“Be kind, for everyone you meet is fighting a hard battle” - Often attributed to Plato but likely from Ian McLaren (pseudonym of Reverend John Watson)

Friday, April 30, 2010

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?

7 comments:

ed-davies said...

There's something badly wrong with this argument but it's too late at night here for me put it clearly straight away so, for now, I'll just say:

1) Your top speed in such a turn is 1/3rd normal, is it?

2) A satellite in orbit needs to be fed energy to keep it going round, does it?

King of the Road said...

1. Yes, it's much slower although there are many freeway curves I take at 55 m.p.h. with radii smaller than 0.9 kilometers. I'll actually be adding another post to compare going straight for, say, 30 meters at, say, 10 m/s vs. slowing for a corner (to, say, 5 m/s), using force to change the direction of the vehicle, and accelerating back to speed. But turning requires force.

2. As to the satellite, no, it doesn't need energy added (well, at least for a while), but that clearly doesn't mean that there is no centripetal force exerted on the satellite. Energy wouldn't be added by gravity, though, since no work is done on the satellite because the force vector dotted with the displacement vector at any instant is zero.

When I bank my airplane to turn, I add a horizontal component to my lift vector. the tighter or faster I turn, the more I must bank in order fornthe lift vector to have a sufficient horizontal component. The lift vector is a result of my forward motion, i.e., the engine's thrust. I must also increase my angle of attack and, for steep turns, add power. Thus, I'm flying at a constant altitude at lower speed and a higher power setting in order to turn.

I will have to give it some hard thought. The centripetal force calculation is pretty straightforward though..

King of the Road said...

Ed, I've penned (keyed?) a mea culpa in an update. Thanks for rattling my cage. I'm suitably chastened.

Becky McCoy said...

great idea! however, the momentum equation is momentum (p) = mass*velocity.

f=ma is force = mass*acceleration

These posts are great applications for Physics students!

ed-davies said...

Thanks for the update.

I, too, have been thinking about this. What I can't quite get my head round is the idea that a centripetal force is acting and changing the velocity of an object without any (or much) work getting done (i.e., energy converted to some lower grade like heat or noise).

Neglecting air resistance and other minor things, for a satellite in a circular orbit energy is constant, though momentum is transferred back and forth with the primary (planet or whatever). For an elliptical orbit the total energy stays constant but is converted backwards and forwards between kinetic and potential energy.

In the case of a banking air/aeroplane not climbing or descending, as you say, the horizontal component of the lift provides the centripetal force and the vertical component of the lift balances the weight. Because the lift is not vertical it must increase over what it is in level flight to continue to balance the weight. Because the lift is increased the drag is also increased (in proportion to the lift/drag ratio at the speed). To maintain the same speed, therefore, more power will be required.

For a road vehicle going round a curve there will be an analogous effect, I think. I imagine that the tyres/tires will have a sort of lift/drag ratio which means that to generate a given amount of centripetal force they also produce a certain amount of extra drag. Does that sound right? Whatever, the extra drag force will be quite a lot smaller than the centripetal force being generated.

King of the Road said...
This comment has been removed by the author.
King of the Road said...

Becky,it's true that we most commonly use f=m*a but Newton formulated it as f=d(mv)/dt which is rate of change of momentum. For a constant mass you can pull the "m" out for f=m*dv/dt and dv/dt=a so the two are equivalent