**The Mark Ortiz Automotive**

**CHASSIS NEWSLETTER**

**Presented free of charge as a service**

**to the Motorsports Community**

**July 2009**

**Reproduction for sale subject to restrictions. Please
inquire for details.**

** **

** **

** **

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/sec^{2}).
An acceleration in g's is a in ft/sec^{2} divided by the number of
ft/sec^{2} 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/sec^{2}, 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/sec^{2}. 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/sec^{2} (g's times 32.2). F will
be in pounds.

If m is in kilograms, a has to
be in M/sec^{2} (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/sec^{2}, 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_{x}^{2} + (H_{rcr} –
H_{rcf})^{2})) / L_{x} (1a)

r_{cgSy
} = H_{cgS} – (H_{pcR}
+ (H_{pcL} – H_{pcR})(m_{SL}/m_{S})) (√(L_{y}^{2} + (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.

** **