A Refined Analytical Approach to Calculating Balloon Burst Altitude Using the International Standard Atmosphere

Table of Contents

Introduction

In order to determine the altitude

H_b

at which the burst volume

V_b

of a latex balloon is reached, we can use one of two methods:

Use a reference model for the atmosphere and solve analytically for $H_b$ .
Simulate balloon ascent using a weather forecast model (e.g. GFS) or real atmospheric sounding data and iterate vertically until balloon volume exceeds $V_b$ .

For a simple burst calculator in 1-dimension (i.e. up-down), method 1 will provide a reasonable estimate and is preferred due to the ease of implementation and speed of calculation. For a more advanced balloon flight simulator in 3D, method 2 would most likely be implemented and would give the most accurate results. One of the most popular balloon burst calculators is hosted online by habhub (now sondehub) here. It uses a simple analytical method which I have explained on my page here, but its limitation is that it assumes an isothermal atmosphere. We can refine this analytical approach by using the ISA reference atmosphere in the graph below, which shows temperature (x-axis) as a function of altitude above sea level (y-axis).

Refined burst height calculation

This section will outline the refined method to find burst height for a given initial volume $V_i$ and burst volume $V_b$ , and will work through an example. We will first derive equations for volume as a function of height (and vice-versa) for each of the discrete layers demarcated by the different temperature gradients (L1, 0, L2,…). Then we will calculate the volume ( $V_n$ ) at the top of current ( $n^{th}$ ) altitude window and check if $V_n$ is less than $V_b$ , starting from the troposphere (n=1) and working our way up. If so, then we can use a formula (shown below) for height as a function of volume to determine $h_b$ . Admittedly, this technique requires more steps than the simplistic approach, but aims to provide a more realistic and rigorous solution.

We will define the lapse rate as the rate at which temperature reduces with altitude. Thus, the lapse rates for the different altitude windows will be the negative of the temperature gradients shown in the figure above (e.g. the lapse rate in the troposphere is k=0.0065 K/m).

Heights measured from the base of a window will be denoted using a lowercase $h$ , whilst altitudes measured from sea level (above sea level or ASL) will be denoted using uppercase $H$ .

(Derivations for the following isothermal and non-isothermal equations will be provided soon)

Non-isothermal equations

For non-isothermal altitude windows (e.g. the troposphere and stratosphere) with a constant lapse rate, we can derive the following formula for volume as a function of height (measured from the base of the window):

$\Large V=V_{base}[1-\frac{kh}{T_{base}}]^{^{1-\frac{g}{Rk}}}$

$\begin{array}{ll} &\text{where:} \\[10pt] h & \text{is a height value between 0 and }h_{max} \\ h_{max} & \text{is the size of the window, or } H_{upper} – H_{lower} \\ H_{lower} &\text{is ASL altitude of the bottom of the window} \\ H_{upper} &\text{is ASL altitude of the top of the window} \\ T_{base} &\text{is temperature at the base of the altitude window} \\ k &\text{is the lapse rate of the altitude window} \\ V &\text{is the volume of the balloon} \\ V_{base} &\text{is the volume of the balloon at the base of the altitude window} \\ g &\text{is accl. due to gravity }(9.81\:m/s^2) \\ R &\text{is the specific gas constant for air }(287\:J/kgK)\end{array}$

Rearranging, we can get height $h$ as a function of volume:

$\Large h=\frac{T_{base}}{k} [1-(\frac{V}{V_{base}})^{^{\frac{1}{1-\frac{g}{Rk}}}}]$

Isothermal equations

For isothermal altitude windows (e.g. the tropopause and stratopause), we can similarly derive the following functions:

$\Large V=V_{base}\exp \frac{gh}{RT}$

$\Large h=\frac{RT}{g} \ln (\frac{V}{V_{base}})$

Example calculation

Let us begin by setting our

V_i

and

V_b

. We will fill an initial volume of gas of 4.2

m^3

, assuming standard conditions of 15

\degree C

and 1 atm at 0 m elevation, which will form

V_{base}

of our first analysis window. For the balloon, we will choose the HY-1600 from Hwoyee, whose bursting diameter

D_b

is stated to be at the minimum 10 m from their website. Converting this to a burst volume, assuming a spherical balloon:

$V_b = \frac{\pi}{6} (D_b)^3=\frac{\pi}{6} (10)^3=523.6 \rm m^3$

Test Troposphere (0-11 km)

This window is non-isothermal with the following properties:

$\begin{array}{lcll} k &=& 0.0065 &\rm K/m \\ T_{base} &=& 288.15 &\text{K} \\ V_{base}&=&4.2 &\rm m^3 \\ H_{lower}&=&0&\rm m\\ H_{upper}&=&11000&\rm m \\ h_{max} &=& 11000 & \rm m \end{array}$

Let us calculate the volume at the top of the window using our non-isothermal volume formula:

$\begin{array}{rcl}V_1&=&4.2[1-\frac{0.0065 \times 11000}{288.15}]^{^{1-\frac{9.81}{287 \times 0.0065}}} \\[10pt] &=&14.15 \: \rm m^3\end{array}$

We can see that $V_1<V_b$ , and so it is evident that burst does not occur in this window. We will use $V_1$ as $V_{base}$ in our next window.

Test Tropopause (11-20 km)

This window is isothermal with the following properties:

$\begin{array}{ccll} T &=& 216.65 &\text{K} \\ V_{base} &=& 15 &\rm m^3 \\ H_{lower}&=&11000&\rm m\\ H_{upper}&=&20000&\rm m\\ h_{max} &=& 9000 & \rm m\end{array}$

Test for burst, as before, by finding the ceiling volume of the window:

$\begin{array}{ccl}V_2&=&14.15\exp(\frac{9.81\times 9000}{287\times 216.65}) \\[10pt] &=&58.53 \rm m^3 \\[10pt] &<&V_b \end{array}$

Again,

V_2<V_b

, so we can advance to the next altitude window.

Test Stratosphere 1 (20-32 km)

This window is non-isothermal with the following properties:

$\begin{array}{rcll} k &=& -0.001 &\rm K/m \\ T_{base} &=& 216.65 &\text{K} \\ V_{base}&=& 58.53 &\rm m^3 \\ H_{lower}&=&20000&\rm m\\ H_{upper}&=&32000&\rm m \\ h_{max} &=& 12000 & \rm m \end{array}$

Test for burst:

$\begin{array}{ccl}V_3&=&58.53[1-\frac{-0.001 \times 12000}{216.65}]^{^{1-\frac{9.81}{287 \times -0.001}}}\\[10pt]&=&389.99 \: \rm m^3 \\[10pt] &<&V_b\end{array}$

Test Stratosphere 2 (32-47 km)

This window is non-isothermal with the following properties:

$\begin{array}{rcll} k &=& -0.0028 &\rm K/m \\ T_{base} &=& 228.65 &\text{K} \\ V_{base}&=& 389.99 &\rm m^3 \\ H_{lower}&=&32000&\rm m\\ H_{upper}&=&47000&\rm m \\ h_{max} &=& 15000 & \rm m \end{array}$

Test for burst:

$\begin{array}{ccl}V_4&=&389.99[1-\frac{-0.0028 \times 15000}{228.65}]^{^{1-\frac{9.81}{287 \times -0.0028}}}\\[10pt]&=&3616.88 \: \rm m^3 \\[10pt] &>&V_b\end{array}$

We see that burst will occur in this window, and we can determine the burst height from the base of the window using our non-isothermal height formula:

$\begin{array}{ccl}h_b &=& \frac{228.65}{-0.0028}[1-(\frac{523.6}{389.99})^{^\frac{1}{1-\frac{9.81}{287\times -0.0028}}}] \\[10pt] &=& 1841.98\: \rm m \end{array}$

Hence, the balloon will burst at height above sea level $H_b$ :

$\begin{array}{ccl}H_b&=&H_{lower} + h_b \\[10pt] &=&32000 + 1841.98 \\[10pt] &=& 33842 \:\rm m \\[10pt] &=& 33.8 \:\rm km \end{array}$

Comparison against the simplistic model

Starting with an initial balloon volume of 1 m³, we can plot the volume vs height within each altitude window for comparison. These graphs below were generated using matplotlib and the formulas explored in previous sections. It can be seen that the refined method gives a lower volume than the simple method in the first 2 windows, but surpasses it at an altitude of roughly 24 km and above. Hence, the refined technique will provide a more conservative result for burst volumes in the stratosphere than a simple burst calculation.