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

Read from the scan by GLM-OCR; expect the odd misread word.

This relationship between loss of lift and loss of superheat is substantiated by analysis of Flight 94. From the rate of descent the unbalance (using the equation of Clarke and Korff, see page 33) is in the neighborhood of 5 kilograms. Although there was no temperature measurement on this flight, a previous flight of this type indicated a superheat of approximately $ 4 0^{\circ} \mathrm{C} $ . By equation (10), with a gross load of 52 kg., the unbalance caused by loss of all of this superheat would be 9.7 kg. It is believed that ventilation past the balloon during a low velocity descent before operation of the ballast mechanism caused loss of superheat. Since this loss caused greater descent, and thus more ventilation, superheat was lost. An enormous rate of ballast flow would have been required to check descent.

## D. Adiabatic Lapse Rate

One of the causes of temperature difference between the lifting gas and free air during rise or descent of balloon systems is the difference in lapse rates of air and the lifting gas. The adiabatic lapse rate is that temperature change caused by adiabatic expansion or compression of a gas during ascent or descent through a given vertical distance. The actual lapse rate of the lifting gas is the adiabatic lapse rate plus the effects of conduction and radiation. The adiabatic lapse rate is defined as:

$$
L R = \frac {A g}{C _ {p}}
$$

where:

A = 2.39 x $ 1 0^{-8} $ cal / erg

$ C_{p} $ specific heat at constant pressure

g = acceleration caused by gravity

In the metric system for helium, ( $ C_{p}=1.25 \frac{\mathrm{cal}}{\mathrm{^\circ C / g m}} $ ):

$$
L R = - \frac {9 8 0 \times 2 3 9 \times 1 0 ^ {- 3}}{1 . 2 5} = - 1. 8 7 ^ {\circ} C / k m
$$

or:

$$
L R = -. 5 7 ^ {\circ} C / 1 0 0 0 f t
$$

The adiabatic lapse rate for air, $ C_{p}=0.239\frac{\mathrm{cal}}{\mathrm{^\circ C / gm}} $ :

$$
L R = - \frac {9 8 0 \times 2 3 9 \times 1 0 ^ {- 3}}{0 . 2 3 9} = - 9. 8 ^ {\circ} C / k m
$$

or:

$$
L R = - 2. 9 8 ^ {\circ} C / 1 0 0 0 f t
$$