Cars Power Torque

Overview

Engineering reference data for Cars Power Torque in dynamics.

Key Formulas

Newton's Second Law

$F = ma$

Force = mass × acceleration.

Kinetic Energy

$E_k = \frac{1}{2}mv^2$

Energy of motion.

Momentum

$p = mv$

Mass × velocity.

Work

$W = Fd\cos\theta$

Force × displacement × cos(angle).

Variables

Detailed Formulas and Explanations

Engine Power at Constant Speed

The power required from an engine to maintain a constant car velocity is given by: $P = \frac{F_T \cdot v}{\eta}$ where $F_T$ represents the total resistive forces (rolling resistance, gradient resistance, and aerodynamic drag), $v$ is the velocity in m/s, and $\eta$ is the transmission efficiency (typically 0.85–0.9).

Example Calculation:
For a car on a flat surface at 90 km/h (25 m/s) with $F_T = 250\, \text{N} + 400\, \text{N} = 650\, \text{N}$ and $\eta = 0.85$: $P = \frac{650 \cdot 25}{0.85} \approx 19{,}118\, \text{W} = 19.1\, \text{kW}$

Engine Torque

Torque $T$ relates engine power $P$ to rotational speed: $T = \frac{P}{2\pi n_{\text{rps}}} = \frac{9.55 P}{n_{\text{rpm}}}$ where $n_{\text{rps}}$ is engine speed in rev/s and $n_{\text{rpm}}$ in rev/min.

Example Calculation:
For $P = 19{,}118\, \text{W}$ and $n_{\text{rpm}} = 1500$: $T = \frac{9.55 \cdot 19{,}118}{1500} \approx 121\, \text{Nm}$

Wheel Force (Traction)

The force $F_w$ between driving wheels and road equals the total tractive force $F_T$, expressed via engine torque: $F_w = \frac{T \cdot \eta}{r} \cdot \frac{n_{\text{rpm}}}{n_{w,\text{rpm}}} = \frac{2 T \cdot \eta}{d} \cdot \frac{n_{\text{rpm}}}{n_{w,\text{rpm}}}$ where $r$ is wheel radius (m), $d$ is wheel diameter (m), and $n_{w,\text{rpm}}$ is wheel rotational speed (rpm).
Note: During cornering, an additional centripetal force acts between wheels and road.

Practical Examples and Applications

Engine Power at Constant Speed

The engine power required to maintain constant speed is calculated as:

$P = \frac{F_T v}{\eta}$

Where:

• $F_T = F_{aero} + F_{rolling}$ (total resistance force in N)
• $v$ is velocity in m/s
• $\eta$ is overall transmission efficiency (typically 0.85 to 0.9)

Calculation Example:

Given:

• $v = 90 \, \text{km/h} = 25 \, \text{m/s}$ (using $1 \, \text{km/h} = \frac{1}{3.6} \, \text{m/s}$)
• $F_{aero} = 250 \, \text{N}$
• $F_{rolling} = 400 \, \text{N}$
• $\eta = 0.85$

$P = \frac{(250 + 400) \times 25}{0.85} = \frac{650 \times 25}{0.85} \approx 19118 \, \text{W} = 19 \, \text{kW}$

Engine Torque Calculation

Torque is derived from power and engine speed:

$T = \frac{9.55 \, P}{n_{rpm}}$

Example: For $P = 19118 \, \text{W}$ and $n_{rpm} = 1500$:

$T = \frac{9.55 \times 19118}{1500} \approx 121 \, \text{Nm}$

Wheel Traction Force

The traction force at the driving wheels links engine torque to motion:

$F_w = \frac{2 T \eta}{d} \times \frac{n_{rpm}}{n_{w,rpm}}$

Where $d$ is wheel diameter (m) and $n_{w,rpm}$ is wheel rotational speed (rpm).

Additional Driving Considerations

• Acceleration: Include the acceleration force $F_a = m \cdot a$ (mass × acceleration) in $F_T$ during non-constant speed.
• Curved Paths: Centripetal force $F_c = \frac{m v^2}{r}$ (where $r$ is curve radius) adds to the total wheel-road force.
• Inclined Surfaces: Gradient resistance $F_g = m g \sin(\theta)$ (where $\theta$ is slope angle) modifies $F_T$ for uphill/downhill driving.

Drivetrain Power and Torque Relationships

Engine Power at Constant Speed

The required power to maintain constant velocity accounts for all resistive forces:

$P = \frac{F_T \cdot v}{\eta}$

Where the total resistive force $F_T$ combines rolling resistance, gradient resistance, and aerodynamic drag.

Efficiency Range: $\eta$ typically ranges from *0.85 (low gear, more losses) to *0.90 (direct drive, fewer losses).

Power-Torque-RPM Relationship

Engine torque relates to power and rotational speed:

$T = \frac{P}{2\pi n_{rps}} = \frac{9.55 \cdot P}{n_{rpm}}$

This inverse relationship between torque and RPM (at constant power) is fundamental to understanding engine characteristics and transmission design.

Wheel Force from Engine Torque

The traction force at the driving wheels incorporates gear ratios and wheel dimensions:

$F_w = \frac{T \eta}{r} \cdot \frac{n_{rpm}}{n_{w,rpm}} = \frac{2T \eta}{d} \cdot \frac{n_{rpm}}{n_{w,rpm}}$

Note: During curved driving, centripetal force adds to the total force acting between the wheels and road surface, affecting traction requirements and stability limits.

Additional Practical Examples

Engine Power Calculation Example

Consider a car maintaining a constant speed of 90 km/h on a flat surface. The aerodynamic drag force is 250 N, and the rolling resistance force is 400 N. The overall transmission efficiency (η) is 0.85.

Step-by-step calculation:

1. Convert velocity from km/h to m/s: $v = 90 \, \text{km/h} \times \frac{1000 \, \text{m/km}}{3600 \, \text{s/h}} = 25 \, \text{m/s}$
2. Calculate total traction force ($F_T$): $F_T = F_{\text{aero}} + F_{\text{roll}} = 250 \, \text{N} + 400 \, \text{N} = 650 \, \text{N}$
3. Calculate required engine power ($P$): $P = \frac{F_T \cdot v}{\eta} = \frac{650 \, \text{N} \cdot 25 \, \text{m/s}}{0.85} = 19,118 \, \text{W} \approx 19.1 \, \text{kW}$

Engine Torque Calculation Example

Using the power calculated above (19.1 kW) for an engine speed of 1500 rpm:

$T = 9.55 \times \frac{P}{n_{\text{rpm}}} = 9.55 \times \frac{19,118 \, \text{W}}{1500 \, \text{rpm}} \approx 121 \, \text{Nm}$

Key Clarifications and Notes

• Transmission Efficiency Range: The overall efficiency ($\eta$) typically varies from *0.85 (in lower gears with more meshing losses) to *0.90 (in direct-drive or higher gears). This range significantly affects the calculated required power.
• Unit Consistency in Formulas: The standard power formula $P = F_T \cdot v / \eta$ requires force in Newtons (N) and velocity in meters per second (m/s). Using other units (e.g., km/h) requires explicit conversion factors, as demonstrated in the example.
• Wheel Force and Drivetrain Relationship: The traction force at the wheel ($F_W$) is directly related to engine torque through the gear ratio and wheel size. The relationship $F_W = (2 T \eta / d) (n_{\text{rpm}} / n_{\text{w,rpm}})$ highlights how changing tire diameter ($d$) or gear ratios alters the force delivered to the road.
• Curved Driving Consideration: During cornering, the total force acting between the tires and the road includes not only longitudinal forces (for acceleration/braking) but also a centripetal force component to maintain the curved path. This increases the total force demand on the tire contact patches.

References

• Original Source