# Gear ratio and power output in a car

I'm a bit confused about power output and gear ratios. I think it's best to explain by going by my example (imperical units, sorry Americas & Brits):

Mass: 1300kg
max. power: 128 kW @ 5500 rpm
max. torque: 245 Nm
Gear ratios (from a slightly different car type, but should be similar):
1st 3.55
2nd 2.11
3rd 1.30
4th 0.94
5th 0.79
Reverse 3.50
Final Drive 4.11

Torque = F.s (F and s perpendicular)
Now, in general, power = work / time = F*s / t = 2*Pi*Torque*(rpm / 60)
In this case, filling in that formula with the torque value gives: power P = 2*Pi* 240 * 5500 / 60 = 138 kW

=>Not exactly 128kW, but okay, that's not my problem.

Officially, the car goes from 0 to 100 km/h in 8.0 seconds.

The kinetic energy difference between 0 and 100 km/h is:
.5 * 1300 * 27.8^2 =~ 500 kJ. Of course, this doesn't include roll and air friction. As a rough estimation, let's add 20% to get the total energy, required to 100 km/h:
500 * 1.2 = 600 kJ

So, divide this by 8 seconds, and you get an average power output of 75 kW - 59% of the maximum.

This number seems to be too high?
It is my understanding that power output at the wheel = engine power / gear ratio. So, in the first gear, you have a max. power of 128 / 3.55 = 36 kW. In second gear, this is 60 kW. Which doesn't make sense to me, because in going from 0-100, even if you constantly drive in 2nd gear at maximum power output, you'd accelerate according to 60kW (that is not even taking into account shifting, being outside of the power band, transmission losses, etc), while the 8.0s requires an average of 75kW.

Where do i go wrong?

Also, what is the "final drive" number from the gear ratios? An additional transmission that is always there?

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brewnog
Gold Member
The only difference a gearbox makes to power transmitted is with drive inefficiencies. Your formula (wheel power = engine power / gear ratio) is incorrect. The equation you need to understand is:

Power = Torque x Angular Speed

In words, a gearbox will decrease its output speed by the same ratio the output torque is increased. The power transmitted remains constant (less a bit for friction).

Final drive is indeed an additional transmission that is always there; generally on a RWD car this is housed within the differential/back axle. For a FWD car it's part of the transaxle unit.

The only difference a gearbox makes to power transmitted is with drive inefficiencies. Your formula (wheel power = engine power / gear ratio) is incorrect. The equation you need to understand is:

Power = Torque x Angular Speed

In words, a gearbox will decrease its output speed by the same ratio the output torque is increased. The power transmitted remains constant (less a bit for friction).

Final drive is indeed an additional transmission that is always there; generally on a RWD car this is housed within the differential/back axle. For a FWD car it's part of the transaxle unit.
I'm not sure if i understand. Intuitively i get what you're saying - lower gears give higher torque but lower final output speed, with constant power (more or less) over all gears.

But i don't quite get your formula: Power = Torque x Angular Speed.

Angular speed here is the speed of the wheel itself? That is: rpm / (60 * gear ratio * final drive ratio) ? Most engines deliver maximum torque from 2000 to 5500 rpm, but this maximum is only reached in first gear, and decreases proportionally in higher gears?

Basically, i don't know how to reproduce 128 kW using 240 Nm of maximum torque and those gear ratios.

brewnog
Gold Member
The product of angular speed and torque into the gearbox (i.e. engine speed x engine torque) is equal to the product of angular speed and torque out of the gearbox (i.e. at the propshaft), minus losses. The same goes for the final drive.

Most engines deliver maximum torque somewhere in the range you quoted. Now, this is irrelevant to the gear you're in. Maximum torque at the wheels is produced in first gear, but that's also where minimum speed is produced.

You need to realise that the performance figures quoted are usually for the engine, and are peak. That is to say, your engine may have a peak power of 128kW and a peak torque of 240Nm, but peak power is delivered at 5,500rpm and peak torque somewhere below that (let's say 4,000rpm) so you won't be able to get the two to match up!

I see. I guess most of the confusion came from the fact that there's a big difference between the torque delivered by the engine, and the torque delivered by the wheels.

One more question, about manual shifted cars. When turning off the engine, some people leave the car in 1st gear instead of using the handbreak. Would it (technically) be safer to leave the car in 5th gear, because more force (or torque, if you will) on the wheels is needed to make one revolution? Of course, in practice, the first gear easily does the trick, but would it help?

Thanks!

brewnog
Gold Member
Nope, first. Think about it - if the car rolls forwards a low gear, the engine will have to rotate many more times than if it were in a higher gear. Therefore you get much more of an engine braking effect in a lower ger.

Ahh, you are right. Just like in Lego, where the smaller but higher torque gear was harder to spin than than the faster, big one, when turned off.

brewnog
Gold Member
Exactly. Lego's a great learning tool for this kind of stuff, dig it back out and have a play!

What about fuel consumption per gear?

Basic physics tells me that if the power delivered is roughly equal for each gear (at constant RPM), then fuel consumption should be roughly equal as well. But does it take just as much gas driving 120 km/h with 3000 rpm in the 5th, as it does driving 25 km/h with 3000 rpm in the 1st?

brewnog
Gold Member
Good reasoning. However, when you're driving at 25kph in 1st gear, you're not producing the same power (or using the same amount of fuel) as you are to drive at 120kph (because you won't have pushed the accelerator down particularly hard). Think about how far you have to push the accelerator to achieve a certain condition and that gives a mental approximation of fuel use.

At wide open throttle, the power produced by the engine at a given engine speed is the same in every gear.

Yeah, going 25kph in 1st you would probably be hardly pressing on the gas pedal at all. Maybe you never noticed because nobody just goes 25kph in 1st. You either accelerate right through it or you change gears.

Going 120kph you would need to give it much more gas, and much more fuel would be burned.

Acceleration is higher at low velocities, up to the point that a≈0.4g for a 2-wheel drive car and ≈0.8 g for a four-wheel drive car, when the drive wheels start slipping:

g= 9.81 m/s2 = 0 to 35 km/hour in 1 second.

E = ½·m·v2

So power is

P = dE/dt = m·v·dv/dt = m·v·a

where v = velocity and a = acceleration. If the static (non-slipping) friction coefficient is 0.8, and the car is 2-wheel drive, with the effective mass m/2 and effective weight mg/2 on the drive wheels, then the maximum acceleration is 0.4g.

So the maximum power that can be used without slipping is then

P ≤ 0.4·m·v·g

which may be less than the power available, especially at low velocity.

Power is related to torque and RPM by

P = torque (Newton-meters) x 2·π·RPM /60

This is the power anywhere in the drive train, including rear axle.

Bob S

Good reasoning. However, when you're driving at 25kph in 1st gear, you're not producing the same power (or using the same amount of fuel) as you are to drive at 120kph (because you won't have pushed the accelerator down particularly hard). Think about how far you have to push the accelerator to achieve a certain condition and that gives a mental approximation of fuel use.

At wide open throttle, the power produced by the engine at a given engine speed is the same in every gear.
Thanks. From an intuitive point of view, it definitely makes sense.

The reason i asked is because my brothers car has a "life fuel consumption" gauge, and it always goes through the roof when in first gear. However, i think that's because in first gear, you go from 800 to 2000 rpm in a matter of seconds (meaning a high acceleration and thus a lot of power), while it takes 5+ seconds to go from 90 to 120 km/h, unless you're driving a Ferrari :)

brewnog
Gold Member
Exactly right. You're in first gear for a reason - to accelerate! As you say, you do this in a matter of seconds (with your foot flat to the floor, producing lots of power, and using lots of fuel, though momentarily). When you're cruising at 60mph on the motorway, you're backed off the throttle, not using too much power, and not using too much fuel.

You'll see on the gauge that with your foot flat to the floor, for a given engine speed the fuel consumption is the same regardless of gear. It's the duty cycle (what power, what engine speed, what vehicle speed, for what duration) which is the variable causing the difference you're seeing.

sophiecentaur
Gold Member
When I am pootling along at almost any speed in 5th gear and then drop down to 4th, the indicated instantaneous mpg drops significantly. This is despite having an expensive engine management system. It always was advised to keep in a high gear to improve fuel consumption, when cars had carburettors and the same thing seems to apply.

brewnog
Gold Member
This is because engine speed increases instantaneously, and on the 'over-run' modern systems will cut fuelling and report an instantaneous MPG increase accordingly.

To prove this to yourself, brim the tank, do a 10 mile cruise in 5th, brim the tank again (measure what you've used), do the same journey at the same cruising speed in 4th, brim the tank again (measure what you've used). Over a drive cycle, a higher gear (provided it's appropriate for the conditions; i.e. not doing 20mph in 5th) will use less fuel.

sophiecentaur
Gold Member
I'm not referring to the transient effect, which lasts less than a second. The efficiency just seems to be less in the lower gear under the (apparently) same load.

brewnog
Gold Member
Ah, I see.

Do you mean the efficiency is less in a lower gear (which would be correct), or the fuel consumption is less (i.e. higher efficiency)?

If it's the former, it's either a fuel flow map error (the ECM doesn't actually measure fuel flow, it infers it from speed, throttle position, manifold pressure and a few other things), and if the calibration engineer hasn't got the map right then it won't read correctly!) or you're genuinely using too high a gear for the conditions.

sophiecentaur
Gold Member
Lower efficiency in a lower gear was what I meant.
I somehow had the naive idea that fuel consumption was measured by what the pump was delivering. Why don't they do it that way?

brewnog
Gold Member
Well a lower efficiency in a lower gear is what you'd normally expect. Engine speed is higher, you're probably further away from peak torque, and the friction and pumping losses are related to engine speed so these are higher.

There are a few reasons fuel flow isn't measured. An electric fuel pump delivers fuel at a constant rate to the engine, so if you put a meter here your flow rate would remain constant! To actually measure fuel consumption, you need to measure both fuel into the engine, and fuel being let back to the tank.

In either event, an inferred fuel flow is an easy parameter to get from the engine control system; this is monitoring/controlling (typically) fuel pressure, injector opening duration, plus engine speed//temperature/manifold pressure/temperature/throttle position so with some calculations within the ECM it's easy to provide a calculated fuel flow value.

Flow meters are also too expensive to fit to a car just to provide a more accurate fuel consumption reading.

russ_watters
Mentor
Why are you dropping back to 4th gear anyway? Just for a test or because you're slowing down or going uphill...?