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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.
“The Pentagon”12 pages
## (2) Solution, Migration and Evaporation through Film
A very slight amount of lift is lost through solution of the gas into the balloon film, migration through the film and evaporation into the atmosphere. The rate of this type of diffusion is a function of the characteristics of the lifting gas and the partial pressure involved. Since the lifting gas is assumed to be very nearly pure, the partial pressure is merely the pressure of the atmosphere in which the balloon is floating. This method of diffusion need not be considered when examining the loss of a balloon's lifting gas since it is of a low enough value to be insignificant as compared with the loss of gas by leakage through openings in the film.
Tests have indicated that this type of diffusion through .001" polyethylene has a value of approximately 4 liters/meter2/day. At sea level this is equivalent to 5.32 gm/hr. for a 20-foot diameter balloon. At 40,000 feet MSL the value would be approximately 1 gm/hr.
## (3) Diffusion through Appendix
We have seen that there is no pressure difference across the open appendix of the balloon during floating. Therefore, the loss of lifting gas through this appendix (except when the balloon is rising and gas is being valved out of the appendix) can be only by true intermolecular diffusion of the gas into the atmosphere and air into the lifting gas. The expression for loss of lifting gas by diffusion is similar in form to the expression for transfer of heat through a given distance by conduction:
$$
\frac {d N}{d t} = - D \frac {d N}{d z} d y d x
$$
where:
$ \frac{dN}{dt}= $ time rate of transfer of molecules of gas across the area dy dx in direction 2
D = a coefficient of diffusion, dependent upon viscosity and density of the gas involved (D=c $ \frac{7}{9} $)
$ \frac{dN}{dz}= $ variation of molecular concentration with variation in direction z
dydx = the differential term for area.
Then, since a molecule of lifting gas has a given weight, we may state that: 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.