<p>You have done multiple things wrong here. One is that you have used a linear interpolation when you should have used a logarithmic interpolation. But that's a different question, one that <a href="https://earthscience.stackexchange.com/questions/18266">you have already asked</a>. I'll address that there. That said, a logarithmic interpolation would indicate that the warming due to a 73% increase would be 2.37 °C rather than 2.2 °C.</p>

<p>But that math is incorrect. We have not seen a 73% increase in <span class="math-container">$\text{CO}_2$</span> levels over preindustrial times. We have seen a 46.4% increase. The second thing you have done wrong is to incorrectly apply a linear interpolation. The correct value, assuming a linear interpolation is correct, is <span class="math-container">$\left(\frac{410-280}{560-280}\right)3^{\circ}$</span>, or 1.39 °C.</p>

<p>We have seen more than a 1.39 °C increase from preindustrial times due to the 46.4% increase in <span class="math-container">$\text{CO}_2$</span> levels over preindustrial times. The correct math is a logarithmic interpolation: <span class="math-container">$\log_2\left(1+\frac{410-280}{560-280}\right)3^{\circ}$</span>, or 1.65 °C.</p>

<p>We have not quite seen a 1.65 °C increase. So what's going on? Another thing you have done wrong is to implicitly assume that the current climate represents the steady-state response to the current <span class="math-container">$\text{CO}_2$</span> level. It does not.</p>

<p>Climate instead changes rather slowly. Suppose we stopped adding ever more <span class="math-container">$\text{CO}_2$</span> to the atmosphere but instead maintained <span class="math-container">$\text{CO}_2$</span> at the current level of 410 ppm. Temperatures will continue to rise for quite some time because the climate has not yet reached its steady-state response to the rather sudden rise from 280 ppm <span class="math-container">$\text{CO}_2$</span> to the current level of 410 ppm.</p>

CC BY-SA 4.0
2020-01-06T11:32:28.023Z
@Fred - I've reverted your changes. Just as to-day used to need a hyphen but no longer does, preindustrial no longer needs a hyphen, look it up. The change in how I wrote $\text{CO}_2$ versus how you wrote it is a bit trivial.
David Hammen
CC BY-SA 4.0
2020-01-06T12:06:04.933Z
I only included the hyphen to be consistent with the way it was written in the third paragraph.
Fred
CC BY-SA 4.0
2020-01-06T12:14:52.833Z
@Fred - Thanks.
David Hammen
CC BY-SA 4.0
2023-09-12T21:52:53.167Z
I recalculated it for current 420 PPM CO2 and got 1.755 degrees C. This means that we already overshot the 1.5 degrees C goal (Paris accord) right?
polcott
CC BY-SA 4.0
2023-09-26T15:02:39.927Z
log2(1+(420−280)/(560−280)) * 3 = 
log2(1 + (140 / 280)) * 3 = 
log2(1.5) * 3 = 1.7548875021634685443612168318434 degees C
polcott
CC BY-SA 4.0
2019-10-29T05:37:00.847Z
2020-01-06T12:14:41.597Z
2020-01-06T12:14:41.597Z
David Hammen
<p><a href="https://en.wikipedia.org/wiki/Climate_sensitivity#Estimating_climate_sensitivity" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Climate_sensitivity#Estimating_climate_sensitivity</a></p>

<p><strong>This indicates that current <span class="math-container">$\small\mathsf{CO_2}$</span> levels have a temperature equilibrium of 2.2 <span class="math-container">$\small\mathsf{^oC}$</span> disagreeing with statements that we are still below 1.5 <span class="math-container">$\small\mathsf{^o}$</span>C. What did I do wrong?</strong><br>
280 PPM * 2 = 560 ppm<br>
410 PPM / 560 ppm = 73%<br>
73% * 3 <span class="math-container">$\small\mathsf{^o}$</span>C = 2.2 <span class="math-container">$\small\mathsf{^o}$</span>C<br></p>

<p><a href="https://skepticalscience.com/climate-sensitivity-advanced.htm" rel="nofollow noreferrer">https://skepticalscience.com/climate-sensitivity-advanced.htm</a></p>

CC BY-SA 4.0
2019-10-28T14:51:20.010Z
Please always provide significant content of links within your post, since links are object to change, eventually leaving your question a stub. Also, we haven't reached equilibrium yet, have we?
Erik
CC BY-SA 4.0
2019-10-28T15:10:22.287Z
@Erik The key content is in the question itself. The links merely provide the reference to this content. Yes it would seem that we have not yet reached equilibrium, yet when we apply this answer for the 1.5C claim the 1.5C claim seems to be refuted. It would certainly not be my first guess that my math correctly refutes a consensus of climate scientists. I would instead assume that I made a mistake.
polcott
<climate-change>
How do we apply CO2 sensitivity (3C ± 1.5C for every doubling) to current CO2 levels
polcott
# How do we apply CO2 sensitivity (3C ± 1.5C for every doubling) to current CO2 levels

(Asked by: username_0 on 2019-10-28)

<https://en.wikipedia.org/wiki/Climate_sensitivity#Estimating_climate_sensitivity>

**This indicates that current $\small\mathsf{CO\_2}$ levels have a temperature equilibrium of 2.2 $\small\mathsf{^oC}$ disagreeing with statements that we are still below 1.5 $\small\mathsf{^o}$C. What did I do wrong?**  
280 PPM \* 2 = 560 ppm  
410 PPM / 560 ppm = 73%  
73% \* 3 $\small\mathsf{^o}$C = 2.2 $\small\mathsf{^o}$C

<https://skepticalscience.com/climate-sensitivity-advanced.htm>

**Tags:** climate-change

> **Comments:**
> **username_1**: Please always provide significant content of links within your post, since links are object to change, eventually leaving your question a stub. Also, we haven't reached equilibrium yet, have we?  
> **username_0**: @username_1 The key content is in the question itself. The links merely provide the reference to this content. Yes it would seem that we have not yet reached equilibrium, yet when we apply this answer for the 1.5C claim the 1.5C claim seems to be refuted. It would certainly not be my first guess that my math correctly refutes a consensus of climate scientists. I would instead assume that I made a mistake.  

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## 1 answers

### Answer by username_2 on 2019-10-29. Score: 2

You have done multiple things wrong here. One is that you have used a linear interpolation when you should have used a logarithmic interpolation. But that's a different question, one that [you have already asked](https://earthscience.stackexchange.com/questions/18266). I'll address that there. That said, a logarithmic interpolation would indicate that the warming due to a 73% increase would be 2.37 °C rather than 2.2 °C.

But that math is incorrect. We have not seen a 73% increase in $\text{CO}\_2$ levels over preindustrial times. We have seen a 46.4% increase. The second thing you have done wrong is to incorrectly apply a linear interpolation. The correct value, assuming a linear interpolation is correct, is $\left(\frac{410-280}{560-280}\right)3^{\circ}$, or 1.39 °C.

We have seen more than a 1.39 °C increase from preindustrial times due to the 46.4% increase in $\text{CO}\_2$ levels over preindustrial times. The correct math is a logarithmic interpolation: $\log\_2\left(1+\frac{410-280}{560-280}\right)3^{\circ}$, or 1.65 °C.

We have not quite seen a 1.65 °C increase. So what's going on? Another thing you have done wrong is to implicitly assume that the current climate represents the steady-state response to the current $\text{CO}\_2$ level. It does not.

Climate instead changes rather slowly. Suppose we stopped adding ever more $\text{CO}\_2$ to the atmosphere but instead maintained $\text{CO}\_2$ at the current level of 410 ppm. Temperatures will continue to rise for quite some time because the climate has not yet reached its steady-state response to the rather sudden rise from 280 ppm $\text{CO}\_2$ to the current level of 410 ppm.

> **Comments:**
> **username_2**: @username_3 - I've reverted your changes. Just as to-day used to need a hyphen but no longer does, preindustrial no longer needs a hyphen, look it up. The change in how I wrote $\text{CO}_2$ versus how you wrote it is a bit trivial.  
> **username_3**: I only included the hyphen to be consistent with the way it was written in the third paragraph.  
> **username_2**: @username_3 - Thanks.  
> **username_0**: I recalculated it for current 420 PPM CO2 and got 1.755 degrees C. This means that we already overshot the 1.5 degrees C goal (Paris accord) right?  
> **username_0**: log2(1+(420−280)/(560−280)) * 3 =  log2(1 + (140 / 280)) * 3 =  log2(1.5) * 3 = 1.7548875021634685443612168318434 degees C  

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