The Mark Ortiz Automotive


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April 2007

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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:  Readers are invited to subscribe to this newsletter by e-mail.  Just e-mail me and request to be added to the list.





As a reader of Racecar Engineering, I would like to know your opinion about a concern I have at the moment. I hope you will consider it and will be able to include it in a future issue of RE, because I think it is quite interesting.


I am studying the suspensions of my rally car at the moment.  Our rally car is a WRC car that we use for both gravel and tarmac events. There are some small differences between tarmac and garvel spec suspensions but the main layout is the same. It is a 4WD, with an active hydraulic center differential and a 50/50 torque split, most of the time.


I have a lot of toe variations during full travel at the rear suspension (pseudo MacPherson type, see attached file).  [Newsletter subscribers will receive the questioner's illustration as a separate attachment.]  In order to solve the problem, I have been testing (real tests and computer simulations) various configurations, with varying mounting point positions.


The results give me the same direction. The toe variations during travel decrease when points B and G are at the same height (vertical axis). We have always used our car with setups where B was around 5 to 10 mm lower than G. Knowing the results from my tests, I would logically change this setup to a line BG nearly horizontal.


However, some people are saying that it will greatly alter the anti-lift/anti-squat feature of the rear suspension. I have the feeling that it's not a problem but I am unsure about that. The problem is that I don't really know how to simply calculate the "anti" amounts of a MacPherson suspension in general and I can't find anything in many books. Could you give me some indications about this?




Readers are referred to the questioner's illustration.  We are discussing the rear suspension, which is shown in the right half of the illustration.  The suspension has a telescoping strut located by a single pivot at its top end, and by three links at its bottom end.  Links CB and FG run transversely and

provide lateral location, controlling camber and toe.  Link AM runs longitudinally and provides longitudinal location and reacts brake torque.  Link FG is partly hidden by the anti-roll bar.  Point C is hidden behind a donut-shaped object whose function is not obvious to me.  Point B is adjustable for height.


Any rear suspension that has a strut of this type, with the lower end of the strut located by some system of arms or links, is sometimes called a MacPherson strut suspension, or sometimes a Chapman strut suspension.  MacPherson was an engineer for Ford in England, and he is credited with inventing this type of suspension.  However, he and Ford only applied it to the front end of the car.  Lotus founder Colin Chapman was the first to apply it at the rear, although arguably his idea is a fairly obvious application of MacPherson's concept.  So there is a rational case for either name when referring to a rear suspension, and I consider them legitimate synonyms.  (I would not, however, consider it correct to call a strut-type front suspension a front Chapman strut.)


To understand the behavior of this suspension, and indeed any suspension, we need to define two more points: the wheel center and the contact patch center or load centroid.  These points are not in the illustration, but they are easily described verbally.  The wheel center is the intersection of the wheel centerplane (the plane midway between the rim flanges) and the wheel's axis of rotation.  I will call this point J.  For the contact patch centroid, I like to use a point directly below point J, even when the wheel has some camber.  Let's call this point K.  Let's also suppose that the wheel has no toe-in or toe-out at the condition we're evaluating.  The plane containing points J and K, and exactly longitudinal to the car, forms a good effective wheel plane to which we can project the suspension's side-view geometry.  Being a vertical, longitudinal plane, it is in true shape in a side view of the car.


All of these points have three coordinates in the car's axis system: x (longitudinal, per SAE convention), y (transverse), and z (vertical).


The distance from point K up to point J is the tire's loaded radius rL.


All we really need to know about any suspension's geometry to determine the suspension's "anti" properties is the motion path of point K.  More precisely, we need to know the instantaneous side-view and end-view (front- or rear-view) slope of that motion path at the suspension position we're evaluating, i.e. the instantaneous rate of change of the point's x or y coordinate with respect to rate of change in the z coordinate in calculus notation, dxK/dzK for anti-squat and anti-lift, and dyK/dzK for anti-roll.


With good enough modeling software, it may be possible to simply move the suspension a very small increment each side of the position being evaluated, and get a ΔxK/ΔzK that very closely approximates a true dxK/dzK.



Note that there are two assumptions we can make when doing this: we can assume that the wheel is free-rolling, meaning that K always stays directly below J, and the motion path of K is J's motion path transposed to the ground; or we can assume that the wheel is locked to the upright, meaning that

any change in the side view angle of the upright as the suspension moves creates a difference between the motion path shapes for J and K.  With independent rear suspension and outboard brakes,

we use the former method for anti-squat (under power), and the latter method for anti-lift (under braking).


We need to know three more things to get a percent anti-squat or anti-lift value: we need to know the percentage Pr of total x-axis force at the contact patch (or rear contact patch pair, if we are assuming the suspension is symmetrical and we are seeking a percent anti for the rear wheel pair); we need to have a sprung mass center of gravity height H (some estimating is normally required to arrive at a reasonable number); and we need to know the car's wheelbase L.


Then, for anti-squat, percent anti-squat = (dxK/dzK)PrL/H.  For the situation where all four wheels are driven and the torque split is 50/50 front/rear, percent anti-squat = (dxK/dzK).50L/H.  Remember that for independent suspension the value for dxK/dzK is taken assuming the wheel to be free-rolling, so that dxK/dzK = dxJ/dzJ.


For anti-lift, we need to know the percentage of brake retardation force, at the ground, that the rear wheels provide.  We use this for Pr .  We use the dxK/dzK value for the condition where the wheel is assumed to be locked to the upright.  That is, dxK/dzK = dxJ/dzJ only if the upright does not rotate at all in side view as the suspension moves.


Many of us do not have computerized animation of our suspension layouts.  We are used to finding the side view instant center, on the drawing board or the CAD tube, and constructing a side-view force line from the instant center to the contact patch center or the wheel center.  To do the job this way with a strut suspension, we need to find the virtual upper and lower control arm planes, and find their intersections with the wheel plane.  The intersection of two planes is a line, so we are talking about two lines, lying in the wheel plane.  These lines are our side-view projected control arms.  Their intersection is our side-view instant center.


A strut-type suspension has a virtual upper control arm plane, which contains the strut's top pivot point (point E, in the questioner's suspension), and is perpendicular to the strut tube axis.


This is not necessarily the same as the effective steering axis.  When the strut has a single ball joint at its lower end, the effective steering axis is a line through the ball joint pivot point and the upper pivot point.


What to use for a lower control arm plane is more of a conundrum.  I think I would start by finding the points where lines BC and FG intercept the axle plane: the transverse, vertical plane containing the wheel centers.  Let's call these intercepts S and T.  A line connecting these, line ST, would be the end-view projected lower control arm.


Once we have line ST, we can establish a plane parallel to that line, and containing line AM.  Once we have that, we can find the intersection of that plane with the wheel plane, and use the line so defined as our side-view projected lower control arm.


I am suggesting using side-view and end-view projected control arms that do not actually lie in a common plane.  However, all things considered, I think this is preferable to any possible alternatives.  The suspension, so modeled, is probably not exactly equivalent to reality, but it is a reasonable approximation.


Once we have the upper and lower side-view projected control arms, we can find their intersection, which is our side-view instant center.  We then draw in the force line for braking, from the contact patch center to the side-view instant center.  We locate a point at the sprung mass center of gravity at height H.  We draw a vertical line in our side view at a distance Pr/L forward of the rear wheel center.  I call this the resolution line for braking.


The force line will intercept the resolution line at some height h1.  The percent anti-lift = h1/H.  Note that if the force line intercepts the resolution line below ground, h1 becomes negative, and we have negative anti-lift, also called pro-lift.  If h1 > H, we have more than 100% anti-lift, meaning the rear suspension will compress in braking.  As we add more rear brake, we get more rear anti-lift (or more pro-lift, if the geometry provides pro-lift), without changing anything else.


For the anti-squat, the side-view projected control arms and instant center are the same as for anti-lift, but the resolution line and force line are different.  The force line is different because with independent suspension, drive torque does not act through the suspension linkage.  The resolution line is different because the rear wheels do not generate the same percentage of the longitudinal force under power that they do in braking.


There are two variations on the basic technique.  Done correctly, they both give the same answer.  The variation I prefer involves constructing a force line that originates from the contact patch center (point K), and using the same equation or rule for calculating the percent anti that we use when the suspension linkage reacts the torque, as with a live axle under power, or with outboard brakes in braking.  The other variation involves constructing a force line that originates at the wheel center (point J), and using a slightly different equation or rule.


In the first method, we construct a line from the wheel center J to the side-view instant center.  Then we construct a parallel line to that one, originating at the contact patch center K.  This is the force line when the car is under power.  We construct the vertical resolution line .50L forward of points J

and K for the questioner's example, or at the front axle for a purely rear-drive car.  The force line intercepts the resolution line at a height h2.  The percent anti-squat = h2/H.


In the second method, we do the same thing, except we omit the second line, and just use the line from point J to the instant center as the force line.  The force line then intercepts the resolution line



at a height h3, and h3 = h2 + rL.  The percent anti-squat = (h3 rL)/H.  That is, of course, the same as h2/H.


Adherents of the second method sometimes insist that this is the right way because the force acts on the car at hub height, and that any force line must pass through the instant center.  I say the first method makes more sense because any force exerted on the car by the road must act where the road

touches the car, and the wheel is part of the car, not a separate entity that acts on the car.  The difference between independent and live axle suspension, or inboard and outboard brakes, is not where the force acts on the car, but how the torque is reacted within the car i.e. through the suspension linkage or not.  I also think it makes sense to construct force lines so that a common rule applies once the force line is constructed, rather than having two different rules.  However, I don't think the whole debate is of much consequence, since both methods give identical answers.  The difference really has more to do with visualization and semantics than actual natural laws.  It is, however, important to understand both methods, so that one doesn't apply the rule or equation from one method when using the other.


The questioner wants to know what effect bump steer adjustment will have on the anti properties.  It is customary to model our antis on the assumption that there is no bump steer, and treat bump steer as a separate issue, a steering geometry issue rather than a suspension geometry issue.  However, if we really want to be rigorous, in fact there is some effect, but it is small, unless there are really large amounts of bump steer, in which case the car will be so undriveable that optimizing the antis will not be our main concern.  Unless bump steer causes the wheel to steer about an axis precisely in the wheel plane, there will be a small component in the side-view motion paths of points J and K resulting from the bump steer.


If the questioner's suspension is configured for minimum bump steer, point B will lie in plane CFG.  If B lies below plane CFG, the wheel will toe in as the suspension compresses, and the car will have roll understeer.  If B is above plane CFG, the opposite will occur, and the car will have roll oversteer.  Assuming that the wheel centerplane is outboard of the effective strut axis, roll understeer will increase anti-squat and anti-lift slightly, and roll oversteer will decrease anti-lift and anti-squat.


Note that this is opposite to the effect we would predict if we supposed that the suspension's side-view properties depended on the side view inclination of line BG, as some older suspension texts would suggest.  Raising point B decreases the anti-squat and anti-lift.  This example demonstrates the importance of looking at what happens at the wheel plane, rather than looking at the side-view inclination of the inner pivot axes of the control arms.



One final, somewhat self-serving note: when discussing suspension antis, I generally try to get in a plug for my video "Minding Your Anti", which includes more discussion of the topic, with more pictures.  They are available in US standard VHS cassette only, for US$50.00, payable by check or money order to Mark Ortiz, 155 Wankel Dr. , Kannapolis, NC 28083-8200, USA.  Price includes shipping and handling, worldwide.