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DVD now available 
This DVD is NOT a repeat of the site, but is, rather, an extended
explanation which can only be achieved with pictures and a lot of "hand waving." |
| IF YOU'RE A SERIOUS BRACKET RACER, HERE'S SOME ELECTRONICS YOU'LL NEED |
GETTING STARTED
What I'm attempting to do, on this page, is present just enough of that which is taught the engineering student to allow the non-engineer to easily digest the information in the following pages. I had hoped a page like this would not be necessary, but it's been pointed out that I occasionally slip in some engineering "jargon" which frustrates the user. So, I hope the following will help. I'll try not to provide more than is needed and I'll also attempt to use examples which will keep it interesting.
Attention is directed ONLY to rear wheel drive cars with beam (live) rear axles. When considering modern production cars, this might seem overly restrictive, but the RWD beam axle cars still retain the greatest success at the dragstrip. The techniques can, of course, be applied to other vehicles.
Link preloading is not considered, though the techniques presented will readily indicate the effect.
SOME BASIC DEFINITIONS:
We all use words like "acceleration" and "speed" and "velocity" with the assumption that the other person has the same understanding of the meanings. But, an engineer is forced to use very specific meanings for each word. In the first place, "acceleration" and "velocity" are called "vector" quantities. All this means is that they have both magnitude AND direction. "Speed," on the other hand, has only magnitude and is rarely referenced in an engineering analysis.
A vector, when applied to a diagram or picture, is simply an arrow. The length of the arrow is proportional to the magnitude of the quantity and the arrow's direction...well, that's obviously what it is: A direction.
If we had a picture of the block layout of a city's streets, we could use vectors to describe a car's path. Starting from a certain intersection, the car might travel 3 blocks east. We would then draw a 3 block long arrow to the right with the arrow head at the extreme right end. A 2 block long arrow, pointing upward, would represent travel northward. Further travel might involve a 4 block long arrow pointing westward, a 2 block long arrow pointing southward, and a 1 block long arrow pointing eastward. If you've kept track, you've noticed that this puts the car right back at its starting point. A rigorous way of keeping track of this sort of thing is called "vector addition." In this case, the map grid forces the vector lengths to correspond to the distance travelled and the head of the last vector arrow reaches the tail of the first vector arrow. But, engineers rarely go to the trouble of constructing vector lengths exactly to scale. Rather, they simply write the magnitude adjacent to the vector. To sum the vectors, then, it is necessary to separate each vector into its horizontal and vertical components, arbitrarily assign positive values to one direction, and sum. In this simple case, the vectors are either horizontal or positive. The vectors could, however, be at an angle, in which case the angle must be known and trigonometric functions must be used to determine the horizontal and vertical components. Note, also, that the street map doesn't indicate the presence of hills. In other words, the vector might be three dimensional. While this complicates matters, the same principles apply. It is only necessary to determine the component magnitude and direction in or out of the paper.
In the example, we could arbitrarily say that eastward travel and northward travel are positive. Looking at east/west (horizontal) travel, then, 1 plus 3 (eastward travel) - 4 (westward travel) = 0, meaning that the net east/west change is zero. Looking at north/south (vertical) travel, 2 (northward travel) - 2 (southward travel) = 0, meaning that the net north/south change is zero.
The magnitudes of both "velocity" and "speed" are the time rate of change of location. A car's speedometer indicates the speed (NOT the velocity) of the car in miles per hour. "Acceleration" is the time rate of change of "velocity," and it's here where I encounter a lot of confusion among users. Since acceleration is the rate of change of velocity, the magnitude of the velocity is totally unimportant. It's very common to have very large values of acceleration when the velocity is zero! The illustration commonly used in engineering textbooks is that of a pendulum. I could use the rear end of a car as it squats or rises during launch, but that takes too long to set up and I'll spend a good deal of time there later. For now, a pendulum will do. The pendulum's weight has maximum velocity magnitude at the midpoint of its swing. At that point, the acceleration is zero. When the weight reaches its highest point, the velocity is zero and the acceleration magnitude is at a maximum.
Force is also a vector quantity. The force vector will be covered more fully when "free body diagrams" are considered.
LAWS OF MOTION:
All that's really needed is an understanding that, when a force is applied to a body (that's engineering talk for a lump of anything), an inertial force is developed which is opposite in sense (in other words, the inertial force vector is pointing in the opposite direction to the force vector producing the acceleration) and equal in magnitude.
(Torques and rotating inertias will be covered as the need arises for their consideration.)
FREE BODY DIAGRAMS:
At this points, the textbooks commonly show a picture of a rectangle. Boring! So, this is what I'll use:
No, this is not a car. This is an outline of a car, torched from a thick piece of plate steel. The total weight is 3000 pounds.
Note the black "dot" with the letters "CG" beside it. If you were to place this 3000 pound of plate steel on its side, it would balance on that point. This is called the "center of gravity" (CG) of the body.
Note, also, the horizontal line below the body. This represents the shop floor which supports this 3000 pound piece of steel plate.
The purpose of a "free body diagram" is to isolate or "free" the body under analysis from its surroundings, replacing the effects of those surroundings with the proper vector representations of the forces, moments, velocities, etc. involved. For an understanding of that which is discussed in the other pages of this site, I believe it will only be necessary to consider forces and moments.
But, I haven't defined a "moment." The motorsports enthusiast already understands the meaning of "torque," however, so I could cheat and simply say that a moment is a torque. More precisely, a force, acting at a distance from a point and with a vector direction which does not intersect the point, generates a moment about that point equal to the product of the force magnitude and the distance to the point measured along a line perpendicular to the line of action of the force. The dimensions and units would correspond to those used for torque.
As a first step in "freeing" this body from its surroundings, the shop floor must be removed. The shop floor had been supplying an upward force at the two points, so vertical upward arrows (force vectors) are shown at those two points. The weight of the body acts through the CG and is shown as a vertical downward arrow:
We know that the plate weighs 3000 pounds, but we don't know the magnitudes of f1 and f2. It's rather obvious, however, that the sum of f1 and f2 must equal 3000. Let's be a bit more rigorous, though, and use the SAE coordinate system. This coordinate system has its origin at the CG with the positive "X" extending forward, the positive "y" to the right (driver oriented), and the positive "z" downward. The total weight, then, would have a positive value while the supporting forces would be negative. A force balance requires that vector sums, in each of the three coordinate directions, equal zero. Remember that forces at an angle can always be resolved into forces in 2 or (if necessary) 3 of the coordinate directions.
The preceding, then, is a force balance in the "z" direction. Since there are no forces in the "X" direction, the force balance portion of the free body diagram is completed. There remains, however, a moment balance. The point about which I calculate the moments is completely arbitrary, as is the choice as to whether clockwise (CW) or counter-clockwise (CCW) is positive. Often, the CG is chosen as the point about which moments are calculated, but, with most automotive problems, it is more convenient to use a tire patch. I will use the rear tire patch. And, though it makes no difference, I will follow the popular usage of the "right hand rule" and call CCW moments positive. (The "right hand rule" calls for pointing the thumb of the right hand in the direction of the positive "y" axis. The direction in which the fingers curl indicates the direction of the positive moment.) Before we can go further, we need dimensions:
So, calculating moments about the application point of the force f2, we see that we have a positive moment of Lf1 and a negative moment of 3000 B. The sum of these two moments must equal zero. With this additional relationship, we see that we have two equations and two unknowns. Remembering your high school algebra, you realize that we can now solve for the two supporting forces. Substituting (3000 B / L) for f1 in the previous equation, we see that f2 = 3000 (L-B)/L pounds. f1, then, must equal 3000 B/L pounds.
(Note that we might have had a force balance in "x." This would have provided a third equation. In a more complex situation, force and moment balances might have been necessary in the top and end views. This would have provided a total of 9 equations and the possibility of solving for 9 unknowns. While such a task remains a high school algebra problem, the algebraic manipulations can become quite cumbersome, resulting in a high probability of error. Great care must be taken!)
Note, also, that the force vectors can be positioned anywhere vertically, along their line of action, and the results are unchanged. In other words, the CG could be at ground level and the upward forces above the "roof line" and the answers would be the same. This is a very important characteristic of forces to be remembered. A FORCE CAN BE LOCATED ANYWHERE ALONG ITS LINE OF ACTION AND ITS EFFECTS ARE UNCHANGED.
With this simple example, it's obvious that the forces supplied by the shop floor are directed upward. Often, however, the direction of a force is not at all obvious. In this case, the direction is assumed and the equations are solved. If a negative force magnitude is obtained, it is known that the assumed direction was incorrect.
BODIES UNDER ACCELERATION:
In the previous examples, it was emphasized that the force and moment sums MUST equal zero. This was because the body was not being accelerated. The piece of sheet steel was merely sitting on the shop floor. This is called a STATIC condition. In those cases when the force and/or moment sums do NOT equal zero, we know that the body is being accelerated. This is called a DYNAMIC condition.
Fortunately, when we do a free body analysis of a dynamic situation, we need only consider those forces and moments which are dynamically involved.
We see that an accelerating force has been applied to what looks like the rear tire patch and that the static forces are no longer shown. From the LAWS OF MOTION section above, we know that the body under acceleration responds with an inertial force that is equal in magnitude and opposite in direction. This, then, is the upper arrow shown acting through the CG.
This arrangement of equal magnitude forces constitutes that which is called a "couple." In this case, the couple is attempting to twist the body in a positive (CCW) direction. As in the static analysis, the moment sum must equal, so we must look for other forces which will provide a negative (CW) torque. Since the body is still resting on the shop floor (remember, the static forces are still present, though they're not being shown), the forces would be generated at these two points.
The two vertical forces are, again, equal in magnitude and opposite in direction, so we have a force balance in "z" and a force balance in "x." And, since the vertical force couple provides the necessary negative torque, moment balance is achieved if the magnitude of "L" times a vertical force is equal to the magnitude of "H" times the accelerating force.
This procedure can be used to calculate the weight transfer which occurs on launch.
Note that the front force vector arrow is directed downwards, but the f1 force vector pointed upwards. If the dynamic force equals f1, there is no remaining load at the front. If it is greater, it becomes an accelerating force, tending to rotate the "car" about the rear contact point. Extending this to a consideration of a real car, it can be seen that this is the origin of that which can become complete "blowover."
LINKS IN A FREE BODY DIAGRAM:
Links are commonly encountered in the free body analysis of a suspension system, but, if they're free to rotate on both ends and if there are no forces imposed between the ends, we immediately know a great deal about the disposition of the end forces. Those forces at the pivots must be equal in magnitude, opposite in direction, and must lie on a line between the two pivots. You are most likely familiar with the 4link suspension and the Watts and Panhard methods for lateral location of the rear axle assembly. But, there is another link which often is not so readily identified. I'm referring to the ladder bar suspension.
The front pivot of a ladder bar is readily recognizable, but the rear is not so obvious. The rear "pivot" is the rear tire patch. Like any link pivot, the rear tire patch allows free rotation without opportunity to provide a torque. The net force, then, must act on a line passing through the rear tire patch and the front pivot.
(It is recognized that this arrangement does not strictly meet the requirements for a link, since the mass of the rear axle assembly provides an inertial force between the two pivot points. The mass of the rear axle assembly is considerably less than the mass of the remainder of the car, however, and, in addition, ladder bar adjustment precision is so poor that, as a first approximation, rear axle mass can be ignored.
The steel plate has now been cut and the two pieces positioned to simulate the ladder bar suspension.
SEPARATION INTO FREE BODY DIAGRAMS FOR EACH BODY:
Here is a free body diagram of the axle assembly with a ladder bar. (All right, it's still a piece of steel plate, but it must be evident by now that the procedues I'm developing are perfectly applicable to a real car.)
The forces are seen to be arranged as described in the "LINKS IN A FREE BODY DIAGRAM" section. Since the two forces are equal in magnitude and opposite in sense, the system is balanced and there is no net accelerating force.
If the lower force vector is resolved into its "x" (horizontal) nd "z" (vertical) components, it should be evident that the "x" component is the "ACCELERATING FORCE" and that the "z" component is the weight transfer. The corresponding components of the upper force would be the "INERTIAL FORCE" and the weight removed from the front end during launch.
Here is the "other" body of the ladder bar car, with the force applied through the pivot, the inertial force, and the force which represents the weight removed from the front end during launch.
Since a foce can be moved anywhere along its line of action, I'll move it to where it meets the intersection of the lines of action of the other two forces.
It is now evident that the horizontal component of the pivot force perfectly cancels the horizontal force at the CG and the vertical component of the pivot force perfectly cancels the vertical force at the front wheel. This means there is no net accelerating force.
But, what if the pivot point had been lower and the force had been located where its line of action crossed the vertical force line? Its horizontal component, acting with the inertial force, would have caused a positive (CCW) torque on the body. You should be able to relate this to the rear end drop or "squat" which occurs during launch when the ladder bar is located in a low pivot hole.
Note, also, that the vertical component of the pivot force is slightly smaller when the ladder bar is located in a low pivot hole. This means a net vertical accelerating force at the front. Specifically, the front will rise higher on launch, further accentuating the change in pitch (front up, rear down).
The opposite pitch change will occur when the ladder bar is located in an upper pivot hole.
Returning to the situation in the picture, we can see that a pivot force line of action which passes through the intersection of a horizontal line through the CG and a vertical line through the front tire patch is quite unique in that it is the only one that does not cause a pitching of the car. This unique line is called the "100% anti-squat line" and is so labeled in the picture. The slope of this line is equal to the CG height divided by the wheelbase. A line through a lower pivot hole would have less slope. If the slope of the lower pivot hole line was 80% of the slope of the line in the picture, it would be the "80% anti-squat line." So, we can imagine a "fan" of lines, each passing through the rear tire patch and the ladder bar pivot, which would define the line of action for any number of ladder bar pivot holes.
While the results are the same if I had left the pivot force vector at the pivot hole, I don't believe the situation is so obvious. Again, the vertical and horizontal forces are balanced, leaving only a consideration of the moments. If moments are taken about the pivot hole, we see that the positive and negative moments are proportional to those considered with the original body and, therefore, are still equal.
MULTIPLE LINKS:
Instead of a simple ladder bar, the following illustrates the situation with multiple links. Again, the rear axle assembly and links will be considered weightless.
Next, I'll do a free body diagram of the rear axle and links. Note that, while there is more than one "body," the procedure is still valid. It is only necessary that the appropriate vector forces and moments are shown at the attachment points.
Though it might be intuitively obvious, I'll say that I don't know which link will be in compression and which in tension during launch. But, I do know (see "LINKS IN A FREE BODY DIAGRAM") the locations of the lines of action of the link forces. I have chosen to show both link forces pointing to the right, meaning that I'm assuming both links are in tension. As I've indicated, if one or both assumptions are incorrect, the final answer will have a negative value for the force magnitude.
As it turns out, my assumptions are unimportant, since the lines of action intersect on the line of action of the force at the tire patch. This means that I can move both forces to that intersection and sum them. Further, it means that the sum must equal the tire patch force in magnitude but be opposite in direction.
The intersection of the link lines is called the suspension "instant center" (IC). In this example, I have caused the IC's location to be at the pivot point for the ladder bar, so, from this point on, the analysis would be identical to that for the ladder bar.
If, instead of the IC falling on the 100% anti-squat line, it had fallen on, say, the 80% anti-squat line, the situation would be the same as that for a ladder bar with the pivot located on the 80% anti-squat line.
Note, however...and this is very important..., that, with multiple links, it is possible to locate the IC at any one of an infinite number of points on a given percent anti-squat line and the final result will be EXACTLY the same.
PARALLEL LINKS:
The special case of parallel links needs to be considered.
To analyze this situation, it is only necessary to remember that parallel lines meet at infinity. Since the links shown are parallel to the 100% anti-squat line, this means that the IC falls on the 100% anti-squat line and launch characteristics will be the same as for the previous example. For other values of percent anti-squat, it is only necessary to position the parallel link pair at the corresponding slope. Some suppliers provide that which is called a "4bar" kit, wherein the links are maintained in parallel while the end brackets are adjusted vertically to provide the desired percent anti-squat.
(It is recognized that, while launch characteristics are the same, street driveability is not. The angled links shown will result in that which is called "roll oversteer." While roll oversteer...unlike the oversteer caused by severe rearward weight bias, for instance...does not result in inherent instability problems, it does cause steering input changes which might be annoying to the average driver. This is why, of course, automobile manufacturers keep the IC somewhere near, in height, to that of the centerline of the rear wheel. For cars used only in dragstrip competition, however, the roll oversteer should not be a problem.)
A ROTATABLE BODY (THE "BLOWOVER")
Consider a hockey puck with an accelerating force acting upon it.
Now, consider the same hockey puck as it receives a "glancing" blow.
The accelerating force has been resolved into 2 components, one of which is directed toward the CG and which would accelerate the puck in that direction and the other, which would tend to rotate the puck counterclockwise about its center of gravity. The inertial reactions, then, would include an inertial force AND an inertial torque. The inertial force is proportional to the mass and the inertial torque is proportional to the moment of inertia. Dimensions for moment of inertia are mass-length squared. Typical units would be pounds mass inches squared, slug feet squared, or gram centimeter squared.
I would hope that you noticed that the point of application of the force is now significant. The previous emphasized statement...that the force can be moved anywhere along its line of action without a change in effect...appears to be in question. This is because the body under analysis is no longer constrained from rotation. If the body is free to rotate, the point of force application determines the accelerating moment and is, therefore, most significant.
Now, consider a car when the accelerating force is sufficient to generate enough weight transfer that the front wheels are no longer loaded at the track surface. The car is no longer constrained from rotation about the rear tire patch and the possibility of a "blowover" exists.
Note that the line of action of the sum of the weight transfer and forward thrust passes below the CG. If the CG were on or below that line, blowover would be impossible. On the other hand, with the CG above the line, it takes only a small unbalance to initiate blowover, for, once it begins, the magnitude of the blowover torque increases as the CG rises. The resulting blowover can become, as I'm sure you're quite aware, very violent!
(The moment of inertia of a typical 3000 pound sedan is about 6.5 million pound mass inches squared in pitch, 7 million in yaw, and 1.5 million in roll. The pitch moment of inertia about the rear tire patch would be 6.5 million plus the product of the car weight and the square of the distance between the CG and the rear tire patch.)
IRS AND THE ANTI-SQUAT LINES
When the free body diagram for the entire car was considered (above), it was not necessary to specify the type of rear suspension. Regardless of the configuration "internal" to the diagram, the force and moment balances were still valid.
If, however, we now consider a free body diagram for that part of the car after the driving wheels and related components are removed, the rear suspension design can influence the force and moment balances and, of specific importance to the racer, can influence the positions of the anti-squat lines.
If, with a beam axle car, it were possible to apply the patch forces directly to this new free body diagram at a point corresponding, in location, to the tire patch, the resulting forces would correspond exactly to those in the preceding illustration. And, of course, the locations of the anti-squat lines would be exactly as determined previously. As has been shown, this means that the vertical and horizontal forces, at the tire patch, could be moved to a point anywhere on the line passing through the tire patch and point A and the force and moment balances would be unchanged.
But, if the rear suspension was either totally independent or of the DeDion design, a moment...equal to the product of the tractive force and the rear tire radius...is added to the diagram. To regain a moment balance, this means that the forces at the rear must be raised a distance equal to the rear tire radius.
Since the force ratio is unchanged and, hence, the slope of the 100% anti-squat line is unchanged, the vertical distance between points A and B would also be equal to the tire radius. The "Japanese fan" appearance of the lines of constant percent anti-squat would simply be raised a distance equal to the tire radius.
It is apparent, then, that...with an IRS or DeDion...the IC will be much higher for a given percent anti-squat. This can create excessive roll oversteer. The launch squat common with production IRS cars indicates that the designers have chosen to sacrifice some anti-squat to avoid the roll oversteer.
OTHER SUSPENSION SYSTEMS:
There are, of course, other suspension systems, some of which...like the torque arm...are discussed elsewhere in this site. I believe the preceding should help you in understanding their capabilities.