Calculation of Brake Horsepower Saved by Drafting

Introduction

In motorsports, "drafting" or "slipstreaming" is a technique where a car follows closely behind another vehicle to reduce aerodynamic drag. This decrease in drag reduces the power required to maintain a certain speed, allowing the car to save brake horsepower (BHP). The car in front disrupts the airflow, creating a low-pressure area directly behind it. This low-pressure region benefits the trailing car by reducing the air resistance, or drag force, that it would normally encounter.

Problem Statement

Calculate the brake horsepower (BHP) saved by a car when drafting another car in front of it. Consider the drag reduction due to the low-pressure wake created by the lead car, which the following car exploits by remaining close behind. We will develop the mathematical relations to calculate the savings in BHP.

Assumptions and Given Data

The speed of both cars, $ V $, is constant and equal.
The coefficient of drag of the trailing car, $ C_d $, is reduced while drafting.
The frontal area of the trailing car is $ A $.
The air density is $ \rho $.

Variables

Define the following variables:

Drag force without drafting, $ F_{d0} $: The drag force experienced by the trailing car without drafting.
Drag force with drafting, $ F_d $: The drag force experienced by the trailing car while drafting.
Power without drafting, $ P_{0} $: The power needed to overcome the drag force when there is no drafting.
Power with drafting, $ P $: The power needed to overcome the reduced drag force while drafting.
Brake Horsepower Saved, $ BHP_{saved} $: The amount of brake horsepower saved by drafting.

Mathematical Model

The drag force $ F_d $ on a car moving at a constant speed $ V $ can be expressed as:

$$ F_d = \frac{1}{2} C_d \rho A V^2 $$

where:

$ C_d $ = Coefficient of drag
$ \rho $ = Density of air (typically around 1.225 kg/m³ at sea level)
$ A $ = Frontal area of the car
$ V $ = Speed of the car

When the car is drafting, the drag coefficient $ C_d $ is reduced by a certain factor due to the low-pressure wake. Let the reduction factor be represented by $ k $ (where $ 0 < k < 1 $), so the effective drag coefficient while drafting is $ kC_d $.

Therefore, the drag force while drafting, $ F_d $, becomes:

$$ F_d = \frac{1}{2} k C_d \rho A V^2 $$

Calculating Brake Horsepower

The power required to overcome drag (in watts) is the product of the drag force and the velocity of the car. Thus:

$$ P_0 = F_{d0} \cdot V = \left( \frac{1}{2} C_d \rho A V^2 \right) \cdot V = \frac{1}{2} C_d \rho A V^3 $$

Similarly, the power required while drafting is:

$$ P = F_d \cdot V = \left( \frac{1}{2} k C_d \rho A V^2 \right) \cdot V = \frac{1}{2} k C_d \rho A V^3 $$

The brake horsepower saved, $ BHP_{saved} $, due to drafting can be calculated as:

$$ BHP_{saved} = \frac{P_0 - P}{745.7} = \frac{\left( \frac{1}{2} C_d \rho A V^3 \right) - \left( \frac{1}{2} k C_d \rho A V^3 \right)}{745.7} $$

Simplifying, we get:

$$ BHP_{saved} = \frac{\frac{1}{2} C_d \rho A V^3 (1 - k)}{745.7} $$

Example Calculation

Let's calculate the brake horsepower saved for a specific example where:

$ C_d = 0.3 $
$ A = 2.5 \, \text{m}^2 $
$ \rho = 1.225 \, \text{kg/m}^3 $
$ V = 30 \, \text{m/s} $ (approximately 108 km/h)
$ k = 0.6 $ (i.e., a 40% reduction in drag coefficient due to drafting)

Substituting into the formula:

$$ BHP_{saved} = \frac{\frac{1}{2} \times 0.3 \times 1.225 \times 2.5 \times (30)^3 \times (1 - 0.6)}{745.7} $$

Calculating step-by-step:

$ F_d = 0.5 \times 0.3 \times 1.225 \times 2.5 \times 30^2 $
Power saved calculation here...