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Diminishing logarithmic relationships?
There've been a number of posts lately along the lines of "CO2's warming effect logarithmically diminishes with concentration, it is the first 0 to 100ppm that has done nearly all the warming, double our CO2 now would hardly have a noticeable effect, fractions of a degree C only" (quoting from one such answer)
Can you show a logarithmic equation in which successive doublings don't have exactly the same effect?
Jeff- a logarithmic equation has the form y = k log x. Your y = kx does not fit the definition, and has nothing to do with what happens each time you double x in a logarithmic equation.
Now try to answer the actual question.
7 Answers
- TrevorLv 78 years agoFavorite Answer
I can see where you’re coming from and also where the skeptics are coming from, but the skeptics just aren’t doing the maths. It seems they’ve latched onto an idea that sounds like it may be plausible and are running with it, without first checking the validity of their claims.
The relationship between CO2 and temperatures is not logarithmic as some people claim, but if you were to plot a graph of temp vs CO2 it would have that sort of shape to it.
Let’s for a moment assume that the skeptics are correct and the relationship between temps and CO2 is indeed logarithmic, and compare this with the claim that very little warming occurs beyond 100ppmv.
We know that natural warming heats the planet by 33°C so, according to the skeptics, almost all of this must occur at concentrations below 100ppmv.
The log of 100 is 2.000, let’s call that 2.000 ‘heating units’. Current CO2 ppmv is just below 400, the log of 400 is 2.602. If 2.000 heating units = 33°C then 2.602 units = 42.933°C.
42.933K above the baseline figure of 254K would mean that if the skeptics were correct the average global temperature would be 294.933K or 23.783°C. This is nearly 10°C above the real AGT of 14.691°C (30 yr RA, all temp records).
Further, prior to the onset of industrialisation CO2 was at 280ppmv, log 2.447, in ‘heating units’ this is 40.376°C, at 560ppmv (double the baseline) the log is 2.748 or 45.342°C. This means that, using the skeptics own argument, equilibrium climate sensitivity would be almost 5°C, and yet they claim it’s negligible. How can that be?
Of course, we know that the skeptics are talking nonsense and we only need look at Venus to see what happens when you get a runaway greenhouse effect. If the skeptics were correct then the temperature of Venus should no higher than 295K, but in reality it’s 740K.
The effective temperature of Venus is 231.6K (Earth is 254.3K), the dense layer of Venusian clouds would reflect most sunlight back into space and Venus would have an average surface temperature colder than that of Earth, despite being nearer to the Sun (-42°C).
If we now apply the “warming occurs at less than 100ppmv” argument then we could add a generous and maximum 63.1K to the Te of Venus (solar constant Earth = 1366Wm² = 33°C, Venus = 2611Wm², proportionately = 63.1K).
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EDIT: TO JIM (1)
I used Venus to highlight what happens when CO2 is at ridiculously high levels and because it totally destroys the 100ppmv argument.
You’ve linked to an article by Steve Goddard writing about the climate of Venus – he hasn’t got a clue what he’s on about. The whole premise of his argument is that Venus and Earth have very similar DALR’s and concludes “This tells us that the primary factor affecting the temperature is the thickness of the atmosphere, not the composition”. Good grief.
If there were no greenhouse gases then the energy budget would be equilibrated, no warming would occur and Venus would be at Te (colder than Earth). Of course, Goddard completely misses this point – was it deliberate deception or just plain ignorance?
Here’s what real physicists who work on the climate of planets other than Earth have to say, probably more reliable than an electrical engineer who is so bad that even Anthony Watts has had to apologise for Goddard’s mistakes
http://www.realclimate.org/index.php/archives/2006...
EDIT: TO JIM (2)
Are you seriously suggesting that Venus could be so hot due to something other than the runaway greenhouse effect? Surely not. I’d love to hear your theory, I think NASA would as well because according to them Venus is as hot as it is because of the runaway greenhouse effect – nothing else, just that.
http://solarsystem.nasa.gov/planets/profile.cfm?Ob...
As for pressure, of course it affects temperature but if no mechanism exists for the retention of the solar energy and the energy budget is in equilibrium then how would pressure cause Venus to be the hottest place in the solar system other than the Sun itself.
- JimZLv 78 years ago
According to this post, it is logarithmic. Forcing = 2.94 Log2(CO2) + 233.6 (R^2 = .997)
I thought I got that link from Jeff Eng. a while back
http://wattsupwiththat.com/2010/03/08/the-logarith...
If it were linear, it would theoretically warm more as CO2 was added. In the slope of the line that was presented in the link, it appears to be near asymptotic conditions where adding more isn't going to have that much of an effect. Most of the effect appears to be in the first 20 ppmV.
http://wattsupwiththat.files.wordpress.com/2010/03...
Did Trevor seriously try to compare Earth with Venus? There is no relationship except one planet has a much larger atmosphere (nearly a hundred times as massive) with much greater pressure at the surface which accounts for very high temperature of Venus.
http://wattsupwiththat.com/2010/05/08/venus-envy/
Note The one thing that I really have a problem with is people that don't recognize theory and fact. I tend to believe that GHG probably account for the warming of the earth's atmosphere. Suggesting that runaway water vapor greenhouse effect on Venus accounts for its current CO2 atmosphere is a theory and it is a theory that has a very good change to be overturned as new evidence is discovered. I notice there was very little in Trevor's link that discussed pressure, yet the difference between Mars and Venus is probably mostly about the mass and secondly about the distance to the sun. Mars' atmosphere is also almost all CO2 yet it is a 100 below or more. GHGs aren't the only thing warming the surface. It is almost like they suggest that Death Valley is warmer than Mount Wilson due to greater GHGs in Death Valley.
- pegminerLv 78 years ago
Well, Jeff Engr's answer has demonstrated one thing: "Engr" does not stand for "Engineer."
Any more attempts at math from the other deniers should be equally as amusing.
EDIT: It would be interesting if jim z was correct and he did find out about that formula in his link from Jeff Engr. That formula (unlike what Jeff Engr writes) certainly IS logarithmic. All you have to do is plug that formula into Excel and calculate the values for 100 ppm, 200 ppm, 400 ppm, etc. and you can easily demonstrate that repeated doublings DO have the same effect, just as the asker asserted.
By the way, there is a problem with a general logarithmic dependence, I wonder if the deniers can spot it?
EDIT for Jeff Engr: You said "You should have read my whole post", well, I read your whole post and none of it made any sense. You are actually part of the motivation for my open question--and I have to say, of all the answers to my question, yours is the only one that I don't believe.
- KanoLv 78 years ago
CO2 is logarithmic even wikipedia accepts that, Trevors answer is ridiculous He mentions 33C caused by our atmosphere and then does calculations based on CO2 being the only part of the atmosphere that keeps us warm. if there were no CO2 our atmosphere would warms us approx 30C compared to no atmosphere at all.
Venus has no greenhouse effect which is obvious due to day temperatures being 737K or 836F and night time temps 737K 836F (a venusian day is 118 earth days)
All the inner planets were balls of molten rock when they formed, mars being smaller, with the smallest atmosphere cooled fastest, Earth being the perfect size cooled enough for life to form. Venus being large with a dense atmosphere 60 times greater than earth, did not cool.
What puzzles me is how come the IPCC models show exponential rise when based on a gas of diminishing returns.
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- Anonymous8 years ago
In a true logarithmic relationship, doublings will always have the same effect. To get the same effect, the input increases exponentially. Nevertheless, log(infinity) = infinity
The claim, "Carbon dioxide effect is logarithmic, therefore adding more will have no effect," is a denialist fantasy.
Mister Zedd
<much greater pressure at the surface which accounts for very high temperature of Venus>
The surface of Venus is not the oulet of a compressor. If the atmosphere of Venus were pure nitrogen, it's pressure would have no effect on temperature.
http://en.wikipedia.org/wiki/Zeroth_law_of_thermod...
Edit
I owe an apology to Paul's Alias 2. I now see what your issue with Trevor and differential equations is. Trevor thinks that you can use computers instead of differential equations. The truth is that you can use computers to solve differential equations, not instead of differential equations.
- Jeff EngrLv 68 years ago
Successive doublings are NOT logarythmic relationships. successive doubling are linear.
i.e. x = 2y
for CO2, the agruement would be that x = Degrees C and Y = CO2 concentration in the atmosphere.
This is NOT an alogrythmic equaiton. CO2's ability to act as a GHG gas is alogrythmic NOT linear. VERY few things in nature are linear.
Logrythic equialtions are VERY different. they are curves and not lines. They can be very extreme curves. In the case of CO2 acting as a GHG gas, the curve is VERY extreme.
Compounding interest is a geometric progression. If you look that up you can get a good idea, however logarythmic curves are generally much steeper.
EDIT: You should have read my whole post. I clearly stated that the equation x = 2y is a linear equation. I also cleearly stated that this equation represents what some AGW proponants have been stating. i.e. for every doubling of CO2 you get an increase of X degrees C.
- Paul's Alias 2Lv 48 years ago
The original concentration is N. So the effect of the first doubling is to have the effect go from being a ln (b)N to a ln(2bN), where a and b are boring constants. The effect of a second doubling is to go from a ln 2bN to a ln 4bN.
So what we want to test is whether (a ln 4bN)/(a ln2bn) is equal to (a ln2bN)/(a ln bN). We can use well-known simple formulas, but I will derive it directly.
If (a ln 4bN)/(a ln2bn) is equal to (a ln2bN)/(a ln bN). then e^ [(a ln 4bN)/(a ln2bn) ]would be equal to e^ (a ln2bN)/(a ln bN).. e^ [(a ln 4bN)/(a ln2bn) ] is b4N/b2n, and e^ (a ln2bN)/(a ln bN).. is b2N/bN, so they are indeed equal.