# Drivetrain losses (efficiency)

The drivetrain (also called driveline) is the sum of components which are delivering the engine power to the wheels. For example, on a rear-wheel drive (RWD) vehicle, the drivetrain consists of: clutch (or torque converter), gearbox (manual or automatic), propeller shaft, differential and drive shafts.

The efficiency of the drivetrain has a significant impact on the overall efficiency of the vehicle. The higher the efficiency of the drivetrain, the lower the fuel consumption of the vehicle (also lower CO2).

Image: Drivetrain architecture (Audi A6 quattro) and main components
Credit: Audi

In the article What is efficiency ? it is explained in detail how mechanical efficiency is calculated.

The current article is split in two main parts. In the first part, to understand the concept of efficiency, we’ll calculate the efficiency of a simple gear, function of the input/output power and torque, and in the second part we’ll calculate the efficiency of each drivetrain component and the overall drivetrain efficiency.

### Gear mesh efficiency

A simple gear mechanism has an input gear and an output gear meshed together. The input torque and angular speed are converted through the gear ratio in output torque and angular speed.

Image: Simple gear input output

where:

Tin [Nm] – input torque
i [-] – gear ratio
Tout [Nm] – output torque

We can calculate the input power Pin [W] and output power Pout [W] as:

$P_{in} = \omega_{in} \cdot T_{in} \tag{1}$
$P_{out} = \omega_{out} \cdot T_{out} \tag{2}$

The efficiency is defined as the ratio between the output power and input power:

$\bbox[#FFFF9D]{\eta = \frac{P_{out}}{P_{in}}} \tag{3}$

Any mechanical component/system, which has moving parts, has friction. The friction is converting part of the energy into heat, which is dissipated in the surroundings environment, therefore lost. The overall friction can be captured as power loss of the components/system. The output power is the difference between the input power and the power losses Ploss [W]:

$P_{out} = P_{in} – P_{loss} \tag{4}$

Replacing (4) in (3):

$\eta = \frac{P_{out}}{P_{in}} = \frac{P_{in} – P_{loss}}{P_{in}}=\frac{P_{in}}{P_{in}} – \frac{P_{loss}}{P_{in}} = 1 – \frac{P_{loss}}{P_{in}}$

gives the expression of the efficiency function of the input power and power losses:

$\bbox[#FFFF9D]{\eta = 1 – \frac{P_{loss}}{P_{in}}} \tag{5}$

The mechanical efficiency of the simple gear can also be calculated function of input and output torque.

The output speed is equal with the input speed divided by the gear ratio:

$\omega_{out} = \frac{\omega_{in}}{i} \tag{6}$

Replacing (1) and (2) in (3) gives the expression of the efficiency function of input and output torque and speed:

$\eta = \frac{P_{out}}{P_{in}} = \frac{\omega_{out} \cdot T_{out}}{\omega_{in} \cdot T_{in}} \tag{7}$

Replacing (6) in (7) gives:

$\eta = \frac{\frac{\omega_{in}}{i} \cdot T_{out}}{\omega_{in} \cdot T_{in}}$

from which we can write the final expression of the efficiency function of input/output torque and gear ratio:

$\bbox[#FFFF9D]{\eta = \frac{T_{out}}{i \cdot T_{in}}} \tag{8}$

### Gearbox efficiency

The moving parts of a gearbox consists of gears (simple or planetary), synchronizes, shafts and bearings. The overall efficiency of a gearbox depends mainly on the gear mesh and bearings efficiency.

Image: Six-speed manual gearbox components
Credit: ZF

Depending on the architecture, a gearbox has at least two shafts (input and output) and several simple gears. Each shaft is sustained in at least two ball bearings, one in each end. Therefore, when a gear is engaged, there are 4 bearings and at least 1 gear mesh as sources of power losses.

The overall efficiency of the gearbox can be calculated as:

$\eta_{gbx}= \eta_{brg}^{N_{brg}} \cdot \eta_{grm}^{N_{grm}} \tag{9}$

where:

ηgbx [-] – gearbox efficiency
ηbrg [-] – bearing efficiency
ηgrm [-] – gear mesh efficiency
Nbrg [-] – number of bearings
Ngrm [-] – number of gear meshes

The efficiency of a ball-bearing is around 0.99 and of a helical gear mesh around 0.98. With these numbers, we can calculate the overall efficiency of the gearbox.

$\eta_{gbx}= 0.99^{4} \cdot 0.98^{1} = 0.941$

In reality the efficiency of the gearbox is not constant but it depends on the temperature and shaft speed. The minimum efficiency is usually obtained at low temperature (high oil viscosity) and high shaft speed. The maximum efficiency is obtained at high temperature (low oil viscosity) and low shaft speed.

### Propeller shaft efficiency

The propeller shaft is transmitting torque from the gearbox to the rear axle. Since the gearbox and rear axle have to move relative to each other while transmitting torque, the propeller shaft needs at least 2 universal (“U”) joints, one at each end.

Image: Propeller shaft components (universal joints)

The efficiency of the propeller shaft depends on the number and efficiency of the U-joints and holding bearings. If the propeller shaft is made up from two pieces, it needs at least one center bearing and four U-joints.

The overall efficiency of the propeller shaft can be calculated as:

$\eta_{prs}= \eta_{brg}^{N_{brg}} \cdot \eta_{uj}^{N_{uj}} \tag{10}$

where:

ηprs [-] – propeller shaft efficiency
ηbrg [-] – bearing efficiency
ηuj [-] – universal joint efficiency
Nbrg [-] – number of bearings
Nuj [-] – number of universal joints

For our example, we are going to consider that the propeller shaft is one-piece, has 2 universal joints and no center bearing. The efficiency of an universal joint is around 0.99. With these numbers, we can calculate the overall efficiency of the propeller shaft.

$\eta_{prs}= 0.99^{0} \cdot 0.99^{2} = 0.98$

In reality, the efficiency of the universal joint is not constant but it depends mainly on the offset (angle) between the front and rear axle. The lower the offset, the higher the efficiency.

### Differential efficiency

The differential does the final gear reduction and the torque split between the right and left wheels. If the vehicle is driving on a straight line, only the final gear and bearings are adding power losses. There are 3 bearings (one on the input pinion, one on the left output shaft and one on the right output shaft) and 1 spiral bevel gear.

Image: Rear differential components

The overall efficiency of the differential can be calculated as:

$\eta_{dif}= \eta_{brg}^{N_{brg}} \cdot \eta_{grm} \tag{11}$

where:

ηdif [-] – differential efficiency
ηbrg [-] – bearing efficiency
ηgrm [-] – gear mesh efficiency
Nbrg [-] – number of bearings

The efficiency of a ball-bearing is around 0.99 and of a spiral bevel gear mesh around 0.96. With these numbers, we can calculate the overall efficiency of the differential.

$\eta_{dif}= 0.99^{3} \cdot 0.96 = 0.931$

In reality the efficiency of the differential is not constant but it depends on the temperature and shaft speed. The minimum efficiency is usually obtained at low temperature (high oil viscosity) and high shaft speed. The maximum efficiency is obtained at high temperature (low oil viscosity) and low shaft speed.

### Driveshaft efficiency

The driveshaft is transmitting the torque from the differential to the wheel. Each wheel has it’s own driveshaft. At each end of the driveshaft there are constant-velocity joints (CVJ), which are needed due to the relative motion between differential and wheel.

Image: Driveshaft components (constant-velocity joints)

The overall efficiency of the driveshaft can be calculated as:

$\eta_{drs}= \eta_{trp} \cdot \eta_{rzp} \tag{12}$

where:

ηdrs [-] – driveshaft efficiency
ηtrp [-] – tripod joint efficiency
ηrzp [-] – rzeppa joint efficiency

The inner CVJ (on differential side) usually is a Tripod type joint, while the outer CVJ is a Rzeppa type joint. The efficiency of these joints is around 0.99. With these numbers, we can calculate the overall efficiency of the driveshaft.

$\eta_{drs}= 0.99 \cdot 0.99 = 0.98$

In reality, the efficiency of the constant-velocity joint is not constant but it depends mainly on the offset (angle) between the differential and wheel. The lower the offset, the higher the efficiency.

### Overall efficiency of the drivetrain

Now that we have the overall efficiency of each component, we can calculate the overall efficiency of the drivetrain (driveline) as:

$\bbox[#FFFF9D]{\eta_{drv} = \eta_{gbx} \cdot \eta_{prs} \cdot \eta_{dif} \cdot \eta_{drs}} \tag{13}$

Replacing the values obtained for each components, gives:

$\eta_{drv} = 0.941 \cdot 0.98 \cdot 0.931 \cdot 0.98 = 0.841$

From our parameters and methodology we got an overall efficiency of the drivetrain of 84.1 %. This means that around 15.9 % of the engine power is lost through the drivetrain. The efficiency could be even lower for four-wheel drive (4WD) vehicles which have a central differential.

Let’s see how much we get at the wheels Pout and what are the drivetrain power losses Ploss, if the engine power at the clutch Pin is 150 kW and the drivetrain efficiency is 0.841.

From (3) we can calculate the power at the wheels (output power):

$P_{out} = \eta_{drv} \cdot P_{in} = 0.841 \cdot 150 = 126.15 \text{ kW}$

From (4) we can calculated the power lost in the drivetrain:

$P_{loss}=P_{in} – P_{out} = 150 – 126.15 = 23.85 \text{ kW}$

The numbers show that the overall drivetrain efficiency has a significant impact on the dynamic performance of the vehicle since a significant part of the engine power is lost. Also, the lower the drivetrain efficiency, the higher the engine fuel consumption.

Front-wheel drive (FWD) vehicles usually have the highest drivetrain efficiency, mainly because they don’t contain a propeller shaft. At the opposite end are the all-wheel drive (AWD) and four-wheel drive (4WD) vehicles, with the lowest drivetrain efficiency (due to higher number of components).

You can also check your results using the calculator below.

### Drivetrain Efficiency Calculator

 ηgbx [-] ηprs [-] ηdif [-] ηdrs [-] Pin [kW] Drivetrain efficiency, ηdrv [-] = Output power, Pout [kW] = Loss of power, Ploss [kW] =

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