Jacking force

[Claude Rouelle]

Bill,

You wrote "....what you are faced with as far as Mz influences on a suspension."

Mx or Mz?

Claude

[BillCobb]

Yes, that should be Mx. At least someone is reading these posts for content ! I was able to redit the post to correct it. This is all about overturning moment: Mx. On current production passenger cars, this increases the roll gradient roughly 10%. May not sound like a lot, but at incipient rollover, you'll pay anything to get 2 wheels back on the ground.

[Z]

Tim,

The interesting thing ... the equivalent stiffness at the point of application of the action force is NOT simply the spring stiffness multiplied by the motion ratio squared like the books and the seminars say. ... the actual wheel rate appears higher than that predicted by the formula Kwhl = Kspring x MR^2.

You can see this for yourself by doing a virtual work analysis on a hypothetical suspension which has a non constant motion ratio. The result is that the wheel rate is equal to:
Kwhl = Kspring x MR^2 + Fspring x dMR/dz,
where dMR/dz is the slope of the motion ratio curve with [respect to] wheel centre vertical travel and Fspring is the force in the spring at any instant.

This extra component ... is often around 10-15% for a direct acting double wishbone suspension. So it's far from negligible.

Yes, your second formula (bold) is indeed the correct one.

While Virtual-Work analysis does give the correct answer, it can be obtained more directly as follows.

Define:
Motion-ratio MR = ds/dz,
(ie. MR = axial-deflection-of-spring/z-deflection-of-wheel, ... so big MR implies long spring),
Wheel-rate Kw = d(Fw)/dz,
Spring-rate Ks = d(Fs)/ds.

Forces at wheel and spring are related by,
Fw = Fs x MR ..... (Law of Levers).

(Edit: Actually, both Law-of-Levers and Virtual-Work can be written as Fw.dz = Fs.ds, so pretty much the same thing, with same derivation below.)

So wheel-rate is given by,
Kw = d(Fw)/dz = d(Fs)/dz x MR + Fs x d(MR)/dz ..... (Product Rule),
Kw = d(Fs)/dz x ds/ds x MR + Fs x d(MR)/dz ..... (1st-term-RHS x 1),
Kw = d(Fs)/ds x ds/dz x MR + Fs x d(MR)/dz ..... (rearrange...).

So,
Kw = Ks x MR^2 + Fs x d(MR)/dz.
~o0o~

An interesting feature of above is that you can use a spring that exerts a CONSTANT force at any extension, namely a spring with ZERO-RATE (ie. Ks = 0), and then develop whatever wheel-rate you want by using an appropriate Motion-Ratio curve. That is, you can eliminate the first term on RHS, and use only the second term to get all the wheel-rate.

Is this practical?

Sure! I would argue that the very effective rising-rate suspensions on the rear of almost all off-road bikes for the last 30+ years, comes mostly from the rising-MR-linkage they use, rather than from their spring's intrinsic "spring-rate". In other words, if you fit a zero-rate spring to those suspensions, and tweak the MR-curve a pinch, then the end result is the same. (Warning to FSers! DO NOT fit such rising-rate linkages to your corner-springs! Good for MotoX-bikes, BAD for FS-cars!)

And where do you get "zero-rate" springs?

One of the simplest approximations is a pneumatic cylinder connected to a largish volume pressure tank. A cylinder with constant cross-sectional-area, multiplied by approximately constant pressure, gives a nearly constant force. Old idea, used with much success.

Another version uses the approximately constant "weight"-force of a given mass near Earth's surface. A simple pendulum works like this, with constant-weight-force x variable-MR = variable-horizontal-restoring-force. A similar approach uses a car's downward weight-force, together with a suitable linkage, to act as the Pitch and Roll springs.

Z

[Claude Rouelle]

Bill.

At least I am reading your posts.

The understating I have is that the contribution of the tire overturning moments Mx to the total roll moment could "only" be about 10 % but that is slope of Mx Vs slip angle is usually "quicker" : Mx peaks at much smaller slip angle than Fy, all other conditions being the same, and that creates a driver feeling that "the car falls on the outside tire and that can only get worse" and that from there some drivers start to do irrational things with the steering and brakes for example and the car effectively rolls over. In other words if the 10 % of tire Mx contribution to roll moment at peak lateral acceleration could be 20 or 25 % at low slip angle and 0.2 or 0.3 G.

In other words the tires Mx doesn't really rolls the car but send the driver at the beginning of the corner a message that "if it continues like that the car will roll over"

Do you have the same perspective and experience? If you have then the tire you showed the graphs of does notillustrate my understanding as you can see the slip angle of peak Mx bigger than the slip angle of peak Fy

Claude

[BillCobb]

Yes, my feelings are the same. Often you will hear drivers express that "the car is rolling over on its tires". It sorta is, actually. In the sample tire I showed, the load sensitivity (derivative) is linear, so I can normalize the Fy, Mz and Mx traces. (That's a nice feature of race tires). The Mx is late as they say.

I'll look around for a tire that has less pressure and a narrower rim and bet I can show the Mx being earlier. That's the current explanation for the driver comment.

The plot title automatically shows the coefficients used to produce all the plots using my Pacejka4_Model tire function. By adjusting the 3rd and 4th Mx coefficient (B & C in Pacejka world), you can move the peak point earlier or later while maintaing the same initial stiffness. A library of these coefficients from all the TTC tires could hand you the 'best' tire without having to plow through all the hand waving. That's how it sometimes goes in Production Engineering.

[Tim.Wright]

The effects of Mx etc on jacking can be seen from the components of the trajectories of the contact patch. Considering a suspension's wheel travel movement (i.e. along it's degree of freedom), if your "force application point" (i.e. contact patch) moves along or rotates about a certain vehicle axes (X,Y,Z) then the application of a force or moment to that same axes will in turn cause a vertical movement (or a jacking force is the movement is constrained).

Therefore reason that Mx about the contact patch creates a jacking effect is because practically every suspension has camber gain (which is a rotation about the X axis). If you do a force analysis on a suspension which has zero camber gain, then the Mx will have no jacking effect. On the other hand, racing suspensions typically have very high camber gains and wide tyres (potentially producing a lot of Mx) so the effect could be non negligible.

This logic be applied to every translation and rotation that the contact patch makes in its wheel travel degree of freedom. It also explains why you see a small jacking effect from Mz on any suspension which has some bumpsteer and (less obviously) why changing your bumpsteer curve also slightly changes your anti dive curve.

For what it's worth, this driver description of a tyre "rolling over on itself" is an interesting one. Having felt it myself too and noted its quite a strong feeling I always suspected it must be connected to something a lot less subtle than an Mx component at the tyres. Particularly on road cars which have narrow tyres very little camber gain. I found it particularly prevalent in vehicles which have the blade control arm suspension on the rear.

[Flight909]

For what it's worth, this driver description of a tyre "rolling over on itself" is an interesting one. Having felt it myself too and noted its quite a strong feeling I always suspected it must be connected to something a lot less subtle than an Mx component at the tyres. Particularly on road cars which have narrow tyres very little camber gain. I found it particularly prevalent in vehicles which have the blade control arm suspension on the rear.

I would feel that the added roll angle of Mx acting around IC will small on race car with stiff suspension. I think decrease in vertical spring stiffness of tire when Fy is high makes a bigger affect of roll angle, Tim maybe this you felt?

[BillCobb]

OK, so what's your feeling about two other dynamic vehicle properties that are affected by the Mx's ?

[Z]

WHERE IS THE ROAD-TO-TYRE FORCE-SCREW?
===================================

Let's start this simple.

Consider a 2-D end-view of a car-body, its wheel, its tyre, and the road. Look at the forces coming from the road and ultimately acting on the car-body, via the tyre/wheel/axle/etc. The overall effect of the multitude of little road-particles acting against the multitude of little tyre-particles can be seen in many different ways. Here are two obvious ways:

1. A specific point is chosen in a cartesian coordinate system fixed to the wheelhub, with that point perhaps called the "notional wheelprint (or contact-patch) centre W". In this 2-D case, two linear force components, say Fy and Fz, from-road -> to-wheelhub, are drawn (or measured, or simulated, or assumed to be) acting on mutually perpendicular Lines-of-Action through the point W, together with a couple, say Mx, that acts on the wheelhub as a whole (remember, couples are "free" vectors!).

2. Alternatively, a SINGLE linear force vector, call it Fr, is drawn to represent the resultant of the sum of all the multitude of little road-to-tyre-particle interactions (ie. the sum of Fy, Fz, and Mx above). In general, this resultant force Fr is at a different angle to either of the cartesian axes given above, and it does NOT pass through the point W.

Take as an example the tyre-curves in Bill's post near bottom of previous page 7. (But note that the numbers on those curves seem odd, with a maximum Fz of 300 N, ... = ~30 kg? See*.).

Anyway, with Fz = 300 N, and Fy high enough to give about 6 degrees of Slip-Angle, the road-to-wheelhub couple Mx = ~50 Nm. So, according to the curves, the Fy, Fz, and Mx force components can be replaced with a single resultant force Fr that lies on a LoA that passes through the road-plane about 17 cm to the side of the notional wheelprint centre W (... because 50Nm/300N = ~0.17 m).

(*Edit: The curves make more sense if they are in pounds and feet. Then the offset of the Fr force is 50 lb.ft/300 lb = 2 inches, which is more reasonable.)

In other words, "Fz + Mx" is the same thing as having ONLY the Fz force acting upwards on a LoA that is 17 cm (or 2"?) closer to the centreline of the car than the "wheelprint centre" W. This inwards shift of Fz is hardly surprising given the video in Bill's post at bottom p7, which shows most of the tyreprint (= "tyre contact-patch") moved similarly inwards. It is also hardly surprising that tall SUVs are prone to falling over as a result of this (large!) reduction in effective half-track dimension.

It is worth asking (rhetorically);
How can "Fz + Mx", both of which by definition are considered to act AT the point W, actually be exerting any sort of force at all, given that there might be NO contact at all between road and tyre, AT the point W!?

And similar comments can be made about "Fy + Mz", which amounts to the same thing as "Fy moved backward by distance = pneumatic-trail".
~o0o~

Moving to more realistic 3-D, it is well known (or should be) that in general the total effect of all those little road-to-tyre-particle forces CANNOT be reduced to a SINGLE, linear, resultant force, acting along some arbitrary LoA. In general 3-D systems of forces, the "resultant" is always a dual-vector quantity.

It is also well known (or used to be), that the best (= most reasonable, most sensible, most useful...) way to specify this 3-D dual-vector quantity is to specify the location of its central axis, together with the magnitudes of its force and couple components. That is, it is very useful to know, or "see", the location of the road-to-tyre "Force-Screw" (aka "Wrench"), and to have an idea of the relative magnitudes of its two components.

So..., does anyone in the Tyre/VD-industry ever do this?

If not, then why not?

If so, then does anyone have animations showing how the Force-Screw moves about with different tyre loading conditions, and which of its two components (force or couple) dominate during the high load conditions of interest in racing?

I have a good idea how this road-to-tyre Force-Screw behaves, but I wonder if anyone else has bothered looking at it?

(Note: All the data needed to see this behaviour should be available to TTC members. It is simply a matter of displaying it appropriately.)

Z

(PS: The jacking effect of "Mx + camber-gain" can be seen, IMO most simply, as the Force-Screw moving closer to the car's centreline, where the n-lines of a positive-camber-gain suspension are steeper. Similar to reducing track-width, and again obvious from watching Bill's tyre-testing video.)

[DougMilliken]

..... So..., does anyone in the Tyre/VD-industry ever do this?

It's been a long time since I read this book, but I have some memory that he does this? John R. Ellis, "Road vehicle dynamics" 1988 & 1989.

http://www.worldcat.org/title/road-v...=di&ht=edition