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Location: UFOUpDatesList.Com > 2012 > Aug > Aug 25

Re: Chasing Mexican UFOs

From: Michael Tarbell <mtarbell.nul>
Date: Fri, 24 Aug 2012 13:39:53 -0600
Archived: Sat, 25 Aug 2012 05:34:26 -0400
Subject: Re: Chasing Mexican UFOs

>From: Bruce Maccabee<brumac.nul>
>To: post.nul
>Date: Tue, 21 Aug 2012 21:59:34 -0400 (EDT)
>Subject: Re: Chasing Mexican UFOs


>As my article shows, the FLIR system was pointed toward the area
>of the oil field, about 100 miles away, as the plane flew
>eastward for many miles. I denonstrated reasonably good,
>although not perfect, agreement between the FLIR lights and the
>oil burnoff fires that were picked up by satellite that day. At
>the very least I was able to demonstrate that the oil fire
>explanation was plausible. The biggest question was, could the
>mid-IR (3-5 micron) radiation travel that far and be detected. I
>proposed to the MExican Air Force an experiment using their FLIR
>equipped airplane to test the oil field hypothesis, but they
>never did it - or at least never told me they had. Their main
>argument against the fires was "we never saw them before or
>since." This of course, does not rule out the possibility that
>unusual atmospheric conditions ocurred on the day of the

Hi Bruce,

I know little about FLIR, but perhaps you can comment on whether this
first-order analysis, which doesn't much depend on the FLIR details,
holds any water:

Whatever the sensitivity of the FLIR may be (in terms of minimum
detectable wattage), we must consider whether IR radiation from
the atmosphere (essentially a ~300 deg-K black body) would swamp
whatever signal might be coming from the oil fields 100 mi.

The mean spectral radiance of the atmosphere in the 3-5 micron
region is ~100 microwatts/sq.cm per steradian per micron of
wavelength [Ref.1]. So for an angular field of view of 3 degrees
(~.002 steradian), which is roughly the 'medium' field of view
setting for the FLIR in question, there will be ~0.4
microwatts/sq.cm of 3-5 micron radiation being received from the
atmosphere alone. In order to be detectable, presumably the
target of interest must itself produce a non-negligible fraction
of this, let's say at least 4 nanowatts/sq.cm (i.e., 1% of the
background value).

Now consider the two principal mechanisms of radiant intensity
loss between the target and camera: geometric (1/R^2)
divergence, and atmospheric extinction (scattering and

For geometric divergence: A spherically isotropic source in a
vacuum delivering 4 nanowatts/sq.cm from a range of 100 miles
would have a total radiant power output of (4 nanowatts) x (4pi
x [100mi x 1.61x10^5cm/mi]^2) or ~13 megawatts in the 3-5 micron
band. Let's say the earth's surface is a perfect reflector and
reduce that by a factor of 2, to ~6.5 megawatts.

For atmospheric extinction: The extinction coefficient in
maritime air in the 3-5 micron band varies considerably, but the
value at 4 microns is characteristic, with a value ~0.055/km
[Ref.2]. Applying this value across the entire wavelength band,
at 100 miles range the transmitted power is reduced by a factor
of EXP(.055/km x 100mi x 1.61km/mi), or roughly a factor of

Hence the total 3-5 micron radiant power at 100 mi. range
required to noticeably exceed the atmospheric background
radiance at the camera is at least ~7000 x 6.5 megawatts or ~46
gigawatts, which is some 7 times the hydroelectric power output
of the Grand Coulee Dam.

While it would clearly rule out the oil field flare theory, this
number is so enormous that I'm seriously doubting my own
calculation. Can you point out where/if I went astray? Not
really my area of expertise, so
please set me straight if necessary.



[1] Wolfe L. W., & Zissis, J. G., The Infrared Handbook, Office
of Naval Research, Department of Navy, Washington, DC, 1978

[2] Yates, H. W., & Taylor, J. H., "Infrared Transmission of the
Atmosphere", U.S. Naval Research Laboratory, 08 Jun 1960

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