Documents / Report
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.
“R.G.”5 pages
Read from the scan by GLM-OCR; expect the odd misread word.
$$
\text {S i n c e a t T}, \left(\frac {\mathrm {d p}}{\mathrm {d} Z}\right) _ {3} = \left(\frac {\mathrm {d p}}{\mathrm {d} Z}\right) _ {5}
$$
$$
m = - \frac {K}{2} \frac {(D - 2 \Delta Z)}{(D \Delta Z - \Delta Z ^ {2}) ^ {3 / 2}}
$$
and, since at T, $ \Delta p_{z3}=\Delta p_{z5} $
$$
m \Delta z _ {T} = - \frac {K \Delta z _ {T}}{2} \frac {(D - 2 \Delta z _ {T})}{(D \Delta z _ {T} - \Delta Z _ {T} ^ {2}) ^ {3 / 2}} = \frac {K}{(D \Delta z _ {T} - \Delta Z _ {T} ^ {2}) ^ {1 / 2}}
$$
and:
$$
\Delta Z _ {T} \left(2 \Delta Z _ {T} - D\right) = 2 \left(D \Delta Z _ {T} - \Delta Z _ {T} ^ {2}\right)
$$
$$
\Delta Z = \frac {3}{4} D
$$
Then:
$$
\Delta p _ {T} = \frac {K}{\left(\frac {3}{4} D ^ {2} - \frac {9}{1 6} D ^ {2}\right) ^ {1 / 2}} = \frac {K}{\sqrt {\frac {3}{4}} D}
$$
$$
m = \left(\frac {d p}{d z}\right) _ {\mathrm {a i r}} (1 - B) = \frac {K \left(2 \cdot \frac {3}{4} D - D\right)}{2 \left(\frac {3}{4} D ^ {2} - \frac {9}{1 6} D ^ {2}\right) ^ {3 / 2}} = \frac {1 6 K}{3 \sqrt {3} D ^ {2}}
$$
Allowable :
$$
\left(\frac {d p}{d z}\right) _ {\mathrm {a i r}} = \frac {1 6 K}{3 \sqrt {3} D ^ {2}} \cdot \frac {1}{1 - B}
$$
For the example above,
$$
D = 3 0 ^ {\prime}, \quad S _ {f} = \frac {9 0 0}{2}, \quad t = . 0 0 1 \mathrm {i n .}, \quad B = \frac {5 3 . 3}{3 8 6} = 0. 1 3 8
$$
Then:
$$
\left(\frac {\mathrm {d p}}{\mathrm {d z}}\right) _ {\mathrm {a i r}} = \frac {1 6}{3 \sqrt {3}} \cdot \frac {4}{2} \cdot \frac {9 0 0}{2} \cdot \frac {. 0 0 1}{(3 0) ^ {2} \cdot 1 2} \cdot \frac {1}{(1 - 0 . 1 3 8)} \quad \mathrm {p s i} / \mathrm {f t}
$$
Allowable
$$
\begin{array}{l} \left(\frac {d p}{d z}\right) _ {\mathrm {a i r}} = 0. 2 9 8 \cdot 1 0 ^ {- 3} \mathrm {p s i} / \mathrm {f t} \\ = 2 0. 5 5 \mathrm {m b} / \mathrm {f t} \\ \end{array}
$$
This is comparable to an altitude of approximately 18,200 ft. Thus the maximum allowable buoyancy for a 30' diameter, .001" thick polyethylene balloon filled with helium is 440 lb.
## (2) Appendix-Opening Considerations
As an open-apperdix, constant-volume balloon ascends the lifting gas will expand due to the decrease in the pressure Report, cited by the archive. The PDF is mirrored here; the original link is above. The text was read from the page images by GLM-OCR; expect the odd misread word. 882 pages are in the text index: search them above, or from the library's search.