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The Roswell Report: Fact versus Fiction in the New Mexico Desert was published by Headquarters United States Air Force in 1995. The Black Vault distributes this copy. It reproduces the report by Col. Richard L. Weaver and the synopsis by 1st Lt. James McAndrew, both written after a General Accounting Office inquiry requested by Representative Steven Schiff. The Air Force search found no evidence of an extraterrestrial craft or crew. It concluded that the Roswell debris most likely came from NYU Flight No. 4, a Project MOGUL balloon train.
“Fort Worth”30 pages
(8)
$$
= \frac {p V _ {1}}{R _ {a}} \left(\frac {1}{T _ {g _ {1}}} - \frac {1}{T _ {2}}\right)
$$
Then:
$$
\frac {\Delta L}{L _ {2}} = \frac {\frac {1}{R _ {a}} \left(\frac {1}{T _ {g _ {1}}} - \frac {1}{T _ {2}}\right)}{\frac {1}{T _ {2}} \left(\frac {1}{R _ {a}} - \frac {1}{R _ {g}}\right)}
$$
(9)
$$
= \frac {1}{1 - B} \left(\frac {T _ {2} - T _ {g _ {1}}}{T _ {g _ {1}}}\right)
$$
or for small temperature differences:
$$
\frac {\Delta L}{L} = - \frac {1}{1 - B} \left(\frac {\Delta T}{T}\right)
$$
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 $ 4 0^{\circ} \mathrm{C}. $ * Polyethylene balloons, on the other hand, show superheat of approximately $ 1 0^{\circ} \mathrm{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:
$$
\frac {\Delta L}{L} = \frac {1}{1 - \frac {1}{7}} \left(\frac {4 0 ^ {\circ}}{2 5 0 ^ {\circ}}\right) = 1 8. 7 \%
$$
Polyethylene:
$$
\frac {\Delta L}{L} = \frac {1}{1 - \frac {1}{7}} \left(\frac {1 0 ^ {\circ}}{2 5 0 9}\right) = 4. 7 \%
$$
*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). Report, cited by the archive. The PDF is mirrored here; the original link is under it. The text was read from the page images by an OCR model; expect the odd misread word. 882 pages are in the text index: search them above, or from the library's search.