Adjusting a (linear) graph to adjust for its area

banjo89

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Hi everyone,

So I am creating a simple CO2 emissions reduction scenario. The Y axis is in the case the total carbon emissions in GtC. The X axis is the time in years.

Y axis: from 0 to 36.6 (in GtC)
X axis: from 2017 to 2070 (in years)

Lets say the remaining carbon emission budget in this time-frame (53 years) is 1000 GtC.

So if I make a linear reduction from (2017 , 36.6) to (2070 , 0) the cumulative emissions (so the area under the graph) is smaller than 1000 GtC.
Hence, there is still a remaining budget I want to divide over those years. I want to do this without increasing the starting point of the Y-axis.
So for example, a linear graph with a bend in it or a more rounded graph. I want to prevent any extremely steep decreases in any given year though to make it as realistic as possible.

Thanks in advance!

Kind regards,

Tom
 

Dr.Peterson

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Is it a requirement that Y = 0 in 2070? If not, just make a linear function that will be positive then.

Or, you could keep Y at 36.6 until some time X=a, and then start decreasing linearly to zero.

Alternatively, it might be possible to make a quadratic function Y = -aX^2 + bX + c that passes through (2017 , 36.6) and (2070 , 0), with the additional constraint that the definite integral is 1000.
 

banjo89

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Thank you very much for your repply!

"Or, you could keep Y at 36.6 until some time X=a, and then start decreasing linearly to zero."
I was indeed trying this but I did not succeed at it..

When I decrease from (2017 , 36.6) to (2070 , 0) the formula is:

Y = -0.69 X + 36.6
You say keep Y at 36.6 unill X=a. But "a" is expressed in emissions and "X" in years so I am a bit confused..
 

Dr.Peterson

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I wasn't aware that you already had a variable called "a". All I meant by "a" was "some value of X, which you have to determine by the condition on area".

So if Y is defined piecewise, Y = 36.6 for 0<X<a and linearly decreasing from there to 0 until 2070, what would the area be (as a function of a)? Set that to 1000, and solve for a.
 

banjo89

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Thanks you for your help. I am still stuck on how I would have to do this.. I have made a picture of what I had in mind (the dotted lines are the methods i want to use) I just cant figure out how to do it. Escpecially on what kind of formula to use.. or how to find untill which year I can keep emissions at 36.6 before decreasing linearly to 0 (with remaining at 36.6 for the shortest possible time).
 

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Dr.Peterson

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For the piecewise linear function, I told you what to do.

Suppose we stay at 36.6 from 2017 to A, and then linearly decreases until 2070. (Incidentally, I would have defined x as "years since 2017", so it would be going from 0 to 53 and the numbers would be smaller, but I'm sticking with your approach.)

Then the area under the horizontal line will be 36.6A, and the rest will be a triangle with height 36.6 and base 2070-A, so its area is 36.6(2070-A)/2. Add those areas and set the sum equal to 1000. Solve for A.

As for other curves, that's entirely up to you. I previously suggested Y = -aX^2 + bX + c, a quadratic passing through the two endpoints (but probably not with its vertex at the starting point as you appear to show. Do you know how to find the area under that (integration)? You haven't yet shown where you are having trouble carrying out my suggestions.
 

banjo89

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Thank you very much! Yes sorry indeed I meant to write 2017 at the origin instead of 0 (so indeed 53 years total).
I now understand the first method I feel so stupid now haha (math is just too long ago). For the second method I am not yet sure. I think I mainly ran into trouble with deciding which kind of formula to use. Integration and then calculating the area is a long time ago for me but I can try figuring it out. I will get to it and let you know how it went.
 

Dr.Peterson

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Good. The integral won't be too hard; I wanted to be sure you were able (and willing) to try it at all. If you have trouble, show how far you got and I can guide you through.

Once you've worked out the quadratic, you may want to think about other curves. Looking at your graph, you may notice that the quadratic seems to be steeper at the end than the linear, which I think has to be true of any curve; that could be a reason to prefer the linear (based on what you said in the OP). But I don't know the realities behind your application, so the decision will ultimately be yours.
 
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