I have a 2005 Saturn Vue 5 speed manual. Recently, I drove from Buffalo, NY to Pittsburgh, PA. The trip down took me 3 hours and 20 minutes at a speed of about 73mph. I was revving at about 3500rpm in 5th gear, and used about 3/4 of a tank. On the way back, I got caught in a blinding snowstorm, and drove 35mph the whole way back, revving at about 3500rpm, this time for about 7 hours! I still managed to make it back having only used 3/4 of a tank. How, since the rpms were the same, did I use the same amount of gas when driving time was doubled on the way back?
There are many factors that will cause a change in the mileage. The fact that the two drives had about the same mpg only means the plus factors and the negative factors averaged each other out.
Your car motor may have turned over more coming back, but even though it was in a lower gear it didn’t have to work as much with less wind resistance. You were using fewer gallons per hour running at lower speeds with less resistance making it a wash. At least in this situation. You can bet that if you had traveled at much less than 73 w/o the snow but fast enough to keep it in 5th gear, your mileage would have been better.
The variables seem to offset each other with the change in situation
Stepping back from the fact there’s a car involved, and it’s been years since I took physics in high school, is it possible that moving the car the same distance at either of two constant speeds takes the same amount of work, regardless of speed? Since gasoline gets turned into work, I suspect this is the explanation.
Work = Force x Distance
Force = Mass x Acceleration
You were going at a constant speed each time, so we can consideration acceleration neglible.
How much would a device which reads out fuel consumption/mile cost?
Don’t forget to include friction in your force diagram. That will be the variable that explains this.
Some cars have them as a standard feature. To quote Woody Harelson as Larry Flynt “Ask the manufacturer.”
I proudly decline so to do, as friction is included in the Force calculation.
“Don’t forget to include friction in your force diagram. That will be the variable that explains this.” I rather think not. Don’t you agree?
Friction is not included in F=MA. F is Force, M is Mass, and A is Acceleration, as you noted. That equation taken by itself is useful only in high school physics classes or in a vacuum while sliding on a frictionless superfluid. Otherwise you must calculate not only the friction of the tires and the roadway, but also the air resistance.
It is, in fact, that friction that makes your F=MA equation have any meaning at all in this case, else going at a constant speed means Acceleration is 0 and therefore the force exerted would be 0, or in other words you could get the car up to speed and then put it in neutral and shut off the engine until the end of your trip.
“Otherwise you must calculate not only the friction of the tires and the roadway, but also the air resistance.”
No, I don’t, because I’m suggesting a real world explanation for the phenomenon described, and the degree of accuracy you’re suggesting by taking into account all possible factors just isn’t necessary since we don’t have either a) precise information about speed (0 to 73 in how much time? no stops? no variation in speed?) or b) precise information about fuel consumption (“3/4 of a tank”? really? exactly? filled precisely the same way each time? measured how?) or c)interest in every little factor; what was the temperature on each trip, for example. If you want to do the math, have at it, but for this question it’s not necessary.
“It is, in fact, that friction that makes your F=MA equation have any meaning at all in this case, else going at a constant speed means Acceleration is 0 and therefore the force exerted would be 0, or in other words you could get the car up to speed and then put it in neutral and shut off the engine until the end of your trip.”
Acceleration means change in speed or direction. Let’s assume a constant direction to keep things simple. For there to be 0 force exerted to move a car in a straight line it would absurdly follow that it took 0 work to move the car.
So, assuming a car moving at constant speed, where is the force? Look to the wheels: any point on the wheel is constantly changing direction (accelerating), while interestingly enough, the point on the wheel in contact with the ground is always motionless relative to the ground. Turning the wheels is 0 work but it is the turning of the wheels that does the work of moving the car. Remember I am decades away from my physics class, but I’m pretty sure that “That equation [F=MA] taken by itself is useful only in high school physics classes or in a vacuum while sliding on a frictionless superfluid” is a little condescending. It doesn’t just disappear the summer before college.
Either you remembered it wrong, or your teacher taught it wrong. The complete equation is “sum of all forces acting on an object” = ma, where forces and acceleration are vectors (meaning they have magnitude and direction). Asking high school kids to draw vectors diagrams and integrating varying forces is a bit cruel; they do that to masochistic college freshmen. Assuming a one dimensional case here, forces generally include the engine’s motive force, rolling resistance, and wind resistance, assuming the vehicle is traveling on flat ground with no gravity involved(1D case). If you’re traveling at a constant speed, the engine provide the motive force to overcome both, resulting in zero acceleration.
Wind resistance can be huge. Look at the original Honda Insight and a Smart Fortwo. They both weigh about the same. One is long and thin, designed to slip thru the air, while the other is tall and upright, designed to be parked anywhere. One does 70 mpg on the highway and the other does 1/2 that.
Tires are not perfectly round at the bottom. That little flat part in front of the wheel center acts as a negative torque, creating rolling resistance. That’s why you’re always told to check your air pressure, to make sure that it doesn’t get too flat. Also, that’s the reason why trains are more efficient, their wheels don’t deform nearly as much as rubber.
Again, you don’t need all that detail for the problem at hand. Plus, “the sum of all forces” okay, but then each of those F’s is also “the sum of all forces” a nice infinite room of mirrors for a relatively straightforward question. Let us not forget the Coriolis force.
“Tires are not perfectly round at the bottom.” Thank you for that. It is my understanding that the sky is generally blue.
Force = Mass * Acceleration gets the job done.
What you seem to not be grasping here is that at different speeds, air resistance changes and therefore does become an important part of the energy equation. The faster you go, the harder it is to get through the air. That’s why they had to add 199 horsepower to the Veyron Super Sport to get it to go 15 mph faster than the standard version.
So assuming you don’t add aerodynamics to the car in order to compensate for the increased drag, you do indeed need to take air resistance into account when figuring out how much energy a car will use at different constant speeds.
You also don’t seem to understand the concept of the F=MA equation. You’re correct in that A = change in speed, but where you fall off is that you therefore assume that a car that does not change speed actually requires energy in order to keep going at that speed. If we strip all of what you feel are “unnecessary details” from the problem - that being road friction and air resistance, the car would indeed not require any energy to remain at a constant speed. That’s why a spacecraft such as the Voyager probe can keep moving through space without firing an engine. Friction in space is negligible (only occurring on the rare occasion that you actually run in to a particle) and therefore there is no need to keep firing a rocket in order to keep the spacecraft at the proper speed.
Here on earth, friction is very much necessary to take in to account, and so the ONLY reason that a car requires energy to remain at a constant speed and direction is because of the road friction and air resistance that you dismiss as unimportant detail.