Vehicle Mechanics and Performance Analysis

Introduction

Vehicle mechanics explains how engine torque becomes motion, how the road and atmosphere oppose that motion, and how vehicle mass, geometry, tyres, and operating conditions determine performance. Performance analysis provides the quantitative basis for selecting powertrains, gear ratios, brakes, tyres, suspension characteristics, and control strategies.

A vehicle travelling on a road must overcome rolling resistance, aerodynamic drag, gradient resistance, and sometimes acceleration resistance. The available tractive force must exceed these demands if the vehicle is to accelerate. However, the available force is limited by both the powertrain and tyre–road adhesion. A satisfactory engineering analysis therefore combines force balance, power balance, energy conversion, load transfer, and experimental data.

Learning Objectives

By the end of this chapter, a student should be able to:

1. Identify and model the principal resistances acting on a road vehicle.
2. Calculate wheel torque, tractive force, power requirement, acceleration, and gradeability.
3. Construct and interpret a vehicle force–speed diagram.
4. Explain the effects of gear ratios, drivetrain efficiency, mass, tyres, and road conditions.
5. Analyse longitudinal load transfer during acceleration and braking.
6. Estimate fuel economy and identify its principal influencing factors.
7. Apply vehicle-performance models to practical design and diagnostic problems.
8. Recognise the assumptions and limitations of simplified models.

1. Fundamental Vehicle Forces

Consider a vehicle of mass m moving at speed v on a road with gradient angle θ. The longitudinal force balance is

where F_t is tyre-road tractive force, F_rr is rolling resistance, F_ad is aerodynamic drag, F_g is grade resistance, F_a is acceleration resistance, and a is longitudinal acceleration. In the basic model, F_a is represented by m a; an equivalent-mass correction may be added to account for rotating components.

For small road angles, sin θ ≈ grade/100 and cos θ ≈ 1. The exact expressions should be retained for steep grades or high accuracy.

1.1 Vehicle Weight and Normal Reactions

The vehicle weight is

where W is weight in newtons, m is mass in kilograms, and g is gravitational acceleration, approximately 9.81 m/s². On a level road, the total normal reaction is approximately W. On a grade, the normal reaction is W cos θ.

The static front and rear axle loads depend on the centre-of-gravity location. Let L be wheelbase, h be centre-of-gravity height, and b be the horizontal distance from the front axle to the centre of gravity. On a level road, the static rear-axle load is

and the static front-axle load is

These equations assume a rigid vehicle, negligible suspension compliance, and no aerodynamic downforce.

1.2 Rolling Resistance

Rolling resistance results mainly from tyre hysteresis, deformation of the tyre and road, bearing losses, and small surface irregularities. It is commonly represented by

where C_rr is the coefficient of rolling resistance. For a level road, F_rr ≈ C_rr W. Typical values may range from approximately 0.006 to 0.015 for well-inflated passenger-car tyres on paved roads, although tyre construction, pressure, temperature, load, and speed cause substantial variation.

Rolling resistance is approximately proportional to load in a simple model, but real tyres show speed, temperature, and pressure dependence. Underinflation generally increases energy loss and tyre temperature.

1.3 Aerodynamic Drag

Aerodynamic drag is

where ρ is air density in kg/m³, C_d is drag coefficient, A is frontal area in m², and v is vehicle speed relative to the air in m/s. The equation assumes steady external flow and uses an effective drag coefficient that incorporates body shape and flow effects.

The corresponding aerodynamic power is

Thus, drag force increases with the square of speed, while the power required to overcome it increases approximately with the cube of speed. Wind changes the relative air speed; a headwind can therefore produce a large increase in consumption.

1.4 Gradient Resistance

The component of weight acting down a road slope is

For a road gradient G expressed as a decimal, tan θ = G. For ordinary road grades, sin θ is close to G. A 10% grade therefore requires approximately 0.10mg of additional tractive force.

1.5 Acceleration Resistance and Rotating Mass

The translational acceleration force is m a. Wheels, shafts, gears, and motor rotors also rotate and require torque to accelerate. Their effect may be represented by an equivalent mass:

where m_rot is the equivalent translational mass of rotating components. The more complete force balance is therefore

This approximation is useful for longitudinal performance calculations, but the exact contribution depends on gear ratio and rotational inertias.

2. Powertrain Tractive Effort

If engine or motor torque is T_e, transmission ratio is i_g, final-drive ratio is i_0, and drivetrain efficiency is η_d, wheel torque is

The tractive force at the tyre circumference is

where r_w is effective loaded tyre radius. The equation assumes that torque is transmitted without clutch slip and that η_d represents all transmission losses.

Vehicle speed and rotational speed are related by

where n_e is engine speed in revolutions per minute. A larger gear ratio increases wheel torque but reduces vehicle speed at a given engine speed.

Power at the wheels is

where P_e is engine or motor output power. Since F_t = P_w/v, available tractive force decreases with speed when maximum power is constant. At low speed, torque and traction limits are more important; at high speed, power and aerodynamic drag dominate.

Traction Limit

The tyre-road interface cannot transmit unlimited force. The maximum force without longitudinal slip is approximately

where μ is the coefficient of adhesion and N_d is the normal load on the driven axle or axles. Therefore,

The value of μ depends strongly on road surface, tyre condition, water, temperature, and slip ratio. Electronic traction control attempts to regulate slip near a favourable operating value.

3. Performance Measures

3.1 Maximum Speed

At maximum speed on a level road, acceleration is zero. Consequently,

or, in power form,

The maximum speed is found by solving this nonlinear equation using the powertrain characteristic and aerodynamic parameters. A graphical intersection of available and required tractive force is often the clearest method.

3.2 Gradeability

Gradeability is the steepest gradient a vehicle can climb at a specified speed. If acceleration is zero,

Hence,

The result must be physically meaningful: the right-hand side cannot exceed one, and the available force must also satisfy the tyre-adhesion limit. At low speed, aerodynamic drag is often small, but at high speed it may substantially reduce gradeability.

3.3 Acceleration

For a level road,

If force varies with speed, acceleration must be integrated numerically:

and distance is obtained from

In practical testing, acceleration is measured over a defined speed interval, such as 0–100 km/h. Results depend on payload, ambient conditions, tyre state, battery state of charge, gear-shift strategy, and test procedure.

Worked Example 1: Tractive Force and Gradeability

A 1,500 kg car has C_rr = 0.012, C_d = 0.30, frontal area A = 2.2 m², drivetrain efficiency η_d = 0.90, tyre radius r_w = 0.30 m, and is travelling at 20 m/s on a 5% grade. In second gear, engine torque is 180 N·m, the gear ratio is 2.1, and final-drive ratio is 3.7. Take ρ = 1.2 kg/m³ and g = 9.81 m/s².

Wheel torque is

T_w = 180 × 2.1 × 3.7 × 0.90 = 1,259 N·m.

Thus,

F_t = 1,259/0.30 = 4,197 N.

Weight is W = 1,500 × 9.81 = 14,715 N. For a 5% grade, sin θ ≈ 0.05 and cos θ ≈ 0.999.

Rolling resistance is

F_rr ≈ 0.012 × 14,715 = 177 N.

Aerodynamic drag is

F_ad = ½ × 1.2 × 0.30 × 2.2 × 20² = 158 N.

Grade resistance is approximately

F_g = 14,715 × 0.05 = 736 N.

The net accelerating force is

F_net = 4,197 − 177 − 158 − 736 = 3,126 N.

Therefore,

a ≈ 3,126/1,500 = 2.08 m/s²,

neglecting rotating mass. The result is valid only if tyre adhesion permits 4,197 N. If μN_d is less than this value, actual acceleration will be lower.

4. Vehicle Force–Speed Analysis

A force–speed diagram plots available tractive force against vehicle speed together with resistance curves. Each gear produces a separate available-force curve. The required-force curve generally rises with speed because aerodynamic drag rises as v². The vertical difference between available and required force indicates acceleration capability.

At gear-shift points, the tractive-force curve may drop because the higher gear has a smaller ratio. Good ratio selection keeps the engine near its useful torque or power range after a shift. A continuously variable transmission can maintain a more favourable operating point, although efficiency, cost, thermal capacity, and control complexity must be considered.

A related power–speed diagram compares wheel power with required road-load power. It is especially useful for high-speed analysis, where aerodynamic power dominates.

5. Fuel Economy and Energy Use

Fuel economy may be expressed as distance per unit fuel, such as km/L, or fuel consumption, such as L/100 km. The two are reciprocals after appropriate unit conversion.

The chemical energy rate supplied by fuel is

where ṁ_f is fuel mass flow rate in kg/s and LHV is the lower heating value in J/kg. If overall fuel-to-wheel efficiency is η_fw, wheel power is

For a steady operating condition, fuel consumption per distance is approximately

Real driving includes idling, transient enrichment, gear changes, accessories, braking, traffic delays, and cold-start operation. Therefore, fuel economy cannot be predicted reliably from road load alone unless the driving cycle and control strategy are also specified.

For battery-electric vehicles, an analogous relation uses electrical energy from the battery. Regenerative braking can recover part of kinetic and potential energy, but recovery is limited by motor power, battery acceptance, tyre adhesion, and control strategy.

Factors Affecting Fuel Economy

Important factors include vehicle mass, C_dA, tyre rolling resistance, speed, gradient, powertrain efficiency, gear selection, auxiliary loads, tyre pressure, traffic, and driver behaviour. Reducing speed on a motorway often provides a larger benefit than modest reductions in rolling resistance because aerodynamic power varies approximately with v³.

6. Longitudinal Load Transfer

During acceleration, inertial effects shift normal load from the front axle to the rear axle. For a vehicle on a level road with acceleration a,

where ∆N is the magnitude of load transfer, h is centre-of-gravity height, and L is wheelbase. During acceleration,

During braking, the signs reverse. These equations assume rigid suspension, level ground, negligible aerodynamic forces, and a concentrated centre of gravity.

Load transfer does not change the total normal load on a level road, but it changes the load available to each axle. Since tyre adhesion is approximately proportional to normal load, acceleration may be traction-limited in a front-wheel-drive car when load shifts rearward. Braking may be limited at the rear axle when load shifts forward.

For a front-wheel-drive vehicle, the approximate maximum acceleration before front-wheel slip can be found by combining F_t = μN_f with the force balance. For a simplified level-road model with rolling and aerodynamic resistances neglected,

where b is the distance from the front axle to the centre of gravity. This expression assumes front-wheel drive, a rigid body, and constant μ. It illustrates why a lower centre of gravity and suitable static weight distribution improve traction.

During braking, brake-force distribution must be designed so that front and rear tyres approach their adhesion limits appropriately. Anti-lock braking systems modulate brake pressure to prevent sustained wheel lock and preserve steering control.

7. Worked Example 2: Load Transfer

A 1,600 kg rear-wheel-drive car has wheelbase L = 2.8 m, centre-of-gravity height h = 0.55 m, and the centre of gravity is 1.2 m behind the front axle. It accelerates at 0.40g on level ground. The static rear load is

N_{r0} = m g b/L = 1,600 × 9.81 × 1.2/2.8 = 6,727 N.

The load transfer is

∆N = m a h/L = 1,600 × (0.40 × 9.81) × 0.55/2.8 = 1,234 N.

Therefore, the dynamic rear load is

N_r = 6,727 + 1,234 = 7,961 N.

The corresponding front load is approximately

N_f = 1,600 × 9.81 − 7,961 = 7,735 N.

The calculation demonstrates that acceleration increases rear-axle normal force and reduces front-axle normal force. It does not imply that the rear axle automatically receives all driving force; suspension geometry and differential design also matter.

8. Illustrative Case Study: Selecting a Powertrain for a Delivery Vehicle

Background

A company is evaluating a 2,200 kg light delivery vehicle intended for urban routes and occasional 8% grades. The design team must determine whether a proposed 90 kW electric drivetrain can meet the performance requirement without excessive energy consumption.

Problem

The vehicle has C_dA = 1.0 m², C_rr = 0.010, tyre radius 0.34 m, and drivetrain efficiency 0.90. At 25 m/s, air density is 1.2 kg/m³. The required level-road speed is 25 m/s, and the vehicle must climb the grade at 15 m/s.

Analysis

At 25 m/s, rolling resistance is approximately

F_rr = 0.010 × 2,200 × 9.81 = 216 N.

Aerodynamic drag is

F_ad = ½ × 1.2 × 1.0 × 25² = 375 N.

Road-load force is 591 N, requiring wheel power

P_w = 591 × 25 = 14.8 kW.

Thus, the 90 kW motor has ample steady-state power at this speed, although acceleration, thermal limits, and accessory loads must also be considered.

At 15 m/s on an 8% grade, rolling resistance is about 216 N, aerodynamic drag is

F_ad = ½ × 1.2 × 1.0 × 15² = 135 N,

and grade resistance is approximately

F_g = 2,200 × 9.81 × 0.08 = 1,726 N.

Total required force is approximately 2,077 N, and wheel power is

P_w = 2,077 × 15 = 31.2 kW.

The corresponding motor power, neglecting auxiliary loads, is approximately 31.2/0.90 = 34.7 kW. The selected motor therefore satisfies the steady grade requirement. Engineers must still verify peak torque at low speed, tyre adhesion, battery voltage, cooling, sustained-grade operation, regenerative braking capacity, and performance at the maximum payload.

Outcome and Learning Points

The analysis shows that rated motor power alone is insufficient as a selection criterion. Low-speed gradeability depends strongly on wheel torque and adhesion, whereas high-speed performance depends strongly on power and aerodynamic drag. A complete design must test the full torque–speed map over realistic payloads and operating conditions.

9. Model Limitations and Professional Practice

The equations in this chapter are engineering approximations. Important limitations include:

• C_rr is not truly constant and varies with tyre pressure, temperature, speed, and load.
• C_d may change with yaw angle, ride height, open windows, roof equipment, and crosswind.
• Drivetrain efficiency varies with torque, speed, lubricant temperature, and gear selection.
• Tyre adhesion depends on slip ratio, slip angle, surface contamination, and temperature.
• Suspension compliance and kinematic effects alter actual load transfer.
• Wind and road roughness make real operation transient.
• Battery state of charge and temperature affect electric-vehicle power availability.

Best practice is to combine first-principles equations with instrumented testing, validated simulation, uncertainty analysis, and clearly stated boundary conditions. Regulatory tests must follow the applicable jurisdictional procedures for emissions, fuel consumption, braking, noise, and safety. Engineers should not report a performance figure without specifying payload, tyre pressure, ambient conditions, test speed range, road grade, and measurement uncertainty.

Modern vehicles use sensors, electronic control units, GPS, inertial measurement units, and data logging to estimate road gradient, wheel slip, energy consumption, and driver demand. Artificial intelligence can assist in drive-cycle prediction, fault detection, eco-driving advice, and energy management. However, AI models require representative data, physical validation, cybersecurity protection, and fail-safe behaviour. Safety-critical decisions should not rely on an unverified data-driven model alone.

Chapter Summary

Vehicle performance is governed by the balance between available tractive force and road-load resistance. Rolling resistance is approximately proportional to normal load, aerodynamic drag varies with the square of relative air speed, and aerodynamic power varies approximately with the cube of speed. Gradient resistance depends on vehicle weight and road angle. Powertrain ratios multiply engine torque, while drivetrain losses reduce wheel power.

Acceleration is obtained from the net longitudinal force divided by effective mass. Maximum speed occurs where available tractive force equals resistance. Gradeability is determined by the surplus tractive force after rolling and aerodynamic losses. Tyre adhesion can impose a stricter limit than the powertrain, especially at low speed.

Fuel economy depends on the energy required at the wheels and the efficiency of converting fuel or stored electrical energy into wheel work. Longitudinal load transfer changes axle normal loads during acceleration and braking, affecting traction, brake-force distribution, and stability. Simplified analytical models are valuable for design screening, but professional decisions require validated vehicle data, realistic operating conditions, and explicit assumptions.

Key Terms

Rolling resistance: Resistance caused mainly by tyre and bearing deformation.

Aerodynamic drag: Resistance produced by air flow around a moving vehicle.

Grade resistance: Component of vehicle weight acting down a road slope.

Tractive force: Longitudinal force transmitted from the driven tyres to the road.

Wheel torque: Torque available at the driven wheels after gearing and losses.

Drivetrain efficiency: Ratio of wheel power to input power.

Adhesion coefficient: Measure of the maximum tyre–road friction capability.

Gradeability: Maximum road gradient a vehicle can climb under specified conditions.

Road load: Combined resistance due to rolling, aerodynamic, and gradient effects.

Equivalent mass: Vehicle mass adjusted to represent the inertia of rotating components.

Centre of gravity: Point at which the vehicle mass may be considered concentrated.

Load transfer: Change in axle normal loads caused by acceleration, braking, or cornering.

Wheelbase: Distance between the front and rear axle centres.

Drag coefficient: Dimensionless measure of aerodynamic resistance shape effects.

Frontal area: Projected vehicle area normal to the relative air stream.

Fuel economy: Distance travelled per unit quantity of fuel.

Regenerative braking: Recovery of kinetic energy through operation of an electric machine as a generator.

Traction control: System that limits excessive driven-wheel slip.

Force–speed diagram: Plot comparing available tractive force and resistance over speed.

Power-to-weight ratio: Available power divided by vehicle mass or weight, used as a performance indicator.

Review Questions

Short Answer Questions

1. What are the principal components of road load for a vehicle?
2. Why does aerodynamic power increase more rapidly with speed than rolling-resistance power?
3. Define gradeability and state the condition used to calculate it.
4. What is meant by drivetrain efficiency?
5. How does acceleration affect front- and rear-axle normal loads?

Long Answer Questions

1. Derive the longitudinal force balance for a vehicle travelling up a gradient and explain every term.
2. Explain how engine torque, gear ratios, final-drive ratio, tyre radius, and drivetrain efficiency determine wheel tractive force.
3. Discuss the construction and interpretation of a vehicle force–speed diagram.
4. Explain the principal physical and operational factors influencing fuel economy.
5. Discuss the assumptions and limitations of simplified load-transfer and traction models.

Analytical / Application Questions

1. A 1,800 kg vehicle travels at 22 m/s on a level road. Given C_rr = 0.011, C_d = 0.32, A = 2.4 m², and ρ = 1.2 kg/m³, calculate rolling resistance, aerodynamic drag, and total road-load power.
2. A vehicle has wheel torque of 1,100 N·m and effective tyre radius of 0.31 m. If road resistance is 900 N, calculate its ideal acceleration for a mass of 1,500 kg, neglecting rotating inertia and tyre slip.
3. Compare two design changes—reducing vehicle mass by 8% and reducing C_dA by 10%—for a vehicle operating at 30 m/s. Explain which change is more beneficial for steady high-speed energy consumption and why.

Multiple Choice Questions with Answers

Question 1

Which resistance varies approximately with the square of vehicle speed?

A. Rolling resistance

B. Aerodynamic drag

C. Static axle load

D. Weight on a level road

Correct Answer: B. Aerodynamic drag

Question 2

A 1,000 kg vehicle has C_rr = 0.01 on a level road. Using g = 9.81 m/s², its approximate rolling resistance is:

A. 9.81 N

B. 98.1 N

C. 981 N

D. 9,810 N

Correct Answer: B. 98.1 N

Question 3

If wheel torque is 900 N·m and effective tyre radius is 0.30 m, the ideal tractive force is:

A. 270 N

B. 600 N

C. 2,700 N

D. 3,000 N

Correct Answer: C. 2,700 N

Question 4

At maximum speed on a level road, assuming steady motion, available tractive force equals:

A. Vehicle weight

B. Acceleration resistance only

C. Rolling resistance plus aerodynamic drag

D. Tyre adhesion multiplied by vehicle mass

Correct Answer: C. Rolling resistance plus aerodynamic drag

Question 5

During acceleration on a level road, longitudinal load transfer generally:

A. Moves load from the rear axle to the front axle

B. Moves load from the front axle to the rear axle

C. Changes total vehicle weight

D. Eliminates tyre normal reactions

Correct Answer: B. Moves load from the front axle to the rear axle

Question 6

Aerodynamic power is proportional approximately to:

A. v

B. v²

C. v³

D. 1/v

Correct Answer: C. v³

Question 7

A vehicle weighs 14,715 N and travels on a 10% grade. Using the small-angle approximation, its grade resistance is approximately:

A. 147 N

B. 981 N

C. 1,472 N

D. 14,715 N

Correct Answer: C. 1,472 N

Question 8

If engine torque is multiplied by a gear ratio and final-drive ratio, but drivetrain losses occur, wheel torque is:

A. Always greater than the ideal multiplied torque

B. Equal to engine torque regardless of gearing

C. The multiplied torque reduced by drivetrain efficiency

D. Independent of tyre radius

Correct Answer: C. The multiplied torque reduced by drivetrain efficiency

Question 9

Which condition most directly limits tractive force at low speed on a slippery road?

A. Tyre-road adhesion

B. Frontal area only

C. Air density only

D. Fuel lower heating value only

Correct Answer: A. Tyre-road adhesion

Question 10

Regenerative braking primarily attempts to:

A. Increase aerodynamic drag

B. Recover part of vehicle kinetic energy

C. Increase rolling resistance

D. Prevent all load transfer

Correct Answer: B. Recover part of vehicle kinetic energy