$$
\Delta G = G \times \frac{\Delta T}{T} \times (1 + K) \, K
$$
where $\Delta G$ = loss of lift
$G$ = gross load (balloon weight plus equipment load)
$\Delta T$ = mean temperature difference in lifting gas before and after sunset
$T$ = free air temperature
$K$ = specific gravity of lifting gas, relative to air
The specific gravity of 98% helium, diluted with air, and with respect to air, is 0.157. It may be noted that with a lower specific gravity of a gas, lower ballast corrections are required. Hydrogen, for example, requires half the ballast which helium requires for the same temperature differential. At high altitudes, a difference of 40°C may be expected in the temperature of the lifting helium from day to night. This would correspond to a loss of lift at sunset, on a General Mills 20-foot balloon, of about 550 grams.
F. Internal Pressure
The maximum internal pressure which can be held within a spherical container is given by Timoshenko⁸:
$$
P = \frac{2 S_u \times t}{r}
$$
where $S_u$ is the ultimate strength of the material in tension, $t$ is thickness of the material and $r$ is the radius of the spherical shape. Applying this equation to a polyethylene film, such as used in the General Mills 20-foot balloons, $S_u$ at room temperature = 1900 psi., $t = 0.001"$ , and $r = 10$ ft., giving the maximum pressure, $P = 0.032$ psi. This pressure is equivalent to about 1.1 inches of water, or 2.5 mb. This small bursting pressure necessitates proper inflation and load values to prevent the balloon's
- 20 -