AGW is myth- its all for the money!

Discussion in 'Earth Science' started by mello, Oct 31, 2013.

  1. Trippy ALEA IACTA EST Staff Member

    I agreed that there was a factor of four in the Stefan-Boltzman equations sure, however, apparently I lost you after that.
    First, consider a patch of ground of unspecified size, thermal properties or optical properties, on a surface of unspecified size and shape. The only thing we know about the body is that it has a diurnal cycle of some unspecified length and this is the only assumption we make about it (well, that and that our ground isn't shaded).
    As our patch of ground rotates across the sunward face, the insolation received by it varies as \(k_l \frac {1+sin( \theta )}{2}\)
    Where \(k_l = k_s \frac{1+cos( \phi )}{2}\)
    Where theta is the angle of the sun measured from where it rose (0 at dawn, 90 at noon, 180 at dusk).
    Where Phi is the angle of longitude.
    And ks is the power density at the planets orbit.

    Under these conditions, the average power density during the day is half the peak power density.
    but this is only half the cycle.

    The average power of insolation across the night side of the planet is zero. Taking the average of the day side and night side insolations combined we find that the average power density across the entire diurnal cycle is 1/4 that of the peak. This means that any equilibrium temperature calculated will be the average equilibrium temperature across the entire day-night cycle.

    So you see, simply by taking the full diurnal cycle into account we arrive at the same 1/4 figure that comes about asa result of assuming ideal behaviour, but we do so without having to assume that the thermal behaviour of the material is ideal.

    Do you follow now? Or do you want me to graph it for you when I get home. Do you understand what it means for at least so.e of your objections?
  2. Google AdSense Guest Advertisement

    to hide all adverts.
  3. Trippy ALEA IACTA EST Staff Member

    apparently my trig is rusty, this doesn't give quite the right shape. Everything else I said still applies though.
  4. Google AdSense Guest Advertisement

    to hide all adverts.
  5. Trippy ALEA IACTA EST Staff Member

    Correct functions are:
    \( k_l\frac{1-cos(2 \theta)}{2}\)
    \(k_l = k_s \frac{1+cos (2 \phi)}{2}\)

    where everything has the same meaning as before.
  6. Google AdSense Guest Advertisement

    to hide all adverts.
  7. Trippy ALEA IACTA EST Staff Member

    To illustrate my point:

    Please Register or Log in to view the hidden image!


    So the factor of 1/4 allows us to account for diurnal variations as long as we keep in mind that any numbers derived from it will not be peak numbers, but will be average numbers, averaged across the entire diurnal cycle.

  8. Andre Registered Senior Member

    Yes, Thanks much appreciated.

    Now I took the liberty to add the Stefan Bolzmann equilibrium temperatures and how the temperature of the surface actually would react if it were a basic first order delay response. That needs more thought though since obviously the forcing for the rate of warming is the momentary difference in insolation and outradiation. But it gives a rough idea, IMO, of the delay that occurs in a non-black body.

    Please Register or Log in to view the hidden image!


    Obviously you can play with the parameters to see the effects.

    Sorry I used degrees on the x-axis, but Myrthe is crying, Babysit priority. Will change later
  9. Trippy ALEA IACTA EST Staff Member

    So then you're in agreement that the factor of 1/4 is actually valid?

    I do actually have more, however, I'm not actually interested in proceeding further until I have your explicit agreement on that point.

    We'll get to the next step soon enough, let us finish with this first one first, although I would point out that your system isn't quite in equilibrium - and note that while your peak temp is 54°C, your average temperature across the entire cycle is -128°C.

    I have a slightly different approach in mind, which I will discuss in due course, however, I'm not quite there yet.
  10. Andre Registered Senior Member

    I agree that it is valid for a perfect black body. I see issues elsewhere. If you divide radiation energy over a black body surface then obviously it divides by 4. However if you divide temperature over the surface then there are issues with the fourth power in the S-B law.

    Thats likely due to the non valid assumptions, however something similar is seen on the Apollo 17 site, max temp 384 K; min temp 102 K; Mean temp 216 K; a bit below the median, and also clearly below "black body" temperature, which is given as 270.7K

    At the Apollo 17 pos, 20.1653 North lattitude, the cos factor is obviously some 0.94

    Edit: In addition:

    This plot of moon temps shows much the same pattern as our graphs with mean values optically estimated in the order of magnitude of 200K.

    Please Register or Log in to view the hidden image!

  11. Trippy ALEA IACTA EST Staff Member

    See, now this is why I insist on consensus before we proceed further.

    This graph:

    Please Register or Log in to view the hidden image!

    Assumes nothing about the properties of the body in question other than that it is rotating about some central axis.

    I thought I made that clear. It assumes nothing about albedo, it assumes nothing about the thermal behaviour of the material. The only thing that that graph illustrates is the amount of radiation recieved by the surface as it rotates.

    None of this has anything to do with the graph I presented or what I am asking you to agree to.

    I haven't considered shape or size, only rotation to represent the diurnal cycle, and what I am trying to get consensus on is the observation that when we consider power over the whole diurnal cycle, before we get as far as considering albedo, shape and surface area (which I will soon contend are irrelevant to the discussion), because power varies as \(f(\theta) = \left{ { \begin{array}{lll} \frac{ 1-\cos (2 \theta) }{2} & \quad \quad \quad & \textrm{if} \; 0 \leq \theta \leq \pi \\ 0 & & \textrm{otherwise} \end{\array} } \right.\)* then the average power received by the surface is \(\frac{1}{4}\) the peak power received at the surface.

    For the purposes of that graph:
    Albedo is irrelevant.
    Blackbody behaviour is irrelevant
    Surface area is irrelevant
    Temperature is irrelevant

    The only thing this graph considers is the rotation of a a point of unspecified size, shape, character, and albedo, rotating about the central axis of a body of unspecified shape.**
    The only thing this graph shows is the power received by the surface as it varies between 0% and 100% of some undefined maximum value.
    What the graph demonstrates is that the average power recieved at the surface, considered across the full diurnal cycle, before any is absorbed or any is reflected, is exactly 25% of the peak power when the sun is at it's zenith.

    It, combined with the fact that, as I shall demonstrate, knowledge of the surface area of the body we are dealing with is un-neccssary, suggests that the source of the \(\frac{1}{4}\) applied to the solar constant has nothing to do with the stefan-boltzman law as you have asserted, but is instead a consideration of the diurnal cycle.

    *Thankyou rpenner for reminding me of the correct notation.
    **Within reason, obviously a peanut shaped body or a point in significant concavity is going to be among the exceptions.
  12. rpenner Fully Wired Registered Senior Member

    That the average influx of radiation from a distant point source over a globe, rotating or not, averages to \(\frac{1}{4}\) of the peak intensity is easily seen.

    The surface area of a sphere of radius r is \(A_{\tiny \textrm{Total}} \; = \; 4 \pi r^2\)
    The cross-section of a sphere of radius r which can been illuminated by a point-like source at distance D from is the same as the area of a circle of radius h, the height of a right-triangle with base/hypotenuse of D and one leg r. \(h \; = \; \frac{r \sqrt{D^2 - r^2}}{D}, \quad A_{\tiny \textrm{Total}} \; \approx \; \pi h^2 \; = \; \pi r^2 \left( 1 - \frac{r^2}{D^2} \right) \)
    Since the ratio is \(\frac{1}{4} \left( 1 - \frac{r^2}{D^2} \right)\) which differs from \(\frac{1}{4}\) by one part in about 550 million, \(\frac{1}{4}\) is a good approximation.

    Alternatively we could start with the assumption of isotropically emitted radiation with power P and an inverse square law giving peak intensity \(I_0 = \frac{P_0}{4 \pi \left( D - r \right)^2}\) at the nearest part of the planet. Then total power received would be the same fraction of P that the shadow of the planet casts of a sphere. Thus \(P = \frac{2 \pi D \left( D - \sqrt{ D^2 - r^2 } \right) }{4 \pi D^2} P_0 = 2 \pi I_0 D^2 \left( 1 - \frac{r}{D} \right)^2 \left( 1 - \sqrt{ 1 - \frac{r^2}{D^2} } \right) \) and ratio of average intensity to peak intensity of \( \frac{I}{I_0} = \frac{1}{I_0} \times \frac{P}{A_{\tiny \textrm{Total}}} = \frac{D^2 }{2 r^2} \left( 1 - \frac{r}{D} \right)^2 \left( 1 - \sqrt{ 1 - \frac{r^2}{D^2} } \right) = \frac{ \left( 1 - \frac{r}{D} \right)^2 }{2 + 2 \sqrt{ 1 - \frac{r^2}{D^2} } } \approx \frac{1}{4}\)
  13. Trippy ALEA IACTA EST Staff Member

    This is part of the problem though. Andre objects to this:
    Because he feels it requires the inate assumption that the sphere in question behaves as a perfect blackbody with perfect thermal conduction but the real earth doesn't behave that way.

    My point, or one of the points, is that we arrive at the same figure for average power (being 25% of the peak power) by a completely different means, by considering only a planet that rotates and nothing else, thus rendering his objections in that regard irrelevant.

    Which I suppose is the point that you're illustrating, demonstrating that there are several different ways of coming to the same conclusion.
  14. Aqueous Id flat Earth skeptic Valued Senior Member

    The cross section of the incident beam is πR². That's the amount eclipsed, which is the amount distributed over the surface of Earth (4πR²) diurnally. The average incident flux is reduced by πR² ÷ 4πR² = 1/4. That is, it's 1366 W/m² ÷ 4 which is approx 340 W/m², the number you introduced. So you already had a factor of 1/4 in mind from the outset.

    That's not only wrong, but you appear to be arguing with yourself. As I said, it was you who posted 340 W/m² which already concedes that 1/4 was correct.

    No it's valid regardless of that. The question of deriving temperature from flux is a separate question.

    It divides by 4 in all cases as matter of geometry, not physics.

    This seems to say you're not certain what you meant by radiative equilibrium temp., which further confuses your meaning overall.

    (back to the prior post)

    If you think that N[sub]2[/sub] accounts for global warming, then you're way off track. Besides the fact that the greenhouse effect is a consequence of the conversion of short wave energy into long wave by surface processes -- which does not relate to the absorption spectrum of N[sub]2[/sub], but rather CO[sub]2[/sub] and water vapor -- climate forcing can not be attributed to N[sub]2[/sub] since N[sub]2[/sub] levels are not increasing. Further, N[sub]2[/sub] is pegged out at about 78% due to other kinds of equilibria (such as with the biomass), with little or no room to shrink or grow, and it certainly can't double. Here is the smoking gun you seem to be looking for: only trace amounts of CO[sub]2[/sub] are controlling our climate, therefore the system is sensitive to anthropogenic contributions.

    But feel free to dismiss the observed data and to try to replace working principles with unworkable ones. You just won't be talking about more than myth. You do realize that this is fairly characterized as pseudoscience, don't you?

    Yet there is only one explanation for global warming, and that is the greenhouse effect. And among the various gases involved, CO[sub]2[/sub] is the main one forcing the climate in trace amounts.

    None of this alters the statement I just made. Curry endorses the greenhouse effect.

    So far the explanations you've been giving don't work.

    Whether we need it or not, nature is producing it, so we're stuck with it. And now that we understand the sensitivity of natural systems to anthropogenic CO[sub]2[/sub], and how trace amounts affect climate, the onus is on us to either mend our ways or else to inflict more irreparable harm on the planet than we already have.

    No, which is why, during her testimony to Congrees, your source (Curry) said:

    If all other things remain equal, it is clear that adding more carbon dioxide to the atmosphere will warm the planet.
  15. Trippy ALEA IACTA EST Staff Member

    This thread has been unsurprisingly quiet...
  16. wellwisher Banned Banned

    My question is this. The earth has shown warming and cooling cycles, many times in the past. The last ice age and the warming from the last ice age to the present is one example. These all occurred without man. How do we factor out, natural, to make sure what we are seeing is not part of another natural trend?

    If we ignore these natural trends (nothing to see here), the naive and easily duped could be made to believe what we see all human doing. So how do we factor natural out 100%, so we can conclusively say the natural trends are 0% in effect at this time? Where is the science for this proof to make sure we assign the correct percent.
  17. origin In a democracy you deserve the leaders you elect. Valued Senior Member

    We do not ignore the natural cycles those are taken into account in the models. This fact is not discussed on Fox News so you may have missed that.

    Please Register or Log in to view the hidden image!

  18. Andre Registered Senior Member

    This is probably not helping to falsify the title of the thread.

  19. origin In a democracy you deserve the leaders you elect. Valued Senior Member

    It is irrelevant. The fact that this guy has some problems that are going to most likely result in him going to prison does not have a bearing on global warming one way or the other.
  20. billvon Valued Senior Member

    In that case this is really going to cripple the cause of the pro-fossil-fuel deniers:

    3 BP executives indicted over Gulf oil spill
    Rick Jervis and Kevin Johnson
    USA TODAY 11:28 p.m. EST November 15, 2012
    BP agrees to pay $4.5 billion in fines in the largest such settlement in U.S. history.

    NEW ORLEANS -- Two employees of the British oil giant BP have been indicted on manslaughter charges in connection with the 2010 Gulf oil rig blowout that killed 11 workers.

    A third executive was charged with lying to authorities about his work estimating the rate oil was flowing during the disaster.
  21. Andre Registered Senior Member

  22. Trippy ALEA IACTA EST Staff Member

    Gosh, well, I guess that's that then. Some dude lied about being in the CIA and some other dudes with green interests invested in green companies. Clearly the conservation of mass and energy, the stefan-boltzmann law, the beer-lambert law, and quantum mechanics are just further examples of fraudulent liberal/green science. I shall simply have to throw my computer away because obviously it will never work...
  23. Andre Registered Senior Member

    Isn't that a bit of a slippery slope? But anyway, you can have your conservation law, Stefan Boltzman, Beer Lambert as building blocks and quantum mechanics as kit for them to glue them together and then you can build your model, anyway you like. But it becomes very questionable, if you say: see I could build a prison of my building blocks, so we're all doomed. Because that's not science, when nature made a palace out of those, using many more different building blocks that we can comprehend.

    And the models don't seem to do so well.

    Now what exactly did Carl Sagan (who diseased 17 years ago) mean with:

Share This Page