Beam Axles - Front, Rear or both.

Quote:

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Originally posted by Gruntguru:
Z, not sure if you noticed, ...
...
This is a very clever and innovative car with a very low suspension component count and one major suspension component doubling as an aero undertray. This design and future iterations are capable of great things.

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GG,

Indeed it is!

Yes, I recognised it straight away. I would really like to see UWA finish this car and perhaps get it over to some Northern hemisphere comps next year. Any chance? Certainly, I hope future UWA teams continue with the concept.

To restress what you said above, it is a VERY SIMPLE CONCEPT, with a minimal part count. So much so that I doubt anyone would have believed it could work if it was only suggested as an "idea" on these pages... http://fsae.com/groupee_common/emoticons/icon_smile.gif

Proof that the concept works, to some degree at least, is one of Rex's (?) photos on the Facebook page that shows the car parked with one front wheel on top of a couple of red and blue boxes, about 10-15 cm high. All other wheels are still firmly on the ground!
~~~o0o~~~

To any interested teams currently running conventional suspensions.

Some of the UWA concept and the Z-Bar sketch above might be combined to retrofit a fairly simple soft twist-mode suspension to a conventional wishbone car. This is something you might do to a previous year's car as a research project.

First, remove all normal springs (dampers can stay). Then, looking at bottom right of sketch, at front and rear of car fit lateral Z-bars to control heave and pitch only. These might be lateral steel or glass/carbonfibre leafsprings as shown, or else any conventional "third spring" arrangement. These do the job of UWA's "W" springs (silly me, thinking they looked like "E's"!), although the W's also provide axle location (clever!). Next fit an unsprung undertray, somewhat like UWA's.

The chassis is now only supported like a bicycle (with single spring at front and rear), so wants to fall over whenever going around corners. So fit a SINGLE lateral U-bar at about mid-chassis with its outer ends connected to the undertray tunnels, thus preventing body-roll (search Rex's photos). During cornering the body leans on this single ARB, which in turn pushes down on the outside tunnel, and lifts the inside tunnel.

These roll forces (up on inner, and down on outer tunnel) are passed on to their respective wheels by the tunnels acting as "balance beams". So LLTD (= ERMD in above sketch) is determined entirely by the geometry of the linkage (specifically, the ARB/tunnel attachment points), and not by any spring rates, or by any bumps or twist in the road.

Note that in the above sketch (bottom-right) there are two longitudinal torsion bars, acting as Z-bars, that control both heave and roll. This layout is well suited to production cars because of the easy packaging of the torsion bars, which are conveniently the right size to carry most of the car's weight (ie. heave loads), as well as the roll loads. The UWA single lateral U-bar is a simpler solution, although it does require the undertray, or some sort of side balance-beams to work.

Z

I am new to the fsae forum and have found a wealth of information here.

To clarify, I am not involved in fsae as a participant, judge or consultant.

I am a Tool and Model designer in industry by profession who spends his free time working on suspension analysis and tuning for lower level oval track race teams in the northeastern region of the United States.

I have been perplexed for quite some time on one particular point of the classical or the static analysis of a live beam axle suspension in common use in the arena I am working in and was hoping someone on this forum may enlighten me.

The problem:

A race car with beam axle suspensions both front and rear.

The main sticking point is the analysis of the live rear axle laterally restrained by a very short (approx. 18 in.) panhard bar offset to the right of the vehicle centerline as viewed from the rear. The panhard bar attaches to the axle just to the right of the axles centerline and to the sprung body just inside the right rear wheel as viewed from the rear. The overall height of the bar is adjustable relative to the axle centerline or ground as you prefer. The bars angularity is cockpit adjustable by the driver while the car is in motion. If we assume the bar to be set level at axle centerline height the range of adjustability is 10 degrees up from axle to chassis to 10 degrees down from axle centerline to chassis as viewed from the rear. The adjustment takes place at the chassis mount of the panhard bar via a vertical 'lead screw' and captive block assembly.

The remaining rear axle degrees of restraint are two parallel trailing links mounted solidly to the rear axle below axle centerline height (approx. 6 in.). The final restraint needed is for axle housing rotation about the y-axis which controlled by a torque arm from the live axle center section (gear housing) to a linear bearing 'sled' with large heim joint being used to attach the torque arm to the chassis i.e. one degree of restraint of the arm rotation about the lateral axle centerline.

The constraints in this form of racing:

There is no data acquisition.
There is no tire data.
The track surface may be pavement or dirt.
There is so little testing time it might as well be considered negligible.
There is no practice time to speak of.

You are left with driver feedback, observation, and possible video and thought to analyze possible setup changes.

My approach to date has been a simplistic 'classical' roll center based model to evaluate front and rear wheel pair loads to have a look at what limit behavior might be at assumed steady state lateral and longitudinal acceleration levels i.e. make spring changes or suspension link geometry changes (IC position changes) assume a 'g' level and calculate (spreadsheet) the front and rear tire pair loads.

The problem I have had with the 'roll center', 'shear point', control point type of analysis, name your favorite author here, IS the location of that point with the short offset panhard bar with angularity described above.

All of the classic texts I have (RCVD, Dixon, Olley etc.) say that with a basically planer linkage as I have described is that the roll center (for lack of a better term) is located where the rear axles axis of rotation as defined by the axles locational linkage pierces the lateral rear axle wheel pairs vertical plane. For the linkage I describe the rear 'control point' is always taken where the panhard bar crosses the vehicle centerline plane.

I see no kinematic nor force based reason for this to be the case and as Dixon points out the rear axles axis of rotation is a piece of engineering fiction useful in locating the notional roll center or force coupling point.

Well, the panhard bar does not cross the vehicle centerline in this case, what to do?

Consult Mark Ortiz.

In Mr. Ortiz's view the roll center can be located in this situation by one of two methods. The first he describes as the simple method of finding the intersection of the panhard bars centerline with either the vehicles centerline or longitudinal CG center plane if the car is not symmetrical and to neglect the internal jacking force and use that point as the roll center. The second he describes as the more rigorous method and uses the panhard bars mid-point to fix the roll center height but also says to include the internal jacking force caused by axle to chassis bar angularity in your calculations and to make the mid-point of the panhard bar the point for this jacking forces vertical point of application.

I have a great deal of respect for Mr. Ortiz's openness in answering any and all questions and agree with much of what he has written about asymmetric race cars but again I can find no kinematic or statics (force based) reason that makes the panhard bar mid-point any more 'special' then any other point.

When looking at this problem from a 2D statics point of view you come to realize that the panhard bar is simply a two force link attaching the beam axle to the body and if you free body diagram the body then the panhard bar force line of action 'on the body' is simply the bars angle and you could decompose that force anywhere along that line of action. (Yes, a Z n-line) So no point is 'special' from this point of view.

Two questions to the forum if I may:

1. In the situation as described where would you take the roll center height for a classical treatment and more importantly why?

2. What would be a better approach to getting a high level view of the effects of the short panhard bars position and angularity with respect to wheel pair loading with an assumed steady state lateral loading?

As one last point of observation I have worked with six different drivers two of which have been multi-time champions in this form of racing and only one of the six can describe the effects he feels when he has adjusted the bar past the point of a positive effect.

Thank you,

Ralph

Ralph,

This is an interesting question, as I haven’t dealt with asymmetric cars before. The following is based solely on my own reasoning, so corrections are welcome.

Normally for most suspension types the roll centre is found by making a few simplifying assumptions: the linkage is reduced to a purely planar linkage, and the tyres are pin jointed to the ground at their contact patches. The only way the chassis can move when these assumptions hold true is by rotating about the roll centre. When considered in plane; a beam axle located by a Panhard bar does not restrict the chassis to rotate about a single point (Panhard bar can rotate relative to the beam, and the chassis can rotate relative to the bar), so the concept of a geometric roll centre becomes grey.

There is also a different force based definition of the roll centre as a point where lateral forces can be applied to the chassis without causing roll. After drawing up a quick FBD I think that the point found from the intersection of a line through the Panhard bar with a vertical line dropped from the CG satisfies this definition if you assume the car is sprung such that the chassis will not roll if a vertical load is applied at the CG. Note though that this definition accounts only for a roll centre height, and does not consider a lateral position for the centre.

This would be easier to explain with a picture but bear with me. We want to apply a lateral force to the chassis, this force will be reacted by the horizontal component of the Panhard bar force. There will also be a vertical component to the bar force if the bar is not horizontal because the bar force must be aligned with the centreline of the bar.
This component will either increase or decrease the load carried by the suspension springs; the change in the spring loads can be represented as a single vertical force passing through the CG because we have assumed that the applied force does not cause roll, and that vertical loads passing through the CG do not cause roll.

Now we have 3 forces introduced by applying the lateral force. These must sum to zero and produce no roll moment for equilibrium. The bar force and the spring change force can be slid along their lines of action to the point where they intersect. We can see now that the lateral force must be applied at the height of this intersection in order for the 3 forces to cancel without introducing a roll moment. Therefore this intersection defines the roll centre height.

If for some reason the position where vertical loads could be applied without introducing roll was not aligned with the CG; then the roll centre height would be determined by the intersection of the bar axis and the line where vertical loads can be applied without causing roll.

The roll centre height depends only on the location of the chassis Panhard bar point, the angle of the bar, and the location of the line where vertical loads can be applied without introducing roll. The length of the bar will only influence how this changes with suspension movement, shorter bar = more change in bar angle for the same movement.

Nathan,

I would say your line of thinking very much parallels my own and is correct when using the classical roll center based model. I arrived at the same conclusions a while ago after exhausting a purely kinematic approach.

The only difference would be that I would say that with a FBD of the body the angle of the panhard bar sets the line of action of the combined vertical internal (to the system) jacking force and the horizontal force at the body mount point form a resultant force that can slide anywhere along the LOA set by the panhard bar at any instant in time. This resultant force can be again decomposed at any point along the line of action and have the same effect on the body. At any point chosen. Thank you Z.

I agree with your reasoning that for the roll center based model that this point should be at the point where the vertical component produces pure heave. For all text book models the examples all place the CoG on the vehicle centerline and assume equal left and right spring rates equidistant from the centerline. I also agree with your reasoning that if we were to add, let's say, spring split that the pure heave point decomposition point would move to the vertical line of the spring center, I think.

I just wish the most regarded texts on this subject would make it clear that this is a force based concept and not obscure this fact by mixing in the kinematic notional concept of the live axles roll axis. A useful concept I guess for picturing roll over or under steer but confuses the issue of a force coupling point for a panhard bar system in particular. Dixon comes closest to a force based description if you read his words very closely but is still making a strong reference to link geometry analysis and the notional axle roll axis.

I am now using a purely FBD and force based approach with a lot of help from correspondence with Z. I can tell opinions vary on this forum, but you must admit the man knows his classical mechanics and his insistence on rigor is correct in my view.

Thank you for your reply,

Ralph

Ralph and I exchanged some PMs prior to Ralph's first post here (on previous page). Nathan seems to have grasped the gist of the problem in his answer to Ralph's question (also previous page). That is, DON'T FORGET the jacking force (Fz) component of the Panhard-Bar force!

For any students interested in these sorts of problems, here are some more general comments, from an "Old-School Mechanics" perspective.
~o0o~

First, a summary of the problem.

Quote:

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Originally posted by Ralph:
... The main sticking point is the analysis of the live rear axle laterally restrained by a very short (approx. 18 in.) panhard bar offset to the right of the vehicle centerline as viewed from the rear. The panhard bar attaches to the axle just to the right of the axles centerline and to the sprung body just inside the right rear wheel as viewed from the rear. The overall height of the bar is adjustable relative to the axle centerline or ground as you prefer. The bars angularity is cockpit adjustable by the driver while the car is in motion. If we assume the bar to be set level at axle centerline height the range of adjustability is 10 degrees up from axle to chassis to 10 degrees down from axle centerline to chassis as viewed from the rear. The adjustment takes place at the chassis mount of the panhard bar via a vertical 'lead screw' and captive block assembly.
...
The problem I have had with the 'roll center', 'shear point', control point type of analysis, name your favorite author here, IS THE LOCATION OF THAT POINT...
All of the classic texts I have (RCVD, Dixon, Olley etc.) say that with a basically planer linkage as I have described is that the roll center ... is located ... where the panhard bar crosses the vehicle centerline plane.

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(My emphasis added.)
~o0o~

One of the keys to solving this type of problem is to appreciate that FORCES and MOTIONS are entirely separate things. A SINGLE, given problem (say, the one above) has TWO entirely separate answers, depending on whether FORCES or MOTIONS are sought.

Unfortunately, the suggestion in the "classic texts" that a single "The One-True-Roll-Centre" point can be used to solve both types of problems is shear nonsense. Even worse is to suggest that the "forces problem" can be solved by finding "The One-True-Force-Application-Point" (as always ... groooaann!!! ... education down the crapper!!! ... blah, blah, blah... :)).

However, this problem, being one of the Dynamics of a very simple mechanical linkage, is very easy to solve using the methods of Old-School Mechanics. Briefly, the problem can be divided into 3 parts, corresponding to the 3 sub-fields of Mechanics.
1. STATICS - Analysis of Forces acting on bodies.
2. DYNAMICS - Here, by using D'Alembert's Principle, showing the centrifugal Inertial-Force (say, Fi) as just-another-force that acts on the car-body in the Statics FBD.
3. KINEMATICS - Based on the changes in Spring-Forces (due to changes in Fi) in the above Statics diagram, determining the subsequent Motion of the car-body wrt live-rear-axle.

To quote Ralph (again with my emphasis):

Quote:

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From Ralph in a PM:
My biggest Aha moment in all of this so far has been the tearing down of the wall I had built up in my mind between my knowledge of classical Newtonian Mechanics and what I thought I was learning from the vehicle dynamics texts. For some reason I was treating VD methods as something new and different.

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The following notes do not go through the full solution of the problem above, but only re-stress a few points that "all school-boys should know", but currently do not seem to be taught very well, if AT ALL... :(
~~~~~~~~~~o0o~~~~~~~~~~

FORCES.
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Key point here is that forces are "SLIDING" VECTORS that can be considered to ACT ANYWHERE ALONG THEIR LINES-OF-ACTION. (Because of the failed education system, this requires endless re-stressing...)

So, as pointed out by both Ralph and Nathan, the Panhard-Bar is simply a link that transmits equal-and-opposite forces between car-body and live-rear-axle. The LoA of these forces is the centreline of the PB (or more accurately, the line through the centres of the Ball-Joints at each end of the PB, neglecting friction in the BJs). So the PB force from-axle->to-car-body, say Fpb, can be considered to act ANYWHERE along this LoA.

Nathan suggested that a good point to do calculations for this force is on the LoA directly below the CG. This way the "Heave" of the car-body is easier to calculate, because the vertical component of Fpb, say Fpb.z, acts directly through the CG. Quite true.

But it is very important to realise that calculations at ANY point along the LoA ALWAYS give the same answers (when done correctly!). So the answers are the same whether the "FAP" is ten miles to the left of the car and one mile above ground, or a squillion miles underground and ten squillion miles to the right of the car.

As just one example, it is worth considering the effect of Fpb on the car body when it "acts at a point" on its LoA that is at the same height as the CG, but offset somewhat sideways. Now the horizontal component of the PB force, say Fpb.y, directly counteracts the Inertial-Force Fi, while the vertical component Fpb.z acts on its horizontal lever-arm to both Heave and Roll the car-body.

Putting all this another way, beware of anyone who suggests that their particular "Force Application Point" is better than any other (ie. don't trust them!).

All the above should be obvious from the definition of, and understanding of, a "Moment". Namely, a Moment is the vector you get when you take the Vector-(Cross)-Product of a Force-Vector and Displacement-Vector.

But nowadays students are not even taught the difference between Couples and Moments.

So ..... more ranting in next post... :)

Z

COUPLES and MOMENTS.
======================
These are two very different things. Here is how I described them in a PM to Ralph (slightly editted).

"The important difference between "Couple" and "Moment" is (IMO, and also in the older teachings):
* A "Couple" is a forceful ACTION applied to a body, namely a "twisting or rotating force" (hence often with "T" (Greek Tau) for Torque).
* By contrast, a "Moment" is an EFFECT of a system of forces, capable of being calculated or measured at a particular Point, but it is NOT THE ACTION itself.

Note that it is easy to have "the Moment M, of a single Force F, at a particular Point P". However, it is impossible to generate a purely rotating force by using only a single linear force. Instead, for that "purely rotating force", you need at least a "Couple of Forces" (ie. two or more forces such that they amount to "two equal magnitude and oppositely directed forces on non-coincident LoAs")."

This is illustrated in this pic.
https://lh4.googleusercontent.com/-I...00/MAccFg6.jpg

"Also the Moment of a single linear Force is, in general, different at different Points (although with axisymetric distribution, with greater Moment the further away P is radially from the Force's LoA). By contrast, the Moment of a Couple is ALWAYS the same, no matter at which Point you measure it. So Forces are "sliding vectors" (can be drawn anywhere ALONG their LoAs, but NOT OFF the LoA), and Couples are "free vectors" (can be drawn ANYWHERE).

There are other subtle details, and it all requires a lot of practice, and more rigour than is common these days... I admit that sometimes I get sloppy in my writing and use the wrong word/phrase(s). But for better understanding I think it important in examples such as yours directly above to think more in terms of Couples, and less in terms of Moments.

So, (briefly) if you want to move all your linear Forces to the CG, then you calculate their Moments at the CG (calc'd with the Forces on their original LoAs), THEN move the Forces to CG, and THEN also ADD A COUPLE to the system that is equal to the sum of all the Moments. (Subtle point, but the Couple can be drawn acting anywhere, although drawing it at the CG is just as good a point as any other.)"

This last paragraph about moving forces AWAY FROM their LoAs is illustrated in this pic.
https://lh4.googleusercontent.com/-E...00/MAccFg3.jpg

In Figure 3a, a force F1 acts on a body. We want to know what changes to the picture must be made if a SIMILAR force to F1, say F2, acts at point P on the body (ie. F2 has the same magnitude and direction to F1, but is on a LoA passing through P).

In Figure 3b, "NOTHING" is added to the picture, in the form of two equal magnitude but oppositely directed forces F2 and -F2, with both acting along the same LoA that passes through P.

In Figure 3c, the two forces F1 and -F2 are replaced by the Couple T. The magnitude and direction of T are found as shown, and as noted above it can be drawn anywhere (ie. it is a "free vector").

Note that when doing such "geometrical calculations" the manipulations can go from a->c, or from c->a. Also, adding "nothings" to the picture, in the form of two equal and opposite forces, is very useful (eg. it can help keep all the calcs on the page). The whole process is very similar to algebraic manipulations, with, for example, "add +X and -X to LHS of equation, then ...", etc.

But the geometric approach gives a much better visual feeling for what is happening. The map is much closer to the terrain. The more abstract algebraic approach, developed by Descartes in his "Analytic Geometry", used to be called "cogitatio caeca". Literally, it is "thinking blind". It is the use of a bowl of alphabet soup to represent a spatial, physical, problem.
~~~~~o0o~~~~~

Here is another pic showing how one Couple of Forces with large magnitude but close spacing in Figure 4a, can be transformed into another EQUIVALENT Couple of Forces with smaller magnitude and wider spacing in Figure 4c.
https://lh5.googleusercontent.com/-O...00/MAccFg4.jpg

Note the arbitrariness of the positions and directions of the two Couples, and hence their "freeness". Also note the repeated sliding of forces along their LoAs, and the addition of "nothing" to move the calculation along. Lots of practice with these types of geometrical vector manipulations makes VD a very simple subject.
~~~~~o0o~~~~~

As a final note, for anyone who thinks that all this "let's draw play-school pictures" stuff is too childish for grown-up Engineers, then NOT SO. The above geometrical approach was developed hundreds of years ago (1600s+) along the same Axiomatic-Deductive lines as Euclid's Elements. And so too, originally, was Descartes' algebraic approach. So both are equally rigorous approaches to solving physical problems.

But nowadays very little of the foundational stuff of the algebraic approach is taught (except, perhaps, some "Set Theory" waffle). So the students effectively enter the movie theatre half-way through the show, and come out thinking that "Couple" and "Moment" are the same person. All the rigour has been tossed out the window, and instead it's "like, ...whatever...".

And then they look at the car in front of them, and at their bowl of alphabet soup, and they can't make sense of any of it...
~o0o~

Still a bit more ... :)

Z

Last bit regarding Ralph's post on Panhard-Bars/Speedway-Cars on previous page, and solutions using Classical Mechanics.
~o0o~

DYNAMICS.
============
The main point I want to stress here is that the use of D'Alembert's Principle can turn an apparently difficult problem in Dynamics into a very simple to understand picture in Statics.

As per Newton's Axioms of Motion, the resultant of a system of forces F, acting on a massive body m, for a short period of time dt, CAUSES a change in "the quantity of motion" (<- Newton's wording, but nowadays called Momentum P) of the body dP.
So, F.dt causes dP,
or F -> dP/dt,
(or, very sloppily, F = m.dV/Dt = mA).

Since Momentum is a vector, it can be drawn in the same way as a Force. So the arrow representing the resultant F of the system of forces acting on the body (for the short period of time dt) can be over-drawn with an arrow representing the "change in Momentum" dP.

(And BTW, I suggest that this is how your VD simulation programs should be structured. Namely, a time-stepping process with forces acting on the body for time dt. Then consequent "changes in quantity of motion" at the end of each dt. Then adjust forces based on new motions, and repeat.)

Anyway, a disadvantage of the above approach is that when it is drawn as a FBD it shows a bunch of forces pushing on the body, say the road-to-tyre forces pushing leftward, and then only a dP vector at the CG that points in the SAME direction. How do these arrows, all pointing leftward, cause the suspension springs to be squashed out of their original (static) positions?

The way to get better understanding is simply to replace the dP vector with the equal sized, but OPPOSITELY directed, Inertial-Force Fi. Now the FBD shows the road-to-tyre forces pushing leftward (and upward), and the Inertial-Force at the CG pushing rightward (and gravitational force Fg acting downward). Now it is quite obvious, visually, why the springs compress or extend as they do, and the body Rolls and Heaves as it does.

Note that in this picture (ie. FBD) the spring forces are internal to the system, acting between car-body and live-rear-axle. The inclusion of Fi in the diagram provides the last bit of external information needed to solve for the internal spring forces. But if no Fi in the FBD, then no solution possible.
~~~~~~~~~~o0o~~~~~~~~~~

KINEMATICS.
================
(Again, for simplicity, consider this only a 2-D, rear-view problem of Ralph's car with Panhard-Bar.)

During cornering, where is the Kinematic "Roll Centre", or "IC", of the car-body's MOTION wrt the live-rear-axle?

Note that this problem is a lot simpler than the case with independent suspensions. As covered elsewhere (Search for "Where is the Motion Centre?" sketch) it is futile trying to find an independent suspension's "Kinematic RC/IC" of the car-body, either wrt the ground (because it depends on how much the tyres slide), or wrt the wheelprints (there are two of them!).

But here there are the two bodies (ie. car-body and live-rear-axle), so an IC can always be found. Conveniently, the two bodies are connected with the roughly horizontal, rigid but BJ-ended link, the PB, and the two lengthwise-flexible Spring-Dampers, taken to be mounted vertically between the L and R sides of the body and axle (they are not actually vertical SDs, but let's assume so...).

So the driver turns into the corner, the road pushes the tyres leftward, inertia pushes the car-body rightward, and the L and R SDs extend or compress a bit. Where is the IC for the resulting motion of the car-body wrt live-rear-axle (either taken "instantaneously", or averaged over a short time period)?

The first half of the solution is very easy. The IC must always lie somewhere along the PB's centreline, or n-line. This comes from the definition of an IC, namely, the point in either body's reference frame that has NO MOTION wrt the other body's reference frame. The PB's n-line includes all the points in BOTH bodies that have NO MOTION in the DIRECTION of the n-line, wrt the other body.

The second half of the solution is a bit trickier. Here is a hint. Consider the tops of the SDs where they attach to the car-body. Draw a line through these two points (take it as horizontal for now). Consider the extension or compression of the SDs, as calculated in your earlier Statics+d'Alembert FBD. Based on these two vertical movements of the body-mounting-points of the two SDs, find the points of the car-body that have NO VERTICAL MOVEMENT.

Problem solved!

And as should be apparent, depending on how the two SDs extend or compress, the car-body's "Roll Centre" (ie. its Kinematic IC wrt the axle) can be ANYWHERE along the PB n-line. Yep, anywhere from left infinity, to right infinity. Or, by chance, it might be under the car...
~~~~~~~~~~o0o~~~~~~~~~~

Last point.

Quote:

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Originally posted by Ralph in PM (with a hint of sarcasm :)):
... Now you KNOW real race cars can't have beam axles! I would love to take the purists on a 140 mph ride around a one mile dirt oval in one of our cars. Which by the way we do on a regular basis.

---

(My emphasis.)

Yep, 140 mph on a rather bumpy track, with BEAM-AXLES!!! :)

Z

z,

Thank you very much for your assistance and for re-kindling my interest in reviewing my classical mechanics. Being a grey beard I did learn all of the above many, many years ago around 1977 to 1982 but my skills had atrophied from lack of use. :(

In my view the best references I have in my library are Mechanics by Den Hartog and Statics by J.L. Meriam. I have many, many other texts concerning general physics, kinematics, machine design, descriptive and analytic geometry etc. etc. but these two works are by far the clearest. Hartog is a little unclear in describing your figure 3 above concerning the relocation of a force by the use of a force and couple but Meriam cites it almost verbatim as you have presented it. It is included in Hartog of course but presented with some subtlety that doesn't reach out and grab you.

Another good resource for a general review is MIT's Open Courseware 8.01 classical mechanics video lectures by physicist Walter Lewin. This is a video archive of his freshman fall lectures on the subject. While the students on here might view this as very trivial it is a great review of the very essence of classical mechanics and if you don't find Prof. Lewy, as he is known, as one of the most entertaining and informative lecturers you've ever seen I would be surprised. To have a single professor like him in your college careers is a very rare thing, at least in my experience.

Link: http://ocw.mit.edu/courses/physics/8...1999/index.htm

To the students who may read this thread and wonder 'why in the heck would you design a beam axle like that??' all I can say is 'I' didn't. :)

The cars as described have evolved over the course of some sixty years. They started as modified U.S. passenger cars of the 1930's and are mostly the product of 'trial and error' garage engineering. Don't laugh, it might be a slow process, but cut and try has produced many beneficial pieces of machinery.

The biggest advances in these cars over those sixty years have been in the area of construction materials available, fabrication techniques, tire size & compounding and as in all things racing aerodynamics. However, to my knowledge at least, only one car has made it to a wind tunnel test which yielded some hard data and which was published in an SAE paper by Eric Koster. The rest, cut and try.

Why did I get involved?

Well, I was actually perplexed that after all these years of evolution that all of the cars produced by basically four builders wound up in the same tiny corner of the box of design possibilities. The beam axles are a product of the rulebook so that is not a design choice. But how to suspend a beam axle car has many possibilities and many have been used over the years on these car.

So, why is a twin trailing link, torque arm, short offset cockpit adjustable panhard bar system the winner for the last ten or so years?

The ever thickening rulebooks are one possibility. Never think that what happens at the upper echelons of automobile racing doesn't eventually trickle down in some form to the grassroots level. It does, if for no other reason, than the illogical reason, of 'I want to be like the 'Big Boys' mentality of the rules makers.

The era of 'credit card' racers may also be a factor as very few participants actually build and develop their own cars anymore. Break it, buy a new shrink wrapped piece from the parts truck.

However, there are some who have asked, who think I might know at least a little bit of engineering, to have a look and explain what and more importantly why the cars are designed as they are. And what might be done to improve them. The driver behind this of course is that like in many current forms of motorsports they have achieved parity and that makes winning frequently very difficult. :)

Why would anyone pick driver adjustability of the panhard bar angle as a design criteria? I don't know but I am working to find out.

Thanks for allowing me to participate in this forum even though my questions and thoughts may not pertain directly to formula SAE.

Thank you again Z for the great illustrations and explanations. (Expect more PM's :) )

Ralph

Just thought I would post a picture of the type of car I have been discussing.

Attachment 250

And then there are the extremes:

Attachment 252