Warmest March on record according to the Japanese Meteorological Agency

[Wuwei]

Not sure why you are having a hard time with this...the resonance frequency is not CMB...CMB is not being captured by the radio telescope...a weak resonance frequency is being captured and interpolated into CMB which could not be captured by the radio telescope.

That's wrong. The entire CMB BB energy is hitting the maser. The maser amplifies one narrow band of one of the BB radiation frequencies at a time. You are still confusing the CMB input to the maser output.

I am not having trouble with this...I understand that the 2.75K radiation is not hitting the dish....a resonance signal is hitting the dish and being amplified into a signal that can then be interpolated into the 2.75K signal.

That is totally wrong! When the CMB hits the dish how can it possibly be a "resonance frequency" when it hasn't even reflected to the maser yet! All frequencies hit the dish. The maser amplifies just one of them after the fact. You have cause and effect turned around.

It is a clear example of getting a signal that can not be received by the telescope...a resonance signal that can be picked up being amplified and vi a mathematical model interpreted into the signal (artificially) that is to weak to be actually received.

You have an amazing incapability for understanding the simplest things. Let me try to make it simpler.

A radio telescope antenna picks up a band of wavelengths. (BB spectrum) and transfers it to a tuned amplifier whose tuning can be changed to select a very small segment of wavelengths (one specific BB sample).

An FM radio antenna picks up a band of wavelengths (several stations), transfers it to a tuned amplifier whose tuning can be changed to select a very small segment of wavelengths (one station).

The concepts are exactly the same, albeit with different technologies. It can't be made simple than that.

Any electrical engineer will tell you that an FM antenna sticking out of your car receives a wide band of EM waves, and not a “resonance frequency”.

You are continually coming up with non-scientific terms or concepts that you don't clearly define. If you disagree, you will have to clearly define exactly what is the “resonance frequency” the CMB is supposedly sending and how it differs from black body radiation.

You are continually coming up with non-scientific terms or concepts that you don't clearly define.

His epicycles take a lot of work.

Yeah it's like talking to a child that makes stuff up as you go along.

 

[Toddsterpatriot]

Not sure why you are having a hard time with this...the resonance frequency is not CMB...CMB is not being captured by the radio telescope...a weak resonance frequency is being captured and interpolated into CMB which could not be captured by the radio telescope.

That's wrong. The entire CMB BB energy is hitting the maser. The maser amplifies one narrow band of one of the BB radiation frequencies at a time. You are still confusing the CMB input to the maser output.

I am not having trouble with this...I understand that the 2.75K radiation is not hitting the dish....a resonance signal is hitting the dish and being amplified into a signal that can then be interpolated into the 2.75K signal.

That is totally wrong! When the CMB hits the dish how can it possibly be a "resonance frequency" when it hasn't even reflected to the maser yet! All frequencies hit the dish. The maser amplifies just one of them after the fact. You have cause and effect turned around.

It is a clear example of getting a signal that can not be received by the telescope...a resonance signal that can be picked up being amplified and vi a mathematical model interpreted into the signal (artificially) that is to weak to be actually received.

You have an amazing incapability for understanding the simplest things. Let me try to make it simpler.

A radio telescope antenna picks up a band of wavelengths. (BB spectrum) and transfers it to a tuned amplifier whose tuning can be changed to select a very small segment of wavelengths (one specific BB sample).

An FM radio antenna picks up a band of wavelengths (several stations), transfers it to a tuned amplifier whose tuning can be changed to select a very small segment of wavelengths (one station).

The concepts are exactly the same, albeit with different technologies. It can't be made simple than that.

Any electrical engineer will tell you that an FM antenna sticking out of your car receives a wide band of EM waves, and not a “resonance frequency”.

You are continually coming up with non-scientific terms or concepts that you don't clearly define. If you disagree, you will have to clearly define exactly what is the “resonance frequency” the CMB is supposedly sending and how it differs from black body radiation.

You are continually coming up with non-scientific terms or concepts that you don't clearly define.

His epicycles take a lot of work.

Yeah it's like talking to a child that makes stuff up as you go along.

Radiation from cold space can't hit the warm antenna, that would violate the 2nd Law.
It magically resonates with the antenna and only "looks like" CMB.
Just silly.

 

[Wuwei]

Radiation from cold space can't hit the warm antenna, that would violate the 2nd Law.
It magically resonates with the antenna and only "looks like" CMB.
Just silly.

SSDD is going to fight that tooth and nail. He has a maniacal obsession against the fact that EM radiation freely goes anywhere no matter the temperatures. If there is any type of counter-example the only recourse for him is to make stuff up.

 

[IanC]

All I was pointing out is that CO2 retains energy as thermal energy consistent with the temperature of the surrounding atmosphere. Not that it retains all the energy it absorbs.

It doesn't retain any of the energy it absorbs....that is why I say that so called greenhouse gas molecules are like holes in the blanket covering earth...If we only had oxygen and nitrogen and depended entirely on convection and conduction to move IR to the upper atmosphere, the earth would be a warmer place. It is just silly to suggest that the presence of a radiative gas would hinder the atmosphere's ability to radiatively cool itself.

Interesting statement. If there were no GHGs would IR simply leave at the speed of light?

Via radiation? How would it radiate if there were no radiative gasses to facilitate the movement? Radiation is the smallest part of energy transport from the surface to the upper atmosphere...GHG's are what facilitate that small percentage of the total movement of heat out into space...take away the GHG's and you are left with nothing but conduction and convection.

???? That statement makes no sense. If there were no GHGs to create the bottleneck at the surface boundary then most of the surface IR would simply fly away to space, unimpeded, at the spped of light.

The surface at 15C radiates 400w. The solar input now is 160w, even if we consider all the albedo due to clouds it would still only rise to less than 250w. The 150w surplus of radiation leaving would cool the surface PDQ don't you think?

You can't have it both ways. Without GHGs forcing energy into alternate pathways you have no energy powering convection and little conduction either.

 

[IanC]

Radiation from cold space can't hit the warm antenna, that would violate the 2nd Law.
It magically resonates with the antenna and only "looks like" CMB.
Just silly.

SSDD is going to fight that tooth and nail. He has a maniacal obsession against the fact that EM radiation freely goes anywhere no matter the temperatures. If there is any type of counter-example the only recourse for him is to make stuff up.

Yup. He thinks that because you have to do work to make the actual measurements then the thing being measured doesn't exist. Or something like that. But it makes sense to him in his bizarro land universe.

 

[Wuwei]

???? That statement makes no sense. If there were no GHGs to create the bottleneck at the surface boundary then most of the surface IR would simply fly away to space, unimpeded, at the spped of light.

The surface at 15C radiates 400w. The solar input now is 160w, even if we consider all the albedo due to clouds it would still only rise to less than 250w. The 150w surplus of radiation leaving would cool the surface PDQ don't you think?

You can't have it both ways. Without GHGs forcing energy into alternate pathways you have no energy powering convection and little conduction either.

Excellent point! I checked S-B law and at 15C the surface of the earth does radiate 400 W/m^2. No possible mechanism can counter all that radiation if there is no back scatter.

 

[mamooth]

Backradiation demonstrated a different way, using myself as an example.

The average surface area of an adult male human is 1.9 m^2.

Since some parts of your body radiate right back into the other parts, let's say the surface area radiating to the world is 1.5 m^2. You can set it at a little more or less, it won't matter for purposes of this example.

Body temperature is 37C, or 310K.

IR power radiated out of me = (surface area) * (S-B constant) * (T^4)

If you run the numbers, that comes out around 800 watts. Over a day, that's 19 kw*hrs, or 16,000 Calories.

If backradiation didn't exist, all of that 16,000 Calories would have to fueled by the food I ate. Even more than that, since heat is also conducting away from me, but we'll just forget about conduction for this example. So according to SSDD, we all must be eating that much.

However, backradiation does exist. The environment, almost all of which is cooler than 37C, is constantly radiating into me. I absorb that radiation and then radiate it back out. I'm absorbing 14,000 Calories of backradiation each day, so that I only have to eat 2,000 Calories to make up the balance.

Another example ...

Inside at 68F, you're comfortable.

Outside at 68F, in the shade (so no direct solar radiation), you'll start getting chilly.

Why? Inside, the walls and ceiling are radiating at 68F. Outside, half of your radiating environment is a colder sky. You're getting less backradiation.

 

[RKMBrown]

Backradiation demonstrated a different way, using myself as an example.

The average surface area of an adult male human is 1.9 m^2.

Since some parts of your body radiate right back into the other parts, let's say the surface area radiating to the world is 1.5 m^2. You can set it at a little more or less, it won't matter for purposes of this example.

Body temperature is 37C, or 310K.

IR power radiated out of me = (surface area) * (S-B constant) * (T^4)

If you run the numbers, that comes out around 800 watts. Over a day, that's 19 kw*hrs, or 16,000 Calories.

If backradiation didn't exist, all of that 16,000 Calories would have to fueled by the food I ate. Even more than that, since heat is also conducting away from me, but we'll just forget about conduction for this example. So according to SSDD, we all must be eating that much.

However, backradiation does exist. The environment, almost all of which is cooler than 37C, is constantly radiating into me. I absorb that radiation and then radiate it back out. I'm absorbing 14,000 Calories of backradiation each day, so that I only have to eat 2,000 Calories to make up the balance.

So basically you are saying warmer is better. Ok, glad we can stop talking about warming as a bad thing now.

 

[Toddsterpatriot]

Backradiation demonstrated a different way, using myself as an example.

The average surface area of an adult male human is 1.9 m^2.

Since some parts of your body radiate right back into the other parts, let's say the surface area radiating to the world is 1.5 m^2. You can set it at a little more or less, it won't matter for purposes of this example.

Body temperature is 37C, or 310K.

IR power radiated out of me = (surface area) * (S-B constant) * (T^4)

If you run the numbers, that comes out around 800 watts. Over a day, that's 19 kw*hrs, or 16,000 Calories.

If backradiation didn't exist, all of that 16,000 Calories would have to fueled by the food I ate. Even more than that, since heat is also conducting away from me, but we'll just forget about conduction for this example. So according to SSDD, we all must be eating that much.

However, backradiation does exist. The environment, almost all of which is cooler than 37C, is constantly radiating into me. I absorb that radiation and then radiate it back out. I'm absorbing 14,000 Calories of backradiation each day, so that I only have to eat 2,000 Calories to make up the balance.

Science 24 May 1963:
Vol. 140 no. 3569 pp. 870-877
DOI: 10.1126/science.140.3569.870

In a practical situation and room-temperature setting, humans lose considerable energy due to thermal radiation. However, the energy lost by emitting infrared light is partially regained by absorbing the heat flow due to conduction from surrounding objects, and the remainder resulting from generated heat through metabolism. Human skin has an emissivity of very close to 1.0 . Using the formulas below shows a human, having roughly 2 square meter in surface area, and a temperature of about 307 K, continuously radiates approximately 1000 watts. However, if people are indoors, surrounded by surfaces at 296 K, they receive back about 900 watts from the wall, ceiling, and other surroundings, so the net loss is only about 100 watts.

Physicist Offers 10 000 To Anyone Who Can Disprove Climate Change Page 30 US Message Board - Political Discussion Forum

SSDD will now show that back in 1963, they didn't understand the 2nd Law.

 

[Wuwei]

I hate taking the side of SSDD, but this is what he would reply to your inane human body example: Mamooth and Todster, you retards are totally wrong. The 2nd law says that colder air outside can't radiate to your warmer body. Period. What actually happens is that the outside environment inhibits your body from radiating anything but 100 Watts. QED.

Furthermore the reason your body behaves that way is due to an anti-phlogiston vortex of dark matter located in your sweat glands.

On a lighter note, that's why I thought it was so important to show the CMB example. It shows explicitly a case of super-cold radiation hitting a hotter antenna in a measurable, repeatable experiment.

That one counterexample illustrates that his version of the second law cannot be,
"Energy cannot flow from cold to hot",
but must be
"Thermal energy cannot flow from cold to hot.

That wording allows backscatter, and also preserves the original intent of the second law. But I expect you already know that. So this post is mostly for other's that may have been lost or bored by the interminable back and forth arguments and digressions.

 

[Toddsterpatriot]

I hate taking the side of SSDD, but this is what he would reply to your inane human body example: Mamooth and Todster, you retards are totally wrong. The 2nd law says that colder air outside can't radiate to your warmer body. Period. What actually happens is that the outside environment inhibits your body from radiating anything but 100 Watts. QED.

Furthermore the reason your body behaves that way is due to an anti-phlogiston vortex of dark matter located in your sweat glands.

On a lighter note, that's why I thought it was so important to show the CMB example. It shows explicitly a case of super-cold radiation hitting a hotter antenna in a measurable, repeatable experiment.

That one counterexample illustrates that his version of the second law cannot be,
"Energy cannot flow from cold to hot",
but must be
"Thermal energy cannot flow from cold to hot.

That wording allows backscatter, and also preserves the original intent of the second law. But I expect you already know that. So this post is mostly for other's that may have been lost or bored by the interminable back and forth arguments and digressions.

What actually happens is that the outside environment inhibits your body from radiating anything but 100 Watts. QED.

His magic theory has warmer bodies slowing their emission, as they reach equilibrium, without ever knowing the temp of their target, because, after all, the colder target can't radiate back.
Maybe the photon (or wave, he doesn't believe in photons) sends info back, as it hits?
It's all very complicated.

 

[jc456]

Backradiation demonstrated a different way, using myself as an example.

The average surface area of an adult male human is 1.9 m^2.

Since some parts of your body radiate right back into the other parts, let's say the surface area radiating to the world is 1.5 m^2. You can set it at a little more or less, it won't matter for purposes of this example.

Body temperature is 37C, or 310K.

IR power radiated out of me = (surface area) * (S-B constant) * (T^4)

If you run the numbers, that comes out around 800 watts. Over a day, that's 19 kw*hrs, or 16,000 Calories.

If backradiation didn't exist, all of that 16,000 Calories would have to fueled by the food I ate. Even more than that, since heat is also conducting away from me, but we'll just forget about conduction for this example. So according to SSDD, we all must be eating that much.

However, backradiation does exist. The environment, almost all of which is cooler than 37C, is constantly radiating into me. I absorb that radiation and then radiate it back out. I'm absorbing 14,000 Calories of backradiation each day, so that I only have to eat 2,000 Calories to make up the balance.

Science 24 May 1963:
Vol. 140 no. 3569 pp. 870-877
DOI: 10.1126/science.140.3569.870

In a practical situation and room-temperature setting, humans lose considerable energy due to thermal radiation. However, the energy lost by emitting infrared light is partially regained by absorbing the heat flow due to conduction from surrounding objects, and the remainder resulting from generated heat through metabolism. Human skin has an emissivity of very close to 1.0 . Using the formulas below shows a human, having roughly 2 square meter in surface area, and a temperature of about 307 K, continuously radiates approximately 1000 watts. However, if people are indoors, surrounded by surfaces at 296 K, they receive back about 900 watts from the wall, ceiling, and other surroundings, so the net loss is only about 100 watts.

Physicist Offers 10 000 To Anyone Who Can Disprove Climate Change Page 30 US Message Board - Political Discussion Forum

SSDD will now show that back in 1963, they didn't understand the 2nd Law.

So, let's take this analogy and, lets say in the room the temperature is 68 F, what is the temperature of the wall absorbing, reflecting and re-emmiting the radiated waves? So when you point the infrared meter at it, what will it read? And if the wall is emitting IR back into the room, does the room get warmer?

 

[Wuwei]

What actually happens is that the outside environment inhibits your body from radiating anything but 100 Watts. QED.

His magic theory has warmer bodies slowing their emission, as they reach equilibrium, without ever knowing the temp of their target, because, after all, the colder target can't radiate back.
Maybe the photon (or wave, he doesn't believe in photons) sends info back, as it hits?
It's all very complicated.

I have not seen his answer to that. I don't think I want to. But as far as photons he already had a theory of photons and zero distance and time. That's odd since he does not believe in photons.

 

[Toddsterpatriot]

Backradiation demonstrated a different way, using myself as an example.

The average surface area of an adult male human is 1.9 m^2.

Since some parts of your body radiate right back into the other parts, let's say the surface area radiating to the world is 1.5 m^2. You can set it at a little more or less, it won't matter for purposes of this example.

Body temperature is 37C, or 310K.

IR power radiated out of me = (surface area) * (S-B constant) * (T^4)

If you run the numbers, that comes out around 800 watts. Over a day, that's 19 kw*hrs, or 16,000 Calories.

If backradiation didn't exist, all of that 16,000 Calories would have to fueled by the food I ate. Even more than that, since heat is also conducting away from me, but we'll just forget about conduction for this example. So according to SSDD, we all must be eating that much.

However, backradiation does exist. The environment, almost all of which is cooler than 37C, is constantly radiating into me. I absorb that radiation and then radiate it back out. I'm absorbing 14,000 Calories of backradiation each day, so that I only have to eat 2,000 Calories to make up the balance.

Science 24 May 1963:
Vol. 140 no. 3569 pp. 870-877
DOI: 10.1126/science.140.3569.870

In a practical situation and room-temperature setting, humans lose considerable energy due to thermal radiation. However, the energy lost by emitting infrared light is partially regained by absorbing the heat flow due to conduction from surrounding objects, and the remainder resulting from generated heat through metabolism. Human skin has an emissivity of very close to 1.0 . Using the formulas below shows a human, having roughly 2 square meter in surface area, and a temperature of about 307 K, continuously radiates approximately 1000 watts. However, if people are indoors, surrounded by surfaces at 296 K, they receive back about 900 watts from the wall, ceiling, and other surroundings, so the net loss is only about 100 watts.

Physicist Offers 10 000 To Anyone Who Can Disprove Climate Change Page 30 US Message Board - Political Discussion Forum

SSDD will now show that back in 1963, they didn't understand the 2nd Law.

So, let's take this analogy and, lets say in the room the temperature is 68 F, what is the temperature of the wall absorbing, reflecting and re-emmiting the radiated waves? So when you point the infrared meter at it, what will it read? And if the wall is emitting IR back into the room, does the room get warmer?

So, let's take this analogy and, lets say in the room the temperature is 68 F,

Air temp, inside wall temp, outside wall temp?

So when you point the infrared meter at it, what will it read?

If the wall temp is 68 F, it'll read 68 F.

And if the wall is emitting IR back into the room, does the room get warmer?

Is there an energy source in the room? What's its temp?

 

[Toddsterpatriot]

What actually happens is that the outside environment inhibits your body from radiating anything but 100 Watts. QED.

His magic theory has warmer bodies slowing their emission, as they reach equilibrium, without ever knowing the temp of their target, because, after all, the colder target can't radiate back.
Maybe the photon (or wave, he doesn't believe in photons) sends info back, as it hits?
It's all very complicated.

I have not seen his answer to that. I don't think I want to. But as far as photons he already had a theory of photons and zero distance and time. That's odd since he does not believe in photons.

It must work with waves as well and it is definitely odd.

 

[Wuwei]

So, let's take this analogy and, lets say in the room the temperature is 68 F, what is the temperature of the wall absorbing, reflecting and re-emmiting the radiated waves? So when you point the infrared meter at it, what will it read? And if the wall is emitting IR back into the room, does the room get warmer?

Objects at the same temperature radiate equal amounts to each other. The net flow of energy is zero.

 

[jc456]

Backradiation demonstrated a different way, using myself as an example.

The average surface area of an adult male human is 1.9 m^2.

Since some parts of your body radiate right back into the other parts, let's say the surface area radiating to the world is 1.5 m^2. You can set it at a little more or less, it won't matter for purposes of this example.

Body temperature is 37C, or 310K.

IR power radiated out of me = (surface area) * (S-B constant) * (T^4)

If you run the numbers, that comes out around 800 watts. Over a day, that's 19 kw*hrs, or 16,000 Calories.

If backradiation didn't exist, all of that 16,000 Calories would have to fueled by the food I ate. Even more than that, since heat is also conducting away from me, but we'll just forget about conduction for this example. So according to SSDD, we all must be eating that much.

However, backradiation does exist. The environment, almost all of which is cooler than 37C, is constantly radiating into me. I absorb that radiation and then radiate it back out. I'm absorbing 14,000 Calories of backradiation each day, so that I only have to eat 2,000 Calories to make up the balance.

Science 24 May 1963:
Vol. 140 no. 3569 pp. 870-877
DOI: 10.1126/science.140.3569.870

In a practical situation and room-temperature setting, humans lose considerable energy due to thermal radiation. However, the energy lost by emitting infrared light is partially regained by absorbing the heat flow due to conduction from surrounding objects, and the remainder resulting from generated heat through metabolism. Human skin has an emissivity of very close to 1.0 . Using the formulas below shows a human, having roughly 2 square meter in surface area, and a temperature of about 307 K, continuously radiates approximately 1000 watts. However, if people are indoors, surrounded by surfaces at 296 K, they receive back about 900 watts from the wall, ceiling, and other surroundings, so the net loss is only about 100 watts.

Physicist Offers 10 000 To Anyone Who Can Disprove Climate Change Page 30 US Message Board - Political Discussion Forum

SSDD will now show that back in 1963, they didn't understand the 2nd Law.

So, let's take this analogy and, lets say in the room the temperature is 68 F, what is the temperature of the wall absorbing, reflecting and re-emmiting the radiated waves? So when you point the infrared meter at it, what will it read? And if the wall is emitting IR back into the room, does the room get warmer?

So, let's take this analogy and, lets say in the room the temperature is 68 F,

Air temp, inside wall temp, outside wall temp?

So when you point the infrared meter at it, what will it read?

If the wall temp is 68 F, it'll read 68 F.

And if the wall is emitting IR back into the room, does the room get warmer?

Is there an energy source in the room? What's its temp?

thermometer in the room reads 68.

Does the reflected IR off the wall warm the room?

 

[Toddsterpatriot]

So, let's take this analogy and, lets say in the room the temperature is 68 F, what is the temperature of the wall absorbing, reflecting and re-emmiting the radiated waves? So when you point the infrared meter at it, what will it read? And if the wall is emitting IR back into the room, does the room get warmer?

Objects at the same temperature radiate equal amounts to each other. The net flow of energy is zero.

I know. He has said that they simply stop radiating.

 

[Toddsterpatriot]

Backradiation demonstrated a different way, using myself as an example.

The average surface area of an adult male human is 1.9 m^2.

Since some parts of your body radiate right back into the other parts, let's say the surface area radiating to the world is 1.5 m^2. You can set it at a little more or less, it won't matter for purposes of this example.

Body temperature is 37C, or 310K.

IR power radiated out of me = (surface area) * (S-B constant) * (T^4)

If you run the numbers, that comes out around 800 watts. Over a day, that's 19 kw*hrs, or 16,000 Calories.

If backradiation didn't exist, all of that 16,000 Calories would have to fueled by the food I ate. Even more than that, since heat is also conducting away from me, but we'll just forget about conduction for this example. So according to SSDD, we all must be eating that much.

However, backradiation does exist. The environment, almost all of which is cooler than 37C, is constantly radiating into me. I absorb that radiation and then radiate it back out. I'm absorbing 14,000 Calories of backradiation each day, so that I only have to eat 2,000 Calories to make up the balance.

Science 24 May 1963:
Vol. 140 no. 3569 pp. 870-877
DOI: 10.1126/science.140.3569.870

In a practical situation and room-temperature setting, humans lose considerable energy due to thermal radiation. However, the energy lost by emitting infrared light is partially regained by absorbing the heat flow due to conduction from surrounding objects, and the remainder resulting from generated heat through metabolism. Human skin has an emissivity of very close to 1.0 . Using the formulas below shows a human, having roughly 2 square meter in surface area, and a temperature of about 307 K, continuously radiates approximately 1000 watts. However, if people are indoors, surrounded by surfaces at 296 K, they receive back about 900 watts from the wall, ceiling, and other surroundings, so the net loss is only about 100 watts.

Physicist Offers 10 000 To Anyone Who Can Disprove Climate Change Page 30 US Message Board - Political Discussion Forum

SSDD will now show that back in 1963, they didn't understand the 2nd Law.

So, let's take this analogy and, lets say in the room the temperature is 68 F, what is the temperature of the wall absorbing, reflecting and re-emmiting the radiated waves? So when you point the infrared meter at it, what will it read? And if the wall is emitting IR back into the room, does the room get warmer?

So, let's take this analogy and, lets say in the room the temperature is 68 F,

Air temp, inside wall temp, outside wall temp?

So when you point the infrared meter at it, what will it read?

If the wall temp is 68 F, it'll read 68 F.

And if the wall is emitting IR back into the room, does the room get warmer?

Is there an energy source in the room? What's its temp?

thermometer in the room reads 68.

Does the reflected IR off the wall warm the room?

No energy source in the room, no energy loss thru the outside walls, sounds like equilibrium.

 

[Wuwei]

I know. He has said that they simply stop radiating

Sorry. I was addressing that to JC.