What explanation of the Greenhouse Effect do you use?

[Wuwei]

May I make a suggestion that this thread get off the definition of "collimated". It's too bad it's turning into a battleground of egos.

 

[IanC]

I wonder why you ignore the actions of methane, water vapor, dust, aerosols, and other gases, etc., in the roles they all play in converting sunlight to IR energy and re-radiating it back into their environment? Will you tell me because CO2 holds a unique place in that it uniquely matches some special resonant place in the mechanics of heat storage and transfer?

Right CO2 is not the only GHG, but it is, in a sense, a case study on how GHG's act. Once you understand CO2, then you can understand the others, and the similarities and differences.

Further, it seems a distinction without a difference that having timed the EXACT time it takes for an atom to re-release an IR photon vs. an atomic collision (ie: kinetic transfer) of energy, the energy is still released, transferred, and in a random way. Either way you cook it, all kinds of stuff up there is catching sunlight, trapping it as heat and then releasing it, some of which is locally transferred kinetically, some of it radiated.

Either way, the air is still collecting and storing energy from the Sun and I'm much more interested in the broad picture than purely just CO2, which is clearly just as fraction of the entire greenhouse process essential to life on Earth.

I thought you were interested in the physics of the GHE, but now it seems you aren't. The broad picture cannot be handled without some of the underlying basics.

.

We have another blowhard on our hands. Tubesucker may be mentally unstable but at least he isnt full blown retarded like jc. Or is he?

.

 

[toobfreak]

Physics is like an onion, with layer after layer making smaller and smaller contributions.

Another great insight from someone with only a high school education who makes up technobabble! Where is that collimated sunlight, fraud?

You have to make a decision on how many factors are enough to give a reasonable answer. Ignoring the amount of sunlight absorbed by CO2 is more than reasonable, it would in fact be a meaningless complication.

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!

 

[IanC]

May I make a suggestion that this thread get off the definition of "collimated". It's too bad it's turning into a battleground of egos.

Never!!!!!!!

Actually I am the only person who gave a partial definition of the word collimated. It fits.

And I dont have a big ego. Im just condescending

 

[Wuwei]

We have another blowhard on our hands. Tubesucker may be mentally unstable but at least he isnt full blown retarded like jc. Or is he?

I am not going to make any judgments until the smoke settles.

.

 

[Wuwei]

And I dont have a big ego. Im just condescending

LOL. An interesting self analysis. I try not to be condescending but I often fail.

.

 

[IanC]

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.

 

[toobfreak]

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%

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!

 

[toobfreak]

I gotta ask you, Ian The Science Guy, you keep trying to discredit me even though I'm the only one here whose offered any info on my background and even been willing to prove it, yet you say so much that is so wrong or at least easily gotten from some common climate change article that I wonder: we used to have another troll from Russia who thought he was Mr. Science always telling us the great inventions of the USSR until one day I finally decided to test him on his knowledge. He ran like a scolded dog, so I will make the same offer to you. If you know so much about science, the Sun, light, physics, et al., certainly you can solve a high school level math problem?

Here it is. Tell us what the solution to this is?

And please, if anyone else knows the answer, keep it to yourself until IanC answers first:

It's a simple enough problem.

Answer this, and at least you'll have a little bit of credibility.

 

[Olde Europe]

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

Oh my...

At the surface of the Sun the intensity of the solar radiation is about 6.33×10^7 W/m^2​

 

[Wuwei]

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!

The amount of energy from the sun that hits the earth has many contexts.

• Is it normal incidence at the top of the atmosphere (noon at equator)?
• Normal incidence at earth surface?
• Is it the total radiation on one side of the earth with the cosine reduction at latitudes off normal?
• Is it the full day average. Divided by two because of the dark side?
• Is it full radiation or radiation per square meter?
• Etc.

These all give different figures. You will have to check the context from your sources.


.

 

[jc456]

I wonder why you ignore the actions of methane, water vapor, dust, aerosols, and other gases, etc., in the roles they all play in converting sunlight to IR energy and re-radiating it back into their environment? Will you tell me because CO2 holds a unique place in that it uniquely matches some special resonant place in the mechanics of heat storage and transfer?

Right CO2 is not the only GHG, but it is, in a sense, a case study on how GHG's act. Once you understand CO2, then you can understand the others, and the similarities and differences.

Further, it seems a distinction without a difference that having timed the EXACT time it takes for an atom to re-release an IR photon vs. an atomic collision (ie: kinetic transfer) of energy, the energy is still released, transferred, and in a random way. Either way you cook it, all kinds of stuff up there is catching sunlight, trapping it as heat and then releasing it, some of which is locally transferred kinetically, some of it radiated.

Either way, the air is still collecting and storing energy from the Sun and I'm much more interested in the broad picture than purely just CO2, which is clearly just as fraction of the entire greenhouse process essential to life on Earth.

I thought you were interested in the physics of the GHE, but now it seems you aren't. The broad picture cannot be handled without some of the underlying basics.

.

We have another blowhard on our hands. Tubesucker may be mentally unstable but at least he isnt full blown retarded like jc. Or is he?

.

still waiting on that back radiation evidence from a gas that transfers 99.9999% of its energy in collisions yep. I'd say, you be looking into mirrors.

 

[IanC]

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?

 

[toobfreak]

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

Oh my...

At the surface of the Sun the intensity of the solar radiation is about 6.33×10^7 W/m^2​

Good catch. I was looking for articles that simply gave a simple estimate of the POWER hitting the Earth near the equator per sq meter on a clear day. Not some long winded analysis. Still, I find it interesting when they finally gave their estimate CLEAR at the bottom (something like 5.78 kw hours per day per sq meter (still differeing from another site above), they too claimed that the atmosphere absorbed 30% of the incoming solar energy which clearly IT DOES NOT.

 

[toobfreak]

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.

 

[toobfreak]

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!

The amount of energy from the sun that hits the earth has many contexts.

• Is it normal incidence at the top of the atmosphere (noon at equator)?
• Normal incidence at earth surface?
• Is it the total radiation on one side of the earth with the cosine reduction at latitudes off normal?
• Is it the full day average. Divided by two because of the dark side?
• Is it full radiation or radiation per square meter?
• Etc.

These all give different figures. You will have to check the context from your sources.
.

THAT WAS MY POINT, Wu. Every place I read, they give a different set of parameters and units. To my mind, I was simply looking for a common sense figure, which to me suggests a standard reference, around noon, around the equator, 0° incidence, at the surface, average albedo, on a clear day. I've seen figures varying between 1000 and 1300 watts per square meter. That seems empirically reasonable.

 

[Billy_Bob]

Never said it was. Ask your OP. He is the one who brought up distant stars, color temperature, collimated light and so much more totally unrelated to greenhouse theories! ALL of this extraneous dialog was generated by several of YOU simply because I gave a very simple definition as asked for by the OP and qualified my background as being in the physics sciences.

I've yet to hear that any of you even have a high school diploma.

That doesn't matter anymore. Sunsetommy finally understood what I was talking about. It's the vibration mode that is important in 15 micron absorption and emission.

But his animation is wrong in one important aspect. The CO2 that is excited by an incoming photon most likely will NOT emit a photon as the animation suggests. A collision with another gas molecule is much more likely to happen. In that way the capture of a photon by CO2 will transfer it's energy to linear kinetic energy of a different molecule and go to it's ground state without emission. The gain in the other molecule of course is random and manifests as local heat.

.

Actually, no. If a CO2 atom is excited by a photon, it has taken some part of that photon and converted it to heat, the source of the excitation. It will then re-radiate that heat in the form of another photon returning to its ground state. Gas molecules OTOH are rather far apart and collisions comparatively less likely.

Your trying to teach a person about the atmosphere and the fact that energy in the troposphere is primarily through conduction and convection. 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.

The entire climate thing is a fraud.

Entire climate thing? Are you thinking the physical basis of the GHE is fraudulent? If so, can you be more specific?

.

sure, 99% of the CO2 absorbed IR is handed off on collisions resulting in a conduction. If 99% hand their energy off, then how does downward IR from CO2 occur?

If 99% hand their energy off, then how does downward IR from CO2 occur?

If 99% hand their energy off, then how does spaceward IR from CO2 occur?

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.

 

[Toddsterpatriot]

Never said it was. Ask your OP. He is the one who brought up distant stars, color temperature, collimated light and so much more totally unrelated to greenhouse theories! ALL of this extraneous dialog was generated by several of YOU simply because I gave a very simple definition as asked for by the OP and qualified my background as being in the physics sciences.

I've yet to hear that any of you even have a high school diploma.

That doesn't matter anymore. Sunsetommy finally understood what I was talking about. It's the vibration mode that is important in 15 micron absorption and emission.

But his animation is wrong in one important aspect. The CO2 that is excited by an incoming photon most likely will NOT emit a photon as the animation suggests. A collision with another gas molecule is much more likely to happen. In that way the capture of a photon by CO2 will transfer it's energy to linear kinetic energy of a different molecule and go to it's ground state without emission. The gain in the other molecule of course is random and manifests as local heat.

.

Actually, no. If a CO2 atom is excited by a photon, it has taken some part of that photon and converted it to heat, the source of the excitation. It will then re-radiate that heat in the form of another photon returning to its ground state. Gas molecules OTOH are rather far apart and collisions comparatively less likely.

Your trying to teach a person about the atmosphere and the fact that energy in the troposphere is primarily through conduction and convection. CO2 collides 30,000 times with other molecules in our atmosphere during the time a photon resides in a

The entire climate thing is a fraud.

Entire climate thing? Are you thinking the physical basis of the GHE is fraudulent? If so, can you be more specific?

.

sure, 99% of the CO2 absorbed IR is handed off on collisions resulting in a conduction. If 99% hand their energy off, then how does downward IR from CO2 occur?

If 99% hand their energy off, then how does downward IR from CO2 occur?

If 99% hand their energy off, then how does spaceward IR from CO2 occur?

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.

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

The mass of the atmosphere restricts the emission of CO2 near the surface?
Tell me more.

 

[Toddsterpatriot]

May I make a suggestion that this thread get off the definition of "collimated". It's too bad it's turning into a battleground of egos.

Never!!!!!!!

Actually I am the only person who gave a partial definition of the word collimated. It fits.

And I dont have a big ego. Im just condescending

Actually I am the only person who gave a partial definition of the word collimated. It fits.​

Part 2: Solar Energy Reaching The Earth’s Surface | ITACA

Seems to fit toobefreak's source as well.

 

[IanC]

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?