What explanation of the Greenhouse Effect do you use?

[jc456]

Funny. When others give reasonable answers and close approximations, they are nitwits. When he gives approximations, he calls it "good science." With Sunlight generating between 1000 and 1300 watts/meter square and CO2 only absorbing 1/10,000th watt, you'd wonder how it effects climate change at all!

As a 'trained physicist ' with an interest in astronomy you should be familiar with blackbody radiation.

The green band of the Sun's spectrum puts out 10^5 more power than the 15 micron band. The average solar insolation reaching the Earth is 340w, 100w is reflected. 240w divided by 10^5 is an amount that can be safely ignored.

If you disagree, explain why.

Edit- it should actually be the amount of visible green light that gets divided by 10^5, a much smaller amount. But CO2 has other absorbance bands. Is this extra complexity necessary? I dont think so. Either amount is insignificant.

Funny, one source claims: 6.33×107 W/m2
Part 2: Solar Energy Reaching The Earth’s Surface | ITACA

Another source says: the average is 6.11 kWhr/m^2 per day.
https://www.quora.com/How-much-ener...nd-per-square-meter-per-minute-solar-constant

This source claims: 1,000 W/m2
The Sun's Energy

YOU CLAIM: The average solar insolation reaching the Earth is 340w, 100w is reflected.
So you just claimed that a full 30% of all sunlight is reflected back into space when I've already shown that figure is closer to 3%

View attachment 262108

Then you have the audacity to suggest this only leaves 240 watts across the entire face of the Earth? You are an incredible accident of fraud and bumbling inexactness, while jumping on the SLIGHTEST inaccuracies of others!

AND STILL, WITHOUT A SHRED of credible independent supporting research to back up a single thing you say! Too funny!

Are you sure that you want keep humiliating yourself?

Only humiliation here is looking you in the mirror.

Ok then Mr 'trained physicist '.

Roughly 1360 w/m2 of solar radiation reaches the Earth orbit distance. The amount intercepted by the Earth would be equal to a disk with the same diameter as the Earth. The Earths surface has four times the area as that disk. 1360 divided by four is 340w/m2.

The Earth has an albedo of 0.3. That means it reflects 30 percent of the solar radiation back into space. That drops the solar insolation to 240w/m2. If you are complaining that I subtracted all of the albedo at once rather than cumulatively through the different layers, then I plead quilty. Tell you what. You can have the whole 340w outside the atmosphere figure to calculate tha amount of solar produced CO2 IR reaching the Earth. It is still insignificant.

Anyone educated in physics and astronomy should already know all of this. Why don't you?

the earth isn't a disk, it is a spheroid, and it is constantly rotating, it's where you loose your argument, you attempt to stabilize it which isn't what's happening. anyone with any kind of science knowledge should know this.

 

[Toddsterpatriot]

As a 'trained physicist ' with an interest in astronomy you should be familiar with blackbody radiation.

The green band of the Sun's spectrum puts out 10^5 more power than the 15 micron band. The average solar insolation reaching the Earth is 340w, 100w is reflected. 240w divided by 10^5 is an amount that can be safely ignored.

If you disagree, explain why.

Edit- it should actually be the amount of visible green light that gets divided by 10^5, a much smaller amount. But CO2 has other absorbance bands. Is this extra complexity necessary? I dont think so. Either amount is insignificant.

Funny, one source claims: 6.33×107 W/m2
Part 2: Solar Energy Reaching The Earth’s Surface | ITACA

Another source says: the average is 6.11 kWhr/m^2 per day.
https://www.quora.com/How-much-ener...nd-per-square-meter-per-minute-solar-constant

This source claims: 1,000 W/m2
The Sun's Energy

YOU CLAIM: The average solar insolation reaching the Earth is 340w, 100w is reflected.
So you just claimed that a full 30% of all sunlight is reflected back into space when I've already shown that figure is closer to 3%

View attachment 262108

Then you have the audacity to suggest this only leaves 240 watts across the entire face of the Earth? You are an incredible accident of fraud and bumbling inexactness, while jumping on the SLIGHTEST inaccuracies of others!

AND STILL, WITHOUT A SHRED of credible independent supporting research to back up a single thing you say! Too funny!

Are you sure that you want keep humiliating yourself?

Only humiliation here is looking you in the mirror.

Ok then Mr 'trained physicist '.

Roughly 1360 w/m2 of solar radiation reaches the Earth orbit distance. The amount intercepted by the Earth would be equal to a disk with the same diameter as the Earth. The Earths surface has four times the area as that disk. 1360 divided by four is 340w/m2.

The Earth has an albedo of 0.3. That means it reflects 30 percent of the solar radiation back into space. That drops the solar insolation to 240w/m2. If you are complaining that I subtracted all of the albedo at once rather than cumulatively through the different layers, then I plead quilty. Tell you what. You can have the whole 340w outside the atmosphere figure to calculate tha amount of solar produced CO2 IR reaching the Earth. It is still insignificant.

Anyone educated in physics and astronomy should already know all of this. Why don't you?

the earth isn't a disk, it is a spheroid, and it is constantly rotating, it's where you loose your argument, you attempt to stabilize it which isn't what's happening. anyone with any kind of science knowledge should know this.

the earth isn't a disk, it is a spheroid, and it is constantly rotating,

Would that make the number more than 340w/m2 or less than 340w/m2?

anyone with any kind of science knowledge should know this

Irony is ironic.

 

[IanC]

CO2 very rarely re-emits near surface due to the mass of the atmosphere and collision energy transfer at ground level.

At altitude the mass is such that LWIR CAN EASILY ESCAPE as N2 and water vapor are non existent and collisions are rare above cloud boundary where re-nucleation of water vapor releases its energy, due to the decreased mass of the atmosphere.

Thanks Todd. I missed this astoundingly stupid post by BillyBoob.

It doesnt even meet the level of coherence to be considered wrong. It is just nonsense made with sciencey words

 

[jc456]

Funny, one source claims: 6.33×107 W/m2
Part 2: Solar Energy Reaching The Earth’s Surface | ITACA

Another source says: the average is 6.11 kWhr/m^2 per day.
https://www.quora.com/How-much-ener...nd-per-square-meter-per-minute-solar-constant

This source claims: 1,000 W/m2
The Sun's Energy

YOU CLAIM: The average solar insolation reaching the Earth is 340w, 100w is reflected.
So you just claimed that a full 30% of all sunlight is reflected back into space when I've already shown that figure is closer to 3%

View attachment 262108

Then you have the audacity to suggest this only leaves 240 watts across the entire face of the Earth? You are an incredible accident of fraud and bumbling inexactness, while jumping on the SLIGHTEST inaccuracies of others!

AND STILL, WITHOUT A SHRED of credible independent supporting research to back up a single thing you say! Too funny!

Are you sure that you want keep humiliating yourself?

Only humiliation here is looking you in the mirror.

Ok then Mr 'trained physicist '.

Roughly 1360 w/m2 of solar radiation reaches the Earth orbit distance. The amount intercepted by the Earth would be equal to a disk with the same diameter as the Earth. The Earths surface has four times the area as that disk. 1360 divided by four is 340w/m2.

The Earth has an albedo of 0.3. That means it reflects 30 percent of the solar radiation back into space. That drops the solar insolation to 240w/m2. If you are complaining that I subtracted all of the albedo at once rather than cumulatively through the different layers, then I plead quilty. Tell you what. You can have the whole 340w outside the atmosphere figure to calculate tha amount of solar produced CO2 IR reaching the Earth. It is still insignificant.

Anyone educated in physics and astronomy should already know all of this. Why don't you?

the earth isn't a disk, it is a spheroid, and it is constantly rotating, it's where you loose your argument, you attempt to stabilize it which isn't what's happening. anyone with any kind of science knowledge should know this.

the earth isn't a disk, it is a spheroid, and it is constantly rotating,

Would that make the number more than 340w/m2 or less than 340w/m2?

anyone with any kind of science knowledge should know this

Irony is ironic.

The Spherical Shape of the Earth

"Because the Earth is a sphere, the surface gets much more intense sunlight, hence heat, at the equator than at the poles. On the equinox, the Sun passes directly overhead at noon on the equator and a square centimeter of ground receives about 1 calorie of heat energy (see solar constant). On the same day, at 60°N, the latitude of Anchorage, Alaska, or Oslo, Norway, or St. Petersburg, Russia, the Sun rises no higher than 30° above the horizon at noon and heats a given parcel of ground with only a half the intensity as at the equator. At the poles, the Sun appears to sit on the horizon for periods upwards of 24 hours, and its rays skim horizontally over the surface."

You tell me, it's your crazy equation made up as if the earth was a disk fully exposed to the sun flat side? hahahahaha, you crack me up dude.

 

[IanC]

Are you sure that you want keep humiliating yourself?

Only humiliation here is looking you in the mirror.

Ok then Mr 'trained physicist '.

Roughly 1360 w/m2 of solar radiation reaches the Earth orbit distance. The amount intercepted by the Earth would be equal to a disk with the same diameter as the Earth. The Earths surface has four times the area as that disk. 1360 divided by four is 340w/m2.

The Earth has an albedo of 0.3. That means it reflects 30 percent of the solar radiation back into space. That drops the solar insolation to 240w/m2. If you are complaining that I subtracted all of the albedo at once rather than cumulatively through the different layers, then I plead quilty. Tell you what. You can have the whole 340w outside the atmosphere figure to calculate tha amount of solar produced CO2 IR reaching the Earth. It is still insignificant.

Anyone educated in physics and astronomy should already know all of this. Why don't you?

the earth isn't a disk, it is a spheroid, and it is constantly rotating, it's where you loose your argument, you attempt to stabilize it which isn't what's happening. anyone with any kind of science knowledge should know this.

the earth isn't a disk, it is a spheroid, and it is constantly rotating,

Would that make the number more than 340w/m2 or less than 340w/m2?

anyone with any kind of science knowledge should know this

Irony is ironic.

The Spherical Shape of the Earth

"Because the Earth is a sphere, the surface gets much more intense sunlight, hence heat, at the equator than at the poles. On the equinox, the Sun passes directly overhead at noon on the equator and a square centimeter of ground receives about 1 calorie of heat energy (see solar constant). On the same day, at 60°N, the latitude of Anchorage, Alaska, or Oslo, Norway, or St. Petersburg, Russia, the Sun rises no higher than 30° above the horizon at noon and heats a given parcel of ground with only a half the intensity as at the equator. At the poles, the Sun appears to sit on the horizon for periods upwards of 24 hours, and its rays skim horizontally over the surface."

You tell me, it's your crazy equation made up as if the earth was a disk fully exposed to the sun flat side? hahahahaha, you crack me up dude.

Answer Todd's question. Does a spinning spheroid collect more incoming solar radiation than a flat disk pointed at the Sun.

If you answer is anything but 'they are equal' then explain.

 

[SSDD]

As per usual, I make a point and you say it is refuted by adding a non sequitur.

How much energy does the Earth lose to space by conduction and convection? 0%

How much of the claimed radiative greenhouse effect takes place above the troposphere?

You seem to think that just because almost all energy is transported to the top of the troposphere via conduction and convection, that somehow it can't radiate out once an altitude is reached in which molecules are so far apart that energy movement via conduction is no longer efficient.

How much energy does the Earth lose to space by radiation? 100%

How much radiation happens above the troposphere is irrelevant to the workings of the claimed greenhouse effect...

I am not saying comductiom and convection have no part in our system. But if I was to act like you then I would say that gravity was the dominant factor. And unlike you, I could built a strong case for it. You just run away.

You don't build a strong case for your beliefs...in fact, you can produce no empirical evidence at all to support them.

 

[SSDD]

My interpretation of you is that you are an unequivocal idiot. Always has been. Always will. Just one more unqualified voice in the dark. Your opinion means nothing.

Fair enough. I will take that assertion as seriously as The atmosphere serves as the MECHANISM by which the greenhouse takes effect. It first takes sunlight and converts it into IR radiation, then acts as an insulating layer to hold it in from escaping back out into space.

Can you show us a physical law which predicts that a body can be warmed by reabsorbing its own radiation?

See? You're just being stupid again. Radiation leaving an object travels in a straight line. It cannot hit the original emitter.

Picking fly specks out of pepper again? Which physical law predicts that if you redirect an objects radiation back to it that it can be warmed?

 

[SSDD]

I dont like Trenberth's Cartoon either but at least it is a starting point. Guys like SSDD just sputter when asked for a better energy budget and say look it up yourself.

It's fun to tease him with it.

.

You believe it wholeheartedly because it is the basis for your religion...

 

[Olde Europe]

I dont like Trenberth's Cartoon either but at least it is a starting point. Guys like SSDD just sputter when asked for a better energy budget and say look it up yourself.

Care to explain what it is you dislike about Trenberth's energy budget diagram? TIA.

 

[IanC]

You seem to think that just because almost all energy is transported to the top of the troposphere via conduction and convection, that somehow it can't radiate out once an altitude is reached in which molecules are so far apart that energy movement via conduction is no longer efficient

GHGs lose radiation to space once they are high enough that an emitted photon is not just reabsorbed by another molecule of the same GHG.

The amount of radiation produced by a GHG that can escape is predicated on the temperature of the emission height.

The amount of radiation that a GHG absorbs from the surface is predicated by the temperature of the surface.

The amount of radiation available to be stored in the atmosphere is absorbed radiation less emitted radiation.

Because the surface is warmer than the emission height there will always be surplus radiation available to be returned to the surface.

I have explained this to you dozens of times. And asked you at what point in this line of logic do you disagree.

I have given you hints even. Like solid and liquid phases of water are not under the same radiation constraints as water vapour. Yet you refuse to step up and describe the pathways energy takes on its way out to space

 

[IanC]

My interpretation of you is that you are an unequivocal idiot. Always has been. Always will. Just one more unqualified voice in the dark. Your opinion means nothing.

Fair enough. I will take that assertion as seriously as The atmosphere serves as the MECHANISM by which the greenhouse takes effect. It first takes sunlight and converts it into IR radiation, then acts as an insulating layer to hold it in from escaping back out into space.

Can you show us a physical law which predicts that a body can be warmed by reabsorbing its own radiation?

See? You're just being stupid again. Radiation leaving an object travels in a straight line. It cannot hit the original emitter.

Picking fly specks out of pepper again? Which physical law predicts that if you redirect an objects radiation back to it that it can be warmed?

All objects radiate according to their temperature and emissivity. If they are in proximity then the net movement of energy will always be from warm to cool but both objects continue to radiate according to their temperature and emissivity.

While you can calculate how much radiation is flowing in opposite directions, you cannot calculate an effect without considering the net flow. Two objects at the same temperature don't stop radiating, there is just no net movement to make a change. And two objects at 100C make a lot more radiation than two at 10C. Even though in both cases thete is no net flow.

 

[IanC]

I dont like Trenberth's Cartoon either but at least it is a starting point. Guys like SSDD just sputter when asked for a better energy budget and say look it up yourself.

Care to explain what it is you dislike about Trenberth's energy budget diagram? TIA.

I dont feel like going over it yet again.

As so often happens, there is actually an article on energy budgets over at WUWT. You should check it out.

 

[Wuwei]

You believe it wholeheartedly because it is the basis for your religion...

I have absolutely no idea how realistic it is.

.

 

[cnm]

By the OP's definitions then, if sunlight is collimated, so then is an ordinary light bulb, which is ridiculous.

No, light from a light bulb is not considered parallel. Ffs, this is secondary school physics.

 

[Olde Europe]

I dont feel like going over it yet again.

As so often happens, there is actually an article on energy budgets over at WUWT. You should check it out.

That's okay. I found your thread (2011), and your understanding sure has improved since, because back then you didn't understand what the GHE is, and misrepresented its magnitude.

You'd have to pay me ever to go to WUWT again. That's time I'll never get back, with nothing of any value in return.

 

[SSDD]

GHGs lose radiation to space once they are high enough that an emitted photon is not just reabsorbed by another molecule of the same GHG.

So called GHG's almost never get to emit a photon...they lose the energy they absorbed energy via collision in damned near every instance...

The amount of radiation produced by a GHG that can escape is predicated on the temperature of the emission height.

So says the model..in reality, GHG's produce almost no radiation because the vast bulk of energy is transported via conduction..not radiation.

The amount of radiation that a GHG absorbs from the surface is predicated by the temperature of the surface.

Irrelevant since in nine hundred 999,999,999 times out of a billion they lose the energy they absorb via a collision...

The amount of radiation available to be stored in the atmosphere is absorbed radiation less emitted radiation.

That has nothing to do with the mode in which the energy is transported through the troposphere where the claimed greenhouse effect happens...energy is being conducted and convected through he troposphere with radiation being barely a bit player..

Because the surface is warmer than the emission height there will always be surplus radiation available to be returned to the surface.

No radiation is returned to the surface except in rare instances of temperature inversion where the atmosphere is warmer than the surface.

I have explained this to you dozens of times. And asked you at what point in this line of logic do you disagree.

Every single point because radiation is not the main mode of energy transport through the troposphere...radiation hardly plays a part in moving energy to the top of the troposphere.

I have given you hints even.

I don't need your hints because they are as wrong as the terribly flawed physics you believe a radiative greenhouse effect works in a troposphere completely dominated by pressure, conduction and convection.

 

[cnm]

STP means standard temperature and pressure, ie roughly what it is at sea level.

I find it hard to believe a trained physicist has not encountered the term STP before.

 

[SSDD]

All objects radiate according to their temperature and emissivity. If they are in proximity then the net movement of energy will always be from warm to cool but both objects continue to radiate according to their temperature and emissivity.

All perfect black bodies in perfect vacuums radiate according to their temperature and emissivity...all other objects radiate according to their area, their temperature, their emissivity and the difference between their own temperature and the temperature of their surroundings... Your belief that the surroundings a radiator finds itself in have no bearing on how it radiates is a terrible flaw in your line of thinking...

 

[Wuwei]

CO2 collides 30,000 times with other molecules in our atmosphere during the time a photon resides in a CO2 molecule, making emission nearly impossible.

How many times have we heard that CO2 is such a small portion of air it can't have much of an effect on climate. I decided to do a calculation of the density of 15 micron radiation of CO2 in a cubic meter of air at room temperature and pressure.

Number of air molecules per m^3 = 2.53 10^25 (Use Avogadro's number)

CO2 density: 400 ppm

Number of CO2 molecules per m^3 @ 400ppm = 1.01 10^22

Number in excitation state 1.01 10^22 x 2/9 = 0.244 10^22
(From Equipartition Principle)

Relaxation time for CO2 vibration 6 microSec.

There will be 0.244 10^22 molecules emitting 15 micron radiation every 6 microSec

Probability of CO2 emission from excited state (with no collision) = 1 / 30,000

No. of photons per sec from CO2 = number x probability / 6.0 10^-6
= 0.244 10^22 / ( 30,000 x 6.0 10^-6 ) = 1.35 10^22 emissions / sec

Energy of 15 micron photon = 1.3 10^-20 J

Joules per second of all 15 micron photons = 1.35 10^22 x 1.3 10^-20 J/s = 1.75 x 10^2 W

Conclusion:
So even though there are only 400 ppm of CO2 the energy radiation density is 175 Watts (J/s) radiating within a cubic meter.
Does that sound large? Not compared to the total energy of air at STP 156,000 J
( = number of air molecules times their average kinetic energy.)
(Someone might want to double check the calculations,)

 

[SSDD]

All fine and good, I won't dispute any of that, my interests have never been in gases but rather solid materials in a laboratory setting, but it remains curious to me you people seem fixated on the exact properties and relationships of ONE atom, CO2 in relation to the atmosphere when CO2 is:

1. Not even the strongest or only GHG.
2. Only one of MANY items in the atmosphere which absorb sunlight as infrared energy.

My point is that there is no radiative greenhouse effect as described by climate science...it simply is not possible in a troposphere so completely dominated by pressure, conduction and convection. CO2 has no effect on the temperature and no other so called greenhouse gas does either except for water vapor.