, -2.8, 1.2, 0.6 1.5, -6.0, -3.0, 1.0, 0.5 1.6, -6.4, -3.2, 0.8, 0.4 1.7, -6.8, -3.4, 0.6, 0.3 1.8, -7.2, -3.6, 0.4, 0.2 1.9, -7.6, -3.8, 0.2, 0.1 2.0, -8.0, -4.0, 0.0, 0.0 The rate of change remains proportional to Teq-T, but per unit Teq-T is half what it was when b+f had twice its value and Teq was only 1 K. _____________________________
129365. dnipper at 03:43 AM on 31 May 2009
        Climate's changed before
        Everyone is talking about GGs and other "causes" of GW. I am a service tech for an electrical company and have learned that you can either fix the problem, or fix the symptoms. From what I have read, there is evidence that GG's are indeed caused by GW, which would make them a symptom, much like a fever to a viral infection. Has any research been done to determine what caused the sudden and dramatic reversals in temperatures in the past? If so, where are we in relation to the trigger(s) today? Are we doing any good what so ever by treating the symptoms, or are we actually doing harm?
        Response: CO2 is both the symptom and the cause. We are emitting CO2 into the atmosphere, which warms the temperature, which causes the warming ocean to outgas more CO2. This is the feedback loop which caused the dramatic warmings in the past that took the planet out of the ice ages into interglacial periods. Eg - warming in the Southern Ocean (due to changes in the Earth's orbit) led to outgassing of CO2 which amplified the warming and spread it through the rest of the planet as the CO2 mixed through the atmosphere. Your question is addressed in detail at the CO2 lags temperature page.
129366. Dan Pangburn at 03:16 AM on 31 May 2009
        It's the sun
        Patrick 027 468 Thermalized means that the energy is mostly conducted to adjacent non-GHG molecules that do not significantly re-radiate. "When a radiant flux is absorbed, it is assumed to be mostly thermalized" Where then does all that radiation back to the surface come from? If the K&E graphic was not misleading there would be no need for all the clarification/explanation/rationalization.
129367. Patrick 027 at 03:06 AM on 31 May 2009
        It's the sun
        Important correction to comment: http://www.skepticalscience.com/argument.php?p=19&t=474&&a=18#3338 I used the same symbol for two different variables (transmissivity (unitless value) and temperature (units K or some other unit on an absolute scale), which caused confusion: P = (e*ec*T)*BC*A(T^4 - Tc^4) What variable shall we use for transmissivity? How about N: P = (e*ec*N)*BC*A(T^4 - Tc^4) Then let U = e*ec*N*BC*A The rest was correct: P = U * (T^4 - Tc^4) This can be rewritten as: P = U*T^4 - U*Tc^4 Original point still valid. Units are correct. Inteligent people sometimes make simple slip-ups like that. This had the significance of a typo - I think I had meant to use a different variable and must have absent-mindedly used called transmissivity T. However, anyone reading that could have figured out what I was trying to communicate and made the correction for their self. And yes, I have been known to substitute 'their' for 'there', as might be noticed in prior comments.
129368. Dan Pangburn at 02:54 AM on 31 May 2009
        It's the sun
        Patrick 027 464 Earth’s climate can be evaluated as a dynamic system for which the science of Control Theory applies. That means that all of the minutia of climate and weather, all of that stuff; forcings, Climate Science type feedbacks, GHG spectral absorption, ocean turnover, plant respiration, atmospheric carbon dioxide level, methane level, etc. etc. everything that Climatologists attend to (including all that you have described in excruciating detail) and even those things that haven’t been discovered yet; everything gets lumped together in a block in the Control Theory block diagram called ‘all internal factors that can alter average global temperature’. The only requirement is that none of them can add any energy to the block. The output from this block is average global temperature (agt). In Control Theory, feedback is a dimensionless number that is proportional to a change in output. In global climate it is a measure of the effect that change to agt has on how effective change to the energy entering the global climate system is at influencing agt. Feedback is a number that is added to 1.000 and the sum is multiplied by the input to the global climate system. Thus a feedback of 0.01 means that the input to the global climate system is multiplied by 1.01. If the feedback is -0.01 the input to the global climate system is multiplied by 0.99, etc. Note that feedback in Control Theory is dimensionless while feedbacks in Climate Science have units. Temperature data are readily available for the last and previous glacial periods (see, e.g. the few numerical data at 414 and graphs in my pdf linked from http://climaterealists.com/index.php?tid=145&linkbox=true . Since the agt trend changes the direction of slope from down to up and vice versa, Control Theory determines that the temperature can not be controlled by temperature feedback. A familiar example of a feedback system is the cruise control for a car. When working properly, it controls on speed and the speed of the vehicle is held fairly constant. Now if the speed changes substantially one must conclude that the unit is not controlling on speed. That is, there is no feedback from speed. In this analogy, agt is analogous to vehicle speed. Since earth’s temperature has changed as shown by the accepted temperature data from the last glacial period, it shows that there can be no significant positive (Control Theory) feedback from temperature in earth’s climate. Since there can be no significant Control Theory feedback, there can be no significant net Climate Science feedback either because that would influence temperature. Without Climate Science net feedback none of the twenty or so Global Climate Models that the IPCC uses predict significant global warming.
129369. Gord at 19:02 PM on 30 May 2009
        It's the sun
        Mizimi - RE: Your Post #467 Your "simplified" model violated two fundamental Laws of Science and actual measurements! Enough said.
129370. Gord at 18:51 PM on 30 May 2009
        It's the sun
        Patrick - Re: your posts #461 and #472: "2. Heat Radiation between hot and colder objects" "P = e*BC*A(T^4 - Tc^4)" When both the hot object and colder object have emissivities e and ec that could be less than 1, and there is a layer between with transmissivity T: P = (e*ec*T)*BC*A(T^4 - Tc^4)" ---------------------- e*BC*A(T^4 - Tc^4) does not equal (e*ec*T)*BC*A(T^4 - Tc^4) PROOF: e does NOT equal e*ec*T and the units of P (watts) have now been changed...by Patrick !!! All justified by Patrick's delusional OPINION. ------------ Mindless drivel that ANY algebra student could OBVIOUSLY see. Only a total "incompentent" and/or someone with "delusions of grandeur" would consider "modifying" the Stefan-Boltzmann Law and the units of Watts. Absolutely Hilarious! Adios, Patrick....you have "QUIT"...finally!
129371. Patrick 027 at 14:46 PM on 30 May 2009
        It's the sun
        " I might as well be your science tutor." By the way, I QUIT.
129372. Patrick 027 at 14:45 PM on 30 May 2009
        It's the sun
        Gord - Let's get this straight - Considering the approx. 33 K difference between the approx. 255 K surface temperature with no greenhouse effect and the approx. 288 K surface temperature with the greenhouse effect about as much as it is (or recently has been): 1. I correctly note that this 33 K warming is from the greenhouse effect, without any change in albedo, and I agree with you (as if I didn't know before) that albedo would change if the atmosphere in it's entirety is removed. 2. NASA makes a mistake in their explanation, implying that removal of the atmosphere would leave the albedo unchanged, or that the 33 K warming comes from the greenhouse effect minus the cooling from the albedo increase, which is of course not true. I know that NO atmosphere means NO atmosphere, but I, assuming I am not more knowledgable than NASA, give them the benifit of the doubt, and choose to think that they did not actually intend what they mistakenly wrote. That should be quite clear to you, and you should have no reason to think I don't understand words. 3. You seem to hold NASA's error against me or against climatology in general - as if it were up to me (or Kiehl and Trenberth (okay, maybe more likely), or...) what NASA says - and at the same time, tell me I have delusions of grandeur, when I said that I assumed that I was not more knowledgable on the subject than the people at NASA, and so chose to assume that they actually know that the 33 K warming is from the greenhouse effect with no change in albedo and that they were just sloppy in their writing. (I still do not think I am smarter or more knowledgable than the people at NASA in general - however I do think I much more knowledgable than you - you keep giving me evidence for that.) 4. For NASA's mistake, you seem to suggest that I am wrong about the greenhouse effect. I have offered over the last several pages of comments and references therein a very good description of the greenhouse effect, but you look for erroneous descriptions from other sources to argue against me, to argue against things - things that I also agree are erroneous. 5. Your entire point seems to be that there is no greenhouse effect. You were arguing against Kiehl and Trenberth's energy budget, and they do not make any mistake in assuming that the atmosphere contributes to albedo. I have made no mistake in that regard either. You also seem to be aware that there is some significant albedo. Yet you offer a calculation of temperature with zero albedo as support for your argument that there is no need to use the greenhouse effect to explain the temperature of the surface of the Earth. BOTTOM LINE: The albedo of the Earth, including atmospheric effects - is about 0.3. With that albedo, but in the absence of the greenhouse effect, the average temperature of the surface would be about 255 K, or possibly slightly colder. The average temperature of the surface is actually near 288 K. Meanwhile, the emission of radiation to space can be observed and it is less than the radiation that the surface would emit to space if it were completely exposed. The difference is especially associated with high cold cloud tops, and at wavelengths where absorption by atmospheric gases - H2O and CO2 in particular, is expected to be sizable based on known optical properties. ---- "It is very, very tedious explaining things to you and backing up my statements with links." Ha! I (perhaps quite unwisely) took the time to go over your mistaken ignorant babbling more or less point by point; I've used as pedantic an argument I ever would and then used another, I've explained the microscopic basis of macroscopic phenomena such as the second law of thermodynamics. I've explained here and offered references to other comments explaining just how the greenhouse effect and radiative energy transfer actually work (in a way that is easy to visualize, hence my occasional anthropomorphism, as in 'what the surface can see'), at times using the examples you offered. I've gone over the math of blackbody radiation. I might as well be your science tutor. And for the most part, all you ever do is repeat a bunch of ill-thought out, ill-informed assumptions that most serious physicists would laugh at.
129373. Patrick 027 at 14:32 PM on 30 May 2009
        It's the sun
        Gord - Let's get this straight - Considering the approx. 33 K difference between the approx. 255 K surface temperature with no greenhouse effect and the approx. 288 K surface temperature with the greenhouse effect about as much as it is (or recently has been): 1. I correctly note that this 33 K warming is from the greenhouse effect, without any change in albedo, and I agree with you (as if I didn't know before) that albedo would change if the atmosphere in it's entirety is removed. 2. NASA makes a mistake in their explanation, implying that removal of the atmosphere would leave the albedo unchanged, or that the 33 K warming comes from the greenhouse effect minus the cooling from the albedo increase, which is of course not true. I know that NO atmosphere means NO atmosphere, but I, assuming I am not more knowledgable than NASA, give them the benifit of the doubt, and choose to think that they did not actually intend what they mistakenly wrote. That should be quite clear to you, and you should have no reason to think I don't understand words. 3. You seem to hold NASA's error against me or against climatology in general - as if it were up to me (or Kiehl and Trenberth (okay, maybe more likely), or...) what NASA says - and at the same time, tell me I have delusions of grandeur, when I said that I assumed that I was not more knowledgable on the subject than the people at NASA, and so chose to assume that they actually know that the 33 K warming is from the greenhouse effect with no change in albedo and that they were just sloppy in their writing. (I still do not think I am smarter or more knowledgable than the people at NASA in general - however I do think I much more knowledgable than you - you keep giving me evidence for that.) 4. For NASA's mistake, you seem to suggest that I am wrong about the greenhouse effect. I have offered over the last several pages of comments and references therein a very good description of the greenhouse effect, but you look for erroneous descriptions from other sources to argue against me, to argue against things - things that I also agree are erroneous. 5. Your entire point seems to be that there is no greenhouse effect. You were arguing against Kiehl and Trenberth's energy budget, and they do not make any mistake in assuming that the atmosphere contributes to albedo. I have made no mistake in that regard either. You also seem to be aware that there is some significant albedo. Yet you offer a calculation of temperature with zero albedo as support for your argument that there is no need to use the greenhouse effect to explain the temperature of the surface of the Earth. BOTTOM LINE: The albedo of the Earth, including atmospheric effects - is about 0.3. With that albedo, but in the absence of the greenhouse effect, the average temperature of the surface would be about 255 K, or possibly slightly colder. The average temperature of the surface is actually near 288 K. Meanwhile, the emission of radiation to space can be observed and it is less than the radiation that the surface would emit to space if it were completely exposed. The difference is especially associated with high cold cloud tops, and at wavelengths where absorption by atmospheric gases - H2O and CO2 in particular, is expected to be sizable based on known optical properties.
129374. Gord at 13:24 PM on 30 May 2009
        It's the sun
        Patrick - Re: Your Post #459 Right from the NASA website: The Greenhouse Effect "Scientists have long known that the presence of an atmosphere keeps the surface of the planet warmer than it would be without an atmosphere. In fact, without an atmosphere, the surface of the earth would be about 30 degrees Celsius cooler than it is now!" http://earthobservatory.nasa.gov/Experiments/PlanetEarthScience/GlobalWarming/GW_Movie3.php ------------------ The Earth with an atmosphere has an albedo of 0.3, without an atmosphere it will be less than 0.3. I chose an albedo of zero and like I have I have REPEATELY said this number is as valid as the 0.3 assumption in the calculation. --------------- Patrick said... "...but I would think the people at NASA are at least as knowledgable on the subject as I am..." Obviously you have "delusions of grandeur" considering your total lack of scientific knowledge and your complete inability to understand what "NO ATMOSPHERE" means. It is very, very tedious explaining things to you and backing up my statements with links. Your posts are all rife with obvious errors and hilarious opinions that are totally unsupported. In short, your posts show a total ignorance of the subjects you babble about.
129375. Patrick 027 at 03:05 AM on 30 May 2009
        It's the sun
        RE 466 Dan - "I agree that some is. K&E doesn't show any. " When a radiant flux is absorbed, it is assumed to be mostly thermalized; K&E implies thermalization whereever they show absorption. It is not misleading. In as far as the fluxes going long distances - this is not a diagram that is meant to be to scale - it is a schematic diagram. (that should be readily apparent - the atmosphere is shown as a layer that is detached from the surface, which appears flat. The solar flux is shown bending and bouncing off of things in an unrealistic manner. But it is correct (or approximately correct) for what it is.) It is only misleading if you misinterpret it.
129376. George Curzon at 01:25 AM on 30 May 2009
        It's cosmic rays
        The point is how much heat arrives on the surface of the earth for what duration. If you leave water in a pot on the stove, and do with the the heat control a mock-up of the suns behaviour over the past 1100 years, the water will get hottest when you turn the heat up and leave it up. The water will not get hot hot when you give irregular bursts of heat, then turn it down again, or no heat. That temperatures continue to rise when sunspots level off is analogous to the water continuing to heat when you leave the stove turned on. Perhaps a pig on a spit might be a sizzling analogy for some. Until our planet reaches the temperature of the surface of the sun as it would be at this solar radius, we are going to continue heating up. Hopefully, sunspots will drop off before then. The 'exact' set interacting mechanisms of cloud formation would be interesting to know, and must affect the amount of the suns energy arriving at what the oil-well-country-occupying military types refer to as 'ground zero', eh?
129377. Mizimi at 22:59 PM on 29 May 2009
        Water vapor is the most powerful greenhouse gas
        AldousH: I have read the article summary and in general terms agree with the conclusions. However it does seem to concentrate on deforestation as the prime mover and disregards other matters such as WV emissions from industrial processes, increased evaporation from man-made dams, lakes etc. Whilst these may be (relatively) small, they are not inconsequential; neither is the increase in heat and WV from animal life.
129378. Mizimi at 22:32 PM on 29 May 2009
        It's the sun
        I am aware of the complexities of radiative energy transfer...my 'model' was, as stated, simplified to demonstrate the principle that GG's simply affect the rate at which heat is transferred from the earth to space. Semantics possibly, but it is important we all ( and those who come here seeking clarification of a comlex subject) understand the terms expressions and concepts used.
129379. AldousH at 19:31 PM on 29 May 2009
        Water vapor is the most powerful greenhouse gas
        Hello guys. Has anyone considered the basis of a biological understanding of global warming? For instance your assumptions about the behaviour of water vapour are explained in terms of temperature, evaporation and condesation. I have read an interesting article from CSIRO SUSTAINABILITY NETWORK UPDATE – No. 64E is title "The biology of global warming and its profitable mitigation" and written by Dr Walter Jehne. I have a MSWord copy of the article though do not know how to added it to the Articles list.
129380. Dan Pangburn at 14:59 PM on 29 May 2009
        It's the sun
        Patrick 027 458 "Generally, almost all radiation that is absorbed is thermalized." I agree that some is. K&E doesn't show any. That is part of what is misleading. The rest of what is misleading is they show big numbers going from surface to clouds and clouds to surface. It misleads by showing all radiation going a long distance instead of being absorbed close to the surface by greenhouse gases.
129381. Patrick 027 at 14:34 PM on 29 May 2009
        It's the sun
        "So what if there were some unforced temperature increase of 2 K? Water vapor would increase, but it would not be strong enough to prevent the temperature from falling back" To be more clear on the matter: Using the assumption of constant climate sensitivity for the sake of simplifying the argument: If a radiative forcing is able to sustain a 1 K increase without feedback, and the water vapor feedback increases the equilibrium change to 2 K, then that implies that the water vapor response to a 2 K increase is sufficient to sustain 1 K of the 2 K increase in temperature on it's own. Thus, if there is an unforced 2 K increase, the water vapor level will increase so as to support 1 K of that increase. Thust the temperature drops perhaps half as fast as it would without the water vapor feedback, but it drops because the water vapor feedback cannot sustain it fully. As the temperature drops, the water vapor drops back, so the temperature doesn't stop dropping as it approaches 1 K - by that point, the water vapor can only sustain 0.5 K, so the temperature keeps dropping. The temperature and water vapor eventually return (or approach) there initial values.
129382. Patrick 027 at 14:26 PM on 29 May 2009
        It's the sun
        Dan - the climate models that simulate longer-term warming also produce some short term flat and cooling periods such as now. Variations in PDO, ENSO, AMO, etc (modes of internal (unforced) variability), could be contributing to these variations, as could some solar effects, though the evidence and physics backing such things, or non-TSI related solar effects, as major causes of the longer-term temperature increase is not at all as strong and solid as that which supports the known forcings in their relative proportions - anthropogenic greenhouse forcings being the biggest warming contribution and anthropogenic aerosols in total being the biggest cooling contribution, but with the total (net) forcing being positive (causing warming) and significant, and increasing. "In Control Theory, a positive feedback means that the trend will continue in whichever direction it is going (but at a declining rate because of the fourth power of absolute temperature)."..." That is, if it is going down, it will keep going down approaching a new level asymptotically and if it is going up, it will keep going up approaching a new level asymptotically." The second part of that actually fits with climatological terms; some change in forcing pushes the climate into a different state; feedbacks amplify or reduce the change, but there is some new equilibrium state the climate tends to approach. The first part, however, is different. Feedbacks in climate do not cause continuing change (in the long-term state, as opposed to weather, interannual variability, etc.) just by existing (they can cause continuing change by reacting to continuing change such as in internal variability). The equilibrium state the climate system finds is a new constant global average temperature - there will be fluctuations about that temperature but the longer term average is a constant. "The only way that the trend direction can change is if there is an external influence that is strong enough to overpower the feedback." A forcing shifts the equilibrium that the climate tends to approach; feedbacks modify that shift but still result in an equilibrium that the climate tends to approach. The long-term equilibrium is not itself a trend but a fixed state. Any forcing at any time can shift that equilibrium and if that forcing plus the feedbacks result in a shift that is in the opposite direction as some previous shift, the climate's tendency could reverse if the climate has already shifted far enough in response to a previous change. Climate sensitivity is the change in equilibrium climate per unit change in forcing and is thus modified by feedbacks. However, unforced changes always reverse due to negative feedbacks - unless their are multiple equilibria, or the climatic equilibrium is a strange attractor, etc, - but in those cases, the short term variability that results can still be encompassed within a full description of a yet longer-term equilibrium climate state. Maybe that is the distinction we need to go over. For example, water vapor: Water vapor is most likely a strong positive feedback (PS I may have been wrong earlier when I identified surface albedo via snow and ice changes as also being a quite strong positive feedback - it is strong regionally and is one of the more obviously observed, but it's contribution to the global average ... I'm not sure offhand). What is meant by that? The amount of water vapor in the atmosphere tends to approach an equilibrium value over a period of several days. The equilibrium value increases with increasing temperature. If there is a climate forcing - say, an increase in solar TSI or an increase in CO2 greenhouse forcing, as this starts to increase the temperature, the temperature will rise, tending to exponentially approach a new equilibrium temperature in which the change in temperature causes a change in LW radiative cooling that balances the imposed radiative forcing. The time it takes for this to occur is proportional to the heat capacity per unit area of the climate system (PS that varies depending on time scale but let's set that issue aside for now), and the equilibrium temperature increase per unit radiative forcing without feedbacks - because the radiative forcing is an energy supply rate per unit area, and the temperature change times the heat capacity per unit area is the amount of energy that needs to be gained in order to reach equilibrium. But as the temperature rises, the water vapor content also starts to rise, and this is a positive radiative feedback - it has a radiative 'forcing' that adds to the initial radiative forcing, so the total temperature change necessary to reach equilibrium increases. Interestingly, this also increases the time it takes to reach equilibrium, because the water vapor feedback to the change is initially zero and increases as the temperature increases. (With some calculus, I could show that the time constant to exponentially approach equilibrium is proportional to the climate sensitivity, which is the equilibrium change per unit externally imposed forcing). But what happens when the climate is at equilibrium? Remember, water vapor tends to be a function of temperature. Thus, an unforced change in water vapor decays to zero over a short time. There is some radiative effect that raises the temperature, but the time it takes for a significant temperature increase is too long compared to the time that water vapor itself changes in response to temperature. And if there is an unforced temperature change - well, that would be like having some forcing and then ending it. In other words, if CO2 is added to cause a temperature increase of 1 K and the water vapor feedback causes an additional increase of 1 K (not sure offhand if that is ther right proportion - this is just to illustrate the concept) - then what happens when the CO2 change is reversed? Well, there is a roughly 1 K decrease in temperature, and the water vapor feedback results in an additional 1 K decrease, returning the climate to where it was before the CO2 increase. Climate sensitivity could vary over temperature, but even if the water vapor feedback is stronger or weaker at the 2 K higher temperature, the senstivity will go through the same range on the way down as it did on the way up; the result is still a return to the initial value. So what if there were some unforced temperature increase of 2 K? Water vapor would increase, but it would not be strong enough to prevent the temperature from falling back - it would only delay the return to the initial temperature. In that sense, the climate, when not being forced to change, could be dominated by negative feedbacks - mainly the negative feedack of the increased radiative cooling in response to an increase in temperature. This is often not called a feedback because it is part of the climate response to forcings and other radiative feedbacks (you have to watch for context - sometimes feedbacks are described as having forcings, etc...). Of course, when the full complexity of climate is considered, there is a possibility of hysteresis loops, and things that do not respond fast to temperature over short periods of time may do so over longer periods of time. Externally-imposed forcings are boundary conditions. What can be approximated as a boundary condition in the short term may shift to being part of system behavior in the long term...
129383. Patrick 027 at 12:57 PM on 29 May 2009
        It's the sun
        "In climatic equilibrium, the average of each rate of energy gain is zero." I mean the average of one is zero and the average of the other is zero, not just the average of both.
129384. Dan Pangburn at 12:57 PM on 29 May 2009
        It's the sun
        Patrick 027 453 The assessment is presented at my pdf linked from http://climaterealists.com/index.php?tid=145&linkbox=true . This is entirely different from the usual Control Theory application in that it shows only that the feedback (as defined in Control Theory) from average global temperature can not be significantly positive. Wikipedia has fairly acceptable definitions for feedback as used in Global Warming and for feedback as used in Control Theory. In Control Theory, a positive feedback means that the trend will continue in whichever direction it is going (but at a declining rate because of the fourth power of absolute temperature). That is, if it is going down, it will keep going down approaching a new level asymptotically and if it is going up, it will keep going up approaching a new level asymptotically. The only way that the trend direction can change is if there is an external influence that is strong enough to overpower the feedback. Since in the paleo data the temperature trends are seen to change direction, it means that there had to be an external influence that was greater than the feedback and the feedback could not be significantly positive. Thus the climate sensitivity predicted by the GCMs should be the value that they calculate for zero net (Climate Science) feedback (1.2°C). I personally think that it will turn out to be less than 1.0. An explanation for the 20th century temperature rise was presented at 371 except I said Maunder Minimum (which refers to sunspot count) when I should have said Little Ice Age (which refers to temperature). The end result corroborated qualitatively the Grand Solar Maximum (which appears to have ended) that others talk about. The comparatively weak in effect on temperature GSM combined with a comparatively strong in effect on temperature PDO (they typically last about 30 years) uptrend to produce the temperature run up from about 1975 to about 2005. The last decade has seen no significant change in average global temperature. There are five agencies that report average global temperature. I averaged them all each year from 1998 through 2008 and the trend is flat. The trend from 2002 through 2008 is significantly down (1.88 °C/century) but it probably won’t stay that steep.
129385. Patrick 027 at 12:55 PM on 29 May 2009
        It's the sun
        Gord - your comment 457: "1. The 2nd Law of Thermodynamics" "Energy will not flow spontaneously from a low temperature object to a higher temperature object." Net energy. Otherwise the website contradicts itself with statements about net radiation loss. "2. Heat Radiation between hot and colder objects" "P = e*BC*A(T^4 - Tc^4)" When both the hot object and colder object have emissivities e and ec that could be less than 1, and there is a layer between with transmissivity T: P = (e*ec*T)*BC*A(T^4 - Tc^4)" Or let U = e*ec*T*BC*A, so that P = U * (T^4 - Tc^4) This can be rewritten as: P = U*T^4 - U*Tc^4 (More generally, e, ec, and U could all vary with wavelength, and wavelength-dependent emission is a more complicated function of temperature, thout at all wavelenght an increase in T results in an increase in emission if e is constant.) The two terms on the right are opposing energy fluxes - you would call them opposing fields and not consider their energy fluxes to be real in their own right. However, in restating this formula, you have essentially admitted that you have no mathematical basis for denying the greenhouse effect, and in particular, backradiation from the atmosphere, which corresponds to the Tc term above in the fluxes between the surface and atmosphere. Your primary disagreement is a matter of labeling and, except for the microscopic basis of macroscopic phenomena, philosophical; You would have to accept that, at least mathematically, the greenhouse effect works as I and others say it does, even if you disagree about whether two opposing radiation fields can have two opposing energy fluxes. "3. The Law of Conservation of Energy (Energy cannot be created or destroyed)" "The atmosphere was heated by the Earth's radiation. The energy used to heat the atmosphere CANNOT be used to heat the Earth because it came from the Earth. If it did heat the Earth, energy would have been created....a violation of The Law of Conservation of Energy."..."It would also violate Electromagnetic Physics, Vector Mathematics etc. and ALL existing measurements." 1. Perhaps you've been confused by your use of the word 'heat' as a verb. Do you think that the temperature of the atmosphere rises when the atmosphere is 'heated' by the surface and the temperature of the surface rises when the surface is 'heated' by the atmosphere, but the temperature of either does not fall when it acts as the heat supply for the other? When radiation is absorbed by the atmsophere from the surface - if this happened in isolation, then aside from phase changes, the temperature of the surface would fall and the temperature of the atmosphere would rise. Energy is conserved. If the atmosphere radiates some energy that is absorbed by the surface, then if this happens in isolation, the temperature of the surface would rise, the temperature of the atmosphere would fall, and energy would be conserved. Of course, both, and other processes, are generally happening at the same time: Surface absorbs RSe from the sun and Rae from the atmosphere. Surface loses heat by Rea to the atmosphere, Res to space, and Cea (convection) to the atmosphere. The atmosphere absorbs RSa from the sun and Rea from the surface, and gains Cea from the surface, and loses energy by Rae to the surface and Ras to space. THESE EQUATIONS ARE BASED ON THE CONSERVATION OF ENERGY: Rate of energy gain by the surface = RSe + Rae - Res - Rea - Cea Rate of energy gain by the atmosphere = RSa + Rea + Cea - Ras In climatic equilibrium, the average of each rate of energy gain is zero. Where is energy being created or destroyed? ------- Oh, how tedious it is to explain anything to you, Gord.
129386. Patrick 027 at 12:23 PM on 29 May 2009
        It's the sun
        Gord - "First, if the atmosphere absorbed 360 w/m^2, it would radiate 360 w/m^2 in all directions...not half up and half down. (see Stefan-Boltzmann Law) http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html" First, I haven't checked it just today but I presume that portion of the "hyperphysics" website is accurate ( I can't say the same of what you think you learned from it), and would also encourage Mizimi to explore that website. True that Mizimi is incorrect in the assumption that the upward and downward radiative fluxes are equal (though it's an okay first guess) - they could be under some circumstances, but the upward flux from the atmosphere at the tropopause and the top of the atmosphere will generally be less than the downward flux at the bottom of the atmosphere to the surface. But why would the atmopshere emit 360 W/m2 up from it's top and down from it's base, given 1. emissions are a function of temperature, and not directly on how much radiation is absorbed, though their is certainly a causal link - and 2. there are variations in temperature within the atmosphere.
129387. Patrick 027 at 12:13 PM on 29 May 2009
        It's the sun
        Gord - "No, the thing that is very obvious is that they all say NO ATMOSPHERE!" I was giving them the benifit of the doubt - maybe I shouldn't do that for the wikipedia source, but I would think the people at NASA are at least as knowledgable on the subject as I am, so I figured they must have meant to refer to the atmospheric greenhouse effect, and not the effects of changing any atmospheric contribution to global albedo. I know that the 33 K warming effect is calculated for a greenhouse effect vs no greenhouse effect with the albedo being invariant at 0.3 in both cases. But you seem to be grasping at straws, because - you can calculate the temperature for no greenhouse effect for whatever albedo you like, but you haven't made any case that the albedo is significantly different than 0.3 in actuality. No, I'm not saying that you said that you did. My point is that you were trying to argue that there is no greenhouse effect. But you KNOW the albedo is 0.3. So what is the point of a calculation with an albedo of 0 or 0.1?
129388. Patrick 027 at 11:34 AM on 29 May 2009
        It's the sun
        Dan - I did notice the numbers had been adjusted, I just used the older numbers that are nearly the same to illustrate my points. I didn't mean to cause confusion. Surface LW emissivity - I had inferred a value of about 0.96 (or maybe 0.95) from another source (Hartmann, "Global Physical Climatology", 1994, p.28) - specifically, this source states a surface emission of about 376.2 W/m2, which, if the temperature is 288 K (at which, blackbody radiation ~= 390.1 W/m2 (1 place beyond significant figures; I am using sigma = 5.67e-8)), implies a LW emissivity of 0.964; - on the other hand, the global and temporal variation of surface temperature might boost the global average surface emission so that it would correspond to a temperature 1 K higher than the actual global average (because radiation varies with the fourth power of temperature, so the areas with T>average T add more to the global average emission than the areas with T
129389. Gord at 09:48 AM on 29 May 2009
        It's the sun
        Mizimi - You said.... "The atmosphere absorbs 360w And radiates 180w to space And radiates 180 back to earth Which absorbs this radiation and rises in temperaure." --- First, if the atmosphere absorbed 360 w/m^2, it would radiate 360 w/m^2 in all directions...not half up and half down. (see Stefan-Boltzmann Law) http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html Second, it is impossible for the colder atmosphere to transfer any heat energy back to the warmer Earth. It violates: 1. The 2nd Law of Thermodynamics "Second Law of Thermodynamics: It is not possible for heat to flow from a colder body to a warmer body without any work having been done to accomplish this flow. Energy will not flow spontaneously from a low temperature object to a higher temperature object." http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/seclaw.html#c3 2. Heat Radiation between hot and colder objects P = e*BC*A(T^4 - Tc^4) Where P = net radiated power (Watts), e = emissivity, BC = Stefan's constant, A = area, T = temperature of radiator in Kelvin and Tc = temperature of the surroundings or another body. http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html 3. The Law of Conservation of Energy (Energy cannot be created or destroyed) The atmosphere was heated by the Earth's radiation. The energy used to heat the atmosphere CANNOT be used to heat the Earth because it came from the Earth. If it did heat the Earth, energy would have been created....a violation of The Law of Conservation of Energy. It would also violate Electromagnetic Physics, Vector Mathematics etc. and ALL existing measurements. --------------------- Mizimi, yes there is a website that you can visit regarding the Solar Oven experiment. Here is a partial re-post of mine: --- Solar Cookers and Other Cooking Alternatives "The second area of solar cookers I looked at was their potential use for cooling. I tested to see how effective they are at cooling both at night and during the day. During both times, the solar cooker needs to be aimed away from buildings, and trees. These objects have thermal radiation and will reduce the cooling effects. At night the solar cooker needs to also be aimed straight up towards the cold sky. During the day the solar cooker needs to be turned so that it does not face the Sun and also points towards the sky. For both time periods cooling should be possible because all bodies emit thermal radiation by virtue of their temperature. So the heat should be radiated outward. Cooling should occur because of the second law of thermodynamics which states that heat will flow naturally from a hot object to a cold object. The sky and upper atmosphere will be at a lower temperature then the cooking vessel. The average high-atmosphere temperature is approximately -20 °C. So the heat should be radiated from the cooking vessel to the atmosphere." http://solarcooking.org/research/McGuire-Jones.mht --- This above link shows that heating cannot occur from the atmosphere. In fact, the article shows how to COOL items placed in the Solar Oven at NIGHT AND DAY! All you have to do is point the Oven away from the Sun during the Day and the Oven will transfer heat from the WARM object in the Oven to the COOLER atmosphere! It can even be used to produce ICE when the ambient air temp is +6 deg C! "If at night the temperature was within 6 °C or 10°F of freezing, nighttime cooling could be used to create ice. Previous tests at BYU (in the autumn and with less water)achieved ice formation by 8 a.m. when the minimum ambient night-time temperature was about 48 °F." This confirms the validity of 2nd Law of Thermodynamics....heat energy CANNOT flow from Cold to Warm objects. --- PS. This method is also used to measure "Back-radiation" indirectly. Ex. Indirect measurements measure the loss of energy (eg.Thermistor) to the cooler atmosphere.
129390. Gord at 09:46 AM on 29 May 2009
        It's the sun
        Patrick - You said.. "So they were sloppy in their explanations to the public on that point. It is obvious from the context that they are refering to the effect of the atmospheric greenhouse effect, seperate from atmospheric albedo contributions." --- No, the thing that is very obvious is that they all say NO ATMOSPHERE! "In the absence of the greenhouse effect and an atmosphere.." http://en.wikipedia.org/wiki/Greenhouse_effect "....without atmosphere." http://www.kowoma.de/en/gps/additional/atmosphere.htm "...it had no atmosphere." http://www.astronomy.ohio-state.edu/~pogge/Ast161/Unit5/atmos.html "...it didn’t have an atmosphere." http://earthobservatory.nasa.gov/Features/EnergyBalance/page6.php See how tedious it becomes to demonstate an obvious point to you? Anyone capable of understanding the English language and old enough to know what the atmosphere is would easily understand what these links said. Yet, for Patrick, the words "NO ATMOSPHERE" somehow has a different meaning. ---------- Patrick, as I have said numerous times, you just don't have clue about any "established science". But, that certainly does not prevent you from trying to continually disprove or re-write Laws of Science....all based on your hilarious opinions. For you 1 + 1 = anything but 2. It's a good thing that you are an amateur (and that's being very generous in your case). At least you are not a threat to public safety.
129391. Dan Pangburn at 08:52 AM on 29 May 2009
        It's the sun
        Patrick 027 447 Apparently you did not notice that nearly all of the numbers on the revised K&T chart are different from the numbers on their 1997 chart. For those who did not pursue the link regarding the K&T update, these are the comments that I made there. “Are they unaware that most absorption takes place close the emitting surface? The graphic is misleading. Still.” “The graphic shows the 356 going all the way to the clouds and from the clouds 333 all the way back to the ground. Since GHGs absorb the IR, the intensity has to decline along the way. Barrett calculates 72.9% is absorbed within 100 meters, http://www.warwickhughes.com/papers/barrett_ee05.pdf using the HITRAN database.” I am surprised that no one had challenged this chart before. Climatologists discovered long ago that some of the absorbed IR radiation is thermalized, that is, it raises the temperature of the air. Climate models didn’t work at all until convection (heat rises) was included in the mid 60s. “…in the real world by the upward convection of heat.” Spencer Weart at http://www.aip.org/history/climate/simple.htm . The chart clearly shows Surface Radiation 396 with 356 of this going clear to the clouds. It is misleading because first it’s closer to 382 than 396 and second there is no indication that some of this (I calculate 59) must get thermalized. The rest, except for the 40 that go all the way out, gets radiated back. The graphic also shows 333 coming from the clouds. This too is misleading since only a fraction of the radiation that leaves the clouds (my guess about 35) gets all the way from the clouds to the ground.
129392. Mizimi at 04:36 AM on 29 May 2009
        It's the sun
        A simplistic model often suffices better to demonstrate a point: At night the earth emits (say) 400w in radiation and its' temperature falls as a result. 10% goes straight out to space (40w) The atmosphere absorbs 360w And radiates 180w to space And radiates 180 back to earth Which absorbs this radiation and rises in temperaure. The net effect is that in that portion of time the earth disposed of 220w and thus the rate at which it cools is decreased by the presence of GG's. I don't know the exact details of the experiment to determine 'backradiation at night' but there are obviously some difficulties like - how did they ensure the dish wasn't warmer than the sky? Is there a website describing the experiment?
129393. paravantis at 04:11 AM on 29 May 2009
        It's Urban Heat Island effect
        Karl, Diaz and Kukla (1988), one of Peterson's (2003) references, carried out an extensive analysis of the effect of urban growth on temperature measures and found that "urbanization has influenced the climate records of even small towns in the US", especially for diurnal minima, means and range. Although from an econometric/modeling point of view I am not convinced that their analysis is top notch, I am equally unimpressed with the rather superficial analysis of the article at the top of this page. I think it is unlikely that there is NO Urban Heat Island effect with a detectable influence on temperature records. Luckily, I am currently supervising a pertinent doctoral dissertation at the University of Piraeus in Greece. With my very best regards to all, John Paravantis Assistant Professor University of Piraeus Greece
129394. Patrick 027 at 11:41 AM on 28 May 2009
        It's the sun
        Dan - "in not being aware that added atmospheric carbon dioxide has no significant effect on average global temperature" - the point is that I don't see how Control Theory justifies such a conclusion. It would help if you explained what you would expect to see from a system that has positive feedbacks; I see nothing to indicate that the climate sensitivity is not in the range where it is thought to be. (It would also help if you explained how, whatever the cause, the temperature rise of the last century and especially the last few decades can be explained without a positive feedback.)
129395. Dan Pangburn at 10:27 AM on 28 May 2009
        It's the sun
        Patrick 027 444 The point is that Control Theory shows that Climate Scientists, in claiming AGW, have made a mistake in not being aware that added atmospheric carbon dioxide has no significant effect on average global temperature. As a result, they have misled a lot of people. The important ones are the politicians who have been crippling the world economy and are contemplating crippling the US economy to solve a non-problem.
129396. Patrick 027 at 03:54 AM on 28 May 2009
        It's the sun
        Gord - --- " You quoted me as saying: "There is absolutely no evidence that the Earth had an albedo of 0.3 before the Earth had an atmosphere." What I actually said was: "There is absolutely no evidence that the Earth had an albedo of 0.3 before the Earth had an atmosphere, so my assumption of zero albedo is just as valid." " --- ... just as valid as some other erroneous assumption - speaking of which: About your quotes from: http://en.wikipedia.org/wiki/Greenhouse_effect http://www.kowoma.de/en/gps/additional/atmosphere.htm http://www.astronomy.ohio-state.edu/~pogge/Ast161/Unit5/atmos.html http://earthobservatory.nasa.gov/Features/EnergyBalance/page6.php So they were sloppy in their explanations to the public on that point. It is obvious from the context that they are refering to the effect of the atmospheric greenhouse effect, seperate from atmospheric albedo contributions. Re 450 - So any physics that is inconvenient for YOU is "rambling", full of "errors and unsupported opinions"? What else is new. (Go look it up in textbooks and encyclopedias, or "hyperphysics" - I'm really just telling you common knowledge. You *almost* might as well criticize me for stating 1+1=2 without backing it up with examples and/or proof.) If you find it so tedious, maybe you aren't qualified to make those judgements. One possible error I made was in not considering that interference patterns among different photons - whose wavefronts will tend to spread sideways as they propagate (diffraction) could affect the absorption (or emission?) of an individual photon even when the other photons continue unaffected; however, if/when this is the case (I presume it is possible, in part because of how a photons' own interference pattern with itself in a double slit experiment affects the probability distribution of where it is absorbed) would still occur via effects on local conditions. It also still doesn't offer a possible mechanism that might somehow someway prevent the absorption by a warmer object of photons from a cooler object, because a hot object with lower emissivitiy, especially at shorter wavelengths, would produce an identical photon population. But again, the second law of thermodynamics is saved because the emissivity of the cooler object, if in local thermodynamic equilibrium, is equal to its absorptivity (at each individual wavelength), so that however much it emits to a a warmer object that is absorbed by the warmer object, it still must absorb more from the emission by the warmer object. Now, if they are seperated by a distance and alternately covered and exposed, and/or change temperature, at some relatively fast rate compared to the time for radiation to travel the distance, then it's more complicated, but... ------------- You're tilting at windmills. It would be comedic if it weren't tragic and perhaps just a bit scary.
129397. Gord at 20:29 PM on 27 May 2009
        It's the sun
        Patrick - Your rambling post #448 is also rife with errors and a real hoot! Far too many errors and unsupported opinions to go into right now (and very tedious). Comedy is your strong point.
129398. Gord at 20:27 PM on 27 May 2009
        It's the sun
        Patrick - It appears that you still do not understand my post. The calculations are for an Earth WITHOUT AN ATMOSPHERE. --- First: You quoted me as saying: "There is absolutely no evidence that the Earth had an albedo of 0.3 before the Earth had an atmosphere." What I actually said was: "There is absolutely no evidence that the Earth had an albedo of 0.3 before the Earth had an atmosphere, so my assumption of zero albedo is just as valid." --- You asked: "And who ever said otherwise? (Without an atmosphere, offhand I think it's somewhere around 0.1)." The AGW'ers who did this calculation, used 0.3! This calculation is the one the AGW'ers use for an Earth without a "Greenhouse Effect", ie. no atmosphere: "Average Temperature of Earth = 255K or -18 deg C (240 w/m^2) due to the Sun and Earth's albedo = 0.3" ---- You said.. "NO, that is NOT what 'AGW's' are assuming at all. The albedo is known to be about 0.3 (PS this is a global time and area average weighted by TOA insolation, hence equal to the global time average reflection/backscattering to space of all solar radiation intercepted by the Earth.). Since the albedo is known seperately from how the greenhouse effect is known, it can be used in an equation to estimate the greenhouse effect. IN OTHER WORDS, The approx. 33 K warming effect is not from the difference between the atmosphere as it is and no atmosphere, it is the difference between the atmosphere as it is and the atmosphere as it is except for the greenhouse effect!" Answer: Greenhouse effect "In the absence of the greenhouse effect and an atmosphere, the Earth's average surface temperature of 14 deg C (57 deg F) could be as low as −18 deg C (−0.4 deg F), the black body temperature of the Earth." http://en.wikipedia.org/wiki/Greenhouse_effect Earth's Atmosphere Functions of the Atmosphere "Makes possible a mean temperature on Earth's surface of +15 deg C instead of -18 deg C as would be without atmosphere." http://www.kowoma.de/en/gps/additional/atmosphere.htm Why is the Earth so warm? "If there was no atmosphere, the Earth's temperature could be calculated by balancing: The energy in sunlight absorbed by the Earth The energy radiated as infrared photons by the warm Earth. Equilibrium Temperature should be T=260 K " "The Greenhouse Effect is responsible for making the Earth about 35K warmer than it would be if there it had no atmosphere." http://www.astronomy.ohio-state.edu/~pogge/Ast161/Unit5/atmos.html Effect on Surface Temperature "The natural greenhouse effect raises the Earth’s surface temperature to about 15 degrees Celsius on average—more than 30 degrees warmer than it would be if it didn’t have an atmosphere." http://earthobservatory.nasa.gov/Features/EnergyBalance/page6.php -------- Your post primarily deals with an Earth albedo of 0.3 in the presence of an atmosphere, which is not what was calculated. The rest of your rambling post covering your opinions on everything from the "big bang" to "chickens wraped in aluminum foil" are good comedy as well.
129399. David Wells at 20:09 PM on 27 May 2009
        Why is Antarctic sea ice increasing?
        Hi, isnt everyone missing the point, I thought that the big scare was according to God (Al Gore) that because the ice caps were melting we would need to buy stilts to get around and turn our gardens into rice paddies. As the ice caps are not melting but increasing in size - British Antarctic Survey Report, being released soon - shows that East Antarctica 4 times the size of West has been growing in size (100,000 sqkms per decade since 1970) and that temperatures there are falling not rising. Also and maybe more important the Oxford professor who designs satellites says that data received by satellite observation should be taken with a pinch of salt because they are too remote to be taken as explicitly accurate. It really doesnt matter whether the sea is a little warmer or a little colder 0.17c so long as the variation is within acceptable limits as to keep most things becalmed and liveable. Remember that Australias recent bushfires were caused because the local pacific sea area was colder than usual meaning less precipitation so less rain in Australia causing drought and extensive bush fires. My feeling is that these insignificant variations in temperatures have been occuring since the beginning of time and will continue whether we are here or not, what supposed scientists and pseudo scientists are trying to do is explain the inexplicable and using flawed computer models to try and predict the future when you would have more change of predicting the 4.30 at Cheltenham and they cant even do that when you have more than enough - more than enough data - to be able to achieve the objective. We burned more animals than was necessary in the last big foot and mouth epidemic in the UK that was necessary because of computer modelling, most of the financial meltdown was caused by people relying on computer models in fact if you really dig we seem to be becoming obsessed with anything to do with computers as though once a computer has made its decision it must be right even though the individual who designed the model was a human being who by default are more stupid than clever. No one ever took Al Gore seriously when he pretended to be a politician precisely why people take him seriously when he pretends to be a scientist is and will remain a complete mystery. My overall impression is that there are far too many people on this planet who are unemployed or unemployable and they fill in their time by trying to resolve the mysteries of the universe, by tryin to apply logic to a system that doesnt even understand the word. A scientist reporting on genetics said this morning that we would like to believe that we understand genetics but what most people either fail to understand or dont want to understand is just how ignorant we are. If that is the case and it most likely is then the same logic should be applied to the supposed science surrounding the idea that less than half of one percent of Co2 in the atmosphere is actually going to cause absolute catastrophe when in fact it wont. It maybe the case that a sea rise in certain places will cause some to have to move but the real and absolute disaster will be caused if the planet cools, it will only take the same temperature drop as the predicted rose for everything to stop dead. The last time in cooled in Europe parents were eating their children to survive, those who are captivated by the idea that a rise in temperature would be an absolute disaster need to chew on that piece of historical fact and go somewhere and get a life, because in anycase most of you will have died from old age before the predicted misery even begins to happen!! David Wells
129400. NewYorkJ at 15:04 PM on 27 May 2009
        It's cosmic rays
        More recent studies: http://www.sciencedaily.com/releases/2008/12/081217075138.htm http://www.sciencedaily.com/releases/2009/05/090511122425.htm

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