The Physics of Motorsport

This course looks at the physics involved in designing and driving fast racing cars. It will concentrate mainly on four wheeled, petrol driven vehicles, but the same physical problems apply to all kinds of vehicles. By taking this course, I hope you’ll learn from:

• applying physical principles to a real-world problem.
• figuring out how to build a better racing car than those guys at McLaren.

There are five parts to the course:

• Power Talk. We’ll discuss the meaning of a car’s performance statistics. For example, what do we mean when we claim our car has 150 bhp? How does it relate to its top speed?
• Power or Torque? Engine performance is governed by the two numbers: maximum power and maxium torque. Together with the gear box, these determine how fast it accelerates; but should the designer concentrate on increasing power or torque, or simply making the car lighter?
• The Straights. Tyres are the secret of motorsport sucess. We’ll discuss what limits a tyre’s ability to grip the road and propel a racing car forwards. `Weight transfer’ will explain why F1 cars have their engines at the rear.
• The Racing Line. What is the quickest route around a corner? The secret is that larger radius turns require the tyres to generate less centripetal force.
• Skids and Why They Happen. We’ll discuss the way the weight of a racing car is balanced between the wheels. Braking or accelerating in a corner shifts the weight distribution and the force a tyre is expected to generate. This leads to a skid.

Additional material:

• Tipler. 1st year Physics course text book. Covers all the physics you need to know. Read this if you’re feeling lost.
• Brian Beckman’s web page: http://members.home.net/rck/phor/ This is the inspiration for much of the material in this course.
• “Tune to Win” by Carroll Smith. Excellent introduction to race car development.

Aims of this Part

• Translate a car’s performance statistics into numbers we’re familiar with in physics.
• Understand what limits top speed.
• Make some simple calculations to compare maximum speeds.

Units

The units in which we talk about cars often differ from the SI units we conventionally use in Physics. This table gives some examples. The last column gives the typical range of values for a normal road car.

• Velocity has the same magnitude as speed, but is associated with a direction. A car moving in a circle may have a constant speed, but it has a changing velocity.
• Mass is often referred to as weight. Weight is the force exerted by a mass due to the earth’s gravitational pull.
• Power (and Torque) are usually quoted as the maximum values. The engine produces less power (torque) if the accelerator pedal is not pressed fully to the floor, or if it is not spinning at the optimal speed (see Part II).
• In SI units, torque is quoted in Nm, but this must not be confused with Joules. They mean completely different things (see below)
• 2π radians = 360^o

Power and Maximum Speed

“Power” is a measure of the rate of doing work. “Work” is the product of a force and the distance through which the force moves (Tipler, p. 135). Combining these shows that the force that is accelerating our race car is:

\(\textrm{Force} = \frac{\textrm{Power}}{\textrm{Speed}}  \quad (1)\)

For a given amount of power, the faster we’re moving, the less acceleration is produced. The maximum speed of a car is usually determined by its (maximum) power. For instance, when accelerating up hill, the car stops accelerating when the force produced at the wheels matches the component of the car’s weight acting down the hill.

Example: A Fiat 500 is entered in a production car trial and is expected to climb a 45⁰ slope. The engine produces only 30 bhp, which it can hope to reach the top of the slope?
Answer: 14.6 mph

Pushing through the Air

On level ground and for a given power, the maximum speed is usually set by the drag produced by the air flowing past the car. This is given by

\(F_{drag} = \frac{1}{2} C_{d}\Lambda\rho_{air}v^{2}   \quad (2)\)

where v is its speed, A its frontal cross-section area. \(\rho_{air}\) is the density of air (~1.3Kgm^-3),and \(C_{d}\) is a coefficient that measures how slippery the car is. Typically, \(C_{d}\) = 0.4 – 0.6

Example. Audi are proud of the A4’s drag coeficient of 0.4. However, the car is still pretty big having a frontal area of 2.2m². If the turbo engine produces 200 bhp(max) at the wheels, what is the maximum top speed ?

Answer : The maximum speed is about 146 kmph. The force required to push through the air at this speed is 2.1X10³N.

In these calculations, we’ve assumed that the maximum power is delivered at any speed we like. In part II, we’ll see that the power produced by the engine in fact depends on how fast it spins, so these calculations only work if we choose the ideal set of gear ratios to match the maxium speed and engine revs.

F1 racing cars tend to have very high values of \(C_{d}\)_, because they use spoilers to push the car down onto the track and increase its cornering speed. We’ll cover cornering speed in Part IV.

Aims of this Part

• Revision of Torque, Couples and rotational motion.
• Explain the purpose of the gear box.
• Understand the relation between power and torque.
• Estimate 0-60 time from gear ratios and engine power.

Torque – making things rotate

Imagine an object that’s fixed at one point but free to rotate. Give it a push. It can’t move laterally, since it’s fixed at the pivot, but it can begin to spin round and round. Analogous to Newton’s laws of linear motion, there’s a set of equations governing how fast the spin rate of the object increases. Check out Tipler (p. 227) if this is new to you.

In rotational motion, Torque plays the same role as force in linear motion. The torque (Γ) is defined as the force multiplied by the perpendicular distance between the force and the pivot.

\(\Gamma = F \times \tau_{perp}  \quad (3)\)

The Gear Box

The torque produced by the car’s engine is not applied to the road directly; first, it is passed through a set of gears in order to optimise the engine speed for the road speed. The effect of the gearing is to change the number of times the engine rotates for each rotation of the road wheel. In addition to the gearbox itself, the differential (or `final drive’) may alter the gear ratio as well as dividing the torque between the two front wheels. The `drive train’ is illustrated in Fig. 1.

If the engine rotates many times for each rotation of the road wheel, the torque on the road wheel will be larger than that produced by the engine.

\(\Gamma_{wheel} = x_{d}\cdot x_{g}\cdot \Gamma_{engine}.  \quad (4)\)

with the gear ratios of the differential and gear-box (\(x_{d}\) and \(x_{g}\)) defined as shown in Fig. 1

The aim of the gearbox is that, by changing the gear ratio \(x_{g}\), we keep the engine turning close to its maximum efficiency (or power) even though the speed of the vehicle may change.

Figure 1: The torque produced by the engine is fed through the gearbox and differential before being converted into a force at the road surface.

The following table gives the gear ratios for an Alfa Romeo 145 QV.

wheel dia inc. tyre 22in (radius = 11in)

To calculate the force that is applied to the road in order to accelerate the car, we need to multiply the torque at the wheel by the radius of the road wheel (including the tyre). This gives:

\(\textrm{Accelerating Force} = \frac{\Gamma_{engine}\cdot x_{g}\cdot x_{d}}{\textrm{tyre radius}}  \quad (5)\)

What’s the difference between Power and Torque?

The performance of an engine is usually quoted by quoting the maximum power and torque and the engine rotation speeds at which they occur.

Why quote both numbers? and how do these two engines compare?

Figure 2: The power curve of an idealised petrol engine. The solid curve illustrates that expected for a road car. A racing car requires less torque at low rpm, but is optimised to work at as high rpm as possible.

These are maximum values. The power curve of a typical petrol engine is shown in Figure 2. The power grows roughly proportionally to road speed while the torque is roughly constant. This is because they are both related to the force applied to the road to accelerate the vehicle.

Compare equations 1 and 5, and you’ll find

\(\textrm{Torque} \propto \frac{\textrm{Power}}{\textrm{engine rpm}}   \quad (6)\)

So why is it useful to quote both? It’s simplest to concentrate on the torque curve. This is what actually accelerates the car. The whole of the torque plateau is usually only 20% below the max torque, so this figure gives us the level of the plateau. In a racing car, we are not too worried about where the plateau starts, but the lower that it begins, the easier it is to drive the car without continually changing gear to keep the engine spinning fast.

The maximum power is useful for calculating maximum speeds without knowing the gear ratios, but it also serves to tell us where the power and torque curves drop. There is no point reving the engine much beyond the max. power rpm. Once the peak power is reached, you soon need to change gear; this will reduce the engine rpm so that you can again accelerate, but the force applied to the road will now be smaller because of the smaller \(x_{g}\).

The 0–60 Time

You should by now have the feeling that we know how to interpret the performance figures given in `Mega Power’ magazine. We know how to interpret the power and torque figures for the engine, and how these translate to the force accelerating the car. If we understand it, the we should be able to make a reasonable estimate of the 0-60 time that the car is capable of. From equation 5

\(\textrm{acceleration} = \frac{\textrm{force}}{\textrm{mass}} = 0.7 \times \frac{\textrm{engine torque} \, \times \, \textrm{gear ratio}}{\textrm{mass} \, \times \, \textrm{tyre radius}}  \quad  (7)\)

Notice the factor of 0.7 that has appeared? This is to allow for the fact only a part of the torque produced by the engine becomes useable force for accelerating the car. Approximately 30% of the power of the engine is ‘lost’ due to friction in the gear box and in spinning up the engine and gearbox components to make them rotate faster. It’s a pity such a large fraction is used up, but there you go.

If we stay in a particular gear, the acceleration is roughly constant (for our idealised engine) until we reach the maximum power rpm. We can therefore use the standard equation of motion:

\(v = u + at  \quad (8)\)

to calculate the time between two speeds (Tipler, p. 30).

Figure 3: Speed as a function of time during a 0 to 60 dash. The idealised engine gives a constant rate of acceleration in each gear.

To reach 60 in the Alfa Romeo, we will have to use 1st, second and third gears, so the speed will increase with time as shown in Figure 3. It takes about 1 sec to change gears. The estimated 0–60 time is 8.0 sec. That’s in reasonable agreement with the manufacturer’s quoted time of 8.4 sec. We’ve so far made no allowance for the limited grip between the tyres and the road.

Notice that we can make a faster 0-60 time in three ways.

1. Increase the maximum power of the engine by making it rev to higher rpm.
2. Increase the torque of the engine by making the cylinders larger (eg., replace the 2.0 l engine with a 2.5 l engine).
3. Make the car lighter.

Aims of this Part

• Introduce a simple model for how a tyre grips.
• Understand the concept of weight transfer during acceleration and braking.
• Calculate the maximum rate at which a racing car can accelerate.
• Explain the basic design of an F1 car.

A Simple Model for the Tyre

The torque from the engine is used to turn the wheel. Friction between the tyre and the road surface is responsible for preventing the tyre slipping across the road surface and hence for propelling the car forwards (Fig. 4).

Figure 4: A schematic diagram showing the forces acting on a tyre and the road

Figure 5: The traction force that a tyre generates as a function of the torque applied to the wheel.

The tyre is pushed onto the road by the weight of the vehicle supported by that wheel. There is a maximum force that can be generated by the friction. If this is exceeded, the tyre `breaks loose’ and spins. A simple model for an ideal tyre is shown in Fig. 5. As the applied torque rises, the force increases, but if too much torque is applied, the tyre begins to slip and no extra accelerating force is generated. If we go on increasing the torque, the tyre spins and the force generated drops rapidly.

The maximum force that can be applied is approximately given by

\(F_{max} = \mu W \quad (9)\)

where W is the reaction force pressing the tyre onto the road (Tipler, p. 107). If the car is stationary, W=Mg where M is the mass of the car supported on the particular wheel. For a road tyre, μ ~ 1.0, for a race tyre,  μ ~ 1.5,

This is an oversimplified description, but it will be adequate for this course. In Part 5, we’ll briefly introduce ‘slip angle’.

Weight Transfer

Most racing cars are driven by their rear wheels. As the car accelerates, more of the weight is supported on the rear wheels. This process is called `weight transfer’.

Figure 6: The forces acting on the front and rear wheels of an accelerating car.

We first need to introduce the `Center of Gravity’, or CoG. This is a useful point of reference. It is defined so that: any force that acts through the CoG has no tendency to make the car rotate. If you apply the force in a direction that does not pass through the CoG, it will make the car rotate unless it is balanced by another force (Tipler, p. 278).

For a racing car, the CoG is located between the wheels and as close to the ground as physically possible.

When the car is at rest (\(F_{f}=F_{r}=0\)), the up and down forces must balance:

\(M_{g} = W_{f} + W_{r}   \quad  (10)\)

and the torque on the car’s body due to the reaction forces must also be zero (since the reaction forces do not act through the center of gravity.)

\(L_{f}W_{f} = L_{r}W_{r} \quad \textrm{(at rest)} \quad (11)\)

Figure 6 shows the forces acting on the car as it accelerates. To make it general, I’ve sketched in a propelling force at the front wheel, although this is zero for a rear wheel drive car, like F1. As before, the weight of the car is balanced by the front and rear reaction forces (eq 10), but equation 11 must be modified to allow for the driving force at the rear wheels, which creates an additional torque about the CoG.

\(L_{f}W_{f} + hF_{r} = L_{r}W_{r} \quad (12)\)

Equations 10 and 12 are a pair of simultaneous equations that can be solved to determine the reaction forces

\(W_{f} = \frac{L_{r}Mg – hF_{r}}{L} \quad (13)\)

\(W_{r} = \frac{L_{f}Mg – hF_{r}}{L} \quad (14)\)

where \(L = L_{f} + L_{r}\)  is the ‘wheelbase’ of the car.

Maximum Acceleration

Figure 7: The effect of increasing the accelerating force on the weight distribution of a racing car. At rest the weight of the car is biased to the rear.

Upper panel: the effect of acceleration. Solid lines show the reaction forces at front and rear wheels, and how these change with acceleration. The dashed line shows the traction limit; ie., the weight that must be placed on front the wheels for them to grip.

Lower panel: the effect of braking, assuming that 70% of the braking force is directed to the front wheels. There are now two dashed lines showing the weight that must be placed on the front and rear wheels respectively.

As \(F_{r}\) gets larger, the reaction force at the front gets smaller while that at the rear gets larger. Since the maximum traction force depends on the weight on the tyre, this increases as the car accelerates harder. The limit to the acceleration is set either by the point at which the traction limit is reached, or the point at which the front wheels lift off the ground and directional control is lost. Figure 7 (upper panel) shows the situation for a typical race car.

Hard on the Brakes

Braking has the opposite effect to acceleration, increasing the reaction force at the front and reducing that at the back. We may generate braking force at both wheels. This leads to replace \(F_{r}\) by \(F_{r} + F_{f}\) in eqns 13 and 14.

Figure 7 (lower panel) illustrates how the weight is transferred at the braking force is increased. This time the braking force is limited either by the front wheels losing traction, or by the rear wheels lifting off the ground. If the front and rear wheels contributed to the braking force equally, we would be limited by the point at which the rear wheels lost traction. We need to limit the braking force at the rear wheels so that it is less than that applied at the front.

Basic Racing Car Design

We now understand enough to design an F1 racing car. Here are some beginner’s guidelines.

We will choose to drive the rear wheels so that the weight transfer during acceleration helps to improve the traction.

The engine is the heaviest part of the car, so we need to place it somewhere near the middle of the car. If we put most of the weight on the rear wheels, we will not be able to accelerate because the front wheels lift off the ground. If we put the engine too far forward, we will not be able to use the rear wheels to brake.

We can help a lot by making the wheel-base long thus reducing the amount of weight transfer. If we make it too long, however, the car will not steer well.

Finally, we can help a lot by making the car as low as possible. This will reduce weight transfer as well as making the car more aerodynamic.

Aims of this Part

• Understand forces acting on a car turning a circle.
• Explain what limits cornering speed.
• Estimate the fastest route through a corner (`The Racing Line’).

Motion on a Circle

In order to keep this section simple, we’ll only try to do the calculations for objects on a turn of constant radius.

In order for a mass (M) to move around a circle of constant radius (r) it must be subject to a centripetal force of magnitude

\(F = \frac{Mv^{2}}{r} \quad (15)\)

acting perpendicular to the direction of motion (Tipler, p. 65).

This force is generated by the tyres, however, the total force that a tyre can generate is limited. The art of race driving is to balance the accelerating (braking) and cornering forces so that the total force stays within the `traction limit’.

Figure 8: The traction circle showing a car accelerating out of a right hand bend, close to the limit of traction.

The Traction Circle

A convenient way to show the forces acting on a vehicle (or individual tyre) is the `traction circle’ (Figure 8). The arrow shows the vector sum of the acceleration and turning forces being exerted on the car. The circle shows the maximum force that can be exerted by the tyre. This depends on the force pressing the tyre onto the ground as discussed in Part III.

Unless the engine is exceptionally powerful, there will not be enough torque to reach all of the forward traction circle. In this case the acceleration is limited by the engine torque.

In this lecture I’ll consider the combined traction circle of all four wheels. We discuss the balance of the weight of the car between the wheels in Part V.

Cornering Speed — Constant Radius Turns

The maximum combined cornering force that can be generated by the tyres is \(F_{max} = \mu M g\) equating this to the centripetal force shows that the maximum cornering speed is

\(v = \sqrt{\mu r g} \quad (16)\)

We can drive around large radius corners more quickly (Tipler, p. 114).

Figure 9: Three possible routes around a 90⁰ corner connected by two long straights. Although route A is shorter, the cornering speed is slower. Route C has largest radius and is fastest.

Figure 9 shows three possible lines around a 90⁰ corner. Consider the time taken to complete the curved parts of routes A and B.

\(t_{A} = \frac{\pi}{2}\sqrt{\frac{r_{A}}{\mu g}} \quad (17)\)

\(t_{B} = \frac{\pi}{2}\sqrt{\frac{r_{B}}{\mu g}} \quad (18)\)

Although she takes the longer route, it takes driver B only a factor \(\sqrt{\frac{r_{B}}{r_{A}}}\) longer to cover the curved part. Further more, she needs to brake less before the corner, and will start travelling down the straight with a larger initial speed.

A Timed Corner

Figure 10: Illustration of Driver A’s speed through the corner. After the time-gate, he is able to maintain speed briefly before braking, turning the corner at a relatively low speed, and then accelerating once onto the straight. Driver C takes a longer path, but does not need to slow for the corner.

Example Drivers A and B have identical cars that can brake at 1g and accelerate at 0.5g, and have tyres with coefficient of friction of 1.1. They are timed over section of the circuit, starting 12m before a 90⁰ corner and finishing 12m after it. They pass through the first time-gate travelling at 41.8 mph. Assuming the inner and outer radii of the corner are 15 and 20m, which driver will make the fastest time?

Comparison of cornering times of Drivers A, B, C.

Notes: max. cornering accel. 10.78; accel. during braking -9.8; accel. 4.9; distance from timing gates to corner 12m.

This is how I calculated the table entries.

Calculate max cornering speed and time to turn corner.

Calculate time to slow to this speed (t=(v-u)/a) and distance travelled (s=(v²-u²)/2a).

Calculate time to cover remaining distance from first timing point at constant speed.

Calculate time to accelerate towards the second timing point (solve s= ut+½at² for t) Final speed is v=u+at.

After allowing for acceleration and braking times, B is only 0.15 sec (4%) slower despite travelling 30% further. Also, B passes the last timing post moving faster.

The Racing Line

You can guess that the fastest route through the corner will be the one with the largest radius. We will start with our wheels just touching the outside of the track, turn into the corner so as to just clip the inside kerb halfway through the corner and then unwind back to the outside of the track at the exit. This path is a good approximation to the ‘racing line’.

If the inner and outer radii of the curve are \(r_{i}\) and \(r_{o}\) this path has radius

\(r_{r} = \frac{\sqrt{2}\cdot r_{o} – r_{i}}{\sqrt{2} – 1} \quad (19)\)

and the driver must start the curve a distance \(x = r_{i} – r_{o}\) before the actual curve begins.

This just happens to be the position of the first time post in the previous example, and the speed at which the drivers pass it is exactly the maximum cornering speed corresponding to \(r_{r}\) in this example. Driver C, using the racing line, is 0.7 sec faster than driver A. Considering that most races are won by a margin of a few sec, this is a huge advantage. Taking the wrong line is very costly.

The Fastest Possible Corner

So far we’ve only considered paths through the corner in which the traction of the tyres are used to either brake, accelerate or corner. The fastest approach to a corner will involve a little combined braking and steering at the entry to the corner, followed by acceleration combined with steering as the corner unwinds. While braking/accelerating the corner must have larger radius than when the tyres are providing cornering force alone, otherwise the total traction vector will exceed the traction limit, and the car slides into the gravel trap.

Ideally we want to keep the traction force always at the limit of the traction circle, making a smooth, continuous transition between braking cornering and accelerating. This is the ability that makes a great driver.

Aims of this Part

• Compare the cornering forces generated at the front and back wheels.
• Outline the terms `oversteer’ and `understeer’ and explain how they relate to front and rear-wheel skids.
• Understand physically what happens when a car skids and how it can be corrected.

The Traction Circle at each Wheel

In Part IV, we considered the traction circle as applying to the car as a whole. To understand why a car skids, we need to look at the traction circles of each of the wheels individually. This is because weight transfer affects the grip offered by each of the wheels.

For example, consider a car that shares its weight equally between the wheels when it is stationary. During braking, the weight of the car shifts towards the front, and the traction circle of the front wheels grows at the expense of the rears’. As we discussed in Part III, we need to make sure that there is less braking effort at the rear than the front.

Now consider what happens if the car is cornering. During cornering the weight shifts to the outer wheels. With our equally weighted car the outside front and rear tyres will have equally large traction circles, and will provide most of the cornering force.

Figure 11: The effect of braking during a fast corner. The braking force transfers weight from the rear wheels to the front, resulting in a rear wheel skid.

But what happens if we panic and brake part way though a fast corner? In addition to the lateral (side-to-side) weight transfer, there is now also weight transfer to the front (Fig. 11). The traction circle at the rear shrinks and suddenly the cornering force that is required at the back of the car lies outside the available traction. If this happens quickly, the rear wheels will lose their grip, and the car is launched into a rear wheel skid.

This is just one example. Depending on how much weight is transfered, and how much braking/accelerating force is applied, we might have cause the front wheels to lose grip. The nature of the skid depends heavily on whether the front or rear wheels provide the driving force, and the distribution of the wheels’ traction when the car is static.

Oversteer and Understeer

Fortunately, a well designed car does not suddenly break into a skid. As the tyre approach their traction limits, they tend to slip sideways across the road. The angle between the tyres actual path and its natural path is called its slip angle. This gives the driver advance warning that the front/rear of the car is in danger of breaking loose and starting a skid.

If the rear tyres approach their traction limit more rapidly than the front, then the effect is for the rear of the car to steer a wider path than the front wheels. This rotates the car more than the driver intended and, if nothing is done, leads to the car turning a smaller radius corner. When this occurs the car is said to oversteer.

If the front tyre approach the traction limit more rapidly, the effect is that the front of the car takes a wider radius curve than the driver intended. The car is said to understeer.

Understeer is safer than oversteer. If the car understeers, and no correction is made the result is a wider corner than intended, but the car remains stable. If the car oversteers, the turn made has smaller radius than intended. The smaller radius produces higher cornering forces bring the required traction even closer to the limit of the rear wheels, and thus causing even more oversteer. The situation becomes worse until the rear wheels lose grip completely; the car spins and all directional control is lost.

If we start with a neutral steering car — one with the front and rear traction limits equal — we can create a table of the effects of accelerating and braking. For acceleration, the effect depends on whether the front or rear wheels are driven. Most modern road cars driven through the front wheels. The results also depend on whether or not the weight transfer compensates for the additional traction required for braking/accelerating.

The likely effects of acceleration and braking are summarised in the following table.

To understand the entries in this table, draw the traction circles for the balanced car, and add the cornering forces. Now redraw the traction circle allowing for the weight transfer, add the vectors showing the driving/braking forces, and determine whether the vector sum of the forces is closer to the traction limit at the front or rear of the car (closer at front leads to understeer, at rear leads to oversteer). The entry marked (*) assumes that most of the braking force is applied at the front wheels.

Correcting Skids

This discussion would not be complete without a discussion of how the racing driver should react to excessive over/understeer.

Because of its unstable nature, oversteer requires quick reactions to control. Assuming the situation has been caused by acceleration in a car driven by its rear wheels, there are two remedies. Firstly, reduce the accelerating force so that the traction is further within the limit. Secondly, to deliberately steer a wider circle than the car is intended to take (`apply opposite lock’).

The situation is different for a car driven by its front wheels since oversteer is only caused by excess weight transfer during braking. Applying the driving force of the engine will reassert the balance of the car.

Understeer cannot be corrected by increasing the steering angle of the front wheels. This simply increases the requested cornering force, making the front wheels lose traction completely. The correct procedure is to steer less, and brake to slow the car. Travelling more slowly, the cornering forces are considerably reduced, and even though a tighter corner must now be turned, it is probable that enough traction will be available.

A Numerical Example

This section has been mainly descriptive, but it is possible to account for the weight transfer to solve numerical problems. To make matters simple we will treat the car as resting on only its outer two wheels (this is really a motorbike approximation).

Example An F1 car has a 60/40 front-rear weight distribution. It is cornering at 1.4g but the rear tyres are capable of producing a total of 1.5 M_rg  of traction (where M_rg is the weight on the rear wheels). The design is such that 20% of the weight is transfered from rear to front per 1g of braking force. What braking deceleration can the driver apply before the rear tyres break loose? [To make the problem simpler, assume that the front wheels do all of the braking!]

Ans The maximum weight transfer is 6% from the rear to front. This corresponds to 0.3g of breaking (Part III). Play with the calculation, and you’ll see how critical it is to ensure that the weight transfer is as small as possible. This means getting the Centre of Gravity as low as possible.