Orders of Magnitude and Estimation
At 05{:}29 on the morning of 16 July 1945, in the New Mexico desert, the
first atomic bomb went off. Watching from base camp about 16\ \text{km}
away was the physicist Enrico Fermi. As the blast
wave rolled past, he let a handful of small paper scraps flutter from his hand and watched how far the
gust blew them: roughly 2.5\ \text{m}. From that single crude number —
scribbled on the back of an envelope, before any instrument had been read — he announced the bomb's
energy: about ten kilotons of TNT. The instruments, analysed over the following
weeks, put the true yield near 20 kilotons. Fermi, with a fistful of
confetti, had landed within a factor of two of the right answer.
That is the power we are after on this page. You will almost never be asked, in real physics, "what is
the exact answer?" before you first ask "roughly how big is it — a hundred, a million, a
billion?" Knowing the answer is 10^{9} and not
10^{3} or 10^{15} is often the whole battle: it
tells you whether an idea is even possible, whether your calculator slipped a decimal, and where to
point your effort. This skill — thinking in orders of magnitude and making
Fermi estimates — is the physicist's most portable tool. It travels from the lab to
the exam hall to the dinner-table argument, and it needs nothing but the back of an envelope.
What "order of magnitude" means
The order of magnitude of a positive quantity is the nearest power of ten to it. Write
the quantity in scientific notation, a \times 10^{n} with the mantissa
1 \le a < 10, and the order of magnitude is essentially that exponent
n. The one subtlety is rounding: because we are rounding on a
logarithmic scale, the dividing line between "rounds down to 10^{n}"
and "rounds up to 10^{n+1}" is not 5 but
\sqrt{10} \approx 3.16 — the geometric middle of a decade.
-
For x = a \times 10^{n} with 1 \le a < 10,
the order of magnitude is n if a < \sqrt{10}
and n+1 if a \ge \sqrt{10}. Equivalently it
is \operatorname{round}\big(\log_{10} x\big).
-
To multiply, add the exponents. An estimate built from factors of orders
10^{p} and 10^{q} has order
10^{p+q}; dividing subtracts, 10^{p-q}. All
the arithmetic of estimation is just adding and subtracting small whole numbers.
-
Two things are "the same order of magnitude" when their ratio is less than about
ten — i.e. they sit in the same rung of the ladder of tens.
So 2.9 \times 10^{4} is order 10^{4} (since
2.9 < 3.16), while 4.1 \times 10^{4} is order
10^{5}. The mass of a person (\sim 70\ \text{kg})
is order 10^{2}\ \text{kg}; the mass of the Earth
(6 \times 10^{24}\ \text{kg}) is order 10^{25}\ \text{kg}.
They differ by 23 orders of magnitude — twenty-three rungs apart.
A picture: the ladder of tens
A logarithmic scale is the natural home of order-of-magnitude thinking. On an ordinary ruler equal
steps add the same amount; on a log scale equal steps multiply by the same
factor. Marking off powers of ten, each rung is ten times the one before, so the almost unimaginable
span from the proton to the cosmos fits comfortably on one line. Notice how objects we think of as
wildly different — a virus and a galaxy — are just a fixed number of evenly-spaced rungs apart.
This is why a scientist can say "atoms are about an ångström
(10^{-10}\ \text{m})" and "the Sun is about a billion metres across
(10^{9}\ \text{m})" and immediately know they are 19
rungs — nineteen powers of ten — apart, without a single digit of precision. The log scale turns
multiplication into geometry.
Fermi estimation: guess big things by breaking them small
A Fermi problem asks for a quantity that looks hopeless to know — how many piano
tuners work in Chicago, how many atoms are in a sand grain, how many breaths you will take in your
life. The trick, which Fermi taught his students relentlessly, is never to guess the big number
directly. Instead you decompose it into a chain of smaller quantities, each of which
you can guess to within a factor of a few, and then multiply. The magic is that the individual
errors, being partly independent, tend to cancel rather than pile up: your
over-guesses and under-guesses partially undo one another, and the product lands remarkably close —
usually within an order of magnitude of the truth.
The recipe is always the same:
- Decompose the target into a product of factors you can estimate.
- Estimate each factor to the nearest power of ten (round mantissas hard — the
precision does not matter).
- Multiply — which means adding the exponents.
- Sanity-check the result: is a number of that size even sensible?
Worked example 1 — how many piano tuners in Chicago?
This is the classic Fermi problem, the one he set at Chicago. We want the number of people
who earn a living tuning pianos in the city. Decompose:
-
People in Chicago. A big city: call it
3 \times 10^{6} people, order 10^{6}.
-
Households. At roughly 3 people each, that is about
10^{6} households.
-
Pianos. Say 1 household in 20
owns a piano (plus schools, churches, bars — round it in): about
5 \times 10^{4} pianos.
-
Tunings per year. A piano is tuned about once a year, so
\sim 5 \times 10^{4} tunings are needed each year.
-
Tunings one tuner can do. Say 4 pianos a day,
5 days a week, 50 weeks a year:
4 \times 5 \times 50 = 1000 \approx 10^{3} tunings per tuner per year.
Now divide the work needed by the work one person supplies:
\text{tuners} \approx \frac{5 \times 10^{4}\ \text{tunings/yr}}{10^{3}\ \text{tunings/tuner/yr}}
= 5 \times 10^{1} \approx 50.
So an order of magnitude of about 10^{2} — somewhere around fifty to a
hundred piano tuners. Directory counts for mid-century Chicago land right in that band. We reached it
knowing not one exact figure, only a chain of confident guesses whose exponents we added up.
Worked example 2 — how many atoms in a grain of sand?
A physics-flavoured one. Take a grain of sand of diameter about half a millimetre,
d \approx 5 \times 10^{-4}\ \text{m}. Sand is silica,
\mathrm{SiO_2}. Decompose towards the atom count.
Volume. Treating the grain as a little cube (a sphere is the same order):
V \approx d^{3} \approx (5 \times 10^{-4})^{3} \approx 1 \times 10^{-10}\ \text{m}^{3}.
Mass. Silica has density about \rho \approx 2.6 \times 10^{3}\ \text{kg/m}^3:
m = \rho V \approx (2.6 \times 10^{3})(1 \times 10^{-10}) \approx 3 \times 10^{-7}\ \text{kg} = 3 \times 10^{-4}\ \text{g}.
Formula units. One mole of \mathrm{SiO_2} weighs about
60\ \text{g}, so the grain holds about
\tfrac{3 \times 10^{-4}}{60} \approx 5 \times 10^{-6} mol. With
Avogadro's number
N_A \approx 6 \times 10^{23} per mole, that is about
3 \times 10^{18} molecules. Each \mathrm{SiO_2}
carries 3 atoms, so
N_{\text{atoms}} \approx 3 \times (3 \times 10^{18}) \approx 1 \times 10^{19}\ \text{atoms}.
About 10^{19} atoms in a single grain of sand. Ten billion
billion — more atoms in that speck than there are grains of sand on all the world's beaches. And notice
we never needed a real grain, a real balance, or a real number: just a chain of order-of-magnitude
guesses.
Two quick ones — and why the exponents are what matter
Breaths in a lifetime. About 15 breaths a minute,
60 minutes an hour, 24 hours,
365 days, 80 years:
15 \times 60 \times 24 \times 365 \times 80 \approx 6 \times 10^{8} \;\Rightarrow\; \text{order } 10^{9}\ \text{breaths}.
Energy to boil a cup of tea. A cup is about
0.25\ \text{kg} of water, warmed by roughly
80\ \text{K}, with specific heat
c \approx 4200\ \text{J/kg/K}:
E = mc\,\Delta T \approx 0.25 \times 4200 \times 80 \approx 8 \times 10^{4} \;\Rightarrow\; \text{order } 10^{5}\ \text{J}.
In every case the answer we actually keep is the exponent. The leading digit is throwaway; the
power of ten is the physics.
Why does this work? Errors cancel, they don't compound
It feels like it shouldn't work. If every one of five factors could be off by a factor of
three, surely the product could be off by 3^{5} \approx 240 — more than two
orders of magnitude? That is the worst case, and it only happens if every single guess
errs in the same direction. In practice your guesses are roughly independent: some are too high, some
too low, and on a logarithmic scale those errors partially cancel. The statistics of
adding independent errors says the typical spread grows not like N but like
\sqrt{N} — far slower.
The chart makes this concrete. The slider sets how badly each individual factor might be wrong (a
factor f, i.e. \log_{10} f decades of error per
factor). The upper curve is the paranoid worst case, N \log_{10} f decades;
the lower curve is the realistic typical error, \sqrt{N}\,\log_{10} f. Even
with a dozen shaky factors, the typical result stays within about one power of ten of the truth.
This is the deep reason Fermi estimation is trustworthy and not mere hand-waving: on a log scale,
random errors behave like a random walk, and a random walk of N steps only
wanders about \sqrt{N} steps from home.
The paper trick is pure order-of-magnitude physics. The blast's pressure pulse gives the falling scraps
a sideways kick; from how far they blow (\sim 2.5\ \text{m}) Fermi could
estimate the overpressure of the shock wave. Knowing his distance from ground zero
(\sim 16\ \text{km}) and how a blast wave's strength falls off with distance,
he could run the chain backwards to the energy released at the centre — arriving at
\sim 10 kilotons against a true \sim 20. A factor
of two. For a first-of-its-kind measurement made with litter, in seconds, without instruments, that is
astonishing — and it is exactly the accuracy order-of-magnitude reasoning promises: right rung of the
ladder, maybe one step either way.
No — and reaching for either is usually the mistake. The commonest misconception about
estimation is that a "better" estimate means more decimal places, so students freeze, hunt for the
exact population of Chicago, and never start. But an order-of-magnitude estimate is deliberately
imprecise: rounding every factor to the nearest power of ten is not sloppiness, it is the method. Punch
3.17 \times 10^{6} into a calculator instead of writing
10^{6} and you have wasted effort on digits that will be swamped by the
factor-of-a-few uncertainty in your next guess anyway.
Two related traps. First, do not confuse "order of magnitude" with "exact": an answer of
10^{2} tuners does not claim there are exactly a hundred — it claims the
truth is nearer 100 than 10 or
1000. Second, do not fear that many uncertain factors compound into garbage:
as we saw, independent errors cancel, growing like \sqrt{N}, not
like N. The whole point of a Fermi estimate is to be approximately
right rather than precisely lost.
Sanity-checking: the everyday superpower
The same reflex that makes estimates also checks them. Before you trust any
result — your own, a textbook's, a headline's — ask what order of magnitude it should be. A
quick Fermi estimate is a smoke detector for slipped decimal points and wrong units. If you compute a
car's kinetic energy as 10^{40}\ \text{J}, you do not need to recheck the
algebra line by line: 10^{40}\ \text{J} is more than the Sun radiates in a
century, so something is off by dozens of powers of ten. Order-of-magnitude thinking is how a
physicist catches nonsense in one glance.
Keep a few anchor values in your head — the mass of a person
(10^{2}\ \text{kg}), the size of an atom
(10^{-10}\ \text{m}), a year in seconds
(3 \times 10^{7}\ \text{s}), Avogadro's number
(10^{24}) — and you can bracket almost any physical quantity to the right
rung of the ladder in your head.