Welcome to the third installment of our series on dirt late model vehicle dynamics. In part two, we looked at some of the basic concepts of engineering analysis to use as a procedural baseline. We also discussed the eight unknown values of force and displacement that we will be solving for. With this understanding, we can now begin to look at the equations that will make up the mathematical model defining our race car.

As discussed in part two, the first subgroup of equations that we will be developing are the load transfer equations. The equations in this group define how the chassis will react to external inputs of force, aerodynamics, etc. applied to the vehicle.

It is worth taking a few moments here to discuss the concept of a resultant vector. A race car is made up of a large number of independent masses that are all physically connected. For example, the fuel cell can be thought of as an independent mass with its own center of mass, moment of inertia, etc. The engine is another mass with its own center of mass, moment of inertia, etc. The driver, the transmission, each of the four wheels are all independent masses that could have a force analysis performed on them independently. In fact, if you wanted to get real detailed, you could break the car down into every nut, bolt, washer, and rivet on the car having its own center of mass, moment of inertia, etc. Although in theory this is possible, and is actually how a CAD system would work, it isn’t a very realistic thing to do when developing our model. To simplify things, we can leverage the concept of a resultant vector, or in our case the resultant force vector. When doing a force analysis, it is common practice to combine all of the independent applied forces on a body, each with its own magnitude and direction, into one equivalent resultant force with its own magnitude and direction. When this resultant force is applied to the body, the body will react exactly the same as if each of the separate forces were applied independently. Similarly, we also define an equivalent center of mass. The equivalent center of mass is the center of mass of the body that is made up of all the independent masses (i.e. the fuel cell, the engine, the driver, the ballast, etc.). The resultant force vector is applied at the equivalent center of mass, which yields the same result that we would get if we evaluated each of the independent masses separately. What does this mean in the real world? Think about what happens when you move a piece of ballast, a battery, or a fuel cell around in the chassis. You are not changing the applied forces; you are simply moving the location of the center of mass. Keep this in the back of your mind as we develop our system of equations!

The resultant force vector applied to the center of mass is a combination of the gravitational force, the centrifugal force, and the acceleration and/or braking forces acting on the chassis. Note that all of these forces can be thought of as “external” forces acting on the chassis. To understand this concept, let’s think about the unrealistic scenario of a car driving in a circle on a flat surface with no body roll and no yaw angle relative to the tangential line of the circle’s curve, and with a moderate degree of forward acceleration. The gravitational force will act directly on the center of mass straight down, or in the negative “z” direction. This will be the “z” component of the resultant force vector. As the car drives around the curve of the circle, a centrifugal force is developed along the “y” axis of the car. Assuming that the car is turning left, this force will point towards the right of the car, or will be in the negative “y” direction. In response to this centrifugal force, there is a centripetal force developed, which is distributed among the four tire contact patches, that acts to hold the car on the path of the circle. This centripetal force is equal and opposite to the centrifugal force, assuming a constant radius curve. This is the case since we are traveling along a circular path. The centrifugal force will be the “y” component of the resultant vector. As the car accelerates forward, the rear tires are acting to push the car. At this point, we will assume the forward thrust is distributed equally between the left and right rear tires, and that there is no rear steer; however, neither would be the case in the real world. As the tires apply a force forward at the tire contact patches in the positive “x” direction, this force is reacted at the center of mass in the opposite direction, or in the negative “x” direction. This reaction force is the “x” component of the resultant force vector.

Under the scenario described above, we would have a resultant force vector going through the center of mass and pointing somewhere towards the right rear tire contact patch. In the real world, this would be a much more complicated scenario. There would be banking, the chassis would have body roll, and there would be some degree of chassis yaw angle from the tangential of the curve path. If these values are known, components of the resultant force vector can be modified accordingly. However, it is quite difficult to predict all of these inputs. How often do you know a tracks banking angle, how often do you know the radius of curvature, how do you know how much yaw the car has? All of these can be estimated to some degree, but there will inevitably be some amount of error involved. Obtaining acceleration data from a DAQ system is a much easier way of calculating the component forces. Simply mount the DAQ system in line with the chassis coordinate system, and the three axis accelerometer will give you all the accelerations at any given point around the track. Forces can then be calculated using the mass of the vehicle. Note that smoothing of the DAQ data will generally be required to get acceleration data. Throughout this series, we will assume that DAQ data is available and we will not worry about estimating these forces using track geometry and velocity.

For our system, let’s assume that the DAQ system is telling us that at the point in question, the chassis is experiencing the following:

z-axis: -1.1g
y-axis: -0.8g
x-axis: -0.5g

Let’s also assume that our car weighs 2350 lbs.

The term “g” is telling us that the acceleration is some multiple of one unit of gravity. One “g” is equal to approximately 32.2 ft/sec2, so 1.1g is 1.1 times the acceleration due to gravity, 0.8g is 0.8 times the acceleration due to gravity, so on and so forth. To find the applied force, we can simply use Newton’s equation of F=ma. We are using our “g” value as a scaling factor for the acceleration. We can also use it for a scaling factor for the force since force is equal to the mass (or a constant) times the acceleration. There is a linear relation between the force and the acceleration. For our example, the vertical force exerted on the chassis will be -1.1g x |2350 lbs.|, or -2585 lbs. You may be asking yourself where this extra 235 lbs. came from, and why is the weight negative. The extra force is more than likely induced force from the bank angle of the track. There will also be additional forces due to aerodynamic effects; however, the accelerometer will not detect aerodynamic forces. Aerodynamic forces, in general, can be thought of as an additional external force acting at the center of pressure. The force is negative because it is pointing in the negative “z” direction. Remember that a force is a vector, so it has magnitude and direction. We use the magnitude of the weight as an input to the equation because we are only interested in the quantity of the weight at this point. The acceleration vector gives the direction, and the output of the equation is the resulting effective weight, with direction.

This leads us to our first equation. For equation number one, we are going to sum the forces in the “z” direction. We will be solving the system of equations using a quazi-static approach, so we will set the equations in our system equal to zero. Equation one looks like this:


This equation is adding all the forces in the “z” direction. This is the applied force of the vehicles weight multiplied times the g-factor plus the load induced by aerodynamic effects plus the reaction forces at each of the four contact patches (Pn). Note that if the weight of the car and the aerodynamic forces are pointing down, or are negative, the sum of the four contact patches must be positive to satisfy the equation and make it equal to zero. Since our tires are not rails, although some of us like to think we are running on rails, they cannot create a negative force holding the car down (unless you are using some really good tire dope); therefore, the “Pn” must always be positive or zero. This means that the sum of the tire forces will always support the weight of the car plus the induced aerodynamic forces. This, of course, assumes we are generating aerodynamic down-force and not aerodynamic lift greater than the effective weight of the car sending the car into flight. The figure below shows a free body diagram for equation number one. Notice that the force vectors are represented by arrows, and that the arrows are all pointing in the positive direction. Drawing the vectors in the positive direction is standard practice when developing free body diagrams. This helps to keep the signs straight when writing the equations. We let the vector go positive or negative in the equation as needed depending on the inputs to the equation. A negative quantity indicates that the vector direction is opposite the arrow shown, so in our case the weight will have a negative value meaning that the force will be in the down direction, opposite the drawn arrow.


For the second and third equations in our system of equations, we will be summing the moments about the x-axis and the y-axis respectively. In order to do this, we will utilize the cross product to calculate the moments. The cross product is a 3D vector mathematics tool used to calculate the moments about a point around the x, the y, and the z axes at the same time. We will not spend a lot of time discussing the tool as it can be found in any appropriate math or engineering text. The cross product is:

undefinedThe value “Mo“ is the moment and is a vector. The “r” is the positioning vector. Since we are taking moments about the axes, we will find the three moments about the origin point, so in our case the tail of the positioning vector will be the origin, and the head will be the point of force application. The “F” is the force vector that is applied at the given point. For simplicity, we will find the moment about the origin from each applied force vector, then we will sum the respective component from each of the cross product results. A free body diagram is shown in the picture below for the external forces applied at the center of mass, and the reaction forces applied at the left rear tire contact patch. Only the two forces are shown to keep the picture uncluttered; however, in practice you would do this for all of the forces in the system that the chassis will see. The following is a list of forces that would most likely be encountered in a race car application. At this point, we are only looking at forces that act on the vehicle as a whole, not internal forces such as springs, shocks, etc. You can imagine drawing a boundary box around the car and looking at things that would act on the exterior of this box.

Force Examples
• Applied force at the center of mass
• Reaction force at the left front tire contact patch
• Reaction force at the right front tire contact patch
• Reaction force at the left rear tire contact patch
• Reaction force at the right rear tire contact patch
• Applied vertical aerodynamic force at top center of pressure
• Applied lateral aerodynamic force at side center of pressure
• Applied longitudinal aerodynamic force at front center of pressure



For our second equation, we are going to sum the moments about the x-axis. There is one more step we need to do before we write our equation. We need to observe the fact that at each of the four tire contact patches, we only know two of the three components. We know the lateral reaction force and the longitudinal braking/acceleration force, but we know nothing about the vertical component, so we will need to break it out from the cross product calculation. We will still perform the cross product at the four contact patches, but we will only use the “x” and “y” components and leave the “z” component equal to zero. We will account for the “z” component separately. Let’s build equation two in multiple steps. First, we will sum all of the known moments about the x-axis from our cross product calculation results. We will call this sum “MXa”.

Next, we will use this result to write equation two. Equation two looks like this.


In this equation, “ti” is the half-track position from the X-Z plane to each of the four tire contact patches. As you can see in the equation, we are starting to add some definition to our mathematical model that is telling us how lateral load is transferred from one side of the car to the other in response to some force inputs. We can start to see how much load is transferred laterally, and how this load transfer is distributed between the front and rear wheels of the car. Remember, however, that this is only load transfer as a result of a moment about the x-axis, or the longitudinal axis, of the car.

For our third equation, we will replicate what we did for equation two, but we will do it about the y-axis. We will sum all of the known moments about the y-axis from our cross product calculations, and we will call this sum “Mya”. 


We will use this result to write equation three. Equation three looks like this. 


In equation three, “li” is the half-wheelbase position from the Y-Z plane to each of the four tire contact patches. Noticed that we are using a negative sign as opposed to the positive sign that we used in equation two. This is to stay consistent with the form of the cross product to keep the moments going in the right direction. With this equation, similar to equation two, we can now start to see how much load is transferred longitudinally, and how this load transfer is distributed between the left and the right wheels.

We now have the first three equations in our system of equations.


What exactly can we learn from looking at the equations? How have you thought about weight transfer in the past? What things did you think impacted weight transfer? Was it influenced by springs or shocks? Do we see springs or shocks in our system of equations? No, we don’t. As we can see from the equations, springs and shocks have nothing to do with the amount of weight transfer. In fact, the only factors that impact weight transfer are the externally applied forces, static weight of the car, the center of mass location, the half-wheel bases, and the half-wheel tracks (widths). With this in mind, start thinking about how different adjustments made to a race car impact these factors. What does moving the rear end to the left or right do? What does moving ballast do? What does moving the right front out an inch do? What happens to weight transfer when we induce rear steer into the car?

In part 4 of this series, we will start to look at the equations in our model that govern how this load transfer is distributed throughout the elements (i.e. shock, springs, links, etc) of our race car. We will also look at how we can use this information to predict the suspension travel.

As always, if you find this blog helpful, then you can help me in return by thinking of Bartlett Motorsport Engineering next time you need to buy parts for your race car.


Joe Bartlett