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From: Joel Henry <jhenry@visi.com>
Date: Tue, 24 Aug 1999 15:01:35 -0500
Fwd Date: Tue, 24 Aug 1999 16:57:16 -0400
Subject: Re: Socorro: The Zamora 'Insignia'


 >Date: Sun, 22 Aug 1999 23:49:51 -0400
 >From: Bruce Maccabee <brumac@compuserve.com>
 >Subject: Re: Socorro: The Zamora 'Insignia'
 >To: UFO UpDates - Toronto <updates@globalserve.net>

 >I thank Dave Rudiak for providing a quantitative analysis of the
 >balloon "problem."   It is a further illustration of th fact
 >that a phenomenon which forms the core idea of a "prosaic
 >explanation" must be a known physical phenomenon and must
 >therefor obey known physical laws.

 >>>>From the fundamental gas law

 >>PV = nRT,

 >P = DRT would have been an easier starting point, where D
 >is the Density.

 >As Dave pointed out, pressure equilibrium at the opening of the
 >balloon (when the hot air is no being injected) means that P
 >inside = DiRTi = P ountside = DoRTo. If the inside temperature
 >is 20% higher than the outside, then Ti = 1.2 To and Di =
 >(1/1.2)Do. (These are nominal numbers, of course.)  The buoyancy
 >force is proportional to the difference in mass which is the
 >difference in the product DV: m supported by buoyancy = V(Do -
 >Di) = V(Do - 0.8Do) = VDo(.2), For a 500 m^3 volume and air
 >density as 1.3 Kg/m^3 the mass supported is 500x1.3x0.2 = 130
 >kg. Multiply by 2.2 to get "pounds of mass " :  286 lbs.

 >As Dave pointed out, even with lightweight balloonists, like 100
 >lbs each, there is not enough supported weight left to allow for
 >the weight of the balloon and the weight of the the gondola
 >,fuel, fuel burner (to heat the air) etc. Moreover, to get this
 >130 Kg difference required V = 500m^3 and that crorresponds to
 >about 5 m in radius or 10 m in diameter.... abou 30 feet.  At a
 >distance of 50 feet the angle subtanded at the eyes of Zamorra
 >by the balloon this big would be about 30 degrees.   This is
 >HUGE...... By way of comparison, an 8 ft car at 50 ft subtends
 >an angle of only about 9 degrees. One may well expect that
 >Zamorra, being a policeman, was well familiar with the
 >appearance and angular size of a car at 50 ft. Hence if he
 >compared it to a car in size we might conclude that the angular
 >size was much closer to 10 degrees than to 30 degrees.

 >Yes, he compared it to a balloon. Probably because of shape and
 >being in the air. A child's balloon? Or perhaps he was thinking
 >of a weather balloon (ca 6 ft in diameter). I'm sure he wasn't
 >thinking of a 30 (or more) diameter of a hot air balloon with
 >basket/gondola, etc.

Dr. Maccabee and the rest,

You have totally missed the most important fact that makes this
whole math issue a moot one. The tripod landing holes were made
by an object weighing TONS, compressing and compacting the sand
and earth below it. It would take a balloon the size of the
freaking Hindenburg to get that much weight into the sky!! I
don't even need to do the math to figure that one out! BEFORE
you start belching pointless theories, please go over ALL the
facts, then you won't be wasting our time or yours. More square
pegs pounded into round holes..... :(

Joel Henry Minnesota MUFON

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