The impact of wheel weight on the propulsion energy consumption of a vehicle

Reducing CO2 emissions is a very big challenge in the transport sector. Engineers around the world are working on solutions to reduce propulsion energy consumption of vehicles because such solutions, either directly or indirectly, help to reduce CO2 emissions. In the following analysis we will make focus on the impact of wheel weight on vehicle energy consumption.

The weight of wheels has an impact on any vehicle because it is a part of a vehicle’s overall weight. There is a question if the weight of the wheels has the same impact on vehicle performance as the weight inside the vehicle? Many people do believe that the weight of the wheels has the same effect on the vehicle as when the weight of the wheels is placed inside of the vehicle. They're obviously mistaken because the wheels are part of the unsprung mass of the vehicle. It is important to know that not all weight on the vehicle is equal. The weight that sits on top of the suspension is referred to as "sprung weight” because it is supported by the springs. The weight that is connected to the springs but does not sit atop the suspension, is called "unsprung" weight (Figure 1). The unsprung weight of a vehicle is the mass of the suspension, wheels, and other components directly connected to them.

Figure 1.

The wheel is exposed to two different kinetic energies (Figure 2):

• energy of rotation that spins the wheel.
• energy of translation that moves the wheel forward.

Figure 2.

The rotational kinetic energy of the wheel can be expressed as:

Er = ? I ω2

Where ω is the angular velocity and I is the moment of inertia around the axis of rotation.

Moments of inertia can be expressed as:

I = ? m r2

Where r is the radius of the wheel and m is the mass of the wheel.

Angular velocity can be expressed as:

ω = v/r

Where v is the velocity and r is the radius of the wheel.

The results resulting from rotational kinetic energy of the wheel will be as follows:

Er = ? I ω2 = ? (? m r2) (v/r)2 = ? mv2

Translational kinetic energy of the wheel can be expressed as:

Et = ? m v2

Where m is the mass of the wheel and v is the velocity (speed).

Total kinetic energy of one wheel can be expressed as:

Etv = Er + Ek = ? m v2 + ? m v2 = ? m v2

Total kinetic energy of four wheels can be expressed as (Figure 3):

E4tv = 4 Et = 4 ? m v2 = 3 m v2

Figure 3.

Total kinetic energy of the moving vehicle and four wheels can be expressed as:

Etotal = Ecar + E4tv = ? M v2 + ?3 m v2

Where M is the mass of the vehicle, m is the mass of the wheel and v is the velocity (Figure 4).

Figure 4.

Let's put our calculations to the test with a real-world example, because that is the easiest way we can quantify the share of wheel weight in fuel consumption.

The new Volkswagen ID.3 has summer tires 215 55 R18 95T Goodyear Efficiency Grip Performance installed in the factory. The weight of the tire is ~9.2 kg. Let’s have an aluminum rim suitable for ID.3 that weighs ~12.7 kg. The combined weight of the aluminum rim and tire is 22 kg.

Let's use ID.3 as an example, with a curb weight of 1800 kg and a speed of 20 meters per second (72 kilometers per hour). We have to subtract the weight of the tires from the curb weight of the vehicle and the weight of the vehicle is 1712 kg.

We do know that the formula for the total kinetic energy of the vehicles looks like this:

Etotal = Ecar + E4tv = ? M v2 + ?3 m v2

When we substitute the values into the formula, we get the resulting total kinetic energy.

Etotal = ? 1712 202 + 3 22 202 = 342400 + 26400 = 368800 J

The total kinetic energy of the vehicle moving with velocity 20 m/s is 368800 Joules.

Let’s find out how many percent of total kinetic energy made up total kinetic energy of four wheels.

%E4tv = 100 E4tv / Etotal ?= 100 26400/368800 = 7.15%

Total kinetic energy of the four wheels made up 7.15 % of total kinetic energy of the vehicle.

When the vehicle is driving at low speeds in urban traffic, the acceleration resistance accounts for about 40% of the total energy consumption (Figure 5), (1). An acceleration resistance force Facc is the force which resists to the total kinetic force Ftotal created by moving vehicle plus its wheels and the resultant force is equal to zero when equilibrium is reached, what should we write as Ftotal - Facc = 0. To avoid confusion, total kinetic force Ftotal does not equal the total tractive force, which propels the vehicle forward. Traction force of the propulsion system of the vehicle must also overcome the rolling resistance force, aerodynamic resistance force, beside acceleration resistance force Facc, as shown in the Figure 5.

Figure 5.

The acceleration resistance force Facc can be expressed as:

Facc = mt dv/dt

Where mt is the mass of the vehicle with wheels and v is the velocity of the vehicle.

The second derivative with respect to time of the total kinetic energy of moving vehicle and all four wheels Etotal ?gives us following result:

?Ftotal? = (Etotal )d2/dt2?

As the vehicle accelerates and decelerates, the ratio of Facc ?and Ftotal forces changes, of course. When the vehicle accelerates the kinetic force Ftotal of the vehicle and the wheels prevails over the acceleration resistance force Facc and vice versa when the vehicle slows down.

Since we do know that acceleration resistance force Facc equals to the total kinetic force Ftotal of vehicle when equilibrium is reached. This is achieved when the vehicle does not move or moves at a constant speed.

Facc = Ftotal?

When both forces are in balance, both energies must be in balance as well. Of course, the vehicle accelerates and decelerates while driving in the city, so the more correct description should look like this:

∑Eacc = ∑Etotal?

We were able to prove that acceleration resistive energy really equals to kinetic energy of the vehicle and its four wheels.

Since we do know that the total kinetic energy of the four wheels makes up 7.15 % of the total kinetic energy of the vehicle, let us calculate this share of the total fuel consumption in a city. Acceleration resistance energy accounts for ~40% of total energy consumption in a city (1). The share of wheels on total kinetic energy is 7.15% and it means that the share of kinetic energy of wheels on total acceleration resistance energy is 2.86% (7.15 x 40/100). The share of the kinetic energy of wheels on the total energy consumption of the vehicle is 1.144% (2.86 x 40/100) when driven in a city.

This share could be reduced by using lighter rims. For example, using magnesium or carbon fiber rims. Forged magnesium wheels are manufactured by mechanically forging a refabricated rod using a powerful forging press (Figure 6) and they are very light. Their disadvantage is that they are relatively expensive.

There is produced a magnesium rim that weighs ~6 kg (2).

Figure 6.

Using forged magnesium rims (6 kg/pc)? instead of aluminum cast rims (12.7 kg/pc) would reduce share of wheels on total consumption from 1.144% to 0.79% and that is, a reduction of 0.35%.

A pure carbon fiber composite rim (Figure 7) with a weight of only 5 kg is also possible (3).

Figure 7.

Using such pure carbon fiber composite rims would reduce share of wheels on total consumption from 1.144% to 0.74%. There is a 0.4% reduction.

As it was possible to see above, the share of kinetic energy (or acceleration resistance energy) of wheels on the total energy consumption of vehicle driving in the city is not very significant. As a result, the claim that using lighter rims has a significant impact on energy consumption is not based on the truth.

The use of magnesium or carbon rims may reduce energy losses in a slight way, but in general, the use of these rims has very little effect on overall efficiency.

References:

1, https://www.sciencedirect.com/book/9780750650540/the-automotive-chassis

2, https://mbdesign.shop/de/felgen/mf1-mg/

3,https://exclutec.com/en/news-carbon-jewelry-details/carbon-wheels-rims-audi-porsche-lamborghini.html