Analysis of the Cecconi silver photograph
by Antoine Cousyn (myself) and Francois Louange
This thread is the continuation of this one:
An in-depth review of the Cecconi Case
Photos were I begun to be interested in the case and to search for the original unmodified photo.
It tooks me more than two months efforts to finally succeed in finding the original B&W 6x6 plate, with the help of E. Russo from the Italian CISU.
(Many thanks to him BTW)
Here's the original scanned photography, resized (640x480) to fit within the forum screen:
Photographic parameters
• Picture taken on June 18th 1979, above the Treviso airport (Italy), with a Vinten F-95 camera
• Film format 70x70 mm
• Unknown focal length, with 4 possible preset values: 76 mm, 100 mm, 152 mm or 300 mm.
Geometric parameters
Consider the following points:
O : camera’s lens
H : vertical projection of O on the ground
A et B : extremities of a known parcel on the ground
B’ : projection of point A on OB
Assuming points H, A and B are aligned (approximately), we have the following sketch in the vertical plane that contains H, A and B:
• Known flying altitude:
OH = 7000 ft = 2134 m
• Known length of the AB parcel spotted on the ground:
AB = 381 m
• Unknown alpha angle between the OH vertical and the OA line
Attempt to determine the focal length
After inputting the film’s dimensions (70 mm x 70 mm) and the focal length (76 mm, 100 mm, 152 mm or 300 mm) into the Camera menu of our favorite
analysis software, one may measure the alpha angle and infer the value of the r ratio, thanks to the Length/Distance function:
r = AB’ / OA
Knowing that:
AB’ = AB cos alpha
OA = OH / cos alpha
One infers:
r = (AB / OH) cos2 alpha
Thus:
alpha = arccos alpha (r / 0,17857)
Measurements with our favorite analysis software (see diagram below) give the following results:
In order to determine the most probable value for the focal length, we shall make use of the horizon line, visible on the picture.
Let us compute first the angle value between the line of sight towards the horizon and the horizontal from the camera, which is at an altitude of 7000
feet (2134 m).
The following sketch shows the Earth, with its R radius, the previously defined points O, H and A, the horizon line and the horizontal passing through
O.
The wanted angle is delta:
Knowing the values of the altitude OH = h'c' and of the Earth’s radius R:
h'c' = 2134 m
R = 6371000 m
One computes:
Delta = arccos [R / (R + hc)]
Delta = 1,5 °
The angle gamma between the camera’s line of sight towards point A and the camera’s line of sight towards horizon is inferred from alpha,
previously estimated for each of the possible focal lengths, and from the delta angle:
For each possible value of the focal length, we shall measure gamma with the Angle function of our favorite analysis software, and compare the result
to that given by application of the previous formula.
Measurement for f = 100 mm
The most coherent result, expressed by a minimum difference in absolute value (taking into account simplifications in computations and uncertainties
in measurements), corresponds to:
f = 100 mm
It should be noted that values of « gamma computed» and of « difference », indicated in the table above, increase or decrease if the flying
altitude increases or decreases as compared to 7000 feet. More precisely, calculation shows that the “difference” would be
null if we
assumed:
f = 76 mm
and h = 7800 ft
OR
f = 100 mm
and h = 6400 ft
Without finally discarding the value of 76 mm, for which coherence is not very significantly worse, we shall therefore retain the value of
100
mm for the following study, bearing in mind that the residual uncertainty will not call into question the orders of magnitude which will be
evaluated.
Study of the unidentified object
The object appears approximately in the form of a “cigar” (cylinder), the nearest extremity of which is dark. Flying at an altitude slightly lower
than that of the plane, it is lighted from above, which is compatible with natural lighting by the sun, and its main axis is parallel to the
ground.
The angular measurement of its diameter gives:
3,473 °
To this angular size corresponds a ratio between the diameter of the object and its distance d from the camera:
Attempt to estimate the distance from the unidentified object
If the nearest extremity of the object is really black or very dark, comparison of its apparent luminance with that of the darkest elements of the
ground should provide information on the ratio between camera-to-object and camera-to-ground distances, taking into consideration atmospheric
diffusion.
The following illustration displays the measurement of the darkest pixel, respectively in 3 strips of the image and at the dark extremity of the
object.
It appears that dark points on the ground are far darker than the object. This would imply, if the object was really black, that it would be farther
away than the ground in the strips, i.e. farther than 10000 feet.
If it had been 10000 feet away from the camera, the object would have had a diameter in the order of 200 m. On the other hand, the value of its h'ufo'
altitude would have been inferred from the h'c' value of the camera’s altitude through the following simplified formula:
h'ufo' = h'c' – d sin ( phi + gamma)
where gamma is the above-defined angle and phi the angle between the camera’s line of sight towards the object and the camera’s line of sight
towards the horizon line.
h'ufo' = 7000 – 10000 sin (5,49 + 1,5)
h'ufo' = 5800 ft
Now, an object with a 200 m diameter, flying at an altitude of 5800 ft (1800 m), could certainly not have gone unnoticed from the ground, in
particular in an urban area and on a relatively clear day.
This only proves that the original assumption was wrong: the object was not completely dark, and therefore could be at any distance from the camera
(less than 1000 ft).
edit on 20-10-2012 by elevenaugust because: typo
edit on 20-10-2012 by elevenaugust because: (no reason
given)
edit on 20-10-2012 by elevenaugust because: (no reason given)
extra DIV