The effect of horsepower on gas mileage

[The Toecutter]

Torque is a function of how much fuel is being combusted and some efficiency, and horsepower is torque at some rpm. Gasoline engine efficiency is, among other things, a function of displacement. A corvette and Insight are not that far off in terms of CdA and weight, but the vette get's crap for mileage due to it's large displacement engine having much more in the way of pumping losses at some given output.

The Insight has a CdA of 5.125 square feet. This is from a published Cd of .25 and published frontal area of 20.5 square feet.

The Corvette has a CdA of 6.132 square feet, from published .28 Cd and estimated 21.9 square foot frontal area(Car and Driver).

Not only is the Insight 1,100 pounds lighter, it also uses LRR tires with a Cr around .006. The Corvette's tires are very sticky, Cr around .012.

The difference in forces to overcome for the Vette versus the Insight are actually going to be quite large. If you go to fueleconomy.gov, and note the differences in fuel economy between a V6 and V8 Mustang, or the L4 and V6 Camry, the differences for combined mpg are around 20%. The highway mpg difference between the V6 Camry and the hybrid Camry is only 18%!


Were that Vette to have the same CdA as an Insight, same weight, same tires, and keep the V8, it would probably do around 40 mpg highway. As is, the Vette does 29 mpg highway. Likewise, would the Insight have the same CdA as the Vette, were 1,100 pounds heavier, and had sticky tires with the same hybrid drivetrain, it would probably get around 45 mpg highway.



By adressing aero drag and rolling losses, we could have 40-50 mpg hwy V8 musclecars. Say, take a GM Precept concept car with a .16 Cd, and hypothetically shove a Corvette engine into it as opposed to the hybrid-diesel drive. I'd lay down money on the table that it would get better than 35 combined MPG. Having the Corvette V8 engine with a 'mild hybrid' powertrain like Citroen uses to kill idling at a stop would probably bring that number over 40 mpg combined.

[omgwtfbyobbq]

Run your numbers yourself, sum up the rolling and fluid friction @48mph for each vehicle. There is ~27% difference in energy required to move the Vette @48mph EPA highway, but the difference in mpg is ~136%! Where does this extra inefficiency come from? Primarily, the fact that the Vette has an engine six times the size of the Insight's, and pumping losses at 50mph are huge!

Hybrids aside, since we're talking about pumping losses, so any tech like electric motors, idle shut-off, mds, that minimizes those, and masks the impact of pumping losses should probably not be looked at...
The Mustang you mentioned (mt) has a difference in displacement of 4.6L/4L=~15%, and a difference in highway efficiency of 28mpg/25mpg=~12%. The Camry you mentioned, not the hybrid version, get's 28mpg for the the 3.3L 4sp auto, and 34mpg for the 2.4L 4sp auto. 3.3L/2.4L=~38% difference in displacement, and 34mpg/28mpg=~21% difference in mpg. So, the drop in efficiency is not linear, probably because friction losses are greater in smaller displacement engines. But, the point I'm making is that pumping losses are a big part of gasoline FE! And the Vette would still get ~48~50mpg (maybe more, maybe less, ballpark) if it had the Insight's engine, since there is only a ~27% difference in glider energy requirements. The only way to get the vette to 50mpg would be to drop the force needed at 48mph by ~67%, LRR tires only give about 8%, so the remaining must be done via aero, and like you said, if we improve that by 50%, we'll be there.
Here I go off topic again!
Anyway, pumping losses>friction losses. *EV motors>ICE engines!

*I've been thinking about a li-ion powered velomobile.

[The Toecutter]

You also have to account for the fact that the Corvette's alignment isn't tuned for efficiency but for traction. Same with its brakes, which are likely the cause of much drag. For that matter, there's also its transmission.

When it comes down to it, the difference between a V8 and a L3/electric drive would be more closer to 50% than 100%. Lets run some numbers, a set for the Corvette, and a set for the Hybrid.

2007 Chevrolet Corvette Coupe:

Mass(W): 1,442 kilograms
Drag Coefficient(Cd): .28
Frontal Area(A): 2.08 square meters
Rolling Resistance Coefficient(Cr): .012
Transmission Efficiency(TE): .90

Velocity(V): expressed in meters per second
Force Drag(FD): expressed in newtons
Force Rolling(FR): expressed in newtons
Stray Force(SF): 40 newtons
Wheel Power(WP): expressed in watts
Engine Power(EP): expressed in watts

Air Density(Rho): 1.25 kg/m^3
Gravitational Constant(G): 9.8 N/kg


Equations used:

FD = .5 * Rho * Cd * A * V^2
FR = Cr * W * G

WP = (FD + FR + FS) * V
MP = WP / TE

Results:

At 21.6 m/s(48 mph):

FD = 170
FR = 170
SF = 40
WP = 8,208
MP = 9,120

At 26.8 m/s(60 mph):

FD = 261
FR = 170
SF = 40
WP = 12,623
MP = 14,025

2006 Honda Insight:

Mass(W): 840 kilograms
Drag Coefficient(Cd): .25
Frontal Area(A): 1.9 square meters
Rolling Resistance Coefficient(Cr): .006
Transmission Efficiency(TE): .94

Velocity(V): expressed in meters per second
Force Drag(FD): expressed in newtons
Force Rolling(FR): expressed in newtons
Stray Force(SF): 20 newtons
Wheel Power(WP): expressed in watts
Engine Power(EP): expressed in watts

Results:

At 21.6 m/s(48 mph):

FD = 139
FR = 49
SF = 20
WP = 4,493
MP = 4,780

At 26.8 m/s(60 mph):

FD = 213
FR = 49
SF = 20
WP = 7,558
MP = 8,040



At 60 mph, the power at the motor of the Honda Insight would be about 57% that of the Vette. At 48 mph, the power required by the Insight is about 55% that of the Vette!


That's where the bulk of the mileage differences are at. Put the V8 in the Insight, and fuel economy would probably drop around 40% or so. And it would still embarass most anything else on the road when it comes to fuel economy.

I chose a higher amount of stray friction for the Vette over the Insight due to the Insight being designed with efficiency in mind, as opposed to maximizing traction and other performance aspects. The Insight also has a more efficient transmission and the numbers were chosen accordingly.

[omgwtfbyobbq]

I think your FR's are way, way high because the manufacturer curb weight already includes gravity, since they weight them on scales. If the vette weighs 1.442kg than we're already including gravity, and in this context weight is in the N=kg as opposed to the actual mass kg. If you divide your FR's by 9.8, than I think you're at the force needed to overcome rolling resistance. As per the rolling resistance expression, Crr is dimensionless, so the W must be in N already.

[The Toecutter]

Kg is a unit of mass, not weight. At the Earth's level of gravity, there are ~2.2035 pounds of force for every kilogram of mass. In space, mass and weight would basically have no relation, as mass would stay the same as it was on Earth and weight would basically be 0 pounds.

To find the total force gravity in Newtons, you take the mass in kilograms and multiply by the gravitational constant.

With pounds as opposed to kilograms, the gravitational constant is already included and it's a simple conversion. Newtons and pounds measure the same thing: weight. Kilograms is not a unit of weight.

It is not unusual for a car to have 30 pounds of rolling force. That equates to a 3,000 pound car with a .010 Cr. 1 pound is equal to 4.45 Newtons. Convert accordingly.

[omgwtfbyobbq]

I hate standard units.
This sucks beause I was seriously underestimating rolling friction, but it's very very cool because if my tires have a Crr=.012, getting LRR tires Crr=.006 and aligning everything 0,0,0 should bring my mileage up to almost 70mpg@50mph! Anyway, as we've seen before, I hate units!
Not that pumping losses aren't significant, there just not as significant as my ignorance of the imperial units, crazy *** things.

Engineering Challenges: Adding another technology to a vehicle does not necessarily improve fuel efficiency. For example, variable valve timing and cylinder deactivation both target the energy ?pumping loss? that occurs as the engine ?breathes? (that is, operates at a lower intensity). Once the first technology is applied, that pumping loss has been largely reduced so the second technology will provide less benefit.

[The Toecutter]

If I had my way, the auto industry would be bringing on the 40+ mpg musclecars, 80+ mpg biodiesel-powered family sedans, and 250 mile range pure electric cars.

My design philosophy is that performance and fuel economy need not be mutually exclusive. Both can be had in a properly designed car that retains much of the amenities of the cars seen today.

[omgwtfbyobbq]

Originally Posted by The Toecutter

Kg is a unit of mass, not weight. At the Earth's level of gravity, there are ~2.2035 pounds of force for every kilogram of mass. In space, mass and weight would basically have no relation, as mass would stay the same as it was on Earth and weight would basically be 0 pounds.

To find the total force gravity in Newtons, you take the mass in kilograms and multiply by the gravitational constant.

With pounds as opposed to kilograms, the gravitational constant is already included and it's a simple conversion. Newtons and pounds measure the same thing: weight. Kilograms is not a unit of weight.

It is not unusual for a car to have 30 pounds of rolling force. That equates to a 3,000 pound car with a .010 Cr. 1 pound is equal to 4.45 Newtons. Convert accordingly.

I think I see where you and I both went wrong! If like you stated, with pounds, the gravitational constant (~32ft/s^2) is already included, to convert to the mass in kilograms, we need to first divide the weight by the acceleration due to gravity to arrive at the mass in pounds, then we convert that mass in pounds to mass in kilograms, then multiply that by 9.8m/s^2, and finally multiply by the dimensionless Cr.

In other words, if the weight in pounds is 3174pdl, then we divide by 32ft/s^2 to arrive at a mass of 99lbs. Converting to kg we get 45kg, and the normal force is 45kg multiplied by 9.8m/s^2, which is 441N, so the rolling friction is the product of this and .012, which is roughly 5N, which converts to ~38pdl, which is what you get if you multiply 3174pdl by .012.

So, it must be that rolling friction is much small than either one of us though, or, in fact, the weight in pounds is actually the mass in lbs and we must multiply by 32ft/s^2, then multiply by the Cr to get the rolling friction, and the same goes for the mass in kg. So, the question is, provided that the units convert correctly in both cases, is weight generally a measurement of mass, or of force?

[Ted Hart]

Originally Posted by Matt Timion

At least that's what I just read on the internet.

Torque helps gas mileage. Horsepower does not. This actually makes sense considering diesel engines get much better gas mileage than gasoline engines.

Now, can anyone give me the physics telling WHY torque increases gas mileage and horsepower does not?

Are there any tricks to increase torque?

Hi, again!
"Tricks to increase torque?" Now you're on the right track! It's pressure on each piston which produces the twist on the crank journal. This pressure (from the burning gases above each piston) is called the BMEP ( brake mean effective pressure). Increase this, and we have increased the torque (maybe not the HP!). How can we do this? Lots of ways! Change camshaft profiles, change CR (pistons? Mill the head? Both?), change this, change that. But be wary of the FED! Lots of changes create emission rate changes! Some good, most bad, some illegal! Welcome to the world of the "snake-oil" salesman! "He" will lead you down every primrose path known to man...to get your money!
I will now tell you a perfect way to change (increase) your BMEP; change the gasoline! Today's gas is slop! I've read "horsewhizz", I've read " Tiger piss"....But, the fact remains- until someone figures out a way to run a gasoline engine on diesel fuel- pump gas ain't real gas! Too much oil in it! If we can't get the oil out, we can do something to burn this "gass" better!
Interested?

[The Toecutter]

I think I see where you and I both went wrong! If like you stated, with pounds, the gravitational constant (~32ft/s^2) is already included, to convert to the mass in kilograms, we need to first divide the weight by the acceleration due to gravity to arrive at the mass in pounds, then we convert that mass in pounds to mass in kilograms, then multiply that by 9.8m/s^2, and finally multiply by the dimensionless Cr.

In other words, if the weight in pounds is 3174pdl, then we divide by 32ft/s^2 to arrive at a mass of 99lbs. Converting to kg we get 45kg, and the normal force is 45kg multiplied by 9.8m/s^2, which is 441N, so the rolling friction is the product of this and .012, which is roughly 5N, which converts to ~38pdl, which is what you get if you multiply 3174pdl by .012.

So, it must be that rolling friction is much small than either one of us though, or, in fact, the weight in pounds is actually the mass in lbs and we must multiply by 32ft/s^2, then multiply by the Cr to get the rolling friction, and the same goes for the mass in kg. So, the question is, provided that the units convert correctly in both cases, is weight generally a measurement of mass, or of force?

You don't divide the weight of an object in pounds over 32ft/s^2 unless you want to measure mass. What we're concerned with is force, not mass. Pounds are already a measure of force. Kg is mass, and we must multiply by 9.8 N/kg to figure out the force given Earth's gravity.

For a 99 pound object with a .012 Cr, 5N of rolling force sounds about right(Actually 5.287). Scale that up for a 3,200 pound car using the decimal value, and you get 171 N of rolling force.

171N of rolling force is equal to ~38 pounds of rolling force.


Go to google.com, type in Newtons per pound. 1 newtons = 0.224808943 pounds force Both Newtons and pounds measure force. Force is essentially weight.