<Oil Shale and the futureThe Oil DrumDrumBeat: July 7, 2006>

80 comments on Linearize this...

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DuncanK on July 7, 2006 - 7:40am Permalink | Subthread | Parent | Comments top
In Hubbert Linearisation graphs, P is the rate of Production (i.e. X million barrels per day or year) and Q is Cumulative Production (i.e. the total amount of Oil produced to date).

The X axis is Q (not time as I first thought) and the y-axis is P/Q, or in other words, the percentage that the Production rate is of the Cumulative Production.

Take an example:
By the end of the first year of production from your oil well, you are producing at a rate of 1 million barrels a year and have cumulatively produced half a million barrels (you started  with no production at the beginning of the year). Thus, Q=0.5 and P=1, so P/Q=2 (or 200%), and a point at (0.5, 2) is plotted to mark the end of the first year.

By the end of the second year, you are producing at a rate of 1.5 million barrels a year and have produced a cumulative total of (say) 1.7 million barrels.  Thus, the P/Q value would be about 0.88 or 88%, and another point at (1.7, 0.88) is plotted.

This carries on and each point defined as (Q, P/Q) is plotted for each year.

It has been found that if production follows the usual Hubbert curve, then the Hubbert Linearisation graph tends to become linear after a certain period of production, and the linear extrapolation projects down to the x-axis to give the URR.

Of course, if the production curve does not fit a Hubbert curve exactly, then the linearisation also doesn't work very well either (as can be seen from Stuart's graphs above)

I've played around with fictional production curves and their linearisations and I've found that you can make graphs that look like the field has a higher URR. For example if the production keeps ramping up and up and then suddenly collapses, then the linearisation essentially 'drops off'.

I believe a lot of the current production is like that, and even areas like the North Sea will decline faster than the linearisation currently predicts.

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alistairC on July 7, 2006 - 9:46am Permalink | Subthread | Parent | Comments top
Thank you Dunk... I believe I understood that a few weeks ago, or thought I did, but my mind is shutting down for the holidays.

In any case, that's an excellent nutshell.

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alan on July 7, 2006 - 11:12am Permalink | Subthread | Parent | Comments top
Plotting the P/Q vs Q and linearizing part of it was easy once I had the data. How do you create the logistic curve from the URR and the linearization?
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Stuart Staniford on July 7, 2006 - 3:11pm Permalink | Subthread | Parent | Comments top
The logistic equation is Q(t) = URR/(1 + exp(-K(T-Tpeak)).  You can use that to build a series of Q(t)s.  Adjust Tpeak to match the model Q to some actual Q (eg at the end of 2005).  Then get P as the difference between successive Q's.
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IowaBoy on July 7, 2006 - 3:34pm Permalink | Subthread | Parent | Comments top
Solve the differential equation. P=dQ/dt, so dlogQ/dt=a(URR-Q). It has an analytic (that is, closed form) solution in spite of its non-linearity. Change dependent variable to Q=1/S (defn of S).
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WebHubbleTelescope on July 8, 2006 - 7:01pm Permalink | Subthread | Parent | Comments top
IowaBoy,
You seem to understand the calculus. Now, can you give it a try and explain the physical meaning of the equation in the context of oil depletion? So far no one has been able to do this without punting on the question and then relying on the heuristics of the fit as a rationale.
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Don Sailorman on July 8, 2006 - 11:52pm Permalink | Subthread | Parent | Comments top
I am sorry to inform you that you cannot differentiate a discontinuous function.

For all intents and purposes, the consensus here on TOD seems to be that we are at the cusp of an age of discontinuity.

Thus I question the relevance of calculous, which was, after all, first invented by Newton to help describe and understand gravity and the motion of the planets.

Solar system astronomy is an atrocious model for trying to figure out petroleum geology, IMO.

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WebHubbleTelescope on July 9, 2006 - 11:14am Permalink | Subthread | Parent | Comments top
You can describe a certain class of deternministic problems using calculus, such as trajectories, etc.  This enables us to follow the temporal behavior of a single particle acted on by physical processes.

Then we have stochastic calculus and strochastic differential equations which applies basically the same formulations to solve problems in probabalistic terms.  In simple terms this approach applies basic laws to aggregated sets of particles using mean values for rate determining factors. I don't think anyone studying the oil depletion problem seriously is trying to apply deterministic calculus to generate a solution.

And I think that is a huge problem in our greater understanding.  For one, I think we have a much better chance of creating simple formulations for oil depletion (and NG depletion, etc) than we do for other econometrics areas of study, which apply some of the same stochastic principles.  

However, if the proverbial "we" do give up, I will keep plugging away on this problem because it is certainly an interesting hobby and I have a tin-foil-free niche market to toil away in.

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IowaBoy on July 12, 2006 - 6:56pm Permalink | Subthread | Parent | Comments top
Sailorman's comment is a bit puzzling; the logistic curve (the Hubble linearization) is a continuous, differentiable function which may or may not match the measured (estimated?) production and production rates. It's no more arbitrary than satellite orbits, except we do know the law of gravity. (I believe that general relativity corrections to satellite orbits are measurable, so it's not just Newtonian, though.)

Anyway, the Hubble formula says when production begins, it grows exponentially (compound interest): growth is proportional to the amount produced. I could make up a story about why that might be true, but it would just be a story. Of course, this part doesn't fit the very early data, which show oscillations before P/Q settles down.

The model also says that the production rate is proportional to the amount remaining (URR-Q). Makes sense, I guess, when you're nearly out of unproduced oil. That's the whole model: rate= aQ(*URR-Q)

It's a very simple pair of assumptions, and it's surprising to me that it fits so well so many situations. (The Fermi function of describing the energy of electrons in metals at finite temperatures is a form of this function.) I hope this helps.

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IowaBoy on July 13, 2006 - 11:09am Permalink | Subthread | Parent | Comments top
Excuse me! I meant "Hubbert" not "Hubble," of course. Senior moment, I guess.
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