Fermi Questions
Larry Weinstein, Column Editor
Old Dominion University, Norfolk, VA 23529;
[email protected].
Solutions for Fermi Questions, September 2015
w Question 1: Shaving time
How much total time does a typical American spend
shaving during his or her life? How much total time
per year do Americans spend shaving?
Answer: To answer this question we need to estimate the
average time spent shaving each day and then multiply
by the appropriate factors. The typical American male
spends more than one and fewer than 10 minutes shaving his face daily so we’ll take the geometric mean and
estimate three minutes per day. The typical American
female spends a similar amount of time shaving her
armpits and legs. The fraction of men who do not shave
is obvious and small. The fraction of women who do not
shave is less obvious and will also be neglected.
The average American lifespan is 80 years, 60 of which
we spend shaving (not full time). The total time devoted
by each us to shaving is then:
Tshave = (3 min/day)(43102 day/yr)(60 yr)
= 63104 min = 103 hr.
One thousand hours per lifetime. Think of what we
could each accomplish with an extra thousand hours!
Now let’s consider the total yearly time spent shaving. If
the average person spends three minutes per day shaving, then the total time Americans spend shaving each
year is:
T total = (3 min/day)(43102 day/yr)(33108)
shave
=
331011
=
53105
min/yr =
53109
hr/yr
yr/yr
= 53103 lifetimes/yr.
Americans spend an aggregate total of five thousand lifetimes every year just shaving. This seems like a stupendous waste.
Actually, this just demonstrates that almost any number
multiplied by 33108 will give a large number. When we
determine an estimate, we should compare it to a relevant quantity. We should compare estimates for individuals to individual quantities and estimates for countries
to national quantities.
Let’s compare the shaving time to the total time available.
For a person, the time available is one lifetime or
Tavail = (104 hr/yr)(80 yr)
< 106 hr,
THE PHYSICS TEACHER ◆ Vol. 52, 2014
and so the 103 hr spent shaving is only one part in a
thousand.
For Americans, the total time available in one year is
T total = (1 yr)(33108)
shave
= 33108 yr.
and so the 53105 yr spent shaving each year is still
only about one part in a thousand. Spending 10–3 of
our available time on one aspect of personal grooming
seems quite reasonable.
In other words, while small numbers can add up over a
lifetime or over 33108 people, they’re still small numbers.
Copyright 2015, Lawrence Weinstein.
w Question 2: Half filling the gas tank
How much money can you save by only half-filling
your car’s gas tank to reduce weight?
Answer: Our cars lose mechanical energy through air
resistance, rolling resistance and braking (except for
hybrid cars which recover much of the kinetic energy
otherwise discarded during braking). If we only fill the
gas tank half full, then we reduce the average mass of
the car. This does not affect air resistance but reduces
rolling resistance and the mechanical energy discarded
during braking.
Typical cars mass between one and two tons. Larger cars
will have larger gas tanks so let’s drive a two-ton SUV
with a 20 gallon (80 liter) gas tank. If we fill the tank
when it is almost empty, then on average the tank will
be half-full and contain 40 l of gasoline. If we fill the
tank only half-full, then on average the tank will contain
only 20 l of gasoline. Filling the tank only half-full will
decrease the average mass of the car by 20 kg (using the
density of water r = 103 kg/m3).
Braking discards mechanical energy equal to the kinetic
energy of the car. In a typical 20-mile trip (e.g., my
morning commute) we will accelerate and brake many
times. Driving on the highway, we will accelerate only
once (ignoring traffic jams) to about 30 m/s. Driving in
the city, we will accelerate about 10 times (more than 1
and less than 100) to about 15 m/s.
By only filling our gas tank half-full, we could reduce
the mechanical energy discarded in braking during this
stop the car. If the car is travelling at a fast walking
pace (2 m/s or 4 mph), then it will roll about 30 m
(more than 10 and less than 100 m). This gives
trip by
.
At a gasoline energy density of 33107 J/l (see
Guesstimation for details), this corresponds to the
chemical energy in about 10–3 l of gasoline. Accounting
for engine efficiency, we would save 4310–3 l (10–3 gal)
of gasoline. Compared to the one gallon of gasoline
typically consumed to drive 20 miles, this is negligible.
Like static and kinetic friction, the force of rolling
resistance equals the coefficient of friction times the
normal force (F = mr mg). We can estimate the coefficient of friction by giving the car some initial velocity v
and observing the distance d that it rolls (assuming the
engine is in neutral). Then W = DKE or
Having pushed a disabled car, I know it is very important to have someone in the car to apply the brakes and
214
THE PHYSICS TEACHER ◆ Vol. 53, 2015
Rolling resistance consumes energy
E = mr mgd
so for our 20-mile (30 km) trip, we will need an extra
DE = 10–2 (20 kg)(10 m/s2)(33104 m) = 63104 J
to overcome rolling resistance. This is twice the
mechanical energy discarded by braking, but still
quite small.
Thus, the total gasoline saved in a 20-mile trip by
only half-filling the gas tank is 3310–3 gal out of the
one gallon otherwise needed. This is not surprising.
By only half-filling the gas tank, we reduce the mass
of the car by an average of 1%. This reduces the rolling resistance and energy needed to accelerate the car
by 1%, but leaves the air resistance unchanged. Thus,
it must decrease gas consumption by less than 1%.
Copyright 2015, Lawrence Weinstein.