Hot Air Balloon Lifting Force

Reference data and engineering information about hot air balloon lifting force for mechanics applications.

hotairballoonliftingCalculator

Overview

Engineering reference data for Hot Air Balloon Lifting Force in mechanics.

Key Formulas

Work

$W = Fd\cos\theta$

Work = force × displacement × cos(angle).

Kinetic Energy

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

Energy of motion.

Potential Energy

$E_p = mgh$

Gravitational potential energy.

Variables

Symbol	Description	Unit
$F$	Force	N
$m$	Mass	kg
$a$	Acceleration	m/s²
$v$	Velocity	m/s

The lifting force of a hot air balloon can be computed using the primary formula. For a balloon with a volume of $V = 10 \, \text{m}^3$ ( $353 \, \text{ft}^3$ ), heated to $100^\circ \text{C}$ ( $212^\circ \text{F}$ ) in surrounding air at $20^\circ \text{C}$ ( $68^\circ \text{F}$ ), the air densities are $\rho_h = 0.946 \, \text{kg/m}^3$ ( $0.00184 \, \text{slugs/ft}^3$ ) and $\rho_c = 1.205 \, \text{kg/m}^3$ ( $0.00234 \, \text{slugs/ft}^3$ ).

The lifting force $F_l$ is calculated as:

F_l = V (\rho_c - \rho_h) a_g

Substituting the values in SI units:

F_l = (10 \, \text{m}^3) \times (1.205 \, \text{kg/m}^3 - 0.946 \, \text{kg/m}^3) \times 9.81 \, \text{m/s}^2 = 25.4 \, \text{N}

In Imperial units:

F_l = (353 \, \text{ft}^3) \times (0.00234 \, \text{slugs/ft}^3 - 0.00184 \, \text{slugs/ft}^3) \times 32.174 \, \text{ft/s}^2 = 5.7 \, \text{lbf}

The mass $m$ that can be lifted, when the lifting force equals the weight force $F_g = m a_g$ , is derived by combining the formulas:

m = \frac{F_l}{a_g} = \frac{25.4 \, \text{N}}{9.81 \, \text{m/s}^2} = 2.6 \, \text{kg}

Specific Lifting Force

Specific lifting force refers to the lifting force per unit volume of the balloon, which depends on the temperature difference between the hot air inside and the cold air outside. This parameter is useful for comparing performance under different conditions. For instance, if the balloon air temperature is $60^\circ \text{C}$ and the surrounding air temperature is $-20^\circ \text{C}$ , the specific lifting force is approximately $3.3 \, \text{N/m}^3$ . Charts are available to visualize this relationship for both SI and Imperial units.