Impossible For CO2 to Heat Oceans
in [Physics]
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From: Retief on 17 Sep 2006 02:16 On 16 Sep 2006 04:48:26 -0700, kdthrge(a)yahoo.com wrote: >Albedo has very little to do with the amount of energy recieved in 1 sq >meter at the equator at solar noon. This is 1000 W m-2. The solar Actually, 1000 W/m^2 is usually considered to be the _maximum_ that the surface can recieve (based on experimental observations). Actual measurements will show that it is almost always less than that amount, due to clouds, etc... >constant is 1370 W m-2. This leaves 370 W m-2 that doesn't make it >through the atmosphere. So that "missing" power is either absorbed, or rejected... What's your point? >So you don't get confused the question is, 'what is the distribution of >frequencies in the 1000 W m-2 collected in a solar collector at the >equator at solar noon.' It's rarely 1000 W/m^2, and the actual spectrum varies with cloud cover, atmospheric water load, dust, etc. Retief
From: kdthrge on 17 Sep 2006 04:26 Retief wrote: > On 16 Sep 2006 04:48:26 -0700, kdthrge(a)yahoo.com wrote: > > >Albedo has very little to do with the amount of energy recieved in 1 sq > >meter at the equator at solar noon. This is 1000 W m-2. The solar > > Actually, 1000 W/m^2 is usually considered to be the _maximum_ that > the surface can recieve (based on experimental observations). Actual > measurements will show that it is almost always less than that amount, > due to clouds, etc... > > >constant is 1370 W m-2. This leaves 370 W m-2 that doesn't make it > >through the atmosphere. > > So that "missing" power is either absorbed, or rejected... What's > your point? > > >So you don't get confused the question is, 'what is the distribution of > >frequencies in the 1000 W m-2 collected in a solar collector at the > >equator at solar noon.' ..... > It's rarely 1000 W/m^2, and the actual spectrum varies with cloud > cover, atmospheric water load, dust, etc. > > Retief The point is that this is the maximum energy collected in actually 1 square meter of solar collector. Not any kind of average for determining overall delivered heat energy. If you take the solar collector 200 miles up it collects 1370 W m-2. So there is missing 370 W m-2. The original 1370 is 50% visible, 9% ultraviolet, 41% infrared and thermal. We know that very little of the visible is blocked in this ideal condition. 1% of Sun's energy is aborbed as ultraviolet in the stratosphere. The visible and ultraviolet amounts to roughly 800 W m-2. The other 200 W m-2 that make it through is at the high end of the infrareds. Almost all of it shorter wavelengths than 2.8 microns whcih can be confirmed by analyses of the Sun's spectrum through the atmosphere. This shows the opacity of the atmosphere to the thermal frequencies longer thatn one micron. Sure with highly sensitive equipment, imaging can be done in some of these wavelengths. In the references these chumps give to "prove" that the air is transparent to these frequencies, they never break down the sensitivity of the equipment or the actual energy that is being transmitted in these frequencies. This is a fundamental flaw in grenhous gas theory. If the atmophere is already opaque to the thermal frequencies that carry the radiated energy away from the earth, no increased CO2 will cause any warming at all, which is the fact. Also their initial model of the earth being 30C cooler without the presence of specific grenhouse gases, including water is completely false.. The fact is without the atmosphere that sheilds a considerable portion of the Sun's radiated energy, the mean temperature would be considerably higher. I was taught very specifically by a very proper astrophysicist, that the Earth is normal for it's mean temperature and recieved radiation. Only Venus and Jupiter have deviations of their mean temperature. The assertion that CO2 is responsible for the deviation in Venus is speculation and not confirmed by scientific method. And certainly is not an applicable example of properties of CO2 to be reapplied to the Earth. Kent Deatherage
From: Orator on 18 Sep 2006 00:22 Lloyd Parker wrote: > In article <D3TNg.28287$rP1.8461(a)news-server.bigpond.net.au>, > Orator <Orator(a)troll.bridge.net> wrote: > >>Lloyd Parker wrote: >> >> >>>In article <XzOMg.26528$rP1.1438(a)news-server.bigpond.net.au>, >>> Orator <Orator(a)troll.bridge.net> wrote: >>> >>> >>>>Lloyd Parker wrote: >>>> >>>> >>>>>And even IR radiated upwards is >>>>>going to encounter other CO2 molecules unless the molecule emitting is at >>>>>the top of the atmosphere. >>>>> >>>> >>**** >> >>>>BRAVO! Now that you have figured that out that above the molecule >>>>radiation "upward" there is another molecule. Then it is bleeding >>>>obvious that it too radiates "upward", and then another molecule is >>>>encountered and it too radiates "upward", and then another molecule is >>>>encountered and it too radiates "upward", and then another molecule is >>>>encountered and it too radiates "upward", and then another molecule is >>>>encountered....... >> >>**** >> >>> >>>Only a small portion of the surface area of a sphere points "upward." Most >>>point sideways or downwards. >> >>The surface that radiates to earth is the minor portion no matter what. >> >>> > > The surface which radiates to the _atmosphere_ is the major portion. Are you > claiming only photons which hit the earth's surface cause warming? > > >>>>Get the picture yet? Each and every one of them radiates more out into >>>>space than they do to earth. >>> >>> >>>That is totally false. >> >>Are you really serious? > > > By "earth" we mean the earth's atmosphere too. "Earth" doesn't just mean the > solid part, you know. Do we? Specially when we are talking about the atmosphere radiating to solid bit -v- space? I don't think so! >>>Even spheres at the top of the atmosphere don't have >>>the majority of their area pointing out to space. >>> >> >>Ahh, another nice one for the Lloyd collection of clangers :-) >> >>If you place a ping-pong ball, touching on top of an inflated beach >>ball, the majority of the ping-pong ball will face away from the beach >>ball. > > > Are you claiming all CO2 molecules sit on top of the atmosphere? > Did I suggest CO2 was NOT part of the atmosphere? No I didn't but we see Lloyd make that claim! That's another for the Lloyd Parker clanger collection :-) >>The further away from the beach ball the ping-pong ball is, the >>less of the surface will point to the beach ball. >> >>This is childishly simple and if you can't even get this right, how the >>hell can anyone expect, or believe, you can get anything right that is >>slightly more complex? Lloyd didn't dare respond to the rest :-)
From: Orator on 18 Sep 2006 00:38 Lloyd Parker wrote: > In article <5ameg2ppdrib5rsqdvt3fboofopde8f42i(a)4ax.com>, > Retief <nospam(a)invalid.invalid> wrote: > >>You have failed to prove that increasing CO2 is the _cause_ of global >>warming. How do you know it's not Methane? > > Because methane hasn't increased enough to be the cause. There is one or more statements embedded in that answer that needs explaining. 1 - Methane doesn't cause GW. 2 - Methane must be at a certain specific (but unstated) amount before it has any effect. These are tm-Lloyd Parker statements. I think (1) is rejected out of hand. Then (2) becomes the issue. So an example will be justified to simplify the issue. It there is a nail, hammered 1/2 way into the wall, and tests have shown that 1 kg of weight acting on the nail to cause it to bend. Would it have any effect at all on the nail if one fly sat on that nail? Would it take the last fly that makes up 1 kg of flies to sit on the nail before it bends?
From: Phil. on 18 Sep 2006 09:13
Retief wrote: > On 12 Sep 2006 22:38:00 -0700, "Phil." <felton(a)princeton.edu> wrote: > > > > >Collisional deactivation by surrounding > > > >molecules will be the predominant mode of energy exchange at STP. > > > > > > Until the energy has been sufficiently "spread around" (i.e. the > > > molecules approach the same level of excitation) throughout the > > > ensemble, then radiative process will take over as the primary > > > exchange mechanism. > > > > Which it never does, you'll have a Boltzmann distribution where energy > > is being continuously being exchanged between molecules via collisions. > > You apparently wish to assert that the collisional process PREVENTS > the radiative process from occurring. This is false. The two > processes occur simultaneously (each with their respective lifetimes). It does, once energy has been lost via collisions with N2 and O2 molecules it's no longer available to be lost via emission. > > The mean lifetime (similar to a half-life) is statistically > determined, and is not a limit on how soon a molecule or atom may > radiate after excitation. Some may radiate immediately upon > excitation, some may take time (far longer than the "mean lifetime"). Quite so, which I have said previously. > > That is, if a certain gas had a mean radiative lifetime of 100 > seconds, after 100 seconds, approximately 2/3 of the excited molecules > would have radiatively decayed, leaving approximately 1/3 in the > excited state. > Right, but in the situation we are discussing the analogy would have a collision occuring every 0.1 sec which would mean that deactivation would 'predominantly' be via collisions, exactly as I said above. |