tag:blogger.com,1999:blog-28837843.post2111376059011652240..comments2024-03-19T07:10:27.303-07:00Comments on Quark Soup by David Appell: Yes, Salem, Oregon is WarmingDavid Appellhttp://www.blogger.com/profile/03318269033139447591noreply@blogger.comBlogger75125tag:blogger.com,1999:blog-28837843.post-67657946326749551892017-03-08T20:14:46.061-08:002017-03-08T20:14:46.061-08:00David Appell
Since you will block this as previou...David Appell<br /><br />Since you will block this as previous post. The fact is obvious you are not a science minded person. You do not possess any natural curiosity. You really do not care about the Truth. You are like a disciple of a cult. If the Authority makes a claim it is unquestionable truth and only truth that you allow to exist in your thoughts. <br /><br />I am very glad science is not composed of your mental type as it would go nowhere. Science is generated by those who love the truth. Never accept blindly anything from anyone. Challenge everything, even established ideas. <br /><br />Even though I think you make a very bad scientist I still thank you for allowing me the privilege to post on your blog. It has not been at all productive. I will gladly return to Roy's site. Much more intelligence and thought on his site. Agree or disagree people will engage in thought provoking ideas. Normanhttps://www.blogger.com/profile/01561705541102129449noreply@blogger.comtag:blogger.com,1999:blog-28837843.post-2194219780262562232017-03-07T21:31:40.141-08:002017-03-07T21:31:40.141-08:00Norman says:
"You can do a simple calculation...Norman says:<br />"You can do a simple calculation to show how they got it wrong."<br /><br />If you think that, then write to the authors or write to the journal.<br /><br />Let us know what you find out.David Appellhttps://www.blogger.com/profile/03318269033139447591noreply@blogger.comtag:blogger.com,1999:blog-28837843.post-2908400741558806302017-03-07T21:29:14.357-08:002017-03-07T21:29:14.357-08:00Norman, T = 288 K and F = 240 K ==> emissivity ...Norman, T = 288 K and F = 240 K ==> emissivity = 0.62<br /><br />Unless you have something new to say here, your future comments will be declined. You have been off the topic of this post all along.<br /><br />David Appellhttps://www.blogger.com/profile/03318269033139447591noreply@blogger.comtag:blogger.com,1999:blog-28837843.post-61213168827395183222017-03-07T21:26:02.890-08:002017-03-07T21:26:02.890-08:00David Appell
If you wanted to raise the same plat...David Appell<br /><br />If you wanted to raise the same plate from 288 K to 289 K how much radiant energy would you have to add to the surface?<br /><br />A 288 K plate with an emissivity of 0.62 has an absorbitivity of 0.62. That means it would need to have a radiant flux of 390 W/m^2 striking its surface to maintain a temperature of 288 K. (390 W/m^2)(0.62) = 241.8 W/m^2 <br /><br />An object with an emissivity of 0.62 only has an absorbitivity of 0.62 so it takes 390 watts to maintain its 288 K temp and it will emit 241.8 W/m^2 same as it absorbs (First Law of Thermodynamics).<br /><br />To go to 289 K if will now be emitting 245.23 W/m^2 and will need the same energy input to maintain a temperature of 289 K (equilibrium condition).<br /><br />How much input energy? 245.23/0.62 = 395.53 W/m^2 which is the same amount if the emissivity and absorbitvity are equal to one. Changing the emissivity does not change the energy balance, first law still dominatesNormanhttps://www.blogger.com/profile/01561705541102129449noreply@blogger.comtag:blogger.com,1999:blog-28837843.post-81972891973955956212017-03-07T21:14:40.490-08:002017-03-07T21:14:40.490-08:00David Appell
When have I been wrong all these tim...David Appell<br /><br />When have I been wrong all these times before. You keeping making this statement with no evidence. Did you look at the emissivity problem?<br /><br />You can do a simple calculation to show how they got it wrong.<br /><br />If you have a 1 m^2 plate in outer space alone and by itself. Set the emissivity of the plate at 0.62.<br /><br />You send radiant energy at it until the surface reaches 288 K. It is emitting 240 W/m^2. How much energy must send to the plate's surface to maintain a 288 K temperature? Would you suggest 240 W/m^2 because that is what it is losing?<br /><br />Normanhttps://www.blogger.com/profile/01561705541102129449noreply@blogger.comtag:blogger.com,1999:blog-28837843.post-37851559363143808212017-03-07T20:55:09.440-08:002017-03-07T20:55:09.440-08:00Norman says:
"I can see the flaw in the paper...Norman says:<br />"I can see the flaw in the paper and I hope you will look into it."<br /><br />I won't. I'm not interested -- you've been wrong too many times before. Just like you were wrong here about emissivity. Figure it out for yourself. <br /><br />Have fun.<br /><br />David Appellhttps://www.blogger.com/profile/03318269033139447591noreply@blogger.comtag:blogger.com,1999:blog-28837843.post-35909738171225558932017-03-07T20:49:11.843-08:002017-03-07T20:49:11.843-08:00No you're not blocked, asshole, I put moderati...No you're not blocked, asshole, I put moderation on comments on posts > 15 days old to prevent comment spam. I upped that to 30 days for your benefit, but the spam is getting too much and I reduced it back to 15 days. David Appellhttps://www.blogger.com/profile/03318269033139447591noreply@blogger.comtag:blogger.com,1999:blog-28837843.post-20460028114917227012017-03-07T20:36:46.438-08:002017-03-07T20:36:46.438-08:00am I blocked?
am I blocked?<br />Normanhttps://www.blogger.com/profile/01561705541102129449noreply@blogger.comtag:blogger.com,1999:blog-28837843.post-32812695376101536562017-03-07T20:36:00.151-08:002017-03-07T20:36:00.151-08:00David Appell
I can see the flaw in the paper and ...David Appell<br /><br />I can see the flaw in the paper and I hope you will look into it. If you do not believe me an think me an idiot, fine I can't change your opinion but do look at Joel Norris paper linked above.<br /><br />http://kiwi.atmos.colostate.edu/pubs/Cessetal-1990.pdf<br /><br />They use the emissivity of 0.62 for the Earth/atmosphere system. If they want to do it this way they have to follow through and not then use a unity of one for the emissivity in the calculation later. <br /><br />If you take their mistaken math. Climate Sensitivity = 4(240 W/m^2)/288 K<br />You get the 3.33 W/m^2-K<br /><br />However you need to divide this by the emissivity of 0.62 because now this becomes the absorbitivity of the system. So you now need to take 3.33/0.62 = 5.37 W/m^2-K which is the same as Joel Norris caculated.<br /><br />If you use 0.62 at the Surface/atmosphere system then that becomes the absorbitivity of the same system. That means it takes 5.37 additional watts added to the system to be able to absorb 3.33 W/m^2.<br />Normanhttps://www.blogger.com/profile/01561705541102129449noreply@blogger.comtag:blogger.com,1999:blog-28837843.post-11198359699891655542017-03-06T18:38:26.359-08:002017-03-06T18:38:26.359-08:00The paper you cited used an emissivity of 0.62. Yo...The paper you cited used an emissivity of 0.62. You agree? David Appellhttps://www.blogger.com/profile/03318269033139447591noreply@blogger.comtag:blogger.com,1999:blog-28837843.post-14144272428383499582017-03-06T18:35:14.545-08:002017-03-06T18:35:14.545-08:00David Appell
One last shot, I know you are not in...David Appell<br /><br />One last shot, I know you are not interested but this scientist (a PhD) Joel Norris is a Climate Scientist.<br /><br />http://scrippsscholars.ucsd.edu/jnorris<br /><br />He has the same value I do with a zero-feedback response.<br /><br />Above I posted: "So instead of doing the scrambled math of mixing surface flux with TOA flux you would use<br />4(390 W/m^2)/288 = 5.42 Watt/m^2-K which gives a zero feedback sensitivity of 0.185 K(W/m^2) so an increase in downwelling IR of 3.7 W/m^2 will increase surface temperature 0.68 K."<br /><br />Joel Norris comes up with the same number. Does he lack physics understanding and is clearly wrong as I am? Hmmmm? <br /><br />http://scrippsscholars.ucsd.edu/jnorris/files/caltechweb.pdf<br /><br /><br />Page 14 Title: Zero-Feedback Climate Responses<br /><br />He rounds it to 5.4 I have 5.42. So now who lacks understanding? Normanhttps://www.blogger.com/profile/01561705541102129449noreply@blogger.comtag:blogger.com,1999:blog-28837843.post-38483692971058925092017-03-06T17:54:46.629-08:002017-03-06T17:54:46.629-08:00Wrong, Norman -- again you show your ignorance of ...Wrong, Norman -- again you show your ignorance of physics. E=m*c2 (m* does not equal mass) only holds in Lorentzian reference frames. For a general reference frame, general relativity is required.David Appellhttps://www.blogger.com/profile/03318269033139447591noreply@blogger.comtag:blogger.com,1999:blog-28837843.post-90978829683288037072017-03-06T17:50:18.169-08:002017-03-06T17:50:18.169-08:00David Appell
http://www.emc2-explained.info/Emc2/...David Appell<br /><br />http://www.emc2-explained.info/Emc2/Deriving.htm#.WL4O0W8rKAY<br /><br />First General relativity is not the same as the relationship between mass and energy. It is a new understanding of gravity and how it works. <br /><br />You do not need an elaborate set of equations to determine the amount of energy released by the conversion of a given mass to a given amount of energy. Any difference would violate the Energy Conservation.<br /><br />Normanhttps://www.blogger.com/profile/01561705541102129449noreply@blogger.comtag:blogger.com,1999:blog-28837843.post-30128394486910318692017-03-06T17:09:10.662-08:002017-03-06T17:09:10.662-08:00Look Norman: Top right-hand column of Cess+ pg 166...Look Norman: Top right-hand column of Cess+ pg 16602 says:<br /><br />"It then follows that Delta(F)/Delta(T) = 4F/T• = 3.3 W m -2 K -• for conditions typical of<br />Earth (F = 240 W m -2 and T• = 288 K)"<br /><br />If F = 240 W/m2 and T = 288 K, then emissivity = emissivity of "surface atmosphere system" = 0.62<br /><br />David Appellhttps://www.blogger.com/profile/03318269033139447591noreply@blogger.comtag:blogger.com,1999:blog-28837843.post-34329616515935974432017-03-06T16:18:04.006-08:002017-03-06T16:18:04.006-08:00Norman, I'm done with your questions, that are...Norman, I'm done with your questions, that are always based on a misunderstanding. Every time. You haven't disproved climate science. I'm simply not interested in whatever you are writing about here -- your comments are off the topic of the blog post, and you have cried wolf too many times.<br /><br />David Appellhttps://www.blogger.com/profile/03318269033139447591noreply@blogger.comtag:blogger.com,1999:blog-28837843.post-26590159122833129362017-03-06T16:15:23.626-08:002017-03-06T16:15:23.626-08:00Norman says:
"Einstein did not have to add mu...Norman says:<br />"Einstein did not have to add multiple variables and differentials to describe the relationship between mass and energy. A very simple equation."<br /><br />Clearly you have never encountered Einstein's equations or studied general relativity.<br /><br />--<br /><br />Both Joe (above) and I (on Spencer's site) have pointed out your many errors. <br /><br />Did you write to one of the paper's co-authors with your disproof of all of 20th century climate science? Or at least ask them a question? <br /><br />Why not?David Appellhttps://www.blogger.com/profile/03318269033139447591noreply@blogger.comtag:blogger.com,1999:blog-28837843.post-69968317561272827822017-03-06T16:13:02.419-08:002017-03-06T16:13:02.419-08:00David Appell
There is no "Boy who cried Wolf...David Appell<br /><br />There is no "Boy who cried Wolf" <br /><br />The flaw in the equation is obvious and simple to see. They use the radiative flux of 240 W/m^2 (which is the TOA flux away from Earth) as the energy forcing the surface to a 288 K temperature. It could not be worse phyiscs and you won't even look at it. Makes you the sad sack of physics. Stick the head in the ground and ignore. <br /><br />I ask you a very simple question that you should easily be able to answer and you attack my knowledge of physics, ignore completely the question, pretend I am this dumb hick that knows nothing. Worthless science. <br /><br />Again make is easy for me. Why do they use a flux of 240 W/m^2 with an Earth surface temperature of 288 K. You have seen global energy budgets. How much radiant energy does the surface have to receive to maintain a temperature of 288 K?<br /><br />https://en.wikipedia.org/wiki/Earth's_energy_budget#/media/File:The-NASA-Earth%27s-Energy-Budget-Poster-Radiant-Energy-System-satellite-infrared-radiation-fluxes.jpg<br /><br />How about 163.3 W/m^2 Solar and 340.3 W/m^2 Downwelling IR? That comes up with 503.6 W/m^2.<br /><br />If the surface needs a whopping 503.6 W/m^2 to maintain and equilibrium temperature (because of surface losses from evaporation and thermals) how do they think using 240 W/m^2 is correct math? What rational math is this concept based upon? Magic?Normanhttps://www.blogger.com/profile/01561705541102129449noreply@blogger.comtag:blogger.com,1999:blog-28837843.post-51076732157444324362017-03-06T16:03:16.292-08:002017-03-06T16:03:16.292-08:00David Appell
YOU: "And, as Joe said above, a...David Appell<br /><br />YOU: "And, as Joe said above, and as I've pointed out to you many times on Spencer's site, you understanding of physics is very weak." <br /><br />That is a weak opinion, based on nothing. You have not pointed out great flaws in my understanding of physics many times. This is not a true or verified statement.<br /><br />Again basically you do not have an answer nor does Joe so you just think the First Law of Thermodynamics is now bad physics. A<br /><br />The paper itself uses a very simplistic calculation (also if you are at equilibrium state you do not need complex math...where did you study physics again? A Doctor degree, that must have been a very easy school). You use the simplest explanation possible and only complicate matters when the problem requires. Occam's Razor, you know. Einstein did not have to add multiple variables and differentials to describe the relationship between mass and energy. A very simple equation.<br /><br /><br />Normanhttps://www.blogger.com/profile/01561705541102129449noreply@blogger.comtag:blogger.com,1999:blog-28837843.post-36812098309782181342017-03-06T14:34:49.172-08:002017-03-06T14:34:49.172-08:00Norman said...
"No you really do not have an ...Norman said...<br />"No you really do not have an answer but do not want to show your lack of knowledge so you hide behind comments like you are too busy."<br /><br />I'm not too busy. I'm just done with you. Much like Trump, you misinterpret your lack of understanding for conspiracy. <br /><br />You think every time your (usually simplistic) calculation doesn't agree with a paper's, it means there is a 100-yr conspiracy hiding the truth that only now you have discovered and can reveal as a fatal flaw in climate sciencew.<br /><br />You're the boy who cried wolf. <br /><br />And, as Joe said above, and as I've pointed out to you many times on Spencer's site, you understanding of physics is very weak. <br /><br />I am tired of explaining this to you time and time again. I'm tired of having to find the flaws in your simplistic arguments. I'm tired of your insults. I'm tired of your ignorance and that you're happy to remain so.<br /><br />Shove off. David Appellhttps://www.blogger.com/profile/03318269033139447591noreply@blogger.comtag:blogger.com,1999:blog-28837843.post-70458479828373186502017-03-06T14:29:41.760-08:002017-03-06T14:29:41.760-08:00Norman says:
"The Earth's surface emits c...Norman says:<br />"The Earth's surface emits close to unity (1) at wavelengths absorbed by atmosphere."<br /><br />What value does YOUR PAPER use for emissivity?<br /><br />David Appellhttps://www.blogger.com/profile/03318269033139447591noreply@blogger.comtag:blogger.com,1999:blog-28837843.post-58730036187896112282017-03-01T18:39:59.865-08:002017-03-01T18:39:59.865-08:00David Appell
From what you wrote and what Joe T w...David Appell<br /><br />From what you wrote and what Joe T wrote:<br /><br />YOU: "Norman, no one is obligated to spend time explaining anything to you. Reread what Joe wrote. Figure it out for yourself. Ask the paper's author. I have other things I wish to spend my time on instead of doing homework problems you assign me."<br /><br />Basically it is most obvious you do not have an answer and may even fear I am correct so you would choose the safe path of ignoring what I post and pretending you are far too busy to answer a simple question. Sending me back to what Joe T wrote indicates you are not reading my return comments which he chooses not to respond to .<br /><br />No you really do not have an answer but do not want to show your lack of knowledge so you hide behind comments like you are too busy. <br /><br />Same with Joe T. Calls all my understanding flawed and then drops out of the debate with no answer to my question. <br /><br />From Joe T "Pretty much everything you wrote is wrong. It's clear to me now that you don't understand the greenhouse effect and don't understand simple concepts in physics."<br /><br />Yet he can't explain why it is wrong or how I do not understand simple concepts in physics. He can't explain nor are you willing to even attempt to explain how 393.7 W/m^2 energy flux can sustain a temperature of 289 K when that surface is emitting at the rate of 395.5 W/m^2.<br /><br />Less energy in means more energy out. I would hope that you could spend some time in explaining why this is possible since it is completely against the First Law of Thermodynamics. Normanhttps://www.blogger.com/profile/01561705541102129449noreply@blogger.comtag:blogger.com,1999:blog-28837843.post-77540384819331442842017-03-01T18:30:06.597-08:002017-03-01T18:30:06.597-08:00David Appell
From your friend at Science of Doom
...David Appell<br /><br />From your friend at Science of Doom<br />https://scienceofdoom.com/2010/04/07/the-dull-case-of-emissivity-and-average-temperatures/<br /><br />The Earth's surface emits close to unity (1) at wavelengths absorbed by atmosphere.<br /><br />The surface emissivity is not 0.62. It is how much energy reaches the TOA from the 288 K surface but it is NOT the emissivity of the surface. The surface radiates 398 W/m^2 according to the latest global energy budgets. Hopefully you know that if you have a radiant flux of 398 W/m^2 emitting from a surface and the emissivity is low like 0.62 it means you must have a really hot surface, much hotter than the Earth's surface currently measures at. <br /><br />T =(398 W/m^2)/(o.62 emissivity you and Joe T seem to want to use)(5.67x10^-8...Stefan-Boltzmann constant)^-4 = 326 K!! or about 53 C! Much warmer than if you use a 1 for the emissivity which would give you a 289.45 K temperature.<br /><br />Normanhttps://www.blogger.com/profile/01561705541102129449noreply@blogger.comtag:blogger.com,1999:blog-28837843.post-36604932811734271312017-03-01T00:33:21.588-08:002017-03-01T00:33:21.588-08:00Norman, against my better judgement, I'm going...Norman, against my better judgement, I'm going to ask: what value are you using for the surface's emissivity?<br /><br />Joe asked you this too. Did you answer him?<br /><br />The Cess et al paper is clearly using 0.62. Are you? <br /><br />PS: You do realize, I hope, that the no feedback numbers in the paper are not measurable or observable, and do not appear anywhere in climate models. Right? <br />David Appellhttps://www.blogger.com/profile/03318269033139447591noreply@blogger.comtag:blogger.com,1999:blog-28837843.post-43024381735137273132017-02-28T21:09:12.032-08:002017-02-28T21:09:12.032-08:00Norman, no one is obligated to spend time explaini...Norman, no one is obligated to spend time explaining anything to you. Reread what Joe wrote. Figure it out for yourself. Ask the paper's author. I have other things I wish to spend my time on instead of doing homework problems you assign me.<br />David Appellhttps://www.blogger.com/profile/03318269033139447591noreply@blogger.comtag:blogger.com,1999:blog-28837843.post-78249669920434474762017-02-28T19:50:42.008-08:002017-02-28T19:50:42.008-08:00David Appell
So to make it easy explain the flaw....David Appell<br /><br />So to make it easy explain the flaw. I explained Joe T's flaw and he quit posting. <br /><br />So rather than spending time telling me I am wrong because of numbers of people involved or what Joe T thinks. Spend a brief time and science and explain exactly how you think the 1st Law of energy Conservation can be violated? <br /><br />With not other effects, a plate in outer space...In order to raise a 1 m^2 plate with near blackbody emissivity 1C you have to add the same amount of energy to the plate that is leaving it. At 288 K the plate will be emitting 390 W/m^2 so to maintain a temperature of 288 K it must receive 390 W/m^2. If the plate was 289 K it would be emitting 395.5 W/m^2. In order to remain at 289 K it must receive 395.5 W/m^2.<br /><br />If you added 3.7 W/m^2 to the flux of 390 W/m^2 to now add 393.7 W/m^2 to the surface it would emit 393.7 W/m^2 at equilibrium. Its temperature would be 288.67 K not 289.2 K (which is what you and Joe T are saying what would happen). <br /><br />It would violate energy conservation if 393.7 watts could raise a surface to 289.2 K because the 289.2 K surface would be emitting more radiation than it is receiving. How do you explain this? I would like more than an appeal to authority. Actually explain it. Thanks.Normanhttps://www.blogger.com/profile/01561705541102129449noreply@blogger.com