The consequences of turning

In my post on Highway m.p.g. I recommended always travelling in such a way that your destination is at a lower elevation than your starting point. And in my post on headwinds I believe I showed conclusively that one should always travel with, rather than against, the wind. But to these, I'd like to add another recommendation: always travel in such a way so as to have no need to turn.

Turning a vehicle uses extra fuel in an obvious way, i.e., it increases the distance traveled. But it uses extra fuel in another way as well. As noted by Sir Isaac Newton, changing the momentum of an object requires a force - this is his famous "second law of motion," commonly summarized as "F=m*a." But momentum is a vector quantity (it has both magnitude and direction) so changing direction at a constant speed means momentum is changing. This requires a force which the road applies to the vehicle through the tires (in reaction to the tires applying this force to the road - Newton's third law of motion). This force must ultimately originate in the prime mover of the vehicle and, for internal combustion driven vehicles, must come from the burning of fossil fuels.

So how much fuel are we talking about here? I've calculated the forces involved in traveling in a straight and level line at 55 m.p.h. before, in my vehicle this takes a force of about 805 newtons. If I look at a circle with a circumference of 3.667 miles (this was chosen only to make it a four minute circle at 55 m.p.h.) and hence a radius of 939.2 meters (slightly less than a kilometer) then, utilizing the equation for centripetal force (F=m*(v^2/r)) I can determine that the force required to turn the vehicle is 1,777 newtons. This force must be developed in addition to the 805 newtons required to overcome aerodynamic drag and rolling resistance. Thus, driving in this particular circle more than triples the amount of force required at 55 m.p.h. and hence more than  triples the fuel used to travel any given distance in this manner.

Now, obviously, unless you have a number and a bunch of corporate logos on your car, you won't be driving continuously in a circle but this does illustrate the effect of having to apply the force required to make the car turn. Thus Rob's rules of fuel efficient driving are as follows:

1. Travel downhill only
2. Travel downwind only
3. Do not turn

Follow these three simple rules and I promise you dramatically improved gas mileage!

Update: DON'T LET THIS HAPPEN TO YOU! In my rush to get a post done in April, I made the simplest of freshman physics errors. Thanks to Ed Davies (see comments) for causing me to give it enough thought to see the error.  There's no question about the centripetal force, however, no energy is added to the car. This makes sense - its mass doesn't change (reduces slightly, actually, as fuel is burned); it's speed doesn't change; and it doesn't change its position in a field. So both its kinetic and potential energy are constant. What gives? Well, energy is added by work being done on an object. It's true that the road exerts a force on my vehicle but that force is perpendicular to the instantaneous displacement of the vehicle at all times. The centripetal (road on car) force is directed to the center of the circular path, the displacement is tangential. Work is the product of force times displacement IN THE DIRECTION OF THE FORCE (defined as the vector dot product of the force and the displacement). Since there is never any displacement in the direction of the force, no work is done on the vehicle and thus no energy is added and the engine needn't do any more work.

As to my comment about the airplane, it stays at a constant altitude and slows down, thus its energy is reduced. So, does that mean that it's doing work on its environment? It does, reflected in the motion of the air displaced.

The rules will still work - a straight path is the shortest so you'll use less fuel by always travelling in a straight line. But your miles per gallon won't go up. And I could quibble by saying that taking turns deforms the tires and point out the hysteresis losses in going into and out of turns. But really, I just goofed.

How does one blush online?

Not to pooh pooh NASA, MIT, et al, but...

In my blog list is a link to a site entitled "Altternative Energy. It's an interesting site and frequently has items well worth reading and other times more silly or trivial. But today I saw an article about an aircraft called the "Puffin. This is a concept aircraft designed by one Mark Moore, a NASA aerospace engineer. The article quotes briefly from and links to the NASA site about the Puffin.

It's a single person vehicle ostensibly capable of vertical take offs and landings and a top speed of 150 m.p.h. (thought it's stated that the Puffin is more efficient at lower speeds). It's 12 feet long with a wingspan of 14.5 feet. It's stated to weigh 300 pounds empty, with 100 pounds for batteries and 200 pounds of payload (pilot and baggage). It's electrically powered and its motors develop 60 horsepower. Its range "with current battery technology" is stated to be about 50 miles.

Well. First, it doesn't exist, even as a prototype. But Analytical Mechanics Associates has produced an animation that can be seen in a YouTube video. According to the video, a one third scale validation model was to have been tested in March for hover capability, with transition to forward flight demonstrated after that. I have been unable to determine if these test flights have actually taken place.

But what about the plausibility of such a craft? Let's start with battery capacity. I'm going to assume the Puffin will achieve its 50 mile range at 100 m.p.h. I'll also assume that the 60 horsepower are required to achieve the top speed of 150 m.p.h. Since power required varies generally with the cube of speed for forces, such as aerodynamic drag, that vary with the square of speed, I can roughly estimate that the Puffin will require 60/1.5^3 (1.5 is the ratio of 150 m.p.h. to 100 m.p.h.) or about 17.8 horsepower. I'll be generous to the claims and assume zero reserves of energy. Thus, the batteries must supply 17.8 horsepower for 30 minutes to go 50 miles at 100 m.p.h. Googling (17.8 horsepwer)*30 minutes in kilowatt hours returns the conversion to the 6.64 kilowatt hours that are required.

Moving on to this excellent battery site I find that Lithium-ion batteries have the highest energy density available currently at 128 watt-hours/kilogram. Thus, I'll need 6,640/128 or 51.9 kilograms of batteries. This mass weighs 114 pounds here on Earth. This 6,640 watt-hour battery pack will cost an estimated $28,000.

For fun, I ran through the same calculation at 50 m.p.h. (the stalling speed in forward flight is not indicated, nor are such aerodynamic characteristics as flat plate area, propeller efficiency, lift/drag ratios at various speeds, etc.) and determined that, with no energy to spare the batteries must store 1,660 watt-hours of energy weighing 28.6 pounds and costing $7,090. But this battery pack could only maintain 150 m.p.h. for a little over two minutes.

OK, let's finally figure a 100 pound Li-ion battery pack. This 45.4 kilogram pack should provide about 5,810 watt hours. That capacity will provide 60 horsepower for just slightly under eight minutes. And remember, this is to "dry tanks" and includes no increment for such things as vertical takeoff and transition to level flight (very energy intensive phases for aerial vehicles with which I'm familiar). Finally, let's realistically assume that it would be nice to have a 10% reserve. How fast can you fly to have a 50 mile range with the 100 pound battery pack and have such a reserve? It's kind of a nitpicky problem in algebra and units, but the answer is about 89 m.p.h.

So my conclusion is that, while the claims may not actually be false, they seem quite misleading. You may be able to go 50 miles (and glide, powerless, to a landing); you may be able to go 150 m.p.h.; you may be able to take off vertically with a payload of 200 pounds; but you won't be able to load up 200 pounds, take off vertically, and fly 50 miles at 150 m.p.h. to your destination.

PA32R-301T

That's the model number of my Piper Saratoga, a 1981 model. Now, I know what you're thinking (at least if you haven't followed my blog for long and maybe even if you have). What's a guy who writes about energy savings, fuel efficiency, and peak oil doing in a private airplane?

The answer is that I've wanted to fly since being a small child and  thus I acquired my Airman Certificate ("pilot's license") in 1981. In 2001, well before I began to concern myself with energy related matters, I acquired the airplane. I've been all over the country in it, and have flown it for around 500 hours.

As transportation security measures have increased, the attractiveness of driving to the hangar, performing the preflight inspection, taxiing out and flying to the airport nearest my destination (as opposed to an air carrier airport) becomes more and more attractive. This increases the range I'm willing to fly myself. There are drawbacks as well: I don't have the weather capabilities of the airlines; outside of about 650 (nautical or 748 statute) miles I'll need to stop for fuel; adverse winds have a huge effect on ground speed; and it's quite costly. For very long distances, the airlines are significantly faster though with the hub and spoke system, one hour early arrivals, waiting for baggage, and other delays, the distance at which airlines become faster door to door is longer than one might think. Also, for business meetings, having my transportation ready when I am means that I don't have to leave early to catch my flight.

Parsing the model number, PA stands for Piper Aircraft, it's the 32 series (no significance to that), R indicates that its landing gear is retractable (chicks don't dig fixed gear), 301 is the maximum continuous horsepower of the Lycoming TIO540-S1AD engine (actually, it's one more for some obscure reason), and T indicates turbocharging. As to the engine, T is turbocharged, I is injected, O is (horizontally) opposed (cylinders), 540 cubic inches is the displacement. It's a "dash S1AD" variant, the nomenclature has no obvious significance.

So I have a turbocharged aircraft capable of developing about 300 horsepower. This power is only used on takeoff and initial climb, typical cruise power is about 70% of this number or about 210 horsepower. Just for fun, if I assume that the engine efficiency at this power setting is 25% and that there are about 125*10^6 joules in a gallon of avgas, my airplane should burn 18.04 gallons/hour. I actually run at a flow of about 18 gallons/hour as shown on my fuel flow meter (pilots, obviously continuously concerned with fuel, think in terms of pounds or gallons per hour). As I've repeatedly stated, I just love it when calculations and data or two methods of calculation agree.

These 18 gallons at 13,000 feet will typically take me about 168 nautical miles or 193 statue miles (the "knot" is one nautical mile or 6080 feet per hour). So, in terrestrial terms, I get 193/18 or 10.7 m.p.g. Not really so good. The Saratoga is known as a bit of a sky-borne SUV, it has six seats and three baggage areas. It's a little draggy and a bit heavy in comparison to piston single engine airplanes.

For a given amount of shaft power, the airspeed achieved (at any particular "density altitude") is controlled by propeller efficiency and total drag. I can increase my range and consequently my "specific range" (miles per gallon) by flying at slower airspeeds. At about 140 knots (a bit over 160 m.p.h) I can reduce the fuel flow to around 10 gallons/hour. This works out to 16 m.p.g., more along the lines of what you might expect for a fuel thirsty road vehicle.

But one of the main reasons people buy airplanes is to go fast, and so I rarely use this method. It can, however, result in a quicker trip if the extended range results in the elimination of a fuel stop. Such a stop typically adds about 45 minutes to a trip. But really, by the time I've been in the plane over five hours, I'm ready for a stop.

Finally, if I put passengers in three of the five seats remaining (more than that results in the need to fly with partial fuel loads, thus reducing range) I can achieve a reasonable figure for "passenger miles per gallon" of 42.8. To be candid, that rarely happens though. No amount of mental gymnastics can make N8409Y a fuel efficient way to travel.