It seems that vehicle MPG would be best on flat ground rather than up and down hills. That is, you’d burn so much fuel going up hill, that the “free ride” going back down hill would not be worth it.
However, according to my car’s dashboard read out, this is not the case (I get instanaeous and average MPG readouts on my Passat).
Here’s what happened - Driving up a fairly steep mountain road at Mt. Rainier (2000 foot gain in 10 miles) I averaged 13 MPG from base to top. The way back down was probably 75% coasting and braking, I averaged about 67 MPG. That would be an average of 40 mpg…right?
The car usually gets about 28 MPG highway.
Am I doing bad math? Should I quit spending so much time reading my dashboard and spend more time with my eyes on the road?
Note that Hoodoo was using a theoretical distance, but it doesn’t matter since it ends up dividing out at the end.
Your problem was that you averaged the Miles per Gallon, but this is incorrect. What you can do is average the Gallons per Mile, and then invert that to get your answer. So:
I believe the best MPG comes when you have a series of short & rather steep uphill stretches, each followed by a downhill stretch that’s quite gradual (though steep enough to allow the car to roll). You drive uphill at a speed that corresponds to an efficient engine RPM, then shut off the engine and coast downhill.
Suppose the OP’s case were altered to be a 10-mile, 2000’ climb followed by a 20-mile descent. Shutting the engine off at the top and coasting down would then produce 39mpg.
Ignore your dashboard. The car’s computer simply doesn’t have the information available to make a meaningful calculation. For example, the fuel gauge is only approximate, and there is no fuel flow sensor anywhere.
Now, back to your basic question. Yes, your mileage suffers going uphill. Downhill, you can’t really re-use all the energy you stored going uphill. Even if you were foolhardy enough to coast and use no brakes on the downhill scream, you’d have to keep the engine running for power steering.
Remember, the energy you used to gain forward motion is converted into heat when you touch the brakes. You use energy by touching either the go or stop pedals. Think of dropping a stream of dimes on the street when you touch the gas, and a stream of nickels when you touch the brake. The longer you go without touching either pedal, the more fuel you save.
Disc brakes and tires drag all the time, but that’s the same whether you’re uphill or downhill.
Your answer is superficially persuasive. Asknott, however, refutes it on thermodynamic grounds at a “whole of car” level:
AskNott is correct here.
Your (Xema’s) argument fails at the “engine” level, as well as at the “whole of car” level.
You’ve assumed that there’s “an efficient engine RPM” going uphill which is more efficient than the engine RPM required when travelling along a flat surface. There’s no evidence to support this assumption.
In fact, the evidence is to the contrary: travelling in high gear on a flat surface with low engine RPM is more efficient than travelling uphill in lower gear with higher engine RPM (setting aside the energy being stored as potential energy, which AskNott has already dealt with). The higher engine RPM necessarily implies increased frictional losses in the engine, and the lower gearing necessarily implies increased frictional losses in the gearbox. Those frictional losses cannot be reclaimed.
Actually, you can use the 40 mpg number, but just divide it into the product of 13 and 67. So the average mpg is (1367)/40 = 21.775, as already pointed out by Hoodoo and flight. So assuming that the two mpg’s are over the same distance, the average mpg is just (mpg1mpg2)/(average of mpg1 and mpg2). This is something you can do in your head.
Given a hill of the right steepness, no brakes are required.
In no car that I’ve owned has it been true that you lose steering when the engine is off. It is true that you lose the power steering boost, so steering forces become noticeably higher. But a running engine is not required. (And I’d expect this to be true with most or all models, since otherwise failure of the belt that drives the power steering pump would lead to an uncontrollable vehicle.)
I don’t believe I have assumed that. I’ve simply assumed that for a given steep hill, there’s a most efficient speed at which to climb it - driving either slower or faster will give fewer mpgs. I propose that you climb at that speed, then turn off the engine and use no fuel at all while you coast down a shallower slope that takes you substantially further than the distance you traveled up the hill.
I think they can be nicely relaimed in the coasting phase, because they have earned you the opportunity to travel a substantial distance without burning any fuel.
It’s worth noting that your average speed over the whole trip will be rather low. If a real-world car on a level surface matched this speed (perhaps out of a desire to suffer a very low total air drag penalty) it would have to travel at a speed well below what’s efficient (given the normal size of an auto engine) and so could not match the “climb and coast” scheme. If you could equip the car with an efficient engine just big enough to move it at the desired low speed, it would probably become quite competitive. In short, the climb and coast scheme takes advantage of the fact that car engines are substantially more powerful than what’s needed for low speeds on level roads.
I disagree. Xema’s “rather short but steep uphill stretches” would require the engine to operate at a higher load going up the hill. IC engines are, in general, more efficient at higher loads than at lower loads – in some cases, multiple times more efficient.
What Xema is proposing is essentially how a series hybrid automobile works. The engine is turned on to work in a high power, high efficiency mode (going uphill, in this case); energy is then stored for later use (as height energy instead of battery energy); finally, this stored energy is drawn on to power the car when the engine is turned off (coasting downhill).
It’s true that you won’t get back all the energy “stored” in the form of car height, but the question is whether or not that inefficiency will trump the efficiency gain from using the engine at higher efficiency points. The answer to that would depend on the size of the hills (equivalent to the design strategy of a hybrid) and how the car is driven, but it seems likely to me that an up-and-down, engine-off strategy would wind up being more efficient.
I had always heard that your engine would not use any gasoline on a downhill with your foot off the gas. Am I missing something, or are we talking about throwing the car in neutral and killing the engine in order to avoid engine braking on the downhill?
This is incorrect. Your engine is still firing, so gas is being used. In any car you can put it in neutral, but the engine will still use some gas at idle. Something is coming out of the tailpipe.
Granted, in neutral something still needs to come out of the tailpipe.
I wasn’t certain that the engine needed to fire if I left the car in gear and then took my foot off the gas on a steep downhill. It had seemed to me that at least in theory, gravity would handle the job of turning the engine for me. I realize that the nature of my carburator (sp?) or EFI system/and/or PCM may determine what actually happens in this scenario.
Mild hi-jack: is there anywhere on the planet I could find an engine efficiency map for the 3 liter Duratec in my Mercury Sable like the one posted for a previous earlier in this thread?
Actually Jonathan Woodall had it right the first time. On a modern EFI car the fuel to the injectors is cut off when the throttle is closed. This is done for emission purposes. There is no, none zip reason to piss fuel out the tail pipe when the car is coasting down. Running the injectors during this period would only increase the total emissions that the car produces. Now obviously the fuel cannot remain turned off or the car would stall, so there is a point where the fuel is turned back on, on the cars I teach on that is between about 1400- 1800 RPM depending on other factors. So if I head down a hill and take my foot off the gas, no fuel will be injected into the engine until the RPM drop below the 14-1800 RPM level.
And while I am picking nits
Modern trip computers (at least the ones on the cars I teach on) don’t use the fuel level sensor to calculate fuel mileage. The have an input direct from the Engine Control Module about the amount of fuel injected into the engine by the injectors, and the engine RPM. This is a very accurate signal. In the interest of full disclosure the trip computer does in fact use the fuel level sensor signal for the miles to empty calculation, which can be somewhat in error. (or wildly inaccurate if you prefer)
Jonathon Woodall, let me fill in some of the spaces.
When you go downhill with the transmission engaged, it does sorta push the engine, and the engine acts as a brake (the drivetrain has to work to spin the engine.) However, the engine still sips a little fuel. Without any fuel, the engine would turn into a big air pump, and the braking effect would be much greater. (In the days of carburetors, the “idle circuit” ran pretty rich, and engine braking used a lot of fuel. Computerized fuel injection is much stingier for that.)
Yes, we are talking about coasting in neutral; Xema is talking about doing it with the engine off, and I assert that running downhill with no power steering or power brakes is too risky in case of emergency maneuvers.
Disclaimer! This is hypothetical, folks. Howling downhill into Nashville or Madison in neutral, with or without the engine, is not safe. Professional Dopers on a closed course. Do not attempt. This intellectual exercise ignores the hazards of overheating the brakes on a steep downhill. In the Mobil Fuel Economy competition, (before EPA gas mileage ratings) drivers actually used techniques like these, but they were paid to be crazy.
What you’re saying is true, but highly misleading. The engine will have a frictional drag that depends upon engine rpm, but not upon throttle opening. Let’s say we have a 100 kW engine that has a constant drag of 10 kW at a fixed speed, say 4,000 rpm. As we vary the throttle opening, we observe the following:
The relative efficiency (gross engine output/net engine output) does, indeed, increase with engine load. However, that doesn’t mean we can save fuel by running with a higher engine load.
Let’s try an example. We’re going to run the car at a constant speed, and it takes, say, 20 kW to overcome wind and rolling resistance at that speed. We’re going to run it for 50 minutes.
For our engine example above, if we run on the flat, that’d require a net engine output of 20 kW, a throttle opening of 20%, and a gross engine output of 30 kW for 50 minutes, for a total gross energy output of 30 kW x 50 minutes = 1,500 kW.minutes.
Let’s also assume that we’ve found a good hill, where we can climb for 10 minutes, and then descend for 40 minutes with the throttle closed. During the descent we’re extracting potential energy at a rate of 20 kW (to overcome wind and rolling resistance). During the climb we have to store potential energy at a rate of 80 kW, plus overcome the 20 kW wind and rolling resistance, so we need 100 kW net engine output, a throttle opening of 100%, and a gross engine output of 110 kW. Total gross energy output is 10 kW x 40 minutes + 110 kW x 10 minutes = 400 + 1,100 = 1,500 kW.minutes. That is, there’s no fuel saving.
In this example, we’ve assumed perfect energy recovery, and we know that that’s not possible. We’ve also assumed a prefect hill, but if in reality we have to do any one of the following:
a) touch the gas pedal
b) touch the brake pedal
c) change down a gear either going up or down the hill,
I suppose I should add in the detail of why it isn’t possible to fully recover the potential energy that’s been stored in the car, since my contrived example obscures that bit.
I made the assumption in the example that the trip would take exactly 50 minutes at exactly the same speed, regardless of whether the trip was on the flat or over the hill. In reality, it’s not possible to hold both the time and the speed constant, because the distance travelled when going over the hill is longer than that travelled on the flat, even though the end points are exactly the same distance apart on the map.
If we hold the speed constant, it takes longer to go over the hill, so the engine is running for a longer time, and uses more fuel.
If we hold the time constant, we have to go faster over the hill, so the wind and rolling resistance is higher, and uses more fuel.
I’ve assumed that engine drag remains constant with increasing engine loading, but drag actually increases with increasing load, even at constant speed, because of increasing forces on bearings and cylinder walls.
I’ve ignored the increased frictional losses in the transmission that occur with increased forces between the teeth, and the increased slip in the torque convertor (if there is one), both of which occur with increased load, even at constant speed.
Part of the engine drag can be reduced by shutting down the engine. Another part can’t (or at least, shouldn’t). The friction of the engine bearings, the pistons, the cams and the valves, can be cut out. The energy required to drive the power steering pump, maintain power braking vacuum, maintain battery charge, cool the engine, run the air-con compressor, etc, will either be dangerous to eliminate, or will be recovered through higher fuel usage once the engine is restarted.
