=======================Electronic Edition========================
RACHEL’S HAZARDOUS WASTE NEWS #199
—September 19, 1990—
News and resources for environmental justice.
——
Environmental Research Foundation
P.O. Box 5036, Annapolis, MD 21403
Fax (410) 263-8944; Internet: erf@igc.apc.org
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MORE WAYS TO UNDERSTAND GROWTH.
[In RHWN #197, we introduced some tools that allow anyone to
understand things that grow exponentially, which is to say,
things that grow by a constant percentage of the whole in a
constant time period, such as a bank account that grows at the
rate of 6% each year. Most things that affect the environment are
growing exponentially. Tools for understanding exponential growth
are very useful because they allow us to see why we’re in the
situation we’re in today.]
RULE 5: If you know some original amount, called N_sub_o, and
some amount that it has grown to, N_sub_t, at some later time, t,
you can calculate the value of k (the annual rate of change,
expressed as a decimal fraction) by the following equation:
k = (ln(N_sub_t/N_sub_o))/t [Rule 5]
where:
ln means “natural logarithm of” (your $15 Radio Shack scientific
calculator will give you the natural log of any number instantly);
N_sub_o is some original amount;
N_sub_t is the amount that it has grown to at some later time, t;
k = the annual percentage increase expressed as a decimal
fraction (in other words, it’s the value we’ve been calling p,
divided by 100) [for example, if something is growing at 6% per
year, p = 6 and k = 0.06];
t = time (in any units you care to choose).
In other words, take the amount N_sub_t at time t, divide it by
N_sub_o (the original amount), take the natural log of the
result, then divide that natural log by t. That will give you k;
if you then multiply k by 100, you have p. Then you can use k and
p with Rules 1 through 4 (given in RHWN #197) to learn important
characteristics of the thing that’s growing.
For example, the consumer price index grew from 133.1 in 1973 to
246.8 in 1980. What was the annual percentage rate of increase,
p? To find p, first we calculate k. The annual increase rate
(expressed as a decimal fraction), k, was:
(ln(246.8/133.1))/t
Since t = 1980-1973 = 7 years, and 246.8/133.1 = 1.85 and the
natural log of 1.85 = 0.62, the value of k = 0.62/7, or 0.088;
thus p = 0.088 x 100 or 8.8% per year increase. With this number
in hand, we can use rules 1 through 4 to learn its other growth
characteristics.
RULE 6: If a quantity is growing exponentially and you want to
know how rapidly it is increasing by a factor of x (i.e., how
fast it’s increasing by a factor of, say, 5, or, as we might say,
how long it takes to increase 5-fold), use this equation:
x folding time = ln(x)/k [Rule 6]
Example: If a bank account is growing at 6% per year, how long
will it take for the amount in the bank to increase five-fold?
The natural log of 5 = 1.61; 1.61/0.06 = 26.8. Therefore, a bank
account growing at 6% per year will increase 5-fold in a period
of 26.8 years.
RULE 7: If an exponentially-growing quantity has increased by a
factor of x during some period of time, t, then the fractional
increase, k, per unit time can be found by the equation
k = ln(x)/x folding time [Rule 7]
Example: The annual production of a very useful chemical
increased 6-fold from 1949 to 1958. What was the annual
percentage increase?
My scientific calculator tells me that the natural log of 6 =
1.79. The x folding time = 1958-1949 = 9. Thus k = 1.79/9 = 0.198
or 19.8% per year, an impressive annual rate of increase, which,
by Rule 1, would lead to a doubling of that chemical’s production
every 70/19.8 = 3.5 years, and, by Rule 4, would lead to an
increase by a factor of a million during one human lifetime.
Impressive growth, indeed. (See Table 1 in RHWN #197.)
RULE 8: If you want to know the total amount of an
exponentially-growing quantity produced during time-period t, use
this equation:
C = (N_sub_o/k)*(e**((k*t)-1)) [Rule 8]
where:
** means “raise to the power of”
* means “multiply by”
C = total amount produced during time-period t;
N_sub_o = the original amount;
k = the increase (expressed as a decimal fraction) per unit of
time (for example, per year);
e = 2.718, the base of natural logarithms;
t = time (in any units).
For example, if 40,000 automobiles were produced in 1920 and car
production increased 10% per year for the next 15 years, how many
cars were produced, total, during the 15 years? N_sub_o = 40,000;
k = 0.10; t = 15. Therefore, C = (40,000/0.10) (e**((0.10*15)-1))
= 1,392,676.
In writing the equation above, we’ve used an asterisk to indicate
multiplication; in other words, 2*2 = 4, and we’ve used ** to
indicate exponentiation, meaning “raised to the power of;” i.e.,
2**3 = 8.
RULE 9: During one doubling time, the growth in an
exponentially-growing quantity equals all the growth that has
occurred in all previous time. Thus if production of chemical XYZ
is growing at 10% per year (doubling in 7 years), during the next
7 years we will produce an amount of chemical XYZ equal to all of
the chemical XYZ that has been produced up to today.
These 9 rules can help us all understand why environmental
problems have sneaked up on us, and why the growth of some things
must be curbed.
Let’s take an example. Let’s say we learn from a magazine article
that the synthetic organic chemical industry produced 38 billion
pounds of chemicals in 1945 and since then has grown steadily at
6.5% per year. From that small amount of information, we can
learn a lot. For example, from Rule 3 we can learn that annual
synthetic organic chemical production in 1990 was about 708
billion pounds. By Rule 1 we can learn that total production is
doubling every 10.8 years (we’ll round it off and call it 11
years). By Rule 9, we can learn that during the most recent
doubling-time (from 1979 through 1990), which happens to be the
time since Love Canal was discovered, the chemical industry
produced an amount of chemicals equal to the amount it produced
during all time prior to the discovery of Love Canal. By Rule 8,
we can learn that since 1945, the chemical industry has produced
a total of 10.3 million million (or 1.03 x 10**13) pounds of
chemicals, all of which has gone somewhere, and Rule 9 tells us
that during the next 11 years they’ll produce another 10.3
million million pounds. From Rule 4, we can learn that if the
chemical industry keeps growing at 6.5% per year, during one
human lifetime it will increase its size 90-fold; in other words,
if the chemical industry keeps growing at its current rate, for
every chemical factory in existence today, there will be 90
chemical factories just 70 years from now.
These facts help us understand why more and more people are
complaining about chemicals affecting their lives: it isn’t
because of growing “chemophobia,” it is because people are being
exposed to more and more chemicals that are being released into
the air and water that we all breathe and drink. A massive
change–the “chemicalization of the environment”–has occurred
during the past 50 years, and the problem is doubling in size
every 11 years.
These facts can lead us to one more conclusion: since the
chemical industry in all its trade publications constantly
stresses the need for continuous growth at rates that match or
exceed the historical rate of growth, we can conclude that
executives in the chemical industry do not understand the
implications of exponential growth. No one who has seen the
heavily industrialized sections of chemical-producing states like
New Jersey, Ohio, Louisiana, Texas and California can believe
that we could survive 90 chemical factories where each chemical
factory stands today. Growth of this industry has got to be
curbed. By what means? By whatever non-violent means are
necessary. Whatever it takes.
Further reading: Ralph Lapp, THE LOGARITHMIC CENTURY (Englewood
Cliffs, NJ: Prentice-Hall, 1973). And: chapters 1 and 2 of
Donella H. Meadows and others, THE LIMITS TO GROWTH (NY: Universe
Books, 1972). And: Albert A. Bartlett, “The Exponential
Function,” THE PHYSICS TEACHER (October, 1976), pgs. 393-401.
–Peter Montague, Ph.D.
Descriptor terms: mathematics; chemical industry;