### The Most Effective Ways to Improve Your Car's Energy Efficiency

It’s that time of year again: opening weekend at Watkins
Glen, which means last Friday nearly 40 people took to the track to drive
(slowly) around it in the annual Green Grand Prix.

Although, not as slowly as you might think. If you do it right, you won’t touch your brakes; my traction control was cutting in on the back sweeper on the Boot and Turn 1. |

I’ve been making the trip out to New York from Illinois for
a few years now, and I always find myself thinking about the efficiency of my
car not just during the competition but also during the long drive (I removed
the stereo several years ago, so I have a lot of time to think!). This year I
had also read something thought-provoking in Carlo Rovelli’s

*Helgoland*, a book about the discovery of quantum mechanics: “Abandoning assumptions that seem self-evident can lead to greater understanding.”What assumptions have I made about energy efficiency that
might be holding me back? Well, the only way to answer that is to abandon everything
and go back to basics. And going back to basics may shed light on how to
improve the energy efficiency of our cars most effectively.

**Energy**

**One of my professors in my first undergraduate degree program used to say, “The difference between mediocrity and excellence is the constant review of fundamentals.”**

The fundamental concept of energy efficiency in automobiles
is one we are all familiar with but perhaps never think about deeply: the ratio
of distance travelled to energy used. MPG is one metric, which assumes a
constant energy available from a unit volume of gasoline or diesel. Mi/kWh is
another; L/100 km; MPGe. All these show distance per energy or energy per
distance in a metric that can be used to easily compare different cars.

What is energy? Technical definitions are vague, for
example: “Energy is a scalar quantity associated with the state (or condition)
of one or more objects” (Walker et al,

*Fundamentals of Physics 10*[Hoboken: Wiley, 2014], 149). In cars it is often defined as the ability to do work i.e. to move the car. I would define it for our purposes as a quantity associated with a moving car that scales with its various parameters such as speed, mass, and drag. We can refine that definition further using mathematical definitions for the multiple types of energy associated with a moving car from classical physics, which I will do below. Just a warning before we go on: the rest of this post will be math-heavy. I’ll try to explain everything clearly, but it might be more useful to work through these equations yourself.^{th}ed.In my post on fuel efficiency and drag, I mentioned
two basic energies of a moving car,

*kinetic energy*and*work*:…where

*m*= mass,*v*= velocity (really, it’s just speed here since these are scalars and not vectors),*F*= aerodynamic drag force,_{D}*F*= rolling or mechanical drag force, and_{R}*s*= distance. There’s another energy I left out last time,*potential energy*due to gravity:…where

*g*= gravitational acceleration and*h*= height above reference. Assuming a car trip begins and ends at the same place, the total change in potential energy Δ*U*= 0 (remember, since energy is a state function, if conditions at the beginning and end are the same, so is the energy). To keeps thing simpler, I’ll ignore this again—but if you’re doing something like a hill climb event where you*won’t*begin and end at the same elevation, you’ll need to consider it. (Note also that I’m ignoring any potential energy turned into heat by the car’s brakes as it goes down, say, a long hill. That’s energy you won’t get back, and in reality Δ*U*≠ 0 because of it. However, it’s small enough to ignore for this discussion).Within the two resisting forces,

*F*and_{D}*F*, there are more complications. The first is that cars rarely travel on an exactly flat road, and anytime the vehicle is angled with respect to the horizontal, the action of gravity will do some work along its horizontal axis and less along its vertical. In mathematical terms,_{R}…where

*Θ*= the car’s inclination from horizontal, measured as positive angle = rear end up. This affects weight (downward force of the car) as well, along with the second complication: aerodynamic lift.In both of these equations,

*ρ*= air density,*A*= car’s reference area (usually frontal area),*C*= drag coefficient,_{D}*C*= lift coefficient, and_{L}*C*= rolling resistance coefficient._{R}Now, we can put all these together into one equation to
describe the energy of a car as it drives along:

Still with me? Good. To simplify again, let’s assume we’re
driving on a perfectly flat road so that

*Θ*= 0. This gets rid of the trigonometry in the last three equations since sin*Θ*= 0 and cos*Θ*= 1, and our energy equation becomes:Note that we

*can’t*ignore*K*(despite the car starting at rest and ending in the same state) since the car must have kinetic energy if it is moving; while we can simplify and assume a trip with no elevation change to get rid of any change in potential energy, it isn't possible to do this with kinetic energy. You might wonder, if potential energy gained by driving up a hill comes back in the form of work as the car descends the other side, where does the kinetic energy go? It can do work as well, or be turned into heat or another form of energy (via friction braking—or captured as chemical energy through regenerative brakes—or lost to aero and rolling drag). Most commonly it is converted to heat as a car brakes to a stop, but if you're driving for best efficiency you will need to let your car coast as much as possible; coasting lets its kinetic energy do work in moving the car forward, and this will always be more efficient than even capturing the kinetic energy through regenerative braking (and vastly more efficient than using the friction brakes).

**Hypothetical Changes**

There are several variables we have some control over in
this equation, specifically

*m*,*v*,*C*,_{D}*C*,_{L}*C*, and_{R}*A*. Note that some of these are related e.g. changing a car’s drag will affect its lift and vice versa.*A*and*C*aren’t usually changeable by huge amounts; fitting LRR tires, using low-viscosity fluids and lightweight rotating components, and removing side mirrors are the only easy things that come to mind. But_{R}*m*,*v*, and*C*are relatively easy to manipulate. What happens to the total energy required to make the same trip of distance_{D}*s*, but reducing each of these variables by 10% in turn?

**1) Mass,**

*m*It’s easy to prove mathematically that reducing

*m*(or*v*or*C*) will reduce the energy required to move a car. Simply subtract_{D}*E*(energy in the initial state) from_{1}*E*(energy in the final state):_{2}…which means, “the change in energy is proportional to the
difference between Mass 2 and Mass 1.” Thus, if Mass 2 is less than Mass 1, the change in energy will be negative (and vice versa).

You can do the same with

*v*and*C*to see that this relationship holds true for them as well. This makes intuitive sense; reduce speed or drag or mass and the energy needed to move a car from one place to another will go down, or in other words, its energy efficiency will improve. But what I’m interested in goes further than that. Is there a way to predict how_{D}*much*each of these affects energy efficiency so that we can decide how to best modify our cars?To do that, we have to consider

*E*and_{1}*E*in proportion, expressed as a ratio i.e. energy in the second condition as a fraction of the first. This gets thorny because, unfortunately, it doesn’t simplify easily:_{2}…where

*T*= constant. So, we end up with the ratio of energies being equal to the ratio of the masses plus some constant. The presence of the constant will have an “adjusting” effect on the fraction, so that the energy ratio is larger than the mass ratio i.e. a ten percent reduction in mass will result in an energy reduction of less than ten percent.

**2) Velocity,**

*v*I won’t bore you with all that math again (you can work
through these other variables yourself), so suffice it to say that the same
simplification results in the ratio:

As above, the presence of the constant

*U*has a tempering effect; so, a ratio of velocities of 0.81 (accounting for the squares) means that energy will be reduced by something less than 19%.

**3) Drag Coefficient,**

*C*_{D}Finally, the same calculation results in:

Now, the question is: which of these
constants results in the largest change in energy for a change in the
associated parameter? Which results in the smallest?

Looking at a bunch of variables like this isn’t very helpful, so let’s run the calculations for
my car, a 2013 Prius, running at 100 kph for 10 km. Using the stock numbers of

*C*= 0.25,_{D}*A*= 2.20 m^{2},*m*= 1380 kg,*v*= 27.8 m/s, and estimating*C*= 0 and_{L}*C*= 0.012, we get_{R}*T*= 1670,*U*= 400, and*V*= 0.207. Plugging each of these into the equations for energy proportion, the percent energy for a 10% reduction in each of*m*,*v*, and*C*gives:_{D}This means that hands down, the best
method for improving fuel efficiency on my car—especially running at steady
speed on a highway—is to

*slow down*. The same is probably true for yours. The second best? Get rid of unnecessary mass in or on the car or reduce aerodynamic drag; they’re about the same improvement. This is assuming the same percentage reduction in each, which isn’t an outlandish ask; it’s about as hard to reduce drag by 10% as it would be to take 10% mass out of my car. This won’t hold true for every vehicle, though, so run the numbers for your own car. And keep in mind the simplifications we used; in reality, mass might make*more*difference in energy efficiency than these calculations show since our cars actually drive up and down all the time and since we didn't account for acceleration, or aero drag might matter more if your SUV or truck has very high drag to begin, or if you tend to drive faster.Another thing to keep in mind: just because a 10% change in mass or aero drag showed a 4-5% reduction in energy needed to make our hypothetical trip

*doesn't*mean this will result in 4-5% better fuel economy. As I wrote before, that depends on a lot of other factors, and the real change in fuel economy might be more or less than 5% because of them. This exercise just tells us what the relative change in energy needed to make the trip will be while various parameters are changed.These numbers don't really change even if we extend our hypothetical trip to very long distances. At

*s*= 100 km, we get energy proportions of 0.96, 0.88, and 0.95 for the same 10% reduction in*m*,*v*, and*C*respectively. And at_{D}*s*= 1,000 km, the percentages are the same as at 100 km.

**Conclusions**

So, the best things I can do to improve my car’s energy
efficiency appear to be, in order:

1) Drive more slowly.

2) Reduce mass and/or reduce aerodynamic drag.

This is

*not*the order I thought these should go; you’ve probably been indoctrinated, as I was, into believing that aerodynamic drag “matters more” on a highway trip than weight when in reality they can be about the same benefit, or that reducing drag means you can drive at the same (or even higher) speed and get the same fuel economy as at a lower speed when in reality you’d have to reduce drag a*lot*more to get the benefit of even a small reduction in speed. Those ideas rest on assumptions that might not be true. Only by abandoning those assumptions can we see, mathematically, what will make the most difference in the energy consumption of our cars and what we should do if our goal is to improve it.
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