- #1

- 285

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from 280 ppm to 560 ppm.

http://www.grida.no/climate/ipcc_tar/wg1/pdf/tar-01.pdf

I came up with an equation of 4(log 560)-4(log 280)= 1.2041199.

Is this the best way to fit this curve?

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- Thread starter johnbbahm
- Start date

- #1

- 285

- 26

from 280 ppm to 560 ppm.

http://www.grida.no/climate/ipcc_tar/wg1/pdf/tar-01.pdf

I came up with an equation of 4(log 560)-4(log 280)= 1.2041199.

Is this the best way to fit this curve?

- #2

- 2,055

- 319

from 280 ppm to 560 ppm.

http://www.grida.no/climate/ipcc_tar/wg1/pdf/tar-01.pdf

I came up with an equation of 4(log 560)-4(log 280)= 1.2041199.

Is this the best way to fit this curve?

This doesn't seem to be more than a coincidence. According to you the direct response for carbon doubling is [itex] 4 (^{10}log 2) [/itex]. Why a base 10 logarithm?

To compute the direct response for a doubling of CO2, you have to calculate the radiative forcing first, and then the temperature increment using the fact that the radiaton is proportional to the 4th power of the temperature

The second calculation isn't too hard, but the first calculation involves detailed calculations, involving

- different wavelengths

- the temperature and pressure at different heights.

- averaging seasons

- averaging different latitudes

- other greenhouse gases, espescially water vapour.

- clouds.

A really simple model of this process doesn't seem to work.

A program used for the radiation calculations is called MODTRAN.

According to chapter 6 of the IPCC report this comes to 3.7 W/m^2

Since the total outgoing IR radiation is 240 W/m^2, this means that the outgoing radiation has to go up by a factor 3.7/240 = 1.0154 and the absolue temperature has to go up by a factor (1.0154)^(1/4) = 1.0038.

Multiplying this with the absolute temperature in kelvin you get

1.0038 * 288 = 289.1 K, so the temperature has to go up by 1.1 K.

- #3

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I will just take Baede et al number as it is, I was just asking if theIf the amount of carbon dioxide were doubled instantaneously,

with everything else remaining the same, the outgoing infrared

radiation would be reduced by about 4 Wm−2. In other words, the

radiative forcing corresponding to a doubling of the CO2 concentration

would be 4 Wm−2. To counteract this imbalance, the

temperature of the surface-troposphere system would have to

increase by 1.2°C (with an accuracy of ±10%)

log equation was the best way to fit the doubling curve.

- #4

- 2,055

- 319

The 1.2°C number is from the IPCC cited Baede et al, and is based their radiative forcing calculation.

I will just take Baede et al number as it is, I was just asking if the

log equation was the best way to fit the doubling curve.

What exactly do you mean with "the doubling curve"?

- #5

- 285

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it would look like this,

increase from 140 to 280 ppm 1.2 °C

from 280 to 560 ppm 1.2 °C

from 560 to 1120 ppm 1.2 °C

The function is not a straight line.

The log function I used seems to fit (within the ±10% anyway)

I am wondering if there is a better way to fit a function to this curve?

- #6

- 2,055

- 319

it would look like this,

increase from 140 to 280 ppm 1.2 °C

from 280 to 560 ppm 1.2 °C

from 560 to 1120 ppm 1.2 °C

The function is not a straight line.

The log function I used seems to fit (within the ±10% anyway)

I am wondering if there is a better way to fit a function to this curve?

This is not curve fitting, it's just exact computation

[tex] \Delta T = (1.2) \frac { log\frac{C}{280}} {log 2}[/tex]

where C is the CO2 concentration in ppm exactly fits all the data points you mentioned. You can even use logarithms to any base.

I think the idea of CO2 doubling was only introduced because more detailed calculations show that the radiative forcing is approximately proportional to the log of the CO2 concentration.

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