Better MPG at 62MPH then at 55MPH

[ksa8907]

I would add that even when a road looks level, it may have a very slight grade, which does make a considerable difference. A/C use is also a factor, (even your defroster will cycle the A/C). slight wind, previously mentioned, will make a difference. all these things together can make a big impact, easily explaining your difference.

[jakobnev]

exponentially

Stop using that word!

[AeroModder]

I have the same thing with my car, it gets better cruising MPG at a touch over 60 rather than at 50-55. It's not just a one-time thing, either, it's very consistent.

[Mustang Dave]

Originally Posted by wungun

Exponentially means just that...
And easy way to look at it is, to double your speed requires 4times the power.
So if your increasing your speed from 10mph to 20mph, it'll still require 4 times the power...except at those speeds, you might only be using 3hp to maintain 10mph (or 12hp for 20mph.)
But if you're doing 50mph, and you want to do 100mph, where it might be 50hp to do 50, you'll need 200hp to do 100mph...and exponential jump!

Someone correct me if I'm wrong....

Doubling your speed requires 8 times the power. Drag force is 4X as much at double the speed.
Power is a function of force * velocity.

[wungun]

So your saying, in my example, if it requires 50hp to do 50 mph, you'll need 400hp to do 100??
I think you're wrong....
Are you thinking of kinetic energy an object has at speed perhaps?
Or am I!?
Lol

[mort]

Originally Posted by wungun

So your saying, in my example, if it requires 50hp to do 50 mph, you'll need 400hp to do 100??
I think you're wrong....
Are you thinking of kinetic energy an object has at speed perhaps?
Or am I!?
Lol

Yes, if all your drag was aero. Look at the ecomodder tools
The default is for a small economy car, but the results chart goes up to 200 mph. At 50 mph aero power is 5.79, at 100 mph aero power is 46.36. hp. At 200 mph it would require 370.86.
-mort

[Mustang Dave]

Originally Posted by wungun

So your saying, in my example, if it requires 50hp to do 50 mph, you'll need 400hp to do 100??
I think you're wrong....
Are you thinking of kinetic energy an object has at speed perhaps?
Or am I!?
Lol

You are.

From Wikipedia:

"Power

The power required to overcome the aerodynamic drag is given by:


Note that the power needed to push an object through a fluid increases as the cube of the velocity. A car cruising on a highway at 50 mph (80 km/h) may require only 10 horsepower (7.5 kW) to overcome air drag, but that same car at 100 mph (160 km/h) requires 80 hp (60 kW). With a doubling of speed the drag (force) quadruples per the formula. Exerting four times the force over a fixed distance produces four times as much work. At twice the speed the work (resulting in displacement over a fixed distance) is done twice as fast. Since power is the rate of doing work, four times the work done in half the time requires eight times the power."

This is well known among salt flat racers.

[wungun]

Doubling your speed requires xpower cubed....End of story.
Think about what is being said...
Exerting four times the force over a fixed distance produces four times as much work.
The distance is not fixed...
The distance is the variable in the work calculation...In this case, how many miles is covered in x amount of time. This is how we measure speed.
At twice the speed, you are covering TWICE as much distance over the same time frame, right?
And the last sentence...
Since power is the rate of doing work, four times the work done in half the time requires eight times the power.
Again, this does not add up go me.
Rate means time, correct? The rate at which you do work. Think about it.
Where does "half the time" come from?
Obviously from the "fixed distance" comment...
Again, the time frame in these calculations is based on the speed measurement, and since speed is a product of time and distance , you are not doing 4 times the work in half the time...you are doing 4 times the work in the SAME amount of time. And the timing in question is one hour.
Make sense?

It does to me!

If I'm wrong, and I very well could be, someone please explain it better to me!?

[Mustang Dave]

Horsepower= torque {lb_ft} * RPM/5252. Universal hot rodding formula.

The following assumes the same gear ratio:

Torque required is directly proportional to drag force, and therefore, proportional to the square of the speed.
RPM is directly proportional to speed.

Suppose a car requires 35 lb_ft torque to maintain 50 MPH, and is running 1500 RPM.

35*1500/5252=9.996HP

Same car requires (100/50)^2 or 4times the torque (140 lb_ft) to overcome the drag @ 100MPH. It's also running 3000 RPM.

140*3000/5252=79.969HP (8 times what it takes to run 50 MPH).

Maybe someone else can explain it in simpler terms.

[mort]

Originally Posted by wungun

Since power is the rate of doing work, four times the work done in half the time requires eight times the power.
Again, this does not add up go me.

Hi wungun,
There are 3 engineering properties to consider. Force is pushing on something. If you weigh 150 lbs you exert a force against the earth of that many pounds.
If you want to move an object, like a car, you can push on it with a force, say 90 lbs. If you push and jog along at 5 mph you are using 1.2 hp, that's power. Horsepower is just a conventional way of saying lb*mph (with a conversion factor).
Now to go anywhere you need energy. Burning gasoline releases energy, in units of horsepower hours. A gallon has about 45 hp hr of energy. Energy and work have the same units.

Imagine a car that only has wind resistance. (OK - that's an airplane.)
At 60 mph the wind resistance force on my imaginary car is about 90 lbs.
If my imaginary car had an engine that was 33% efficient at 15 hp, and it takes 15 hp to go 60 mph Then I could go for an hour on 1 gallon of gas and that would move me 60 miles. That's 60 mpg.

Now if I went 120 mph the force needed to overcome wind resistance is four times more than at 60 mph. The power rquired to overcome wind resistance is 8 times higher. The energy in a gallon of gas is not changed.

So in my imaginary car I also have an engine that can produce 120 hp at 33% efficiency. With that engine I can go 120 mph. With 1 gallon of gas delivering 15 hp hr through the engine, I can go for 7.5 minutes. That's 15 miles. Thats 15 mpg.

So at twice the speed I can travel for 1/8 as much time, and go 1/4th the distance on the same amount of fuel. I've done the same amount of work, but the engine needed to be 8 times more powerful. If I wanted to go faster, say 240 mph it would take 960 hp. I'd have just less than a minute of fuel, and I'd move 3.75 miles. (3.75 mpg)
-mort