WEBVTT
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What's the fuel consumption of a car? Say,
on the highway moving at highway speeds? We
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can look it up, but can we understand those
numbers? Can we predict them from our knowledge
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of science and engineering? That's what we're
gonna do in this video.
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This video is part of the problem solving
video series.
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Problem-solving skills, in combination with
an understanding of the natural and human-made
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world, are critical to the design and optimization
of systems and processes.
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Hi, my name is Sanjoy Mahajan, and I'm a professor
of Applied Science and Engineering at Olin
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College. And I excel at streetfighting mathematics.
Before watching this video, you should be
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familiar with free body diagrams and dimensional
analysis.
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After watching this video, you will be able
to: Model drag to predict terminal velocities;
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and Determine the fuel efficiency of a car.
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Fuel is consumed in fighting drag, on the
highway at least. What's the force of drag
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on a car? That's fluid mechanics. We could
try the hard wayâ€”solving the Navier-Stokes,
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the equations of fluid mechanics.
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Now after 10 years of learning mathematics,
you'll discover this is a really hard problem.
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You can only solve it analytically in certain
special cases like a sphere moving at really
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slow speeds and a car is not one of those
special cases; certainly not a car moving
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at any reasonable speed. So we need another
way.
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By applying approximations artfully, we're
gonna find the drag force and the fuel consumption
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of a car with a simple experiment and some
scientific and engineering reasoning.
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First, we need to model drag. To do this,
we're gonna figure out how the drag force
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depends on the quantities that control it.
Drag force F depends on the density of the
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air, to some power which we don't know yet,
times the speed of the car moving through
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the air, to some power we don't know yet.
It depends on how big the car is, which we'll
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represent by the area; traditionally, the
cross- sectional area, to some power. And
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it also can depend on the viscosity here,
the kinematic viscosity of the air to some
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power. And what those powers are, we don't
yet know. To figure them out, let's do an
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experiment.
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For the experiment, we make two cones. First,
by making a large circle and cutting out a
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quarter of it. The large circle has a radius
of seven centimeters. And the small three-quarter
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circle has half the radius. Cut them out and
then tape this edge to that edge and this
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edge to that edge in order to make the 2 cones.
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Next we're gonna race them, the big cone vs.
the small cone. But first, make a prediction.
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The question is: what is the ratio of fall
speeds of the big cone and the small cone
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approximately? Is the big cone roughly twice
as fast, is the small cone twice as fast,
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or do they fall at roughly the same speed?
Pause the video and make your best guess.
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You may want to try the experiment yourself.
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The experimental result is that they both
fall at roughly the same speed. Now what does
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that mean for the exponents in the drag force?
To decide, use a free body diagram. Here is
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a free body diagram of a cone as it falls.
What are the forces on the cone? Pause and
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fill in the diagram.
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There is drag and gravity. Because the cones
are falling at their terminal speed, in other
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words at the constant velocity that they reach
pretty quickly, the drag must equal the force
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due to gravity.
So the drag force can easily be measured just
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by sticking the cones on a scale. We don't
actually have to stick the cones on a scale
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because we know what their relative weights
are and that's what we're gonna use to figure
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out one of the exponents.
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Knowing that the drag equals the force due
to gravity, let's compare the drag forces,
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which are the factors on the right side, of
the small and large cones. The large cone
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was made out of a paper circle with twice
the radius so it has four times the area,
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four times the mass, and four times the force
due to gravity. Therefore, four times the
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drag force.
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Now let's look at the causes one by one. This
density here is the density of the air and
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both cones feel the same air so there's no
difference here, so that's times one no matter
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the exponent. Now, what about the speed? The
speeds were the same; that was the result
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of our experiment, so no matter what the exponent
here, the speed contributes to the effect
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by a factor of one; in other words, nothing.
Now the area. This is traditionally the cross-sectional
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area. The big cone has four times the cross-sectional
area of the small cone so this area factor
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is times four and raised to the unknown exponent.
And here, the kinematic viscosity, that's
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the kinematic viscosity of air and both cones
feel the same kinematic viscosity. Therefore,
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that contributes a factor of one just like
the density does.
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So now we know what the question mark here
has to be. The question mark here must be
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one so that this four to the question mark
is equal to this four. And so does this one.
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Now that we have found one exponent, we can
find the remaining three using dimensional
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analysis.
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We make sure that the exponents chosen on
the right side produce dimensions of force,
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which are the dimensions on the left side.
A force has dimensions of mass, length per
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time squared, that's mass times acceleration.
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Density has dimensions of mass per length
cubed, raised to this unknown exponent. Velocity
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has dimensions of length per time, raised
to this unknown exponent. Area has dimensions
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of length squared, raised to the power 1,
we've already figured that out. And kinematic
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viscosity; now, that's the regular viscosity
divided by the density. It has dimensions
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of length squared per time raised to this
unknown exponent
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Now let's solve for these unknown exponents
one at a time. Looking first at the mass,
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there is mass here and mass here on the left
side, but there is no other mass. So the only
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way to get the mass to work out is to make
this exponent one.
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Now, what do we have left? We have to make
sure the lengths and times work out and we
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have two unknowns, this exponent and this
exponent.
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Let's see how many more lengths we need, given
what we already have. We have mass worked
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out, we have two lengths here over three lengths
so we have mass per length. So we have length
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to the minus one so far.
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And what about time? Well, we have no times
yet except in these question marks so we need
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to get two more times on the bottom two more
lengths on the top. So we need to multiply
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by length squared over time squared, and the
only way to do that is to make this exponent
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a two... and this exponent here a zero.
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You can pause the video and go ahead and try
other combinations of exponents. You'll find
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that none of them will get both length and
time correct.
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So now we have all the exponents, we can actually
enter that into our force formula. Drag force
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is proportional to density to the first power,
speed squared, area to the first power we
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got from our home experiment and viscosity
here to the zeroth power.
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Now let's test our formula with a second demonstration
and see if it predicts this new situation.
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We're gonna race four small cones stacked
on top of each other. One, two, three, four.
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So we stack them to make one small cone that
is four times the mass of the other small
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cone. Next we're going to race them.
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But first, what's the drag force ratio between
the small cone and 4 small cones stacked together.
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Pause the video and use the formula to make
a prediction.
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The four cones weigh four times as much as
one cone so that's times four on the left
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side. What about the right side? Again, both
contestants feel the same air density. The
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velocity ratio, that's what we're trying to
figure out, and let's see what all the other
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pieces are. The area; well here in this case,
the stack of four has the same cross- sectional
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area as the one cone so that's just one. And
the viscosity doesn't matter. So, this factor
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of four on the left side has to be produced
by the v squared; in other words, the v goes
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up by a factor of two so that when it's squared,
you get a factor of four. That means the correct
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answer should be two to one.
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Now, let's test our prediction with a demonstration
to check whether nature behaves as we predict.
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It looks like our prediction was correct!
Now we feel pretty confident that we can use
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this formula to try to figure out the fuel
consumption of a car.
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Let's clean up our formula here just a bit
and erase the viscosity since it comes in
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with a zero exponent so it doesn't matter.
And this is our drag formula, which we're
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gonna use to estimate the fuel consumption
of a car on the highway.
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Fuel efficiency is measured as the distance
that that the car travels on one gallon or
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liter of gasoline.
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So it depends on how much energy you can get
out of a gallon of gasoline divided by the
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drag force, which is the formula we've just
found.
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So let's put in some numbers. What's the energy
from a gallon of gasoline? A gallon is roughly
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4 liters. And gasoline is roughly like water
so, there are 1,000g/liter.
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How much energy do you get out of each gram
of gasoline? Gasoline is like fat and from
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the side of a butter packet, you can read
that 100 kilocalories, 10 to the 2 kilocalories
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come from 11 grams of butter.
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But now we want to convert kilocalories to
some reasonable metric unit. So that's four
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joules/calorie or four kilojoules per kilocalorie.
And that all divided by the drag force so
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let's put in those terms one by one.
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First the density here, one kilogram per meter
cubed is the density of air roughly. Now v
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for highway speeds. Say 100 kilometers an
hour, 65 miles an hour. That's roughly 30
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meters per second. Don't forget to square
it. And then the area of a car. Well, it's
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about... two meters wide by 1.5 meters high.
Now we have to work all the numbers, but before
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you do that, pause the video to check that
the dimensions all cancel appropriately.
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You should have found that the result has
units of meters, which is good! But how many?
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A quick numerical approximation will tell
you that you get 50 kilometers, which is 30
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miles roughly. The fuel efficiency of a car
on the highway is about 30 miles per gallon;
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or 50 kilometers per gallon. In typical European
Union units, that's eight liters per hundred
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kilometers.To Review With our approximation
methods, we've come to pretty reasonable numbers
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for the fuel efficiency of a car, without
having to solve any of the complicated equations
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of fluid mechanics. Instead, we used one simple
home experiment, dropping the cones, and the
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powerful principle of dimensional analysis.
From that, we were able to find all the exponents
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here, in the relationship between drag and
density, speed, and cross-sectional area.
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So the moral of this is that there is an art
to approximations, and reasoning tools such
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as dimensional analysis, they give you an
understanding of how systems behave without
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having to solve every single last bit of the
detailed mathematics. This understanding allows
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us to redesign the world.