Comments by "Killed The Cat" (@killedthecat1034) on "Global climate strike: Millions take to the streets to save the world" video.
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@thebutcherofbenghazi.libya3348
" or else your entire argument is a lost cause. " That would be only on your opinion, correct?
Again, if you have to assume someone has an educated themselves, yet you know nothing about them, to validate your own point, maybe you need to re-evaluate that point.
I am on topic. Just like I was a second ago with you. Both of them and you tried to point out what you thought was faulty logic on my part. Which only turned out to be assumptions on yours. Which I pointed out.
Now you're trying to put words into my mouth in order to support your point. Please point out where I said I was not smart. Feel free to quote me.
What I did say, was that neither of us are climate science experts. Which is far from not being smart. It just means that neither of us have the education background to fully understand the issue with climate science. So to be so arrogantly sure of your opinion on the matter, especially when it goes against what expert climate scientists have said, is reductive and redundant. It's also not very logical.
Neither is you, trying to attack me, for defending myself against somebody who chose sarcasm and childish insults as a form of communication, with no provocation. I don't see where you said that to them. That looks awful bias. Maybe it's because you agree with their point of view on climate change? Talk about hypocrisy...
I'm breathing just fine, sweetheart. 😉✌
As I have already stated, I am not a climate science expert. I do not intend on pretending that I am either. That actually would make me a hypocrite . I can refer you to some good sources of info on the specifics you want. This link goes to NASA and has several links on the page, itself that helps to explain the information you were asking for.
https://climate.nasa.gov/news/2743/the-scientific-method-and-climate-change-how-scientists-know/
I'm pretty sure I've already said this... asking for the source of the data that Pat referenced when speaking to me, is not in any way a negative. I did not ask for this data. They provided it willingly and of their own volition. I merely asked for the source of that data. How is asking for the source of that data a negative or in any way puts the onus on to me to provide it?
I will ask you to prove your claim. You're correct that scientific theory, which is what you're referring to, has a higher standard of probability. ONE of the criteria's for this is to be able to repeat your experiment and receive the same or within a margin of error similar results. That is only one of the standards but it is one of them.
You, on the other hand are making claims like they where. This has not been proven. This is also not a negative. It is verafiable. You just have to be willing to go check the results and scientific method of all 98% of climate scientists in the world who have come to the same conclusion. I will state again, I am not one of them.
Your theory that the onus is on me to prove something that I am not an expert in because I am trying to convince someone to take an action, only holds water if you assume that I am trying to change your mind in this case. I am not. I'm not trying to get either one of you three to do a damn thing. Other than prove the claims that you were making.... Or back up the data, you have willingly provided. If you willingly provide what you are presenting as data and fact to disprove climate change, it has to come from somewhere. Whether you're a climate scientist or not. Asking for the source material that you have take it upon yourself to provide, has nothing to do with an onus on me. it is rather Shady that none of you seem to be able to provide it, though.
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Here! A few good source links! Some equasions and quotes.
https://www.scientificamerican.com/article/science-behind-climate-change/
https://climate-nasa-gov.cdn.ampproject.org/v/s/climate.nasa.gov/evidence.amp?amp_js_v=a2&_gsa=1&usqp=mq331AQCKAE%3D#aoh=15691814703424&referrer=https%3A%2F%2Fwww.google.com&_tf=From%20%251%24s&share=https%3A%2F%2Fclimate.nasa.gov%2Fevidence
"The carbon released worldwide from burning carbon and deforestation has recently been about 7.1 Gt/yr. The number of CO2 molecules released is
NCO2 = (7.1 x 1015 g/yr)(6.02 x 1023/mole)/(12 g/mole) = 3.6 x 1038 molecules CO2/yr. (1)
The mass of the atmosphere is the surface area of Earth times the atmospheric pressure of 105 Pascal divided by g:
Matmos = PA/g = (105 Pa)(4p)(6.4 x 106m)2/(9.8 m/sec2) = 5.3 x 1018 kg. (2)
The number of O2and N2 molecules in the atmosphere is
Natmos = (5.3 x 1021 g)(6.02 x 1023/mole)/(29 g air/mole) = 1.1 x 1044 molecules. (3)
This gives the rate of increase in concentration of CO2 molecules,
cCO2 = NCO2/Natmos = (3.6 x 1038 CO2/yr)/(1.1 x 1044 air) = 3.3 ppm/yr. (4)
This is more than twice the atmospheric CO2 rise of 1.4 ppm/yr (325 ppm in 1970 to 354 ppm in 1990 to 370 ppm in 2000). Thus, about half of the CO2 remains in the atmosphere, the other half goes into sinks in the oceans and on land.
CO2 Before Industrialization. The pre-industrial CO2 level was 280 ppm in 1800. By 1959, the level had grown to 316 ppm. We can estimate total change in concentration by integrating backwards in time. Using a rate of 0.9 ppm/yr in 1959 and a global carbon rate growth rate of aboutl= 3%/year, the increase in CO2 concentration between 1800 and 1959 should be about
ΔcCO2 =  (0.9) eltdt = 0.9(e0 – e)/l = 0.9/0.03 = 30 ppm. (5)
Subtracting this from the 1959 value of 316 ppm gives a pre-industrial CO2 level of 285 ppm, close to the accepted value of 280 ppm.
CO2 in the 21st Century. 2050 CO2 levels may be obtained by projecting 60 years growth onto the 1990 level of 354 ppm. Energy Information Agency estimated a business-as-usual approach will give 2%/yr global growth in fossil fuels, for a 2050 concentration of
cCO2 =  (1.4 ppm/yr) e0.02t dt + 354 ppm = 162 ppm + 354 ppm = 516 ppm. (6)
This figure is consistent with most business-as-usual projections.
Upper-Atmospheric Temperature Ta.Earth's temperature is determined from a heat balance between absorbed energy from solar flux so = 1367 W/m2 and infrared emission to space. The solar power intercepted by the area of Earth's disk (pRE2so) is distributed over the entire spherical area (4pRE2), giving an average solar flux of so/4 = 1367/4 = 342 W/m2. Of this, 70% is absorbed by the Earth, and 30% is reflected (Earth’s albedoa = 0.3 in the visible), giving an average fluxabsorbed by surface and atmosphere,
sabsorbed = (1 – a)(so/4) = (1 – 0.3)(1367/4) = 239 W/m2. (7)
Absorption by clouds and atmosphere reduces solar flux at the surface to an average of about 200 W/m2. The energy absorbed by Earth’s surface is sent upward by infrared, evaporation and air currents, which is captured by the atmosphere or passes directly to space. In our first model, we assume that allthe absorbed energy is reradiated to space as IR from a thin surface at the top of the atmosphere.
The power balance at the top of the Earth's upper atmosphere is
Pin = (1 – a)(pRE2so) = Pout = esTa4(4pRE2), (8)
where temperature at the top of the atmosphere Ta is in Kelvin, s is the Stefan–Boltzmann constant, 5.67 x 10–8 W/m2K4, and e is emissivity (about 1 for 10–micron infrared). Solving for the upper atmosphere temperature,
Ta = [(1 – a)so/4es]1/4 = [239 W/m2/se]1/4 = 255 K = –18oC = 0oF. (9)
The temperature in the middle of the troposphere is 255 K at 5 km above the surface (and at 50 km.) This is 32 K colder than the observed average surface temperature of 287 K (14.0 oC with 1997 averages of 14.6oC in the northern hemisphere and 13.4oC in the southern hemisphere).
As a comparison we calculate Ta-V for Venus, which has a higher solar flux since the radius of its orbit is only 60% that of Earth:
so-V = so(rE/rV)2 = (1367 W/m2)(1.50 x 108 km/1.08 x 108 km)2 = 2610 W/m2. (10)
However, Venus's higher albedo of 0.76 reflects a greater fraction of sunlight, greatly reducing the average absorbed flux to
(1 – a)so-V/4 = (1 – 0.76)(2610 W/m2)/4 = 157 W/m2, (11)
which is smaller than Earth's 239 W/m2. The upper atmospheric temperature of hot Venus,
Ta-V = [157 W-m–2/s] 1/4 = 229 K (12)
is 26 K colder than Earth's 255 K. However, Venus's higher CO2 concentration traps IR, giving it a surface temperature of 750 K, three times Earth's surface temperature of 287 K.
Surface Temperature Ts. Our zero-dimensional box model did not take into account the following variable factors: Reflection, absorption and emission by air, aerosols, clouds and surface; Convection of sensible and latent (evaporation) heat; Coupling to oceans and ice; Variations in three dimensions; and Variable solar flux.
Next, we estimate the surface temperature Ts without considering Ta. We assume that all the solar flux that is not reflected is transmitted through the air and totally absorbed by the Earth's surface fabsorbed = (1 – a)so/4. The warmed surface radiates as a blackbody, and also loses heat through rising in air currents or evaporated moisture. We allow a fraction of the light radiated from the earth, fIR to be absorbed by the atmosphere, which is mostly in the infrared. The atmosphere radiates 50% of the IR absorbed flux to space and 50% to Earth, giving an IR flux downward of (fIR/2)fabsorbed. Again, a fraction fIR of this energy is absorbed in the atmosphere again and 50% of this radiates downward and is absorbed by the surface, (fIR/2)(fIR/2)fabsorbed. This process gives an infinite sum in the energy balance:
fabsorbed + (fIR/2)fabsorbed + (fIR/2)2fabsorbed + (fIR/2)3fabsorbed + ….. + (fIR/2)nfabsorbed = sTs4. (13)
After manipulation, this becomes
fabsorbed/[(1 – (fIR/2)] = sTs4. (14)
We obtain Earth’s surface temperature of Ts = 287 K with fIR = 0.76. For the extreme case of no IR absorption in the atmosphere (fIR = 0), we obtain Ts = 255 K, the temperature of the upper atmosphere. For the other extreme case of 100% IR absorption in the atmosphere (fIR = 1), we obtain Ts = 303 K, consistent with the next calculation.
Ta and Ts Together: Next, we assume that all sunlight is transmitted through the air and absorbed by the Earth's surface. We vary the one free parameter, the emissivity of the atmosphere ea, but retain the surface of Earth as a blackbody with eE = 1. Equation 15 balances heat flow in the single layer of air. The left side doubles the infrared flux emitted by Earth’s atmosphere, since IR goes both up into space and down to Earth's surface, essentially doubling the radiating surface area. This is balanced with IR flux emitted from Earth’s blackbody surface and absorbed by the gray body atmosphere.
2easoTa4 = easTs4 (15)
Equation 16 is an energybalance at Earth’s surface. The left side is the sum of solar energy absorbed at the surface and the absorbed downward flow of IR from the atmosphere, which is balanced with upward IR flux from the surface,
(1 – a)so/4 + easTa4 = sTs4. (16)
Solving equations 15 and 16 gives
Ts = 21/4Ta and sTs4 = (1 – a)so/4(1 – ea/2). (17)
If the air layer is a blackbody (ea = 1, considerable CO2), the atmosphere is Ta = 255 K (as before) and the surface is Ts = 303 K (16 K warmer than actual value of 287 K). If ea = 1/2 (from less CO2), the atmosphere is too cold at Ta = 230 K and the surface is also too cold at Ts = 274 K. By adjusting ea to 0.76, we obtain the “correct” surface temperature, Ts = 287 K.
Multi-Layer Atmosphere. Next we divide the planetary atmosphere into n zones, layered vertically. By using several layers, the temperature gradient in each layer is reduced, smoothing the temperature profile to become more continuous. The thickness of a layer is such that almost all incident IR on a layer is just absorbed in that layer, which then radiates it upwards and downwards. Planets with small amounts of CO2 and H2O have less than one zone, while Venus has many zones. Due to lack of space, we leap to the answer:
T0 = [(1 – a)so/4s]1/4 and Ts = (n + 1)1/4T0, (18)
wheren is the number of IR absorption layers Earth has. Earth’s so = 1367 W/m2 and a = 0.3 gives T0 = 255 K, T1 = 303 K, T10 = 464 K, T20 = 546 K and T75= 753 K. The answer depends greatly on the amount of greenhouse gasses in the atmosphere. Earth's surface temperature of 287 K is somewhat colder than that for one full layer (n = 1) at 303 K. The number of layers for the Earth’s atmosphere is obtained by solving for n, giving
n= (Ts/T0)4 – 1 = (287 K/255 K)4 – 1 = 0.6. (19)
It is not surprising that Earth's atmosphere contains only 60% of an IR layer since O2 and N2 hardly absorb IR, leaving the task of IR absorption to trace amounts of CO2 and H2O. Venus, on the other hand, has a large temperature difference between the upper atmosphere at T0 = 229 K and the surface at Ts = 750 K. These temperatures give 74 IR layers for CO2 rich Venus!
Solar Variations. We might expect solar variations of 0.2% are possible since that is twice the present 11-year solar variation.The 0.2% variation gives a surface temperature variation of
ΔTs = Ts(Δs/so)/4 = (287 K)(2 x 10–3)/4 = 0.14 K. (20)"
https://www.aps.org/units/fps/newsletters/200807/hafemeister.cfm
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