Sunday, November 05, 2017

Green Plate Challenge


Izen has turned the Green Plate Effect and his animation into a video just in time for the bunnies to show it to their uncles at Thanksgiving or Christmas.  Copy this onto your smart phone for future use



and Science of Doom has issued a challenge to the back radiation deniers.


189 comments:

Fernando Leanme said...

A cooler object only warms a warmer object in the Underuniverse. Thats the place where the missing vacuum energy is hiding together with billions of socks, gloves, and Riddick's eye goggles.

Canman said...

I take it that someone saying only 300 w/m**2 is coming out is ignoring the 100 w/m**2 coming out of the green plate and going towards the blue plate. If the blue plate absorbs it, the energy numbers have to be recomputed. If not, it's reflected (or something) as another 100 w/m**2. I suppose whether it is or isn't has something to do with quantum mechanics, which according to Richard Feynman, nobody understands.

Susan Anderson said...

Crickets! Shameless crickets. Ask youself, do you know what you're talking about? Do you have something to contribute? Certainly Fernando and Canman don't. Time for reality to break in.

Betty Pound said...

Wow, the blue plate heats itself by recycling 50% of its sun-warmed energy back to itself. An amazing feat! But only to retarded cranks like Eli.

Eli, why would 67 of 267 not go to the 133 side? Why does the blue plate prefer to heat space and not the green plate?

Why are you an idiot?

Christian Anders said...

Nice video.

It visualizes that all comes down to a simple counting of energy quantities and one don't even need to apply any calculations in the first step. Thus it's enough to own the ability to count to see that 1st law requires a further heating of blue when green enters the game.

It took only 4 comments to show that some even lack the ability to count 1, 2, 3,...

I lost the bet against myself it would take at least ten comments for some well-known expert to show up and demonstrate that even a visualization of this simple counting operation is too much to digest intellectually for some.

I lost and have now to pay a bottle of really good red wine to myself. Well, could've been worse.

EliRabett said...

Member the rules BP. They apply here too.

EliRabett said...


BP loses the thread again. The blue plate does not heat itself. Some of the heat from the external source is recycled back to the blue plate by the green one, resulting in a warmer blue plate.

Bernard J. said...

BP, if the 'red' photon labelled "back radiation" does not go from the green box to the blue box, where does it go?

Betty Pound said...
This comment has been removed by a blog administrator.
Christian Anders said...

Just to make clear for silent readers, who exactly is the one confusing something:

"In your example, it inderd does heat itself beyond then sun power provided. You write self-contradictory nonsense."

It is not heating "beyond sun power provided" [ emphasize added ]. You can heat things up to a temperature. And as long as blue does not have reached the surface temperature of the sun, it can be heated further by reducing its speed of heat loss, until its temperature converges at the temperature of the sun. Then the limit is reached according to the laws of thermodynamics.

That you, BP, of all people confuse energy and temperature, is funny. It was you who insisted on a clear distinction between those two, wasn't it? Right you was, but you forget so fast...

Your statement about power is true in one way, namely for the whole system containing all plates that there might be. This system cannot emit more energy than it receives. The blue plate is only one part of the system blue/green and is not affected by a law that touches the sum of all parts. If you confuse "part" and "whole", your confusion of power and temperature gets a reason.
But you of all people will not make such a banal mistake, won't you?

"Colder objects (green plate) can not rewarm hotter objects because they lack the higher frequency and more intensity of photons."

This simply does not answer, what exactly happens to these photons. According to you they are forbidden to add their energy to blue, when coming from green. According to the 1st law energy is forbidden to be destroyed. Thus, if both, you and the 1st law are right, it leaves us with the very question what happens to the energy of those photons, which are forbidden to act in any known way.

Enlighten us with your answer to this question! You stated clearly enough that we are wrong, now make a step further and tell us what is right instead! We are waiting for 450 comments or so now.

John Garland said...

Re:
-----------
"Some of the heat from the external source is recycled back to the blue plate by the green one, resulting in a warmer blue plate."

That's not physics. Try again, old fool.
------------

If you put a perfectly reflecting mirror behind the blue plate, would its temperature rise? Yes. That's how reflective insulation works. Even a 0K mirror so long as it is perfectly reflecting. If you put nothing behind do you get a rise? No. Put a black body which absorbs 100% and emits 100% and you get an intermediate situation.

Easy peasy, Betts.

Canman said...

Susan A,

"The humorlessness of cultists may be their greatest gift to comedy"--Brad Keyes

https://cliscep.com/2016/05/07/breaking-mann-quits-climate-science/#comment-4083

Tom Dayton said...

It looks like Betty finally made a concrete, specific claim (sort of)! "Colder objects (green plate) can not rewarm hotter objects because they lack the higher frequency and more intensity of photons."

Well, concrete and specific except for the "intensity" part, which is too vague.

So let's address the "frequency" part. But that, too, is still vague, so Betty, please confirm that you mean that a photon can be absorbed by an object only if (a) that photon is greater than or equal to some particular minimum frequency/wavelength, and (b) that minimum frequency/wavelength depends on the temperature of the object the photon hits.

izenmeme said...

This is hopelessly one-sided, BP is obviously incapable of forming a cogent counter-arguement.

So to play 'devils advocate'...

What happens if the Blue plate (or box) is coated in gold?
It absorbes 80% of UV, but reflects 98% of photons with a wavelength longer than ~500um. down into the IR.
Below 300K almost all the energy in a S.B. spectra is below that wavelength.
As emissivity = reflectivity how on Earth can Gold cool any further ?!
Or be warmed by a S.B. spectra from a <300K object !?
izen (grin)

EliRabett said...

That's how cryostats and dewars work, also IR telescopes whose sides are mirrored.

izenmeme said...

@-"That's how cryostats and dewars work. It's also a different issue"

I know, as Science of Doom shows you have to calculate the energy flows as modified by the albedo of the objects/molecule involved in relation to the S.B. spectra.

It is possible to minimise warming from back radiation by judicious choice of emitting/absorbing surface. But unless you include a Maxwell's Demon, it has to balance. Otherwise you have a system that either destroys energy or creates it de novo.

(Its a trap!)

Betty Pound said...
This comment has been removed by a blog administrator.
Tom Dayton said...

Betty is a typical fake skeptic: Ignores my simple questions that are about direct consequences of her assertions.

Betty Pound said...
This comment has been removed by a blog administrator.
John Garland said...

izen...

Classic Maxwell demons and their ability to create perpetual energy machines are basic to Bett's "thinking".

Christian Anders said...

Betty said:

"[...] explain to your cult followers how an object emits 267 and receives 133 at the same time, yet this does not make the system average 200. How does this miracle occur?"

This has been explained uncountable times by far more people than Eli alone. The main reasons have been, are, and will always be:

a) The SB-law claims radiation from every surface according to the temperature of the surface. This mechanism is independent on other radiations that reaches the surface.

b) If the system averages to 200 for each surface, this would violate the 1st law of thermodynamics. This can be shown by simple counting of the energy quantities.

YOU have to tell why a) or b) ( or a) AND b) ) are wrong. Not everyone else why they are right.

CapitalistImperialistPig said...

Trying to explain thermodynamics to the back radiation deniers is a bit like trying to teach a pig to sing. It's a waste of time and it annoys the pig.

If someone actually want to check on back radiation all they have to do is get an IR radiometer and point it at the sky.

CapitalistImperialistPig said...

The fundamental failure of those who think they are invoking the second law of thermodynamics is the failure to understand the difference between radiation flux in one direction and net radiation flux. The latter is always from hotter object to cooler, but that only says that more radiation is flowing from hotter to cooler than vice-versa.

Those who fail to understand the distinction should never try to start a business.

EliRabett said...

As you know, annoying the pig can be a feature CIP

CapitalistImperialistPig said...

@Eli - ;-)

Christian Anders said...

No. In GPE the hottest object is the sun and it is not warmed further by the plates.

Your complete ignorance of the nearby fusion oven is root of your confusion. You can't even identify it as the hottest object of the system.

Betty Pound said...
This comment has been removed by a blog administrator.
Betty Pound said...
This comment has been removed by a blog administrator.
CapitalistImperialistPig said...

BP ought to be banned, of course, but her final comment above ought to be carved in stone, to celebrate the logical and rhetorical incoherence and futility she so eloquently represents.

izenmeme said...

BP gives a good example of back radiation.
When they feel the heat, they return it in lower form.
izen

John Garland said...

Betty: I mean, if I went around saying I was an emperor just because some moistened bint had lobbed a scimitar at me, they'd put me away!...
[obscene statement physically deleted]
Betty: Ah, now we see the violence inherent in the system! Oh! Come and see the violence inherent in the system! Help, help, I'm being repressed!
[second obscene statement physically deleted]
Betty: Ooh, what a giveaway! Did you hear that? Did you hear that, eh? That's what I'm on about! Did you see him repressing me? You saw it, didn't you?...

If only Betty were as .001% as talented as Michael Palin.

Betty Pound said...

Eli,

1) Why would 67 of 267 not heat 133, and instead choose to heat space?

2) How does an object send 267 AND receive 133 on an atomic/sub-atomic basis, without the system average becoming 200?

All you had to do was answer these two questions that I've repeated over 10 times, and you and your guests would not be insulted. But you refused to do so. Because you can't?

Betty Pound said...

John, you sure like to gaslight. All Eli had to do was answer my questions. If he didn't like my obscenities he could have quoted the non-obscene physics parts and answered them, while deleting the rest.

I told Eli that I will keep insulting him and his guests when he refuses to tell the truth.

I don't think I'm being "repressed". I'm repressing Eli with Truth, and he can't take it. I gave him an easy way out with my obscenities, and like a sucker, he fell for it - he now has an excuse to ban me. The wise will see through this.

marcoclimate said...

Betty, your questions don't make any sense. Literally. For your question 1, photons don't "choose" anything, they are not sentient. The radiation has no choice to go where it goes, it just goes where it goes because of the geometric and physical limitations put on the system: a blackbody, constructed as an essentially infinitely thin plate will radiate equal amounts in both directions. Period. It will not 'decide' to radiate less to one side, just because it receives more radiation from that side.

Your second question is in principle answered by a simple counter-question: why would it? It can't! If it would do what you think it does, it would create a situation where one plate (blue) receives MORE radiation than it emits, but remains isothermal, whereas the other plate (green) emits MORE than it receives and ALSO remains isothermal - isothermal with the blue plate at that. This is thermodynamically impossible. But that is related to the fact that you seem to believe in sentient photons, or photons that magically turn back when they realize they are going towards something that is hotter...

John Garland said...
This comment has been removed by the author.
John Garland said...

In case you're not caught up, my best understanding of his "reasoning" is that photons are something like perfectly conducting heat pipes and have a temperature rather than an energy. That is, that the blue plate radiates 240K photons that warm up the green plate. While the green plate is colder it emits cold photons which have no effect on the blue plate because, er, well, they are cold photons and "obviously" can have no effect. Once the green plate warms to 240K, the 240K photons are swapped back and forth in harmony as 240K photons are now emitted by both sides of the green plate. The ones hitting the blue plate just maintain the blue plate at 240K because "obviously" that's all 240K photons can do.

He has stated in another thread that this process can be extended to any number of successive green plates in a row. Steady state would be that all warm to 240K as the 240K photons from the Sun's heat are radiated/(conducted?) down the line.

Of course this doesn't work, but there it is. That is what he really is saying. And it does kinda make one wonder why the blue plate doesn't warm up to the same temperature as the surface of the Sun as the sun is emitting 5778K photons at the blue plate which somehow only heat it up to 240K. But that point never enters his thinking.

Bernard J. said...

"All Eli had to do was answer my questions. "

That's rich. I seem to recall that I asked you above what happened to the 'red' photon labelled "back radiation" and not getting a coherent answer.

Many people ahve pressed you on this, but you don't respond. Perhpas it's an awkward question for you, so let me reframe it. Imagine a laser that emits radiation of the same wavelength as the photons emitted by CO₂. Point it at the ground and pull the trigger. What happens to the photons? Do they pass through the Earth, or do they bounce off the ground and into space, every single one of them?

Take your time 'Betty Pound', but do tell. Can a CO₂ laser heat the ground?

Christian Anders said...

John Garland,

I think it was me who brought in additional plates.

That was because I think the easiest way is a simple bookkeeping of energy quantities, to show if 1st law is violated or not.

This leads to a geometric series, bringing blue further and further to the limit of 400 emittance from each surface.

But, indeed you're right, Betty's approach must lead to plates all at 200 emittance per surface, regardless of how many there are.

Tom Dayton said...

Betty, you still have not answered my two simple questions about your assertions.

Betty Pound said...

Some people knocked down their own strawmen and threw in some red herrings.

Some people still don't understand that photons are not all the same, and that a hotter object emits photons of a higher frequency. A hotter object sends all the same photons that a colder object sends to it PLUS more. A colder object will never heat a warmer object, i.e. the green plate will not recycle blue's photons to make blue warmer.

Betty Pound said...

continuing...

the green plate will not recycle blue's photons to make blue warmer. why? because it lacks the higher frequency photons to do that.

Eli is not doing physics. He created a pseudophysical simultaneous math equation where objects can boost their own temperatuee by recycling 50% of their provided energy off of a passive object.

It's getting very tiring dealing with people who do not want to join reality.

Christian Anders said...

So the very basic question: What is the energy carried by the lower frequency photons doing when the those photons reach the warmer plate.

Remember that energy is not to be destroyed because of 1st law.

Betty Pound said...

Christian,
The photons from the cold object travel to the warm object at the same time as the warm object sends its "subset of photons matching the colder object" to the cold object. It's a swap! If that wasn't the case the colder object would be cooling further.

How are you still not understanding this?

Timothy Folkerts said...

Betty, just out of curiosity, suppose you WANTED to make one plate 262 K (ie P/A = sigma T^4 = 266.7 W/m^2) and make a second plate 220 K (ie P/A = sigma T^4 = 133.3 W/m^2). The two plates are floating in deep space and each has an area of 1 m^2 on each face. Each plate has an adjustable heater. What power would you supply to each plate to achieve this result?

Christian Anders said...

Betty,

I understand that "swap" well.

Can you provide a textbook where this principle is described?

EliRabett said...


A while ago Eli worked out the emission spectra from each plate using Planck's black body law. It is posted at

https://docs.google.com/spreadsheets/d/1RsCR_f9CnTHQNZfplwTrCjGsLVG8lcyv_Tl6wjf-l_Y/edit#gid=0

BP is quite confused tho, the green plate emits fewer photons, and the spectrum is a bit red shifted compared to the emission from the blue plate, but all the frequencies are there.

Betty Pound said...

You put one plate at a distance from the sun that will make it 262K and the other plate at a distance that will make it 220K. Not very difficult.

Betty Pound said...

Eli, I see you're playing with the word "all". All the frquencies are there? LOL. Yes, the green plate has all the frequencies of the green plate - a tautology that explains nothing. Very pathetic.

Betty Pound said...

Christian, how many times did you have to repeat the 9th grade?

Betty Pound said...

Bernard J,
Does the co2 laser have an energy source? Are you claiming its powered by co2 alone? Didn't think so. What a stupid example.

Christian Anders said...

How many times will you refuse to answer a simple question?


I asked you what happens to the energy of those photons, who travel from green to blue, when they "collide" with blue.

Your answer: They are "swapped" with those leaving blue at the same time. Correct?

Then it should be easy for you to specify if the energy of photons from green has "entered" blue in this case, or not. What does "swap" mean, if they don't?

A simple swap sounds in terms of emission and absorption like that what everybody here is claiming since long.

marcoclimate said...

"The photons from the cold object travel to the warm object at the same time as the warm object sends its "subset of photons matching the colder object" to the cold object. It's a swap! If that wasn't the case the colder object would be cooling further."

Well, at least now Betty acknowledges that the Green Plate (GP) is sending radiation to the Blue Plate (BP), and that the BP takes up this radiation.

It's only a small additional step for Betty to realize that this means the BP in the presence of the GP is receiving more energy per time unit than in its absence, and thus *must* warm up.

Unfortunately, this likely is also the one impossible step for Betty to take...somehow Betty is stuck in thinking that photons can only be transferred into work when they strike an object that has a lower temperature than the source.

So, a simple question, Betty, and one I know you will ignore or handwave away:

If in your reasoning the GP will heat up to the same temperature as its source (the BP), why does the BP *not* heat up to the same temperature as *its* heat source: the sun!

izenmeme said...

One wonders how BP can look at themselves in a mirror.

Especially if it is colder than them.
:) izen

John Garland said...

-----
Well, at least now Betty acknowledges that the Green Plate (GP) is sending radiation to the Blue Plate (BP), and that the BP takes up this radiation.

It's only a small additional step for Betty to realize that this means the BP in the presence of the GP is receiving more energy per time unit than in its absence, and thus *must* warm up.
-----
Sadly, marco, no. You see, in BP's universe photons have a temperature not an energy. Cold photons just go on in and do nothing. Only photons which are greater than a plate's temperature can do anything there. The energy from cold photons simply disappears, apparently.

Our universe operates a bit differently and doesn't require drugs to maintain an awareness of.

Bernard J. said...

"Does the co2 laser have an energy source? Are you claiming its powered by co2 alone? Didn't think so. What a stupid example. "

I'm simply asking you what happens to photons emitted by a CO₂ laser, when it's pointed toward the Earth. The salient issue in this particular question is what happens to them, and not what energy source generated them - that's a different question...

So, can a CO₂ laser warm the surface of the planet, or not?

Tom Dayton said...

Betty, you still have not answered my two simple questions that are direct outgrowths of your claims. Yes or no to each, please. Do you believe a photon can be absorbed by an object only if

(a) that photon is greater than or equal to some particular minimum frequency/wavelength, and

(b) that minimum frequency/wavelength depends on the temperature of the object the photon hits.

Bob Loblaw said...

Betty's thought pattern reminds me of a post I made (tongue-in-cheek) over at Skeptical Science years ago.

https://skepticalscience.com/news.php?p=1&t=101&&n=1617#84925

"Thermodynamics, schmermodynamics. You've got the physics entirely wrong. They're not even "light" bulbs, they're dark bulbs. They don't give off light, they suck out dark. Haven't you ever heard of blackbody radiation?

The natural state of the world is to be filled with dark. A dark bulb lets us see by removing the dark, so it doesn't block the view any more. Proof? What does a bulb look like when it stops working? It's black - because it's full of dark. There's no more room for it to suck more dark."


This is an internally self-consistent explanation. The only problem is that it is completely contrary to our knowledge of physics and electricity.

Betty's explanations are internally self-consistent, as long as you follow a highly-compartmentalized method of thought. You can't collect several comments together and try to see consistency amongst them - you're only allowed to look at comments one at a time, in isolation. That appears to be how Betty's mind works. each comment, internally consistent, only to be forgotten ("outside the compartment") when another thought comes about. Betty can even switch compartments mid-sentence.

It's a scary world.

Betty Pound said...

A warmer object receiving photons from a colder object can not *warm up*, because the colder object doesn't have the higher frequency photons to do so.

All the photons that the colder object sends to the warmer ... the warmer sends to the cooler, PLUS more.

Keep trying guys. You can't seem to follow directions in physics. You have to resort to strawmen. Pathetic human beings.

Betty Pound said...

I wonder why izen thinks that a mirror warms you up just because it reflects your photons back to you.

Tom Dayton said...

Betty, you wrote "because the colder object doesn't have the higher frequency photons to do so." That implies (actually, requires) you believe the following. Please answer "yes" or "no" to the following two simple questions.

A photon can be absorbed by an object only if

(a) that photon is greater than or equal to some particular minimum frequency/wavelength, and

(b) that minimum frequency/wavelength depends on the temperature of the object the photon hits.

Christian Anders said...

Betty

"All the photons that the colder object sends to the warmer ... the warmer sends to the cooler, PLUS more."

Yes. That is, so far, what everybody else here also thinks.

You simply don't answer the question: What happens to the energy of the photons sent by green when arriving at blue?

- absorbed ( and reemitted maybe )

- reflected

- destroyed

Betty Pound said...

Tom, how many times did you repreat the 9th grade?

A warmer object will absorb all the photons from a coldect object but this will NEVER raise its temperature because you are missing the higher frequency photons necessary for that.

http://www.sun.org/uploads/images/mainimage_BlackbodySpectrum_2.png

You see, the 300K object will send all its photons to the 500K, and this will do nothing to raise 500K's temperature.

The 300K photons is swapped with the 300-subset-of-500K. People on this forum like to pretend that I'm claiming thr 300K is LOST to the 500K, but that is a strawman.

You get it now?

Christian Anders said...

Betty,

"I wonder why izen thinks that a mirror warms you up just because it reflects your photons back to you."

I wonder why first aid measures claim to wrap an undercooled human body in reflective foil?

Tom Dayton said...

Betty, you wrote "A warmer object will absorb all the photons from a coldect object but this will NEVER raise its temperature because you are missing the higher frequency photons necessary for that." Thank you for answering my question. That raises another question:

Do you believe that the warmer object's absorption of those "lower frequency" photons entails the warmer object acquiring energy (some energy, greater than zero) from each of those photons?

Betty Pound said...

No, Tom, no energy is gained, because the warmer object is draining its energy (sending photons) to the colder object.

You want to eat your cake and have it too.

Tom Dayton said...

Betty, I should clarify my previous question to you: I am asking about your definition of "absorption"--whether absorption of an individual photon by matter, by definition means that matter acquires a non-zero amount of energy from that individual photon.

I'm not asking how much energy is absorbed, other than whether it is greater than zero. I'm not asking whether the absorbing matter later emits photons. I'm not asking about the cumulative effect of multiple such individual photons. I'm not asking what happens even a femtosecond after absorption, only what happens literally in that moment.

Betty Pound said...

Buckets, Tom, Buckets. To fill a bucket you need to empty it out. Want a bucket of "photons"? You need to empty out the "photons" in the bucket. Except you have no extra disposable container.

I've asked Eli repeatedly how it's possible for an object to send 267 AND receive 133 at the same time without the system average going to 200 W/m2.

It seems you want to have your cake and eat it too. Sorry poor guy, it can't happen.

Betty Pound said...

Christian, and what temperature would reflective foil raise your body temperature to? 100F? 110F? 120F?

Christian, were you one of those boys that liked to torture small animals?

Tom Dayton said...

Betty, you did not straightforwardly answer my question. Here is my interpretation of your answer. I'm sorry if I misinterpreted.

When an individual photon is absorbed by matter, that matter may or may not acquire a non-zero amount of energy from that individual photon:

a. If the matter already is "full" of energy, then the matter will acquire zero energy from the photon.

b. If the matter is not already "full" of energy, then the matter will acquire more than zero energy from the photon.

Please say whether I interpreted your belief correctly.

Betty Pound said...

Tom, does the warmer object send photons to the colder object, or does the colder object perpetually cool to 0K?

I'm arguing with jerks that just repeat the same nonsense without ABSORBING what they're told.

Tom Dayton said...

Betty, please answer my most recent question. Your response is irrelevant to that question.

Betty Pound said...

I already answered your question. The hotter object is "full" of itself, therefore a colder object can't raise the hotter object's temperature.

John Garland said...

Your problem as you have been repeatedly told is you think the green plate is warming the blue plate. It isn't. I repeat, the green plate is not heating up the blue plate.

As you have been repeatedly told, the Sun is doing all the heating. All the relationships of the plates--i.e., black bodies--do is define how the how the Sun's energy flows through subsequent black bodies.

The max temp for the blue plate is not 240K as you seem to be thinking. 240K is the temp when 400 W/m2 are received and half emitted back towards the Sun and half emitted to space.

Let's put a perfect reflective plate behind the blue plate such that the blue plate can only radiate from its sunward face, what temp does the blue plate reach now at steady state with 400W/m^2 in and all radiation from the back side reflected back into the blue plate? Hint: It is not 240K by a long shot. Yet the reflector is not heating the blue plate in the least, the Sun is. What temperature do you suggest the blue plate would reach in this new setup? I'm guessing you think 240K. You'd be wrong.

BTW, re. your wrong-headed blanket example, 37C is the body's temp with internal heating and cooling mechanisms in balance. Disallow cooling with a perfect insulating blanket and the human body will very much warm to more than 37C--it's called heatstroke. This is why astronauts in EVA suits wear a liquid cooled undergarment that can remove up to 2 million joules per hour, for example.

Tom Dayton said...

Betty: No, you did not answer my question, because I did not ask you about the temperature of the object, nor about the whole object. I asked you specifically and narrowly:

When the mass is "full" of energy (that's your term and concept), all photons hitting it are absorbed (that's what you wrote). When an individual one of those photons is absorbed, does that absorption contribute energy to (transfer energy to) the mass?

Christian Anders said...

If you wait long enough, due to the foil your body temperature can in fact rise above the normal 37.

Not because the foil warms you, but because the heat produced by you is not allowed to leave properly. This may result in higher final temperature.

This heat stroke by insulation is an issue for astronaut suits. Maybe NASA engineers simply don't know physics and waste money and effort?

Christian Anders said...

Again you are right and just refer to the problem in the way everybody else here does.

You are not gaining energy in blue due to photons from green, because more photons leave green and so in total no energy is added to blue from the direction of green.

Correct?

But is this loss towards right not of lower value than it would be without photons from green? It is still a loss and thus no warming, but is this loss slower with green than without grein?

What is your statement to that?

EliRabett said...

BP scribbles: Christian, and what temperature would reflective foil raise your body temperature to? 100F? 110F? 120F?

Survival or space blankets can raise your body temperature. However, if it raises it too high, you will be sick or die, which is why people throw off blankets when they get too hot.

The general principle is the same for IR reflective coatings in halogen lamps which reflect IR from the colder envelope back onto the filament, raising its temperature and thus increasing the amount of visible light "Recycling the radiant heat, less energy is required to raise the filament temperature to the optimum level" Bulbs using this technology last twice as long
https://www.topbulb.com/blog/ir-mean-halogen-bulb-description-codes/

Christian, were you one of those boys that liked to torture small animals?

More likely one of those who enjoys twitting the ignorant.

Betty Pound said...

^ canonization of the cranks ^

Betty Pound said...

"Your problem as you have been repeatedly told is you think the green plate is warming the blue plate. It isn't. I repeat, the green plate is not heating up the blue plate.

As you have been repeatedly told, the Sun is doing all the heating. All the relationships of the plates--i.e., black bodies--do is define how the how the Sun's energy flows through subsequent black bodies."

That's not what Eli's diagrams show. He is in fact saying that the Green recycles 50% extra of sun's energy from Blue, back to Blue.

You can write all the gibberish you want, it means nothing. The diagram and math show what it shows.

Don't be a liar. The added Green plate magically turns the sun's 400 into 533. So if you're not claiming the Green plate is a heater, then you are left with an equally stupid position of claiming the Green plate tells the sun to send more radiation to Blue. Such a thing does not exist in physics.

Your words are nothing more than cheap rhetoric, not physics.

Betty Pound said...

Eli, you didn't answer my question. How much will space blankets raise your temp? Show me a temp vs. time diagram.

Next time quote scientific or engineering papers, not blogs.

Again, you confuse IR loss prevention with energy generation. Pathetic.

Betty Pound said...

Tom, it is absorbed ... the hotter object emitted a photon and made space for a photon from the colder object.

Tom Dayton said...

Betty, thank you for answering my question with your answer that the photon from the colder object is absorbed by the warmer object via the mechanism of "the hotter object emitted a photon and made space for a photon from the colder object."

Your answer requires the hotter object to emit a photon in one of these ways:

(a) Before the colder object's photon arrives at the hotter object. That violates temporal causality.

(b) After the colder object's photon arrives, but before that photon is absorbed. That requires the hotter object to keep the photon intact inside the hotter object's matter until the hotter object emits a photon. That mechanism has never been described by anyone ever in the history of physics as far as I know.

(c) Simultaneous with the photon's absorption. That, too, violates temporal causality.

Have I missed your explanation of the mechanism?

Tom Dayton said...

Betty, so are you saying that the hotter object and colder object coordinate with each other to each emit a photon of the same frequency, at the exact same time?

marcoclimate said...

I think Tom understands quite well what he is being told: he's being told fantasy physics by you, Betty.

Betty Pound said...

They both emit photons spontaneously which travel at the speed of light. Physics. There is no backed-up traffic. Ever heard of Planck time and speed of light? They are moving in DISCRETE time chunks.

In any case, if there was backed-up traffic it would violate ideal blackbody curve. Eli said the plates are ideal blackbodies.

Betty Pound said...

Typical leftist comment. No intellectual content.

winston said...

I'm wondering if Betty Pound is perhaps related to Mark V Shaney? The literary style is remarkably similar, and if true it would provide an excellent demonstartion of the futility of attempting to reason with a Markov chain.

marcoclimate said...

Winston, I don't think a Markov chain is capable of maintaining the same ignorance of basic physics throughout so many comments.

In the meantime, Betty still deliberately evades the question why Betty believes the Blue and Green plate will equilibrate at the same temperature...which is much lower than the temperature of the sun. Surely, in the strange world of Betty, the hotter sun must continue to warm the blue plate until it is as warm as the sun. If not, it makes no sense that Betty is adamant that the green plate must warm up to the same temperature as the blue plate, because the blue plate is the heat source for the green plate, just like the sun is the heat source for the blue plate.




izenmeme said...

@-BP
" and what temperature would reflective foil raise your body temperature to? 100F? 110F? 120F?"

If you weight around 200 pounds then because your metabolism provides about 100W you will heat up if enclosed in a (completely) reflective blanket at just less than 2F per hour. So if you are covered for a day your body temp would reach over 120F.

Except you would have died before this and your metabolism would have stopped producing that much energy. Decomposition produces less energy so after death you would heat up more slowly.
izen

(there are numerous plumbing/water heater sites that provide these sorts of calculations)

Betty Pound said...

That's a lie.

Betty Pound said...

The sun at that distance can only provide 400 W/m2. Because the sun is a radial point source both plates will warm to 200 W/m2, or 244K. There is no reason why the blue plate must send 267 to space, when the green plate is 133.

Betty Pound said...

You forgot to mention that in Eli's imaginary world the blue plate ends up heating itself off of the green plate. There is no denying this fakery.

John Garland said...

No Betty...The green plate insulates to blue plate from the cooling it would experience if it were not there. It heats nothing. The Sun does all the heating.

John Garland said...

And don't forget to tell us how a highly/perfectly insulating suit will not cause heat stroke.

Betty Pound said...

Yet this insulator magically raises Blue's temperature beyond what the sun alone can do. That makes it a heater, regardless of your gibberish.


Heater = adds heat.

Blah blah blah, the diagrams show blue heating itself off of green. There is no extra arrow coming from the sun because of green's presence.

A suit will not raise your temperature.

John Garland said...

Yup. Just like a blanket can raise your temperature beyond what your body can do alone lying on a blank mattress in a cold room.

Slowly you're starting to get it, Betts! Good job!

BTW, NASA, if not you, knows that yes a space suit with no cooling can kill. They are not ignorant of thermodynamics unlike you.

Betty Pound said...

That's a lie.

Christian Anders said...

Betty,

the blue plate is much cooler than the sun when emitting 400 ( which is 200 for each surface ). It is still much cooler when emitting 533 ( which is 266 for each surface.

Equilibrium with the temperature of the sun is at 400 for each surface. Then blue has the same temperature as the sun.

So there is much potential by what the sun can do.

Well, and as to suits and human body temperature: Tell your theory to the millions of medic aid assistents that proof otherwise.

If they wrap an cooled human with a body temperature of 32C in a thin reflective foil, his temperature rises again, even in the 20C envoirenment of the ambulance car. If they don't use the foil, the human might die of cold.

Tell them they are successfully doing something impossible every day.

Your comments to this point "It's a lie" only show how helpless you are facing every day physics at practical work with your approach.

Christian Anders said...

Funny you have nothing more to say to experimentally proven physics.

If you encapsulate yourself in a very good insulator you are going to die of heat shock after a time, even if you are out in a winter night.

But please feel free to try and find out.

izenmeme said...

@-BP
"There is no extra arrow coming from the sun because of green's presence.
A suit will not raise your temperature."

No point in wearing a coat, or a suit, or any clothes when outside on a cold day then ?
izen

Betty Pound said...

On a cold day your body is heating the environment. If you want to keep warm you will wear a coat. The coat will never raise your skin or internal temperature above 98.6F. Your coat just blocks covection and reduces cooling.

Betty Pound said...

Link us to the experiment, filthy liar, or shut up.

John Garland said...

You are completely free to wear a space suit with no cooling any time you want! You will die. Even on a cool day. But at least you might learn.

winston said...

This is all ripping fun, although feeding the troll is probably unwise, but I have a small request of Betty. When saying "that's a lie" or calling some one a "filthy liar" could you add a bit of context so we know ehich previous comment you are referring to? Thanks in advance.

marcoclimate said...

"The sun at that distance can only provide 400 W/m2. Because the sun is a radial point source both plates will warm to 200 W/m2, or 244K. There is no reason why the blue plate must send 267 to space, when the green plate is 133."

In this Gedankenxperiment it doesn't matter at what distance you place the green plate from the blue plate. If the two plates are perfectly flat and perfectly parallel, and there is vacuum in-between, you can place the green plate as far away as you want: it will *still* not heat up to the same temperature as the blue plate, because the blue plate receives MORE energy per time/unit than the green plate.

Also, you now ignore your magical photons, and go back to bad accounting. The Green plate nicely emits all of the 267 W it receives, while the blue plate nicely emits all 533 it received.

Christian Anders said...

Oh Betty,

are you serious with this:

"Link us to the experiment, filthy liar, or shut up."

20 seconds of google leads you to the official NASA statements about the problems as well as the patent from 1969 for the cooling garment. You will also find engineering papers with real measured values.

So you really say that this physic, successfully applied by generations of engineers and working the way it is supposed to every day, is wrong?

Just to give you a very small viewe of what is to be found there on the web:

https://www.nasa.gov/pdf/188969main_Keeping_Your_Cool.pdf
( a veary easy one for you to start with )

https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19710009964.pdf
( The patent of the cooling garment from 1969. Explaining nicely, what cooling is needed for. )

https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20150003483.pdf
( A paper with actual measurements. You might find all the wrong physics measured and applied there highly entertaining. )

It is hard not to see all the evidence, but I'm quite sure that you can manage this trick.

Just because I'm curios: What do you claim is happening to the sun, if it is surrounded by a sphere of a perfectly reflective material, the material to be at 0K?

Christian Anders said...

And what happens if you reduce cooling of something that is constantly heated by a power device at the same time the cooling is reduced?

- Will the power device reduce its output also?

- Will it work on?

If the latter is true: What is the energy doing that is produced as ever before, but now is not transported away as fast as before, because cooling is reduced?

Can you answer this simple question about naked and clothed energy generators like human bodies?

izenmeme said...

@-BP
“Your coat just blocks covection and reduces cooling.”

The green plate just blocks radiation and reduces cooling.
You DO understand it really !

Betty Pound said...

The Blue only received 400 from the sun - the only raw power source.
Yes, it does matter how far away the green plate is from the blue.

Betty Pound said...

You filthy liar.
There is nothing about the space suit heating you up via your own body temperature.
The space suit is for hot (and cold) EXTERNAL environments.

Betty Pound said...

Reduced cooling does not lead to heating. Reduced breaking is not accelerating...

Betty Pound said...

The Blue plate can easily heat the green to thermal equilibrium. In your case, 533 can easily heat 267 to an equilibrium of 400.

Did you know that the sun was responsible for bringing your kitchen to room temperature (20C)?

Does hot warm cold in your sun-warmed kitchen? Or does hot warm itself up further by recycling 50% of its energy off of a cold object?

Betty Pound said...

I'm the one feeding the trolls. If you can't tell what lies they are feeding you, then you are hopeless or just another troll.

Tom Dayton said...

Thank you for answering my most recent question, Betty. Summary of what you wrote:

1. The colder object emits photons independently from the hotter object's emissions. The colder and hotter objects emit photons spontaneously, without coordinating with each other. The hotter object does not emit a photon in anticipation of the arrival of a colder object's photon. The hotter object does not empty its energy bucket in anticipation of the arrival of a photon from the colder object. (You wrote that.)

2. There is no delay in absorption of a colder object's photon by the hotter object, due to the hotter object being "full" of energy and needing to emit a photon to make room for the arriving photon. There is no backed up traffic. (You wrote that.)

3. All photons from the colder object that hit the hotter object are absorbed by the hotter object. None are reflected. None pass through the hotter object and come out the other side. Absorption of the colder object's photons entails all their energy being absorbed by the hotter object. (You wrote that.)

Let's use your above beliefs in the context of a single batch of photons from a colder object, all arriving at the hotter object together--at the same time. There are no more photons behind those. (The colder object was removed after that first batch.)

The necessary logical conclusion from all of what you wrote, is that all the colder object's photons increase the total energy of the hotter object. The only way to prevent that increase in total energy is for the hotter object to lose energy by emitting photons. But you wrote that the hotter object does not do that tit for tat, instead emitting spontaneously. "Spontaneously" means in accord with SB. SB means that the following process unfolds over time:
(1) First, the total energy of the hotter object increases due to energy gained by absorption of photons from the colder object. That happens regardless of how much energy the hotter object already has, because as you wrote, there is no prevention or delay of absorption.
(2) Later in time, the hotter object emits photons. That is not tit for tat regarding the absorbed individual photons. Instead it is in response to the new, larger, total energy content of the hotter object. How much energy is lost by the hotter object is dictated by SB, which depends on the total energy in the hotter object. That energy loss does not happen all at once, but over time, during which the total energy of the hotter object remains larger than it was before the photons were absorbed.

During the time interval in between the hotter object's total energy increasing by absorption and then finishing decreasing by emission, the hotter object has more energy than it did before the colder object's photons arrived. By definition of temperature, during that time interval the hotter object's temperature is higher than it was before the photons arrived. The hotter object's temperature has been increased by the colder object's photons, for a non-zero period of time.

Betty Pound said...

OK, got it. You're a bloviating imbecile.

"Later in time"
There is no later in time. Photons travel in discrete time chunks. There is no backed-up photons from the cold object.

"The hotter object's temperature has been increased by the colder object's photons, for a non-zero period of time."

No, you need higher frequency photons.

Betty Pound said...

When a cold object photon releases a photon, the hot object releases a photon. Like all cranks, you overload one side of the equation.

Tom Dayton said...

It's been fun getting Betty to plainly state flat contradictions of his previous statements.

EliRabett said...

BP: When a cold object photon releases a photon, the hot object releases a photon. Like all cranks, you overload one side of the equation.

So they use Twitter, What's App or Tinder to communicate? Just askin.

Timothy Folkerts said...

Betty says: "Reduced cooling does not lead to heating. Reduced breaking is not accelerating..."

If only you had the insight to understand own analogy! Suppose my car is rolling down a hill in a steady-state condition -- the brakes are applied just enough to keep the car from speeding up or slowing down. If I reduce the braking, then the car accelerates.

Reduced cooling (restricting the energy outflow) -- when the heating remains constant (maintaining the energy inflow) -- does indeed lead to increasing temperatures. This is so blindingly obvious and intuitive that any 10 year old should be able to grasp it!

Christian Anders said...

Betty,

"There is nothing about the space suit heating you up via your own body temperature.
The space suit is for hot (and cold) EXTERNAL environments. "

Yes, that is also one of their uses. But read thesources carefully: NASA engineers believe the cooling garment also prevents overheating due to the power of the own metabolism.

Question: Why?

"Reduced cooling does not lead to heating. Reduced breaking is not accelerating..."

Again you forget that there is a source of energy, or, in your analogy thus has to be, a source of speed.
If the motor delivers constant power in unchanging gear, than reduced breaking leads to what? Heh? Can't hear you?

And yes, that is a really good analogy for teaching the principle. I have never thought of it before, thank you for that.


marcoclimate said...

Betty is telling us that light bulb manufacturers are putting reflective coatings in halogen lamps for no reason whatsoever. After all, the reflected IR cannot make the filament hotter in any way...in Betty's world.

Betty Pound said...

Rolling down a hill, you have the extra effect of gravity. On a flat surface, with only power provided by car, reduced breaking does not lead to acceleration, it leads to less deceleration.

Betty Pound said...

Why don't you find the exact quote in your links which claims a space suit spontaneously heats you up using just your body's emitted energy.

"If the motor delivers constant power in unchanging gear, than reduced breaking leads to what? Heh? Can't hear you?"

Ease up on your brake and see if the car accelerates. It doesn't. Brakes don't speed up your car, imbecile.

Betty Pound said...

Where did I say that "manufacturers are putting reflective coatings in halogen lamps for no reason whatsoever."

Find that quote.

"After all, the reflected IR cannot make the filament hotter in any way."

That's right, it won't make it hotter, but it will dissipate less energy to environment.

Betty Pound said...

Pathetic strawman. They don't need to communicate. Both hot and cold objects emit photons in discrete time chunks.

What university did you go to?
You need to get your degree revoked.

Christian Anders said...

Betty,

"Ease up on your brake and see if the car accelerates. It doesn't. Brakes don't speed up your car, imbecile."

Done.
Experimental setup: Nissan Micra 2014. Gear 1, no use of gas pedal. With slight breaking the speed equilibrated at 5km/h. Losing the breaks accelerated the car to 9km/h.

Now you!

Ken Fabian said...

Betty - are you sure you are not confusing an inert object, that has only the heat that it starts with, and a body - human body in this case - that produces it's own heat (via exothermic chemical reactions)? The inert body won't and can't get warmer by wrapping it in insulation, but the human body keeps making heat and will - if you can't take that perfect, 100% insulating coat off and shed excess heat once you fully warm up - exceed "normal body temperature". You would start to sweat but that won't cool you down and you will keep getting hotter until you start getting heat stress and heat stroke - (hyperthermia - which is defined medically by having a raised core body temperature of 40.6 °C (105.1 °F) or above).

Of course that doesn't make it a good analogy for what the greenhouse effect in atmosphere does, just a good example of what insulation does. The key things with the greenhouse effect are having some solar energy at wavelengths capable of travelling in through the atmosphere without warming it but still warming the ground and water, which radiates some of that heat at different wavelengths that do warm the atmosphere on the way back out.

(Note - removed previous version of comment and replaced it.)

Timothy Folkerts said...

"Rolling down a hill, you have the extra effect of gravity."

YES! Exactly! And in this example you have the 'extra effect' of incoming sunlight! In both cases, a continuous input of energy (either sunlight to the plate or gravity to the car) must be balanced with a continuous siphoning off of energy (either radiation escaping to the rest of the universe or heat generated in the brake pads) if a steady state is to be maintained. If you change EITHER incoming or outgoing, then the steady-state conditions will be disturbed unless/until a new steady-state can be re-established.

izenmeme said...

@-BP
"Reduced cooling does not lead to heating. Reduced breaking is not accelerating..."

So no reason to streamline a car to make it go faster for the same engine power then...
izen

EliRabett said...


Betty's world is a very strange place indeed. Yes BP, the IR reflective coating makes the filament hotter. Marco simply underestimated how invested BP is in lunacy.

Betty Pound said...

A good insulator that reduced cooling can only change the temperature gradient to the external environment. It can never raise the energy emitted by source.

Heat can not be held, and objects can't heat themselves by bouncing their own radiation off of a passive colder object. That fact has been known for 200+ years.

A 10yo will repeat what they are told. An adult can discern between conflicting facts and choose wisely. Someday you'll learn that skill.

Betty Pound said...

Chris, You just made up numbers. Shame on you for lying.

Betty Pound said...

Tim, retard, the sun is analogous to the internal energy of the car. Gravity is an extra thing that you brought up, which doesn't apply to Green plate theory.

Betty Pound said...

No, Eli, the reflective coating does not make the filament hotter. The reflective coating helps scatter light better, so it appears brighter outside.

Betty Pound said...

Eli, I want you to link data comparing two bulbs: regular and reflective coated. I want to see power input and max reached filament temperature for both.

If you can not produce this data, you will have conceded.

Betty Pound said...

Ken, no I'm not confusing an inert object for an energy source.


"your body keeps makig heat"
and losing heat to the environment.

No such thing as a 100% insulator. No object in the universe heats up due to an insulator.

Why do you believe all insulators are heaters?

John Garland said...

-----
No object in the universe heats up due to an insulator.
-----

Which is why you walk outside in the snow naked, right?

-----
Why do you believe all insulators are heaters?
-----

Because personally, I find wearing insulating clothes does, in fact, warm me.

Your reality may vary.

Betty Pound said...

Well you are an imbecile. YOU warm the clothes, the clothes don't warm you. Only poetic license let's one get away with saying such things.

Timothy Folkerts said...

Betty asks: "Eli, I want you to link data comparing two bulbs: regular and reflective coated. I want to see power input and max reached filament temperature for both."

"In 1990, a halogen incandescent light bulb was introduced that used a new technology. An infrared (IR) reflecting, visible light transmitting optical coating, called a hot mirror, was deposited on the outside surface of the inner quartz envelope of a halogen incandescent light bulb. This hot mirror coating reflects a portion of the heat back to the filament where it reheats the filament. This recycling of the former waste heat allows the filament to reach the proper temperature using less electricity when compared to an uncoated halogen incandescent bulb (Figure 1). These early ‘hybrid’ halogen incandescent light bulbs achieved efficiencies of greater than 18 LpW (lumens per watt) for a 70 watt bulb. This gave the 70 watt hybrid halogen incandescent bulb the same light output as the 100 watt incandescent bulb."

https://www.techbriefs.com/component/content/article/1236-lighting-technology/lighting-technology/15641-the-rebirth-of-the-incandescent-light-bulb

There is plenty more out there, Betty. Just open your eyes and look around.

John Garland said...
This comment has been removed by the author.
John Garland said...

I'm an imbecile?

You're the one walking around naked in the snow thinking you--like "every other object in the universe"--will not be warmer if you put on some insulating clothes!!!

Betty Pound said...

The clothes don't raise your temperature above 98.6F. How stupid can you be?

Betty Pound said...

"Just open your eyes and look around."

Yes, let's have a good look see, shall we?

"This hot mirror coating reflects a portion of the heat back to the filament where it reheats the filament."

That's a lie. No supporting evidence.

"This gave the 70 watt hybrid halogen incandescent bulb the same light output as the 100 watt incandescent bulb."

That's completely true. You obviously lack intelligence, so I will explain to you how it works.

The key word is "light"

The coating acts as an IR filter. By blocking IR from leaving, the VISIBLE light is increased. The coating acts as a "converter" from IR to Visible Light.

Furthermore, according to Eli the filament heats up due to backradiation, so it should emit more IR! of course it doesn't.

http://minerva.union.edu/newmanj/Physics100/Light%20Production/LampSpectra.gif

Why would you and Eli believe that an IR filter increases ougoing IR? Because you are both professional dummies.

Furthermore, this device violates blackbody curve, and so is not even applicable to Green Plate Theory.

This device is great at increasing visible light while reducing heat (IR). So you can indeed get more [visible] light with less electricity.

Try using your brain next time, Tim.

Betty Pound said...

Tim, I didn't even read your article before I replied because I already knew what the problem was.

Now that I skimmed it, here's the crux of your embarassment:

https://www.techbriefs.com/images/stories/LT/2013/FEATURES/45599-302_fig1.JPG

You see the Red "Curve" ???

You see, according to you and Eli, the IR coating should have bounced the IR photons back to the filament and made it magically produce more IR photons. But guess what?

EXACTLY THE OPPOSITE HAPPENS!

Wow, Tim, thanks a lot! I doubt you feel shame, but the wise will snicker at you. What a dolt.

marcoclimate said...

"EXACTLY THE OPPOSITE HAPPENS!"

Well, I'd almost say "INDEED!" You'd think this was the concept: to add a coating that prevents IR radiation to get through, and magically, IT WORKS!

But Betty...photons in the visible light have a *higher* energy than those IR photons, so how can that be?

Unless you are now going to tell us that an IR photon *knows* whether it is coming from a warmer or colder object and moving towards a colder or warmer object, despite having the same wavelength...

"No, Eli, the reflective coating does not make the filament hotter. The reflective coating helps scatter light better, so it appears brighter outside."

Sigh. This makes no sense whatsoever. Scatter light better so it appears brighter? Nope.
And yes, Virginia, the filament does get hotter due to the IR reflective coating. And if you believe that is wrong, you'd wonder why no one challenges patents like the one below, which explicitly states that this is how the frikkin' reflective coating works!
http://www.google.ch/patents/US6967443?hl=de&cl=en
"The totally reflecting coating reflects the infrared radiation escaping at acute angles and directs the infrared radiation towards the filament to increase the temperature of the filament and thus increase the efficacy of the lamp."

Christian Anders said...

No. This Nissan is actually the car I own. The numbers are what the analoge display shows when performing in the way described.

Different cars will get to different numbers, but surely no different principle.
You can test this if you have a car on your own or know somebody who has a car and helps you out.

Or are you really claiming that a car with running motor, fixed gear and slightly held breaks will NOT accelerate if the breaking is reduced?

Be aware of how simple it is to prove or disprove such a claim, for everywhere you can test it by experiment.

Christian Anders said...

Nobody claimed insulators to be heaters. Everybody claimed heaters which perform constantly will lead to different states if insulated to the envoirenment than without such an insulation.

You state otherwise. Please wrap the CPU of your PC in some insulating material.
Nothing to fear from this, for the temperature of the CPU cannot rise due to the presence of the colder insulator.

Christian Anders said...

Betty,

yet another question remained unanswered so far:

You claim the GPE to equilibrate at 200 per surface area element of the plate. This you call equilibrium.
Correct?

The sun is in this state much warmer than the plates, emitting 400 per surface element.
According to SB-law the sun is therefore warmer than the plates.

How can you construct equilibrium within a system with parts of the system being at different temperatures?

Has the correct equilibrium not to show evenly distributed temperature of the entire system?

EliRabett said...

Just for the non physicists here, scattering refers to changing the direction of a photon without changing its wavelength. Things like mirrors and droplets can do that. If anybunny wants a demo, stick a laser pointer behind a small glass of milk.

John Garland said...

-----
You state otherwise. Please wrap the CPU of your PC in some insulating material.

Nothing to fear from this, for the temperature of the CPU cannot rise due to the presence of the colder insulator.
------

This is simply the space suit example written much more easily and cheaply done with the added benefit of being something Betts can immediately try as we know he has access to a computer and can guess he may have access to a blanket--his protestations that blankets do not warm him aside. Yet it is no less deadly to the insulated object. So instead, of killing Betts, you simply kill his means of posting more drivel! Much better legally! Good example.

There is a famous real world example of environment suit cooling failure leading to heatstroke in the old Man High project high stratosphere balloon flights of the late 50s. The pilot of Manhigh III developed a temperature of over 108F while the cabin was in the high 90sF. One--well Betts, that is--can only wonder where this body heating came from. Didn't quite kill him but it was a very close run thing, and that was the last flight in the series.

http://stratocat.com.ar/fichas-e/1958/HMN-19581008.htm

Christian Anders said...

Betty,

ah, I nearly missed this nice additional opportunity:

"Why don't you find the exact quote in your links which claims a space suit spontaneously heats you up using just your body's emitted energy."

Oh, they are easy to find, don't worry. Here they come:

One major problem of manned space flight is the proper thermal control of the crewman during intravehicular and extravehicular missions. Heat generated metabolically and also from the external environment must be removed from the crewman's environment in order to protect him from excess thermal loads.

- from the patent of 1969, second link I provided. Page 3 of the pdf, lines 32-38. -

Maintaining sensible cooling as metabolic rates increase requires ever lower coolant and skin temperatures, with the difference between skin and coolant temperature increasing nonlinearly as indicated by typical thermal comfort curves (Figure 4). The sensible cooling approach is not always a good match for human metabolism, which evolved to control core temperature by latent cooling (i.e., perspiration).

- from the paper of 2015, the third linik I provided. Page 4 of the pdf, line 1-4. -

It is not sufficient for the health and well-being of an astronaut just to be protected from the hazards of the environment in which he or she is trying to work. It is also necessary to consider the conditions that are created by the suit itself. One of the most important ofthese conditions is temperature. Suit insulation technologies protect the astronaut from extreme high and low temperatures of the space environment. However, the same insulation technology also works to keep heat released by the astronaut's body inside the suit. To get an idea of what this is like, imagine walking around in summer wearing a plastic bag. For this reason, an active cooling system is employed

- From the tutorial, first link I provided. Beginning with line 1. -

-----

It is some fun, to show the world you are not even able to read sources which are delivered to you for free.

Simple question: Why are you sweating, even if environment temperature is below your body temperature? Why does your biology insists on the need of cooling your body, although your brain knows better?

Bernard J. said...

There's a scene in Erik the Viking where Aud gives Erik a cloak (actually a towel...) of invisibility, which Erik finds to his later detriment works only on Arnulf.

It seems that actual principles of physics are similarly only invisible to 'Betty Pound', and that what is completely apparent to everyone else is absolutely unperceivable by BP. It's a classic example of a cognitive scotoma, where the bit of reality that cannot be seen by the afflicted mind is substituted with a construct of said mind, that is congruent with what the mind believes should be there.

BP's excuses and irrational reasonings on this thread and the previous one are fascinating from a psychological perspective. It matters not whether his beliefs are genuine or feigned - either way they open a telling window into the mind of a science denier.

Betty Pound said...

Marco,

https://www.techbriefs.com/images/stories/LT/2013/FEATURES/45599-302_fig1.JPG

The area under the red curve is smaller than grey curve. That is total less energy - exactly the opposite of what Eli claims.

LOL. There is no requirement that patents be true or work.

The spectrum curves tell the whole story, and fully debunk the idea that the filament gets hotter due to IR backradiation.

Betty Pound said...

Chris,
Not surprised that you could not produce any data on space suit raising your internal temperature.

Yes, perspiration and getting rid of your toxins is important. You can certainly die if wrapped up in your own crap.

Keep trying to find some actual data that is ditrctly relevant.

Betty Pound said...

Bernard J is a giant projector.

Betty Pound said...

Eli,
"Just for the non physicists here, scattering refers to changing the direction of a photon without changing its wavelength."

So why does the sun look yellow in the atmosphere and why is the sky blue?

Same reason the coated halogen shifts IR to visible light.

You failed to prove the filament gets hotter. How embarassing for you.

Betty Pound said...

Chris,
The sun is a radial point source shining on a surface. If the sun was represented as another plate then the system would equilibrate at 400 W/m2, not 200 W/m2.

Betty Pound said...

John,

https://www.thisdayinaviation.com/tag/project-manhigh/

"During the flight, Lieutenant McClure became dehydrated. Later, temperatures inside the gondola rose to 118 °F. (47.8 °C.). The cooling system was unable to dissipate heat from McClure’s body, and his body core temperature rose to 108.6 °F. (42.6 °C.). After twelve hours, it was decidede to end the flight."

marcoclimate said...

"The area under the red curve is smaller than grey curve. That is total less energy - exactly the opposite of what Eli claims."

Sigh. You manage to find an article with a graph that you *think* supports your claim.

Well, what does the article say:
"This hot mirror coating reflects a portion of the heat back to the filament where it reheats the filament. This recycling of the former waste heat allows the filament to reach the proper temperature using less electricity when compared to an uncoated halogen incandescent bulb (Figure 1)."

Eh, what? The reflected IR reheats? Surely you're joking, Mr Feynman? That's what you claimed simply cannot happen, because the IR photons have lower energy than the UV photons that the filament (also) emits!

And yes, patents do not need to be scientifically correct, but not being so makes them easily challenged. You just describe the same invention but with the 'correct' principle, and you can claim you do not use the same principle as the old patent and therefore have made a new invention!
(sounds easier than it is in practice, but it happens - or so my patent lawyer claims, but what would he know, right?)

John Garland said...

If you read the tech report, the 118F reading was from a badly placed sensor that was placed above a chemical reactor. Specifically, on p. 126 of the now declassified final report http://www.dtic.mil/get-tr-doc/pdf?AD=AD0259635 it states:

"Only the panel temperatures read by the pilot were inordinately high, 1180F. It was soon established that this was caused by the fact that the sensing element was accidently left out of reach on top of the C02 and water absorbing air regeneration unit which used a highly exothermic chemical reaction. This reading was readily expLainable, but essentially meaningless." (as is your badly sourced comment)

Also, as this 118F reading was noted early in the flight and eventually went off scale over the 120F limit of the sensor. Had this reading been true, it would have led to an immediate abort and again if true if the mission were not immediately aborted, the pilot most definitely would have died. The dry bulb temperature of the cabin is talked about in several places and is graphed in Figure 50 on page 169. It never gets to body temperature. The lowest line in Fig 50 is from an additional sensor that was also erroneously placed in a cool area. The dry bulb temp was considered the most accurate in the report.

The whole report is 216 pages but makes for fascinating reading. It discusses temperature effects in great detail. McClure was able to take a last reading near landing of his rectal temp as 108.5. (see page 132). As I am an Air Force brat, I read all this long, long ago. I've even pilgrimaged to one of the launch sites.

Christian Anders said...

You think the reason for sweating is to get rid of toxins?

The whole world thinks it's for cooling due to evaporation enthalpy.

But you do see that those NASA engineers also believe in wrong thermodynamics from reading the text, don't you?

EliRabett said...


Go argue with the Patent Office BP

"Incandescent light sources usually generate approximately 10-15% of visible light, and about 85-90% of infrared light. Incandescent infrared coating on the transparent cover a considerable part of the filament by the infrared light emitted reflected back to the filament. Filament absorb some infrared light, thereby reducing the amount of electric power to the heating filament at a given temperature required, and thus increasing the luminous efficiency of the lamp"

http://www.google.mk/patents/CN101529554A?cl=en

Christian Anders said...

Betty,

you like pain, don't you?

"Not surprised that you could not produce any data on space suit raising your internal temperature."

This is overwhelmingly funny. I gave you sources, you seem to ignore them. There is a whole industry doing nothing else than measure the effect of protective clothes on the core temperature of humans. The papers published about that may go to the hundreds, and so do the measurements. They all show this:

Encapsulation in thermal insulators can easily bring up the core temperature of a human. Furthermore different insulations leads to different results under the same environmental conditions. Thus the temperature change cannot be determined by the physical activity alone.
This is measured and plotted in so many publications - do you really insist on being embarassed further by me linking to some of those measurements?
Or is it enough for you to see that everyday's application of thermodynamics performed by engineers all over the world proves your claims wrong?

It was fun enough to see you completely unable to grasp the simple concept the engineers work with. It will be more fun to see you explaining how on earth things work fine millionfold which according to you should be impossible in every single case.

By the way: Will you deliver some more stuff to this amusing "reduced breaking doesn't lead to accelaration" claim you made, which could be disproved within 30 seconds? I told you how to perform the test... running motor, fixed gear, two conditions of breaking.
Can you at least grasp the concept behind this simple everyday experiment, which simply leads to book keeping forces the same way you should book keep energies in proper thermodynamics?

Come on, don't be shy. We are not used to you being shy.

Betty Pound said...

The data trumps falacious statements.

Total energy is determined by two factors: frequencies and intensities.

If you have some IR photons becoming UV, but you reduce the IR photons, you have not gained total energy. To get total energy you integrate under the curve. You would like to pretend that merely raising some frequencies is good enough. It's not.

There is no evidence of heating via IR backradiation. There is however more efficient way to create visible light.

The data trumps idiots' words.

Learn how to read spectral charts and figure out total energy.

Betty Pound said...

Still no data, eh?

Yeah, I remember when you fabricated speeds proving that more breaking makes your car go faster, and less breaking makes it go slower.

Betty Pound said...

I don't believe your false story.

The point is that somewhere in the vehicle the temperature was high. Thus it taints the whole story.

Your link also reports cabin temp of 96 degrees at only one time period, then it talks about his temperature later on, leading an imbecile like you to assume it was 96 thr whole time.

Betty Pound said...

Wow, look how pathetic Eli is!

He's quoting a patent (which is not required to be true or work) as an authority, while neglecting to properly analyze spectral data.

Betty Pound said...

"And yes, patents do not need to be scientifically correct, but not being so makes them easily challenged. You just describe the same invention but with the 'correct' principle, and you can claim you do not use the same principle as the old patent and therefore have made a new invention!"

Completely false. The patent is for special IR filter/coating. No one can use that specific coating. The physics explanation of why and how it works is irrelevant. They can claim the IR filter makes Santa Claus happy, and that is what makes it efficient.

There are even patents for perpetual motion machines. There are also many overlapping patents.

Not that this matters. The physics principle involved here is not IR backradiation, but refraction/scattering.

There is also a reverse filter. It blocks the visible light and enhances IR. The total provided energy is the same, but it's in a different spectral band that is closer to earth-like temperatures. So it feels like more HEAT.

Christian Anders said...

Uhm, no.

YOU claimed reduced breaking will not lead to acceleration. This can be disproved easily.

Now you say the opposite in your above statement: A straightforward lie.

Seems you realized how lost you are.

John Garland said...

Um Betts...The original official report is neither a story to "disbelieve" nor "fake news". Later misreadings of the official report may be stories not to believe or fake news. It is also quite clear you did not read the temperature parts of the report nor even glance at Figure 50 long enough to minimally understand it.

Believe what you will...you certainly have given ample proof of your ability to accept false beliefs here.

Betty Pound said...

John, the retard, science requires experiments to be verified.

You look so desperate trying to find one exotic situation where you could be right, and it's not even for the correct reasons.

From your source, page 132:

"A long 40 minutes later, the capsule landed in the dark with-
in a few miles of the runway from which it had been launched
nearly 12 hours before. A recovery helicopter landed beside it
a few minutes later in time to see Lt McClure remove the upper
hemisphere and crawl out under his own power.
The last setting of the temperature bridge was 108.3F and
Lt McClure remembered that the needle was off zero by an amount
that represented an additional 0.20. The final internal tem-
perature reading at the time of landing was 108.59F."

And you quoted earlier:

"The cooling system was unable to dissipate heat from McClure’s body, and his body core temperature rose to 108.6 °F. (42.6 °C.)"

Hmm, we have a match between external circuit and internal body temp. Funny how the whole report doesn't mention spontaenous self-heating due to insulating backradiation.

You obviously learned nothing about the cause of his heat fatigue.

Figure 50 shows the dry bulb is the least reliable data source: only 3 data points, biggest intervals. It completely missed the extreme rise in the instrumental panel. I find it funny how figure 50 debunks you completely. You're only argument is complaining about the instruments when they rose to 120. They "must" be wrong" only then. How can they be wrong on the way up to 120?

John Garland said...

Read a little closer and don't just pick and choose. The dry bulb temps were taken by the pilot in his actual immediate environment. Possibly there is a reason they were accepted as accurate. Naahhh.

And again, the high temps were immediately identified as erroneous. Of course we also have the much lower temps from the other sensor, but of course you don't even consider them. The tech report did consider them but dismisses them for good reasons.

You lecturing people about what science is and is not with sophomoric depth is pretty funny! Deniers often do this. It's always ludicrous.

Betty Pound said...

I don't say the opposite. Less and reduced are synonyms.

Reduced breaking does not lead to acceleration.

accelerataion means increase in velocity.

breaking means reduction in velocity.

reduced breaking means the rate of reducing velocity is slowed.

A less negative can never lead to a positive.

You are braindead.

Betty Pound said...

Yeah, the 120+ temp are considered "wrong", but not the whole curve up to 120 deg.

John Garland said...

Actually the whole curve was wrong as a cabin temp indicator as the sensor was on the CO2 scrubber which involves an exothermic reaction. As was noted immediately at the time. See quote above from p. 126. Your deep reading must have missed it.

Christian Anders said...

Betty,

"I don't say the opposite. Less and reduced are synonyms.
Reduced breaking does not lead to acceleration."

Yes, that is what you said and that is easily proven wrong by a simple experiment everybody with a car can perform. If you had a car you could even falsifiy yourself by experiment, but you'd rather won't try to find out, will you?

You insinuated I had been fabricating numbers showing more breaking leads to more speed:

"Yeah, I remember when you fabricated speeds proving that more breaking makes your car go faster, and less breaking makes it go slower. "

Which is a straightforward lie, for I have shown numbers that prove reduced breaking leading to accelaration when motor running in fixed gear. That is less breaking = more speed. It's the very thing you claim to be impossible.

Why can everybody owning a car do things which are according to you physically impossible? Any answer to this?

And now data come which are about to ashame you further:

https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3784664/

Figure 2 here shows how identical systems are heating up differently when wrapped in reflective foil. Those with the foil are all warmer than those without.

The systems contain of anesthetized mice ( which are real ammaml bodies like humans ), without and with thermodevices, each setup with and without foil.

This simple measurement shows in all clarity something that should be impossible according to the physics you preach here.

But I'm sure you will not accept this either, although it matches the discussion perfectly ( three bodies + a heat sink, one body at least being a heat source ).

marcoclimate said...

Betty, how do you know how the curves relate to each other and how they should be compared? Just taking an integral without knowing how the different curves relate to each other is outright stupid. Did you notice the red curve abruptly stopping at about 3000 nm? Do you know why?

But that taken aside, you just argued that low energy photons (IR region) can cause the formation of high energy photons (UV-VIS region). How does that fit with your prior argument that photons from colder objects cannot do something with warmer objects, because those photons do not have as much energy as those coming from the warmer object, and just 'replace' the same (but more) photons coming from the warmer object? You now argue *against* that simple exchange, and claim formation of new and higher energy photons. What magic is involved in that?

Betty Pound said...

"Just taking an integral without knowing how the different curves relate to each other is outright stupid."

Uhuh, doing things right is stupid, and doing it wrong is "smart"

http://astronomy.swin.edu.au/cosmos/B/Blackbody+Radiation

"The total energy being radiated (the area under the curve) increases rapidly as the temperature increases (Stefan–Boltzmann Law).

https://atmos.washington.edu/~caseyw8/Ch2_blackbody_radiation.pdf

"The height of any one curve tells you how bright the self-radiation is at each wavelength for a black body at that temperature, and the area under the curve tells you the total energyemitted."

"But that taken aside, you just argued that low energy photons (IR region) can cause the formation of high energy photons (UV-VIS region)."

Nope, never said that. It's the refractory IR coating that makes the "conversion" not photons.

"How does that fit with your prior argument that photons from colder objects cannot do something with warmer objects, because those photons do not have as much energy as those coming from the warmer object, and just 'replace' the same"

The total energy has not changed.

I made a simple statement that I clarified multiple times:

There are two factors: frequencies and intensity at each frequency.

The IR coating does not lead to more IR photons due to backradiation, as Eli claimed. While they got switched to higher frequency, their intensity at IR frequency dropped to compensate.

I'll make it simple for you since you have difficulty understanding two dimensions.

6*2 = 12
2*6 = 12

You can have a short fat rectangle and a tall skinny rectangle be the same area.

You can also warp the rectangle to be 1*12, or 3*4.

Doesn't matter. The total energy, i.e. heating potential remains.

Like all cranks you overload one side of the equation, and neglect the other.

Betty Pound said...

John,
Page 126 doesn't have your quote. Your quote is not in the document at all.

You got it from the website you linked to.

The document has no reference to "co2 scrubbers" or even the subword "scrub". The term "co2" contains irrelevant info.

"Your deep reading must have missed it."

Yup, I missed something that wasn't there.

You pulled page number 126 out of your backside.

Betty Pound said...

Chris,
anesthetized mice? foil to block convection?

I don't see IR backradiation in the study. Funny how the scientists missed that.

My fever once ran to 104.5F. I guess if I was wrapped in tin foil, you'd be claiming the tin foil did it.


"Which is a straightforward lie, for I have shown numbers that prove reduced breaking leading to accelaration when motor running in fixed gear. That is less breaking = more speed. It's the very thing you claim to be impossible."

That's funny. I ran the experiment myself.

I sped up to 20mph, then stepped on the break fully, I came to a quick stop.

I sped up 20mph again, and only stepped on my break half-way, and guess what? I slowed down slower.

My car did not go faster as a result of depressing the breaks gently.

You are a pathological liar.

Good thing you will get cancer and die early.

marcoclimate said...

"The IR coating does not lead to more IR photons due to backradiation, as Eli claimed. While they got switched to higher frequency, their intensity at IR frequency dropped to compensate."

I am sure that Betty will now show us the evidence that the IR reflective coatings in halogen lamps have an anti-Stokes effect.

Insert citations here, Betty:
...

In the meantime you keep on contradicting the people who actually MAKE these lamps. But what would they know?

Betty Pound said...

The spectral diagram is the evidence, you braindead moron.

As any non-braindead human can see, the area under the curve did not increase, i.e. no extra energy is generated by IR coating.

If that's not good enough for you:

https://www.fh-muenster.de/ciw/downloads/personal/juestel/juestel/4-InkohaerenteLichtquellen-Glueh-_und_Halogenlampen_english_-1.pdf

Read section 4.6

Christian Anders said...

Dear Betty,

are you really that unable to remember anything of importance? I told you thrice how the experiment is designed. Since we are talking about systems that are continuously driven by a constant energy input, the car analogy requires constant force on the wheels by the motor.

That is what "fixed gear" means. The car is then driving at a certain speed constantly on even street.
And if you break slightly enough the car will drive on slower, but the motor will not die. However, if you then reduce the breaking the former speed will come back. That is acceleration.

If you break hard enough for a total stop you have proven nothing, for you have just killed the experiment. Everyone can bring the motor to die that way. Can you turn off the sun?

And yes, the mice study was not about radiation at all. It simply shows how reduced cooling leads to higher temperatures if energy sources feed the system.

That is something you claimed to be impossible, but it was measured anyway. How?

Betty Pound said...

"And if you break slightly enough the car will drive on slower, but the motor will not die."

Driving slower is not accelaration.
Thanks for debunking yourself, braindead moron.

"And yes, the mice study was not about radiation at all. It simply shows how reduced cooling leads to higher temperatures if energy sources feed the system."

No, the experiment what happens to mice with anasthesia.

We already know greenhouses block convection which leads to trapping heated air.

This is not backradiation from foil.

Betty Pound said...

"However, if you then reduce the breaking the former speed will come back. That is acceleration."

Not in any real car. You are imagining a brick on the gas pedal.

Betty Pound said...

"I told you thrice how the experiment is designed."

When I make up an analogy, I decide!

You don't get to put a brick on the gas pedal (or assume cruise control) and pretend that's what I must have meant, asshole.

marcoclimate said...

Betty, thank you for that link. Unfortunately, it does not answer my question to you: please provide citations that show the IR coating in halogen lamps work as anti-Stokes refraction compounds, *which is what you claimed*.

That the integral doesn't change is irrelevant, because no one here contradicts this. Of course energy in = energy out, the same as for the Blue Plate - Green Plate system. However, what you continuously ignore is that for the filament to send out more visible light, it *must* have heaten up. Or taken from a different point of view: since a substantial portion of energy no longer is emitted out of the system (much less IR radiation leaves the system due to the coating), more energy must leave through other wavelengths. But how do we get those other wavelengths? Apparently, as you also seem to accept based on the links you gave us, it is primarily the shorter wavelengths that increase in intensity. But how do we get those, Betty? That's the question. *You* claimed it was due to an anti-Stokes effect of the coating, but *you* did not provide any citations for that.

Christian Anders said...

I don't know if you own a car with manual gear control or one that is automatic.

But yes, a brick on the gas pedal is what makes the analogy, if you want so.
Or how else are you going to create a constant force to the wheels that remains while breaking?

And no, you are not free to chose a design of experiment if you want to prove me wrong. You can do it with the same design as me or you can't do it. Comparative studies require same conditions.

And yes, there is a blocking of convection that leads to reduced cooling. But you demanded this will never lead to higher temperatures whatsoever. You didn't say this for pure radiation cases. Are you going to restate your claim?

That with hindered cooling mechanisms the heating due to a heating source simply becomes more effective is what all the fuzz ever was about.
You stated this as untrue. Reality contradicts you ( as do the 1st law, but you never cared for this ).

Betty Pound said...

"That the integral doesn't change is irrelevant, because no one here contradicts this. Of course energy in = energy out, the same as for the Blue Plate - Green Plate system. However, what you continuously ignore is that for the filament to send out more visible light, it *must* have heaten up."

Yes, indeed you do contradict this.

You've redefined heat willy nilly not to be an increase in total energy of the system, but an increase in energy in a subcomponent of the system, while neglecting the compensating decrease elsewhere.

No, it *musn't* heat up, because the total energy has not increased.

"Apparently, as you also seem to accept based on the links you gave us, it is primarily the shorter wavelengths that increase in intensity."

That's what the spectral data shows.

https://en.m.wikipedia.org/wiki/Stokes_shift

"If the emitted photon has more energy, the energy difference is called an anti-Stokes shift;[5] this extra energy comes from dissipation of thermal phonons in a crystal lattice, cooling the crystal in the process. Yttrium oxysulfide doped with gadolinium oxysulfide is a common industrial anti-Stokes pigment, absorbing in the near-infrared and emitting in the visible region of the spectrum. Photon upconversion is another anti-Stokes process."

Hello, retard, you are just complaining about calling something what it is.

Are you claiming there is no photon upconversion due to coating?

You are a pathetic idiot and liar. Good thing you will get cancer and die early. Best of luck in reforming yourself.