Jacking force

[Z]

Sid@TR,

Some comments on your "ROLE OF A ROLL CENTER (Mystery solved or mystery continues?)".
~~~o0o~~~

"Basic definition of roll center is the point about which the sprung mass rolls under the influence of centrifugal force."

This "RC = Motion Centre" definition is perhaps the worst of the many definitions of RCs. Mainly because it does NOT take into account any vertical (= jacking!) motion of the sprung mass, or lateral sliding of the wheels on the ground. I would avoid using it.
~~~o0o~~~

"It is also a point through which the lateral forces transmitted from Tire’s contact patch acts upon achassis."

... together with the jacking forces! It is the point (in this simplified 2-D example) where the "control-arm forces" from both wheelprints, acting along their respective n-lines, can be vectorially combined into a single force acting on the body. In general this force will NOT be purely horizontal.

IMPORTANT NOTE (to restress above). Each control-arm force is, in general, NOT horizontal, because its n-line is not horizontal. So, since the control-arm force has horizontal Fy, and vertical Fjacking, components, BOTH of these must be moved to the RC. That is, each resultant control-arm force (Fr.ca or Fl.ca in previous posts) is SLID along its LoA (= the n-line) to the RC, carrying its two components, Fy & Fj, with it.

Less important note. RCs, namely the intersection point of the two lateral n-lines, frequently zoom off to infinity (or at least out past Pluto...). This happens with any suspension with that has n-lines that remain notionally horizontal wrt the body (= ~ ground level RC), and the body then adopts a roll angle (ie. because parallel lines meet at infinity). There is NOTHING WRONG with these suspensions. They are common, and well behaved. But your computer won't be able to do the calculations for "roll moments due to the forces at the RC" because the numbers are tooooo big!

BTW, the above bold print is a reasonable definition of "the RC".
~~~o0o~~~

"6. Concepts given in “Suspension Geometry And Computation” by J.C.Dixon"

I have Dixon's book "Tires, Suspension and Handling.", but nothing with the above name. Is that a paper?

I have fewer notes in the margins of Dixon's book that in most others (ie. it is "less wrong" ). IMO, Chapter 5 "Suspension Characteristics" is better than most other books on this subject.
~~~o0o~~~

"Steps-

1. Draw vector Fg + Fi=R from COG. Here, Fg=mg i.e. weight and Fi is the inertial force.

2. Assume/get the coefficient of friction of one tire."

There are many ways of taking this step (2), some of which I covered at the top of page 3. Briefly, for equilibrium the wheelprint forces Fl and Fr MUST intersect somewhere along Fg+i's LoA (your "R"). So you can pick, or assume, ANY point along this LOA, and then all results are found.

A good first approximation is to assume that both tyres have the same CoF, so Fl and Fr are parallel to each other, and also to Fg+i (ie. they all intersect "at infinity"). Note, however, that different toe angles, say from "Ackermann" at the front, can greatly affect whether the intersection point is above or below the car.
~~~o0o~~~

"8. Now, resolve Fl and Fr along and perpendicular to n-line-fig(b)

9. Let Fl.ca be the force along n-line and Fl.sd be the force perpendicular to n-line."

Having the control-arm (F.ca) and "virtual" spring-damper (F.sd) forces perpendicular is how Dixon does it, and is perfectly acceptable.

I prefer to have the F.sd force always vertical (wrt body) so that it represents the "virtual wheel-spring-damper-rate" (ie. one that is always vertically above the wheelprint).

IMPORTANT NOTE. When forces are "decomposed", the "components" do NOT have to be perpendicular. They can be at any angle that suits you. That is why the Parallelogram Rule is not called the Rectangle Rule.

For a given total wheelprint force, say Fl, changing the angle of Fl.sd will change the magnitude of Fl.ca. Likewise, changing the angle of a real spring-damper (or pushrod) will change the loads on real control arms. However, for the vertical wheel-rate to remain as before, the newly positioned spring-damper must also have a different rate spring (stiffer or softer). If this is done, then the body roll in corners will be as before, even though the control-arms and spring-damper are differently stressed. This better understood by working through numerous examples...
~~~o0o~~~

"10. Now, consider Fl.ca and Fr.ca. Resolve them into horizontal and vertical component namely Fl.ca.y and Fl.ca.j similarly with right side-fig(d)

11. Net jacking force=Fl.ca.j-Fr.ca."

Strictly speaking, net jacking force = Fl.ca.j + Fr.ca.j (Fr.ca.j has negative magnitude in the sketch).

As I said before, VW Beetles can have their tails jump upwards so much from these forces that both rear wheels leave the ground.
~~~o0o~~~

"CONCLUSION
...
2. Roll center migration should not be a headache because unless the jacking force is undercontrol (if you want so) the roll center migration is irrelevant.

3. The more the inclination of n-line, the more will be the jacking force. So migration of n-lineshould be controlled."

Yes. I suggest you aim for n-line slopes close to horizontal (say, less than 10 degrees up or down), and not suddenly changing their angle, wrt body, with small movements of the suspension.

Another way of saying this same thing, is that the centre-of-wheelprint's "path of motion" should be a straightish line (gentle curve is acceptable) that is close to vertical.
~~~o0o~~~

"DOUBTS
1. In this concept, what constitues a roll couple? Which force acts along with the Centrifugal force and what is the moment arm in this case?"

All force diagrams (that use D'Alembert's Principle) ultimately result in zero force (ie. equilibrium)! Along the way you can arrange the forces in many ways, some of which represent "roll couples" acting on the "sprung mass". One way to do this was touched on above. Namely, slide both Fl.ca and Fr.ca to the RC. Next decompose the resultant of these two into horizontal (Fca.h = Fl.ca.y+Fr.ca.y), and vertical (Fca.v = Fl.ca.j+Fr.ca.j) components.

The couple formed by Fca.h, at RC height, and the centrifugal force Fi, at CG height, typically acts to roll the body outwards during cornering. This couple is resisted by (equilibriated by) a couple formed from changes to the vertical spring forces, namely delta.Fl.sd and delta.Fr.sd. This explains why cars with high RCs, and thus lesser CG-RC "moment arm", roll less than ones with low RCs, all else equal.

The couple formed by the jacking forces Fca.v, at the RC, and a component of the weight force Fg, at the CG (assuming that the RC is not directly under CG), is typically ignored by many people, such as Danny! Including this couple will correct Danny's earlier calculations with RCs, and give the same results as his FAPs. (Edit: Note that this "correction couple" is small when the RC is near the CG, but it is astronomically large when the RC is out near Pluto...) The fact that this couple includes a part of Fg means that Fg is now "diminished", so lesser spring forces are required to carry it. So the car "jacks up", which is also all too often ignored.
~~~o0o~~~

"5. Why in Tune To Win, Carroll Smith has entirely a different approach?"

The section on jacking (page 38) and Figure 18 (p39) is reasonably accurate IMO, and similar to what I have been saying here. (The Figure has some small errors, and could do with more arrows and explanation.) However, some other parts of the book might be taken with a large grain of salt...

Carroll, a largely self-taught race engineer, includes this statement in the Preface (and in most of his other books).

"I am fully aware that much of what I have to say is subjective...
I wish that my knowledge and wisdom were such that this were not so....
I will put forth my personal best shot on the subject at the time of writing.
I reserve my right to change my thinking at any time."

And, indeed, Carroll often did change his thinking when he came upon better explanations.
~~~o0o~~~

"6. So many concepts, which one to follow? Or more specifically, which one is more closer to reality?"

I stress again that the explanation I gave above (post with sketch) is a greatly simplified version of "reality". (Even in 2-D it doesn't include "roll acceleration". This adds an "inertial couple" to the sketch, and changes all the forces quite a bit. Though also easy enough to understand...)

However, the concepts in Euclid's Elements, Newton's Principia, and all the rest of Classical Mechanics, is a good enough approximation of reality to have taken man to the Moon, and beyond! With that track record, and up against the alternatives from the automotive/motorsport cottage industries, I know which one I choose.

Z

[exFSAE]

Discussing all these nice things about the suspension geometry without considering tire force distribution (through a tire model) and overturning moments = quite distant from reality yet.

[Sid@TR]

Originally posted by Z:
.................. With that track record, and up against the alternatives from the automotive/motorsport cottage industries, I know which one I choose.

Z

Z,
Those were some really helpful explanations..
I will try to justify whatever option i will choose....

[Flight909]

exFSAE I was asking about this earlier and Z answer.

I was thinking more on this, I guess I could apply the same thinking for longitudinal tyre forces. If I have a Fx and Fy from the tyre. Can I then use this two components and use the front view n-lines and side view n-lines? Or should I try to find an instant axis (screw axis, yes?) in 3D and project the forces on their? I think I should get the same result, yes?

[Z]

Flight909,

"I was thinking more on this, I guess I could apply the same thinking for longitudinal tyre forces. If I have a Fx and Fy from the tyre. Can I then use this two components and use the front view n-lines and side view n-lines? Or should I try to find an instant axis (screw axis, yes?) in 3D and project the forces on their? I think I should get the same result, yes?'

Yes. The "Screw Axis", and the "Line Complex of Normal-Lines" that surrounds it, are two different ways of looking at the same concept.

Briefly, since any motion of an idealised, rigid body can always be described as a screwing motion about some axis, it follows that all points of the body have, "at the instant of motion", displacement vectors that are tangent to helixes centred on the screw axis. Thus at each point of the body there is also a "planar pencil" of n-lines that is perpendicular to the instantaneous displacement vector.

(A "planar pencil" is all the lines lying in a flat plane and passing through a given point. So the n-lines are like the spokes of a wagon wheel, with the displacement vector being the axle. Easier shown with a small sketch... Also there are "three infinities" of n-lines in the "line complex" surrounding the screw. This is a lot of lines, but much less than the "four infinities" of lines that fill all of 3-D space.)

If the screw axis has zero pitch (like a revolute, or simple hinge, joint), then the helixes become circles centred on the axis and ALL the n-lines intersect the axis somewhere. However, whenever the screw has some finite pitch (ie. the more common case), then ALMOST ALL of the n-lines DO NOT intersect the axis. Again, easier seen with a simple sketch.

For purposes of your longitudinal and lateral "jacking" calculations you only have to determine the instantaneous displacement vector of the wheelprint at each position of the suspension, which some software programs do. The longitudinal and lateral n-lines are simply perpendicular to this vector, in the X and Y directions. The control-arms then carry a force from road to body that is the sum of Fx, Fx.jacking (= Fx.tan(long-n-line-slope)), Fy, and Fy.jacking (=Fy.tan(lat-n-line-slope)). This resultant control-arm force vector acts on a LOA through the wheelprint.

Also acting on the body are the spring-damper forces (which depend on instantaneous position, for spring, and velocity, for damper), and the gravity force. These can then be summed to zero (equilibriated) by inertial force Fi and couple Ti acting on the body (= D'Alembert's Principle). Or you can say that the unbalanced forces produce linear and rotational accelerations of the body.

Note again that the above does NOT take account of the wheel-assembly (unsprung) forces, such as those from linear accelerations or gyroscopic effects. To account for these forces (~10-20% mass of car, so quite big...) you have to consider more n-lines than just those at the wheelprint. And Euler's Equations for Rigid Bodies... The method shown on this thread might be a good place to start (since each suspension link is an n-line). You would still have to add the gyroscopic couples acting on the wheel assembly to the RHS, etc....

Z

[Z]

Recent discussions on this "Anti-dive..." thread cover similar material to this thread, so I link it here for anyone interested in further reading. The rest of this post could equally have gone on the "Anti-dive..." thread, but I put it here because it better relates to earlier posts on this thread.
~~~~~o0o~~~~~

The following text and figure are from a Racecar Engineering article from a few years ago. (The version below is how the article was intended to appear, prior to the RCE sub-editor butchering it.) The point is to show how unnecessary it is to worry about "migrating Roll Centres".

THE ROLL CENTRE - We now briefly consider one last point along the n-line to which we can slide the control-arm force Fca. This is the intersection point of the two lateral n-lines of a pair of wheels in end-view, which is traditionally called the "Roll Centre". Our key argument is that knowing the Roll Centre position for a pair of wheels is of no more help than knowing their n-line slopes.

The usual process for finding the Roll Centre is to first find the wheelprint positions, and then find their n-lines. The next step of finding the intersection of the n-lines can be counterproductive. The problem is that focussing on the Roll Centre's absolute position can be misleading. Many empirical rules have developed regarding this position, such as that it should not be too high because this will lead to too much jacking. Or that it should stay roughly mid-way between the wheels, and woe betide anyone who lets it wander too far sideways. Is this really so?

Figure 13 shows an end-view of a car during spirited cornering. The suspension has very soft springs and no anti-roll bars, giving the kinematics ample opportunity for mischief. The suspension is of the trailing-arm type, widely used on production cars for more than 50 years. Although trailing-arms have some minor disadvantages, they have proved to be well behaved with few quirks. The lateral n-lines are nominally horizontal with respect to the car, and stay that way throughout the range of travel.

The cornering forces produce a body roll of about 6 degrees (=1:10 = ~0.1 radians). Due to manufacturing tolerances the right n-line is very slightly steeper than the left. This puts the intersection of the n-lines, namely the Roll Centre, at an altitude of 16,000 kilometres, well and truly into space! The lateral location of the Roll Centre is at 160,000 kilometres, half way to the Moon! Surely this can't be good, can it?

We slide the two wheels' control-arm forces to the Roll Centre, add them, and consider the moments of the horizontal and vertical force components about the car's CG. As expected, the horizontal force component (Fca.h = 4 kN) creates an astronomical clockwise moment of 63,999,998.24 kNm. The vertical component (Fca.v = 0.4 kN) creates an even larger anti-clockwise moment of 64,000,000.48 kNm, after rounding. We take a stiff drink to steady ourselves.



We add these two moments, by hand, because the calculator keeps overflowing. We get a net moment of very slightly more than 2.24 kNm. This is equal to the horizontal wheelprint forces (equal and opposite to the centrifugal force Fi = 4 kN) multiplied by a little less than CG height (= 0.6 m). So the body rolls as if it had a perfectly normal "Roll Centre" a little above ground level.

The point we are trying to make is that Roll Centres frequently zoom off into space, often out past Pluto, but that is nothing to worry about. There is nothing wrong with the Roll Centre as it is shown in Figure 13, nor with sliding the control-arm forces there.

The problems only come, firstly, if we assume that Roll Centre position is an important indicator of performance, which it clearly is not. And secondly, if we perhaps only move the horizontal components of the wheelprint forces to the Roll Centre, and then forget the vertical components behind under the car. The slopes of the n-lines are important, not their intersection point.

[See note below*] The reader might try finishing the calculations of body Heave and Roll, as started in Figure 13. The calculations can be done in the ground reference frame, or the car body frame, as shown in the sketch. For small roll angles there is not much difference.

So far we have avoided any mention of the detailed kinematics of the suspension. By considering only the mass of the car's body we, quite correctly, have had only to find the n-lines of the wheelprints. So in two-dimensional kinematic terms, we haven't bothered looking for the instant centres of the suspension. This is in contrast to statements often made in suspension design textbooks, such as, "for anti-dive under braking the side-view instant centre must be above ground and behind the front wheel". This is a half-truth, and hence a half-falsehood, because the instant centre can be anywhere along the n-line, including below ground and in front of the wheel, for the same anti-dive.

But so far we haven't considered the mass of the wheel assembly. [... covered in the rest of the article ...]
~~~o0o~~~

[Note*] I have edited Figure 13 to show all the Heave and Roll calcs, and to only show the spring and control-arm forces in the ground-frame (it possibly had too much information before). Note that doing the calcs as in the figure (in ground-frame) is slightly inaccurate because it assumes the spring forces are vertical wrt ground-frame, rather than vertical wrt car-floor. The error here, even at these quite high roll angles, is only a few percent. The more accurate approach (done in the body-frame) is shown in Figure 10, bottom of page 4 of this thread.

Getting back to Migrating RCs, it is worth thinking about what would happen if the right wheel's n-line above had a slope just a teeny-weeny bit LESS than 1:10. Now the RC might be 16,000 kms BELOW ground, and 160,000 kms to the LEFT of the car.

Would that make any difference to the car's dynamic behaviour? Nope ... not even diddley-squat difference!

This should be apparent from going through similar calculations of moments-about-the-CG as above. At a more practical level, it would be bizzarre in the extreme if a tiny manufacturing tolerance of the angle of a Trailing-Arm's pivot-axis wrt car-floor, could make an astronomical difference to the car's cornering behaviour.

Or looking at it another way, both wheelprint's control-arm forces are close to ground level under the car, and both are almost horizontal and only sloping up-to-right by a few degrees. So we should expect that the VECTOR SUM of these two forces should also be close to ground level under the car, and also only sloping up-to-right by a few degrees. Now, whether we take this SINGLE summed force and slide it along its Line-of-Action until it is a million miles to the left of the car, or a million miles to the right of the car, makes NO DIFFERENCE to this force's affect on the car, whatsoever. This is one of the fundamental Axioms of Statics!

And yet another way of looking at Migrating RCs. Based on the above, and providing the n-line slopes are close to horizontal, we should also expect that a car that has its front-RC a million miles to the left of the car, and its rear-RC a million miles to the right, will behave very similarly to a car with similar n-lines slopes and both RCs close to the car centreline.

So, once again, the RC position, especially when ONLY its height, OR its lateral position is mentioned, is not a good indicator of the car's potential behaviour, as it can be very misleading. N-line slopes are a much better indicator.

Z

[Tim.Wright]

I have had a read through the entire thread since I missed it the first time.

One of the points disagreement between Danny and Erik was where the jacking forces are actually applied. Since my own investigation of rollcentres/jacking etc in the last few years, I have visualised the jacking force as something which occurs in parallel to the spring and damper of the suspension rather than something which is applied directly at the contact patch.

Sure, the jacking force does make its way down to the contact patch but in my opinion its more correct to think of it as originating in parallel to the spring and damper. The reason for this is if you increase the jacking force (by increasing the n-line slope) then you decrease the amount of force going through the spring. It is somewhat analgous to a situation of 2 springs in parallel. If you increase the stiffness of one spring, then it will carry a larger share of the force and the other spring will see less.



The other point of contention was why are the FAP's used by Danny and Mitchell sitting directly under the CG. My understanding is that if you shift the total contact patch force (vert and lat components) to this FAP, then only the lateral component is creating a moment about the CG. This moment is equal to the lateral contact patch force multiplied by the lever arm from the FAP to the CG.

Then the assumption is made that:

• Any moments about the CG are reacted by the suspension elastic elements
• Any forces at the CG are reacted by the links

So then this split of forces/moments gives you the split of geometric/elastic load transfer. I think this is a fair approximation. It also answers Eriks question of why Danny has not put the vertical forces in his analysis. However, I think it would have been more complete to add them in but show how they have a zero moment arm about the CG at the FAP.

[Z]

Tim,

One of the points disagreement between Danny and Erik was where the jacking forces are actually applied. Since my own investigation of rollcentres/jacking etc in the last few years, I have visualised the jacking force as something which occurs in parallel to the spring and damper of the suspension rather than something which is applied directly at the contact patch.

Sure, the jacking force does make its way down to the contact patch but in my opinion its more correct to think of it as originating in parallel to the spring and damper. The reason for this is if you increase the jacking force (by increasing the n-line slope) then you decrease the amount of force going through the spring.

That is pretty much they way I have done it in Figure 13 above (ie. very similar to your sketch). Looking at the car in Figure 13, the three white-arrows are the major forces acting on the FBD of the whole car (ie. left-wheel total force Fl, right-wheel total force Fr, and CG total force Fg+i, all in equilibrium). The left-wheel force Fl is also shown as its black-arrow horizontal (3 kN) and vertical (4.5 kN) components, in a RECTANGULAR parallelogram-of-forces.

At bottom left of the figure is the same left-wheel total force Fl, but this time it is decomposed, in a SKEW PARALLELOGRAM, into a sloping control-arm force Fl.ca, and a NECESSARILY SHORTER vertical component representintg the spring-damper force Fl.s. Naturally, the steeper the n-line, then the more skew the parallelogram, and the shorter the vertical Fspring-damper component.

So in Figure 13 the "calculation" of the spring-damper force is done by the simple act of drawing that skew parallelogram at the bottom left. This then gives the deflection of the left spring (from delta-Fl.s = 4.2 - 3 kN = 1.2 kN, and so Deflection = 1.2/17.5 = 0.068 m). A similar calculation at the right-wheel gives its deflection, and then "Roll and Heave" deflections are also easily found.

Note that the vertical component of the control-arm force Fl.ca (= 3.015 kN), namely its "jacking" component, is easily seen to be 0.3 kN, and this plus the Fl.spring force of 4.2 kN equals the original total vertical force at the left-wheel. And also note that any component of force "FROM-road-TO-car-body" has an equal and opposite force "FROM-wheelprint-TO-road".

I repost Figure 10 here (was on page 4) because it helps show, I hope, how the "SAME total wheelprint force" (= "Fres[ultant]") can be decomposed in many different ways to suit the goal of your calculations.


~~~~~o0o~~~~~

The other point of contention was why are the FAP's used by Danny and Mitchell sitting directly under the CG. My understanding is that if you shift the total contact patch force (vert and lat components) to this FAP, then only the lateral component is creating a moment about the CG. This moment is equal to the lateral contact patch force multiplied by the lever arm from the FAP to the CG.

Then the assumption is made that:

Any moments about the CG are reacted by the suspension elastic elements
Any forces at the CG are reacted by the links

So then this split of forces/moments gives you the split of geometric/elastic load transfer. I think this is a fair approximation. It also answers Eriks question of why Danny has not put the vertical forces in his analysis. However, I think it would have been more complete to add them in but show how they have a zero moment arm about the CG at the FAP.

I stress again that I am happy to slide ANY force ANYWHERE along its Line-of-Action, as long as ALL OF THE FORCE is taken there. My problem with Danny's, Mitchell's, and also Ortiz's, various explanations, is that they only show the horizontal component of the control-arm force at its new FAP location. What has happened to the control-arm's vertical (= "jacking") component?

Is it simply forgotten? Has it somehow been hidden in the Sigma-Fz equations. Is its "jacking" effect on the Heave motion of the car calculated via these Fz equations? Will we ever know? And can we trust these people's calculations (and computer programs) if they keep this simple matter a secret? (I do not suggest that this secrecy is deliberate. Rather, I wonder if they actually know where the "jacking force" has gone?)

As an example of "sliding forces about", if, in Figure 13 above, we slide the two control-arm forces to under the CG, then:
Left-wheel: Fl.ca is now decomposed at a "FAP" 8 cm ABOVE ground, with Fl.ca.h = 3 kN, and Fl.ca.v = 0.3 kN.
Fl.ca.h creates an ACW moment at the CG = 3 x (0.6 - 0.08) = 1.56 kNm.
Right-wheel: Fr.ca is now decomposed at a "FAP" ~8 cm BELOW ground, with Fr.ca.h = 1 kN, and Fr.ca.v = 0.1 kN.
Fr.ca.h creates an ACW moment at the CG = 1 x (0.6 + ~0.08) = ~0.68 kNm.

The sum of the above two moments is 1.56 + ~0.68 = ~2.24 kNm, which is just the same as when it was calculated at the RC "half way to the Moon", so the amount of Roll is as calculated before.

The sum of the two jacking components is 0.3 + 0.1 = 0.4 kN, which acts directly towards the CG and gives a Heave deflection = 0.4/(2 x 17.5) = ~0.012 m, which is, again, as before.
~~~~~o0o~~~~~

As another example, students might try sliding the two control-arm forces (Fl.ca and Fr.ca) along their LoAs, towards the right, until they are at the same HEIGHT as the CG (ie. to "FAPs" at 5.2 m and ~6.8 m to RIGHT of CG).

What are the moments-at-the-CG from the horizontal and vertical (= jacking) components of these forces now?
What are the resulting body Roll and Heave deflections?
Is there anything wrong with using these points as the magical FAPs??? (<- Answer to this one, as before, is "NO, no problems at all!")
~~~~~o0o~~~~~

To sum up, I think "the problem of the missing jacking forces" is the result of spending too little time drawing FBDs, and instead diving too quickly into the equations, such as the Sigma-Fy/Fz/Mx's = 0. These equations are an alphabet soup of a,b,c's and x,y,z's, and important physical concepts can easily get lost in there. The equations are quite literally "cogitatio caeca" = "thinking blind". However, geometric FBDs are a much closer representation of the physical reality, so, IMO at least, it is a lot harder to "lose" forces in them.

Z

[Z]

(Ben, I just replied to your PM, but your "Inbox is full" message came up (I had the same problem .. and lost all my PMs... ). I will try again tomorrow...

Z)

[Tim.Wright]

I have spent a lot of time this morning thinking about the decomposition of the forces in the contact patch...

That is pretty much they way I have done it in Figure 13 above (ie. very similar to your sketch). Looking at the car in Figure 13, the three white-arrows are the major forces acting on the FBD of the whole car (ie. left-wheel total force Fl, right-wheel total force Fr, and CG total force Fg+i, all in equilibrium). The left-wheel force Fl is also shown as its black-arrow horizontal (3 kN) and vertical (4.5 kN) components, in a RECTANGULAR parallelogram-of-forces.

At bottom left of the figure is the same left-wheel total force Fl, but this time it is decomposed, in a SKEW PARALLELOGRAM, into a sloping control-arm force Fl.ca, and a NECESSARILY SHORTER vertical component representintg the spring-damper force Fl.s. Naturally, the steeper the n-line, then the more skew the parallelogram, and the shorter the vertical Fspring-damper component.

I think this is (almost) valid to decompose the contact patch forces into a skew parallelogram (or n-z coords from my point of view) to get the correct vertical spring force. You have shown that this decomposition reduces the spring force as the n line slope increases. Additionally, wheel rates are given (from a rig or simulation) as vertical force per vertical wheel centre travel. So if you want to calculate the deflection using the wheel rates, you need to know the vertical component of the contact patch force (or more correct the body angle corrected vertical as you showed in Fig 10)

BUT...

I am a bit suspicious of using the skew parallelogram to calculate the control arm force. Yes it's a valid way of representing the resultant contact patch force, but you get a control arm force smaller than it should be. I have taken your same problem and decomposed the LH wheel force into n-t coords which is a recangular parallogram. I get a control arm force of 3.434kN (as opposed to your 3.015kN).

To represent it graphically, below on the right is how you have decomposed the contact patch force into n-z (normal,vertical) coords. On the left is how I did it using n-t (normal,tangent) coords.


The reason that your control arm force is smaller is because your Z (vertical) component contains a component acting along the n-axis. I think this is significant because in this example alone there is a 13% difference.