Trenberth's Energy Budget

[gslack]

Notice the little punk avoids the post reminding him of his outing.. What a coward...

 

[polarbear]

your pretty pictures seem to be incomplete.

First You tried to make Your case by confusing POWER and ENERGY,...then after I pointed this out to You You said it was an
oversight..
Then You tried to make the same erroneous case again using the "extra time" a photon spends in Your phantasy climatology world,..and then I did the math for You and the "extra time" was only 10^(-15) seconds by which You would have had to multiply the "extra photon POWER" You
try to use to get the "extra Energy"...and when You realized, that Your "extra Spencer`s example Power" evaporated by a factor of 10^(-15)
You come back here claiming that all photons are the same...we all know where and from who You got that one from...

So now You are claiming that these "all the same photons" each having been emitted by a radiation source say at 10 C....
that 10 of these photons stack up to a new and higher temperature of 10 times X ("climatology photon degrees")...

You got that nonsense from this idiot who told You that the photons coming from a 2000 degree black body are the same
as 10 photons coming from a 200 degree source and therefore that enough these are able to raise the temperature of a 2000 K black body

Then I pointed out the spectral energy shift an increased temperature causes (Wien`s law)

And now You claim that I don`t understand the laws of thermo-dynamcis, Plank`s, Wien`s and Kirchoff`s laws...( but You do..??? )

Let`s take a look at Kirchoffs laws again:

Kirchhoff's law of thermal radiation - Wikipedia, the free encyclopedia

>>>if we consider an ideal situation in which an enclosure with perfectly reflecting walls contains radiation with a certain amount of energy, then at equilibrium, this "photon gas" will have a Planck distribution of energies.[1] This will be true even though the walls are perfectly reflecting due to the very small amount of interaction between the photons themselves. The equilibration process will take a considerable amount of time, but the distribution of energies and radiation density will ultimately approach a Planck distribution. The next step is to realize that, as a result of the second law of thermodynamics, any enclosure at thermal equilibrium must also have a Planck distribution of radiation.[1] If this were not true, then we could bring that system in contact with the above ideal system, both at the same temperature, and by connecting them through an optical filter, we can have a net amount of radiation pass from one body to the other. For example, suppose in the second system, the density of photons at narrow frequency band around wavelength &#955; were higher than that of a black body at that temperature. If a filter that passed only that frequency band was inserted in an opening that connected the two bodies, then there would be a net transfer of photons, and their energy, from the second system to the first. This is in violation of the second law of thermodynamics, which states that there can be no net transfer of energy between two bodies at the same temperature.<<<<

So let`s see when You start violating it...shall we...:
You say that there is an energy transfer from a -10 C body to a +10 C.
So does this energy transfer increase when this colder energy donor gets colder...or does it increase
when the colder energy donor is warmer, say -9 C instead of -10 C.

Let`s see how fucked up You get with Your climatology energy transfer from a colder to a hotter body as we increase the temp. of the colder body..
How far can You go up using this idiotic climatology cold====> hot "energy transfer"
without violating all the existing laws of thermodynamics...

Can You go up from -10C on the colder body to 9.9999 C ....?
Remember Your theory has it that there are now even more photons coming from the
"colder body" as did when it was -10 C....

So, are You disputing all laws of thermodynamics...or are You claiming
that You understand these, but I don`t...?



So who`s picture is not complete...Your`s or mine.
If You want to play quantum physics with me make sure You are playing with a full deck Ian

 

[IanC]

IanC you have just shown how immature and ignorant you are..



No you little lowlife you have gone back and forth insulting and the pretending we insult you over and again for weeks now. You were condescending again and the fact you try and deny this is all the more pathetic..



Your words your previous post to me:

"the electricity is decreased to keep the heated bar at 150F, by 75w the last time I explained it, and the container is cooled because it is only receiving 675w instead of 750w."

Your words above:

I am somewhat amused that you think Spencer would change the power input into the heated bar according to a thermostat set for 150F when the whole concept of the experiment is to find the change in temperature when the second bar is added.

Recognize them? You better dumbass they are your words...

So...whats your excuse this time? You just said one thing and then tried to make it seem ridiculous I would think that one thing..So, which is this time dancing bear does he change the input level or not? Or does he change it then not change to suit your silly posturing ass?

Got an explanation didpshit? NO?????? Didn't think you would.. Now dance boy!



Really? Hows that work? What planet does something get hotter with less energy input?

the container temp is cooled (at least less energy is needed to keep it at 0F than when the heater was receiving full power) when the power is reduced to keep the bar at 150F.

Yes yes and what exactly does this address regarding my post or contention, other than the fact your are once again talking out of your ass? nice way to dodge and divert dancing bear..

do you understand the concept of input=output? I think it is funny that you insulted me for bringing up this scenario previously (and explaining it), while you think you are brilliant and original for bringing it up yourself! I suppose you didnt understand it the first time and you probably dont understand it now. hahahahahaha

Yes I do it seems you don't though because you made a contention then denied making it or was it I made it then you denied I made it? Does it matter? After all you knew that too didnt ya phony.. Sure ya did because you know everything don't ya... You even talked about this before but didn't if its wrong... Sure pal sure..

IanC, you want to explain to me how it is a self proclaimed expert on this didn't recognize fermats theorem? How about the fact you didn't know the two-slit experiment? Remember what you told me about that experiment genius? Oh I sure do.. YOU told me you can't learn about physics watching waves in a liquid... LOL..

Now dance, and squirm and google up terms to try and save face phony, but from here on out I will make sure I remind you how, when and where you got outed as a fake..

you said-

Really? Hows that work? What planet does something get hotter with less energy input?

you guys were the ones who insisted that the bar would not heat up. I simply told you what would happen if the bar was kept at the same temperature. in that case the power input would have to be lowered, which in turn would cool the container by the same amount that the input was lowered. to answer your question that was so badly phrased. in Spencer's experiment, I showed that the electrical power would have to be decreased by 75w to stay the same temperature, therefore if you decrease it by any value from 1-74w the bar will still increase in temperature. increased temp with decreased power when the second bar is placed next to the heated bar. on this planet or any other.

 

[polarbear]

ho hum, more ad homs

your pretty pictures seem to be incomplete.

I couldnt find the same flourescent light as in polarbear's picture. perhaps the unlabelled y axis is part of the problem.

flourescent lights do not emit like a blackbody. the curve is a different shape because it uses specific atomic bands of emission.

sodium lights are even less like a blackbody

Just a minute here IanC...
Where exactly did I say that sodium lights are a black body...???


Let me remind You that it was You and Your 7th grade "teacher" that said :

all photons are the same...
Nobody can tell them apart
photons coming from a 2000 K source are the same as photons coming from a 200 K source

Let me get this straight,...and now You are saying that we have to distinguish between "light bulb photons" and "Spencer`s metal bar example photons"...?

Are You in fact getting Your physics ideas from Archie comics books like wirebender remarked?

And by the way I did not post the light spectrum emitted by a sodium light, had I You should have seen several very bright yellow lines.

The 2 picture composite unfortunately had on the right side a fluorescent light...but I posted it as it was @ upload.wikimedia.org/wikipedia/commons/b/b0/Spectral_Power_Distributions.png because the left side has a 2800 K incandescent light...

what was I supposed to do here...use MSPAINT and clip this picture, then upload it to photobucket and put another url between the {img} tags so that You can`t nit-pick
And why would that bother You if I did not bother to waste even more time responding to these outlandish claims You continue to make here...after all according to You all photons are supposed to be the same, so why would "fluorescent light photons" not fit into Your "climatology billiard ball photon" scheme...?
Having a little trouble finding those in a -10 C Planck curve have You...according to you and the crap You quote they should be there..should they not..?


I also noticed that You rather play word games with wirebender than answering my question about the energy flow that You say flows from the cooler body to the hotter one...
Does in Your world of climatology flow more energy from the cooler (-10C) body to the hotter (+10C) one if the cooler body is only 0 C,...and even more if it is +5C...and more again if it is 9.99999 C...??

Then it follows according to your "climatology physics" that even more energy is transferred if this second body is @ +10 C....

...which as You should know by now, if You read a real physics book is in fact in gross violation of Kirchoff`s law..You know that pesty physics law, that Your grade 7 teacher, who turned "climatologist" called "imaginary"...

well You directed this @ wirebender, not me,...but I can`t resist:

I showed that the electrical power would have to be decreased by 75w to stay the same temperature, therefore if you decrease it by any value from 1-74w the bar will still increase in temperature. increased temp with decreased power when the second bar is placed next to the heated bar. on this planet or any other. __________________
"Nothing shows a lack of mathematical understanding more than an exaggeratedly accurate calculation.&#8221;
Carl Friedrich Gauss (1777-1855)

Tell me now how did You "calculate" the power to increase the temperature of the second bar...if You don`t even know the mass or the distance of the second bar....and exactly what temperature did You get for either bar in this calculation You pretend to have done...

In Missouri they say "show me...!"

Let me ask You another question about Your 2-bar calculation, regarding this:

increased temp with decreased power when the second bar is placed next to the heated bar. on this planet or any other.

Okay, let`s use a mass of 1 kg, the unspecified metal of your "metal bar" shall be lead as in Pb with an oxidized surface...looks pretty black...
the distance between the 2 is 10 cm,...and each cylindrical bar is say 5 cm thick...one bar = @ +10 C, the other @ -10 C

Allright IanC...do the math...because ""Nothing shows a lack of mathematical understanding more than faking it"

Oh and please do tell me also what happens when the bars mover ever closer...
and what happens when they make contact...does the -10 C bar still heat the +10 C bar..???
as "contact each other" lets use the distance at which they would bridge to make an electrical contact at 1 Volt...
According to You it still should...after all atoms can`t really touch each other...or do "climatology atoms" touch each other ...?

Hey lets use Carbon instead of Lead...Carbon rods are as close as You can get to a "black body"...and to make it even simpler let`s use square graphite rods, 10 cm long and 1 cm sides...and if You look it up the atom/atom distance for Carbon is 1.42 Ångstroms...

So Your "climatology back-radiation photons" coming from the -10 C rod flat surface should have no problem "heating" the + 10 C carbon rod flat surface...as opposed to doing so over a distance of 10 cm..

There is nothing unfair about this problem...I have seen You write the word "simple" many many times...and after all the back-radiation coming from the CO2 in the air that climatologists claim heats our planet does not come from a distant bar in a vacuum...it comes from the air, that undeniably touches the "balck body climatology earth"...so do the math and use the Avogadro gas constant & the Loschmidt number to calculate the Molecular distance at - 10C and 760 Torr ...and use that distance
If You can show, as Your "climatology" explains it with their "simple examples..." like Spencer etc.. that -10 C air can "warm" +10 C water over this very short "back-radiation" distance, then all the ice we see on our lakes must be just as imaginary as Kirchoff`s law the way established physics "mis-understood" it to this day...to quote Your source

 

[IanC]

Then You tried to make the same erroneous case again using the "extra time" a photon spends in Your phantasy climatology world,..and then I did the math for You and the "extra time" was only 10^(-15) seconds by which You would have had to multiply the "extra photon POWER" You
try to use to get the "extra Energy"...and when You realized, that Your "extra Spencer`s example Power" evaporated by a factor of 10^(-15)

when you jumped into the thread I was excited that there was finally someone who was going to discuss ideas without all the distortion and unfounded denial of wirebender and gslack. unfortunately that is not the case. I dont know if it because you couldnt be bothered to read anything I write or if you just like being an ignorant asshole like those other two.

with respect to the extra time for escape of radiation- I gave three basic options to consider. 1.the photon escapes at the speed of light 2.the photon is absorbed and reemitted by the CO2 molecule and escapes, at less than the speed of light 3. the photon is absorbed and reemitted towards the surface and does not escape.

the first option leads to the coldest surface because the shedding of radiation is the most efficient.
the second option slows the escape of radiation, making the surface warmer. you are implying that is the only mechanism for warming. I am stating that at the very least there is some warming. you agree that it takes time to absorb and reemit the radiation therefore you agree that CO2 warms the earth. the amount doesnt matter in this scenario, just the concept.
the third option stops the loss of radiation completely and returns the energy to the surface. obviously this is where the bulk of the inefficiency of radiation shedding due to CO2 occurs.

I made a point of not putting values on any of the three options, nor did I bring up any of the numerous complexities of multiple absorption/emission, transforming to different wavelengths etc. I brought up real world mechanisms that warm the surface by reducing the radiation loss. I have no problem with you pointing out that the second option doesnt account for much warming, I do have a problem with you implying that it was the only option and that you have 'refuted' my case somehow.

 

[IanC]

polar bear said-

Just a minute here IanC...
Where exactly did I say that sodium lights are a black body...???

you put up a pretty picture of visible light from an incandescent bulb. unfortunately it cut off the rest of the radiation, more than 80% from the blackbody curve.

flourescent lights and sodium lights are not blackbody curves.

 

[IanC]

polar bear said-

Just a minute here IanC...
Where exactly did I say that sodium lights are a black body...???

you put up a pretty picture of visible light from an incandescent bulb. unfortunately it cut off the rest of the radiation, more than 80% from the blackbody curve.

flourescent lights and sodium lights are not blackbody curves.

 

[polarbear]

when you jumped into the thread I was excited that there was finally someone who was going to discuss ideas without all the distortion and unfounded denial of wirebender and gslack. unfortunately that is not the case. I dont know if it because you couldnt be bothered to read anything I write or if you just like being an ignorant asshole like those other two.

I made a point of not putting values

on any of the three options, nor did I bring up any of the numerous complexities of multiple absorption/emission, transforming to different wavelengths etc.

I brought up real world mechanisms that warm the surface by reducing the radiation loss. I have no problem with you pointing out that the second option doesnt account for much warming, I do have a problem with you implying that it was the only option and that you have 'refuted' my case somehow.

So You won`t or can`t do the math, ...?
And You "excused" Yourself from not doing the math by saying:

I made a point of not putting values ..

and claim:

I brought up real world mechanisms that warm the surface by reducing the radiation loss.

Well haven`t You noticed, that in the real world this second "Spencer example balck body imaginary bar" is air containing CO2 and that they do make a point to put numerical values on this "extra heat"...and post these all over the internet and news papers

Not just with Your "Trenberth`s energy budget" but with every temperature graph with which "climatologists" claim that the temperature increase is caused by 380 ppm CO2...

I`m not asking You to do the math for an entire energy budget...
Do the math then for 2 bars in a chamber, just like Spencer`s "simple example" but use 2 carbon bars, 10 X 1 X 1 cm at a distance where they could make low voltage electrical contact and 1 bar is at +10 C, the other one is -10 C...

If You know math & physics then You should not have any problems with that...

Because any Kindergarten kid could figure out that if You don`t supply any more (extra) heat other than what You have in calories or watt seconds in these 2 bars that they will both stabilize at 0 C

Now please do use Your "climatology math & physics" for Your "climatology photons" and assign numbers and do the math...and tell us what temperature You come up with

Unless of course You want to continue to dodge this simple problem with "intelligent" answers like (me) :

being an ignorant asshole like those other two.

 

[wirebender]

Okay, let`s use a mass of 1 kg, the unspecified metal of your "metal bar" shall be lead as in Pb with an oxidized surface...looks pretty black...
the distance between the 2 is 10 cm,...and each cylindrical bar is say 5 cm thick...one bar = @ +10 C, the other @ -10 C

Allright IanC...do the math...because ""Nothing shows a lack of mathematical understanding more than faking it"

Oh and please do tell me also what happens when the bars mover ever closer...
and what happens when they make contact...does the -10 C bar still heat the +10 C bar..???
as "contact each other" lets use the distance at which they would bridge to make an electrical contact at 1 Volt...
According to You it still should...after all atoms can`t really touch each other...or do "climatology atoms" touch each other ...?

You had me holding my breath there for a second when you laid out the parameters using cylindrical bars. Ian seems to believe that adding sides to one bar or the other, or both will cause the cooler bar to warm the heated bar. I didn't know how he would handle a cylindrical bar.

Don't expect any set of rational calculations that actually incorporate any known physical law. The one pseudomathematical response he has gave me in this whole long discussion appeared to suggest that if you increase the number of faces an object was radiating from, you could increase the amount of power it was radiating. I don't know which physical law that is based on but I wish I did. Add enough faces and you could power a city from the output of a AAA battery.

For that matter, if you could harness the energy created by backradiation, you could power the earth. I never could get ian to name a physical law that supported and predicted backradiation. It is a matter of faith for him. Those free agent photons are just zipping around the universe carrying their power till they run into something solid and phenomena like interference and cancellation which result in the subtraction of EM fields don't actually decrease the strength of the fields by reducing the number of photons that make up the fields.

Apparently the fields are diminished or cancelled, but either retain the same number of photons or the number of photons by which the field was reduced break ranks with the rest of the photons in the field and continue on their merry way across the universe till they hit a brick wall or some other suitably solid bit of matter.

In ian's eyes, the subtraction of EM fields is magical. With that, I believe I have said enough. Good luck, but I don't think you will get anywhere. What you are doing is far more complicated than what I was doing and he didn't even get what I was doing. He thought it was a description of a magical process.

You guys have fun and let me know if he manages to disprove the SB law. Till that happens, my proof that the heated bar will not be further warmed by the addition of a second unheated bar stands.

 

[gslack]

IanC you have just shown how immature and ignorant you are..



No you little lowlife you have gone back and forth insulting and the pretending we insult you over and again for weeks now. You were condescending again and the fact you try and deny this is all the more pathetic..



Your words your previous post to me:

"the electricity is decreased to keep the heated bar at 150F, by 75w the last time I explained it, and the container is cooled because it is only receiving 675w instead of 750w."

Your words above:

I am somewhat amused that you think Spencer would change the power input into the heated bar according to a thermostat set for 150F when the whole concept of the experiment is to find the change in temperature when the second bar is added.

Recognize them? You better dumbass they are your words...

So...whats your excuse this time? You just said one thing and then tried to make it seem ridiculous I would think that one thing..So, which is this time dancing bear does he change the input level or not? Or does he change it then not change to suit your silly posturing ass?

Got an explanation didpshit? NO?????? Didn't think you would.. Now dance boy!



Really? Hows that work? What planet does something get hotter with less energy input?

the container temp is cooled (at least less energy is needed to keep it at 0F than when the heater was receiving full power) when the power is reduced to keep the bar at 150F.

Yes yes and what exactly does this address regarding my post or contention, other than the fact your are once again talking out of your ass? nice way to dodge and divert dancing bear..



Yes I do it seems you don't though because you made a contention then denied making it or was it I made it then you denied I made it? Does it matter? After all you knew that too didnt ya phony.. Sure ya did because you know everything don't ya... You even talked about this before but didn't if its wrong... Sure pal sure..

IanC, you want to explain to me how it is a self proclaimed expert on this didn't recognize fermats theorem? How about the fact you didn't know the two-slit experiment? Remember what you told me about that experiment genius? Oh I sure do.. YOU told me you can't learn about physics watching waves in a liquid... LOL..

Now dance, and squirm and google up terms to try and save face phony, but from here on out I will make sure I remind you how, when and where you got outed as a fake..

you said-

Really? Hows that work? What planet does something get hotter with less energy input?

you guys were the ones who insisted that the bar would not heat up. I simply told you what would happen if the bar was kept at the same temperature. in that case the power input would have to be lowered, which in turn would cool the container by the same amount that the input was lowered. to answer your question that was so badly phrased. in Spencer's experiment, I showed that the electrical power would have to be decreased by 75w to stay the same temperature, therefore if you decrease it by any value from 1-74w the bar will still increase in temperature. increased temp with decreased power when the second bar is placed next to the heated bar. on this planet or any other.

IanC do you really think you can talk in a circle and I won't call you on it?

You didn't respond to my question, you talked in a circle and pretended to...

You do that a lot dancing bear....

You are such a fake ..

 

[konradv]

You are such a fake ..

Expert on fake, eh? I guess you go with what you know!!!

 

[gslack]

You are such a fake ..

Expert on fake, eh? I guess you go with what you know!!!

Sorry junior this thread has an age limit.. Now run along and play with the other kids..

 

[konradv]

You are such a fake ..

Expert on fake, eh? I guess you go with what you know!!!

Sorry junior this thread has an age limit.. Now run along and play with the other kids..

Fake, fake, fake, fake fake. Nothing you post makes much sense. Your only value is as an object of ridicule.

 

[wirebender]

Expert on fake, eh? I guess you go with what you know!!!

Sorry junior this thread has an age limit.. Now run along and play with the other kids..

Fake, fake, fake, fake fake. Nothing you post makes much sense. Your only value is as an object of ridicule.

Do you want to prove that mathematically or are you just here to wave the pom poms for ian?

 

[westwall]

Then You tried to make the same erroneous case again using the "extra time" a photon spends in Your phantasy climatology world,..and then I did the math for You and the "extra time" was only 10^(-15) seconds by which You would have had to multiply the "extra photon POWER" You
try to use to get the "extra Energy"...and when You realized, that Your "extra Spencer`s example Power" evaporated by a factor of 10^(-15)

when you jumped into the thread I was excited that there was finally someone who was going to discuss ideas without all the distortion and unfounded denial of wirebender and gslack. unfortunately that is not the case. I dont know if it because you couldnt be bothered to read anything I write or if you just like being an ignorant asshole like those other two.

with respect to the extra time for escape of radiation- I gave three basic options to consider. 1.the photon escapes at the speed of light 2.the photon is absorbed and reemitted by the CO2 molecule and escapes, at less than the speed of light 3. the photon is absorbed and reemitted towards the surface and does not escape.

the first option leads to the coldest surface because the shedding of radiation is the most efficient.
the second option slows the escape of radiation, making the surface warmer. you are implying that is the only mechanism for warming. I am stating that at the very least there is some warming. you agree that it takes time to absorb and reemit the radiation therefore you agree that CO2 warms the earth. the amount doesnt matter in this scenario, just the concept.
the third option stops the loss of radiation completely and returns the energy to the surface. obviously this is where the bulk of the inefficiency of radiation shedding due to CO2 occurs.

I made a point of not putting values on any of the three options, nor did I bring up any of the numerous complexities of multiple absorption/emission, transforming to different wavelengths etc. I brought up real world mechanisms that warm the surface by reducing the radiation loss. I have no problem with you pointing out that the second option doesnt account for much warming, I do have a problem with you implying that it was the only option and that you have 'refuted' my case somehow.

Ian, I have been reading this and the various other threads for awhile now and the one thing that has become very plain is you are terribly ignorant of real world physics and the mathematical skills one needs to express the world of physics. I suggest you get yourself a CRC and a good book on math and study, study, study.

You are far, far out of your depth here and need a good deal of study to catch up.

 

[gslack]

Expert on fake, eh? I guess you go with what you know!!!

Sorry junior this thread has an age limit.. Now run along and play with the other kids..

Fake, fake, fake, fake fake. Nothing you post makes much sense. Your only value is as an object of ridicule.

Come back with one of your socks.. You know one of them that's older..or something...

 

[polarbear]

Then You tried to make the same erroneous case again using the "extra time" a photon spends in Your phantasy climatology world,..and then I did the math for You and the "extra time" was only 10^(-15) seconds by which You would have had to multiply the "extra photon POWER" You
try to use to get the "extra Energy"...and when You realized, that Your "extra Spencer`s example Power" evaporated by a factor of 10^(-15)

when you jumped into the thread I was excited that there was finally someone who was going to discuss ideas without all the distortion and unfounded denial of wirebender and gslack. unfortunately that is not the case. I dont know if it because you couldnt be bothered to read anything I write or if you just like being an ignorant asshole like those other two.

with respect to the extra time for escape of radiation- I gave three basic options to consider. 1.the photon escapes at the speed of light 2.the photon is absorbed and reemitted by the CO2 molecule and escapes, at less than the speed of light 3. the photon is absorbed and reemitted towards the surface and does not escape.

the first option leads to the coldest surface because the shedding of radiation is the most efficient.
the second option slows the escape of radiation, making the surface warmer. you are implying that is the only mechanism for warming. I am stating that at the very least there is some warming. you agree that it takes time to absorb and reemit the radiation therefore you agree that CO2 warms the earth. the amount doesnt matter in this scenario, just the concept.
the third option stops the loss of radiation completely and returns the energy to the surface. obviously this is where the bulk of the inefficiency of radiation shedding due to CO2 occurs.

I made a point of not putting values on any of the three options, nor did I bring up any of the numerous complexities of multiple absorption/emission, transforming to different wavelengths etc. I brought up real world mechanisms that warm the surface by reducing the radiation loss. I have no problem with you pointing out that the second option doesnt account for much warming, I do have a problem with you implying that it was the only option and that you have 'refuted' my case somehow.

Ian, I have been reading this and the various other threads for awhile now and the one thing that has become very plain is you are terribly ignorant of real world physics and the mathematical skills one needs to express the world of physics. I suggest you get yourself a CRC and a good book on math and study, study, study.

You are far, far out of your depth here and need a good deal of study to catch up.

Hi there old friend...!!! How are You today...??...By the way guys, my daughter was released from the hospital now...and we thank You all for Your concern and well wishing
@ Westwall...:
Have You noticed how the "lefties" are joining in, just as soon as You corner a pro-warmer or the author he quoted with the math..:

konradv
post number 312
Quote: Originally Posted by gslack
You are such a fake ..


Expert on fake, eh? I guess you go with what you know!!!

post # 314
Fake, fake, fake, fake fake. Nothing you post makes much sense. Your only value is as an object of ridicule.

Well he did not direct that at me, but at wirebender and Gslack...
All that`s missing now is "OldRocks" and his sidekick "Thunderfart"

But You know I don`t really blame IanC so much as the authors who spread this kind of garbage and have the nerve to pass themselves off as scientists...IanC does not really do that...:
The Amazing Case of &#8220;Back Radiation&#8221; &#8211; Part Three « The Science of Doom

The major reason that people give for thinking that DLR can&#8217;t affect the temperature is (a mistaken understanding of) the second law of thermodynamics, and they might say something like:

A colder atmosphere can&#8217;t heat a warmer surface
​

There are semantics which can confuse those less familiar with thermal radiation.

If we consider the specific terminology of heat we can all agree and say that heat flows from the warmer to the colder. In the case of radiation, this means that more is emitted by the hotter surface (and absorbed by the colder surface) than the reverse.
However, what many people have come to believe is that the colder surface can have no effect at all on the hotter surface. This is clearly wrong. And just to try and avoid upsetting the purists but without making the terminology too obscure I will say that the radiation from the colder surface can have an effect on the warmer surface and can change the temperature of the warmer surface.

Therefore, if your current belief is that radiation from a colder atmosphere cannot &#8220;change the temperature&#8221; of the hotter surface then you have to believe that all of the radiation from the atmosphere is reflected.

They all look the same to me&#8221; &#8211; The Energy of a Photon

This part is very simple. The energy of a photon, E:
E = h&#957; = hc/&#955;
where &#957; = frequency, &#955; = wavelength, c = speed of light, h = 6.6 ×10&#8722;34 J.s (Planck&#8217;s constant).
You can find this in any basic textbook and even in Wikipedia. So, for example, the energy of a 10&#956;m photon = 2 x 10&#8722;20 J.
Notice that there is no dependence on the temperature of the source. Think of individual photons as anonymous &#8211; a 10&#956;m photon from a 2,000K source has exactly the same energy as a 10&#956;m photon from a 200K source.

This guy can`t do the math any more than IanC...actually I think if IanC had really tried to do the math as I asked him now how many times...he would have done better than this fraud.
Anybody can make a mistake, make a typo or fuck up by accident with a decimal place...it can happen to all of us...but I`m certain this is not the case in this "science of doom back-radiation"...
It seem s to happen everywhere You look in "climatology"...we all remember how they fucked up with their Himalayan glacier predictions and demanded they should be cut some slack for this decimal place slip-up...but they have been doing so all over in their statistics,....everywhere You look...especially in their hockey stick "average temperature graphs " and just a little while ago again with the amount of ice in the arctic...it`s a pattern, they have these "accidental/mistakenly" placed decimal points, which produces numbers which they then all use and publish,....knowing full well that not very often somebody checks up on that...and when It is discovered then it was merely a slip of the pen..and the next fraud is already concocted again, using the same method

But let`s see how many SERIOUS mistakes this copy&paste "author" has made in just these few lines...
If You actually bother doing the math for his example You get for a 10 &#956;m wavelength an energy quantum of:
1.98 ^ (-21)

and he not only came up with a number 10 times higher than what it really comes out at if You do the math for E = hc/&#955;, he commits much more gross errors than that...totally disregarding what the "h" in the equation hc/&#955; really represents:


So what`s the dimensional difference You may ask Ian...well let`s take a look, it happens to be the difference what an Energy quantum really is, as opposed to the "photon energy" in "all photons are the same"....(as stipulated in "climatology physics")..:
Joule-second - Wikipedia, the free encyclopedia

The joule-second is a unit equal to a joule multiplied by a second, used to measure action or angular momentum. The joule-second is the unit used for Planck's constant.
In SI base units, the joule-second is

.

And the guy that said :

You can find this in any basic textbook and even in Wikipedia. So, for example, the energy of a 10&#956;m photon = 2 x 10&#8722;20 J.

has been using photons as if they were little pebbles that You can add up into a bigger one...

I am 100 % certain that this fraud has no idea what an angular momentum means in quantum (wave) mechanics..or that he would be able to comprehend it...

Maybe You will IanC, if You let me explain it to You:

Any rotating mass has an angular momentum which resists a change...
it is a product of angular speed in radians per second and a mass...in classical mechanical physics..
In wave quantum physics the "angular speed" is the FREQUENCY of the wave ...which resists a change in frequency and orientation the same way a mechanical angular momentum does...it is also the reason why the speed of light is constant unless You change the medium,...and then it only changes very very little..the energy quantum of a wave at any frequency is a discrete quantity ...You cannot take a fraction of this quantity, ... somehow divide it and use these fractional components or other whole photons of a lower energy to add up and to merge these to a new photon that carries a higher energy quantum and thus increase the angular speed or the frequency.







So who is the asshole here...this usurp who pretends to understand science or the guys who quoted and have been relying on him...:
The Amazing Case of &#8220;Back Radiation&#8221; &#8211; Part Three « The Science of Doom

There are semantics which can confuse those less familiar with thermal radiation.

If we consider the specific terminology of heat we can all agree and say that heat flows from the warmer to the colder. In the case of radiation, this means that more is emitted by the hotter surface (and absorbed by the colder surface) than the reverse.


However, what many people have come to believe is that the colder surface can have no effect at all on the hotter surface. This is clearly wrong. And just to try and avoid upsetting the purists but without making the terminology too obscure I will say that the radiation from the colder surface can have an effect on the warmer surface and can change the temperature of the warmer surface.

Therefore, if your current belief is that radiation from a colder atmosphere cannot &#8220;change the temperature&#8221; of the hotter surface then you have to believe that all of the radiation from the atmosphere is reflected.

At least IanC never PRETENDED to have done his own math...he actually told me, that he did not want to assign any specific numbers when I pressed him for it...

I have been trying in vain to explain the most elemental parts of quantum physics to IanC, like the wave duality of a photon,...and that no not all photons are the same as this quack keeps saying...and I did already mention to IanC, that a photon has no "resting mass" and what in quantum physics the energy of a wave is

Unfortunately he keeps going back to the same fodder troth...
In the future I suggest to You Ian, to keep an eye out who Your PC is connecting to when You go Googling for science, ...
and if it connects to something like "wordpress.com"....which is a favorite server for climatology propaganda, like what You have been defending here...don`t expect to find reliable information on a server like that,... as You would in a real & serious physics book for quantum physics.
and by the way what this "scientist has told You here was also utter bullshit:

a 10&#956;m photon from a 2,000K source has exactly the same energy as a 10&#956;m photon from a 200K source.

the Energy for a photon at a 10 &#956;m wavelength =1.98 * 10^ (-21)
and for a photon where the Planck curve peaks according to the Wien`s displacement law is at a wavelength of ~ 1.2 &#956;m ...
which has a quantum angular momentum energy of ~ 2 * 10^(-20)
which is in turn what the author of "the science of doom" had for :

So, for example, the energy of a 10&#956;m photon = 2 x 10&#8722;20 J

See it was not just a simple oversight like You made, when You confused POWER and ENERGY...
He never did do his own math...he did a copy&paste...of somebody else`s numbers but he did a typical "OldRocks" and a copy&paste on another "climatology science" garbage page and also confused the numbers what the Planck equation has for 1 &#956;m and 10 &#956;m...!!!

If that doesn`t expose him as a fraud, what will it take to (finally) open Your eyes InaC....??
Wirebender told Yu more than once that this "physics expert" is nothing more than a (fired) grade 7 school teacher, who is now cashing in as a "climatology" P.R. asshole.

And what I really would like to see is a "climatology" equation for "back-radiation" this time let`s settle on graphite rods, 10 by 1 by 1 cm with a 0.5 cm groove milled in the 2 sides of these rods as they face each other...now You have a perfect cavity and with graphite the best thing You can get as a "black body"..:
Black body - Wikipedia, the free encyclopedia

Blackbody radiation becomes a visible glow of light if the temperature of the object is high enough. The Draper point is the temperature at which all solids glow a dim red, about 798 K.[4][5] At 1000 K, the opening in the oven looks red; at 6000 K, it looks white. No matter how the oven is constructed, or of what material,
A closed box of graphite walls at a constant temperature with a small hole on one side produces a good approximation to ideal blackbody radiation emanating from the opening

One rod is at +10 C, the other at -10 C...then we move them together till the rods make contact and the optical cavity is 10 by 1 cm with the 2 channels milled 0.5 cm deep.
I can assure You IanC, that both rods if they have the same mass, will settle at 0 deg C and that You will not wind up with the +10 C rod being heated by "back-radiation" or any other means to more than +10 from the -10 C rod...
Has it not occurred to You, that if indeed the colder rod "back-radiates heat energy" then it just got also even colder...now it will emit less and less high quantum energy photons as this "climatology" Kirchoff`s laws defiant heat gain of the hot rod getting even hotter is supposed to progress the way these quacks are describing it...even when the 2 rods don`t touch with nothing but a channel for space between the 2,... there is no way You can gain energy from the colder rod...and the colder rod will gain energy from the hotter rod till You have a perfect thermodynamic equilibrium, and at the same time and only then the energy distribution as shown in Planck`s curves...and it will be exactly the same in both rods.

If You were to heat either rod, You would not even have an energy distribution over the spectral range as in Planck`s curves till the new & extra heat energy You are feeding to one of these rods has been distributed over the entire mass of the black body...and that takes a lot longer than You think.!!!!


You can get graphite rods from almost any welding supply store, scrape off the copper coating, square them off with a file and scratch or file a grove into each rod....
In Germany we say "studieren macht schlau, studieren und probieren noch schlauer"
{ study can educate, to study and to try it out yourself makes You even more educated and smarter}
That`s why in real science we don`t just use "examples" that exist only on paper, like "Spencers example" with his 2 rods, no mass, distances or temperature specified...and he managed to convince You that the hotter rod can get even hotter from the "back-radiation heat energy" that is supposed to come from the colder rod...in real science we have labs and confirm with experiments..
So go & buy 2 graphite rods, heat one and not the other one, use some thin wire to bind them together and then suspend them in a $20 thermos bottle ..





Well I had enough of this...I`m renovating my house at the same time as I`m reading & writing here...the putty is dry..I can start sanding and painting the next wall now...
Have a good time guys and good bye again for quite some time

 

[IanC]

You come back here claiming that all photons are the same...we all know where and from who You got that one from...

So now You are claiming that these "all the same photons" each having been emitted by a radiation source say at 10 C....
that 10 of these photons stack up to a new and higher temperature of 10 times X ("climatology photon degrees")...

You got that nonsense from this idiot who told You that the photons coming from a 2000 degree black body are the same
as 10 photons coming from a 200 degree source and therefore that enough these are able to raise the temperature of a 2000 K black body

this is the type of nonsense that pisses me off. you read the article so you know he didnt say any of those things.

1. he said a photon of a specific wavelength is exactly the same whether it is emitted from a cooler body or a warmer one. do you disagree? do you think there is some sort of brand labels on photons with warmer surfaces turning their noses up at Levis and only accepting Calvin Kleins?

2. he said that a warmer blackbody emits a set of radiation that totally contains the set of radiation from the lower blackbody. do you disagree? every Planck curve diagram for blackbodies that I have seen shows the lower curve totally under the higher curve, do you have some evidence otherwise?

3. the radiation graph of the lower temp blackbody is perfectly matched by the warmer body. these are the two sets of radiation involved with backradiation. each body is radiating into the other body this set of radiation, so there is no gain and no loss. these two set exactly cancel each other out but that does not mean the radiation magically disappears. it is still there, going back and forth. just like if there were two bodies at the same temperature, the radiation would cancel out. this is where you obviously disagree but where is your proof that the radiation of these two equal sets disappears?

4. the left over part of the warmer Planck curve, above the balanced out lower portion, is the reason why heat always flows from hot to cold. the warmer body ALWAYS radiates more, as can easily be seen in the graph of +10C and -10C. or in any other Planck curve for blackbodies of differing temperatures. this excess set of radiation is considerably shifted to lower wavelength which is more energetic.

5. the last concept he brought up is more counterintuative. the range of wavelengths radiated by the two bodies only 20C different is almost exactly the same. the only difference is at the far left, probably off the graph. the warmer body can give off a rare photon of higher energy that the cooler body cannot. but 99+% of the range is identical although the amount of radiation at each wavelength is always higher for the warmer body and the proportional excess continues to get higher as you move left into higher energy photons. when you compare larger temperature differences it is much easier to see how the average wavelength and intensity deliver more energy.

please note that the area of back radiation is the area under the pink curve which perfectly balances out radiation going back and forth between the two bodies. the area between the pink and blue curves is the radiation capable of transfering heat that only goes in one direction. a visual desciption of how and why the 2nd law of thermodynamics works.

please note that even with a 2000K difference in temperature the range of wavelengths for emitted photons almost completely overlaps. there is no difference in a 500nm photon emitted from a 3500, 4500 or 5500K source. there is no difference in a 15micron photon emitted from a 263, 283 or 5500K source.

 

[IanC]

http://spie.org/documents/Newsroom/Imported/1199/1199_4082_0_2008-07-08.pdf

Incandescent bulbs emit light in a manner closely resembling
Plank&#8217;s law of blackbody radiation. The law describes how a
body capable of absorbing all radiation contacting it (a blackbody)
will emit at a given range of wavelengths dependant on its
temperature. The inefficiency inherent in an incandescent bulb
is due to the fact that it emits both infrared and visible light at
temperatures between 2000 and 3000K. Specifically, the infrared
portion of the radiation consumes about 88% of the input electric
energy and becomes wasted heat (see Fig.1). Hence, recycling infrared
light into useful visible light would improve incandescent
efficiency.

The filter acts as a perfect transmitter for the useful visible
light and a perfect reflector for the undesirable infrared light.
The reflected light is re-absorbed which, in turn, helps to heat
up the filament. This infrared recycling process has two major
energy consequences. First, it reduces the amount of electricity
required to maintain a hot filament and thus improves electricto-
optical conversion efficiency. Second, it reduces the thermal radiation of the bulb as infrared photons cannot escape. Withthis approach, the energy efficacy of an incandescent light bulb
can be improved by as much as eight times.

first off, this paper demolishes wirebender's claim that no photons can be reflected back at the filament of a flashlight.

second, it demolishes the claim that back radiation can have no effect on the warmer body. if the 'generic' heat is not needed to produce the IR because it is already there, then that heat is freed up to produce higher energy radiation.


to answer polarbear's request for an explanation of my figures for Spencer's thought experiment-

I used wirebender's value of 1 for emissivity. I used wirebender's value of 1m2 area for each bar. I used wirebender's calculation of ~750w for 150F heated bar. I constructed a rectanglular block of proportions 2H,1D,1W for ease of calculations. I am also ignoring the temperature differential due to the shape. the face of the heated bar which will be towards the second bar is therefore 0.2m2 and radiates 150w.

the second bar is placed in the container. I have placed it as close to the first bar as possible to reduce stray radiation but not touching so there is no conduction. Spencer said the second bar reached 100F which works out to ~75w for 0.2m2 but did not point out that there would be a temperature differential across the second bar because the radiation is only coming in through the inside face.

the heated bar radiates into the second bar until equilibrium is reached and the inside face of the second bar is 100F and therefore radiating 75w. it is time to recalculate the heated bar. it is recieving 750w from the heater, and 75w from the second bar for a total of 825w. that means the 0.2m2 inside heated face is now radiating 165w.

time to check the inputs and outputs. the two inside faces are not exposed to the container so they do not output. 825w times the 0.8m2 effective radiating surface of the heated block is 660w. what about the second bar? it has 165w coming in, and 75w going back for a net gain of 90w. that 90w is unevenly divided between the 0.8m2 remaing of effective radiating surface. 660w and 90w =750w. input equals output.

I have not done the calculation for the temperature of 825w but I bet it is ~160F

my model is vastly oversimplified and only meant to give a rough idea of how the heat flows through the system. but it is vastly more sophisticated than wirebender's claim that both bars will be the same temperature with the full 2.0m2 area radiating equally.

 

[gslack]

IanC I can't wait to see them destroy that bit of garbage you just posted as science..

The "paper" you cited is not a scientific paper at all.. Its sales pitch.. Its selling a conceptual (at least from the way it reads) invention where as they invention attempts to recycle emitted IR into more incandescent visible light using nanostructures... As in a machine doing more work... HAHAHAHAHAHAHAA!

Notice the way it works at all? No of course not all you did was shout eureka I have done it and go with it... Again... LOL it recycles the heat from the light source through a system designed specifically for the task and does so by redirection.. Jesus IanC...WOW...

And one thing you didn't see because once again you didn't read it fully... They say this...

"Photon recycling via a metallic PBG filter is a promising new
route to creating a &#8216;cool&#8217; light bulb. Our next step is to study a
cylindrical filter geometry that is comparable to the commonly
used tungsten-filament configuration."

Ah they say "is promising" as in maybe, or could or theoretically by their calculations should.. So... Yeah...

They also say this:

"To illustrate the validity of our approach, we have employed
an ideal system that has a spherical blackbody filament enclosed
by the filter: see Figure 2(a). "

So they are using and "ideal system" "that has a spherical blackbody filament"... So then its not even based on anything real is it.. A perfect blackbody? Got one lying around do ya? LOL...

I will stop there because I really want to see wire and polarbear trash this further...LOL