I am not the world’s most mechanically-minded person. Nor am I the world’s worst. But I’m definitely closer to the latter than the former. For example, I can’t make head nor tails of the above formula, even with the following guide [via howstuffworks.com]:
The letter v represents the velocity of the car, and the letters a, b and c represent three different constants:
The a component comes mostly from the rolling resistance of the tires, and friction in the car’s components, like drag from the brake pads, or friction in the wheel bearings.
The b component also comes from friction in components, and from the rolling resistance in the tires. But it also comes from the power used by the various pumps in the car.
The c component comes mostly from things that affect aerodynamic drag like the frontal area, drag coefficient and density of the air.
Bottom line? “if you double your speed, this equation says that you will increase the power required by much more than double.” Yes, yes, but the question was “What speed should I drive to get maximum fuel efficiency?” Can TTAC’s Best and Brightest provide the required dose of enlightenment, either specifically or generally?
Someone has to give you the specific a, b, and c coefficients for you to make any sort of evaluation. If the velocity-cubic term ‘c’ has a coefficient of 0.0001 and ‘a’ and ‘b’ are much larger, v-cubed won’t matter a hill beans.
Gave a dumb reply, claim it was due because English isn’t my native language
Interesting topic, Robert.
Those coefficients are nebulous enough to make this calculation inaccurate: Michelin Pilot Sports influence A, a big bearing transmission (like a Ford C6) negatively impacts B, and the front end of a Chrysler 300 is a big demerit on C.
But by how much?
Looks like figuring out the coefficients’ value is flawed enough to make common sense about your car’s design just as helpful. I guess the benefit is for people who understand things better with formulas.
The formula is an integral of force divided by time. We first add up forces:
a=forces that act on the car independent of speed
bv=forces that increase as vehicle speed increases, hence the multiplication of v.
cv^2=air resistance, where C is a function of coefficient of drag, fluid density, and frontal area.
that gives you the sum of forces a+bv+cv^2, which you then integrate to get the amount of work: ad+bvd+cv^2d, where d is displacement. To get power, you divide work by time (or differentiate by it, if you want to get technical), and d/t is v. So that gives you the final formula of: av+bv^2+cv^3.
C is not negligibly small, and at high speeds, that last term makes for most of drag.
Basically, the answer to “what speed is most fuel efficient” is different for every car in every set of conditions. Hell, say the temperature drops a few degrees. The air becomes more dense, making c larger, but then your cooling system doesn’t need to work as hard, decreasing b. An answer this complicated usually signifies that the question was stupid to begin with.
seoultrain –
I agree on the mathematical explanation as it stands, but the verbal guide on “How Stuff Works” sounds like it’s hand-waving, empirical, ‘hey let’s just fit an equation to it’ kind of thing.
Which is fine, except that it doesn’t tell you anything about YOUR car, only about the tested car.
I’ve sometimes thought that this kind of information should be included on the window sticker, in either algebraic or graphical form. The city/highway MPG number just isn’t enough for today’s increasingly geeky fuel miser. The manufacturer already has that information, where’s the harm in publishing it?
Someone correct me if this is wrong, but doesn’t it also depend on the engine’s BSFC, which depends on volumetric efficiency and such? I know that if this is the case, there should be some “correct” engine speed at which the engine will have it’s peak VE because VE decreases when piston speed increases. Of course that point will also depend on the cam (or cams for you OHC guys) and gearing, since you may not be able to keep a reasonable speed at this RPM range if your gearing is too tall or short.
Craigles: it appears that coefficient “B” accounts for BSFC. Of course, this assumes you know everything about every moving part bolted to the chassis. Which is humanly impossible for most folks.
Again, not a very helpful formula in reality.
To get the best efficiency, you should drive at the a speed/load which gives you best brake specific fuel consumption. This operating point gives you the best trade off for power delivered for fuel used at engine output.
So to get the most efficieny in terms of driving speed, it would be driving the engine at the operating point above with the lowest vehicle speed possible. This allows all your engine power to go to moving the vehicle, rather than frictional losses, and aerodynamic losses.
Interestingly though, to get the best fuel economy, you should drive as slow as possible even though you may not be at your best brake specific fuel consumption operating point.
Fuel efficiency and fuel economy are not exactly the same thing!
That equation alone won’t provide the answer you seek since it only shows the power required to maintain any particular velocity. As Craigles pointed out, you need another equation showing the power available to the drive wheels provided by the engine as a function of engine speed (proportional to forward velocity if the wheels are not slipping) and the rate of fuel consumption. At equal power, the equations can be combined and solved for fuel consumption as a function of velocity. A plot of this equation (or the first derivative thereof) will probably show a minimum fuel consumption at some low velocity.
I can’t explain the formula. But I can tell you your most fuel efficient speed is when your car has just shifted to its highest gear. Highest gear @ the lowest rpm possible = MAX FUEL ECONOMY!
Example: I drive currently a 2005 Toyota Tundra. City:15 Highway:18
But my most fuel efficient speed is between 40 and 55 mph.
When I go within that range my truck typically gets between 21 and 25 mpg (depending on hilly or mostly flat terrain)
DOES THIS SOUND CORRECT TO EVERYONE?
Makes sense to me and Click and Clack radio talk show hosts on car talk.
This equation isn’t all that useful. It sounds like they’re trying to tell you how the various HP-suckers reduce your power at the wheels. But to say that tire rolling resistance increases linearly with velocity, your pumps increase load 2nd order exponentially, and so on is a HUGE oversimplification. A valid calculation would require precise power requirements for all components acting on the engine, and even then would be specific to your exact vehicle, road surface, ambient conditions, etc.
At best, this equation just tells you that going faster and/or adding additional loads to the engine requires more power. Duh.
Just drive fast. It’s more fun.
Westhighgoalie:
Pretty much. Long story short, fuel is injected based on engine speed and load; more engine speed and/or more load = more fuel. Odds are you would find more points that use the same amount of fuel (possibly in a lower gear at low RPM) but since you don’t cover as much ground at that combination, your economy isn’t as high.
This is pretty much the way I took it: the whole answer is basically giving the finger to everyone ignorant enough to ask the question expecting a simple solution (like 55mph, which infuriates me).
westhighgoalie has a good, simple rule of thumb: top gear, low rpm. Stick with that, and don’t care so much about eking out that 1 or 2 extra mpg.
Engine efficiency varies with RPM, load, air temp, humidity, etc. It’s not always correct that keeping engine RPM low = best MPG. Gasoline Otto-cycle engines have to deal with pumping losses due to the way they are throttled, unlike Diesel engines, for example.
Which means efficiency can actually increase as RPMs increase, to a point. With perfect volumetric efficiency and no internal friction, assuming constant engine speed, Otto-cycle gasoline engines can be most efficient at wide-open throttle, assuming the load on the engine keeps RPMs within reasonable bounds. At some point, of course, you would start outrunning the expanding gasses in the cylinder, not to mention subject the engine to stresses it would not survive.
All of which is just another reason why the answer to the question “At what speed is my car most efficient?” is “It depends”. If you really want to know how speed effects MPG in your car, just get an ODBCII instantaneous MPG meter and plug it in. They’re pretty cheap.
Ah finally a question that I can answer! Ok, the howstuffworks article is silly and doesn’t really say anything. And as was mentioned before, to optimize on anything you’ll need to know what A, B, and C are for your car.
Here’s the thing, when trying to describe a system by an equation, physicists will take data and then try to fit it to a model that describes the system. If all else fails, a polynomial expansion, particularly a Taylor series expansion is what they fall back on. Basically what this amounts to is that you have some variable that’s a result (power in your example) and some variable that it depends on (velocity in your example).
And you try to find out what transforms V into P. So what the physicists do is drive a car at various velocities and keep track of the power required. Then they go back and try to fit P as a function of V. Since they don’t know what it really depends on, they first say that P = a*v, so there’s some term that depends on how fast you’re going.
They see that that doesn’t fit the data well, so they expand on the v term and add in b*v^2, that doesn’t work perfectly, so they add in a c*v^3 term. After that, the function fits the data well and they stop, but if they wanted they could always add a v^4, v^5, … term until the function fit as well as they wanted it to. Then afterwards they go back and try to assign physical meaning to the a, b, and c constants.
So the values of a, b, and c will depend on your car, driving style, roads you’re driving on, every thing. It’s not a good way (although there’s no real good way) to generalize to all cars and all driving styles.
To answer you’re real question about how to maximize fuel efficiency. Well, there’s no substitute do experimentation and looking at the data. Since you’re probably more into the modern world than I am, you cars probably all support OBD-II.
You can get a cheap reader that can save data to your laptop so you can see how much fuel you’re burning per unit time in various driving conditions and then you can try to minimize that. If you get data, I’m more than happy to fit it to your personal functions and do all the analysis for you, just send me some data.
the universal answer according to Douglas Adams : 42
Your “road load power” equation is only half of what is needed to figure out maximum MPG speed. According to your formula, you would achieve maximum efficiency by going a speed only marginally faster than 0. What is missing is that at that [lack of] speed your engine is generating power at an efficiency only marginally better than 0 because it is spending almost all its effort just keeping itself running.
You additionally need to know how efficiently your engine can generate power at a given car velocity. This will also involve your transmission gearing. Note that “engine efficiency” isn’t MPG… it is: “amount of power at the crankshaft divided by the volume of fuel stream being sucked into the engine.” i.e.: how well your engine turns gasoline into work.
As a rule of thumb for engine efficiency, you want to run your engine fairly hard and at an RPM close to where it would generate maximum torque at full throttle. Let’s say 3/4 throttle and 3500 RPM. Depending on gearing and engine, 3/4 throttle and 3500 RPM would yield either vigorous acceleration (in 1st gear) or it would be at a velocity (in a very tall 6th gear) where “road load power” is already getting high due to c*v^2.
So you compromise by either lightening the throttle or lowering the RPM the get the velocity slower. Usually, engine efficiency falls off faster with reduced throttle than it does with reduced RPM. So, you usually want to use a tall gear to get your RPM low and then give the engine a fair amount of gas but not so much gas that you exacerbate c*v^2.
In conclusion, do what everyone else has said: Use your tallest gear and go at a speed somewhat faster than the slowest speed possible in that gear. That way you are still putting a load on the engine (for efficient engine operation) but not going so fast as to have an inefficient c*v^2.
The equation Robert provided increases polynomially with respect to V, so each additional mph costs more than the previous one. So if internal combustion engines were equally efficient at all speeds and loads, idling along is best. But they are not.
As Sunnyvale and others point out, operating ICEs at wider throttle openings and low rpm is usually the sweet spot of efficiency. This is the operating regime that minimizes pumping losses and friction losses.
Pumping losses are related to a technology which most TTAC readers have heard when driving downhill next to a big rig: the “Jake Brake“. Notice in the graph that the Jake Brake can produce around 500hp braking force in a large diesel engine. Pretty good for just using an engine as an air pump! Do any TTAC readers know how many hp are needed to overcome pumping losses at peak manifold vacuum — I dimly remember about 25hp for a V8?
But my blathering for a few paragraphs pales in comparison to BMW. In the mid 1980s, they produced an entire vehicle to answer Robert’s question : the eta-engined 528e. The WSJ article (plus its link to Wikipedia) describe how BMW engineers matched the engine’s efficiency graph to common driving conditions. Pumping and frictional losses were minimized and the 528e got great mileage for the time.
Like many attempts to “do the right thing”, WSJ notes that this one did not go unpunished: “Innovative or not, Bimmerphiles complained about the engine and the company responded by returning to more conventionally tuned, performance-oriented engines.”
Westhighgoalie is on the right path…it is a zen question, not an absolute. The answer will very depending upon the vehicle you are driving. In whatever vehicle you are driving, the most efficient speed is the lowest speed at which the tallest gear will function without lugging or stalling the engine (in a manual tranny) or downshifting (in a slushbox). If you clustered the data from different cars under different loads and varying road conditions (level, inclined), the data would cluster around a mean, which would be the most common, most efficient speed.
42 miles an hour on a level road, sounds like a good guess….;)
Bozzie, BMW 528es have been my car of choice for the last 12yrs. They make a great commuter car. Best of all, I can maintain them in my driveway.
The key factor is aerodynamic drag. The faster that you go, the more that drag becomes the most important determining factor for fuel economy.
You hear some claim that they get better fuel economy at higher speeds than at lower speeds, but given the laws of physics, this cannot possibly be true. I believe that you will find that optimal fuel economy is achieve with steady cruising at a speed somewhere between 25-40 mph, not at 55, 60 or above.
In practical terms, though, this has to be balanced with how these speeds affect interaction with other drivers. For example, one can’t possibly cruise on an interstate at 40 mph without creating a hazard or a lot of start-stop driving that would harm fuel economy.
You might well be better off driving at higher speeds, not because of the theory, but because of the practice that allows you to drive in traffic more steadily and evenly at a higher speed than at a lower, more erratic one.
Forget the formula; buy a ScanGuage II ($159) which gives you instant mpg readouts on any car with an OBDI port (post 1996).
As others have said, the polynomial equation provided does nothing to help answer the question of the optimum speed for fuel economy. The feds have posted a typical graph on their website which fits with my experience when I have played around with the real time fuel economy indicators in a couple of our cars.
Have a look:
http://www.fueleconomy.gov/feg/images/speedVsMpg3.gif
The text explains that every vehicle is different and that this is a typical graph. Once you get into highway speed operation and above, fuel economy as a function of speed declines as aerodynamic effects dominate.
A bunch more examples and graphs are found here:
http://www.metrompg.com/posts/speed-vs-mpg.htm
Well SImple is the answer.
whatever you do, whatever your car, keep your RPM as low as possible in respect to how much power you need. if you have a torquey engine, it will take care of the power and render early shifts more efficient and less damaging.
when it comes to drag and all that is lost, you can put tape over the gaps and use low resistance tyres, put the tyre pressur a tad over, get a comfortable cushion to make up for the roughness, and yeah dump the spare and any dead weight.
and dont go over the limit, that should save you a bundle.
it should work, i read that somewhere :P
Well, miked is probably closest but misses it by a bit.
The formula is the generic form of the equation that describes the power to maintain a constant speed, we all agree on that. Fuel efficiency does not enter this equation in the least, so let’s tackle that later.
The equation addresses almost all variables that affect drag (aero, rolling, mechanical friction). There should be an additional scalar constant that is not a product of velocity to be mathematically correct, but I guess it’s OK to ignore since it would be small relative to the other products in the equation.
Tests are performed on vehicles and the data are used to calculate the value of the constants for any particular vehicle. This is done using a coast-down test where a vehicle is driven to a high speed, shifted into neutral and shut off. The car’s velocity over time is recorded and the 3rd-order polynomial curve fit will calculate the constants (excel is really good at this). This is a little different than miked’s explanation since you aren’t actually measuring power but calculating an instantaneous rate of change of velocity which can be used to determine actual power required to resist that rate of deceleration.
Aerodynamic drag is a function of the square of velocity, so aerodynamics will play mostly into the “B” constant. But your initial statement is right; power required to increase speed increases much more rapidly than the square of velocity. I always thought the old Bonneville racers’ rule of thumb was that power increased with the cube of speed. Only aerodynamic drag increases with the square of speed.
Now maximizing fuel efficiency at highway cruise can be calculated empirically but it will typically require much more information that any individual typically knows about their vehicle. Not only do you have to know the constant values for the above equation but it would be best also if you had an isoefficiency plot. This is a chart where the islands of thermal efficiency are broken down like a topographical map on a plot where the x-axis is RPM and the y-axis is engine load. Since you need to know your engine’s load capabilities you also need to know the flywheel torque curve output from a dynamometer.
Now that we know the power to maintain any given speed you can calculate the percent torque load required to achieve that power knowing your gear ratios. The gear ratios are the critical conversion between the x-axis of RPM on the isoefficiency plot and the x-axis of speed (velocity) in the road load equation. Overlay the two results on the isoefficiency plot and you can see where, in each gear, your steady cruise load is in the highest area of thermal efficiency. Duplicate that engine RPM in that gear and maintain it and voila, maximum highway fuel efficiency.
Wrong question, Robert.
Gas is cheap. Unless your funds are really constricted, the number you really want to know is how fast you can drive while remaining safe and not damaging your car.
Repairs from overloading the equipment will easily overrun any savings due to mileage that you could build up over a even a single year.
Wow. Nobody got it right. Alot of close, but not right.
Start at the beginning and consider “Road Load”. Load is a force. It’s easy to find your road load driving force by a coast-down test (described above but there’s no reason to turn off the engine). After coast down, you’ll know the driving force (not power, not yet) you need to keep your car moving. The equation is a 2nd (not 3rd) order polynomial that looks like: a + bV + cV^2.
a= rolling drag, it’s a constant that comes mostly from the tires deforming
b= mechanical friction (depends on speed, usually very small compared to the others)
c= wind drag, depends on the square of velocity.
The question is about POWER. Power = force x velocity. Since we know our driving force (road load), then to find the power, multiply by velocity to get the first equation.
aV + bV^2 + cV^3. You can find the a b & c from the EPA website for new cars.
Trouble is, this still doesn’t factor in efficiency of the motor.
Bottom line, the answer comes from experiment or a bunch more calculations if you already know everything about the car.
Then, the pumping loss discussion is another good one. OEM’s are overcoming the pumping losses by opening the throttle valve and controlling fuel intake (off idle) using the intake valves. That’s BMW’s valvetronic and Nissan V-VEL and others. If you can precisely control the valve opening, you can leave the throttle plate open, or eventually, eliminate it like diesels or HCCI engines.
Drop the hammer and have fun! I drive a 4 cylinder 27 Roadster and get 21mpg. Works for me!
Take a look: http://www.MyRideisMe.com/Garage/pikesan
This may be an old thread, but in case anyone is wondering…
You can Roughly Calculate Road Load Coefficients A, B, and C!!
1) Find a very flat road. Flat means no slope AND no bumps.
2) Get a partner with a stopwatch and clipboard
3) Wait until there is NO wind and NO traffic
4) Don’t do anything illegal, e.g., speeding or blowing stop signs.
5) Use a consistent starting place
6) Drive up to the max speed, hold it steady
7) Same Time: start watch, shift to neutral (or release clutch) so engine doesn’t “suck” your speed
8) Let off gas pedal and say “Now” every time the speedometer drops 5 mph (or kph). This may be too fast at very high speeds.
9) Partner jots down time (helps with ‘Lap’ function) every time you say “Now” (Fill in the corresponding speeds later, or before hand)
10) Go all the way until the vehicle stops.
11) Do this as many times as you can, back and forth -> Collect tons of data!
Analyze:
You will notice at lower speeds it takes longer and longer to fall 5 mph. This is because the ACCELERATION decreases as speed decreases. Since Accel = Force / Mass, the force is decreasing as speed decreases. By going on a flat road, with no wind, no bumps, no slope, and disconnecting the engine, you’ll hopefully just have the one force left acting on the vehicle: Road Load!
Convert Speed to meters/second (mps)
Calculate acceleration between time steps.
A = (V(2)-V(1))/(T(2)-T(1))
Calculate “velocity” between time steps.
V = (V(2)+V(1))/2
Estimate the Road Load force by taking A*Mass.
F (N) = A (m/s^2) *Mass (kg)
Plot F as a function of V. In MS Excel you can just Add Trendline. You can add whatever trendline you want, but the best one to describe your car will be a 2nd order polynomial, then display equation on chart!
You’ll have to make your best guess on mass. Lookup your vehicle specs, and add passenger mass. You may also want to try a range of masses to see where they are.
Notes: 1 mph = 0.447 mps and 1 lbs = 0.454 kg
A Midsize Sedan:
A = 93.45 (N)
B = 3.58 (N)/(m/s)
C = 0.858 (N)/(m/s)^2
http://www.ecocarchallenge.org/docs/ecocar_rfp.pdf