$$
\Delta G = G x \frac {\Delta T}{T} x (1 + K) K
$$
where $ \triangle G= $ loss of lift
G = gross load (balloon weight plus equipment load)
$ \triangle 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 $ 4 0^{\circ} \mathrm{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 Timoshenko $ ^{8} $ :
$$
P = \frac {2 S _ {u x t}}{r}
$$
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