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 CO_{2}).

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.

where:

T_{in} [Nm] – input torque

ω_{in} [rad/s] – input speed

i [-] – gear ratio

T_{out} [Nm] – output torque

ω_{out} [rad/s] – output speed

We can calculate the input power P_{in} [W] and output power P_{out} [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:

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 P_{loss} [W]:

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:

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**:

### 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.

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:

where:

η_{gbx} [-] – gearbox efficiency

η_{brg} [-] – bearing efficiency

η_{grm} [-] – gear mesh efficiency

N_{brg} [-] – number of bearings

N_{grm} [-] – 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**.

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.

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:

where:

η_{prs} [-] – propeller shaft efficiency

η_{brg} [-] – bearing efficiency

η_{uj} [-] – universal joint efficiency

N_{brg} [-] – number of bearings

N_{uj} [-] – 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**.

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.

The overall **efficiency of the differential** can be calculated as:

where:

η_{dif} [-] – differential efficiency

η_{brg} [-] – bearing efficiency

η_{grm} [-] – gear mesh efficiency

N_{brg} [-] – 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**.

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.

The overall **efficiency of the driveshaft** can be calculated as:

where:

η_{drs} [-] – driveshaft efficiency

η_{trp} [-] – tripod joint efficiency

η_{rzp} [-] – 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**.

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:

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 P_{out} and what are the drivetrain power losses P_{loss}, if the engine power at the clutch P_{in} 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

η_{gbx} [-] | η_{prs} [-] | η_{dif} [-] | η_{drs} [-] | P_{in} [kW] |

Drivetrain efficiency, η_{drv} [-] = | ||||

Output power, P_{out} [kW] = | ||||

Loss of power, P_{loss} [kW] = |

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## Renato Maniago

So, on a rear-wheel drive, are there 2 or only 1 driveshaft? Some articles consider the driveshaft as 2 halfshafts. What is even more confusing is the driveshaft is also the propeller shaft! So on a rear-wheel drive, is is 0.841 or 0.826, on a front-wheel drive, is it 0.859 or 0.841?

## Nats

So, using Kris’s example, the RWD loss should be 0.826 and 0.841 since it has 2 driveshafts?

## Renato Maniago

Using Kris’s example, the RWD loss should be 0.826 and not 0.841 since there are 2 driveshafts instead of 1? I thought a driveshaft consists of 2 half-shafts, or a driveshaft is also one whole axle, left and right wheels.

## Bill Hooghuis

How do you calculate the losses due to diveline angles of the CV joints and U-Joints? My buddy has a drag car and we changed the subframe to reduce the angles and were curious.

## Ben

I am interested in losses based off of this scenario. You start with an AWD car, but then you decouple the transfer case in which it is now RWD, but still has all the components of the AWD. Now can you neglect the forward drive shaft, transfer case, forward axles and front differential? They are still on the car and are essentially still driven by the front wheels when they turn, but are not directly driven from the engine. Does this indeed transfer more power to the ground through only the rear diff or does the system as a whole still suffer the same losses as if everything were connected? TIA.

## Anthony Stark

If those components still rotate, the losses are still there. The only difference is that the power now comes through the road, via the rear wheels but still from the engine. You can eliminate those losses only if you disconnect them from the wheels. Compared with the AWD situation, now the rear axle has to compensate for the loss of traction of the front axle and also for the friction losses of the front axle. You might save something from the transfer case (some components not rotating) but pretty much, overall, you have same level of losses.

## Johnny Waclosh

They are not coupled and thus not rotating. 4×4 vehicles have freewheeling capability built into the sprockets inside the front wheels (hydraulically, pneumatically or even very old school manually controlled by twisting a rotating centre of a hub cap) which decouple the fron wheels from their driveshafts. On the other end for the drivetrain, inside the transfer case, a teethed clutch completely decouples/separates the shaft which goes from the gearbox to the front differential. So the rotation of the mass is no longer taking place and the only added inefficiency compared to a car with no FWD capability is the dead weight of the drivetrain components, that the RWD system simply carries around. But nothing is being driven in the front part of the car. Neither by the engine, nor by the front wheels turning.

## bhuvan sharma

In reference to the values of gear mesh efficiencies eg. .98 for helical mesh gears, what is the method to calculate these efficiencies? OR Is there any standard databook which shows the variation of gear mesh efficiencies for different types of gears like spur , helical , bevel, etc. with torque, rpm , no of teeth in mesh?

Also the same for bearing efficiencies ?

## Darin Selby

At what point do you abandon one idea for another one, mathematically speaking?

I have a ‘differential friction losses’ question. The ‘RC truck differential’ is what I’m using to build a small model of my invention. How would the friction losses of this be calculated? https://tinyurl.com/yc7uet5x

## Kris

When you are calculating your total loss, how would you correctly (at least, roughly speaking) account for the quantity of each element?

For instance on your rear drive example, you just multiplied everything together which would make enough sense if there was only one of each component, but in reality there is actually two driveshafts. Would you not do something like the calculation below?

0.941*0.98*0.931*[0.98^2] = 0.826

## Anthony Stark

Hi Kris,

Yes, you have to take into account the losses of each element. Since both driveshaft have frictions, both have losses. In the example, the losses value is combined but you can split it in two.

## Johannes

There are two driveshafts, but each transmits only half of the torque. Each half of the torque goes through one driveshaft, therefore the formula given in the article (simply multiplying everything together) is correct.

## Hugo

May I know where do the values of efficiencies involved in calculations of drive-train loss come from? Any reference support?

## Anthony Stark

Hi Hugo,

The value of the efficiencies are correct in terms of order of value but not specific from a real component. The idea was to describe the method of calculation. For reference you can check “Automotive Transmissions – Fundamentals, Selection, Design and Application by Harald Naunheimer et al.”