The Mark Ortiz Automotive

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July 2009

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Mark Ortiz Automotive is a chassis consulting service primarily serving oval track and road racers. This newsletter is a free service intended to benefit racers and enthusiasts by offering useful insights into chassis engineering and answers to questions.  Readers may mail questions to: 155 Wankel Dr., Kannapolis, NC 28083-8200; submit questions by phone at 704-933-8876; or submit questions by    e-mail to: markortizauto@windstream.net .  Note that this is a new e-mail address.  Readers are invited to subscribe to this newsletter by e-mail.  Just e-mail me and request to be added to the list.

THE UNSPRUNG COMPONENT IN LOAD TRANSFER

What is the correct way to treat the unsprung components when analyzing or modeling cornering?  It is customary to treat the unsprung masses as not acting through the springs or the linkages.  But is that really true when the suspension is independent and the wheels lean with the car?  What happens when there is some camber recovery in roll?  What is the influence of roll center location on this?  What is the correct way to treat the unsprung components for braking and acceleration?

This is a subject that I first received some correspondence about several years ago, and have been puzzling about ever since.  I finally feel ready to present some conclusions.

First, a brief introduction to the subject under discussion.  Dynamic load transfer (weight transfer) in response to ground-plane forces (forward, rearward, lateral) at the tire contact patches breaks down into four basic categories:

· Elastic load transfer: load transfer through the springs, anti-roll bars and any other springing devices

· Geometric load transfer: load transfer through support forces within the suspension, induced by linkage geometry

· Frictional load transfer: load transfer resulting from damper forces and friction in suspension pivots

· Unsprung load transfer: load transfer resulting from the overturning moments created by the inertia of the unsprung components, which does not act through the suspension.

In addition to these, there is also load transfer with steer, which results from caster jacking, and load transfer due to z or road-vertical forces, including vehicle accelerations from crests, dips, and turn banking, and aerodynamic lift/downforce.  Load transfer with steer depends on steering geometry and front and rear angular warp stiffness and track widths.  Load transfer with z force depends on relative wheel rates in roll, pitch, and heave, and on mass locations and center of pressure location relative to the contact patches.  Neither of these depends in any direct way on suspension geometry.

Finally, with live axles, we have load transfer due to driveshaft torque.

To simplify discussion of the principles at issue, we will be ignoring effects of tire deflection, even though in some cases these can be significant.

What concerns us here is unsprung load transfer, and its relation to elastic load transfer.

To understand what unsprung load transfer is, imagine an axle with two wheels rolling down the road on its own, with no car sitting on top of it.  The assembly has mass, centered at about hub or wheel center height.  For it to go around a corner, the tires must exert a lateral or centripetal force, which they will do at ground level.  The mass of the assembly will react with an equal and opposite

inertial or centrifugal force, acting at hub height.  The resulting couple will try to turn the assembly over, toward the outside of the turn.  The tires will keep it from overturning, by exerting an increased force against the ground on the outside tire, and a correspondingly decreased force on the inside tire.  This increase or equal decrease is the unsprung load transfer.

The axle does this just the same when it does have a car sitting on top of it.

It has been accepted practice to treat the unsprung masses in an independent suspension as if they were equivalent to an axle.  However, many are now realizing that this isn't really correct, and that different parts of the unsprung masses can act as sprung masses in roll, and therefore a good analysis or model should take this into account.

There are also some interesting subleties to deciding whether some parts of the suspension are sprung or unsprung at all for a given type of analysis, which we will explore.  We will also consider how this aspect relates to common methods of measuring unsprung weight in the garage or on a test rig.

Since we will be needing some equations, it is time to define some variables.  Readers will recognize many of these from last issue.  I am repeating those here, partly for the reader's convenience, and partly for the benefit of new subscribers.

There are also some additions, appropriate to our present discussion.

First, the vehicle axis system:

Per SAE convention, x is longitudinal, y is transverse, and z is normal or vertical.  Unless otherwise stated, sign conventions are: x positive forward; y positive rightward; z positive downward.  Unless otherwise stated, vehicle origin is per SAE aerodynamic axis system: on the ground plane, midway along the wheelbase, and centered with respect to the front and rear tracks.  However, for particular purposes, we may use local origins and sign conventions may deviate, so it is necessary to pay

attention to context.  Even then, x will always be longitudinal, y transverse, and z vertical, in some sense that is appropriate to the context.

SAE vehicle axis convention takes as origin a point on the roll axis, directly below the sprung mass center of mass or c.g.  I don't care for that, because the points in question move and are subject to uncertainties.  Actually, there is no origin that is completely immune to some uncertainties and relative movements, but for our purposes here, we will use the SAE aerodynamic convention.

x, y, and z may denote coordinates in this axis system, or linear displacements in this axis system.

Next, angular quantities about the axes:

φ (phi) is roll, or angular movement about an x axis, conventionally positive rightward, or clockwise as seen looking forward.

M_x or M_φ is a moment in roll, or about an x axis.

K_φ is the elastic angular roll resistance rate.  Depending on context, this may be for only a front or rear wheel pair, or for the entire vehicle.

It is also customary to use φ to denote camber of a wheel.  This can create some confusion.  Where we need to use this convention, φ_t is camber, and φ with no subscript is roll, unless otherwise indicated.

γ (gamma) is tire inclination.  This is the same as camber except for sign convention.  Camber is positive when the top of the tire tilts outboard with respect to the car.  Inclination is positive rightward, or clockwise looking forward, for either wheel.  By this convention, for a right wheel       γ = φ_t, and for a left wheel γ =  – φ_t.  In less formal, more conversational usage, inclination may also be understood to be positive into the turn.  Note also that both camber and inclination refer to the wheel's angle with respect to the road or ground plane or a perpendicular to it, not the body or sprung mass.

θ (theta) is pitch, or angular movement about a y axis, conventionally positive rearward, or clockwise when looking from left to right.

M_y or M_θ is a moment in pitch, or about a y axis.

K_θ is the elastic angular pitch resistance rate.  It is analogous to K_φ, only in the pitch plane.  Accordingly, it may refer to the rate for the whole vehicle, or for a right or left wheel pair, according to context.

θ_u is angular position or displacement of an upright or axle housing, e.g. caster or caster change.  This is somewhat analogous to φ_t, only in side view.  As with φ_t or γ, this is an angle or displacement with respect to the ground plane or a perpendicular to it, not the body or sprung mass.

ψ (psi) is yaw, or angular movement about a z axis, conventionally positive rightward, or clockwise when looking down.

M_z or M_ψ is a moment in yaw, or about a z axis.

L is a length.  In our discussion here, it particularly refers to center spacing between two wheels of a pair under consideration.

L_x is wheelbase.

L_y is track.

r_t is tire loaded radius.  For simplicity, we will ignore distinctions between loaded and effective tire radius.

F is force.  F with an x, y, or z subscript is force in an x, y, or z direction.

m is mass.

a is acceleration.

F = ma.  When m is expressed in pounds, a should be in g's.

Or, to be a bit more scientifically correct, weight W (pounds) is the force mg exerted on a mass m by a gravitational field in which a free-falling body accelerates at g (ft/sec²).  An acceleration in g's is a in ft/sec² divided by the number of ft/sec² per g.  When we substitute mg, or W, for m, then F = ma becomes F = (mg)(a/g) = Wa/g if a is in ft/sec², or F = Wa if a is in g's.

Users of the metric system may be snickering at the use of pounds weight for mass, but weighing things in pounds, and getting accelerations from data acquisition systems in g's is the norm in English-unit vehicle engineering.  But is weighing things in kilograms any more rational?  Using kilograms for mass is correct, and a can then be in M/sec².  However, one could similarly quibble that objects should really be weighed in newtons, since weight is force, not mass.  a should then be in g's, or the weight properly taken in newtons should be divided by gravitational acceleration to obtain mass.

Is this confusing enough yet?  Is it of only academic interest?  I guarantee it will become hugely relevant, just as soon as we start racing on other planets.

Until then, it can at least be simple to apply F = ma for practical engineering, if we use these rules:

If m is in pounds, a has to be in g's.  F will be in pounds.

If m is in slugs (pounds divided by 32.2), a has to be in ft/sec² (g's times 32.2).  F will be in pounds.

If m is in kilograms, a has to be in M/sec² (g's times 9.8)  F will be in newtons.

If m is in newtons (kg times 9.8), a has to be in g's.  F will be in newtons.

Don't use pounds directly with ft/sec², or kilograms directly with g's – unless you want F in kgf.

m_S is sprung mass.

m_U is an unsprung mass.

H is height.

H_cg is height of the center of mass or center of gravity (c.g.).  Unless otherwise indicated, this means the overall c.g. for the whole car.

H_cgS is the height of the sprung mass c.g.

H_cgU is height of an unsprung mass c.g.  This is typically approximated as being equal to r_t, absent better information.  Since we have more than one unsprung mass, further subscripts are applied per below.

H_rc is roll center height.

H_pc is pitch center height.

Subscripts applied to various quantities are as follows:

_L is left.

_R is right.

_r is rear.

_f is front.

For example:

m_Sr is the portion of the sprung mass statically supported by the rear wheels.

m_SR is the portion of the sprung mass statically supported by the right wheels.

dx/dz is the instantaneous rate of change of a point's x coordinate with respect to change in its z coordinate: the first derivative of x displacement with respect to z displacement.

dy/dz, similarly, is the first derivative of y displacement with respect to z displacement.

dx/dz and dy/dz are also the instantaneous slopes or inclinations, from vertical, of the contact patch center's path of motion in side and front view respectively.

A force line is a notional line of action for the vector sum of an x or y ground-plane force at a tire contact patch and the induced  z-direction support force within the suspension system that results from angularities in the suspension linkage.  In front-view geometry, the force line is the line from the contact patch center to the front-view instant center.  The front-view force line is an instantaneous perpendicular to the contact patch center's path of motion in front view, and has a slope or dz/dy, relative to ground plane, equal to the contact patch center's dy/dz.  The side view force line is an analogous construction in side view, whose slope is the contact patch center's dx/dz.

dγ/dφ is the rate of tire inclination change with respect to angular roll.  If dγ/dφ = 1, the wheels lean the same amount as the sprung mass, as with parallel control arms or pure trailing arms.  If dγ/dφ = 0, the wheels don't lean at all with roll, as with a beam axle.  We can say that (1 – dγ/dφ)*100% is our percent camber recovery, or that dγ/dφ is the camber non-recovery, expressed as a decimal.  (I suppose if we really want to be fastidious, we should speak of inclination recovery –  only that isn't common usage.)

Analogously, dθ_u/dθ is the rate of caster or side-view inclination change of an upright or axle, with respect to angular pitch.  If dθ_u/dθ = 0, that means that θ_U does not change in pitch.  This requires a side-view swing arm length x_svsa equal to half the wheelbase.  It also means that dθ_u/dz = – dθ/dz.  That is, the rate of caster change with respect to linear suspension displacement in the absence of pitch is equal in magnitude and opposite in direction to the rate of pitch displacement with respect to magnitude of linear displacement per wheel.  If dθ_u/dθ = 1, that means that θ_u changes in pitch identically with sprung mass angular pitch.  This requires parallel side view projected control arms, an x_svsa that is infinite or undefined, and dθ_u/dz = 0 in the absence of pitch.  dθ_u/dθ can be considered to be the rate of caster non-recovery, expressed as a decimal.

r_cgSx is the moment arm of the sprung mass about the roll axis.  That is, it is the side-view perpendicular distance from the sprung mass c.g. to the roll axis.

r_cgSy, correspondingly, is the moment arm of the sprung mass c.g. about the pitch axis.  That is, it is the front-view perpendicular distance from the sprung mass c.g. to the pitch axis.

Mathematically, we can compute r_cgSx and r_cgSy as follows:

                       r_cgSx  = H_cgS – (H_rcf + (H_rcr – H_rcf)(m_Sr/m_S)) (√(L_x² + (H_rcr –  H_rcf)²)) / L_x               (1a)

                    r_cgSy  = H_cgS – (H_pcR + (H_pcL – H_pcR)(m_SL/m_S)) (√(L_y² + (H_pcL –  H_pcR)²)) / L_y           (1b)

Equation (1b) applies for a case where front and rear track are equal.  Where we have unequal track front and rear L_yf and L_yr, for best accuracy we need to substitute for L_y a weighted average of L_yf and L_yr, which will be (L_yf – (L_yf – L_yr)(m_Sr/m_S)).  A similar correction could even be required for unequal right and left wheelbases, but ordinarily the wheelbases are identical, or differ by so little that a simple average will suffice.

These equations can be simplified in certain cases.  If the front and rear roll centers are similar height, there is little difference between the perpendicular distance from sprung mass c.g. to roll axis and the vertical distance.  Similar reasoning applies regarding the moment arm in pitch, if the pitch

centers are similar heights at the right and left.  For a simple case where front and rear track are equal, the sprung mass c.g. is laterally centered, and the roll and pitch axes have little slope, the equations simplify to:

                                               r_cgSx  = H_cgS – (H_rcf + (H_rcr – H_rcf)(m_Sr/m_S))                                        (1c)

                                                          r_cgSy  = H_cgS – (H_pcL + H_pcR)/2                                                  (1d)

We use r_cgSx to calculate the overall elastic roll moment M_φE – the moment resisted by the springs, anti-roll bars, and any other springing devices that resist roll, for a given lateral acceleration a_y.

                                                                     M_φE = m_S a_y r_cgSx                                                           (2)

Knowing the elastic angular roll resistance rates for front and rear wheel pairs K_φf and K_φr, we then calculate the angular roll displacement φ.

                                                                 φ = M_φE / (K_φf + K_φr)                                                        (3)

We now know the angular roll displacement, and we know the angular rate at which the front and rear suspensions resist roll, so we can calculate how much elastic roll resisting moment we have at each end of the car.

                                                                       M_φEf = φ K_φf                                                               (4a)

                                                                       M_φEr = φ K_φr                                                               (4b)

Finally, knowing the front and rear track widths L_yf and L_yr, we can calculate the front and rear elastic load transfers ΔF_zEf and ΔF_zEr.  (Note that ΔF_z is the load change at one wheel.  The resulting change in load difference between the two wheels is 2ΔF_z.)

                                                                   ΔF_zEf = M_φEf / L_yf                                                           (5a)

                                                                   ΔF_zEr = M_φEr / L_yr                                                           (5b)

Next, we calculate the geometric load transfer at each end, ΔF_zGf and ΔF_zGr.

                                                            ΔF_zGf = (m_Sf a_y H_rcf) / L_yf                                                      (6a)

                                                            ΔF_zGr = (m_Sr a_y H_rcr) / L_yr                                                      (6b)

Frictional load transfer is difficult to calculate with good accuracy.  Ordinarily, suspension bearings are treated as frictionless – not because we really think they are, but because we cannot reliably calculate their frictional loads at a given instant.  For steady-state analysis, we assume that the suspension systems are at zero velocity: the car has assumed a steady attitude, in response to steady ground-plane forces.  When the suspension does have some velocity, the dampers are making forces.  These are customarily modeled as being in a consistent relationship to shaft velocity, often a linear relationship, although behavior of actual dampers is considerably more complex.  At any rate, for

steady-state cornering, or combined cornering and longitudinal acceleration with unchanging a_x and a_y, frictional load transfer is generally considered to be zero.

We now have all the components of load transfer due to ground-plane forces, except the unsprung load transfer.

So far, I have merely been explaining conventional theory.  Now we are going to break some new ground.  Traditionally, the total unsprung mass at the front and rear has been taken as equivalent to an axle assembly, even where the actual suspension is independent, and the front and rear unsprung load transfer ΔF_zUf and ΔF_zUr have been calculated as follows:

                                                        ΔF_zUf = (m_Uf a_y H_cgUf) / L_yf                                                   (7a)  

                                                        ΔF_zUr = (m_Ur a_y H_cgUr) / L_yr                                                   (7b)

In my opinion, this is correct for any beam axle, and also for any independent suspension with 100% camber recovery in roll.  However, for most independent suspensions, it is incorrect.  The equations really should be:

           ΔF_zUf = ((m_URf a_y H_cgURf)(1 – (dγ/dφ)_Rf) + (m_ULf a_y H_cgULf)(1 – (dγ/dφ)_Lf)) / L_yf           (7c)  

           ΔF_zUr = ((m_URr a_y H_cgURr)(1 – (dγ/dφ)_Rr)  + (m_ULr a_y H_cgULr)(1 – (dγ/dφ)_Lr)) / L_yr          (7d)

It will be apparent that when dγ/dφ = 0 for both wheels, and the unsprung mass c.g. height is the same right and left, equations (7c) and (7d) simplify into equations (7a) and (7b).

It will also be apparent that when dγ/dφ > 0, there is now a component of load transfer that is unaccounted for in the equations for the unsprung load transfer.  No matter what the value of dγ/dφ is, the components still have the same mass, and the same c.g. height, so at a given a_y the lateral load transfer from those masses must be a constant regardless of dγ/dφ.  So what happens to the remaining component?

It tries to roll the car on its suspension, and thus becomes a part of the elastic load transfer!  When camber recovery is less than 100%, it really is possible to create roll in the suspension by exerting a y force near the hub of a wheel, resisted by friction at that wheel's contact patch, without applying any other force to the sprung structure.

We therefore have to revise equation (2), M_φE = m_S a_y r_cgSx, to include the elastically reacted component from the unsprung masses, as follows:

M_φE = (m_S a_y r_cgSx) + (m_URf a_y H_cgURf (dγ/dφ)_Rf) + (m_ULf a_y H_cgULf (dγ/dφ)_Lf)

                                           + (m_URr a_y H_cgURr (dγ/dφ)_Rr) + (m_ULr a_y H_cgULr (dγ/dφ)_Lr)                (8a)

If we wish, we may also write that this way:

M_φE = a_y ((m_S r_cgSx) + (m_URf H_cgURf (dγ/dφ)_Rf) + (m_ULf H_cgULf (dγ/dφ)_Lf)

                                           + (m_URr H_cgURr (dγ/dφ)_Rr) + (m_ULr H_cgULr (dγ/dφ)_Lr))                     (8b)

Or:

                                            M_φE = a_y ((m_S r_cgSx) + Σ(m_U H_cgU (dγ/dφ)))                                 (8c)

Note that we do not simply take an effective mass (m_U (dγ/dφ)) at height H_cgU and add it to the sprung mass m_S.  Nor do we add such a mass to m_S when calculating the geometric load transfer ΔF_zG.

One of the correspondents who helped set me to thinking about this whole business suggested that what mattered was the lateral movement of an unsprung mass's c.g. with respect to the origin, or to the sprung mass, as the car rolls.  That would mean that both the x and z coordinates of the front view instant center would matter.  That is, both camber change and track change with suspension motion would matter.  After considerable thought, I have concluded that what really matters, for an unsprung mass or a sprung mass, is the rate of lateral motion of the mass's c.g. with respect to the relevant contact patch or patches, per unit of roll.

For an individual wheel, the relevant contact patch is the wheel's own.  For the sprung mass, it's a weighted average of the contact patches – weighted according to the right/left distribution of cornering force contribution.  This is what we are getting to when we assign roll centers and roll axes the correct way, which I have described in my video and in other documents.

That is why we take a moment about the roll axis for the sprung mass, but take individual moments about a wheel's own contact patch, with a camber non-recovery multiplier, for the unsprung masses, and then sum these get a correct M_φE.

The whole thing can be thought of as an application of force/motion relationships.  The more a particular mass's c.g. moves laterally in response to its own centrifugal inertia, with respect to the relevant contact patch(es) where the centripetal acceleration force is applied, to roll the suspension a degree, the more roll a pound of force applied at that point will produce.  This is directly analogous to a lever working against an elastic resistance.  A longer lever produces more angular deflection for

a given force at the lever end, in direct proportion to how far that lever end moves per degree of deflection.  (For large angular displacements, the relationships start to get non-linear, but for the small angles we normally deal with in analyzing roll and pitch of automobiles, the non-linearities are very small.)

What components are we talking about when we refer to unsprung components?  One might think this would be a simple question, but when we get down to the fine points, it isn't quite so simple.

For the parts with the most mass – the tire, wheel, upright, brake rotor, and caliper – it's easy: they can be treated as entirely unsprung, with a single center of mass as an assembly.  The whole assembly has the same value for dφ_t/dφ, and the above discussion and equations apply.

But what about the spring?  The unsprung portion of the damper?  The control arms or links?  The pushrod or pullrod?  The rocker?  An anti-roll bar or torsion bar arm?  If we take out the spring, support the sprung structure on stands or on a test rig, and note the weight of the unsprung components at a wheel scale or the wheel support pad on a rig, how accurate is that?  Do we add half the weight of the spring?

In my opinion, using weight as measured above, without adding half the spring, will in most cases yield a very good figure for use in the calculations we are considering here.  It will be less accurate if we are seeking a value for use in ride analysis, or modeling of unsprung mass response to road perturbations.  For that, we need to know the effective inertia mass for the components, in wheel z acceleration.

For example, suppose we have a coil spring of total mass in pounds m_spring, in a typical big-spring stock car front end, acting at a motion ratio of 0.50.  Half of m_spring acting at a wheel scale through that 0.50 motion ratio would increase the scale reading by m_spring/4.  However, when the wheel hits a bump, and sees an acceleration in g's a_zt, half of the spring sees an acceleration of a_zt/2.  The inertia force at the spring is then a_zt m_spring /4.  That inertia force at the wheel is a_zt m_spring /4, times the 0.50 motion ratio, or a_zt m_spring /8.  The scale weight contribution of half the spring's mass is that mass times the motion ratio, but the inertia contribution is that m_spring /2 mass times the square of the

motion ratio.  In this example, an ounce on the spring only has 1/8 as much effect on the wheel's ability to ride bumps as an ounce on the wheel itself.

In roll, how does the spring act?  It pretty much acts as part of the sprung mass, and can be treated as such.  That is, we don't need to add any part of it to our unsprung mass for purposes of roll analysis.

What if we have a formula car with pushrod suspension, and a coilover lying horizontal, operated through a rocker?  Here again, parts of this mechanism undergo accelerations when the wheel sees an a_zt, but they pretty much move with the sprung mass in roll, not with the tire and upright.  No part of a horizontal coilover adds gravitational force at the contact patch, but the parts that move when the suspension moves definitely contribute to unsprung mass inertia when the wheel hits a bump.

Or, consider the case of an upper control arm that serves as a rocker operating an upright inboard coilover, as was popular for open-wheel cars in the '60's and '70's.  Here, the moving bits inboard of the fulcrum appear to have negative weight when measuring on a wheel scale.  If we hang enough lead on the inboard end of the arm, we can get a reading of zero at the wheel scale.  That doesn't mean we are reducing unsprung mass or unsprung inertia, or improving the wheel's ability to ride bumps.

It is sometimes said that one measurement is worth a thousand expert opinions.  There is some truth to this, but only if the measurement in question is an appropriate one for the phenomena under consideration.  For most dynamic analysis, we are concerned with inertia, not weight, and a given object exerts different amounts of inertia force in different modes of movement.

To get a valid experimental measurement of unsprung mass inertia in response to an a_zt, we need to subject the wheel to an actual vertical acceleration of known magnitude, with a dummy shock to eliminate damper forces, and measure the resulting inertia force.  To do that, we need a seven-post rig, and the acceleration needs to be measured in terms of wheel or displacement pot motion, not ram or contact patch motion, despite the fact that we will necessarily be measuring the resulting force at the ram or contact patch.  Chances are that we can get at least equal accuracy by weighing the parts individually on scales, and applying some expert opinion in the form of sound mathematics.

Depending on the design of the suspension, the various inboard components may not exactly move as part of the sprung mass in roll.  Fortunately, these components are generally small in mass compared to the rest of the system, so any error resulting from simply treating them as part of the sprung mass will be correspondingly small.  Therefore, I suggest treating them that way as a general-purpose approximation.

Now, how does all of this apply for longitudinal accelerations (a_x)?  Rather similarly, except that the wheel is round in side view, and only generates a ground-plane force when a torque is applied to it.  Also, the wheel is spinning in side view when the car is moving, and it undergoes rotational

accelerations a_yt about its axis of rotation, which create inertial moments M_yt in addition to the linear inertia of the wheel.  These are relatively small, and it is common to ignore them, but they are real.

We will consider the linear inertia effects first.  We know that the unsprung components will react to any a_x with an inertia force F_xU = m_U a_x.

If we have a car on a kinematics and compliance (k&c) rig, and we exert, say, a forward force directly upon a wheel at about hub height, what does the car do?  Does it pitch?  Unless we lock the wheel somehow, the car doesn't pitch; it just moves forward.  To resist the force, we have to apply the brakes or otherwise lock the wheel.  The way the forces react once the wheel is locked depends on the load path for the braking or locking torque, as it does when determining longitudinal anti and pitch center height.  However, as with the unsprung component in roll, we are concerned with the unsprung mass's motions relative to is own contact patch center rather than a weighted average of two contact patches.

When the torque reacts through the linkage, the situation is directly analogous to roll.  The portion of the unsprung mass that creates a pitch moment in the suspension is proportional to the caster non-recovery dθ_u/dθ.  When dθ_u/dθ = 0, or when the torque does not react through the linkage, as with inboard brakes, the unsprung mass acts entirely as unsprung, and creates no suspension pitch. The equations that result are of the same form as those for roll, as follows:

When dθ_u/dθ = 0, or when torque does not react through the linkage,

                                                        ΔF_zUR = (m_UR a_x H_cgUR) / L_xR                                                   (9a)  

                                                        ΔF_zUL = (m_UL a_x H_cgUL) / L_xL                                                   (9b)

Otherwise,

           ΔF_zUR = ((m_URf a_x H_cgURf)(1 – (dθ_u/dθ)_Rf) + (m_URr a_x H_cgURr)(1 – (dθ_u/dθ)_Rr)) / L_xR          (9c)  

           ΔF_zUL = ((m_ULf a_x H_cgULf)(1 – (dθ_u/dθ)_Lf)  + (m_ULr a_x H_cgULr)(1 – (dθ_u/dθ)_Lr)) / L_xL          (9d)

And when a component of the unspung mass acts to create a suspension pitch displacement,

M_θE = (m_S a_x r_cgSy) + (m_URf a_x H_cgURf (dθ_u/dθ)_Rf) + (m_ULf a_x H_cgULf (dθ_u/dθ)_Lf)

                                           + (m_URr a_x H_cgURr (dθ_u/dθ)_Rr) + (m_ULr a_x H_cgULr (dθ_u/dθ)_Lr)                (10a)

Which we may also write:

M_θE = a_x ((m_S r_cgSy) + (m_URf H_cgURf (dθ_u/dθ)_Rf) + (m_ULf H_cgULf (dθ_u/dθ)_Lf)

                                           + (m_URr H_cgURr (dθ_u/dθ)_Rr) + (m_ULr H_cgULr (dθ_u/dθ)_Lr))                     (10b)

Or:

                                            M_θE = a_x ((m_S r_cgSy) + Σ(m_U H_cgU (dθ_u/dθ)))                                  (10c)

It is possible to imagine cases where dγ/dφ < 0 or where dθ_u/dθ < 0 – that is, where camber or caster recovery is greater than 100%.  In such cases, the equations still apply, and the unsprung components actually create an anti-roll or anti-pitch moment in the suspension, reduce elastic load transfer, and create unsprung load transfer as if they had more than their own actual mass.  Such cases are rare, however.

Finally, the rotating components have rotational inertia.  As mentioned earlier, these forces are smaller than the linear unsprung inertias, and it is common to ignore them completely.  They do not add to the unsprung load transfer.  However, they can produce some forces within the suspension linkage, and they present a rather interesting analytical challenge.

The rotating portions of the unsprung assembly – typically the tire, wheel, hub, brake rotor, and driveshaft – can be thought of as a single mass m_r, acting at a radius of gyration k_y about the wheel axis.  At a given vehicle longitudinal acceleration a_x, the inertial moment M_yui about the wheel axis is equal to m_r times the square of the ratio of radius of gyration to tire radius:

                                                                  M_yui = m_r a_x (k_y/r_t)²                                                      (11)

One way to think about this is to compare the effect of an ounce at the tire tread surface to an ounce halfway in toward the hub.  The ounce at the tread surface accelerates circumferentially at a_x, and acts at r_t, so its inertial contribution rotationally is equal to its linear contribution.  An ounce saved there is worth two off the frame, as it were.  The ounce halfway in accelerates circumferentially at only a_x/2, and its circumferential inertia force acts at half the tire radius.  Thus its inertial contribution rotationally is only ¼ of its linear contribution, and an ounce saved there is only worth 1.25 off the frame.

Probably the best way to find k_y is to computer-model the various parts.  Failing that, a better-than-nothing approximation would be to assume k_y/r_t = 0.5 where the assembly includes a brake rotor, or k_y/r_t = 0.7 where the brake rotor is not included.

When looking at anti-pitch geometry last issue, we noted that the big forces acting through the linkage can include both a thrust and a torque (outboard brake) or only a thrust (inboard brake).  With the much smaller forces from the rotating unsprung components, we have some different possibilities.

In an ordinary car with four-wheel outboard brakes, each brake has to exert an extra increment of torque to overcome rotational inertia at its own wheel.  This torque does not show up as a longitudinal force at the contact patch, but it does react through the linkage: a torque without a thrust.

If the additional torque is generated through an inboard brake or transaxle, and transmitted through a jointed shaft to the wheel, the additional torque does not act through the linkage: neither a torque nor a thrust through the linkage.

Finally, we have the unique case of a wheel being neither driven nor braked, such as a front wheel of a rear-drive car under power.  In this case, the torque to overcome rotational inertia comes to the front wheel through its contact patch.  The creates a small rearward inertial thrust, and this does act through the linkage.  Moreover, that thrust ultimately comes from the rear wheel, as an equal forward thrust at the rear contact patch.  That reacts through the rear suspension linkage, and the additional increment of torque may too, if the rear suspension is such that drive torque does that.

In the first case, the torque to overcome rotational inertia creates a pitch moment M_θEui which must be reacted by the suspension's elastic components that resist pitch, whose magnitude depends on the caster non-recovery dθ_u/dθ, and these on the front and rear at each side create an anti-pitch moment as follows:

                                                M_θEui = M_yuif (dθ_uf/dθ) + M_yuir (dθ_ur/dθ)                                           (12)

The only way this small pitch moment can affect wheel loads is if it is different on the right and left sides of the car.  However, it may slightly affect pitch displacement even if it is the same on both sides.

In the second case, the rotational inertias do not affect wheel loads or suspension displacements, but they do affect brake requirements, or power requirements in the case of all-wheel drive.

In the third case, the thrust F_xuif at the undriven wheel creates an F_z that depends on the dx/dz at the wheel center, or at the contact patch of a free-rolling wheel, which is an identical dx/dz.

                                                 F_zuif = F_xuif (dx/dz)_f = M_yuif (dx/dz)_f / r_t                                          (13)

The rear wheel F_x is equal in magnitude to the front one but opposite in direction and hence in sign.

                                                                    F_xuir = – F_xuif                                                                (14)

There is then a F_zuir induced in the rear suspension which depends on the dx/dz that applies for the type of rear suspension the car has.  There is simply a slightly greater propulsion force at the rear contact patches than would be otherwise be needed to produce a_x, and that produces forces in the rear suspension in accordance with its anti-squat properties.

If the rear suspension has anti-squat, and the front wheel has thrust anti-dive (that is, if its hub moves forward when the suspension compresses), then there are small upward forces induced in both the front and rear suspensions, and the resulting jacking force for the pair is the sum of the two, F_zuir + F_zuif. The difference of the two, times half the wheelbase, is the pitch moment resulting from the rotational inertias' reactions through the linkages.