Bartlett Motorsport Engineeringhttp://blog.bartlettmotorsportengineering.com/feed.php2017-07-07T17:59:26+00:00Dirt Late Model Vehicle Dynamics - Part 4<p>Welcome to the fourth installment in our series on dirt late model vehicle dynamics. In part three, we looked at the development of the first three equations in our dynamic system. We are referring to the first three equations as the “load transfer equations”. In part four, we are going to take a closer look at how engine torque reactions are incorporated into the second and third equations.</p>
<p><!-- pagebreak --><br />The first three equations in our system focus on how the vehicle reacts to external inputs of force such as gravity, lateral acceleration, and longitudinal acceleration. When looking at the three sources, we have to understand that they are the reactions to some change in state of the system that cause the accelerations. Newton laws are in effect here. First, we have gravity which is the result of a natural source of acceleration driven by the relative mass between two objects (a topic far beyond the scope of this blog). Second, we have lateral acceleration which is the result of the changing direction a body takes as it traverses a curve, and is a function of the bodies speed and the curve radius. Third, we have longitudinal acceleration which is the reaction to the torque generated by the engine. This torque is the result of a conversion of “external” chemical energy, in the form of liquid fuel, to mechanical energy, in the form of rotational torque. The power generated by the engine can be thought of as an external source of energy acting on the vehicle.</p>
<p><br />In each of the scenarios noted above, the sources of external force must eventually be traced back to a “fixed ground” for each of them to “push” against. You can visualize this by thinking of a hydraulic cylinder. One end of the cylinder needs to be grounded in order for the other end to generate force and motion to move or support a load. Gravity is a source of energy that is constant and does not need to “push” against anything solid to be generated. It is, in a sense, its own source of solid foundation to push against. The change of direction that generates the lateral acceleration is from the slip angle forces generated in the front tires as a result of steering input. In this case, the Earth is the ground that the tire pushes against to change vehicle direction. This force transmits up through the suspension to the chassis to cause yaw rotation. We will talk more about this later. The torque from the engine is slightly different from the other two scenarios. The “fixed ground” in this case is the chassis itself. This means that the reaction from the external energy is “grounded” within the very system we are looking at; therefore, we need to account for it in our dynamic equations.</p>
<p><br />The engine torque has an action and a reaction. We can think of the engine as the “action” side and the axle housing as the “reaction” side. In reality, both the engine and axle housing responses will be reactions to the explosion in the combustion chamber, just in opposite directions, but we will call the axle side the “reaction” and the engine side the “action” to keep things separated verbally. We can actually think of the entire chassis, minus the rear axle, as being on the “action” side since the engine is bolted directly to the chassis frame.</p>
<p><br />Let’s take a look at the “reaction” at the axle. As engine torque is transmitted to the wheel, it follows a path along the driveshaft, makes a 90° turn at the axle ring & pinion, and splits down each axle shaft until it finally reaches each of the wheels. Since there are two paths, the axle housing will react in two directions. The first reaction will be along the axis of the axle pinion shaft, which is roughly in line with the x-axis of our coordinate system. This reaction is in response to the drive shaft torque and is countered and distributed to the rear tire contact patches through the axle housing, and conversely to the front contact patches through the frame and front suspension. The second reaction is in response to the pinion gear in the rear end trying to “climb” the ring gear. This reaction will be along the axis of the axle shafts, which are roughly in line with the y-axis of our coordinate system. This reaction will be countered by a fifth-arm device, a pull-arm device, or directly through the suspension links depending on what type of rear suspension configuration is used. A fifth-arm device is assumed in this example. We will evaluate each of the reactions with respect to the x-axis and to the y-axis and incorporate them into equations two and three.</p>
<p><br />In order to integrate the reactions into our system of equations, we will need to know what type of forces and moments we are dealing with. We can establish this information in various ways. Since we are utilizing information from a data acquisition system, we can determine what our longitudinal accelerations are at various points. We can use this information along with the loaded radius of the tire/wheel assemblies and the gear ratio of the ring and pinion in the rear end to determine what the moments are along the axle and pinion axes. Once we know the moments, we can use them, along with our known suspension geometry points from a positional analysis, to establish what the reaction forces are and where they are applied. They can then be integrated into equations two and three accordingly.</p>
<p><br />Illustration #1 below shows a side planar view of how the fifth-arm forces would be applied when looking at the moments. This is for demonstration purposes only. In reality, we would use the cross product method described in part 3 to ensure that we are accounting for moments about both the x and the y axes. The blue arrows indicate the two ends of the fifth arm. One end is attached to the axle and pushes down on the axle housing, while the other end is attached to a “fifth arm shock” coil over assembly and pushes up on the chassis frame. This action generates what is commonly referred to as anti-squat. If the position of the fifth arm shock is positioned so that the rear chassis lifts up under acceleration, we have greater than 100% anti-squat. If it is positioned so that the rear goes down under acceleration, we have less than 100%. Perfect 100% anti-squat would result in no upward or downward movement under forward acceleration. Note that fully decoupled rear suspensions typical on most dirt late models have other mechanisms that contribute to anti-squat. The left and right four-bar links, along with the J-bar, all contribute to anti-squat when conditions are right. When using a decoupled rear suspension, we typically always have significantly more than 100% anti-squat. The amount of rear lift is controlled by the use of a limiter chain that controls the amount of separation between the frame and the axle. This is a good time for the reader to take a minute and think about what is happening here. What will happen if you put a longer fifth-arm on the car? What will happen if you put a shorter arm on and move the limiter chain left or right along the axle tube? For reference, test conducted by Bartlett Motorsport Engineering show that forces in a fifth-arm spring can be upwards of 800 to 900 lbs. when using a 36 to 38 inch fifth-arm.</p>
<p> </p>
<p><img class="nb-align-center" src="http://blog.bartlettmotorsportengineering.com/content/public/upload/sideview4_0_o.png" alt="undefined" /></p>
<p style="text-align: center;">Illustration #1</p>
<p>Illustration #2 shows the reaction forces with respect to the x-axis. Since most domestic engines rotate CCW as viewed from the rear, the “action” on the frame will be to rotate CW. This, of course, assumes that the crankshaft axis is parallel to the x-axis of our coordinate system. This is seldom the case in real life, and can be corrected for, but to simplify things in our case, we are going to assume that it is parallel. As the engine transmits torque to the rear-end through the drive shaft, the entire axle housing will try to rotate with the drive shaft causing it to rotate CCW. As this action-reaction occurs, force #1 will decrease, force #2 will increase, force #3 will increase, and force #4 will decrease. This can be validated anytime you spin the right rear tire of a car with a standard differential. As engine power is applied, the chassis rotates CW while the axle housing rotates CCW. The imbalance in roll couple between the front and rear causes the axle housing to rotate more than the chassis; consequently, the right rear tire unloads. As a result, power is diverted to the path of least resistance (i.e. the right rear), and the tire spins. The dashed lines, intentionally drawn opposite direction, indicate the input forces from the reaction moments. They can be found using the known geometry of the four tire contact patches and the known reaction moments to develop two equations with two unknowns. For the front, use the “action” from the engine and take a moment about contact patch one for the first equation, then about contact patch two for the second equation. Repeat for the rear accordingly. In reality, if the chassis were symmetrical, you could probably get away without even doing this, and the forces would fall out in equations to be developed in later blog post; however, with an asymmetric chassis, it is best to break them out and define them here for added detail.</p>
<p> </p>
<p><img class="nb-align-center" src="http://blog.bartlettmotorsportengineering.com/content/public/upload/torquereact4_0_o.png" alt="undefined" /></p>
<p style="text-align: center;">Illustration #2</p>
<p>In part five, we will start to look at each of the four corners of the car and determine the path that the forces follow through the suspension, and how much each corner of the suspension will move.</p>
<p><br />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.</p>
<p><br /><a href="http://shop.bartlettmotorsportengineering.com/" target="_blank">http://shop.bartlettmotorsportengineering.com/</a></p>
<p><br />Thanks,<br />Joe Bartlett</p>http://blog.bartlettmotorsportengineering.com/index.php?controller=post&action=view&id_post=92017-07-07T17:59:26+00:00Dirt Late Model Vehicle Dynamics - Part 3<p style="text-align: left;">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.</p>
<p><!-- pagebreak -->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.</p>
<p>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!</p>
<p>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.</p>
<p>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.</p>
<p>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:</p>
<p style="text-align: center;">z-axis: -1.1g<br />y-axis: -0.8g<br />x-axis: -0.5g</p>
<p style="text-align: center;">Let’s also assume that our car weighs 2350 lbs.</p>
<p>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.</p>
<p>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:</p>
<p style="text-align: center;"><br /><img src="http://blog.bartlettmotorsportengineering.com/content/public/upload/eqsumfz-1_0_o.jpg" alt="undefined" /></p>
<p style="text-align: left;"><br />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.</p>
<p style="text-align: left;"><img class="nb-align-center" src="http://blog.bartlettmotorsportengineering.com/content/public/upload/sumfz-2_0_o.jpg" alt="undefined" /></p>
<p style="text-align: left;"><br /> <br />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:</p>
<p style="text-align: left;"><img class="nb-align-center" src="http://blog.bartlettmotorsportengineering.com/content/public/upload/eqcrossproduct_0_o.jpg" alt="undefined" />The 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.</p>
<p style="text-align: left;"><br /><span style="text-decoration: underline;">Force Examples</span><br />• Applied force at the center of mass<br />• Reaction force at the left front tire contact patch <br />• Reaction force at the right front tire contact patch<br />• Reaction force at the left rear tire contact patch<br />• Reaction force at the right rear tire contact patch<br />• Applied vertical aerodynamic force at top center of pressure<br />• Applied lateral aerodynamic force at side center of pressure<br />• Applied longitudinal aerodynamic force at front center of pressure</p>
<p style="text-align: left;"> </p>
<p style="text-align: left;"><img class="nb-align-center" src="http://blog.bartlettmotorsportengineering.com/content/public/upload/summomentsx-yaxes-2_0_o.jpg" alt="undefined" width="693" height="416" /></p>
<p style="text-align: left;">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”.</p>
<p><img class="nb-align-center" src="http://blog.bartlettmotorsportengineering.com/content/public/upload/eqsummxa_1_o.jpg" alt="undefined" /><br />Next, we will use this result to write equation two. Equation two looks like this.</p>
<p style="text-align: left;"><img class="nb-align-center" src="http://blog.bartlettmotorsportengineering.com/content/public/upload/eqsummx-2_0_o.jpg" alt="undefined" /></p>
<p style="text-align: left;">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.</p>
<p style="text-align: left;">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”. </p>
<p style="text-align: left;"> <img class="nb-align-center" src="http://blog.bartlettmotorsportengineering.com/content/public/upload/eqsummya_1_o.jpg" alt="undefined" /></p>
<p style="text-align: left;">We will use this result to write equation three. Equation three looks like this. </p>
<p style="text-align: left;"><img class="nb-align-center" src="http://blog.bartlettmotorsportengineering.com/content/public/upload/eqsummy-3_0_o.jpg" alt="undefined" /></p>
<p style="text-align: left;">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.</p>
<p style="text-align: left;">We now have the first three equations in our system of equations.</p>
<p style="text-align: left;"><img class="nb-align-center" src="http://blog.bartlettmotorsportengineering.com/content/public/upload/eq1-3list_0_o.jpg" alt="undefined" /></p>
<p style="text-align: left;">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 <span style="text-decoration: underline;">amount</span> 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?</p>
<p style="text-align: left;">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.</p>
<p style="text-align: left;">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.</p>
<p style="text-align: left;"><br /><a href="http://shop.bartlettmotorsportengineering.com/" target="_blank">http://shop.bartlettmotorsportengineering.com/</a></p>
<p style="text-align: left;"><br />Thanks,<br />Joe Bartlett</p>http://blog.bartlettmotorsportengineering.com/index.php?controller=post&action=view&id_post=82017-05-25T19:05:11+00:00Dirt Late Model Vehicle Dynamics - Part 2<p>Welcome to the second installment in our series on dirt late model vehicle dynamics. In part 1 of the series, we talked a little about some of the technical resources available on race car vehicle dynamics and the differences between the engineering involved when looking at a pavement race car as compared to dirt track race car. We also talked about the differences between analytical tools used to evaluate a race car, <!-- pagebreak -->and conceptual tools used to visualize the dynamic characteristics of a car. In part 2, we are going to start looking at the analytical tools in more detail.</p>
<p>Before we start looking at the equations that govern the dynamics of a dirt late model, we need to make sure we have a good general understanding of some basic engineering analytical techniques. We won’t try to cram an entire year of statics and dynamics courses into one paragraph, but we will touch on the necessary basics to understand how our system of equations work.</p>
<p>First, let’s look at the concepts of vectors and coordinate systems. In general, there are two types of quantities that we are interested in when developing a mathematical model. One is a “scalar” quantity, and the second is a “vector” quantity. A scalar quantity only gives you information about the magnitude of something. For example, the length of the left upper four-bar link is a scalar quantity. It is the same length no matter how it is oriented in 3D space. Seventeen inches is seventeen inches, no matter how you look at it. A vector quantity, on the other hand, gives you information about the magnitude and direction of a quantity. This directional information is determined by breaking the vector down into its component quantities which are relative to some frame of reference coordinate system. This coordinate system is made up of three axes which are the “x-axis”, the “y-axis”, and the “z-axis”. Each component quantity of the vector will also be a vector itself and be parallel with the associated axis of the coordinate system, so the x component of the vector will be parallel with the x-axis, etc., etc. Looking at our four-bar example, the left upper four bar link will have a magnitude of some value, say seventeen inches, and a direction in 3D space that is defined by the three components of the vector. The components of the vector are defined by the locations of the tail and head points of the vector in the 3D space of the reference coordinate system. We will not go into great detail describing this here as it is a little more complicated for our discussion; however, it can be found in any appropriate text book on engineering and/or mathematics. The point here is to understand that a vector has magnitude and direction, and it is relative to some frame of reference.</p>
<p>Second, let’s look at how systems of equations are derived. In mechanical engineering, it is common practice to develop systems of equations by summing forces along given directions, and/or summing moments (or torques) about given points. Forces along a given coordinate axis are typically summed and made equal to some known quantity. If a system is static, or not moving, they are set to zero. If the system is dynamic, or accelerating in some direction, they are set to some known force. Moments are rotational forces taken about an axis of a given coordinate system relative to some point. Moments can also be static or dynamic. For example, we may take a moment about point A around the x-axis. In vector mechanics, we typically use the cross product to simplify this process. We sum forces and moments until we have a system of equations to define our unknown quantities. In order to solve for a number of unknown quantities, a system of equations must contain at least one independent equation for each unknown quantity. This means if you have five things that you are trying to define, you will need to have five independent equations that define the unknowns. If you do not have much experience in mathematics, some of this may not make much since. That’s okay, it will make more sense as we move along. The key thing to remember is that we sum forces and moments to derive equations, and we need an equation for each unknown quantity we are trying to define. We can then solve the system of equations to find values for our unknowns. It should also be pointed out that forces and moments are always vector quantities.</p>
<p>With this basic understanding of vectors, coordinate systems, and equation derivation, we can now start looking at how to develop the equations that govern dirt late models (or any other race car for that matter). We will cover other mathematical concepts as they arise.</p>
<p>The first step in developing our system of equations is to define our coordinate systems. The first coordinate system is what we will call the world coordinate system. This is the system relative to the flat surface of Earth. In this coordinate system, gravity always points down. The second coordinate system is what we will call the track coordinate system. This coordinate system will move with the flat surface, so as the track surface banks, this coordinate system will move with it. For example, if there is no banking, gravity will point straight down, but if there is 15° banking, the gravity vector will have a vertical component and a component pointing to the left or right depending on the direction of banking. The third coordinate system is what we will call the chassis coordinate system. The chassis coordinate system is relative to the chassis frame. As the chassis moves around in roll, pitch, and yaw, the chassis coordinate system moves with it. You can think of the driver as having a perspective from the chassis coordinate system, provided he/she doesn’t move around inside the cockpit. We will be developing all of our equations in the chassis coordinate system. This means that we will use our chassis as a fixed object, and we will move the world around the chassis instead of moving the chassis around in the world and/or track coordinate systems. This may seem odd at first, but doing it this way simplifies the mathematics. In addition, this method also makes incorporating data acquisition information into our models much easier. DAQ boxes are generally mounted, in some known orientation, to the chassis frame and will give accelerations along all three axes relative to the accelerometer mounted in the DAQ box. The acceleration values can simply be input into the model, which is conveniently already in the chassis coordinate system. If data is wanted in the world coordinate system or track coordinate system, it is easy to do a coordinate transformation of the model results using the roll, pitch, and squat of the car to transform the data into a perspective that we as observers see the race car in. This would be done to obtain things like camber, caster, bird cage indexing, etc. that we would physically measure when setting up a race car in the real world. We also need to define directions within our coordinate systems. For our model, we will define the x-axis as running longitudinally with the chassis with positive x pointing towards the front. We will define the y-axis as running laterally with the car with positive y pointing to the left of the driver. The z-axis will be defined as up and down with the positive direction pointing up from the drivers perspective. As we define point locations within our coordinate system, we will define them by where they lie relative to each of the three coordinate axis. Any given point will have an “x” location, a “y” location, and a “z” location. We denote this by putting the xyz locations in brackets like [x, y, z]. The given xyz locations can be negative or positive depending on where they lie on the given axis. For example, a point in space defined by the 3D coordinates of [25”, -34”, 18”] will lie a distance of 25 inches on the positive side of the origin along the x axis, 34 inches on the negative side of the origin along the y axis, and 18 inches on the positive side of the origin along the z axis. Along with directions, we also need to define where the origin is located. The origin is the center point of the coordinate system and has the coordinates of [0, 0, 0]. When measuring points on a chassis, the head of the positional vector will be at the point [x, y, z] and the tail will be at the origin. In theory, you could choose to place the origin anywhere in space, and as long as you can accurately measure the coordinates of a point in question, the math model will work out. In reality, when working with a race car, it is good to place the origin somewhere in the center of the car near ground level. It is worth noting here that as you begin to understand how a coordinate system is laid out, you can start to see why things like roll center locations are not very useful as mentioned in part 1 of this series. When talking about roll center location, you often hear people say “put the roll center so many inches to one side of center and so far off the ground”. Well, where exactly is center, especially on an asymmetric car like a dirt late model? Depending on where you place the origin, three different people may place the origin in three different locations. If the three different people set their roll centers “so many inches” from their center, they will have three completely different set ups, but set up to the same roll center location target!</p>
<p>The second step in developing our system of equations is to determine what things we want to solve for. For our model, we have eight different things that we are interested in solving for. The first four unknowns are the vertical loads at each of the four tire contact patch centers. Note that when we refer to vertical in this case, it is vertical in the chassis coordinate system, not vertical with respect to the ground. In a more technical sense, we are looking for the z component of the force vector at the tire contact patch center in the chassis coordinate system. We already have an idea of what the lateral component (i.e. the y component), and the longitudinal component (i.e. the x component) of this force vector is based off of either DAQ system accelerometer outputs, or from track geometry and throttle/brake inputs. Note the use of the words “an idea of”. Although we may have an idea of what the lateral and longitudinal components are, we don’t yet know how they are distributed among the four tire contact patch centers. We will get into that later. We do not, however, have a good idea of the vertical component because of the weight transfer that takes place as a result of applying the lateral and longitudinal forces. The vertical loadings on the tires are a combination of the gravitational force, centripetal force from banking, and lateral & longitudinal load transfer forces. The last four unknowns are the vertical displacements of the four contact patch centers from some initial starting position. This starting position is simply some initial condition that is used to give the simulation process somewhere to start from. We will get into this more later, but for now you can think of it as a point where the chassis is setting somewhere just above the static ride height. Don’t worry, this will make sense later. Again, this vertical displacement is in the chassis coordinate system and is the z component of the displacement vector of the tire contact patch. We can translate the force and displacement vectors into the world or track coordinate system later as we see fit in order to give more “real world” useful results. The vertical loads and displacements give us a total of eight unknown quantities that we will build a system of equations to solve; therefore, we will need to develop eight equations to solve for the eight unknown quantities.</p>
<p>The eight equations can be broken down into three subgroups. The first group consist of equations one through three and are the “load transfer” equations. The second group consist of equations four through seven and are the “load distribution” equations. The third group consist of the single last equation eight which is a compatibility equation that ties everything together. In part three of this series, we will start to look at these equations in detail.</p>
<p>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.</p>
<p><a href="http://shop.bartlettmotorsportengineering.com/" target="_blank">http://shop.bartlettmotorsportengineering.com/</a></p>
<p>Thanks,<br />Joe Bartlett</p>http://blog.bartlettmotorsportengineering.com/index.php?controller=post&action=view&id_post=72017-05-16T20:58:38+00:00Dirt Late Model Vehicle Dynamics - Part 1<p style="text-align: left;">Welcome to the first installment in a series on dirt late model vehicle dynamics. The purpose of this series is to provide a mathematical explanation to some conceptual ideas well known throughout the dirt racing community. Many people successfully use chassis engineering concepts to tune their race car without understanding the underlying mathematics. This series aims to close that gap and help the reader obtain a better understanding of the science behind what their race car is doing. No need to worry if math was not your strongest subject in <!-- pagebreak -->school, mathematical concepts will be explained as needed. </p>
<p>Having a good understanding of what is going on with the physics of a race car is a very important tool for any racer to have. Over the past decade, an increased focus on race car engineering has crept into the dirt late model world. The days of just running what the chassis builder provided are gone. Today, the racer must be creative and innovative to come up with ways to get an advantage, and that often means tweaking the OEM chassis to meet specific needs. One should remember that a chassis builder is building a chassis to perform on a wide array of track configurations. Simply running a baseline setup will usually not result in the most optimum performance on a given night of racing. Understanding race car engineering and having the ability to adjust the race car to specific conditions is vital to getting the most performance and best finish in a race.</p>
<p>There are many debates out there today about rather or not an increase of engineering resources in the sport is a good thing or a bad thing. Like it or not, engineering has become a part of the dirt late model world. It is here to stay, and it will only become more and more important as the sport grows. Racers will be wise to get ahead of the wave now and start understanding it rather than fighting it.</p>
<p>The first part of this series is a discussion of technical resources. By no means should the reader regard this series as an all-encompassing set of information. There are many resources available on the topic of race car engineering. One only needs to query Amazon.com for “race car engineering” to start building their library of knowledge. The problem with most race car engineering books, however, is that they focus on pavement racing applications, not dirt. Most books about dirt racing only talk about conceptual ideas for tuning with no analytical topics addressed. The two worlds of racing share many characteristics, but overall, they are very different. This should not deter someone from reading the books, there is much knowledge to be gained by studying them. A listing of suggested readings will be included at the end of this article.</p>
<p>As a dirt racer, you must be conscious of the fact that many of the analytical tools in pavement racing do not apply directly to the dirt racing world. One example of this is the use of tire data to help set up a race car. The engineer working on a pavement race car is very interested in the tire’s slip-angle and slip-ratio performance characteristics to help predict the balance of the race car. In the dirt world, this type of information simply isn’t readily available. Even if information was available, it probably wouldn’t be a very useful tool because of the ever-changing surface of the dirt race track. However, this statement is made a little tongue-in-cheek as it may eventually find its way into the dirt racing world. Who knows.</p>
<p>One should also understand that many conceptual ideas used to tune a race car discussed in various books and articles are not applicable when performing a mathematical analysis of a chassis. Again, this holds true when looking at how to analyze a pavement car versus analyzing a dirt track car. One example of this is the use of the roll-center to analyze a race car. The concept of the roll-center is often talked about in race car engineering books and can be a useful tool for conceptually understanding what a chassis is doing; however, the roll center is not used in any way when analyzing a dirt track racing chassis. This may be contradictory to what many of you reading this have read in articles and books in the past. To understand this, one needs to appreciate a few characteristics of the dirt late model chassis. First, as compared to a pavement race car that has relatively little suspension travel, the dirt late model has a large amount of suspension travel, especially on the right front and left rear of the car. This means that the motion ratios of a dirt late model will change much more as compared to a pavement race car. Often, when analyzing a pavement race car, the engineer may simply assume that the motion ratios are constant throughout the operating range of the suspension. This is not the case on a dirt late model race car, especially on the rear suspension of a live axle decoupled suspension. (i.e. the four-link suspension utilizing birdcages and a lift arm) When analyzing a dirt late model chassis, one needs to accommodate for the changing motion ratios. Secondly, many paved track race car engineers can effectively calculate roll moments by using the moment-arm-method to determine chassis balance. This can be done only if there is minimal suspension motion and if both the front and rear roll centers can be clearly defined. This is not the case on a dirt late model chassis. Many people will approximate the rear roll center of a dirt late model, and this is okay for conceptually trying to understand what is going on, but this is not adequate when mathematically analyzing a race car suspension. Furthermore, the moment-arm-method is a rather archaic engineering tool. Remember that this method was developed during a time when computers were not readily available forcing engineers to use pencil and paper graphical methods to develop suspension systems. Today, there is more computing power available in the cheapest of computers than could have ever been imagined fifty years ago. There are much better ways to utilize our computers to analyze a chassis.</p>
<p>The key point here is to understand the difference between conceptual tools used to help the human brain comprehend what is going on with a chassis, and analytical tools used by a computer to analyze a chassis. The two are very different, and the latter is what will be focused on in this series.</p>
<p>In part 2, we will start diving into the mathematical system of equations that govern the dynamics of our race cars.</p>
<p>If you find this series helpful, then you can help me by thinking of Bartlett Motorsport Engineering next time you need to by parts for your race car.</p>
<p><a href="http://shop.bartlettmotorsportengineering.com/" target="_blank">http://shop.bartlettmotorsportengineering.com/</a></p>
<p>Thanks,</p>
<p>Joe Bartlett</p>
<p> </p>
<p>The following is a list of suggested readings on the topic of race car engineering.</p>
<p>• Race Car Vehicle Dynamics - by William and Douglas Milliken</p>
<p>• Race Car Design - by Derek Seward</p>
<p>• Race Car Engineering and Mechanics - by Paul Van Valkenburgh</p>
<p>• Tune to Win - by Carroll Smith</p>
<p>• Engineer to Win - by Carrol Smith</p>
<p>• Fundamentals of Vehicle Dynamics - by Thomas Gillespie</p>
<p>• Vehicle Dynamics and Damping - by Jan Zuijdijk</p>
<p>• The Multibody Systems Approach to Vehicle Dynamics - by Michael Blundell and Damian Harty</p>
<p>• Chassis Engineering - by Herb Adams</p>
<p>• Tire and Vehicle Dynamics - by Hans Pacejka</p>
<p>• An Introduction to Race Car Engineering - by Warren Rowley</p>http://blog.bartlettmotorsportengineering.com/index.php?controller=post&action=view&id_post=62017-05-05T21:40:33+00:00