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So I understand that a lower gear as advantageous when accelerating a vehicle or when driving up hill (e.g because you are having to accelerate against gravity). I know this is because of the increased torque measured at the gearbox output due to to the leverage ratio provided by the gearbox, so that each rotation of the wheel has more force behind it.

The question is that if a constant speed (say 50mph) is maintained using two different gears (a high one and a low one), then it will be more energy efficient to usually use the higher gear to maintain the speed. This is because the engine has a lower RPM and therefor is firing fewer times per wheel rotation and the amount of energy used per engine revolution is fixed (not precisely true I know).

But if speed is held constant (and other factors like friction/drag on the vehicle) then the amount of energy that "goes in" to moving the car during each WHEEL rotation should be the same, but if the lower gear is using more energy to achieve this, then that implies that there is extra energy being used and that energy must be going somewhere, but what I don't understand is what happens to that energy? I know there is some that is lost due to "losses" inside the engine that are inherently caused by the higher RPM, but I'm not sure if this would account for all of it.

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    $\begingroup$ you have basically constructed a scenario whereby everything after the transmission is kept constant, so must be that the energy input and losses prior to that that must be the ones changing. Why do you think you cannot account for all of it that way? $\endgroup$ Commented Apr 15 at 1:56
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    $\begingroup$ I guess you can, but I'm not sure how it would all be accounted for in a way that "scales" with the change in gear ratio $\endgroup$ Commented Apr 15 at 2:00
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    $\begingroup$ Have you ridden a bicycle with different gear levels? What happens to your energy when keep the speed the same and change gears? The same is true for a car. $\endgroup$
    – Barmar
    Commented Apr 15 at 14:25
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    $\begingroup$ Where does the extra energy go when you sit with your car in neutral and the engine idling? Why can't you sit there forever without using gas? $\endgroup$ Commented Apr 15 at 22:30
  • $\begingroup$ There is both dry friction and fluid friction in the engine. Note that "overdrive" is a gear ratio that is higher than what would give the car it's highest speed. It is used for allowing a lower RPM at cruising speed for higher efficiency; "ecodrive" would be a better term. $\endgroup$ Commented Apr 16 at 18:48

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the amount of energy used per engine revolution is fixed (not precisely true I know)

This is the error. The work per engine revolution is the torque at the engine.

If the energy dissipation per wheel rotation is the same in both cases then the dissipative torque at the wheel is the same. The gear ratio will decrease the engine rpm by the same ratio as it increases the torque at the engine. This changes the work per engine revolution by the gear ratio.

it will be more energy efficient to usually use the higher gear to maintain the speed

Engines, particularly internal combustion engines, do not have a fixed efficiency at all rpm. Where your statement is true, it is due to increased efficiency at lower rpm.

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    $\begingroup$ The important concept that describes efficiency as a function of load (and therefore gear ratio) is called "Brake Specific Fuel Consumption" $\endgroup$
    – Señor O
    Commented Apr 15 at 4:14
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    $\begingroup$ Engines, particularly internal combustion engines, do not have a fixed efficiency at all rpm that's basically the answer. $\endgroup$ Commented Apr 15 at 12:35
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    $\begingroup$ Well, the last sentence is, I believe, exactly what the OP is asking. Clearly, Driving 30 mph in first gear at 6000 rpm guzzles more gasoline than doing it at 1300 in 4th gear -- but both have the same useful energy output: Both are overcoming all resistances (wind, rolling, mechanics) to go 30 mph. Where's the delta !? $\endgroup$ Commented Apr 16 at 14:41
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    $\begingroup$ @Peter-ReinstateMonica the delta is in heat, noise, mechanical wear, and exhaust $\endgroup$
    – OrangeDog
    Commented Apr 17 at 16:53
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    $\begingroup$ @Peter-ReinstateMonica To the first order, for a gasoline engine, variable losses come from the throttle at low RPMs (converted into charge air heat) and at high RPMs, viscous losses from lubrication oil and charge flow dominate. The best specific fuel consumption (g/kWh) comes from high powers (open throttle) at mid RPMs. $\endgroup$
    – user71659
    Commented Apr 17 at 19:26
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Ignoring all losses from the gearbox on to the wheels, let’s just look at the engine.
Internal combustion engines are about 30% efficient. About 65% of the energy is lost as heat, 5% as friction. from
engine losses

and from Wikipedia "An engine has many moving parts that produce friction. Some of these friction forces remain constant (as long as the applied load is constant); some of these friction losses increase as engine speed increases, such as piston side forces and connecting bearing forces (due to increased inertia forces from the oscillating piston.) . . . Along with friction forces, an operating engine has pumping losses, which is the work required to move air into and out of the cylinders.
This pumping loss is minimal at low speed, but increases approximately as the square of the speed, until at rated power an engine is using about 20% of total power production to overcome friction and pumping losses."

When you increase rpm you are increasing all of these losses. That is why many cars have overdrive - the gearing is set up to increase wheel rpm with reduced engine rpm.

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  • $\begingroup$ Re, "many cars have overdrive." I always thought that "overdrive" meant that the car's drive shaft rotated faster than the engine. But, the majority of new cars have a front wheel transaxle. There's no drive shaft. What does "overdrive" mean when there's no drive shaft? $\endgroup$ Commented Apr 15 at 14:56
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    $\begingroup$ @SolomonSlow from that wiki article With the popularity of front wheel drive cars, the separate gearbox and final drive have merged into a single transaxle. There is no longer a propeller shaft and so one meaning of "overdrive" can no longer be applied. However the fundamental meaning, that of an overall ratio higher than the ratio for maximum speed, still applies: higher gears, with greater ratios than 1:1, are described as "overdrive gears".[ $\endgroup$
    – Rich
    Commented Apr 15 at 15:02
  • $\begingroup$ A lot of "pumping losses" dissipate heat in the throttle--a device which is expressly designed to waste energy (every joule of energy wasted in the throttle will, as a side effect, reduce the amount of fuel fed into the engine, and thus reduce by many joules the amount of energy produced by burning such fuel, so throttling an engine wastes less energy than would e.g. driving with the brakes engaged, but the losses are still significant). $\endgroup$
    – supercat
    Commented Apr 17 at 15:14
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If you are in a situation where the energy used to keep the car moving is unchanged, but you are putting more energy into the engine, then the difference must be losses in the engine. Whether you are sure about it or not, physics says the energy needs to go somewhere, and in the constrained situation you give, the only place it can go is losses.

As to how the balance works out, the key is that you constrained the work done by the tires. You stated that it must be constant. The only way for this to be true is if you adjusted the gas input until it was true. Thus, by definition you will increase the amount of fuel put into the engine until the extra fuel offsets the losses.

What might be confusing you is that fuel input does not directly correlate to the accelerator pedal's position. The accelerator positions is closer to gas-per-cycle if anything, so at higher RPMs, you need a lesser accelerator position to provide the same amount of gas.

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At the same vehicle speed, a lower gear requires more revolutions but less force — you don't need to depress the gas pedal all the way in the lower gear but may have to in a higher gear. But in the higher gear you'll have fewer revolutions per minute. Essentially, it's the $E=F*s$ in play with transmissions.

But. Friction.

Most of the friction losses in the drivetrain (motor, clutch, gears) grow with the number of revolutions. All moving parts get hotter. The gears slurp up oil; motor coolant is pumped faster (yes, the cooling produces heat, just like with your laptop!).

This effect can be observed when "motor braking" in low gear going downhill. All the potential energy is converted to heat without any brakes. Just the noise from the motor and gears must eat hundreds of watts of power!

The drivetrain will suffer from exactly the same losses if you hit the gas pedal in flat terrain and revv the motor to the same rpm. You could as well stand on the brake pedal the same way you would have to on a downhill drive without motor braking.

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No matter what causes them, energy "losses" inside an engine go directly to heat. This makes the engine warmer, making to coolant warmer trying to not overheat the engine.

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This can be explained even if you do not account for the extra energy losses at a higher rpm. Take the example of a bicycle. Your legs are equivalent to the pistons that generate the torque. How fast you pedal is your pedal rpm, which is equivalent to the engine rpm.

Now assume you are moving at a gear ratio of 1:1. So essentially your pedal rpm and wheel rpm are equal and that causes you to move at a certain speed. Lets say the force you apply to the pedal is F and are going at a speed S.

Now what happens if you decrease the gear ratio to 2:1 and you have to maintain the same speed S?

You will have to do 2 things. 1st, you will have to increase your pedal rpm to double of what it was previously. This will create the same wheel rpm and maintain the same speed. But in doing so, you will also require only half the pedal force, i.e. F/2,i.e., you have to use less torque to maintain this rpm. If you try to use anymore force than F/2, you will increase the pedal rpm and hence the speed. This is what makes all the difference.

In this specific case, at a higher rpm, the pistons are pushing down at the crankshaft with a lower force and hence generating lower torque, hence generating lesser power in the first place. So to answer your question of where is the excess energy going? The excess energy is not even generated in the first place! (even if we assume no losses at higher rpm). And the reason this excess energy is not generated is because of a lower throttle. When you are at a lower gear, you need less torque to overcome the same amount of load as compared to when you are at a higher gear.

This image shows how the throttle position affects power

When you are maintaining a high rpm at a lower gear, you are giving less throttle, so the pistons are pushed down at a lesser force and hence the torque generated by the engine is low. But if you try to maintain the same rpm at a higher gear, you have to overcome a greater load, hence you have to give more throttle and as a result you are generating a lot more power from the engine.

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Outside of additional parasitic loss of the engine spinning faster there is no extra energy being created. You're creating less energy at a greater rate that's geared down to produce the same tail shaft output. Then there's the thing about the higher the power setting (more open throttle) the more efficiently a recip operates.

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  • $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Commented Apr 17 at 2:32
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When I was at the university, attending the course on Internal Combustion Engines, our professor was explaining the energy balance of an ICE.

When one looks into it, it can be seen that the power output grows with the square of the engine RPM. But don't think that you can get more power just increasing the RPM, the professor warned us, because engine losses due to friction grow with the cube of the engine RPM, thus there is a point where any more energy put into increasing the RPM actually goes lost into friction and thus heat.

And this is where the excess energy goes. Incidentally, this is also why when you want to brake with the engine you shift to a lower gear: to increase the losses and convert kinetic energy into heat.

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The power output of an engine at any gear can be adjusted by pushing the acceleration pedal. When you use a lower gear to increase torque you either accelerate or reduce the pressure on the gas pedal to reduce intake of fuel and air. If you use a higher gear with lower torque you push the gas pedal a little to increase the amount of fuel.

Two different gears can have an overlapping power output using different mixture of fuel and air, but with the equal efficiency.

One can adjust the gas pedal so that both gears have the same efficiency.

Low gear: more torque,less fuel, slower speed. Higher gear: less torque, more fuel, faster speed.

Energy spent to go a constant distance the same.

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