Why start here
Race car vehicle dynamics can look intimidating: tire curves, load transfer, stability derivatives. Underneath, almost all of it is something you already know: Newton's second law applied carefully to a car. This lesson is a quick refresher of the tools the rest of the course leans on: units, \(\Sigma\mathbf{F} = m\mathbf{a}\), circular motion and free-body diagrams. If they feel familiar, move quickly. If anything feels shaky, this is the place to fix it.
Units you will use constantly
| Quantity | SI unit | Conversions you will need |
|---|---|---|
| Speed | m/s | km/h \(\div 3.6\) → m/s; mph \(\times 0.447\) → m/s |
| Acceleration | m/s² | 1 g = 9.81 m/s² — race engineers almost always quote accelerations in g |
| Force | N | Weight \(W = mg\): a 750 kg car weighs 7358 N |
| Angle | rad | 1 rad = 57.3°; tire and steering data are usually quoted in degrees |
| Angular velocity | rad/s | Yaw rate is often logged in deg/s: multiply rad/s by 57.3 |
| Power | W | 1 kW = 1000 W; 1 hp ≈ 746 W |
“The front-left wheel carries 180 kg.” Race shops really do talk like this: “corner weight” scales read in kg or lb. But the load on a tire is a force. Before using it in any equation, convert: \(F_z = 180 \times 9.81 = 1766\) N.
Every answer box in this course accepts calculations, so you can type
180/3.6 or 180*g directly. See the calculator guide.
Newton's second law for a car
For most of this course we treat the car as a rigid body. Its center of gravity (CG) moves according to the total external force, however that force is spread around the car:
The car can also rotate. For rotation about a vertical axis through the CG, called yaw, the matching law is \(\Sigma M_z = I_z \ddot\psi\), where \(I_z\) is the yaw moment of inertia and \(\ddot\psi\) the yaw acceleration. In a steady turn the yaw rate is constant, so \(\ddot\psi = 0\) and the yaw moments must balance. That balance is what decides whether a car understeers or oversteers (Module 5).
The only external forces on a car are gravity, the forces at the four tire contact patches, and aerodynamic forces. Engine torque, brake torque and steering are all internal; they act on the car only by changing what the tires do.
Circular motion: accelerating at constant speed
A car driving around a circle of radius \(R\) at constant speed \(V\) is accelerating, because the direction of its velocity is changing. The velocity vector turns at an angular rate \(\omega = V/R\), so its rate of change has magnitude \(V\omega\), pointing toward the center:
In a steady turn the car's heading rotates at the same rate as its velocity, so the car's yaw rate is \(r = V/R\). Speed, radius, yaw rate and lateral acceleration are tied together. Know any two and you know the other two.
The free-body diagram of a car in a turn
The recipe never changes:
- Isolate the car: draw it on its own, cut free from the road.
- Add every external force where it acts: gravity at the CG, and contact forces wherever the car touched something (the four contact patches).
- Choose axes and write \(\Sigma F = ma\) in each direction (and \(\Sigma M\) if needed).
Viewed from behind, a car in a steady left turn has its weight acting down at the CG, the road pushing up on each tire, and the road pushing sideways on each tire toward the center of the turn. Writing the force balances:
Taking moments about the CG adds a third equation. The sideways tire forces act at ground level, a height \(h\) below the CG, so they try to roll the car outward. Only a difference between the normal forces can balance that moment. With track width \(t\):
That one line is the seed of Module 3, load transfer.
- Build the diagram, check it, and read the note beside every option, including the ones you got right.
- Double the speed. By what factor does the sideways force grow? Halve the radius instead. Which change is “worse” for the tires?
- Change the mass. Why does the “friction coefficient needed” not change? (You will meet this result again in Lesson 1.1.)
- Watch the normal-force arrows as the speed rises. Which tire carries more load, and why?
“Centrifugal force throws the car outward.” In a ground-fixed (inertial) frame there is no outward force at all. The car's natural tendency is to keep going straight, and the tires must keep pulling it inward. The sensation of being pushed outward is your body wanting to go straight while the car turns under you. If you ever see “\(mV^2/R\)” drawn outward on a free-body diagram, it is d'Alembert's trick of moving \(m\mathbf{a}\) to the force side. That is legitimate only if you then don't also write \(m\mathbf{a}\).
A 750 kg car takes a 100 m radius corner at 108 km/h. Find its lateral acceleration, the total sideways tire force and the friction coefficient it needs.
\(V = 108 / 3.6 = 30\ \text{m/s}\), so \(a_y = V^2/R = 900/100 = 9.0\ \text{m/s}^2 = 0.92\ \text{g}\).
\(\Sigma F_y = m a_y = 750 \times 9.0 = 6750\ \text{N}\), and the tires need \(\mu \ge F_y / (mg) = 0.92\). A road car on good tires could just manage this; a race car on slicks would have grip to spare.
Check your understanding
A car's speed is 144 km/h. What is it in m/s?
1 km/h = 1000 m / 3600 s.
\(144 / 3.6 = 40\ \text{m/s}\).
A car drives around a 60 m radius corner at 25 m/s. What is its lateral acceleration, in g?
\(a_y = V^2/R\) gives m/s². Divide by 9.81 to convert to g.
\(a_y = 625/60 = 10.42\ \text{m/s}^2 = 10.42/9.81 = 1.06\ \text{g}\).
A Formula SAE car and driver (300 kg in total) circle a 9 m radius skidpad at 15 m/s. What total sideways force must the four tires provide?
\(\Sigma F_y = m V^2 / R\).
\(F_y = 300 \times 15^2 / 9 = 300 \times 25 = 7500\ \text{N}\), equivalent to 2.55 g. That is a lot to ask of the tires; real skidpad speeds are lower.
In a steady 100 m radius turn at 30 m/s, what is the car's yaw rate, in degrees per second?
In a steady turn, \(r = V/R\) in rad/s.
\(r = 30/100 = 0.30\ \text{rad/s} = 0.30 \times 57.3 = 17.2\ \text{deg/s}\).
A car drives around a roundabout at a perfectly steady 40 km/h. Is it accelerating?
- No: its speed is constant.Speed is constant, but velocity is a vector, and its direction is changing.
- Yes: toward the center of the roundabout, with magnitude \(V^2/R\).Changing direction is acceleration. Without a net inward force the car would go straight.
- Yes: outward, away from the center.The acceleration points toward the center. The outward sensation is your body's inertia.
What external force actually makes a car go around a corner?
- The engine, through the driven wheels.The engine provides forward force. Turning needs a sideways force.
- The steering system.Steering changes the wheel angle, an internal action. The external force still comes from the road.
- Sideways friction forces from the road on the tires.They are the only horizontal external forces available (ignoring aerodynamics), so they must supply all of \(mV^2/R\).
A corner-weight scale under the right-front wheel reads 180 kg. What vertical force does that tire carry?
- 180 N.The scale reports mass. Multiply by \(g\) to get force.
- About 1766 N.\(180 \times 9.81 = 1766\) N. Always convert corner weights to newtons before using them.
- 18.3 N.That divides by \(g\) instead of multiplying.
Going further with Milliken
Read: RCVD Chapter 13, Historical Note on Vehicle Dynamics Development. It's an unusual way to start a course, but it explains where the ideas in this book came from, and why aircraft engineers ended up shaping how we analyze cars.
As you read, look for answers to these questions:
- Which developments in tire testing made modern vehicle dynamics possible?
- What did engineers borrow from aircraft stability and control, and why did it fit cars so well?
- Which ideas from the chapter do you expect to meet later in this course?
Summary
| Idea | Key relation |
|---|---|
| Newton's second law (translation) | \(\Sigma \mathbf{F} = m\,\mathbf{a}_{CG}\) |
| Yaw rotation | \(\Sigma M_z = I_z \ddot\psi\); in a steady turn \(\Sigma M_z = 0\) |
| Circular motion | \(a_y = V^2/R = V r\), with yaw rate \(r = V/R\) |
| Weight | \(W = mg\); corner weights in kg × 9.81 = N |
| FBD of a turning car | Gravity + normal forces + inward tire forces. No “centrifugal force”. |
Next: before we can write equations with signs, we need to agree which way is positive. Lesson 0.2 sets up the vehicle axis system.