(8)
$$
= \frac{pV_1}{R_a} \left( \frac{1}{T_{g1}} - \frac{1}{T_2} \right)
$$
Then:
$$
\frac{\Delta L}{L_2} = \frac{\frac{1}{R_a} \left( \frac{1}{T_{g1}} - \frac{1}{T_2} \right)}{\frac{1}{T_2} \left( \frac{1}{R_a} - \frac{1}{R_g} \right)}
$$
(9)
$$
= \frac{1}{1 - B} \left( \frac{T_2 - T_{g1}}{T_{g1}} \right)
$$
or for small temperature differences:
(10)
$$
\frac{\Delta L}{L} = - \frac{1}{1 - B} \left( \frac{\Delta T}{T} \right)
$$
the negative sign indicating a loss of lift.
From this equation we may approximate the amount of ballast required to compensate for the loss of superheat of the lifting gas. It is apparent, then, that the amount of superheat gained or lost by a balloon's gas is of extreme importance to the control of the flight.
For this reason a transparent film has a definite advantage over a reflecting fabric. For example, aluminum-coated fabric balloons floating at 40,000 feet have exhibited lifting gas superheat in the neighborhood of 40°C.* Polyethylene balloons, on the other hand, show superheat of approximately 10°C under the same conditions.
Assuming a total weight of 30 kilograms in the balloon system, with helium as the lifting gas ( $B \approx \frac{1}{7}$ ), the following compensation at sunset, or when superheat is lost, will be necessary:
Aluminized fabric:
$$
\frac{\Delta L}{L} = \frac{1}{1 - \frac{1}{7}} \left( \frac{40^\circ}{250^\circ} \right) = 18.7\%
$$
Polyethylene:
$$
\frac{\Delta L}{L} = \frac{1}{1 - \frac{1}{7}} \left( \frac{10^\circ}{250^\circ} \right) = 4.7\%
$$
*This will explain the rapid descent of flight with fabric balloons and will show the need for high rates of ballast flow at sunset with polyethylene balloon flights (see Part III, "Summary of Flights," of this report).
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