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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…
The actual atmospheric distribution, however, does not indicate an adiabatic lapse rate for air but rather a lapse rate which varies with altitude. For the troposphere the lapse rate of the atmosphere averages $ - 1.98^{\circ}\mathrm{C} / 1000 $ ft. It may be shown then that in the troposphere a rising balloon will get warm with respect to the air (neglecting ventilation and radiation effects) at a rate of $ 1.98 - .57 = 1.41^{\circ}\mathrm{C} / 1000 $ ft. In the tropopause the lapse rate of the atmosphere is zero. Thus the lifting gas (if helium) will cool relative to the air at a rate of $ .57^{\circ}\mathrm{C} / 1000 $ ft.

Similarly, in the stratosphere, the lifting gas will cool relative to the air at a rate of 2.24 + .57 = 2.81 $ ^{\circ} \mathrm{C} / 1 0 0 0 $ ft. This effect is plotted as Figure 22.

Figure 22. Lapse rate of air and helium.

Here, below point A, the lifting gas will be warmer than the air. Above point A, the lifting gas will be cooler than the air. The effect of this temperature difference on the lift (as shown in the previous section) is approximately $ \Delta L=1\frac{\Delta T}{1} $

$$
\Delta L = L \frac {\Delta T}{T} \frac {1}{(1 - B)}
$$

Thus, as a balloon system passes through point A, it will have less lift than at release. This effect has been observed on several flights, where a balloon system slowed down during ascent through a temperature inversion.

Since the effect of the sun in heating the lifting gas decreases the effect of different lapse rates, the effect is not as noticeable during the day as at night. At night the balloon system may pass through an inversion, lose its lift, and remain at an altitude much below its estimated floating altitude until warmed by the sun's rays at sunrise.

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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.