Driving resistances:
While driving, the vehicle experiences several resistances:
- Rolling resistance
- Grade resistance
- Aerodynamic drag
- Acceleration resistance
These resistances must be overcome to maintain speed. The force required to do so is called Frij; this is the sum of all driving resistances.
Rolling resistance increases at higher speeds due to tire deformation; grade resistance only applies when there is a grade (on a flat road it is therefore 0); and aerodynamic drag is very low at low speeds. As vehicle speed increases, aerodynamic drag rises quadratically. Aerodynamic drag plays the largest role when we look at the total driving resistances.
On this page the driving resistances are calculated up to the total driving resistance (Frij).a0
Rolling resistance:
Rolling resistance arises from various factors such as tire deformation, wheel alignment (and the resulting tire scrub), and the type of road surface. The amount the tires deform depends on the tire type. The more easily the tire can roll over the road surface (i.e., the less resistance it encounters), the less force is needed to keep the wheel in motion and the lower the fuel consumption will be. On a deformable (soft) surface, such as sand or mud, rolling resistance increases further due to additional friction forces between the tire and the surface, and due to the permanent deformation of the surface itself.
In the table alongside we see that the rolling resistance coefficient is low on dry asphalt (0.010) and high on sand (up to 0.3). The table is based on a low to medium vehicle speed (up to 80 km/h) at which the rolling resistance of different tire types is fairly constant.a0
At these speeds, the influence of speed on rolling resistance is negligible.
Factors that affect rolling resistance:
- Slip (during braking or accelerating): slip occurs in the contact patch when the tire transmits forces. With light slip (such as during gentle acceleration) rolling resistance can even temporarily decrease. With heavy slip (hard launches or hard braking) rolling resistance increases because the deformation behavior changes and more energy is lost.
- Wheel alignment (toe and camber): incorrect alignment causes extra lateral forces (for example due to toe or camber). These forces increase rolling resistance. This is because the wheel no longer rolls purely straight ahead but builds a lateral load. The effect becomes stronger with larger alignment deviations.
- Temperature: As the tire begins to roll, it heats up. This changes the material properties of the rubber: damping decreases, stiffness changes, and the tire deforms less. As a result, rolling resistance decreases as the temperature rises. After a few minutes the temperature reaches an equilibrium, where heat production and heat dissipation are balanced.
- Speed: At higher speed more deformations occur per second (more revolutions), which leads to higher temperature and larger deformation forces. Therefore rolling resistance increases, especially with tire types with higher hysteresis such as standard or winter tires.
At low speeds (up to approx. 80 km/h) the rolling resistance coefficient remains largely constant. The table with values per surface shown at the top of this paragraph assumes these speeds. At higher speeds, rolling resistance increases, especially with standard and winter tires. Mathematically, the increase in rolling resistance can be approximated with a quadratic relationship. The graph below shows this effect:
- SR tires (standard tires) have relatively high hysteresis and show the strongest increase in rolling resistance at higher speed.
- M+S tires (winter tires) fall between SR and HR tires in performance. These tires have extra tread and sipes, which increases rolling resistance.
- HR tires (high performance) have reinforced carcasses and low-hysteresis rubber compounds. They are most efficient at high speed and show the smallest increase in rolling resistance.
When the rolling resistance coefficient and the vehicle weight are known, rolling resistance can be calculated. The following data are known:
- BMW X3 with a mass (m) of 1700 kg;
- Acceleration due to gravity (g): 9,81 m/s^2;
- Rolling resistance coefficient (μ): 0,010;
- Level road surface.
First, we multiply the vehicle mass by the acceleration due to gravity to calculate the normal force (Fn):
Next, we multiply the normal force by the rolling resistance coefficient to calculate the force required to overcome the tires rolling resistance on the road surface.
Grade resistance:
When a vehicle drives uphill, grade resistance occurs. This resistance arises because part of the gravitational force acts opposite to the direction of travel. Additional engine force is therefore required to drive the vehicle uphill at constant speed or while accelerating.
When driving uphill the full gravitational force no longer acts perpendicular to the road surface, but partly along the slope. This changes the distribution of forces on the vehicle:
- The component perpendicular to the road surface determines the normal force (Fn), which affects rolling resistance.
- The component parallel to the road surface produces the grade resistance (Fhelling).
Over a distance of 100 meters the vehicle has climbed 5 meters (see image). That means the grade is 5%. We calculate the grade angle with the tangent (tan).
Calculating tan α:
tan α = opposite / adjacent = 5 / 100
α = tan⁻¹ (5/100) = 2,86°
Tip: on the calculator press Shift and then the tan button to get tan ̄ ¹, and put (5/100) in parentheses). The result can be displayed in degrees or radians, depending on your calculators settings. To convert radians to degrees, use the following formula:
Degrees = Radians * (180 / π)
Rolling resistance becomes slightly lower when driving up a grade, because the normal force decreases. A smaller portion of the gravitational force then acts perpendicular to the road surface, so the tires press less firmly on the road. That leads to less deformation and therefore less rolling resistance.
In the equation for rolling resistance this is represented as follows (the cosine of the grade angle determines how much force still acts perpendicular):
The influence on rolling resistance is small, for example only 0.21 N in this example, and is neglected in most practical situations. We can calculate the grade force (Fhelling) by multiplying the normal force (Fn) by the grade angle. We call the angle sine (sin) alpha. The sine of the grade angle determines how much gravitational force acts along the slope.
It takes a force of just over 832 Newtons + the rolling resistance of 166,56 N to drive up the grade. We can also combine the formulas for rolling and grade resistance, because the grade also affects rolling resistance. Note that aerodynamic drag has not yet been included here, so this is not yet the total driving resistance! That follows later on this page.
Aerodynamic drag:
While driving, the vehicle experiences resistance due to airflow. This is called aerodynamic drag. As speed increases, aerodynamic drag increases quadratically. As a result, the vehicle will accelerate less strongly as vehicle speed increases.
When driving on a secondary road, the difference in fuel consumption between 60 and 80 km/h will be minimal. The difference in consumption between 120 and 140 km/h is much greater due to the increasing aerodynamic drag. Fuel consumption is often most favorable around 90 km/h as a result of the ideal engine speed in the highest gear; see the page about specific fuel consumption.
The force required to overcome aerodynamic drag can be calculated as follows:
Explanation of the formula:
½ = one-half, which can be entered on the calculator as 0.5;
ρ = Rho. This denotes density. In this case the density of air, with the m³;
Cw = drag coefficient. For a passenger car the Cw value is between 0.25 and 0.35. For a truck between 0.65 and 0.75;
A = frontal area of the car (determined in the wind tunnel) in m²;
v² = the vehicle speed squared, with the unit m/s;
For this calculation we use the following data:
- ρ = 1.28 kg/m³ (depending on temperature and humidity)
- Cw = 0.35
- A = 1.8 mb2
- v² = 100 km/h = (100 / 3.6) = 27.78 m/s² (meters per second squared because it concerns an acceleration):
With the known data we fill in the formula for Flucht:
Thus a force of 311.11 N is required to overcome aerodynamic drag.
Acceleration resistance:
During acceleration or deceleration, acceleration resistance arises. Force in Newtons is required to overcome this resistance. We again assume the BMW X3 with a vehicle mass of 1700 kg.a0
The force required to overcome acceleration resistance (Facceleration) depends on the vehicle mass (m) and the acceleration or deceleration (a) in m/s^2. In this example we assume a minimal deceleration ofa0
0.2 m/s^2. The required force to overcome acceleration resistance can be calculated with the following formula:
Total driving resistance:
The total driving resistances (Frij) are all of the previously mentioned resistances added together. Rolling resistance + grade resistance + aerodynamic drag together equal Frij:
Conclusion: to drive on a 5% grade at 100 km/h with an acceleration rate of 0.2 m/s² in calm conditions (0 Bft), the total required force is 1,649.78 N.
Not only the driving resistances, but also the efficiencies and reductions in the transmission are important for the designer to calculate in advance.
The transmission and the gear ratios are matched to the characteristics of the engine. This is described on the page about gear ratios.
