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Loz,
Confusing! Upon a re-read it is simply awful and I am surprised you took the time to answer.:p
I think we'll just chalk this one up to a 'brain fart' and leave it at that. I was investigating one thing and starting to ask questions about a completely different topic. Too quick to the keyboard I guess.
Thanks for the attempt at a response and I will pose a well formed question when I have a little more time.
Ralph
Quote:
Originally Posted by rwstevens59
Ralph,Quote:
Originally posted by Loz:
... (unless of course the tyre is rigid and moves in unison with the wheel in which case we only need be concerned with the change in orientation of each wheel and we are back to using the axle housing).
As Loz suggested, "roll steer" is determined by the direction of the wheel-planes, namely the direction perpendicular to the Axle-Housing (AH). So for a Roll motion of the AH, wrt the car-body, you want to determine the relative movements of the wheel-centres, NOT the wheelprints (= contact patches).
BUT (!), unfortunately, you can NOT determine these Roll steer effects purely from a 2-D side-view of the linkage.
Your side-view diagram does let you determine the "Pitch-plane" (or "side-view") "anti-squat/lift" properties of the suspension. Assuming that yours is a live-axle (diff fixed to AH, so not a De-Dion which works differently), and the brakes are also fixed to the AH (not on "birdcages", etc.), then the analysis is as follows.
The two Trailing-Links are n-lines for the motion of AH wrt body. However, in side-view these two n-lines are on top of each other, so appear as one. The Slider on the torque-arm has a "contact normal" which is a vertical line passing through the S contact zone (assuming the sliding surface is horizontal). This vertical line is a third n-line for AH motion wrt body.
Draw the vertical S n-line on your drawing. The point where this S n-line passes through your TL n-lines is the Instant Centre for the 2-D side-view motion of AH wrt body. A line through the IC and wheelprint determines the wheelprint n-line. This in turn determines your anti-squat and anti-lift properties (ie. horizontal = 0%, steeper = more).
BUT (!), your roll-steer is NOT determined by the apparent movement of the wheel-centres in this side-view!!! This is because the Roll motion takes place in the [drum roll...] THIRD-dimension. So you must also do the analysis there. I will definitely get around to doing some sketches of this one day (yes, promises, promises...) but briefly for now.
Your Panhard-Bar is the fourth n-line that constrains AH motion wrt body. Four constraints = two degrees of freedom, which means the left and right wheelprints can move independently up-or-down. Or you can think of the AH moving independently in a Roll motion or a Pitch motion (with this Pitch motion covered above).
Strictly speaking, the "Pitch" motion (of AH wrt body) is about an APPROXIMATELY horizontal-lateral line that intersects the two TLs and the vertical S n-lines, and also intersects the Panhard-Bar-n-line. Only when this "Pitch-axis" intersects ALL four of the AH-to-body n-lines (ie. 2 x TLs, S, and PB) is this line an Instantaneous Screw Axis (ISA) of zero thread pitch (ie. it is a "revolute" joint, like a simple hinge).
The second ISA of zero pitch (there are always two of them) might be called the Roll-ISA, and is roughly longitudinal and (going from the rear of the car forwards) intersects the PB-n-line, then the vertical S-n-line, then the two TL-n-lines. If the two TLs converge towards the front of the car, then the Roll-ISA intersects both of them at their mutual intersection point. If the two TLs are parallel, then this Roll-ISA is also parallel with them, intersecting them "at infinity".
BTW, the generalisation of all this is that ANY four straight lines in 3-D space (eg. 4 x n-lines) can ALWAYS be intersected by 2 other straight lines (eg. 2 x ISAs of zero pitch). The two ISAs of zero pitch then determine the position of the "cylindroid", which in turn determines all possible ISAs for this 2 DoF linkage. Maybe more later...
Lastly, the slope of your Roll-ISA (two paragraphs up) determines your Roll-Steer. If Roll-ISA slopes up-to-front, then Roll-OVERSTEER. If Roll-ISA slopes down-to-front, then Roll-UNDERSTEER. Assuming your TLs are close to parallel, then it is most likely (for typical PB position) that the Roll-ISA will have a similar slope to the TLs, so your sketched car should have Roll-OVERSTEER.
Enough for now...
Z
(PS. The torsion-bars will give variable rate springing from their changing geometry, but that should not directly affect the Roll-Steer kinematics.)
Z,
Thanks for the very clear response to a very poorly worded question. Have both of Jack Phillips books now but I'm still wrestling with those as I guess I've spent too many years in front of a 2D drawing board or CAD station. :)
The original question as posed to me that I was working on when I created this side view was, 'what would be the difference between the torsion bar springing as shown and replacing the torsion bar arrangement with linear coil springs?'. My answer to that was do not try and emulate the varying rate of the torsion bar system but simply calculate the static deflection for the travel needed, pick a pair of coils that will give close to that deflection and start testing/tuning with the coils.
The side view diagram then caused a bit of a ruckus with the local racers who saw it and questioned the axle housing motion as drawn. Hey wait a minute with a trailing link, torque arm, panhard bar suspension we should have roll oversteer meaning in this right side view the housing should be moving back not forward. I tried to explain the if you were to support the car on jackstands and then used a pair of jacks to return the axle housing to a level condition at a measured static ride height WRT chassiss (torsion bar stop removed, i.e. no windup, and shocks removed) and then were to jack the rear axle housing in pure level bump and then let the jacks down in pure level droop that the axle motion as drawn should be what you see/measure.
I then wrote my very hastily and poorly worded question here on the forum.
Two questions I really have are:
1. For a symmetrical left/right car with live solid axle,with spool (no differential) and outboard brakes using parallel trailing links, torque arm and panhard bar would this 2D drawing be reasonable?
2. If we were to plot the contact patch arc produced when the driveline is in a locked condition would a line drawn perpendicular to this arc also be the n-line for the contact patch?
I am working on a true representation of axle housing roll to satisfy the doubters that they do indeed have roll oversteer. :rolleyes:
Thanks again,
Ralph
P.S. Jack Phillips definitely has a unique 'style' in his writing. :)
This can still be done on a drawing board and in 2D views. BUT, it is necessary to also include a top view and rear view with many projection lines, kinematic paths and Euclidian geometrical construction techniques. It would still be palatable for a mechanism such as this, but unnecessarily time consuming compared to other techniques available (especially a simple 3D CAD based linkage model with 1D elements).Quote:
Originally Posted by rwstevens59
Your answer would be the more sensible approach and certainly quicker and ultimately more likely to result in spring selection that works rather than pursuing a lot of needless calculations only to arrive at the same starting point. Either way you still have to test and tune the spring selection to take into account all of the real world variables on that day for those track conditions.Quote:
Originally Posted by rwstevens59
It wouldn't be difficult to determine the torsion bar non-linearity and upper and lower bounds of stiffness. However, there would not be a singular value of non-linear stiffness but multiple values given there are different kinematic positions the suspension could assume during operation - i.e. different combinations of positions the torsion arm rollers could be acting on the arms due to combined modes of pitch, roll, etc...).
The use of a panhard bar for lateral constraint automatically dictates that it is not a symmetrical setup. If the panhard bar is close to horizontal and only moves vertically up and down a small amount (i.e. small n-line slope changes away from horizontal), then it may be approximately symmetrical enough to be considered as symmetrical (this symmetrical condition is the same as required for having a revolute-joint pitch axis.Quote:
Originally Posted by rwstevens59
e.g.
Quote:
Originally Posted by Z
What do you mean by contact patch arc? And when you say "locked condition" do you mean a spool/locked axle or that some kinematic movement in constrained?Quote:
Originally Posted by rwstevens59
It would probably be quicker and more convincing to the skeptics to take the wheels off and springs out, attach a couple of plumb-bobs to the wheel centres, jack the car and wheels up and down in the motions you require and draw some points and lines on the ground in chalk (plus you inadvertently take out any error you introduce into a 2D/3D model from measurement errors of the mechanisms). Despite not using "high-tech" whiz bang technology such as a computer, methods like this can still be extremely precise for the accuracy and resolution required for assessing and developing vehicle dynamics. A prime example is using a string line to do a wheel alignment.Quote:
Originally Posted by rwstevens59
Loz,
My question two is referring to the following example, one of several given to me by Z to locate the longitudinal n-line slope. In the case of my diagram I did not plot the path of the contact patch in the three positions shown, but if I had the locked driveline would cause a point marked on wheel centerline at the ground tire interface to rotate forward and back as the axle was moved from full droop through static to full bump which would, based on only the three points shown, produce an arc.
The question: Would the example stated below be applicable?
Z:
"1. You have a car in front of you, and your job (as Junior Test Engineer) is to determine its "% Anti-Squat" or some similar number. Ie. you want to find the side-view, longitudinal-n-line slopes of the rear wheels. Unfortunately, the car is covered in so much mud that you can barely recognise any suspension links... So, you might lock the transmission somehow (put in gear, or vicegrips, etc.), paint a mark at the 6-o'clock point of the wheel (= wheelprint), somehow move the wheel up-and-down through its full suspension stroke (wrt car-body), and mark on a piece of plywood (again fixed wrt body) the not-quite-vertical Path-of-Motion of the wheelprint. The longitudinal-n-lines are the lines that are perpendicular to this PoM, at different points along the path (ie. PoM is probably curved, so n-lines have different slopes at different points). The road-to-wheelprint force is doing a similar thing to JuniorTE here, in that it has no hope of knowing where the "IC" is. It simply pushes on the wheelprint, and if it can compress/extend the suspension a bit, because it has a component along the PoM, then that is what happens. "
Thanks,
Ralph
Semantically, but pertinent to your description, in a geometric description, a point is a zero-dimensional vector and cannot rotate.Quote:
Originally Posted by rwstevens59
So the contact patch point originally at the ground cannot rotate, but the position vector (i.e. a line joining two points), can rotate. In this case perhaps the points used could be the original point at the contact patch and some point infinitesimally close (but not coincident with the original point) which are both on the path you describe (i.e. the "arc"). They are by nature tangent to the path. The lines perpendicular to any two adjacent infinitesimally close points on the arc are the n-line vectors in the example.
This is for the static case only. Add some vehicle dynamics and variations of tyre loads and tyre-loaded-radius and path of motion of the contact patch "point" will follow a different path with an infinite number of variations possible.
Z and Loz,
Attached a very, very rough lunch time ball point pen rough sketch of what I imagine to be the relevant n-lines as described by Z.
This is the part with which I am struggling:
Quote Z:
"Strictly speaking, the "Pitch" motion (of AH wrt body) is about an APPROXIMATELY horizontal-lateral line that intersects the two TLs and the vertical S n-lines, and also intersects the Panhard-Bar-n-line. Only when this "Pitch-axis" intersects ALL four of the AH-to-body n-lines (ie. 2 x TLs, S, and PB) is this line an Instantaneous Screw Axis (ISA) of zero thread pitch (ie. it is a "revolute" joint, like a simple hinge)."
"APPROXIMATELY horizontal-lateral line that intersects the two TLs and the vertical S n-lines"; I can visualize that.
"and also intersects the Panhard-Bar-n-line."; This I can not visualize. In my mind the line crossing the TLs n-lines and the S n-line would be in a different plane, at least as I picture it for this particular linkage and would never intersect. Unless of course I am not visualizing the correct n-lines particularly for the panhard bar.
Again pardon the very poor blob used while I was thinking and not going to post but...what the heck it would not come out any better if we were all sitting around the table at lunch at the first go.
Any help and clarifications would be welcome.
Attachment 285
Thanks,
Ralph
Ralph,
Your sketch is good. I will try to do a similar one, with a bit more added, in the next week or so. Have to get a few other things out of the way first.
Firstly, it is a truism of Euclidean geometry that ANY four straight lines in 3-D space can ALWAYS be all intersected by another two straight lines. So, it is just a matter of going a-hunting for those other two lines.Quote:
"APPROXIMATELY horizontal-lateral line that intersects the two TLs and the vertical S n-lines"; I can visualize that.
"and also intersects the Panhard-Bar-n-line."; This I can not visualize. In my mind the line crossing the TLs n-lines and the S n-line would be in a different plane, at least as I picture it for this particular linkage and would never intersect. Unless of course I am not visualizing the correct n-lines particularly for the panhard bar.
If, in your sketch, the PB was exactly horizontal-lateral, then the "Pitch-revolute-ISA" (ie. a "simple hinge" for "pitching" motion of AH wrt body) would also be an exactly horizontal-lateral line, parallel to the PB (so intersecting it "at infinity"), and intersecting the 2 x TL and 1 x S n-lines. So roughly along the line you have drawn through the two TL mounting points to body.
However, if the PB slopes down-to-left, as you have drawn, then the Pitch-revolute must be rotated anti-clockwise about the S n-line in plan-view. Since this Pitch-revolute must still intersect the 2 x TL n-lines, its right-end moves up-and-forward, while its left-end moves down-and-backward, as it slides along the two TLs. Eventually it intersects the PB n-line somewhere to the left of the car.
Regarding the "Roll-revolute", you have drawn the two TLs converging slightly to the rear of the car. If the TLs intersect somewhere behind the car, then the Roll-revolute is roughly horizontal-longitudinal and passes through this TL intersection point, then through the PB n-line, then through the S n-line. If the two TLs DO NOT intersect at all (ie. if they are slightly "skew"), then there will still be a roughly horizontal-longitudinal Roll-revolute that intersects all 4 x n-lines. However, this Roll-revolute might not be as close to longitudinal as before.
Jack Phillips used to make lots of 3-D models of these things. Typically, they would be a largish lightweight steel frame, then lots of different coloured strings stretched within the frame, to represent all the different straight lines.
Z
(PS. If the Slider mechanism is to ONLY constrain vertical movement of the end of the torque-arm (via a single vertical "n-line"), then, yes, it must necessarily be a "5 DoF" joint. Ie. 6 DoF of a free body, minus 1 constraint = 5 DoF.)
Ralph
I have seen far worse, less "rough" sketches than that. I would say it is quite reasonable actually.
They key here is the approximately horizontal-lateral line which isn't actually horizontal and is in fact most likely ever so slightly angled in some direction such that it intersects the panhard bar n-line. Where is intersects though is likely to be way off in the distance unless the panhard bar is angled away from the horizontal by large angle.Quote:
Originally Posted by rwstevens59
Where your visualisation probably goes wrong is from assuming the panhard bar n-line and the pitch axis are exactly parallel (not approximately). In which case they don't explicitly intersect, rather they implicitly intersect at infinity.
Realistically, only when the panhard bar is exactly horizontal and the approximately horizontal ISA is parallel to this line, is there likely to be a true revolute ISA (pitch axis). In all other cases, even for small angular angular changes in panhard bar or TLs in roll, there is an ever changing set of instant pitch and roll axes.
Loz
Well the stubborn old man is back with some preliminary drawings for plotting body roll and heave WRT an assumed fixed solid rear axle. i.e The use of descriptive geometry from the 'old days'.
What follows are pictures that I hope are mostly self explanatory.
Assumptions:
The axle is fixed in space and acts as the 'frame' while the body points are moved about it.
In the initial layout all links are co-planer with the knowledge that as soon as movement takes place they will not be.
Right from the start the very first problem I encountered was how to rotate the body about the axle, i.e. what frame of reference should I use?
Right or wrong I have chosen, for pure roll, to rotate the body around a horizontal line connecting the body spring/damper mounting locations. Due to the lateral restraint of the panhard link this will require a rotation and an associated translation of the body WRT the solid axle. Again your input on how correct or incorrect this might be would be appreciated.
Some pictures to go by:
Attachment 309Attachment 310Attachment 311Attachment 312Attachment 313
Looks like I will need to add a second reply????