Does Pulse and Glide only work with manual transmissions and a kill switch?

[Jurgen M]

When car drive uphill, gravity/weight has drag component, that is reason why you must add more throttle compare to ligther car at same constant speed uphill. So logicaly car increase fuel consumption.

But if you drive downhill, gravity/weight has thrust component, so your top speed("terminal velocity") will be higer compare to ligther car.(Do you agre with this?)

If we neglect cars acceleration (somehow) can both cars have same fuel consumption, because downhill and uphill cancel each out?
(If we assume that uphill and downhill section are equal on your trip and both cars has same aero drag)

So is acceleration only reason why heavier car has higher fuel consumption ?
Neglect rolling resitance which increase by weight..

 

[jbriggs444]

Real life is messier than our theories predict. But yes, if we model the engine as having constant efficiency regardless of power setting and we decide that the two cars are both being driven at the same constant speed regardless of slope then the energy consumption over a trip will be the same for both cars and, hence, the fuel consumption will be as well.

In fact, engines tend not to be equally efficient at all power settings. Your mileage may vary.

 

[Jurgen M]

Real life is messier than our theories predict. But yes, if we model the engine as having constant efficiency regardless of power setting and we decide that the two cars are both being driven at constant speed regardless of slope then the energy consumption over a trip will be the same for both cars and, hence, the fuel consumption will be as well.

In fact, engines tend not to be equally efficient at all power settings. Your mileage may vary.

But conclusion is that aero drag affect fuel consupmtion more then weight, if both cars has same average "mxa", especially at EV that can restore energy at downhill section?

 

[russ_watters]

But conclusion is that aero drag affect fuel consupmtion more then weight, if both cars has same average "mxa", especially at EV that can restore energy at downhill section?

Yes.

 

[jbriggs444]

But conclusion is that aero drag affect fuel consupmtion more then weight, if both cars has same average "mxa", especially at EV that can restore energy at downhill section?

Yes. I would conclude that.

Note that the argument I gave implicitly assumes regenerative braking at a physically impossible efficiency if the down slopes are extreme enough to reduce the engine power requirement below zero.

If you have 40% efficient conversion of stored potential energy to vehicle mechanical energy then the regenerative system must operate at 250% efficiency to maintain the same ratio.

If one is driving a hybrid, it could be only the final electric stage that is implicated. So it might be more like 90% efficient electrical propulsion requiring 110% efficient regeneration to match. Numbers pulled from thin air.

 

[Jurgen M]

Yes. I would conclude that.

Note that the argument I gave implicitly assumes regenerative braking (at a physically impossible efficiency) if?

In real life for EVs, what contribute more for better range, low aero drag (CdxA) or low mass?
Aero drag cant be restored.

 

[russ_watters]

...and it doesn't take much slope to reduce the engine power required to zero.

My gas car actually has selectable modes for downhill behavior. You can opt to leave the transmission engaged or to go neutral. It strikes me that neutral is counterproductive since an idling engine is all waste whereas rolling downhill in gear, gravity powers the accessories with zero fuel consumption.

 

[Jurgen M]

It strikes me that neutral is counterproductive since an idling engine is all waste whereas rolling downhill in gear, gravity powers the accessories with zero fuel consumption.

Yes, but cylinder compression and internal engine friction slow you down.
But in practice on downhill you are going too fast so put in gear is better..

 

[jbriggs444]

Yes, but compression and internal engine friction slow you down.

But we are assuming constant speed. Something will be slowing you down regardless. If not the engine, then the braking system. Or the cops.

Engine braking probably reduces the fuel delivered by the injectors below the idle requirements. (Though my last exposure to automobile engines was in the era of carburetors with venturis). @russ_watters makes the point that engine braking also unloads the engine from any load imposed by the air conditioner and other accessories that might require more than just idle power.

 

[Lnewqban]

... But if you drive downhill, gravity/weight has thrust component, so your top speed("terminal velocity") will be higer compare to ligther car.(Do you agre with this?)
...

According to Galileo, do heavier objects fall faster than light ones?
Please, see:
https://en.m.wikipedia.org/wiki/Frictionless_plane

Two identical bicycles; heavy rider and light rider with same leg’s strength: which one climbs faster?
Pedaling dowhill as fast as they can, which one descends faster?

Engines can only rotate as fast as they can properly breath.
Too many rpm will complicate breathing and reduce delivered torque.

They are designed mainly as a one way driving force, to impulse a mass from standing still up to the car’s aerodynamic terminal velocity, and to climb hills at max torque.

With fuel injection, when output torque is not necessary, the supply of fuel to the combustion chambers if shut-off.

 

[russ_watters]

Yes, but cylinder compression and internal engine friction slow you down.
But in practice on downhill you are going too fast so put in gear is better..

Right, as I said earlier it really doesn't take much of a hill to drop fuel consumption to zero. And even if it doesn't it is still a reduction.

 

[russ_watters]

Engine braking probably reduces the fuel delivered by the injectors below the idle requirements. (Though my last exposure to automobile engines was in the era of carburetors with venturis).

I've verified this with an ECM monitor. They report "throttle" and fuel flow at zero and the coolant cools noticeably in just a 30 second downhill run.

 

[pbuk]

According to Galileo, do heavier objects fall faster than light ones?

Does a cannonball fall faster than a balloon?

Please, see:
https://en.m.wikipedia.org/wiki/Frictionless_plane

Why do you think that helps answer this question?

Two identical bicycles; heavy rider and light rider with same leg’s strength: which one climbs faster?

Why do you think that is relevant: here it is stated that both cars travel at the same constant speed (and so clearly the heavier one must have "stronger legs" for this to be true while traveling uphill).

Pedaling dowhill as fast as they can, which one descends faster?

Why do you think that is relevant: here it is stated that both cars travel at the same constant speed?

 

[jack action]

In real life for EVs, what contribute more for better range, low aero drag (CdxA) or low mass?

EV or not, if you use your car mostly at a constant speed, especially high speeds (highway traffic for example), a lower aero drag is what you want since, as you already mentioned, weight only affects fuel consumption when acceleration is present.

Otherwise, it means you are constantly accelerating and decelerating, thus a lower mass is preferred, especially if done at lower speeds (city traffic for example).

 

[Lnewqban]

Does a cannonball fall faster than a balloon?Why do you think that helps answer this question?Why do you think that is relevant: here it is stated that both cars travel at the same constant speed (and so clearly the heavier one must have "stronger legs" for this to be true while traveling uphill).Why do you think that is relevant: here it is stated that both cars travel at the same constant speed?

I don’t know how to answer your questions, but I would love to read the reasons for which you find those statements of mine misleading and irrelevant.

I also hope to learn something for your future post that helps answer the OP’s questions.

 

[jrmichler]

When car drive uphill, gravity/weight has drag component, that is reason why you must add more throttle compare to ligther car at same constant speed uphill. So logicaly car increase fuel consumption.

When I had my previous truck, a GMC Canyon, I regularly drove between two places that were 72 miles horizontally and 800 feet vertically apart. My gas mileage was typically 3 to 4 MPG better driving downhill than uphill even though the average slope was only 11 feet per mile. I once did a theoretical calculation of the difference in energy due to the elevation change, and it agreed with the difference in measured gas mileage. I averaged 35 MPG in that truck, so the difference in MPG was about 10%.

 

[russ_watters]

When I had my previous truck, a GMC Canyon, I regularly drove between two places that were 72 miles horizontally and 800 feet vertically apart. My gas mileage was typically 3 to 4 MPG better driving downhill than uphill even though the average slope was only 11 feet per mile. I once did a theoretical calculation of the difference in energy due to the elevation change, and it agreed with the difference in measured gas mileage. I averaged 35 MPG in that truck, so the difference in MPG was about 10%.

Did you consider wind?

 

[Halc]

When car drive uphill, gravity/weight has drag component

A hill adds a work component. Don't confuse this with drag, the latter being a waste of energy. In theory, in a frictionless environment, a car can go up and down hills with no expenditure of energy at all. The moon certainly does that every month.

But if you drive downhill, gravity/weight has thrust component, so your top speed("terminal velocity") will be higer compare to ligther car.(Do you agre with this?)

For the same reason that a bit rock dropped on the moon hits the ground at the same time as a feather, no, the heavier car doesn't go faster down hills. It's all about the respective mass to coefficient of friction ratio, and larger vehicles invariably have larger coefficients of friction. It's harder to push a truck on level ground at a walking pace than it is to push a little car.

If we neglect cars acceleration (somehow) can both cars have same fuel consumption, because downhill and uphill cancel each out?

A well driven efficient car will get similar fuel efficiency on reasonably hilly terrain as it would on level roads. Big hills are a different story since one probably needs to brake going down them to maintain a reasonable speed. Higher speed decreases fuel efficiency on any terrain.

So is acceleration only reason why heavier car has higher fuel consumption ?

Acceleration doesn't happen at all while cruising. Yea, if there are starts and stops, it takes more work to do that with a bigger mass. But big things have more friction than little ones, so the little cars have less drag, and drag is what causes most fuel consumption.

Neglect rolling resitance which increase by weight..

If you neglect that, and the stop signs and such, then it takes zero fuel for any vehicle. Again, the moon is wonderfully fuel efficient despite massing more than a battleship. But it has no rolling resisitance.

It strikes me that neutral is counterproductive since an idling engine is all waste whereas rolling downhill in gear, gravity powers the accessories with zero fuel consumption.

Not so. A car uses considerable fuel while off the gas while in gear. You can tell because of the significantly increased drag if you actually shut off the fuel injectors. Leaving it in gear is engine braking. There's a lot of friction and associated lost fuel efficiency by keeping your engine RPM high while it isn't doing any work. Most of the mechanical friction is in the engine.

I have one favorite spot where I can get from the top of one (usually) gentle hill all the way to the stop sign 16 km away. I can't do that if I leave the thing in gear but foot off the pedals. I honestly have no way of measuring fuel consumption of 22 minutes of engine idle vs 32 minutes of the wheels pushing a high RPM drive train with theoretically sub-idle fuel consumption. At times the engine would definitely be accelerating the car, which sort of disqualifies it as coasting.

 

[pbuk]

The answers here saying "if you ignore aerodynamic drag and rolling resistance and assume 100% efficiency then the heavy car will use the same amount of fuel as the light car" are wrong.

1. Consider the amount of work done in ascending the hill: the heavy car's engine will do more work and therefore use more fuel.
2. When descending the hill the light car doesn't use any fuel. In order to make up for the additional fuel it used on the way up the heavy car would have to use less than no fuel, which is not possible.
3. Also the constraint that both cars maintain the same constant speed throughout is violated: with no drag both cars will accelerate downhill.
4. If instead the cars travel the downhill leg first, then neither will use any fuel, however the constant speed constraint will still be violated.
5. If instead we consider an electric vehicle with 100% efficiency, and also 100% efficient regenerative braking and make the same drag-less assumption then by conservation of energy both vehicles use no net energy in the round trip, and they can fulfill the constant speed constraint.
6. If we add air resistance (the same for both cars) as in the OP things get more interesting.
7. On the uphill leg the cars' engines will do work equal to $mgh + f_ds$ where $f_d$ is the drag force (which is constant because the speed is constant) and $s$ is the distance travelled: the heavier car does $\Delta_m gh$ more work where $\Delta_m$ is the difference in the mass of the cars.
8. On the downhill leg gravity will do work at the rate of $mg\frac{dh}{dt}$, working against air resistance at $f_d \frac{ds}{dt}$, and in order to maintain speed each cars engine must work at a rate of $f_d \frac{ds}{dt} - mg\frac{dh}{dt}$. Now the heavier car's engine works at a rate $\Delta_m g\frac{dh}{dt}$ less than the lighter car and in the total descent does $\Delta_m gh$ less work.
9. Because the heavier car does the same amount less work in the descent as it did more in the ascent, it uses the same amount of fuel.

We now have enough information to completely answer the questions in the OP.

When car drive uphill, gravity/weight has drag component, that is reason why you must add more throttle compare to ligther car at same constant speed uphill. So logicaly car increase fuel consumption.

Yes a heavier car uses more fuel than a lighter car when traveling uphill (other things being equal).

But if you drive downhill, gravity/weight has thrust component, so your top speed("terminal velocity") will be higer compare to ligther car.(Do you agre with this?)

I am not sure it is helpful to think about "thrust", however it can be seen from the expression for work in 8 above that if $mg\frac{dh}{dt} > f_d \frac{ds}{dt}$ then the car will accelerate. Because (all things being equal) the term on the left hand side is greater for a heavier car then yes, other things being equal, terminal velocity is greater for a vehicle with a greater mass. This is why we have escape lanes (aka runaway truck ramps).

If we neglect cars acceleration (somehow) can both cars have same fuel consumption, because downhill and uphill cancel each out?
(If we assume that uphill and downhill section are equal on your trip and both cars has same aero drag)

Yes, we can see from conservation of energy that fuel consumption is equal because "downhill and uphill cancel out", however the analysis in the numbered points above shows that there is an additional constraint that aerodynamic drag is sufficient to require both cars to use non-negative engine power downhill to maintain speed.

So is acceleration only reason why heavier car has higher fuel consumption ?

No, rolling resistance is the principal reason heavier cars have higher fuel consumption (other things being equal).

Neglect rolling resitance which increase by weight..

You can't neglect rolling resistance because (if aerodynamic drag is assumed equal) this is the principal factor affecting fuel consumption.

 

[jrmichler]

Did you consider wind?

And a few other factors. I made that round trip once per week for 11 years, so got a good idea of what affected the mileage. Headwinds vs tailwinds affected mileage by at least 5 MPG. Temperature had an effect. I averaged 32 MPG in winter, and 38 MPG in summer. The average winter temperature was about 60 deg F lower than the summer average, so 10 deg F was worth 1.0 MPG. A more complete discussion is at: https://ecomodder.com/forum/showthread.php/modding-06-gmc-canyon-17070.html.

The effect of uphill vs downhill on gas mileage is more complex than simple physics energy calculations. Engine efficiency has a large effect. Engine efficiency varies with RPM and the percent torque, and is shown in a performance map. A typical engine performance map is shown below:

The contour lines are lines of constant efficiency. The most efficient operating point is near 2500 RPM at about 125 Nm torque. The red dots are points of 10 hp at various RPM's. Ten hp at 1000 RPM burns about 295 gr/kW-hr, while ten hp at 4000 RPM burns 450 gr/kW-hr, or 50% more fuel to generate the same power. That's only one reason why gas mileage cannot be predicted by only considering the laws of physics.

 

[russ_watters]

The answers here saying "if you ignore aerodynamic drag and rolling resistance and assume 100% efficiency then the heavy car will use the same amount of fuel as the light car" are wrong.

Thank you. Things are starting to get messy here and they started off so well. This is not an ideal situation. These approximations others added are wrong for the real world, and the assumptions that are appropriate here should be fairly obvious or clearly assigned in a way that agrees with reality.

Let's keep our eye on the ball, folks!

 

[russ_watters]

@jrmichler Holy cow! I should have known!

 

[russ_watters]

Not so. A car uses considerable fuel while off the gas while in gear. You can tell because of the significantly increased drag if you actually shut off the fuel injectors. Leaving it in gear is engine braking. There's a lot of friction and associated lost fuel efficiency by keeping your engine RPM high while it isn't doing any work. Most of the mechanical friction is in the engine.

If you're coasting at highway speed you might have a relatively high rpm, but otherwise you won't. And sure, engine and drivetrain drag is a significant fraction of overall engine load/fuel use. At highway speed it can be a quarter, slower it might be half. But all these losses are small compared to the energy gain or loss from a hill. The math on this is straightforward:

Say you have a 4,000 lb car that gets 30 mpg at 60 mph. From prior calculation that's about 17 horsepower or, obviously, 2 gal/hr. At idle a car uses around half a gallon per hour. 17 horsepower on a hill is 2.3 ft/sec of elevation change or at 60 mph a 2.6% grade. That's all it takes to shut off fuel flow to the engine and use engine braking to maintain constant speed. Or if your rpm is 4x idle a 4.6% grade.

 

[jack action]

1. Consider the amount of work done in ascending the hill: the heavy car's engine will do more work and therefore use more fuel.

Agreed.

2. When descending the hill the light car doesn't use any fuel. In order to make up for the additional fuel it used on the way up the heavy car would have to use less than no fuel, which is not possible.

Not true. with no energy input or output, the car will necessarily have increased its velocity at the bottom of the hill. That is "stored fuel" to do extra work. You cannot neglect that.

3. Also the constraint that both cars maintain the same constant speed throughout is violated: with no drag both cars will accelerate downhill.

Agreed. But the velocity doesn't matter: if a vehicle goes from height $h_i$ to $h_f$, and then goes back to $h_i$, considering only that point, the net energy will always be zero.

4. If instead the cars travel the downhill leg first, then neither will use any fuel, however the constant speed constraint will still be violated.

Agreed.

5. If instead we consider an electric vehicle with 100% efficiency, and also 100% efficient regenerative braking and make the same drag-less assumption then by conservation of energy both vehicles use no net energy in the round trip, and they can fulfill the constant speed constraint.

This proves the point made earlier: You restrain the acceleration while going downhill and instead of increasing the kinetic energy of the vehicle, you store it in some battery, spring, or rotating mass and use it to do extra work later on (doesn't have to be going uphill).

6. If we add air resistance (the same for both cars) as in the OP things get more interesting.

No, it doesn't. It is just adding a different problem completely independent of the previous one.

7. On the uphill leg the cars' engines will do work equal to $mgh + f_ds$ where $f_d$ is the drag force (which is constant because the speed is constant) and $s$ is the distance travelled: the heavier car does $\Delta_m gh$ more work where $\Delta_m$ is the difference in the mass of the cars.

8. On the downhill leg gravity will do work at the rate of $mg\frac{dh}{dt}$, working against air resistance at $f_d \frac{ds}{dt}$, and in order to maintain speed each cars engine must work at a rate of $f_d \frac{ds}{dt} - mg\frac{dh}{dt}$. Now the heavier car's engine works at a rate $\Delta_m g\frac{dh}{dt}$ less than the lighter car and in the total descent does $\Delta_m gh$ less work.

Although all of this is true, it is actually hiding what is really going on.

This is a case of superposition of two independent problems:

1. The car will go uphill and downhill and, assuming both initial and final levels are the same, this results in no energy requirement;
2. There is a drag force $F$ acting on a path of length $s$ requiring an energy input of $Fs$.

The total energy required is the sum of both: $0+ Fs = Fs$. In fact, it doesn't matter:

• whether $F$ is constant or not (i.e. velocity is constant);
• whether the resistance force is a drag force, rolling resistance, coming from a big magnet pulling the car back, or the sum of all of these forces;

The total energy required will always be $\int_{s_1}^{s_2} F_{net}ds$, just like assuming there is no uphill or downhill.

9. Because the heavier car does the same amount less work in the descent as it did more in the ascent, it uses the same amount of fuel.

Yes, it does the same amount of work, i.e. drag force times the distance traveled. Your equation in point #8 $f_d \frac{ds}{dt} - mg\frac{dh}{dt}$ only means (assuming constant velocity):
$$f_d \frac{ds}{dt} = mg\frac{dh}{dt}$$
$$f_d\cancelto{}{\frac{ds}{dt}} = mg\frac{dh}{ds}\cancelto{}{\frac{ds}{dt}}$$
$$f_d = mg\sin\theta$$
If we assume the slopes of the hill $\theta$ are the same on both sides (which it doesn't have to be) and a constant drag force (constant velocity), then:
$$E_{up} = f_d\frac{s}{2} + mgh$$
$$E_{down} = f_d\frac{s}{2} - mgh$$
$$E_{total} = E_{up} + E_{down} = f_d s$$
Assuming the slopes are different will only create a different ratio of distance traveled uphill versus downhill, but, in the end, only the total distance traveled $s$ matters.

 

[Jurgen M]

no, the heavier car doesn't go faster down hills.

Yes it does.
Do you understand that parachute with 10tons weight fall down faster the parachute with 100kg weight?

In vacuum and frictionless world ,they acceleration downhill will be the same

 

[pbuk]

Not true. with no energy input or output, the car will necessarily have increased its velocity at the bottom of the hill. That is "stored fuel" to do extra work. You cannot neglect that.

The question is about fuel consumption which is measured in either mpg (miles per gallon, US or Imperial) or l/100km (litres per 100 km). Nowhere does "stored fuel to do extra work" feature in mpg or l/100km.

Also the "stored fuel" argument assumes that you can increase the speed of the car arbitrarily. In order to compensate for any reasonable height of ascent this soon becomes absurd. Note that the final speed after starting a frictionless descent from height $h$ with initial speed $v_0$ is $v = \sqrt{v_0^2 + 2 g h}$, so with an inital speed of 25 m/s (about 55 mph) and height of 100 m (about 330 feet) you end up more than doubling your speed to about 110 mph.

Agreed. But the velocity doesn't matter: if a vehicle goes from height $h_i$ to $h_f$, and then goes back to $h_i$, considering only that point, the net energy will always be zero.

The net energy change may be zero, but if we have converted some of our fuel to kinetic energy then the fuel consumption goes up. Again, the question is about fuel consumption.

This proves the point made earlier: You restrain the acceleration while going downhill and instead of increasing the kinetic energy of the vehicle, you store it in some battery, spring, or rotating mass and use it to do extra work later on (doesn't have to be going uphill).

There is nothing about this in the original question.

 

[Jurgen M]

Problem is IC engine cant store energy when going downhill,computer just kill fuel 0.0L/100km on downwhil leg and that is it.
But on downhill leg you will get higher top speed with heavier car so you can use this momentum to move longer before you again must push gas pedal.

But I think from downhill/uphill trip from point A to B you will allways use more fuel with heavier car, even both car have same average speed.

 

[Devin-M]

In my car I can regularly get 40mpg even though it’s listed at 33mpg highway. (I do check every time I fill up by dividing the miles traveled by the gallons pumped).

In the mountains, on the downhills it’s beneficial to be in neutral coasting as this drops the engine rpm from around 2500rpm while coasting to around 750rpm.

On the flat highway, I note the lowest rpm that can be achieved while in the highest gear. This results in me cruising at about 61mph even though the speed limit is 70mph. Fortunately the speed limit is 55mph for tractor trailers but they usually go about 61mph, and being a short distance behind one of those helps the fuel economy considerably by reducing the air resistance/drag. Also, turning off the air conditioner and making sure the tires are fully inflated helps. The car is a Nissan Versa.

Only if I drive slowly (40-50mph) uphill in the mountains such that my RPMs don’t exceed 2500, and coast in neutral on the downhills will I get similar 40mpg as I do while cruising at 61mph on a flat freeway.

 

[Motore]

In the mountains, on the downhills it’s beneficial to be in neutral coasting as this drops the engine rpm from around 2500rpm while coasting to around 750rpm.

I also thought that back in the time, but it actually consumes more fuel being in neutral than being in gear on downhill (with no foot on the gas).
https://www.popularmechanics.com/cars/hybrid-electric/a5977/coasting-in-neutral-fuel-economy/

 

[anorlunda]

I also thought that back in the time, but it actually consumes more fuel being in neutral than being in gear on downhill (with no foot on the gas).
https://www.popularmechanics.com/cars/hybrid-electric/a5977/coasting-in-neutral-fuel-economy/

Does anyone remember the BMW ad from the 80s that said in bold letters 0 mpg with BMW? The picture showed a BMW going downhill. Of course, the ad was just a clumsy translation from the German version which said 0 l/km. But I think they were advertising free wheeling when in the ECO mode.

Does anyone remember the Saab 96 two stroke with "free wheeling"? Owners of that car loved it and bragged about it.

Ditto the Austin mini. Ditto Volvo V40. The CLS [what does CLS stand for?] goes in neutral and stops the engine when coasting.

I think the key point is fuel injection. Carbureted cars can't achieve zero fuel flow when running.

 

[russ_watters]

In the mountains, on the downhills it’s beneficial to be in neutral coasting as this drops the engine rpm from around 2500rpm while coasting to around 750rpm.

I also thought that back in the time, but it actually consumes more fuel being in neutral than being in gear on downhill (with no foot on the gas).
https://www.popularmechanics.com/cars/hybrid-electric/a5977/coasting-in-neutral-fuel-economy/

Again, this will depend on the slope and what you are trying to achieve speed-wise. If the slope is pretty mild and the speed high it probably makes sense to idle in neutral. But that's probably a pretty narrow range of situations. For my car anyway it doesn't take much slope before it will do a runaway acceleration from highway speed in idle. So maybe the situation can be summed-up: if you're in neutral and on the brakes you're wasting energy that could be used to spin your engine and run your accessories.

I did a quick calc earlier, but it may be possible to calculate exactly the range of speeds and slopes where it makes sense to idle (again, I expect it is very narrow)...

That link, btw, starts off talking about coasting in idle to a stop sign/light. Maybe that's some really lazy manual transmission stuff? At worst I'd just leave it in my top gear until I was going slow enough I had to get out or stall. In my newer automatic with paddle shifters I often downshift while coasting to a stop. Better for fuel economy and better for your brakes than being in neutral.

Carbureted cars can't achieve zero fuel flow when running.

That I wasn't aware of, but then I don't think I've ever driven one. Is that because there's no fuel flow control except via carburetor airflow throttling, and that never goes to zero (otherwise you pull a vacuum at the engine)?

 

[anorlunda]

That I wasn't aware of, but then I don't think I've ever driven one. Is that because there's no fuel flow control except via carburetor airflow throttling, and that never goes to zero (otherwise you pull a vacuum at the engine)?

There's an idle screw that keeps the throttle open enough to set idle speed.

You're no spring chicken Russ. I can't believe that you never drove a car prior to fuel injection. I never drove prior to the invention of the wheel, but it was close.

 

[russ_watters]

You're no spring chicken Russ. I can't believe that you never drove a car prior to fuel injection. I never drove prior to the invention of the wheel, but it was close.

Google tells me fuel injection went mainstream in the '70s. How old do you think I am?

I've had a few carbureted devices, but haven't dug into them enough for that. Thinking back to the RC plane I built as a kid, I remember it had a screw/pin for adjusting mixture, but no idle setting (probably because you want to be able to cut it off with the throttle). I'm flying real ones now, and I guess it probably has one, but it isn't for me to adjust.

 

[anorlunda]

Google tells me fuel injection went mainstream in the '70s. How old do you think I am?

LOL. I didn't mean to suggest ancient. But I did assume that most of us started with a 5-10 year old car.

Also the mainstream takeover of fuel injection was gradual. Wikipedia says:

Manifold injection was phased in through the latter 1970s and 80s at an accelerating rate, with the German, French, and U.S. markets leading and the UK and Commonwealth markets lagging somewhat. Since the early 1990s, almost all petrol passenger cars sold in first world markets are equipped with electronic manifold injection.

So up until 1995-2000 there were still numerous carbureted cars on the roads in the USA.

 

[Devin-M]

I also thought that back in the time, but it actually consumes more fuel being in neutral than being in gear on downhill (with no foot on the gas).
https://www.popularmechanics.com/cars/hybrid-electric/a5977/coasting-in-neutral-fuel-economy/

Again, this will depend on the slope and what you are trying to achieve speed-wise. If the slope is pretty mild and the speed high it probably makes sense to idle in neutral. But that's probably a pretty narrow range of situations. For my car anyway it doesn't take much slope before it will do a runaway acceleration from highway speed in idle. So maybe the situation can be summed-up: if you're in neutral and on the brakes you're wasting energy that could be used to spin your engine and run your accessories.

The article was an interesting read, and I didn't know the fuel injectors would bring the fuel flow to 0% while coasting in gear on a downhill.

That said I don't believe all factors have been considered in the article. When I was talking about coasting downhill in neutral in the mountains, for me, I meant going 60-70mph downhill for 5 miles or so at a time while in neutral on a hill like this:

From what I gathered in the article, even if fuel injectors aren't injecting any fuel at all while the car is in gear coasting downhill, surely keeping the engine turning at 2500rpm with no fuel being consumed is an additional source of drag above and beyond the air resistance and rolling resistance.

If my memory serves, and I do keep the car in gear while coasting downhill, it will lose speed faster than if it was in neutral, and so I might need to push the gas a little bit to maintain speed, whereas if I were to keep the car in neutral, it will maintain speed on its own from the release of gravitational potential energy.