Note:
The following paper was published in the Journal of Scientific
Exploration, Vol. 13, #2, pg. 199, 1999. It was originally presented at
a symposium of the Society for Scientific Exploration in 1984.)
Full original report and images at
searchable text as
follows
OPTICAL POWER OUTPUT OF AN UNIDENTIFIED
HIGH ALTITUDE LIGHT SOURCE
by Bruce Maccabee (c) 2000 by B Maccabee
ABSTRACT
A Royal Canadian Air Force pilot while flying at
an altitude of about 11 km saw and photographed a very
bright, disclike object that was remaining stationary
near a thunderhead. An analysis of the photograph
suggests that it would have been radiating in excess of
a gigawatt of power within the spectral range of the film.
INTRODUCTION
At about 7:20 PM, MDT (about 20 minutes before sunset), on Aug. 27,
1956 a Royal Canadian Air Force pilot was flying nearly due west over
the
Canadian Rockies near Ft. MacCleod, Alberta (49.5 degrees latitude,
113.5
degree longitude). He was flying at 36,000 ft (about 11 m) in the
second
position (far left side) of a formation of four F86 Sabre jet
aircraft.
While approaching a large thunderhead (cumulonimbus) at a ground speed
of
about 400 kts (740 km/hr) he saw, at a much lower altitude, a "bright
light
which was sharply defined and discshaped" or "like a shiny silver
dollar
sitting horizontal" (see Appendix 1 below).
As he continued westward the sighting line to the object rotated
backward to an “eight o’clock low” position before he lost sight of it,
indicating that it was stationary in the lee of the anvil of the
thunderhead
at an altitude considerably below the plane. (Note: his recollection of
the
object being at his left side conflicts with the direction to the sun
as
determined from the photo; see below). It was below the upper layer
of clouds but above the lower layer of clouds which, according to the
pilot’s weather report, were at 10,000 to 13,000 ft (3 to 4 km). The
object, which was viewed against the dark purplish background of the
lower
cloud layer, appeared to be "considerably brighter than the sunlight."
As
he flew past he decided to take a photo of the object. He had to
“quarter
roll” the aircraft in the direction of the object in order to take the
photo. The exact direction to the object at the time of the photo is
not
known. However, it probably was more than 30 degrees north of the
direction
to the sun (276 deg. azimuth, 8 deg. elevation) because the sun was
apparently to the left of the left edge of the film which in turn, was
about
28 degrees to the left of the object. (This latter statement is based
on
the locations of shadows on clouds made by other clouds at the lower
left
side of the photo). The pilot estimated the sighting duration at
between 45
sec (2) and 3 minutes (1). After pointing the object out to the flight
leader the pilot took a photograph of it (Figure 1). (Appendix 1
provides
further details of the sighting.) The photograph, a 35 mm color slide
(Kodachrome), is the subject of the analysis presented here. This is a
scan
of an excellent print made from the slide by the Jet Propulsion
Laboratory.
A few minutes later, because he had his camera out or its container, he
decided to take a picture of his flight. Figure 2 confirms his location
relative to the other planes.
ANALYSIS
The exact nature of the object has not been determined. An initial
suggestion that it was merely a brightly illuminated small cloud (1)
has
been ruled out for two reasons. The first is that it is just as bright
on
the right (east) side as it is on the left side whereas clouds in the
photo
are noticeably darker on their right sides, a fact that is consistent
with
the sunlight coming from the west. The second reason is that portions
of
the object were brighter than the brightest clouds. Klass (2 ) has
suggested that the object was a plasma or something akin to ball
lightning
and Altschuler (3) included a discussion of the object in an article on
ball
lightning. Whatever the nature of the object, it would be of interest
to
have order of magnitude estimates of its radiance (watts/steradian/cm^2
or
w/st/cm^2), radiant emittance (w/cm^2) and total power output within
the
spectral range of the film. (Note: the symbol ^ as used here means
exponentiation or "raised to the power of".) The radiance in the
direction
of the camera can be estimated from the film density of the image
combined
with published film characteristics and with suitable assumptions about
the
camera settings and the range to the object. By also assuming the
object to
be a Lambertian radiator with constant emittance over its surface one
can
estimate the total radiant emittance and the total radiated power
within the
spectral range of the film.
The radiance is found by solving a standard photographic equation (4),
corrected for the effects of atmospheric attenuation (5) as shown in
Appendix 2:
L = (4/p)E(f#)^2 e^[(ba) /cos(A)]/Tcos^4(B) 1)
where
E = H/t. 2)
and where e is the natural base, 2.71828.
In these equations L is the radiance of the object in w/st/cm^2, E is
the irradiance on the focal plane of the camera lens in w/cm^2 and f#
is the
ratio of the focal length to the lens diameter (set by the operator of
the
camera). The factor e^[(ba)/cos(A)] corrects for atmospheric
attenuation
along the slant path, at angle a, relative to vertical from the object
to
the camera. In this correction factor, b is the optical thickness of
the
atmosphere from the ground to the altitude of the plane and a is the
optical
thickness from the ground to the altitude of the object. T is the
transmission of the optics (lens, aircraft window), B is the angle
between
the optic axis of the camera and the direction to the object, i.e., the
angle corresponding to the offset of the image from the center of the
photo,
H is the film exposure level in j/cm^2 at any particular location
within the
image and t is the shutter time in seconds.
The quantities which go into Eq. (1) and (2) are not definitely known.
However, reasonable estimates have been made in order to carry out the
calculations. Based upon the camera settings that are recommended for
the
film (Kodachrome ASA 10) when used under the known lighting conditions
(bright daylight) it is estimated that f# = 8, although it could have
been
one stop above or below this (i.e., either 5.6 or 11). A camera is
designed
so that an increase in f# by one stop increases the aperture area by
two and
a decrease in f# by one stop divides the aperture area by two.
Therefore an
uncertainty of one fstop setting corresponds to a factor of two (or
1/2)
uncertainty in the lens area and ultimately in the calculated radiance.
The values of b and a in the attenuation correction factor depend upon
the particular altitudes and are weighted averages over the spectral
range
of the film. From attenuation data (5) for a clear atmosphere the
difference ba has been determined for the airplane altitude in
combination
with two assumed altitudes of the object. The values of ba and angle A
have been estimated in the following two ways.
The first way of determining ba and the zenith angle makes use of the
statements by the pilot that (a) the object looked like a horizontal
coin
(i.e., a thin disc with its axis vertical) seen from an angle to its
axis
(the zenith angle) and (b) he guessed the distance was about 3 nautical
miles (5.5 km). Statement (a) implies that the image of the object
would
have a roughly elliptical shape. The aspect ratio of the ellipse (minor
/
major axis) would provide an estimate of the angle between the line of
sight
and the plane containing the disc (the complement of the zenith angle).
The brightest central part of the image does have a roughly elliptical
shape which is moderately consistent with the shape of a thin disc
viewed
obliquely (6). A good overall fit to the image is obtained with a 70
degree
ellipse (i.e., a circle viewed at an angle of 70 degrees from the axis
of
the circle or 20 degrees from the plane of the circle) with a major
axis of
the image being 1.28 mm long. Thus the depression angle of the line of
sight from the plane to the disc would be about 20 degrees, assuming
that
the disc itself was horizontal. The pilot stated that he had to roll
the
plane a bit in order to take the picture through the canopy indicating
that
the depression angle may well have been about 20 degrees or greater.
Assuming again that the disc was in a horizontal plane, then the angle
from
the vertical axis to the airplane, A, was about 70 degrees. The 20
degree
depression angle combined with the approximate 6 km range to the object
yields an altitude of (11 km  6 km x sin 20) = 9 km. The difference
ba
is found for an optical path from 9 to 11 km by roughly averaging the
optical thickness of the atmosphere over the sensitive spectral band of
the
film using tables in ref. 5. One finds that (ba) is approximately
0.03.
Combining this with A = 70 degrees, (ba)/cos A = 0.09 and e^0.09 =
1.09.
A second estimate of the attenuation correction is obtained by using
the same angle a as found above but using a lower altitude for the
object.
The justification for using a lower altitude is based on two
considerations:
(a) the distance estimate given by the pilot was only a guess; the
distance
could have been greater than 6 km, and (b) the film imagery seems to
show
that the object illuminated the clouds just below it. (Note: the
illuminated clouds are below the bright horizontal linear structure
which is
just below the elliptical image.) This would place the object
relatively
close to the lower cloud layer, which was at an altitude of about 3 to
4
km. Obviously the exact altitude of the object is unknown, but for the
purposes of this second attenuation calculation it has been set equal
to 4
km. At this altitude the range to the object is about 20 km and
(ba)/cos(70), as found from data in ref. 5, is about 0.29. The
attenuation
correction factor is e^0.29 = 1.34.
Each of the attenuation correction factors quoted above was calculated
for clear air. Although there was water vapor in the nearby clouds, it
is
probable that the vapor between clouds was not sufficient to make the
attenuation correction factor much larger than the largest value listed
above. The optical transmission, T, of the camera lens and airplane
window
is estimated to have been about 0.7 due to glass surface reflection
losses
(about 4% loss at each surface), although it might have been slightly
greater. The shutter time is less certain. It may have been 1/125 sec,
which, along with f/8, is the value recommended by the manufacturer
(Kodak)
for ASA 10, film under daylight conditions. However, it may also have
been
as long as 1/60 sec or as short as 1/250 sec. For these calculations t
=
1/125 = 0.008 sec = 8 x 10^(3) = 8E3 sec (note the definition of
exponential notation, 1E1 = 10, 1E2 = 100. 1E3 = 1/1000). The angle b
in
Eq. (1) is only about 9 degrees so the cos^4(B) factor is about 0.95
The only remaining quantity to be determined is the average exposure,
H, over the image. H is related to the optical densities within the
image
of the three color forming emulsion layers which make up the film.
(Density
is inversely related to the brightness of an image. In the case of
slide
film the density is the negative of the logarithm to base ten of the
transmission factor of the film image.) Since the slide is color
reversal
film, image density decreases as the exposure increases. The image of
interest appears quite overexposed and white (i.e., devoid of color).
The
color of the image of the object is the same as that of the images of
the
brightest white cloud areas which are also overexposed. Since the image
of
the object is white one may assume that all three color forming layers
were
exposed to approximately equal amounts of energy. The minimum neutral
density at several locations within the image is about 0.11, which is
only
slightly larger than the density of completely overexposed film. This
density is also about 0.03 lower than the density of the brightest
(white)
cloud area. Because lower density corresponds to greater exposure,
certain
portions of the object were brighter than the clouds. The average
density
over much of the image is approximately 0.12 (see Figure 3).
An order of magnitude estimate can be made of the value of H which
could create a density as low as 0.12. The estimate is made by
combining
the exposure curves [film density vs log(H)] with the spectral
sensitivity
curves that are published by Kodak. The spectral sensitivity curves
indicate how the monochromatic sensitivity varies with wavelength for
the
three emulsions, where sensitivity is the inverse of the energy density
in
j/cm^2.
The sensitivity values are given for a decrease in film density of 0.3
units from the maximum density which is between 3.2 and 3.5 units.
Using
the integral form of the van Kreveld addition law (7) one can determine
the
sensitivity of a film to a continuous spectrum of light.
Assuming for simplicity a flat spectrum over the spectral range of each
emulsion one finds that the shorter wavelength emulsion (blueviolet to
green, 350500 nm wavelength) is roughly three times more sensitive
than the
mid range (green to redorange, 500600 nm) or long wavelength
(redorange
to deep red, 600700 nm) emulsion. Assuming that the object radiated a
spectrum that would cause equal density changes in the three layers
thus
producing a white image, the approximate energy density needed to
produce a
film density decrease of 0.3 units would be 1E7 j/cm^2. The exposure
curves for the three emulsions indicate that to produce a film density
decrease of about 3.2 units, i.e. to cause the film density to decrease
to a
value comparable to that of the image of the object, the energy density
would have to be about 1000 times greater. Therefore an order of
magnitude
estimate of H is lE4 j/cm^2, corresponding to the average image
density of
about 0.12.
Inserting the above values into Eq.(1) and (2) yields a radiance of
about 1.7 w/st/cm^2 if the object were at 6 km distance and about 2.1
w/st/cm^2 if it were at the 20 km distance. (The difference in the
calculated radiances arises because of the atmospheric attenuation
correction.) These values are so close to each other that the radiance
will
be approximated as 2 w/st/cm^2 regardless of the assumed distance.
If the object were a Lambertian radiator then the radiant emittance, W
= pi L, would be about 6 w/cm^2. This value of W can be produced by a
2450K
blackbody with emissivity of 100% since it radiates about 3% of its
total
200 w/cm^2 emittance within the bandwidth of the film (8). However, the
color balance of the film suggests that at that temperature the image
would
have a distinctly reddish hue. On the other hand, a 3200K greybody with
an
emissivity of about 10% would provide approximately the same radiant
emittance in the bandwidth of the film, but the color temperature would
be
high enough to produce a white image. A 10% emissivity is typical of
some
polished metals (8) which, even though hot enough to radiate, might
give the
visual appearance of a "shiny silver dollar sitting horizontal."
It is possible to estimate the total power output of the object within
the spectral range of the film by assuming a specific shape for the
object
and also that it was a Lambertian emitter with constant emittance over
its
surface. For simplicity the following calculations have been done
assuming
that the object was a cylinder with a length l/l0 of its diameter.
Assumptions of other shapes will yield comparable or higher power
outputs.
The diameter of the cylinder is estimated from the major axis of the
ellipse fitted to the image length (0.00128 m) combined with the focal
length of the camera (50 mm = 0.05 m) and an assumed distance to the
object.
Using the simple imaging relation established by the camera lens, if
the
object had been 6 km distant then its diameter would have been (6000
m/0.05
m) x 0.00128 m or about 150 m. The total radiating surface area (top,
bottom and the cylindrical surface) would have been about 42,000 m^2 =
4.2E+8 cm^2. Multiplying this by the emittance, 6 w/cm^2, yields about
2.5E+9 w within the spectral band of the film. If the object had been
at 20
km distance the diameter would have been about 500 m, the surface area
about
4.7E+9 cm^2 and the power output would have been about 3E+10 w within
the
spectral band of the film. Of course, the total power emitted over all
frequencies might be much greater.
A second method of estimating the total power, a method which treats
the object as a “point” source emitter and which requires an estimate
of the
total energy deposited on the image, is presented in Appendix 2. This
second method yields powers of 1.8E+9 w (instead of 2.5E+9) and 2.4E+10
w
(instead of 3E+10). Since both methods require assumptions about the
data
it is difficult to decide which is more likely to be correct. However,
the
rather close agreement suggests that, at the very least, the numerical
results to be calculated by either method for any chosen set of values
for
the f#, the shutter speed and the distance to the object would be
acceptably
accurate.
The radiation power levels calculated here make it difficult to sustain
the argument that the object was a plasma as ordinarily understood.
Neither
the origin of such a plasma, its size, its duration nor its total power
output seem consistent with known plasma phenomenology. Its location
near a
huge thunderhead suggests that some transient electrical effect of the
storm
(e.g. lightning bolt) might have created it, but this would not explain
its
duration of many tens of seconds. Assuming a minimum estimated sighting
duration of 45 sec and a minimum estimated distance of 6 km the energy
radiated would have been on the order of 1E+11 j (about two orders of
magnitude greater than the energy in a typical lightning stroke3 ) and
the
volume energy density of the assumed (cylindrical) plasma would have
been
about 4E+5 j/m^3. An energy density this great could be attained with
single ionization of a portion of the air at 9 km altitude (3).
However,
the recombination time for a plasma is very short, typically 100 ms or
less.
Therefore energy storage (ionization) during a transient event (e.g., a
lightning bolt) followed by relatively slow energy release during
recombination could not account for the sighting duration. This implies
that there must have been a continuous power source either within or
outside
the plasma. To determine whether or not such a power source exists
would
require theoretical and speculative analyses which go beyond the scope
of
this paper.
An alternative phenomenon which may be considered is ball lightning or
"kugelblitz." Ball lightning has some characteristics of an ordinary
plasma
but it exists for much longer periods of time and therefore must have
some
(unknown) energy storage mechanism "built in" or else it is continually
powered in some (unknown) way by an external power source. According to
independent surveys by McNally and Rayle totalling 627 sightings (see
ref.
1, pgs. 32 and 3538 for details) observers of ball lightning usually
report
sighting durations on the order of seconds. although about 8% have
reported
durations exceeding 1/2 minute and about 2% have reported durations
greater
than two minutes. Therefore, because of the duration of this sighting
one
might consider the possibility that the photographed object was a ball
lightning. However, typical ball lightnings are roughly spherical blobs
with diameters on the order of tens of cm, although about 4% of the
observers have reported sizes greater than 1.5 m and at least one
observer
(a pilot, see ref. 3) has reported diameters up to 30 m. Since the
photographed object was well over 100 m in diameter (depending upon the
assumed distance), at the very least this object was a highly unusual
example of ball lightning. Unfortunately there is no good theory that
explains how even a typical ball lightning could exist. Therefore there
is
no way to scale predictions for ball lightning from the typical size
under
1.5 m up to the size of the photographed object and to thereby
determine
whether or not it was ball lightning.
By way of comparison with the radiation of this object (1E+9 to 1E+13
watts for 45 seconds or more), lightning radiates 1E+13 watts for 1E4
sec,
1 kiloton of TNT “radiates” 1E+13 watts for 1E3 sec, the Grand Coulee
Dam
power station generates 1E+10 watts, a nuclear rocket can generate
exhaust
power at a rate of about 1E+10 watts and the Saturn space rocket
generates
1E+11 watts at maximum thrust.
I thank J.F. Herr for supplying useful technical information and C.
Spring for photographic help. I also thank Dr. Peter Sturrock, Philip
J.
Klass, Dr. David Fryberger and the two reviewers for helpful comments.
This
work was supported in part by the Fund for UFO Research.
NOTE ADDED IN
PROOF
The original version of this paper was submitted to Applied Optics
magazine in 1984. It was rejected because of my claim that the object
was
unidentified. The editor believed that it could be identified as some
natural phenomenon. The editor first rejected the paper because, in his
opinion, the photo showed a sun dog. I then pointed out that there was
a
large angle difference between the direction to the object and the
direction
to the sun and also that the object appeared below the aircraft and
“hidden”
from the sun behind a large thunderhead. The editor then passed the
paper
to a referee who suggested a solar reflection in a lake. I responded
that
such a reflection, should it occur, would appear directly below the sun
and
not way off to the right, and that it would appear reddish because of
passage of the light through a long atmospheric path at sunset. The
referee responded that a rough lake surface could scatter light at
considerable angles from the sun. He also pointed out that the film was
saturated and therefore did not register the correct color of the
image. At
this point the editor ended the debate with a final rejection.
Had I been able to respond I would have pointed out that colored light
sources leave their color “fingerprint” even if the image is saturated.
This is because sideways scattering of light within the image causes
film
exposure outside the geometric boundary of the image (the boundary
given by
the equation, image size = object size times focal length divided by
distance). This has been proved numerous times in experiments and is
especially evident with red light. The image of a red light that is so
bright as to achieve complete overexposure (white) at the center of the
image will have a rather wide redcolored annular region around the
central
overexposed region. Hence, if the pilot had photographed the reflection
from a lake just before sunset, the atmospherically reddened reflection
would have caused reddening of the image around the central overexposed
region. However, there is no such reddening, nor is there evidence of
any
other coloration. Instead, the image is totally consistent with a white
light source. Hence it was not a reflection in a lake.
Had I been able to respond further, I would have pointed out that the
pilot flew past this object as you would drive past a telephone pole or
any
other object fixed to the earth. This object seemed to remain
stationary
relative to the thunderhead. He last saw it dropping behind him as he
flew
on westward, i.e., at a very large angle relative to the direction to
the
sun.
BIBLIOGRAPHY
1) R.J. Childerhose, affadavit written in May 1958 and private
communications with Philip J. Klass (1966), Dr. James McDonald (1968 
1969), Jan F. Herr (1980  1981) and this author (1984  1988). Mr.
Childerhose and another pilot set a transCanada speed record two days
after
the photograph was taken.
2) Klass, P.J., UFOs IDENTIFIED, Random House, NY (1968)
3) Altschuler, M.D. in SCIENTIFIC STUDY OF UNIDENTIFIED FLYING OBJECTS,
E.
Gilmore, Ed., Air Force Office of Aerospace Research (1968); also
Bantam
Books, New York (1968), pg.733
4) Kingslake, R., APPLIED OPTICS AND OPTICAL ENGINEERING, Academic
Press,
New York (1965)
5) HANDBOOK OF GEOPHYSICS AND SPACE ENVIRONMENTS, S.L. Valley, Ed., Air
Force Cambridge Research Laboratories, Office of Aerospace Research
(1965);
also McGrawHill Book Co, New York (1965)
6) A careful study of the image reveals a brightness structure not
completely consistent with the short cylinder model used here. In
particular, there are two distinct bright spots, one at the left (west)
end
of the image and one slightly above the centerline at the right end.
The
structure of the image further suggests that there may have been two
glowing
objects very close to one another.
7) van Kreveld, A., "Standardization of Photographic Sensitometry Based
on
the Addition Law," Journal of the Optical Society of America 29, 327
(1939)
8) "Infrared Radiation Calculator," by Sensors, Inc., Ann Arbor, MI;
published by Perrygraf Div., Nashua Corp., Los Angeles (1976)
APPENDIX 1
Pilot Testimony
The following statement by pilot Robert J. Childerhose was published by
Flying Saucer Review in October, 1958. It was written in the late
spring
of 1958, roughly 1 3/4 years after the event. Note that the date given
here
was not recalled correctly. During the 1968 investigation of this
sighting
by Dr. James McDonald, Mr. Childerhose referred to his log book and
found
the correct date, August 27, which was two days before he and another
Sabre
jet pilot set a speed record flying east over Canada. The time, 1820
hours,
is also incorrect. In other correspondence Mr. Childerhose stated that
he
landed in Vancouver at about 7:20 PDT which corresponds to 8:20 MDT. He
estimated that the sighting took place about an hour before the
landing,
which would be at 7:20 MDT. The one hour estimate is based on the
calculated flight time from a location near Ft. MacCleod to Vancouver
(445
miles, 450 miles per hour estimated ground speed).
“On the afternoon of 23 August 1956 the writer was flying #2
position in a four plane formation of Sabre 6 (F86) aircraft from
Gimli, Manitoba, to Vancouver, B.C. Our flight altitude was 37,000
feet, weather was good with cumulus and cumulonimbus cloud
formations forming an intermittently broken undercast beneath us.
Visibility was unlimited.
At about 1820 hours (local time) and at a point roughly over
the foothills of the Rocky Mountains along a direct path between
Gimli and Vancouver, we encountered a largerthanusual
thunderstorm. The leader of the formation elected to climb over the
storm and called for climb power. The writer, who was shooting 35
mm color pics, decided to get a shot of the dark purple areas
beneath the CB (cumulonimbus).
On looking down, he saw a bright light which was sharply
defined and discshaped. (Like a silver dollar lying on its side.)
The light being emitted from this source was considerably brighter
than the sunlight which was beginning to set (sic). The sunlight
was reflecting on the tops of cumulus formations, and was coming
from our 10 o’clock position. (This is relative to the aircraft’s
heading. ) The white disc was at 1 o’clock.
The writer called the attention of the formation to the light,
asking for an opinion. The leader, F/L Ralph Annis, commented at
that time: “Maybe it’s a shaft of reflected sunlight.”
(Note: Speaking to F/L Annis in May 1958, writer asked if he
recalled the incident. F/L Annis said that he didn’t. )
The writer took a photo of the disc, thinking only that it was
a peculiar phenomena (sic). The idea of it being ‘reflected
sunlight’ did not sound plausible then, or since.
The disc of light appeared to be at a distance 3 miles from our
position and at an altitude of about 20,000 feet. However, since
the size of the object is unknown, the range quoted above is
strictly a fighter pilot’s guess.
On landing at Vancouver a short while later, the members of the
section discussed the light briefly. Everyone agreed that it was an
unusual sight. Nobody had even encountered a similar orb of light ;
nobody had any reasonable explanation to offer; nobody suggested
that it might have been a ‘Flying Saucer.’”
In correspondence dated ten or more years after the above written
testimony (and 12 or more years after the event) Childerhose stated
that he
had made an error in the above testimony: the object was at 10 o’clock
and
the sun was at 1 o’clock. Moreover, he claimed he rolled his aircraft
to
the left in order to take the picture. He said that he doubted that he
rolled his aircraft to the right since that would have taken him toward
the
other planes, an action which “frightened him.” In other correspondence
(see below) he indicated considerable confusion over his recollection
of
the direction to object from the aircraft. On the other hand, he also
recalled (see below) that the flight leader wanted the jets to turn to
the
right while climbing to avoid the anvil of the thunderhead, so perhaps
he
took the picture as he began his right turn at a time when the other
aircraft were also turning to the right. He may have confused the
avoidance maneuvers with his maneuver to photograph the object.
Whatever
may be the explanation for his confusion, the fact is that the original
slide photograph shows light coming from the left (west) which means
the
sighting line to the object was toward the northwest. The photo format
(50
mm focal length, 35 mm film) and the image location in the picture can
be
used to determine the angle between the image and the left edge of the
film.
That angle is about 28 degress. Therefore the sighting line to the
object must have been greater than about 28 degrees to the right
(north) of
the sun, or greater than 304 deg. azimuth. The lighting on the clouds,
as
shown in the original slide photograph, is consistent with this.
In his later correspondence he was more specific with details about the
object and the sighting. From a letter to Philip Klass, September,
1966:
“I had the object in good view for upwards of 45 seconds. It was
stationary, with sharply defined edges. Looked like a shiny silver
dollar
sitting horizontal. The light emitted was much brighter than the
existing
sunlight and overexposed the film causing blurred edges in the
picture....... It neither moved nor changed shape while I had it in
sight..... I remember looking down at the object from 38,000 ft and
thinking
that it was close to about 12,000 feet since it appeared to be close to
the
scattered layer of fluffy cumulus which I recollected was forecast to
be
between 10 and 12,000. (Pretty standard.)”
From a letter to Dr.James McDonald, June 12, 1968: “I don’t know
whether I mentioned this to you but the photo of the bright object
doesn’t
represent quite what appeared to the naked eye. When I first saw the
object it appeared as a very bright, clearly defined discoid, like a
silver
dollar lying on its side. The photo makes it look like a blob of light,
the
result of light intensity. It appeared much brighter than that (sic) of
the
sun which, of course was setting behind the clouds up ahead. What
appears
in the Kodachrome slide is a disappointment, really.”
From a letter to Jim McDonald, March 22, 1969: “It was in good view for
some minutes because I looked at it trying to figure out what I was
seeing
and I called the attention of the formation to it before remembering
that I
had a camera in my leg pocket.”
From a letter to this author, September 19, 1984: “The object remained
perfectly stationary throughout the period that I witnessed it. I
recall
the formation turned starboard and began climbing. The UFO, now at 8
o’clock low position relative to me was lost to view behind a cloud.
This
could have been the low cumulus near the UFO or in the mists of the
scud
roll of cloud which we climbed through to reach our (final altitude).”
APPENDIX 2
Derivation of the Photometric
Equation
Eq. (1) is based on the standard “camera equation” for film exposure as
modified for offaxis light sources (4) and for atmospheric
transmission,
written in the units of power rather than the usual (for photography)
lumens:
E = (pi)LT e^k cos^4(B) /(4f#^2) 3)
where E is the irradiance in w/cm^2 within the image on the focal
plane, L
is the radiance in w/cm^2/st of the source (st = steradians), T is the
transmission of the lens, k is the optical thickness of the atmosphere
between the light and the camera, f# = “fnumber” = f/D, where f is the
focal length and D is the diameter and b is the angle between the lens
axis
and the direction to the source (and image).
The irradiance reaches the focal plane for a time, t, determined by the
shutter setting. Thus a total energy per unit area or “exposure” of the
film, H, is the product Et (pun intended). Inverting the equation for E
to
get L, with E = H/t yields
L = (4/pi)H(f#^2)e^k/tTcos^4(B) 4)
which, except for the form of the atmospheric correction, is the result
of
combining Eq. (1) and (2).
The atmospheric attenuation over a slant path distance from one
altitude to another is most easily found from tables of calculated
attenuation from the ground up to a particular altitude. (Direct
calculation is not easy since the absorption and scattering of light by
the
atmosphere are functions of altitude and light wavelength.) Suppose a
light
beam travels vertically upward from the ground to altitude h1. The
exponent, k, is replaced by the tabulated optical thickness value, a.
The
beam power reaching h1 is therefore ea times the initial power.
Similarly, when the light beam travels vertically upward to h2, k is
replaced by by the tabulated value b so the beam power at h2 is e^(b)
times
the initial power. When the power at h1 is known, the power at h2 can
be
calculated by multiplying the power at h1 by the ratio of these two
attenuation factors, e^b/e^a = e^(ba). This is the ratio for a
vertical path. For a slant path the distance traveled is greater than
the
vertical distance by a factor 1/cos(A), where A is the angle measured
from
vertical. The optical thickness is therefore greater by the same factor
so
e^k in Eq. (3) is replaced by e^(ba)/cos(A). The inverse of this
quantity replaces ek in Eq. (4) and yields the form shown in the
combination of Eq. (1) and (2).
An alternative estimate of the total radiated power can be derived
under the assumption that the light source was effectively a “point”
source
unresolved by the camera. Under this assumption the intensity of the
source, I , in w/st, is calculated from the camera optics, the range to
the
light source, the shutter time and the energy density on the focal
plane
integrated over the image area. The integrated energy is approximated
here
by the product HAi, where Ai is the image area over which H is
approximately
constant. In this case the power, J, which passes through the lens to
the
focal plane is given by
J = I (AL cos(B)/R^2)Te^[(ba)/cos(A)] 5)
where AL is the area of the lens aperture and R is the range. Note that
this expression contains the inverse square law explicitly whereas the
previous expression does not. Also, in this expression the cos(B)
enters
only to the first power to account for the oblique incidence of the
light on
the lens area. This power reaches the film plane for a time t and
deposits
a total energy over the image area, Ai, which is approximated as
HAi = Jt. 6)
Solving Eq. (5) and (6) for intensity yields
I = (HAi/t)R^2 e^[(ba)/cos(A)]/[AL T cos(B)]. 7)
The image area over which the exposure seems to be nearly constant is
approximated as an ellipse with dimensions 1.28 mm by 0.6 mm for which
Ai =
6E3 cm^2. Therefore, with H = 1E4 j/cm^2, AL = 3E5 m^2 (based on f#
= 8
and a 50 mm focal length lens), T = 0.7, t = 8E3 sec, cos (9) = 0.987
and
using the closer distance, R = 6000 m and e^(ba)/cos(A) = 1.09, Eq.
(7)
yields I = 1.4E+8 w/st. Assuming uniform radiation in all directions (a
“4pi” radiator) we find the total power emitted within the bandwidth of
the
film to be P = 4piI = 1.8E+9 w. Similarly, for the 20 km assumed
distance
and the associated atmospheric attenuation factor 1.34, I = 1.9E+9 w/st
and
P = 2.4E+10 w. These values of total power are within a factor of 2 of
the
values calculated from Eq. (1). :
