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

USAF / The Black Vault · 1995 · 882 pages · text by OCR

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.

  • p. 193 …overhead and followed the balloons out to sea. I have no idea about the results that…
  • p. 200 …the regular 334 that we had at sea level. From that they could deduce the temperature…
  • p. 243 …actually located on the jurisdictional lines between Sea Girt and Springlake, New Jersey. It was an…
  • p. 244 …The Sea Girt Inn? A: Exactly. That's where John had his office, and I was…
  • p. 320 …alone is about 24,000 miles at sea level, and about 4500 miles at 45,000…
  • p. 325 …twenty-five (25) feet at their largest sea-level diameter. The sonic unit was a combination…
  • p. 378 …per hour when one-fifth inflated at sea level). One other type of balloon which has…
  • p. 402 …Let us, then, compare the rate of leakage at any given altitude with leakage at sea…
  • p. 404 …The leakage at any altitude may be expressed as a function of leakage at sea level…
  • p. 405 Comparing rate of leakage at 40,000 feet with leakage at sea level: $$ \frac {L _ {4…
  • p. 407 …If a 20-foot diameter balloon $ \frac{1}{1 0} $ full were tested at sea level…
  • p. 408 …At sea level this is equivalent to 5.32 gm/hr. for a 20-foot diameter…
  • p. 414 …Using the rules of subsonic aerodynamics, Picard suggests that air at sea level escaping at 1333…
  • p. 415 …air at sea level (lb./ft. $ ^{3} $ ) 14.7 = pressure of air at sea level (psi…
  • p. 432 …to about 20 millibars and increased to sea-level pressure at different temperatures. The most comprehensive…
  • p. 563 …The height above mean sea level as determined from pressure measurements used in this work with…
  • p. 644 …point at which the radiosonde reaches the sea surface. ## 2. Earlier attempts There have been numerous…
  • p. 645 …The balloons floated between the surface and 30,000 ft above sea level; those which reached…
  • p. 704 …Met Gifford who has 90' sea rescue boat this project is planning to use. Stayed at…
  • p. 719 …Worzel working on gravity at sea. Saw Geo Woollard and the Ryders. Woollard after Guggenheim fellowship…
  • p. 779 …the launching site is markedly different from sea level, a shift in this curve is needed…
  • p. 817 …balloon at all times with respect to sea level. On this curve also it is customary…
  • p. 825 …The height above mean sea level as determined from pressure measurements used in this work with…
If a balloon is teardrop in shape rather than spherical, the curve would be modified so that the value of $ C_{D} $ , for a given Reynolds number would be lower. In this case the sudden drop in $ C_{D} $ as Reynolds number increases (the change from viscous to turbulent flow) would occur at a lower Reynolds number.

We have thus far in our discussion assumed that there is no vertical motion of the air surrounding the balloon system relative to the coordinate z . However, this is not necessarily the case under actual conditions. In many instances vertical air movement is found in the atmosphere. Therefore, we must introduce a term to allow for this vertical air movement. In equation (2) this term was indicated as $ \pm F_{A} $ , the external atmospheric force.

We may consider this vertical air movement in terms of a velocity D $ \zeta $ . Then the vertical velocity of the balloon system relative to the air surrounding the system will be the difference between the velocity of the balloon relative to the absolute altitude Dz and the velocity of the surrounding air relative to the absolute altitude This may be equated as Dz-D $ \zeta $ , where Dz and D $ \zeta $ are both considered positive in the direction of increase of altitude.

The total force due to the drag, or friction will be:

$$
F _ {D} + F _ {A T M} = C _ {D} \frac {\rho}{z} A (D z - D \xi) ^ {2}
$$

where the notations are those used previously, except that now $ N_{R}=\frac{(Dz-D\zeta)d\rho}{\mu} $ . The relationship between $ N_{R} $ and $ C_{D} $ will be those used previously.

The force due to buoyancy of the lifting gas $ F_{b}=V_{b} \left( \dot{\rho}_{a}-\dot{\rho}_{g} \right) $ where:

$$
V _ {b} = \text {b a l l o o n v o l u m e (f t .} ^ {3})
$$

$ \rho_{a}, \rho_{g} $ density of the air and lifting gas, respectively (1b./ft. $ ^{3} $)

This term may also be stated as: F = $ V_{b}\left(\frac{P_{g}}{R_{a}T_{a}}-\frac{P_{g}}{R_{g}T_{g}}\right) $ where:

$ p_{a}, p_{g} = $ pressure of air and lifting gas

$ R_{a}, R_{g} = $ specific gas constant of air and lifting gas

$ T_{a}, T_{g} $ = temperature of air and lifting gas

The changes that will take place in this expression are those due to a temperature difference between the lifting gas and the free air, change in volume of the balloon due to loss of lifting gas, change of the gas constant of the lifting gas due to dilution with air, and (in the case of a balloon that will hold an internal pressure) pressure difference between lifting gas and surrounding air.

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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.