The Roswell Report: Fact vs. Fiction in the New Mexico Desert

Temperature effects were discussed previously in this report. Those discussions on superheat and adiabatic temperature change will apply to the general equation. In general, temperature of the free air and lifting gas can be measured to a fair degree of accuracy.

Balloon volume at any time is a function of original full balloon volume plus the summation of all the changes in volume due to pressure and temperature changes and loss of lifting gas. It will also be affected by loss or gain of air by the balloon through diffusion and intake of air through the appendix. The nonextensible balloon will have a maximum volume and thus any changes tending to increase the gas volume to a value greater than the balloon volume will result in a valving of the excess lifting gas into the air, or (in the case of a balloon which will carry internal pressure) a pressure increase of the lifting gas.

It is for this reason that a non-extensible balloon is said to be in a state of stable equilibrium in a direction of greater altitude when it is full. However, in a direction of lesser altitude, and with the case of a partially full floating balloon, the system is in a state of neutral equilibrium.

Composition of the lifting gas will change due to contamination of the lifting gas by the entry of air into the balloon, either by the flow of air through the appendix opening or by diffusion of air into the balloon. We may then modify our term for density of the lifting gas to include a term for the pure gas and a term for the contaminating air. Using the method of partial volumes, we may equate the density of the lifting gas at any time by:

where:

$$
\rho_ {g} = \frac {P _ {g}}{V _ {b} T _ {g}} \left(\frac {V _ {p}}{R _ {p}} + \frac {V _ {a}}{R _ {a}}\right)
$$

$ P_{g} $ pressure of the lifting gas

$ V_{b} $ = total lifting gas volume

$ V_{\mathbf{p}} $ volume of pure lifting gas in balloon

$ V_{\sigma} $ volume of air in balloon

$ R_{g}= $ specific gas constant of pure lifting gas

$ R_{a} $ specific gas constant of air

Then, calling $ \frac{V_{p}}{V_{b}}=x_{p} $ and $ \frac{V_{a}}{V_{b}}=x_{a} $ (here we see that since

$ V_{p}+V_{a}=V_{b}, $ $ x_{p}+x_{a}=1 $ we may equate:

$$
P _ {g} = \frac {P _ {g}}{T g} \left(\frac {x _ {p}}{R _ {p}} + \frac {x _ {a}}{R _ {a}}\right)
$$