What is Forces Acting on a Rolling Vehicle?

What is Forces Acting on a Rolling Vehicle? – Whatever type of four-wheeled vehicle you are, all need the right concept to build it all. Moreover, electric cars are actually the result of the development of conventional cars. So for that, we will look at Forces Acting on a Rolling Vehicle as one of the basic vehicle concepts.

What is Forces Acting on a Rolling Vehicle?

The picture below shows the major forces at work on a rolling vehicle.

Forces Acting on a Rolling Vehicle
Figure: 1. Force diagram for a rolling vehicle.

F_{ad} is aerodynamic drag F_{rr} is rolling resistance, F_{hc} is hill climbing (a proportion of the
gravitational force
mg as defined by the angle \psi), F_{la} is linear acceleration, F_{\omega a} is angular
acceleration, and F_{te}
, the sum of all of them, is the total tractive effort required to move the
vehicle.

There are a variety of other forces acting on the car, such as lift (the tendency for a fast car to want to take off like an airplane) and wind resistance (above and beyond aerodynamic drag through still air). But the forces illustrated in Fig. 1 are sufficient to form a satisfactory model of the vehicle’s behavior for our purposes.

Based on Fig. 1, the total force (in newtons) required to move a vehicle on wheels is defined by Eq. (3.1):

Forces Acting on a Rolling Vehicle

As shown, the force required to move the car forward, F_{te}, is equal to the sum of all of the forces opposing that forward motion.

Aerodynamic Drag (F_{ad})

One of the primary forces acting on a vehicle is aerodynamic drag, which is the oppositional force imparted on the vehicle by the air that it collides with as it moves forward. The drag force can be calculated using Eq. (3.2):

Forces Acting on a Rolling Vehicle

where \rho is the density of air (in kg/m^3), A is the frontal area of the vehicle that is pushing against the air (in m^2), C_d is the drag coefficient, and v is the velocity of the vehicle (in m/s).

\rho is a measure of how much mass the vehicle has to push out of the way to travel through it. For example, it’s easier to walk through air than it is to walk through water, because air is less dense than water. At sea level, \rho is 1.225 kg/m^3, but as you rise in altitude, this figure decreases. For example, the air density atop Mt. Everest is about a third of what it is at sea level. In general, for our calculations, we will assume our vehicle is driving at sea level.

A is a measure of the area that the vehicle is presenting to the air. If you put the vehicle in front of a backdrop, and looked at it from the front, A is equivalent to the two-dimensional area of the backdrop that would be blocked by the vehicle. The bigger the area, the more air needs to be pushed out of the way to get through. As an analogy, imagine pushing the point of a large umbrella through the air when it’s closed as shown in Fig. 2. Now imagine pushing the top of the umbrella through the air when it’s open: the larger area presented by the open top of the umbrella makes it harder to push through the air.

Forces Acting on a Rolling Vehicle
Fig.2 The second umbrella has larger A.

Aside from how big an object is, C_d is a measure of how “draggy” it is. If you turn the umbrella around and try to pull it through the air with the bottom facing forward, so that it catches the air like a bowl, it will be even more difficult than pushing the top through the air, even though their areas are exactly the same. This means that the concave bottom of the umbrella has a higher C_d than the convex top of it. (Note that in Figs. 2 and 3, the arrows represent the relative flow of air as all of the umbrellas move through it to the right.)

Forces Acting on a Rolling Vehicle
Fig. 3 The second umbrella has larger C_d

Technically, C_d is a function of the circumstances that a vehicle finds itself in, including its velocity. But for simple analyses like the one we’re doing here, C_d is usually treated as a constant for a given vehicle.

Given that C_d and A are both constants relating to a particular vehicle, they are often combined to create a compound constant, C_dA. The following table shows the values of C_d and C_dA for several popular electric vehicles:

Forces Acting on a Rolling Vehicle

Why is F_{ad} proportional to v^2? F_{ad} is the force required for the car to push the air out of the way: in other words, it’s the force required to change the momentum of the air in front of the car. Momentum is equal to mass times velocity ( p=mv), and force is equal to mass times acceleration (F=ma), so force is equal to change in momentum over time F=m\frac{dv}{dt}.

Imagine that the front of the car was able to push all of the air in front of it out of the way at velocity v. The amount of the air being pushed out of the way each second is equal to the density of the air, \rho (in kg/m^3), times the frontal area A (in m^2), times the velocity (in m/s), for final units of kg/s, mass over time. If we multiply by the velocity (m/s) a second time, we get exactly what we’re looking for: mass times velocity (i.e., momentum) over time, which gives us the force required to push the air out of the way.

The reason we multiply \rho Av^2 by a C_d>1 is that the front of the car isn’t actually pushing 100 percent of the air out of the way: a good proportion of it is actually slipping by, so the total force is less than the full \rhoAv^2. The reason we multiply it all by 1/2 is simply an artifact of the way that C_d is defined elsewhere in the field of fluid dynamics. In an alternate universe, we could’ve defined all values of C_d as half of their current value and taken the 1/2 out of this equation, but for reasons unrelated to this equation, it’s more elegant to define C_d as the larger value and add the 1/2 to this equation.

Rolling Resistance (F_{rr})

In addition to overcoming air resistance, the vehicle needs to overcome rolling resistance, which is a loss generated by the wheels in contact with the road surface. While there are many things that can contribute to rolling resistance, the primary component of it is hysteresis: as the wheel rolls along the road, the wheel (and to a lesser extent, the road surface) is constantly being deformed, which causes a loss of energy in the form of heat.

Rolling resistance is calculated by Eq. (3.3):

    \[F_{rr}=\mu_{rr}mg\]

where \mu_{rr} is the coefficient of rolling resistance, m is the mass of the vehicle (in kg), and g is gravity (9.81 m/s^2). Alternatively, mg can be thought of as the weight of the vehicle (in newtons).

The direct proportionality of rolling resistance to weight makes sense: the harder the vehicle is pressing down on its wheels, the more those wheels are likely to deform, resulting in greater rolling resistance. (As an interesting aside, this is also why it’s important to keep your car tires inflated to the recommended pressure: when the pressure is low, the tires deform more, resulting in higher rolling resistance, and as a result, higher energy consumption.)

The direct proportionality of rolling resistance to weight also gives us a helpful way of thinking about what it actually means. If the \mu_{rr} of our tires is 0.01 (a common value), that means that the force (in pounds) required to move a load on those tires is 0.01 times (i.e., 1 percent) of the weight of that load in pounds. Therefore, it takes only 1 lb of force to move a load of 100 lb on wheels with a \mu_{rr} of 0.01.

The following table shows some common values of \mu_{rr}:

Forces Acting on a Rolling Vehicle

And this table shows the value of m for several popular electric vehicles:

Forces Acting on a Rolling Vehicle

Hill Climbing (F_{hc})

Aerodynamic drag and rolling resistance are the two major forces opposing a vehicle traveling at a constant velocity on a flat surface, but if you’re traveling up a hill, you’ll also need to account for gravity. The force required to travel up a hill is calculated by Eq. (3.4):

    \[F_{hc}=mg\sin{\psi}\]

where m  is the mass of the vehicle (in kg), g is gravity (8.91 m/s^2), and \psi is the vertical angle of the road relative to flat (in radians).

Figure 4 illustrates why sine is the appropriate function to use here. The two angles \psi are congruent because they have the same relationship to horizontal and vertical. The force pulling back on the car is mg sin ψ because we’re interested in the opposite side of the vertical triangle relative to the hypotenuse of the vertical triangle, and sine gives us the ratio of opposite to hypotenuse (the SOH of the famous SOH CAH TOA).

Forces Acting on a Rolling Vehicle
Fig. 4 Hill climbing force is equal to mg\sin{\psi}.

In most cases, the hill climbing force equation can be simplified by using the small angle approximation, which states that for small angles (in radians) (Eq. 3.5):

    \[\sin\theta\approx\theta\]

This fact is clearly illustrated in Fig. 5.

Forces Acting on a Rolling Vehicle
The small angle approximation.

The error of the small angle approximation is less than 1 percent until you reach 0.24 rad (14°), so as long as your roads are flatter than that, it’s a perfectly reasonable assumption to make. And your roads are almost certainly flatter than that! For reference, the maximum allowable angle of federally funded highways in the United States is around 3.5°, and the maximum recorded angle of any road in the world, Ffordd Pen Llech in Wales, is 20.5°. Even on Ffordd Pen Llech, the error of using the small angle approximation is just over 1 percent, far less than the error that will be introduced by other aspects of our calculations.

So, in general, we’re free to simplify to (Eq. 3.6):

    \[F_{hc}\approx mg\psi\]

as long as \psi is in radians. In any case, it’s important to note that if the vehicle is going downhill, \psi, and therefore F_{hc}, will be negative, as gravity pulls the vehicle forward down the slope and reduces the force that the motor needs to provide in order to keep the vehicle moving forward (or increases the force required to slow it down when braking).

Linear Acceleration (F_{la})

If the vehicle is traveling at constant velocity, the three forces described above are sufficient to form a basic model of the vehicle’s behavior. But much of the time (and nearly all of the fun times!) a vehicle is also accelerating. The linear acceleration of a vehicle (along the road) is defined by the basic force law (Eq. 3.7):

    \[F_{la}=ma.\]

The faster you accelerate, and the more massive the vehicle, the more force will be required. Note that if the vehicle is slowing down, a, and therefore F_{la}, will be negative. 

Angular Acceleration (F_{\omega a})

While it’s clear that a force is required to accelerate a vehicle down the road, there’s another kind of acceleration going on that’s a bit less obvious. Every time the vehicle accelerates, there are a variety of rotational parts inside the vehicle that need to be spun up, which also requires a force (or more accurately, a torque). Given that the force required for angular acceleration within the vehicle is usually much smaller than the force required for linear acceleration of the vehicle itself, and the calculations involved are significantly more difficult, requiring numbers that are much less readily available, we will follow the suggestion of Larminie & Lowry (2012) and use a fudge factor that assumes F_{\omega a} is around 5 percent of F_{la}.

Therefore, we will say that (Eq. 3.8):

    \[F_{la}+F_{\omega a}\approx 1,05 F_{la}.\]

If the vehicle is slowing down, F_{la}+F_{\omega a} will be negative, as a negative force is required to reduce both the linear momentum of the vehicle and the angular momentum of its rotating parts.

Tractive Force (F_{te})

As we saw at the beginning of the chapter, the total tractive effort required to move the vehicle is simply the sum of all of the forces we have discussed (Eq. 3.9):

    \[F_{te}=F_{ad}+F_{rr}+F_{hc}+F_{la}+F_{\omega a}.\]

Elaborating each term as we have defined them, we get (Eq. 3.10):

    \[F_{te}=\frac{1}{2}\rho AC_d v^2+\rho_{rr}mg+mg\sin\psi+1.05 ma.\]

Putting in our assumptions and approximations for a vehicle driving at sea level on gentle slopes, we get (Eq. 3.11):

    \[F_{te}=0,6125 AC_d v^2+9,81 \rho_{rr}m+9,81m\sin\psi+1.05 ma.\]

To clean things up a bit, let’s divide the last three terms by 9.81 m so we only have to multiply by m once. This leaves us with the following elegant equation for the total force required to move a rolling vehicle on gentle slopes at sea level: 

    \[F_{te}=0,6125 AC_d v^2+9,81m( \rho_{rr}+\sin\psi+1.05 a).\]

In order to calculate the force required to move a rolling vehicle, all we need to know is: its frontal area (A), its drag coefficient ( C_d), its velocity (v), its mass (m), the coefficient of rolling resistance for its wheels (\mu_{rr}), the angle of the road (\psi), and the vehicle’s current rate of acceleration (a), if any.

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