RMCO-Interstate travel is allowed within Malaysia after 10 June 2020. As economy becomes uncertain, how to save more petrol as you drive, let's look from physics prospective

Interstate travel is allowed in Malaysia after 10 June 2020. It was indeed great news for all Malaysians.

As the economy becomes uncertain nowadays due to CoV-19 situation, people start to look for several ways to save money during “rainy days”, as we all are experiencing it now. 

There were a lot of experiments had done over the years that basically shown as below:

1.  The heavier the vehicle, the higher the fuel consumption.

2.  The higher the speed of the vehicle, the higher the fuel consumption.

So today I am going to provide an explanation on how the mass and the speed of the vehicle are impacting the fuel consumption, from physics perspective.

First we need to draw a model for analysis, refer figure 1 below:

Figure 1

A forward force, F_f is needed to move the car into rightward position on the surface of the road. There is a resistance force existed in the model that opposes the movement of the car, which is the rolling resistance. Rolling resistance sometimes called as rolling friction, which is the force resists the motion when a body (such as tire) rolls on a surface (such as road). The equation of the rolling resistance is:

Eq1 : Fr = CrrN

Fr = Rolling resistance force (N) Crr = Dimensionless rolling resistance coefficient N = The normal force (N), the force perpendicular to the surface which the tire is rolling. In the figure 1 model, the force is the weight of the vehicle that is acting perpendicular to the surface of the road

In figure 1, the vehicle has 4 tires that are direct contact with the surface of the road. In reality, the weight that is acting on these 4 tires are slightly different (the weight of the vehicle is not equally shared among these 4 tires, but the summation of these weights that is acting on these 4 tires will be equivalent to the total weight of the vehicle), and it will be very complex to calculate the rolling resistance base on each tire. I have to simplify my figure 1 model as below.

Figure 2

The details of the vehicle information are as below:

1.  Mass of the car = 1145 kg

2.  Rolling resistance coefficient, C_rr = 0.014

From the figure 2 model, I calculate the rolling resistance as below

Fr = CrrN Fr = 0.014 * 11221 Fr = 157.094 N

Fr = Rolling resistance force (N) Crr = 0.014 N = Weight of the vehicle   = 1145 x 9.8 = 11221 N

Recall that the force is a vector quantity. I give a positive vector to forward force F_f, and since the rolling resistance, F_r is in the opposite direction that always acting against the forward force, the rolling resistance is therefore a negative vector force (refer to figure 2).

In order to move the vehicle to the rightward position, the forward force F_f that is generated by the vehicle engine must overcome the rolling resistance force, F_r to produce a resultant force F that is pointing rightward position. The F is a summation of the 2 vector forces.

Eq 2: F = Ff – Fr

F = Resultant force Ff = Forward force (N) generated by vehicle     engine Fr = Rolling resistance (N)

Figure 3

Refer to figure 3, as the vehicle starts to move and accelerate, there is an additional resistance force comes into the picture, which is air resistance.

Air resistance is a force which tends to slow down the movement of an object through air. As a moving object pushes the air out of its way, the air pushes back on the object. The air resistance is always opposite to the object’s motion. In our model, the object is vehicle.

The details of the vehicle information (drag coefficient and area) and air density are as below:

ρ = 1.1644

C_D= 0.28 (given by the vehicle specification)

A = 2.55175

As the air resistance is always opposite direction against the forward force, thus it is a negative vector force. Hence, the F_f needs to overcome the air resistance and rolling resistance in order to continue to accelerate the vehicle. The Eq 2 is revised as below

Eq 4: F = Ff – Fr – Fa

F = Resultant force Ff = Forward force (N) generated by vehicle     engine Fr = Rolling resistance (N) Fa = Air resistance (N)

Figure 4

From Eq 1, the heavier the vehicle, the greater the rolling resistance F_r.

From Eq 3, the higher the speed of the vehicle, the greater the air resistance F_a. Therefore, the higher the rolling resistance and air resistance, the greater the required forward force F_f is needed to overcome them. Thus, more fuel will need to be provided to the engine to produce the required forward force F_f.

Next, we need to investigate how the air resistance and rolling resistance are impacting the fuel usage in quantitative number.

Obviously, the vehicle needs a driver to operate, so the mass of the driver is added into the rolling resistance. Assume that the mass of the driver is 70 kg, so the revised F_r as below.

Fr = CrrN Fr = 0.014 * 11907 Fr = 166.698 N

Fr = Rolling resistance force (N) Crr = 0.014 N = Weight of the vehicle(including     driver)   = (1145 + 70) x 9.8 = 11907 N

As the speed limit on the highway is 110 km/h (or 30.5 m/s), we assume that the vehicle accelerates from 0 km/h to 110 km/h within 30.5 seconds with constant acceleration 1 m/s². It means that after 1 s, the speed is 1 m/s, after 2 s, the speed is 2 m/s, and so forth. The speed-time graph is shown as below:

Figure 5

Recall that the distance travelled is the area under the curve of the speed-time graph. In this case, the area under the curve is triangle shape.

Distance travelled @5 s = 0.5 x (5-0) x (5-0) = 12.5 m

Distance travelled @10 s = 0.5 x (10-0) x (10-0) = 50 m

Repeat the same calculation until 30.5 s. Refer to table below for distance travelled information

Table 1a

Since the air resistance is depending on the speed of the vehicle, we can then compute the air resistance as below:

Repeat the same calculation until 30.5 m/s. Refer to table below for air resistance information.

Table 1b

To compute the resultant force F, we need the acceleration value that is caused by this resultant force and the mass of the object. The acceleration is 1 m/s² and total mass of the vehicle and the driver is 1215 kg.

Eq 5: F = ma      F = 1215 * 1      F = 1215 N

F = Resultant force (N) m = mass of the vehicle (1215 kg) a = acceleration m/s2 (1 m/s2)

Table 1c below shows the air resistance, rolling resistance, resultant force and distance travelled from 0 m until 465.125 m. Notice that the rolling resistance does not change.

Table 1c

Next, we have to find the energy consumption.

Recalled that the energy formula is

Eq 6: E = F * D

E = Energy (J, Joules) F = Force (N, Newton) D = Distance (m, metre)

The total energy that had been spent on overcoming the rolling resistance from 0 m until 465.125 m is shown in the rolling resistance versus distance graph as below.

Refer back to Eq 6, the energy is the area under the curve (rectangle shape) of rolling resistance versus distance graph, therefore:

Er = F * D Er = 166.698 * 465.125 Er = 77535.40725 J

Er = Energy (J, joules) F = 166.698 N D = 465.125 m

The total energy that had been spent on overcoming the air resistance from 0 m until 465.125 m is shown in the air resistance versus distance graph as below.

As the speed of the vehicle is increasing from 0 m until 465.125 m, thus the air resistance is also increasing accordingly (refer to table 1c). Therefore, the area under the curve in air resistance versus distance graph looks like a triangle shape.

So the energy that had been spent on overcoming the air resistance from 0 m until 465.125 m is area under the curve (triangle shape).

Ea = 0.5 * 386.9617466 * 465.125 Ea = 89992.79118 J

Refer from table 1c and air resistance vs distance graph, the air resistance @465.125 m is 386.9617466 N (assume that the gradient is relatively constant, thus the computed area is a triangle shape)

The total energy that had spent to accelerate the vehicle due to the resultant force from 0 m until 465.125 m is shown in the resultant force versus distance graph.

Enet = Area A + Area B Enet = 7593.75 + 549939.375 Enet = 557533.125 J

Enet = Energy consumption on the resultant force Refer to table 1c and resultant resistance vs distance graph, the total energy is the summation of Area A and B Area A = 0.5 x 12.5 x 1215       = 7593.75 Area B = (465.125 – 12.5) x 1215       = 549939.375

The total spent energy from 0 m until 465.125 m is the summation of E_net, E_r and E_a which are 557533.125 J + 77535.40725 J + 89992.79118 J = 725061.3234 J.

Assume that the driver’s destination is 780 km away and the vehicle keeps the constant speed @30.5 m/s (~110 km/h) from 465.125 m to 780 km. Thus, the resultant force F is 0 N since the acceleration is 0 m/s² from 465.125 m to 780 km. It also means that the forward force F_f is equivalent to the summation of air resistance F_a and rolling resistance F_r. Therefore, E_net is 0 J between 465.125 m and 780 km. 

The total energy that had been spent on overcoming the rolling resistance from 465.125 m until 780 km is shown as below.

Er2 = F * D Er2 = 166.698 * 779534.875 Er2 = 129946904.6 J

F = 166.698 N D = 780,000 m - 465.125 m   = 779534.875 m

The total energy that had been spent on overcoming the air resistance from 465.125 m until 780 km is shown as below.

Ea2 = F * D Ea2 = 386.96174 * 779534.875 Ea2 = 301650176.8 J

F = 386.9617466 N D = 780,000 – 465.125 m   = 779534.875 m The air resistance is constant since the speed of the vehicle does not change from 465.125 m until 780,000 m

The total spent energy from 465.125 m until 780,000 m is the summation of E_r2 and E_a2 which are 431,597,081.4 J.

Thus the total energy consumption from 0 m until 780,000 m is

E_total = 725061.3234 J + 431,597,081.4 J = 432,322,142.7 J

Take note that there is some energy wasted due to the internal working of the car, for example thermal heat, internal engine mechanical friction, etc. that are not counted in this analysis.

Table below shows that how the changes of the mass are impacting on the energy consumption.

Table 2

With the mass of 1215 kg, the energy consumption is 432,322,142.7 J.

With the mass of 1225 kg, the energy consumption is 433,396,891.4 J, thus there is a 0.2485 % increase, compare with the energy consumption with the mass of 1215 kg.

The total estimation fuel cost is RM 75 with the mass of 1215 kg. Compare to the mass of 1425 kg, the fuel cost should be at least increase by RM 3.9. (Note, the energy consumption wasted due to internal working of the car is not included into this calculation, so the actual increase of the fuel cost should be higher than this)

Table below shows that how the changes of the speed are impacting the energy consumption.

Table 3(note with the mass of 1215 kg)

With the speed of 30.5 m/s, the energy consumption is 432,322,142.7 J.

With the speed of 36.1 m/s, the energy consumption is 553,473,191.4 J, thus there is a 28.02% increase, compare with the energy consumption with the speed of 30.5 m/s.

The air resistance has increased significantly at higher speed, thus there is a significant increase in energy consumption. The air resistance is 542 N and 722 N at speed 130 km/h and 150 km/h respectively. However, the air resistance is 386 N at speed 110 km/h.

We can see that the speed and the weight of the vehicle have huge impact on the energy consumption, by performing simple physics calculation.

-    It is wise to unload unnecessary things from the car. Keep the vehicle as light as possible.

-    It is unwise to over-speed, it is illegal and dangerous, and the cost of over-speeding is way too high.  

-    Exercise more to lose body weight

Need more example of speed-time graph, force and motion, please refer to

speed-time-graph-of-spacex-falcon9