The Verhulst Function is a good function to use for fitting minerals-depletion data. It is used to model the fact that their is usually an exponential growth in the rate of extraction a mineral from the Earth, followed by a peak, after which the extraction rate declines exponentially. The function is

..

is the amount to be eventually extracted, is the rising exponential time constant, n is the falling exponential time constant and t_{1/2} is the time at which the resource is one-half depleted. The parameter n determines the amount of skewing at large times. For n = 1 the extraction curve is symmetrical and the peak occurs at t_{1/2}. The deviation of the peak time from t_{1/2} is negative for n > 1 (skewed toward large times) and is positive for n < 1 (skewed toward small times).

The maximum of P(t) occurs at , which yields - Note that, for the symmetric case (n=1): and .

When a peak is symmetrical, the Verhulst function simplifies to

.

The asymmetry parameter, n, must be greater than 0. For the case of n=0, the Verhulst function becomes the Gompertz function:

.

The amount left to be extracted at time t is

.

The following graph shows the Verhulst function for = 100, t_{1/2} = 1950 and = 5 with 6 different values of n:

The area under all the curves is = 100.

For a symmetric Verhulst function used to fit data up to a time t and given the definitions A = amount already extracted and R = true reserves = amount left to be extracted:

.

For a known A, t and , the peak position, t_{1/2}, varies logaithmically with 1/R, which is much less than linearly. If enough data are present in the extration exponential rise, the rate, , can be determined by fitting those data by the Verhulst function. A can be determined by adding up the extraction for all years up to year t.

The following graph shows the amount-left Verhulst function for = 100, t_{1/2} = 1950 and = 5 with 6 different values of n:

Of course, the amount already extracted at time t is -Q(t).

It is useful to define a "duration" for the extraction of a mineral by the difference in the times when (f-1)/f of it has been extracted and when 1/f of it has been extracted:

.

This is derived from .

For the symmetric case (n = 1) .

A good choice for f is 10; then the duration would be the time interval for extracting the middle 80% of .

=32422 x 10^{6} tons, t_{1/2} = 1913.5, = 47.50 and n = 0.1 .

=2720 x 10^{6} tons, t_{1/2} = 1977.3, = 24.22 and n = 0.1 .

First peak: =13.52 x 10^{9} barrels, t_{1/2} = 1985.9, = 2.645 and n = 2.316 .

Second peak: =17.07 x 10^{6} tons, t_{1/2} = 2004.6, = 1.876 and n = 7.565 .

L. David Roper interdisciplinary studies

L. David Roper, roperld@vt.edu

2-jul-16