The goal in this and the next instalment of the Physics of Racing is to combine the magic formulae of parts 21 and 22, so that we have a model of tyre forces when turning and braking or turning and accelerating at the same time. In this part, we figure out combination slip, and in the next instalment, we figure out combination grip. Roughly speaking, slip is the input and grip is the output to our model. Slip comes from control inputs on brakes, throttle and wheel, grip comes from reaction forces of the ground on the tyres.
The regular magic formulae apply only to a tyre generating longitudinal or lateral forces in isolation, that is, to a tyre accelerating or braking and not turning, or a tyre turning but not accelerating or braking. In part 7, we approximated the response under combination slip by noting that it follows the circle of traction. A tyre cannot deliver maximal longitudinal grip when it's delivering lateral grip at the same time, and vice versa. According to my sources, modelling of combination slip and grip is an area of active research, which means we are on our own, once again, in the original, risktaking spirit of the Physics of Racing series. In other words, we're going way out on a limb and this could all be totally wrong, but I promise you lots of fun physics on the journey.
From part 21, recall our definition for the longitudinal slip, , the input to the longitudinal magic formula
where V is the forward speed of the hub w.r.t. ground, R_{e} is a constant, the effective radius, for a given tyre, and R_{e} is the backward speed of the CP w.r.t. the hub. Therefore, R_{e } V is the backward speed of the CP w.r.t. EARTH. A slick technique for proving this, and, in fact, for figuring out any combinations of relative velocities (see part 19) is as follows. Write V = HUB  GD, meaning speed of the HUB relative to the ground (GD). Now write R_{e }= (CP  HUB) meaning the backward speed of the CP relative to the HUB; the overall minus sign outside reminds us that we want R_{e} positive when the CP moves backwards w.r.t. the hub. Now, just do arithmetic:
R_{e} V = (CP  HUB)  (HUB  GD)
=CP + HUB  HUB + GD = (CP  GD)
voila, backward speed of the CP w.r.t. the ground. This realization gives us intuition into the sign of : if and only if the CP moves backwards faster than the hub moves forward; the car accelerates forwardvisualize that in your head; in that case, R_{e} is greater than V; R_{e} V is greater than zero; and is positive.
It turns out that we developed this formula only for the case when V is positive, that is, the car is moving forward. And, in fact, the formula only works in that case. To generalize it to cars moving in reverse, we'd best analyse it in excruciating detail. A moment's reflection reveals that there are eight cases: two signs for V, two signs for R_{e}, and two cases for whether the absolute value of V is greater than the absolute value of R_{e}, yielding eight = (2 2 2) possibilities, which have the following physical interpretations:
We've caught all this in the following diagram, in which we have drawn V and R_{e} as arrows, pointing in the actual direction that the hub moves w.r.t. the ground and the CP moves w.r.t. the hub, respectively. Algebraically, V and R_{e} have opposing sign conventions, so R_{e} is negative when its arrow points up. In looking at this table, note that the longitudinal force F_{x} has the same sign as . When F_{x} and are positive, the car is being forced forward by the ground's reacting to the tyres. When they're negative, the car is being forced backwards. So, to figure out which way the car is being forced, just look at the sign of .
Inspection of this table reveals that the following new formula works in all cases:
Where the numerator, R_{e}  V is the signed difference of the two speeds and the denominator is unsigned. It is perhaps surprising that there is so much richness in such a little formula. However, it is precisely this richness that we must maintain as we add steering, that is, lateral slip angle at the same time. The best way to do that is to vectorize the formula so that the algebraic signs of the vector components take the place of the signed quantities V and R_{e}. The approach here parallels the approaches of parts 16 and 19. We want the signed component V_{x} to take the place of the old, signed V, the signed component L_{x} of the slip velocity L to take the place of the old R_{e}, and V now to denote the unsigned magnitude of the vector V, that is . The next table summarizes these changes:
Quantity 
old notation 
new notation 
vector 
signed, forward speed of hub w.r.t. EARTH 
V 
V_{x} 
V 
signed, backward speed of CP w.r.t. hub 
W_{x} 
W 

unsigned magnitude of hub speed 
V 
or V 

signed, backward speed of CP w.r.t. EARTH 
L_{x} 
L = V + W 

signed, longitudinal slip 
? 
Slip velocity, L [L_{x}, L_{y}, L_{z}] = V + W is the plainold vector velocity of the CP w.r.t. EARTH with no secret sign convention to confuse things. As an aside, we note that when the car sticks to the ground on flat road, we may assume L_{z} = 0. W is CP velocity w.r.t. hub. In the TYRE system, W has only a (signed) xcomponent, that is, W_{TYRE} = [W_{x}, 0, 0]. These definitions hold whether the car is moving forward or backward, accelerating or braking.
The big question mark in the table indicates that we do not have a vector for combination slip because we measure its longitudinal and lateral components differently, as a ratio and as an angle, respectively. Note that, since lateral slip is the angle made by V in the TYRE system, it is . Since L = V + W, it's easy to see that
,
which is a most convenient expression, though some attention must be paid to the quadrant in which the angle falls. We resolve this in the next two instalments of PhORS as we stitch together the two magic formulae to make Combination Grip.
But first, let's update the big diagram, showing all eight cases with a little slip angle thrown into the mix, and the vector sum, L = V + W, replacing the ad hoc, signed quantities of the old notation. The sign of the slip angle does not introduce new cases so long as because the righthand and lefthand cases are precisely symmetrical. The nice thing, here, is that we can treat all eight cases the same waythe nature of vector math takes care of it because the magnitude of a vector is always unsigned. Using signed, scalar quantities, we had to dissect the system and introduce absolute value to get everything to work. Absolute value has always struck me as a kind of crock or kludge to use when the math is just not sufficiently expressive. The main contribution of this instalment is to fix that problem.