BASIC DATA AND ANALYSIS OF INFRARED VIDEO TAPE OF … · Web viewThe radar beam is emitted from an...

Mexican DOD Video Maccabee Page

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A STUDY OF THE MARCH 5, 2004 RADAR AND FLIR SIGHTINGS DURING A SURVEILLANCE

FLIGHT BY THE MEXICAN DEPARTMENT OF DEFENSE

byBRUCE MACCABEE

MAIN TEXT: Version 1 Aug.11, 2004 REVISIONS Aug 28; Appendix 4 added Oct 28

NOTE: THIS WRITING HAS BEEN DONE WITHOUT FULL ACCESS TO ALL INFORMATION NEEDED TO MAKE A COMPLETE ANALYSIS. CONSIDER THIS VERSION TO BE A “TEMPLATE” FOR A FINAL VERSION AFTER CALIBRATION EXPERIMENTS WHICH SHOULD BE DONE.………………………………………………………………………………………Preliminary comments added for web release: Dec 27, 2004It is nearly 8 months since the event described herein was first publicized and about 7 months since I requested a special test flight by the Mexican Air Force. Although I was told last summer that such a flight would be done, it has not been. As you can see from the Aug. 11 date above, the initial version of this paper was completed long ago. I sent the early version of this paper to the Mexican Air Force in the hope that it would provide them a reason to carry out the test flight. I have delayed publicizing it while waiting for the test flight. In the meantime others have publicized their solutions to this puzzle. Of particular interest is the oil flame explanation proposed in May, 2004 (see www.alcione.org). Whether or not that is a valid explanation for all of the forward-looking infrared (flir) images could be proved or disproved by the test flight I requested. Arguments for and against are presented in Appendix 4. ………………………………………………………………………………………….

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http://www.alcione.org/


ABSTRACT

This is a history and analysis of the infrared “lights” and radar “targets” that were recorded by the Mexican Department of Defense (DOD) between 4:40 and 5:30 PM (local standard time) on March 5, 2004 during a surveillance flight over the state of Campeche which is in the northern western portion of the Yucatan Peninsula. The story of the event is based on recollections of the crew combined with the audio and video tape recorded at the time by an infra-red viewing system on the plane. The first object detected on radar was quite close to the airplane, yet remains unexplained. Other radar targets seem to have been reflective objects on the ground. Many of the of the radiation sources or reflectors (“lights”) detected by the forward-looking-infra-red (flir) system were much farther away than the crew thought at the time Some of these may have been be distant oil field gas burnoff flames. Others might have been reflections from clouds. Only one flir light was possibly associated with the first radar target.

I. SUMMARY OF THE SIGHTING EVENTS

The sighting began during the afternoon of March 5, 2004 as a C-26 surveillance airplane operated by the Mexican DOD was flying over the Yucatan peninsula. At about 4:42 PM the plane was at an altitude of about 2 miles (2 mi; statute miles are used in this report unless otherwise noted). It was flying at about 230 mph (statute miles per hour), heading approximately northwest and was near the city of Candelaria. It was nice flying weather with a somewhat hazy atmosphere, a 6 mph tailwind from the southeast and scattered cumulus clouds, many with tops at the altitude of the airplane. The radar, which was tilted about 30o down and continually rotating about a vertical axis (thus sweeping out a conical volume beneath the plane), suddenly detected an unexpected target about 5 (?) mi ahead. It was moving approximately northwestward, but more slowly than the C-26 surveillance aircraft. This target was about 30 degrees to the left of straight ahead of the C-26. The immediate assumption by the crew was that they had detected a small airplane that had just taken off, probably from an airfield near Candelaria, so the C-26 turned to follow it. Over the next couple of minutes the C-26 approached to a distance of about 2.3 mi as the object increased speed in a somewhat erratic way. The C-26 followed the object for about 10 minutes, never getting closer than about 2.3 mi. The crew noted four unusual characteristics of this radar target: (a) its sudden appearance at a short distance (ca. 5 mi), (b) somewhat erratic speed changes, according to the radar, (c) invisibility” (it could not be seen visually despite the short distance) and (d) lack of a thermal signature (it was not detected by the “forward looking infrared” (flir) viewing equipment on the airplane).

The operator of the flir device (which can look in all directions, not just forward) began a search for the radar object (“target”) as soon as it was reported by the radar operator. The flir output was recorded on a videotape and provides flir imagery and some normal TV images as well as recordings of the conversations by the air crew. This flir videotape is the primary source

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of information about the sighting events. The flir device can detect not only the hot engines, but also the body of the aircraft itself, making an image which can be recognized as that of an airplane. The flir operator used the variable elevation and azimuth of the flir optical system to search the sky and ground in the general direction of the radar target. Numerous ground level objects, which were either sunlight reflectors or sources of heat, were observed, but no flir object (“light”) seemed to definitely correlate with the radar target. However, just before the plane turned to fly eastward , a bright flir light was detected which might have been in the same direction as the radar target (see below). (Note: in this discussion the bright objects detected by the flir viewing system, described below, are generally called “lights” because “light” is electromagnetic radiation in the invisible or infrared range and “a light” would be a source of such radiation.) Then, because of low fuel, the C-26 had to make a right turn eastward toward the air base. The radar target continued on its west-northwest track, thus moving away from the C-26. It was followed by radar until it arrived in the vicinity of Ciudad del Carmen on the northern coast of the Yucatan Peninsula (see map below), more than 40 mi behind the plane. By that time its speed had increased to over 380 mph, according to the radar system, and it then faded from the radar. The consistency of appearance on the radar scope, the continuous motion with acceleration and the short distance to the radar target during most of the time seem to remove this from the general class of spurious targets known generally as “radar angels.” The radar data raise two important questions: (1) what object large enough to be detected by the (3 cm) radar radiation at a distance of up to 40 mi would be invisible to the human eye and to a flir system when it was only about 2 mi away and, (2) what type of object can travel at hundreds of miles an hour without generating a heat signature of some sort?

Just before the plane began its right turn to the east the flir operator picked up a very bright “light” (heat source or solar reflector) about 12o to the right of straight ahead and at an angular elevation of about -4o (four degree angle downward). The sudden appearance raises the following question: if this light (a hot object or a reflector of sunlight) was on the ground, as suggested by the downward elevation, why wasn’t it detected earlier (by this time the plane had been flying in the direction of this light for many minutes)? The air crew apparently thought that this light was the radar target. Unfortunately it was recorded on the video tape for only a few seconds, not enough time to prove that it was (or wasn’t) the radar target. Although numerous other anomalous flir lights were observed later on (see below), this is the only one which was detected ahead of the aircraft. All the others were 90o or more to the left of straight ahead.

The flir videotape has a “time jump” of 61 seconds (recording turned off) while the plane turned about 148 degrees to the right. When the video recording resumed the plane was leveling out on the new track, flying due east. The video at this time shows various views of the side and top of a cloud. As the flir operator searched for the previous light, a much dimmer version of it or what was, more probably, a completely different light suddenly appeared from behind the right side of a cloud. It seemed to move to the right past the clouds. It appeared to be at the distance of the clouds or beyond the clouds. To its right was a much brighter light that the flir operator apparently did not notice because it is at the right edge of the field of view for only a few frames of the video (see below). The flir operator tracked the dimmer light (not the very bright one). It seemed to be at the level (altitude) of the aircraft and seemed to be traveling along with the plane. (Note: the possibility that the appearance of “traveling” was an illusion caused by

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watching very distant light as the plane passed by relatively nearby clouds, has been considered; see discussion below.)

During this time the radar target continued toward Ciudad del Carmen, finally disappearing at the limit of the range of the radar, somewhat more than 40 miles. At that time the flir light which was first seen after the right turn was at about -135 azimuth (135o to the left of straight ahead) and the radar target, as stated by the radar operator, was at about the 7 o’clock (7:00) direction (-150o) . (In “clock” notation, hours become angles relative to the forward direction of the plane. Thus, 12:00 is straight ahead, 9:00 is 90o to the left, 3:00 is 90o to the right and 6:00 is directly behind. The individual “hours” are 30o apart and each “hour” represents an angular width of +/- 15o centered on the hour or half hour. Thus a target at 7:00 could be anywhere from -135o azimuth to -165o azimuth. ) If the radar target was exactly at 7:00 then it was about 15o to the left of the flir light. If, however, the radar target was at the “upper limit” of 7:00, i.e., at 7:30, then the two could have been in the same direction. However, the radar target traveled almost directly away from the plane. Had the flir light traveled away from the plane it would have gotten dimmer during the observations. Instead, the flir light seemed to maintain its brightness level, suggesting that it did not get farther away. Hence the flir light that was seen and tracked after the turn was not the radar target.

This light was occasionally blocked by small clouds, but then it disappeared behind a

larger cloud. The flir operator searched again as the plane flew along. Suddenly, instead of a single light he found a group of lights that formed a line. The most notable of these are a pair of lights (herein called “the twins”) that were so bright that they saturated the dynamic range of the flir electronics, making large round overexposed images. A few minutes later, as they passed behind a cloud, it became apparent that there were as many as 15 lights accompanying the “twins.” These were lost behind a large cloud. Then new lights (or the same ones) were detected in a different roughly linear arrangement that include two bright “triplet arrays” (herein called “the triplets”) instead of “twins.” One light in the “triplets” split into two parts. It was at this time that the flir operator reported to the crew that he saw 11 lights. (This was featured in the initial press release of the video.) These lights seemed to pass behind clouds that were estimated by parallax and triangulation to be over 20 miles from the aircraft. During this flir sighting the radar operator detected nothing. He thought this was because of a “dead space” in the radar coverage (radar blocked by some portion of the airplane) that happened to be approximately in the direction of these lights, which were, again, about 135o to the left. The analysis below suggest that another possible reason for failure to detect them on radar is that they were beyond the 40 mile limit of the radar.

Eventually this second “family” of lights was lost behind a cloud. The flir tape recorder was turned off for about 4 minutes. When it was turned on again it recorded a very bright light and two not-as-bright lights against a dim, featureless background (no clouds as before). The lights were at the left side of the aircraft at the 9:00 o’clock position (90o to the left). Several minutes later only a single bright light appeared. It was viewed through the rotating propeller and went behind a cloud top that was estimated at about 14 miles away. There was also a new radar target, this time ahead of the aircraft. The new radar target was traveling at a “ground speed” of about 60 mph and the plane was catching up with it.

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At 5:19 PM the flir operator turned the flir forward and obtained images of the rising moon when it was only about a degree above the horizontal. These images provided useful data for calibration of the pointing accuracy of the flir system. Over the next several minutes the flir operator searched the sky but found only a few small, bright cloud tops (reflecting the setting sun). Then the flir operator found a very bright light that appeared against a featureless background. This light was detected almost directly behind the aircraft. Its brightness decreased as the plane traveled along. In the meantime the radar target had disappeared at the right side after the airplane caught up with it. (It may have been a moving radar reflector on the ground, such as a truck.) The videotape ends, showing a dim light almost directly behind the plane, at 5:28 PM.

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II. ANALYSIS OF THE SIGHTING EVENTS

This discussion is based on the information provided in Appendix 1 which contains the technical data and a transcript with translation of the air crew statements as obtained from the officially released flir data tape. Although this author has had discussions with researchers in various countries (France, Brazil, Mexico, USA) who have publicly stated their own opinions, the conclusions presented here are the opinion of this author alone.

II. A DISCUSSION AND ANALYSIS OF THE SIGHTINGSII. A.1 THE SIGHTING EVENTS

The series of events or sightings can be roughly divided as follows: (a) detection of a radar target that increased speed as the plane approached it, reaching a minimum distance (about 2 miles) and failure to detect any corresponding visual or infrared radiation source; (b) detection of a “flir light” (a source or reflector of infrared radiation not visible to the human eye or a TV camera) just before a sharp turn to the right with the direction to the light being possibly the same as the direction to the radar target; (c) after the right turn, loss of the radar target behind the airplane followed by the flir detection of numerous lights behind and to the left of the aircraft which seemed to travel along with the airplane and which were at distances greater than the distances to the clouds (i.e., greater than 10 – 25 miles); (d) loss of the groups of lights but acquisition of a few other lights at the left of the plane; (e) detection of a radar target nearly ahead of the plane; (f) loss of the light at the left and detection of a light nearly behind; and (g) loss of the light behind and the radar target that was ahead. These various sections of the sighting will be considered separately for detailed analysis below.

II. A.2 THE FLIGHT PATH OF THE AIRCRAFTThese sightings occurred as the plane flew along the track illustrated in Figure 1. (The

location of the plane at any time is determined by the latitude and longitude presented by the infra-red viewing system described below.) Initially the plane was headed northwest and was flying near Candelaria when the first radar target was detected. The plane made a turn to the left and followed the target as it traveled toward Ciudad del Carmen. Figure 1 shows the track after this initial left turn that occurred at about 16:43:15 (4:43 PM and 15 seconds). (Times are given in 24 hour notation. For example, 13:00:00 is 1 PM, 14:05:10 is 5 minutes and ten seconds after 2 PM, etc.) However the plane was running out of fuel so, at about 16:51:30 the pilot began a right turn to travel eastward toward the air base. This turn was completed at about 16:53 and the plane traveled directly eastward for several minutes before turning onto a heading of about 82o. (Azimuth angles relative to the ground are based on clockwise rotation from true north. North is 0o , due east is 90o, due south is 180o, due west is 270o, etc. Negative azimuth numbers are rotations to the left: -90o is due west, the same as a rotation to the right of 270o).

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FIGURE 1 THE FLIGHT OF THE C-26

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II. B THE FIRST RADAR TARGETII. B.1 THE RADAR SYSTEM

The C-26 surveillance aircraft carries a 3.2 cm (wavelength) search/weather radar, model AN/PS-143B(V)3 radar manufactured by the Telephonics, Corp. of Farmingdale, NY (www.telephonics.com). The radar antenna is mounted in a radar “dome” at the rear bottom of the aircraft, as indicated in the picture below.

MEXICAN DOD SURVEILLANCE PLANE

FLIR

RADAR

FIGURE 2 THE C-26 WITH THE RADAR AND FLIR DOMES ILLUSTRATED

A radar system (“radar” is an abbreviation of “radio detection and ranging”) operates by repeatedly sending out pulses of electromagnetic radiation in some small range of directions about a central direction (the radar “beam”). If there are radar-reflective objects within the radar beam they reflect some radiation back toward the radar antenna. The radar electronic system detects the reflection, or “echo,” and measures the time it takes for each echo to return. Since the speed of the radar pulse is known, the distance to a radar reflective object (“radar target”) is (1/2) cT, where c is the speed of radiation (speed of light), T is the time for an echo to return and (1/2)cT is the one-way trip distance.

According to the published literature this particular radar system is sensitive and selective enough to be able to detect, on the water surface, a radar target (e.g., a boat) with a 1 square meter “cross section” at 23 (statute) miles in sea state 3 (moderately rough sea). (The “cross section” is a measure of the efficiency of reflecting radar radiation. It is determined by the wavelength of the radar, the actual size of the object, its orientation relative to the radar beam

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and the material from which the object is made. More explicit capabilities of the radar are classified or proprietary information.) One can expect at least comparable capabilities for detecting airborne objects.

The radar beam is emitted from an antenna that rotates around once every ten seconds (6 rpm). The angular elevation (angle up or down) of the beam can be changed but it is generally set at a “depression angle” (angle downward) of about 30o. The beam sweeps out a “thick conical shell” shaped volume of space below the plane. The conical shell intersects the ground making an annular ring, as illustrated in Figure 2. The most intense portion of the beam (the most powerful radiation intensity) is at the geometric center of the beam (conical shell) and it drops off with angular distance from the center making a roughly elliptical beam shape with a half-power height (variation in the vertical direction or angular elevation within the beam) of about 8o and a half-power width (variation in the horizontal direction or azimuth within the beam) of about 2o, as illustrated by the intensity distribution curves in Figure 2. However, it would be incorrect to interpret the illustrations in Figure 2 as implying that objects can only be detected by the radar if they are within several degrees of the center of the beam. It is important to note that there is radiation outside the “main beam” which is defined by the 2 and 8 degree half power points. Thus, the conical volume below the airplane is mostly filled with radiation of lower intensity and low intensity radiation extends upward in elevation to zero (horizontal) so that, although the main portion of the beam spans only 7 miles when the plane flies at a 2 mile altitude (see Figure 2) the actual radar coverage extends far beyond 7 miles.

FIGURE 3 RADAR BEAM DETAILS

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The radar was operated in the “MTI” (moving target indicator) mode which means that it would only display targets which have motion toward or away from the plane. The radar system also provides speed and direction data about radar targets that are detected on successive rotations (“sweeps”) by comparing the most recent position with the preceding position and calculating the distance moved and in what direction (the calculation takes into account the location and motion of the C-26 to provide the target speed and direction relative to the earth).

RADAR DISPLAY

FIGURE 4 THE RADAR DISPLAY

The radar screen includes an immediate indication of the relative speeds of the aircraft and the radar target: the target indication (“blip”) on the radar display (“scope”) is green if the distance between the plane and the target object is increasing and the blip is red if the distance is decreasing. When the first radar target was detected (see below) the blip was initially red as the C-26 approached the object, but then the blip turned green as the object increased its speed to a value greater than the C-26. The radar blip does not give an indication of the relative strength of the radar target. However, the radar system does provide a measure of “quality” which is related to the strength of the radar return as compared to noise and background clutter.

The largest range at which the object or target was detected, as described on the flir audio channel (see below), is on the order of 46 mi. Note that if the “radar cross section” of an object were to remain constant, and if everything except the distance or “range” remained constant, then the strength of the reflection (echo) would vary as the inverse fourth power of the range, i.e., as (1/r)4 . A very crude estimate of the cross section of this radar target can thus be obtained by comparing the 1 m2 cross section detectable at 23 mi with what would be detectable at twice the range, i.e. 46 mi. At twice the distance the return signal from a 1 m2 target would be (1/2)4 = 1/16 as strong. Thus, to get the same amplitude of the return signal (the same echo

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strength as from a 1 m2 target at 23 mi) from a target at 46 mi would require a cross section of 16 times larger or 16 m2. Such a target might have a size somewhat comparable to an object with a width (or diameter), as seen from the perspective of the radar, of about 4 m (12 ft). (Note: any non-spherical object will have a cross-section that varies with aspect ratio or orientation relative to the radar beam.) One would think that a target of such a substantial size would have been seen by the crew when it was only 2.3 mi away (see below).

II. B.2 THE FIRST RADAR TARGETThe first radar target was detected shortly (seconds?) before 16:42:20 (the pertinent

portion of the flir recording tape supplied by the Dept. of Defense of Mexico begins at this time). Hence the distance and direction of the initial detection are not known. (See Appendix 1 for a timeline and translation of the recorded events.) At 16:42:20 the location of the target was stated by the radar operator as being at eleven o’clock (11:00 which is about 30o to the left) at about 4.6 mi (4.6 mi = 4 nautical miles = 4 nm). The radar system indicated that the object speed was about 86 mph (86 miles per hour = 75 knots= 75 kt) and it was traveling in the direction 297o (azimuth measured by clockwise rotation from north; this direction is west-northwest). At this time the direction of the plane was about 327o. (Note: although a tape recorder for the radar was available it was not used. All radar information presented here is based on translated statements by the radar operator as recorded on the flir tape. The exact times of some of the statements may be accurate only to within a couple of seconds. See Appendix 1 for the available transcript.) The plane turned to the left at about 16:43:00. After the turn its direction of travel was about 302o (west-northwest).

The radar operator continued to provide target information:

-16:42:42 The location of the radar target relative to the plane was given as 12:00 at a distance of 3.3 mi and (still) traveling at 86 mph (1.43 mi/min; no information on its travel direction was given at this time). Since the plane was traveling at 230 mph = 3.8 mi/min, it was catching up with the radar object/target.

-16:43:06 It was 2.4 mi (2.1 nm) ahead and traveling more slowly at 75 mph (1.25

mi/min; 65 kt)

-16:43:41 The speed was given as 111 mph (97 kt)

-16:44:04 It was about 2.4 mi (2.1 nm) ahead and was traveling in the direction 298o, almost exactly the same as before, but the speed displayed by the radar was changing rapidly. First the radar displayed a speed of 130 mph, then the speed varied rapidly between 328 mph and 245 mph.

The nearly constant distance of about 2.4 miles during the 58 seconds from 16:43:06 to 16:44:04 seems unusual since, according to the radar speed readings, the average speed over that time was considerably less than that of the plane. How did the radar object manage to stay at a nearly fixed distance from the plane while traveling at a lower speed? This raises a question: could this be an artifact of the radar not giving accurate speed readings because the target was so close, or is it a real characteristic of the unknown radar target? (This could, perhaps, be resolved

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by having the C-26 follow another aircraft, approaching to a distance of 2 nm to find out if the speed readings become erratic.)

It is important to keep in mind, when studying the times at which radar distances and speeds were given, that the radar “updates” the position of the target every 10 seconds. Therefore, it is reasonable to assume that the radar operator gave speeds and locations that were no more than 10 seconds old. If the radar operator gave a location that was 10 seconds old and then the radar updated to the new location immediately afterward, the radar operator would state first the old location and then follow that immediately by the new location, thereby making it appear that the radar target had jumped from the old location to the new location. The same would occur for the speed: giving the new speed immediately after the old speed could make it appear as if the speed had suddenly jumped. Unfortunately the times given for the stated speeds of the target are not sufficiently accurate to prove that the target actually changed speed erratically. Since the radar target was, on the average, increasing its speed, any decrease in speed from one radar reading to the next could be considered unusual.

-16:44:10 The speed readings ranged from 322 mph (280 kt) to 245 mph (213 kt) to 130 mph (113 kt).

-16:44:32 It was at 12:00 at 2.4 mi (2.1 nm)-16:44:52 It was 2.4 mi ahead traveling at about 203 mph (117 kt) in the direction

300o

-16:45:00 It was 2.4 mi ahead-16:45:37 Still at 12:00-16:45:48 The radar operator reported that the radar was repeatedly showing the

target at 12:00 with a speed of 237 mph in the direction 301o -16:46:43 It was 2.4 mi directly ahead, traveling along 302o and the radar operator

stated the speed as (166 kt) 190 mph but then corrected it to (207 kt) 236 mph-16:48:25 It was 2.4 mi ahead with speed of 221 mph (192 kt)-16:48:37 Still 2.4 mi ahead-16:49:18 It was still in front of the C-26

At this time it appeared that the radar target was heading toward Ciudad del Carmen, so the crew called Carmen airfield to find out if they were expecting an airplane. However, Carmen was not expecting any aircraft. At this time the plane was about 48.6 mi from Carmen so the radar target was about 46.2 mi from Carmen.

The videotape was switched off between 16:49:30 and 16:51:24. Then the video was switched on and it ran for about 8 seconds. During this time the first potentially correlated flir light was recorded (see below) but no radar data were recorded. It is unfortunate that the video was not on longer. The video was switched off again from 16:51:32.6 to 15:52:33.8 while the plane was turning to the right. After the C-26 made its right turn (see Figure 1) the radar target was still detected traveling northwestward.

-16:53:09 It was now behind the plane at 7:00, 12 mi (10.5 nm) away and traveling at about 380 mph (334 kt), according to the radar, in the direction 2-8-3.

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Assume the target was 2.4 mi ahead of the plane when the turn began at 16:51:32. The next radar reading, at 16:53:09, showed the target about 12 mi away. Plotting these positions on a map one finds that the target traveled about 9.7 mi during the 1 min and 39 sec = 1.65 min. This corresponds to an average speed of 9.7/1.65 = 5.88 mi/min = 353 mph. Thus it appears that the radar target was accelerating as it traveled northwestward away from the plane.

-16:55:45 It was near the end of the radar range, 37 mi away (31.8 nm)-16:56:09 It was a mile from Carmen-16:56:17 It was one or two miles from Carmen-16:56:22 It was not on the radar screen but the distance readout shows 43 mi (37

nm)-16:56:28 It was over Carmen, which was about 48 mi behind the plane and about

-150o azimuth relative to the direction of the plane. (The flir light detected at this time was at an azimuth of about -130o. Hence it probably was not the radar target. See below.)

-16:56:50 The target was lost.-16:57:03 The radar operator reported “It’s passing by Carmen. It is still there.”

-16:57:19 The radar operator reported “It seems to be turning.” Then radar lost it.

It is possible to make estimates of the speed of the target based on the radar distance data and the location of the airplane. The next - to - last radar reading before the turn was at 16:48:25 when the target was 2.4 mi ahead. The plane, at this time, was about 48.6 mi from Carmen and the target was 2.4 mi closer to Carmen, at 46.2 mi. If the radar operator was correct in saying that the target was “over Carmen” at 16:56:28, then the target covered the 46.2 mi distance in 8 minutes. This corresponds to 5.78 mi/min or 347 mph average speed over that time.

Another speed estimate can be made as in the following way. At 16:53:09 radar put the target at 12 mi behind the plane (which had turned to the east) and heading northwestward at about 380 mph. At 16:55:45 the radar gave the distance to the target as 37 mi which means that the distance between the plane and the target had increased by 37-12 = 25 mi in 2 min and 36 sec or 2.6 min for an average separation speed (the plane and the target were traveling in nearly opposite directions) of 9.6 mi/min = 577 mph. Therefore the average speed of the target over this time interval was 577 – (speed of the plane) = 577 – 230 = 347 mph (the same speed as above).

Yet another speed estimate can be made using the distance at 16:53:09. At this time the plane was 37 mi from Carmen and the radar target, being 12 mi behind the plane in the direction of Carmen, was about 25 mi from Carmen. According to the radar operator it was “over Carmen” at 16:56:28. Thus it traveled the 25 mi in 3 min and 19 sec or 3.32 minutes. Its average speed was 25/3.32 = 7.5 mi/min = 452 mph.

This third calculation yields a higher speed during the latter part of the trip to Carmen than the second calculation. The variation in calculated speed could be a result of imperfect radar distance estimates but it could be also be interpreted as evidence that the radar target continually increased its speed as it headed toward Carmen.

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Aside from the reported erratic speed changes (which may have been a radar artifact caused by the closeness of the target; a radar engineer would have to comment on this), it appears that, shortly after being first detected by the radar, the radar target increased its speed and paced the plane for many minutes, staying about 2.4 mi ahead, and then increased its northwestward speed after the airplane turned to the right. This is surprising: one would think that a conventional radar target that could reach a speed of several hundred mph, perhaps a smuggler’s airplane, would accelerate to its top speed immediately and depart as quickly as possible rather than waiting for the surveillance aircraft to break off the chase (how would it know the surveillance plane would suddenly stop chasing it?) before accelerating. Was this radar target “playing” with the C-26?

The greatest puzzle about this radar target is the failure of the crew to see it despite its probable size (meters? See the discussion above.) and the short distance (2.4 mi). An airplane or helicopter is visible and identifiable at distances much greater than 2.4 mi. They may have briefly detected it for a few seconds with the flir system just before the right turn but there appears to be no conclusive data to prove this. But if it is true, the question becomes, why the flir didn’t detect the target soon after the radar did? It certainly would have detected any ordinary aircraft soon after being alerted to search for one. On the other hand, if the flir didn’t detect the radar target, the question becomes, why didn’t the flir detect it? Any powered flying machine made by earthmen uses energy and creates heat as it flies. “Stealth” attempts are made to reduce the thermal signature, but it cannot be reduced to zero, especially when the plane is seen from the rear.

Detection of any object comes as a result of contrast between the object and the background. If the object radiates or reflects in exactly the same way as the background at the same distance then it will not be detected. The fact that the target was radar reflective indicates that there was some sizeable contrast (in conductivity or polarizability) between it and the air (or ground at the distance of the target). The fact that it was detected when the radar operated in the MTI mode indicates that it was moving in some manner with respect to the airplane or in some way changed the frequency of the reflected radiation by a small amount that was interpreted by the radar system as a Doppler shift caused by relative motion (the distance either shrinking or increasing). The fact that it made a continuous, even logical. track on the radar (flying in basically a straight line while speeding up), indicates that it was not a “weather angel” or some odd atmospheric effect. It’s speed rules out birds, balloons, dirigibles and also objects on the ground.

The available data indicate that this was an anomalous radar target, a “radar UFO.”

II. C THE FLIR LIGHTSII. C.1 FLIR CHARACTERISTICS

The C-26 airplane is equipped with a “forward-looking-infra-red” (flir) viewing system. (Although “forward-looking” is part of the name for historical reasons, it can actually look at any azimuth angle around the airplane.) This viewing system, called the STAR SAFIRE II, was made by FLIR SYSTEMS, INC, Portland, Oregon and the specifications can be found at their

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web site, www.flir.com. It uses a cooled InSb (indium antimonide) CCD focal plane rectangular array of detectors (“picture elements” or “pixels”) 320 pixels wide by 240 pixels high. A wavelength filter restricts the accepted infrared radiation to the bandwidth of about 3 – 5 microns (1 micron = 1/1,000 millimeter). This is the “mid infrared” region that can detect heat from hot objects and also infrared radiation from the sun as reflected from objects, clouds, lakes, etc. The optical portion of the system is contained within a “dome” that is located on the bottom of the airplane, as illustrated below in Figures 2 and 5.

The dome contains a combination of the flir optics and ordinary (visible) TV camera optics. The optics are designed to give the flir several different effective focal lengths which, in turn, provide several fields of view. The fields of view (FOV) of the flir system are given by the manufacturer as: WIDE FOV – 25.2o x 18.8o ; MED FOV – 3.4o x 2.6o ; NRW FOV– 0.8o x 0.6o; and NRWX2 FOV – 0.4o x 0.3o.. To illustrate the meaning of these angular sizes, consider that the width of the moon (as seen from the earth) is 0.55o. Thus at NRW FOV the moon would nearly fill the viewing screen and at NRWX2 the moon would more than fill the viewing screen (one could easily see craters, etc.) Most of the flir imagery of unusual lights was obtained using the MED, NRW and NRWX2 fields of view.

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FLIR INSTALLATION ON C-26

FIGURE 5 FLIR SYSTEM OPTICAL DOME

Even though the flir electronics are basically digital, the SAFIRE II flir system records whatever is seen by either the infrared viewer or by the TV camera (they cannot be recorded simultaneously) in an analogue NTSC format for viewing on a standard TV monitor. (Newer versions use digital recording.) Therefore some fine details of the scene may be lost and instantaneous “gain settings” (amplification factors of the automatic “gain control” circuitry) are not recorded. Also, the recording is limited by dynamic range (ratio of the maximum recordable brightness level to the minimum level). Furthermore, it is susceptible to a basic electronic noise level when the scene is dim and it suffers from saturation when objects in the scene are very radiant, such as reflectors of sunlight or “hot spots” on the earth. One effect of increasing brightness of a radiation source is increased image size. This is illustrated in the diagrams in Figure 6, below.

II. C.2 FLIR IMAGES OF THE GROUND

Image brightness levels in the flir video range from barely above the noise level (barely detectable against the background) to so overexposed that they include dim halos, cause by light scattering within the optical system, around the images. Figure 7 is a typical flir image of the ground showing the streets of a small city. The polarity is set to “white hot” meaning that the hotter or more radiant objects appear “whiter” (brighter) in the video. This heat could be from the objects themselves or it could be reflected heat, such as from the sun. Figure 7 and the following flir pictures presented here have been digitally captured from an S-VHS video copy of a digital video copy of the original VHS copy. (The reason for making an S-VHS copy of the digital copy was that stop-frame VHS equipment was available to this author for analyzing the

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FIGURE 6 THE EFFECTS OF INCREASING BRIGHTNESS AND SATURATION

video. There was no comparable equipment available for direct analysis of the digital copy.) The black and white analogue images have been limited to 256 digitized grey levels which show brightness ranging from 0 brightness (black) to 255 (saturated white). (Basic flir images are not in color, although sometimes they are presented using various colors to represent various temperatures ranges. The movie “Predator”, featuring a creature that could see in the infrared, made effective use of colorized thermal images in which different colors correspond to different temperatures.)

FIGURE 7 FLIR IMAGE OF A CITY

This image was obtained at 16:42:54, during the initial search for the radar target described above. The time was before sunset so the very bright images in Figure 7 could be

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reflections of sunlight from the windows of buildings or even automobiles or they might be hot objects like chimneys. The streets may appear because of heat they accumulated during the day, keeping them hotter than the surrounding ground and foliage.

Useful data are presented at the bottom of the video screen. For this particular frame of the video the data are: the latitude (18o 4.56’) and longitude (90o 57.64’) of the airplane, the pointing (azimuth) angle relative to the forward direction of the plane (-6.1o or 6.1o to the left), the flir pointing elevation angle (-18o or 18o down), the date (05-03-04) and the time (16:42:54L where L means local). The airplane flight path (track), speed and flir pointing angles have been determined from data such as these. The azimuth and elevation are given for the center of the field of view (FOV). At the right of center at the top of the screen are the FOV (MED), and polarity (white hot). Latitude and longitude values are updated every 5 seconds. The latitude and longitude are given with a precision of a hundredth of a minute of arc, e.g., 4.56’. However, because they are only presented every 5 seconds, the accuracy in locating the plane at any instant is only to a few tenths of a minute of arc. (There are 60 minutes per degree of arc. One minute of latitude is a distance on the earth that corresponds to 1 nm or 1.15 mi; 1 minute of longitude is [cos(latitude)] nm. At the equator where the latitude angle is zero a minute of longitude is also 1.15 mi. Since the latitude of the upper Yucatan peninsula is about 18o, 1 minute of longitude is about cos(18) = 0.95 nm or 1.1mi.)

Figure 8 shows the ground and a river that was recorded by the flir while the operator was searching for the radar target. The ground appears “warm” (medium grey), the river appears cold (black). The water may be colder than the land, but also it reflects the cold sky above. Several times sunlight was reflected in the water and the images were white or overexposed. Unfortunately it is difficult to determine the range of radiation intensity or range of temperature represented by the grey scale from black to white because the electronic system uses automatic gain (amplification) controls and other processes to provide a “good” image.

FIGURE 8 FLIR IMAGE OF THE GROUND

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FIGURE 9 FLIR IMAGE OF A PENINSULA (?)

Figure 9 shows another ground scene. This may be a MED FOV image of small peninsula on the southeast side of the Laguna de Terminos,which is the body of water south of Ciudad del Carmen (see Figure 1). If this is true, then the black area above the image of the land is the Lago de Terminos. Note that there are no bright lights even though the general direction is toward the oil field north-northwest of Carmen (see Figure 1).

II. C.3 THE FIRST FLIR LIGHT

After showing the image in Figure 9 for several seconds the video makes a “time jump” from 16:49:30 to 16:51:24. Apparently the flir operator turned the recorder off for nearly two minutes. When the tape resumes it shows the first flir image of an interesting, unknown light, a potential UFO (PUFO). The initial image was obtained at 16:51:24 using the WIDE FOV. The image, shown in Figure 10, is a tiny dot just above the dot that indicates the center of the field of view. There is a cloud to the right of center. The top of the video frame is dark indicating low radiation from the upper atmosphere (mostly cold air). The lower 2/3 of the image shows gradations of “gray level” indicating that the ground is warmer than the upper atmosphere. (Note: ignore the “x” shaped “feature markers” which were generated by the flir electronics to indicate infrared “hot spots” of potential interest.)

Then the flir operator switched to the MED field of view. As Figure 11 shows, the bright unknown appears just below the center of the FOV and the left edge of the cloud that appears in Figure 10 is at the lower right. The cloud appears white, not because the cloud is hot (the water vapor is really very cold), but because of sunlight reflected from the cloud. (Clouds that are not

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FIGURE 10 FIRST IMAGE OF A POSSIBLE UFO (PUFO)

illuminated by the sun…see images below… appear black.) Note that the bright light was at an azimuth relative to the plane of about 11.7o and at an elevation of -6o (which should be corrected to -4o; see Appendix 2).

FIGURE 11 FIRST MED FOV IMAGE OF A PUFO

The bright light image in Figure 11 is quite highly exposed (232 on a brightness scale of 255), with a definite halo probably caused by light scattering within the optical system.

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FIGURE 12 FIRST NARROW FOV IMAGE OF A PUFO

Figure 12 is a “blowup” of the PUFO obtained using the NRW (narrow) FOV (0.8 degrees wide). Of course, the image is larger but there is also a glare or “halo” around the light, suggesting that the light could be either a very intense emitter of infrared radiation or a good reflector of sunlight (or both). There is a smaller “dot” image at the left edge of the bright light/object. According to the azimuth reading this light/object was about 12o to the right of straight ahead. At the time of the sighting the crew apparently thought the light was the radar target last reported to be only about 2.5 mi at 12:00, “directly ahead” of the plane because 12:00 is an angle range from -15o (left) to +15o (right). Therefore it could be that the radar target and the flir light were in the same direction. Whether or not they actually were in exactly the same direction cannot be determined from the available data.

The elevation is given as -6o. However, the “moon calibration” discussed in Appendix 2

shows that elevation readings in the forward direction should have about (+2) degrees added to them. Hence the elevation angle was actually about -4o. Since the plane was 2 mi high, if this were a light on the ground it would be at a horizontal distance of about (2mi)/tan(4) = 29 mi. It would require a considerable amount of radiation to make an overexposed image such as this from a distance of nearly 30 mi. On the other hand, if it were the radar target at 2.3 mi, then its altitude was nearly that of the plane and it could radiate much less power (less than 1/100th of the power needed at 30 mi) to make the same brightness image. Unfortunately there is no way of determining from a few seconds of flir imagery alone how far away it was.

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II. C.4 THE SINGLE LIGHT AFTER THE RIGHT TURN

The timeline in Appendix 1 shows that within seconds after this strangely bright flir light was discovered ahead of the plane, the plane initiated the right turn and there is a time break in the tape which jumps from 16:51:32 .3 sec (20 fields or 10 frames following 16:51:32.00 sec) to 16:52:33.77 (46 fields or 0.77 sec after 16:52:33.00) for a total “break time” of about 61 seconds.

When the tape resumes the flir video shows close-up views of the cloud that appears to the right of center in the previous images (Figures 10 and 11). The video shows various views of the cloud as the operator scans the flir around looking for the light. Apparently the PUFO disappeared into or behind the cloud. [It is to be noted that if the light were far away (much farther than the cloud) and stationary (fixed on the earth), when the plane turned to the right, the sighting line to the light could have moved in such a way as to make the cloud interrupt the sighting line and block the view of the light.]

At 16:52:49.17, in the MED FOV scene, a small “dot-sized” unknown light (UL) suddenly appeared, seeming to move to the right from behind a cloud (the clouds generally move to the left as the plane travels along). Was this the PUFO seen before the turn? Because of the time break in the videotape one cannot be certain. However, there is a much brighter light that appears farther to the right in the video. It seems more likely that this Bright UL was the PUFO. The small, Dim UL by itself is illustrated in Figure 13 which is a frame obtained during the 1 second time period at 16:52:53.

FIGURE 13 UNKNOWN LIGHT THAT APPEARED FROM BEHIND A CLOUD

From the image in Figure 13 the flir operator briefly panned to the right revealing a much brighter light that appears several times in the next few seconds. The operator did not seem to

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notice this brighter light, illustrated Figure 14. (Relative brightness levels in the range 0 to 255 are presented for comparison.)

FIGURE 14 TWO UNKNOWN LIGHTS

A comparison of the MED FOV image of the PUFO in Figure 11 with the bright image in Figure 14 shows some similarities in size, shape and brightness. The only thing missing in Figure 14 is the small “dot” image just to the left of the Bright UL in Figure 11. The elevation in Figure 14 reads 2o. However, as shown in Appendix 2, in the section on “Elevation Correction,” because the flir was pointing toward the rear of the plane (azimuth -136o) and because the plane flies tilted upward (front higher than the rear) by several degrees, the elevation was actually close to 0o or a bit lower. Since the horizon was at an elevation angle of about -1.7o, that could mean that these lights were above the horizon (and therefore not on the earth’s surface). Since the elevation of the previous bright “PUFO” was at an estimated elevation of -4o, this could mean that the PUFO increased its altitude and became the bright UL. On the other hand, if both lights are assumed to have been on the earth’s surface, then the higher the angular elevation in the image, the further away is the light. Hence, since the angular elevation of the bright UL is greater than that of the PUFO, the bright UL could not be the PUFO because the bright UL would have been much farther away.

But, if the brighter light is the PUFO that was recorded before the turn, then what is the dimmer light? Could it be another PUFO above the earth’s surface or some light at a great distance but on the earth’s surface? Notice, by the way, the relative brightness numbers: in Figure 13 the Dim UL peak (maximum) brightness is 152 on a scale of 0 (black cloud, not illuminated by the sun) to 255 (maximum recordable brightness or saturation). The brightest sky, at 223, approaches complete saturation. In Figure 14 the Dim UL brightness is 147 which is very close to the previous value. The brightness of the “Bright UL” image is the same as the cloud, 229, and is close to the 232 of the PUFO in Figures 11 and 12. Therefore, if the Bright UL is the PUFO, then its distance from the plane did not change much during the turn, in

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contradiction to the radar target that definitely moved away from the plane during that time interval (16:51 to 16:53). (The radar target distance increased from 2.5 mi to 12 mi during this time. According to the inverse square law, the ratio of radiation intensity reaching the flir from a 12 mile distance as compared to a 2.5 miles distance would be (2.5/12)2 = 0.04, i.e., at 12 miles one would get only about 1/25 of the brightness at 2.5 mi. Such a large decrease in intensity should show up in the flir imagery but does not, indicating that the radar target was not the flir light.)

FIGURE 15 UL ABOUT TO ENTER OR GO BEHIND A SMALL CLOUD

The flir operator switched to the NRW FOV and followed the “Dim’ UL as it appeared to travel along with the plane, going in and out of, or behind, clouds. A typical cloud-passing event worthy of study occurred at 16:53:12 as illustrated in Figure 15. In this video frame the “Dim UL” image is now quite bright, in fact, so bright that it completely saturates the dynamic range with a brightness of 255. (Why did the image get so much brighter? Apparently, at least in part, because the FOV was switched to NRW (narrow) and this makes the image larger allowing for greater resolution of the “hot spot” at the center of the image. This also happened in going from Figure 10 to Figure 11 when the FOV was switched from WIDE to MED. Perhaps the Dim UL was the PUFO after all?) The cloud brightness is still around the 230 range, which is not complete saturation of the dynamic range.

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FIGURE 16 UL BEHIND A CLOUD

According to the flir, the elevation angle at this time was about 3o. Even allowing for a 2o

tilt of the aircraft (see Appendix 2) this means that the light probably was above the horizon (which was at -1.7o). The light appears nearly at the tops of the cumulus clouds. The weather report (see Appendix 2) indicates that the tops of the highest clouds were in the range 10,000 – 12,000 ft, i.e., about level with the airplane. If the tops of the clouds in Figure 15 were at that altitude, then they provide further evidence that the light was above the horizon.

When the UL went into or behind the cloud it completely “disappeared,” as illustrated in Figure 16. The video frame shown in Figure 16 was obtained at the same time that the radar operator reported the radar target was 12 mi behind the plane at 7:00, which is azimuth -150o. Notice that in Figure 16 the UL is almost exactly at -136 which would be 7:30 in “clock” notation. Figure 17 illustrates the difference between flir and radar directions at 16:52:53 at the time when Figure 13 was obtained and 19 seconds before Figure 16. Allowing for a degree or so of possible error in the azimuth readings and for a possible crabbing angle to the right of about degree (see Appendix 2), the azimuth is (just barely) within the range -150 +/- 15 which would be reported as 7:00. So the question is, did the radar operator mean exactly 7:00 (+/- 15 minutes of arc) or was he including angles up to 7:30?

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FIGURE 17 COMPARISON OF FLIR(F) AND RADAR(R) DIRECTIONS

Figure 18 shows the light as it was emerging from the cloud. The brightness (204) is lower because there was still some cloud in the way of the radiation. It should be noted that as this light appeared to travel along its brightness was not always 255 when not partially obscured by a cloud. In fact, its brightness was generally lower than 255. Furthermore, the brightness fluctuated rapidly, generally on a frame by frame basis (30 frames per second, 1/30 sec to change brightness) as demonstrated below. If the light were very far away (many tens of miles) this might have been caused by atmospheric “twinkling” (as with stars).

FIGURE 18 UL EMERGING FROM (BEHIND) A CLOUD

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FIGURE 19 UL USING THE NARROWEST FOV

The flir operator followed this light using the NRW FOV until 16:53:27.57. At this time there is another time break in the tape while the plane made a slight turn to the left. When it resumes at 16:55:16.03 the flir is showing what is presumed to be the same light, but now using the NRWX2 field of view which is only 0.4o wide, as illustrated in Figure 19. At the time of this frame the radar operator stated that the target was 37 mi (31.8 nm) behind the plane. Note that the peak brightness is again 255. This fact suggests that, if this is the same light as appears in Figure 16, it didn’t move away from the plane (or it was at such a great distance that the motion of the plane didn’t effect the brightness)..

The flir operator followed this light continuously until 16:57:47.7. It was during this time interval, at 16:56:28, that the radar target reached Carmen. This is the one time when the direction from the plane to the radar target can be determined independently of the direction given by the radar operator (assuming it actually was over Carmen). Figure 17 shows the flir direction and the radar target direction at this time. The flir azimuth was fluctuating but it was generally around -133 (although at this specific time the angle was -130). At the same time the angle from the track of the plane to Carmen was -149o or almost exactly 7:00. Hence the two were definitely in different directions at this time. This difference in direction and the approximately constant brightness of the flir light (even though the radar target distance kept increasing), when combined, indicate that the radar target and the flir light recorded after the right turn were not the same objects.

The video didn’t record between 16:57:47;7 and 16:59:34.5. When it resumes the flir shows a single light in the MED FOV such as illustrated in Figure 20.

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FIGURE 20 UL ABOVE CLOUDS

Notice the brightness level of the UL. The very next frame after the one in Figure 20 is shown in Figure 21. It shows how much the brightness changed in just 1/30 sec from 128 to 77, a decrease of about 60%. By way of comparison, the cloud brightness levels changed very slightly. The computer-generated dot that represents the center of the FOV also changed slightly from 103 to 111 (about an 11% increase).

The single UL was heading for a nearly vertical “cloud wall” which appears at the very right edge of Figures 20 and 21. It is shown explicitly in Figure 22 as the irregular vertical section of the cloud with brightness 235. Note that the UL was partially obscured by thin clouds and had a brightness of 118. Initial viewing suggested that this was evidence that the light was

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FIGURE 21 UL ABOVE CLOUDS

between the plane and some more distant cloud. However, subsequent analysis (see below) suggests that the light probably was much farther way than the clouds and the radiation penetrated the thin cloud.

FIGURE 22 UL APROACHING CLOUD WALL

A few seconds after the frame in Figure 22 the UL moved behind the nearly vertical cloud “wall.”

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II. C.5 AZIMUTH VARIATION AND DISTANCE ESTIMATES FOR THE SINGLE LIGHT

From 16:55:16 to 16:57:48 the flir operator followed the single light with no time breaks. During these 152 seconds the plane traveled about 9.7 mi and the azimuth of the light relative to the plane changed with time. If the azimuth had remained substantially constant, the implication would be that the light traveled at the same speed as the aircraft or else it was so far away that the direction to it wouldn’t change much even when the plane traveled 9.7 mi. On the other hand, the azimuth did change so it should be possible to determine how far away the light would be if it were stationary relative to the earth. As the plane flies along, the azimuth (relative to the heading of the airplane) of any stationary object it passes would start at a some value (a small value if it were initially ahead and to one side of the aircraft) and then the azimuth would continually increase in magnitude. For example, the azimuth might start at -30o (30o to the left of straight ahead; the 11:00 o’clock position) or perhaps +30o (the 1:00 position) and increase in magnitude continually, eventually reaching 180o (6:00) when it was directly behind the plane. A graph of the azimuth angle increasing as a function of time would show whether or not the azimuth was changing uniformly and, if so, by how much. The azimuth plotted as a function of time for an object moving with the plane at the same speed would make a horizontal line (constant azimuth). For a fixed object at some distance the graph would be an upward sloping line. The line would be nearly straight for very distant objects. The closer the fixed object is to the plane the greater would be the azimuth change which means that the slope of the line on a graph of azimuth vs. time increases as the distance to the object decreases. Figure 23 illustrates these azimuth effects for the flir light. The wiggly line shows the variation of azimuth for every second during the interval.

FIGURE 23 AZIMUTH VARIATIONS OF THE SINGLE LIGHT (UL) COMPARED TO THEORETICAL VALUES

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The distance calculation by parallax (azimuth change) is based on the assumption that the flight track is straight. The flight path of the aircraft is shown in Figure 24. (Note that the plane was heading east-northeast so the latitude increased as the longitude decreased.) This shows that there was a turn to the left just before 16:56:00 and that after that time the flight track was approximately straight. The actual flight track is “jerky”, which is probably an effect of the time lag in the flir indications of latitude and longitude. The non-uniform flight track was probably not caused by large variations in the “crabbing angle” or “yaw angle ” (as seen from above, a twist or rotation of the main axis of the airplane into the wind; see Appendix 2) because the wind velocity was so low (6 mph) that only a degree of twist was needed to counteract the crosswind component (Appendix 2).

On the other hand, variations in the crabbing angle directly affect the azimuth values to be used in this graph. The azimuth values are measured relative to the direction of the airplane but the calculations of distance from the variation of azimuth with time are based on azimuth angles with respect to the earth. If the crabbing angle is small and constant there will be little effect on the distance calculation based on changes in the azimuth. However, if the crabbing angle varies with time as the crosswind varies with time, then any distance calculated from the azimuth values may be erroneous. Fortunately the wind velocity was only about 6 mph with about 4.8 mph of that being the wind component across the airplane track, so the crabbing angle would be only about a degree (see Appendix 2) and fluctuations in the wind would probably cause variations in the azimuth of only a fraction of a degree, similar to the random variations of the azimuth in Figure 23.

FIGURE 24 TRACK OF THE C-26 DURING THE UL OBSERVATION

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The azimuth data in Figure 23 from 16:55:20 to 16:56:00 fluctuate about a constant value slightly larger than 132o. The implication of this near constancy of azimuth is that, for this time period, either the light moved along with the aircraft or it was so far away that the actual azimuth change was much less than a degree. Which possibility is factual cannot be determined from these data. After 16:56:01, as the plane flew along a slightly different but substantially straight path, the azimuth increased slowly (with lots of fluctuations of (+/-) 1o or less, possibly due to variations in crabbing angle and/or “slop” in the flir mechanical system). The average slow increase is consistent with a distant, stationary light or a moving light that was not as far away as an “azimuth equivalent” stationary light.

In order to quantitatively test the hypothesis that the gradual azimuth variation could correspond to a distant light as viewed from a moving platform (the airplane), the equation for azimuth variation with time (multiplied by velocity = distance along the airplane track) was converted to a computer program. This program calculates the azimuth, A(t), as a function of time for any chosen distance to the fixed object under ideal conditions of straight flight at constant speed with no yaw or crabbing angle and no errors in the azimuth measurements. The method of calculating the distance to the fixed location of a stationary light source, herein called the “fixed point” (see the triangle in Figure 23), is based on parallax (measuring the azimuth angles at two locations along a straight path). Trigonometry is used to calculate distance. The mathematics are presented in Figure 23 and the method is described further in Appendix 3. The calculated values of A(t) for several assumed fixed point distances start at 16:56:01 and continue for every second to 16:57:47, a time interval of 106 seconds. During this time the plane traveled about 6.7 mi. The calculation yields a final azimuth value for each assumed distance, such as A(106 sec) = 137o for the 50 mi distance in Figure 23. The best fit of the calculated azimuth variation to the actual variation corresponds to initial distances, Ro, in the range 135 to 200 mi. The calculated azimuths for fixed point distances of 100 mi or less, e.g., Ro = 50 mi, are not consistent with the azimuth data.

The “optical horizon” is about 135 mi away at the altitude of the airplane (see Appendix 2). Thus the azimuth data indicate the fixed point for the light might be at or beyond the “geometric horizon” (about 127 mi). If the heat source were closer than the fixed point then the light was moving (see below) and it was above the ground. The possibility that it was above ground is consistent with the elevation angles during this time (see Appendix 1). The elevation angles were generally 1 – 3o. The tilt of the aircraft reduces these numbers by a degree or two, so the actual elevation angle may have been zero or slightly negative (see Appendix 2 for “Elevation Correction”). However, the elevation of the optical horizon was about -1.7o, as mentioned previously, so one must consider the possibility that the light was above the horizon and, hence above ground (or else explain why the flir elevation readout was wrong; an experiment could determine an error in the elevation readings).

It is important to not misinterpret the information in Figure 23. The calculations of object distance were performed under the assumption that the light was stationary. If this were true it was over 100 mi away. However, the light could have been closer and still have its azimuth increase with time if the component of its speed parallel to the plane were not as large as the speed of the plane. (For whatever direction the light was traveling in, one can multiply the speed of the light by the cosine of the angle between the plane’s travel direction and the light’s

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travel direction to obtain the “component” of speed parallel to the airplane.) The graph in Figure 23 could be used to estimate the speed of the light if its distance were known. Since the 135 mi graph is roughly consistent with the azimuth data, assume the fixed point is 135 mi away. The 135 mi distance corresponds to zero speed parallel to the plane. Because the length of the plane track, 6.7 mi, and the distant fixed point 135 mi away form a triangle, the distance a moving light would travel parallel to the plane track during the same time period would be proportional to the distance from the fixed point to the plane. Consider the small triangle in Figure 23. Halfway between the fixed point and the airplane track there is a line drawn parallel to the airplane track. The length of this line is ½ that of the airplane track. If the light were traveling along that line it would cover ½ the distance that the airplane covered in the same time (106 seconds) so it’s speed would be ½ that of the airplane. If the light were 1/3 of the distance from the fixed point to the airplane track the speed would be 1/3 that of the airplane.

The azimuth data analyzed as above provides an estimate of the maximum distance of the light. The clouds provide an estimate of the minimum distance. As described in Appendix 3, certain cloud features were viewed for long enough times to allow for distance estimates by parallax. Calculations of cloud distances during the period of UL visibility range from 6 to 34 miles. Calculations for other clouds suggest values 10-25 mi. The range of calculated values is great because the baselines (distances the airplane traveled) are short and the azimuth angles are not sufficiently accurate. Also, the clouds are widely scattered over the area, so there may have been distances of many miles between clouds. Therefore the most one can say is that the light was not “very close” to the airplane. For example, it might have been at, say, 20 miles. If this were the actual distance, then, combining this assumption with the result in Figure 23 that the fixed point distance could be 135 mi, would mean that the light was about 135 – 20 = 115 mi from the fixed point and traveling at a speed of (115/135) x 230 = 196 mph. This calculation is not to be taken as definitive, but only illustrative of the situation. Without knowing the actual distance all one can do is make “educated guesses” as to the speed (if any!) of the light.

II. D THE “TWINS AND FRIENDS”II. D.1 THE FIRST APPEARANCE

After the single light shown in Figure 22 disappeared behind the cloud there is another time jump in the videotape from 16:59:51 to 17:03:34 (see Appendix 1). When the video resumes it shows a WIDE FOV scene of clouds and, above some of the the clouds, a nearly linear array of lights spread over about 2.1 degrees of azimuth or about 0.0365 radians (0.0365 rad; multiply the number of degrees by 0.0174 to convert to radians). If the lights were 10 mi away this would corresponds to a distance measured transverse to the line of sight of about 0.365 mi; if 20 mi away, then 0.73 mi; if 100 mi, then 3.6 mi; etc. (Multiply the number of radians by the [radial] distance to get an estimate of the transverse distance or size. Since the actual distance is not known – unless these are oil field gas burnoff flares - one can only assume a radial distance and then calculate a possible transverse distance or “size.” ) There is a single very faint light at the right end, the “leader,” followed by two extremely bright lights which I have named “the twins.” The “twins” are followed at the left by another single light and then a linear collection of lights as illustrated in Figures 25 and 26. (Note: the brightness of the dim lights fluctuates considerably. For example, the “leader” is too faint to see in Figure 25 but it appears in most of

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the video frames, including the frame in Figure 26. Ignore the “x” feature marker in these pictures.)

FIGURE 25 THE “TWINS AND FRIENDS”

FIGURE 26 THE “TWINS AND FRIENDS”

Recall that, during the first seconds of video following the right turn, there was a “Dim UL” near the center of the frame and, at the right side of the frame, there was a “Bright UL” (Figure 14). Comparing with the array of lights displayed in Figure 26 one might wonder if the

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Dim UL was the light labeled “Follower 1” and the Bright UL was the left hand “twin.” Unfortunately there is no way to confirm this conjecture using only the available data. Furthermore, a combined azimuth variation analysis to be presented indicates that the previous single light may not have been related to the “twins and friends.” So this author assumes that the Bright UL was not a “twin.”

Note that the azimuth is about the same as before and the elevation reading is still high enough for the lights to be above the horizon if the flir elevation is correct. Of course, simply looking at the picture one gets the impression that they are above the horizon since they are above clouds that are at estimated distances of 12 and 16 mi (see Appendix 3). The highest cumulus cloud tops are given as 10,000 to 12,000 ft (see Appendix 2). This suggests that the lights above the cloud tops are at about the elevation of the airplane. Unfortunately there is no way to determine whether or not the tops of these particular clouds were at 10,000 ft or so. On the other hand, having the lights appear just above the cloud tops is consistent with the flir elevation angle which indicates that the lights were above the altitude of the airplane if the flir elevation angle is correct.

The “twins” make striking images, like headlights pointing at the camera. The angular separation is about 0.1o or about 0.00174 rad. Hence if they were 10 mi away (10,500 ft) they were separated center to center by about 18 ft; if 20 mi, then 36 ft, etc.. (Actually, this is the separation as projected onto a line that is perpendicular or transverse to the line of sight. If the line connecting their centers was not oriented perpendicular to the line of sight then their spacing could be greater.) This does raise the question of whether or not the twins could, in fact, be the headlights of a very distant airplane many miles behind the C-26 and heading almost in the same direction. This hypothetical airplane could have been undetected by the C-26 radar by being above the radar beam or beyond its range. But, if there were such a plane, why did not ground control report it to the C-26? (Although headlights of an airplane might explain the twins,, it is unlikely that an airplane can explain the “double triplet” lights that were seen after the twins as described below.)

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FIGURE 27 “CLOSE UP” VIEW OF THE TWINS(Ignore the “x” and the “+” symbols)

The flir operator quickly narrowed in on the twins and followed them until they disappeared behind a cloud. A NRW FOV image is shown in Figure 27. The fuzziness of the cloud edge images suggests that the picture is not perfectly focused. There is a slight vestige of a halo around each light and each light has a smaller light below it. The halo effect, which was much more pronounced in Figure 12, may be explained as stray light reflections within the optical system. This is a result of the extreme brightness and suggests that the twins were actually very much brighter than the cloud because there is no halo around the cloud image even though the digitized relative cloud brightness, 226, is not much less than the twin brightness, 232. There is, as yet, no explanation based on internal optics (e.g., internal reflections) for the fainter lights that are below each of the twins. (Ignore the “feature markers” generated by the flir system, i.e., the + sign at the right of the left hand lower light and the X at the right side of the right hand lower light.)

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FIGURE 28 THE TWINS ABOUT TO GO BEHIND A CLOUD

The sizes (diameters) of the twin images, about 40% of the spacing, suggest large size sources (e.g., 8 ft if 10 mi away). However, a large part of this size (perhaps most of it) is a result of high exposure. Evidence of the effect of brightness on image size is provided in Figures 28 and 29 as the twins were about to be completely obscured by a cloud. Figure 28 shows the right hand twin before entering the cloud. Its diameter is the same as the left hand twin. Figure 29 shows the effect on the image size as light moved behind the cloud and the brightness of the light was diminished by passing through the cloud (which acted as an attenuator of the radiation). As the light moved behind the cloud, the cloud density increased and so did the attenuation of the light. Then, as discussed previously, (see Figure 6), the brightness decrease resulted in a decrease in the image diameter. The amount of decrease is illustrated by the circles around the images in the inset in Figure 29. As the light moved into the cloud even farther than shown in Figure 29 the diameter decreased even more as the brightness decreased to that of the cloud itself (the image could not get dimmer than the cloud brightness since the cloud is “on top of” the image of the twin). However. a careful frame-by frame study of the twin image seems to show that the diameter did not shrink to “zero” (a very small diameter). Another contribution to image size could be slight defocus which could prevent the image sizes from shrinking to zero. Nevertheless, it would not be unreasonable to assume that the lights/objects were several feet in size if 10 miles away and larger if farther away. (Note: if the lights are actually bright glints of sunlight from some reflective objects then the effective source sizes could be much smaller than the reflective objects themselves.)

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FIGURE 29 THE EFFECT OF BRIGHTNESS DECREASE ON IMAGE SIZE

The nearly vertical ragged edge of the cloud (see Figure 28) was used as a fixed point in a distance calculation (see 17:03:58 in Appendix 3). The distance came out to be about 17 mi indicating again that the lights were quite far from the airplane.

After the twins disappeared the flir operator searched the clouds and spotted other lights which entered the cloud after the twins. There were at least 14 other lights, some moderately bright, others very dim, which also went behind the cloud. A cluster of 6 of these is shown in Figure 30. The lights were all dimmer than the twins and one was so dim as to be essentially undetectable in a single frame and becomes barely visible when the video is running (the second from the left in Figure 30). All of these images are rather diffuse, as are the cloud edges, which suggests that the scene is defocused somewhat.

II. D.2 THE TWINS AT A DISTANCE

After all of the lights had disappeared behind a cloud the flir operator searched the clouds. At 17:04:36 he widened the FOV to MED. Then he picked up lights that seemed to emerge from behind a cloud, as illustrated in Figure 31. The visual impression of this scene is

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FIGURE 30 DIM LIGHTS FOLLOWING THE TWINS

FIGURE 31 THE RE-EMERGENCE OF THE TWINS AND FRIENDS

that the lights are silhouetted against a more distant dull grey cloud background. However there are no “dull grey” clouds. Clouds are either bright or somewhat bright, when directly illuminated by the sun, or totally black when shaded from direct solar illumination. The lights are actually seen through a clear space between clouds. The background may be caused by sunlight scattered from the distant atmosphere. Brightness levels 50 and 60 are placed in the

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image to indicate the more or less uniform brightness of the “clear” (hazy?) sky. (Note that the elevation of the center of the FOV is 2o, so, if the flir elevation calibration is correct, the lights were above the horizon.)

At the bottom of Figure 31 are two “cloud pillars,” the left one of which was used for a distance calculation (see 17:04:39.8 to 17:05:12.3 in Appendix 3). The calculated distance is 17 mi. Also in Figure 31, curving from left to right over the clear space where the lights appear, is a black “claw” structure of a cloud. The right end of this “claw,” which is almost exactly above the left hand cloud pillar, was also used for a distance calculation. When the video is running one can see by the changes in relative position (azimuth) that the cloud pillar was closer to the camera than was the cloud claw (the pillar moves to the left or “backward” faster than the claw). The calculated distance to the cloud claw is 25 mi (see 17:04:37.0 to 17:05:16.5 in Appendix 3) which is 8 miles farther then the pillar.

The lights went behind another cloud in a long irregular line of clouds of various sizes and shapes. They emerged again appearing to travel over a thin bumpy line of small clouds as illustrated in Figure 32.

FIGURE 32 TWINS AFTER RE-EMERGENCE

The effect of partial blockage of the light by a thin cloud is apparent: the left hand twin image is much smaller than that of the right hand twin, unlike the brightness a few frames later when neither image is blocked by a cloud, as in Figure 33. Figure 33 also shows “follower 1” which has been used in a speed estimate that is based on an assumed distance of 25 mi (this assumes that the light is actually moving, not stationary). The time it takes for follower 1 to reach the cloud feature (edge) that it passes behind was measured (21.6 sec). The distance from the starting point (the circle in Figure 33) to the cloud edge was also measured in degrees and then converted to miles traveled, 0.625 mi, if 25 mi away. Thus, if it were 25 miles away, then it traveled 0.625 mi in 21.6 sec at a speed of about 104 mph. (Note: this calculation assumes that the line of clouds was perpendicular to the line of sight. It is probable that the line of clouds

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was not perpendicular, in which case the travel distance, and therefore the speed, could have been greater than estimated here.)

FIGURE 33 SPEED ESTIMATE

At 17:05:21 the flir operator switched to the NRW FOV and watched “Follower 1” appear to travel over the line of clouds. At 17:05:28 he increased the magnification by changing to NRWX2. Before the light went behind the cloud feature noted in Figure 31, the flir operator brought the image into good focus. Figure 34 shows the image before focus. The light image is large and quite dim, reading only 95 relative brightness units at the center. Even the clouds have less brightness and are quite diffuse. The small cloud pillar at the right in Figure 34 should be compared with the same cloud pillar in focus in Figure 35.

FIGURE 34 FOLLOWER 1 DEFOCUSED

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When the images are focused, not only are they smaller and have more definite edges,

FIGURE 35 FOLLOWER 1 FOCUSED

they are also brighter.

II. D.3 DISTANCE ESTIMATE FOR THE TWINS

FIGURE 36 TRACK OF THE AIRPLANE

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A distance estimate to the twins has been made using the same azimuth variation method as described before. During the time that the twins were visible the plane flew in a substantially straight line as illustrated in Figure 36. Figure 37 shows the azimuth variations. There is a gap in the data as the twins went behind clouds (see Figures 29 – 33). The data are sparse, consisting of a cluster of azimuth data points from 17:03:35 and 17:04:00 and a smaller cluster of points from 17:05:17 to 17:05:44. Nevertheless, the available data are more consistent with a distance on the order of 100 mi than either 50 or 200 mi.

FIGURE 37 AZIMUTH OF THE TWINS

II. E THE “DOUBLE TRIPLETS AND FRIENDS”II. E.1 THE TRIPLETS BEHIND CLOUDS

FIGURE 38 TRIPLET/QUADRUPLET SEEN THROUGH CLOUDS

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After the “Twins and Friends” disappeared behind the clouds at about 17:05:44, the flir operator continued to search the clouds using the NRWX2 FOV. At 17:06:25 lights were seen again and continued to be seen through breaks in the clouds until 17:06:45 when the lights entered an area free of clouds. During this time it became apparent that this arrangement of the lights was different from arrangement of the “Twins and Friends.” The brightest lights now form a “Double Triplet” and also a “Quadruplet and Triplet.” There is a lead light, then the Double Triplet (or Quadruplet) and finally the followers.

Figure 38 (NRWX2 FOV) shows one frame during the second view of the triplets. The lagging triplet and three followers were first seen briefly through small holes in the clouds at about 17:06:25. This particular image, obtained about 11 seconds later, shows the last member of the leading triplet at a time when the last member had split into two lights (actually making a “quadruplet”). This image also shows the second member of the second triplet; the first and third members are behind clouds. Figure 39 shows the lights as they emerged from behind clouds. Although it may appear to the observer that the cloud is behind the lights, it is more likely that the lights were beyond the clouds and the radiation had penetrated the thin cloud. As is usual, the image sizes changed with brightness. At their dimmest the images shrink to about the minimum possible size of an image based on the “pixilation” of the image (no image can be smaller than 1 picture element or “pixel”). This observation suggests that the lights are, indeed, essentially unresolved point sources, that is, radiation sources that are so far away that they appear as tiny “points.”

FIGURE 39 TRIPLETS EMERGE FROM BEHIND A CLOUD

II. E.2 THE TRIPLETS IN CLEAR SKY

After they emerge from behind the cloud the triplets are seen against a uniform background, perhaps the atmosphere itself rather than ground, since the elevation is given as 2 - 3o. As pointed out before, because of the tilt of the aircraft, the elevation may actually be about

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0o, but probably not as low as the horizon elevation of -1.7o. Figure 40 (NRWX2 FOV) shows the “triplets,” the “leader” and one of the “followers.” Notice that the brighter images are larger than the dimmer ones. Compare brightness levels of the same lights as they appear in different frames (different figures). The lights fluctuate in brightness from frame to frame, in 1/30 sec. The dimmer lights visibly change greatly in brightness, sometimes going from nearly invisible in one frame to quite bright in the next frame. The brightest lights also fluctuate, but not as much. The lack of large fluctuations of the brightest lights may be a result of nonlinearity of the intensity response of the analogue electronic circuitry (high end saturation of the intensity vs voltage response).

FIGURE 40 THE TRIPLETS AND FRIENDS

Figure 41 shows some of the other followers in a “one up, two down” arrangement that is the opposite of the triplets (two up, one down). Again compare image brightnesses. The brightest lights stay about constant but the dim lights, such as the first follower, change greatly (183 in Figure 40 and 85 in Figure 41). The rapid brightness fluctuations are characteristic of distant lights (radiation sources miles away) seen through the atmosphere.

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FIGURE 41 THE TRIPLETS AND FRIENDS

II. E.3 THE QUADRUPLET TRIPLET

The third light in the first triplet in Figure 38 appears as a doublet, making the triplet into a quadruplet. In Figure 39 it is a singlet and in Figure 41 it is again a doublet. When viewing the video one can see this change from single to double occur gradually. Figure 42 shows the leading triplet just before it (and the others) disappeared behind another cloud. The spacing between the smaller lights which make up the doublet in Figure 41 (17:07:09) has increased considerably in Figure 42, which occurred only 7 seconds later (17:07:16). (It was at about this time that the flir operator reported to the air crew that there were 11 lights apparently traveling with the airplane.)

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FIGURE 42 FINAL VIEW OF THE QUADRUPLET

The change from a single light image into two images over a period of seconds could be a result of actual splitting of the light, but it may also have another explanation. This alternate explanation is based on the assumption that there are two lights, one farther from the plane than the other, and that they appear as a single light during the brief time that they are aligned with the airplane. The line joining them would intersect the airplane flight path at a specific location. When at that location the viewer would see only a single light (one in front of the other). Approaching that location the observer would see two lights. Because they are on the left side of the airplane track (-138.8o Az) the farther one would appear to the left of the nearer during the approach to the alignment point. When leaving that location the observer would again see two lights, but this time the farther would be to the right of the nearer. When approaching that location the apparent spacing would decrease to zero and when leaving that location the apparent spacing would increase. Thus one would see two lights approach one another, then become “one” and then become two again, with an ever increasing separation. The latter observation, continual increase in separation after appearing as a single light, is recorded on the tape. (The shrinkage of the spacing, if this actually occurred, is not recorded because the lights were behind clouds).

This hypothesis makes a testable prediction about the data, as illustrated in Figure 43: the spacing should increase linearly with time (make a straight line on a graph of spacing vs time) except over some range of time near zero (zero time is when the lights are aligned) where the sizes of the images make it difficult or impossible to determine the spacing. (If the images were perfect points then the separation would reduce to zero spacing at the time of alignment.)

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FIGURE 43 DOUBLET SPACING THEORYThis is a view from above, as if the lights and the plane were at the same altitude.

The small circles along line R,D are the closer and more distant lights. The distance from the alignment point to the airplane at time t is vt. Az is the

azimuth at alignment and a is the angular spacing between the lights at time t.

The spacings of the images after they separated have been measured each second for 10 seconds. Figure 44 shows that the spacing data, sparse though they are, tend to agree with this theory.

FIGURE 44 DOUBLET DATA AND THEORETICAL TIME VARIATION

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Unfortunately the image sizes change with brightness and their shapes also change somewhat from frame to frame. Therefore it is difficult to determine what amount of shape change or image growth (in width) is caused by random fluctuations and what is caused by an increase in spacing of the lights. Therefore no measurements were made until the spacing became large enough to be definitely measurable. The first measured spacing is 6 mm and it occurred at about 17:07:09. (The zeroes between 0 and 15 sec are not data.) The data lie roughly along a line that projects to zero time 15 sec before the 6 mm point. In terms of time on the videotape this is 15 sec before 17:07:09 or 17:06:54. In other words, this theory combined with the data predict that the alignment occurred at 17:06:54 and at that time the image spot should be minimum. In fact, when observing the videotape as it runs one sees that the single light image (when the two are too close together to resolve) fluctuates about a minimum size (which changes somewhat with brightness variations) from about 17:06:48 to about 17:07:00, which includes the predicted minimum time. Figure 40 shows the doublet as a minimum sized single image.

The previous discussion applies only to the “horizontal alignment” of the two assumed lights, which could occur if the light sources are side by side, at the same elevation as the airplane. It should be obvious that if the two hypothetical lights were both at the altitude of the airplane that they could align horizontally and vertically, i.e. they could, for a short time, appear as one light image rather than two light images, one above the other). The flir elevation readings (2o), after being correct for airplane tilt (subtract about 2o), suggest that the lights were about at the altitude of the airplane. However, it is not immediately obvious that (apparently) perfect alignment of two lights could be consistent with the hypothesis that the lights were at the surface of the earth, many miles away (possibly gas flames associated with oil wells; see Appendix 4). This hypothesis requires that the lights align vertically as well as horizontally. The accuracy of vertical alignment can be checked by assuming a distance to the closer light, S, and a separation between the lights, L. These distances lie along a “straight” line over the curvature of the surface of the earth, starting under the airplane and going to the closer light at distance S and then to the farther light at distance S+L. It is necessary to calculate the (negative) elevation angle, E, first for the light at S and then E for the light at S+L. The difference, ES – ES+L between these would be the vertical angular separation of the images of lights at distances S and S+L. The geometry and equations for E are presented in Appendix 4, Figure A4C (but without showing L or S+L).

Suppose the closer light was 111mi (=S) away and the farther was 114 (=S+L). (These are distances of two oil platforms discussed in Appendix 4.) Then, using the equations presented in Figure A4C, one finds the elevation of the closer light is ES = -1.840o and the angle of the farther light would be ES+L = -1.835o (Note: this calculation ignores the atmospheric refraction which might reduce these angles slightly, but not the difference between them.) The difference between these angle is 0.0053o. At the time of Figure 40 the field of view was NRWX2 so the scale factor to convert degrees to mm on the TV screen used to view this video (see Appendix 2) is 0.0012o/mm. The angle 0.0053o therefore corresponds to 0.0053/0.0012 = 4.4 mm. With this much spacing the images would always appear one above the other; they would never align to become a single image.

The only way to achieve horizontal and vertical alignment of two distant ground lights would be if they weren’t at the same height above the ground. Specifically, the closer would

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have to be higher. The vertical angle 0.0053o = 0.000092 rad corresponds to a height difference at 111 mi of 0.000092 x 111 x 5280 = 54 ft. One wonders, however, at the probability of there being two hot radiation sources or solar radiation reflectors that just happen to be at the correct spacing horizontally and vertically to allow for a chance alignment from the point of view of a (distant) aircraft. (Note: if the lights are oil field fires this postulated alignment could be proved by a flight experiment.)

II. E.4 DISTANCE ESTIMATE FOR THE LAGGING TRIPLET

Once again the airplane track is found to be essentially a straight line (see Figure 45) but this time the azimuth values are so scattered as to make impossible a good distance estimate (Figure 46). Note that two very different sloping lines have been drawn on the azimuth graph corresponding to about 127 mi and about 300 mi, but they are nearly meaningless. However, when the triplet azimuth data are combined with the twin azimuth data there is a significant result, as shown in the next section.

FIGURE 45 TRACK OF THE AIRCRAFT FOR THE LAGGING TRIPLET

II. E.5 TWINS AND TRIPLETS TOGETHER

Because the light arrangements of the twins and triplets are different there is a question as to whether or not they are the same lights seen from somewhat different locations along the track of the plane or if they actually rearranged themselves somehow. On the other hand, if they were the same lights only appearing different because of the different locations of the plane, then, assuming they were either stationary or moving with the plane but more slowly, it should be

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FIGURE 46 AZIMUTH VARIATION FOR THE LAGGING TRIPLET

possible to combine the azimuth data of the twins and triplets and get a better estimate of distance. The first step in this process is to check for straightness of the airplane track. The track is shown in Figure 47. It is substantially a straight line.

FIGURE 47 THE TRACK OF THE AIRCRAFT

Figure 48 shows the result of combing the azimuth data for the twins and triplets. Despite the wide, seemingly random variations in azimuth shown in Figure 46, when these triplet data are plotted on an azimuth scale that includes the azimuth data of the twins, there appears a degree of consistency that suggests a reasonable fixed point distance would be 100 +/- 25 miles.

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Again, a distance as short as 50 mi is not consistent with the data. (Note: isolated data “dots” between 150 sec and 225 sec are based on the initial occasional views of triplets through clouds.)

FIGURE 48 AZIMUTH DATA FOR TWINS AND TRIPLETS

II. E.6 THE GRAND COMBINATION: SINGLE, TWINS AND TRIPLETS

The comparison of lights in the previous discussion raises another question: was the single light which was viewed from 16:55:20 to 16:57:48 one of the “Followers” of the Twins? To answer this question, the single light azimuth data were plotted on a graph along with the azimuth data of the twins and triplets. First, the airplane track was checked. Figure 49 shows that there was a bend in the track of about 1.8o corresponding to a slight turn to the left. This would decrease azimuth values (of some fixed object far from the plane) immediately following the turn by 1.8o as compared to the values just before the turn.

Figure 50 shows the result of combining all these azimuth data. It also includes a small cluster of data points for a single light which were not used before. This graph does make it seem that the single light data are not related to the twins and triplets data. If the single light and the twins and triplets were all at the same fixed location on the earth one would expect a continual increase in azimuth over the whole graph but not a decrease of about 5o between the final single light azimuth values (about 137o) and the initial twins azimuth (about 132o). As mentioned above, the left turn of 1.8o could create an abrupt decrease of 1.8o in the azimuth values at the time of the turn. However, the actual decrease is closer to 5o so the slight left turn of the aircraft cannot explain the abrupt decrease in azimuth. The unfortunate time

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FIGURE 49 AIRPLANE TRACK FOR SINGLE, TWINS AND TRIPLETSLongitude values are 90o + the listed value in minutes of arc

Latitude values are in minutes of arc above 18o 22’.

breaks in the data tape mean that it is impossible to demonstrate a continuous change in azimuth during the times covered by those breaks.

Nevertheless, the combined data increase the probability that the lights were fixed at a considerable distance, perhaps 80 – 120 miles from the airplane or else they were traveling along a track that was roughly parallel to the plane but at a slower speed. Because the elevation angles were continually 2 – 3 degrees it seems possible that, even with the correction for airplane tilt, the lights were above the horizon. The great distances of the fixed point implied by the changes in azimuth (parallax) seem inconsistent with the great brightness of some of the lights. If one accepts the great distances and assumes that the lights were stationary on the ground, then one must explain what IR radiation sources would be of sufficient brightness to appear as bright as or brighter than sunlight on clouds after traveling through 80 – 120 miles of hazy atmosphere. (Note: it has been suggested that all the lights were gas burnoff flares from oil platforms northeast of Cd. del Carmen as illustrated in Figure 1. Such flares could be sufficiently bright for detection under good weather conditions. Whether they could be as bright as the brightest images in the video can only be determined by experiment., The flare hypothesis is discussed in Appendix 4.)

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FIGURE 50 COMBINED AZIMUTH DATA

II. E.7 LACK OF RADAR CONTACT

During the sightings of the single light, the twins and the triplets the radar operator was continually watching for any targets. In particular, he attempted to locate any potential radar target that could be one of lights that formed the “Triplets and Friends.” Just after the lights disappeared behind clouds the following conversation took place on the plane:

17:07:25 Radar Operator (RO): At what position do you have it, Tellez ?FLIR Operator (FO): At 8:30 o’clock. 8 or 9 o’clock.17:07:30 RO: At 9 o’clock? I don't think the radar can catch them. We have a blind angle there, Sir.FO: At 8:30?RO: Yes.17:07:37 RO: We have 20 degrees of blind angle there.FO: But you should see them now.17:07:42 RO: Yes, but it's strange, now I have nothing. I continue searching.Place Commander (PC): Don’t lose them, flir. Now they are going to appear again. They should be visible.,FO: There’s a bunch of clouds.RO: I didn’t catch them, on radar, sir.FO: Maybe a bit later we’ll see them again.Unknown Voice: Is that all the zoom, Tellez?

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FO: Affirmative.PC: At what position (direction) did we lose them, Tellez?FO: At the 8 o’clock position (direction).

Notice that the flir operator gave the position (sighting direction) as 8 o’clock, which, strictly interpreted, would be -120 +/- 15o. Actually the flir direction was about 140o (see Appendix 1) which is closer to 7 o’clock (150 +/- 15o). The flir operator continued searching but the lights did not appear again. The video recorder was turned off at 17:09:08.

Clearly the radar operator and the air crew were puzzled by the fact that there was no radar detection despite the number of objects detected by the flir. The radar operator offered a possible explanation based on a limitation of the radar (a “blind spot”), and this might be the explanation. However, the flir data offer another possible explanation: if the lights were above the horizon and the radar was still pointed downward at about 30o then the objects/lights would have been well outside the main radar beam and so might have been too high for the radar to detect. A related possibility, even if the lights were objects at ground level, is based on the distance estimates. The distance estimates presented above are so great that, if the radar beam was pointing downward, then the objects may have been too far away to detect in the upper “sidelobe” of the radar beam. If this last explanation were true, then the lights were farther than about 40 mi from the plane, because the radar had earlier demonstrated the ability to detect a target above ground at about 40 mi.

The preceding discussion can be summarized as follows: the available data suggest that there might have been bright sources of infrared radiation which were not detected on radar and not seen visually which might have been traveling at some altitude and speed roughly parallel to the airplane at distances greater than, say, 10 mi and perhaps at distances greater than 50 miles. The failure of the air crew to detect the lights visually may be because (a) they did not radiate or reflect in the visible range of light wavelengths (0.4 – 0.7 microns) or (b) they did radiate or reflect in the visible, but the haze in the atmosphere over a long optical path (tens of miles) was sufficient to attenuate the visible portion of the radiation to the point that they could not be distinguished against the glow of the atmospheric haze (“path luminance”). Hypothesis (b) would also apply to the infrared radiation: there would be attenuation, but the attenuation at the infrared wavelengths would have been much less than at the visible wavelengths so the infrared could be detected over the long distance even though the visible was not. (In other words, the long path through the atmosphere acted as a frequency selective filter, allowing the infrared to pass and “blocking” the visible.)

There is, as of this writing, no definitive identification for these “lights.” (See Appendix 4.)

II. F DIMMER TWINS AND THEIR BRIGHT FRIENDII. F.1 FOLLOWING THE LIGHTS

Nearly 6 minutes later, at 17:14:58, the flir operator again turned on the video recorder. The NRWX2 FOV shows two lights side by side, much dimmer than the twins. They appear against a generally featureless dim, electronically noisy background. Then the camera rotates to

55


the left and a very bright single light with irregular shape appears. Some of the background “noise” moves in a correlated way which suggests that the background actually is an image of

FIGURE 51 DIM DOUBLE LIGHTS

the ground that is so dimly lit as to make a seemingly random image barely above the electronic noise level (this would be what is called “snow” on a regular TV). Figure 51 shows the initial image of the two lights. Figure 52 shows the separation between the pair at the right and the bright light at the left and Figure 53 shows the bright light by itself.

The azimuth reading for this light is -90o +/- 0.5o. Hence its direction is about 45o to the right of the previous lights (clockwise rotation from the -135 azimuth). An important difference between the previous lights and these three is that the elevation now reads -1 or -2o which means that the sighting line is essentially downward (if the elevation reading is correct). The negative

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FIGURE 52 DIM DOUBLE AND VERY BRIGHT LIGHTS

FIGURE 53 VERY BRIGHT LIGHT

elevation is consistent with the hypothesis that the motion of the background “structured noise” results from dimly perceived images of the ground scenery. This “structured noise”can only be seen when the video runs; it can’t be visually detected in a still frame.

At 17:16:18 , the flir operator switched to the NRW FOV and the background became less noisy and looks more like the ground. As he searched with the flir, moving it in many directions, a dim pair of lights and a few other single dim lights appear to move with the background, as does the bright light, suggesting that the lights could be on the ground. Then at 17:16:28 he switched to the MED FOV and clouds become visible at the top of the screen, as in Figure 54. The “dimmer twins” no longer appear.

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FIGURE 54 BRIGHT LIGHT

The flir operator continued to follow this light as its azimuth increased. The light was viewed part of the time through the rotating propeller blades. In Figure 54 a propeller blade is beginning to come into the scene. It then moves to the right, as the flir operator rotated the flir to the left to follow the light (as the plane flew past the light). The propeller creates a very annoying flashing effect. However, because of the 1/30 sec time of a video frame it is possible to “look between” the propeller blades. While viewing the video frame by frame one can see the top of a cloud pass between the plane and the light. Figure 55 shows the scene a few frames before that happens (propeller blades are at the left and right sides creating a whitish glow). As the video runs, the cloud top moves to the left faster than the light (evidently it is closer) and momentarily obscures the light. An estimate has been made of the distance to the cloud (see Appendix 3, 17:17:38.5 to 17:18:01.0), about 14 mi.

The operator continued to watch this light until about 17:18:35, at which time he turned the flir to search for a radar target that had been detected ahead of the airplane. The last view of this light is at 17:18:39 as it moves rapidly to the left out of the FOV (as the operator turned the

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FIGURE 55 A CLOUD TOP ABOUT TO OBSCURE THE BRIGHT LIGHT

flir to the right). This light was not seen again, although the flir operator did search for it a short time later.

II. F.2 AZIMUTH VARIATION AND DISTANCE OF THE BRIGHT LIGHT

As with the previous lights, this one was tracked more or less continually for several minutes allowing for a parallax calculation of distance. As Figure 56 shows that the airplane track was straight during this period.

FIGURE 56 AIRPLANE TRACK FOR THE BRIGHT LIGHT AT 9LEFT

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Figure 57 shows the azimuth variation data. Compared to previous data this is remarkably consistent, suggesting that there was very little variation in the yaw or crabbing angle of the airplane and that the flir system did a good job of tracking the light. A straight line can be drawn through the data which means that the data are consistent with a stationary.light at 50 mi of a moving light at a closer distance.

FIGURE 57 AZIMUTH OF THE BRIGHT LIGHT AT THE 90o TO THE LEFT

Figure 58 shows the data compared with predicted azimuth variations for various assumed distances. This time the 50 mi distance is found to provide an excellent fit

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FGURE 58 AZIMUTH VARIATION FOR VARIOUS DISTANCES

to the azimuth data. It was beyond a cloud that was at about 14 mi distance. If it was just on the far side of the cloud then it traveled at about 3/4 the speed of the plane. But if it was 50 miles away then it was probably stationary and on the ground. The elevation angle for an object on the ground at 50 miles would be about -2.3o. The flir system indicated elevation angles in the range 0 to -2o (see Appendix 1).

II. H. SHOOT THE MOON

At 17:15 the radar operator reported a new target at the 1:00 position. The flir operator decided to search for it. He turned the flir to the right but saw nothing. Whilte rotating to the left to find the previous flir light, however, he noticed the moon rising in front of the plane. This provided a useful calibration of the flir. The moon at the time was about 1.3o above horizontal but the flir gave the elevation as -1o . Hence it appears that the plane was tilted upward (front higher than rear) by about 2.3o. The implications of this for flir pointing calibration are discussed in Appendix 2.

After confirming the moon by switching to the normal TV and seeing the craters, the operator then turned the flir back to the left side to search for the previous light but couldn’t find it. He searched the clouds, occasionally concentrating on small bright cloud areas, until 17:21:12 at which time he turned the recording off. He resumed at 17:22:53, taping what was probably a small cloud area until 17:23:42. He again turned off the recording and resumed at 17:26:30. He then recorded a bright light behind the plane until the end of the tape. This is described below.

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II. I THE SECOND RADAR TARGET

At 17:15:05 the radar operator reported another target, this time ahead of the plane.

17:15:05 RO: No, almost at 1 o’clock. (refers to a new radar target)PC: Well that's one, but we have another at 9. (refers to the flir light)17:05:08 RO: Ah, I have one here. Let me check the one at 9 o’clock.17:15:12 FO: There are two others ahead of that one.PC: And they're the same ones, eh?FO: We have them in front, in the middle and at 8 o’clock.

The crew reacted to this discovery with some dismay because there was a flir target at the left which couldn’t be detected on radar and a radar target in front that couldn’t be detected on the flir (see II. H. above).

17:15:22 Unknown Voice: [Laughing] Fasten your seat belts!UV: Let's see...UV: [nervous laughter] Don't scare me.17:15:26 FO: Can anyone check through the window? We have one going along at approximately 1 and the rest are still going along at our 9 o’clock17:15:33 UV: Does that means we are surrounded?FO: Well... I don't know.UV: At what distance?17:15:40 UV:[in a loud voice, as if making an announcement] Boys, today is your lucky day!![Laughter]UV: At what distance?UV: And that one is getting near to the other one.UV: Do you mark the distance?FO: I don’t get distance readings.17:15:49 RO: Okay, the one that is at 1 o’clock, I have it at 22 mi (19 nm) going at 60 mph (52 kt).

This radar target is 30o +/- 15o to the right of straight ahead and traveling at a typical “ground speed.” Most aircraft have to fly at 70 mph or faster to maintain lift. Projecting the radar target location onto the airplane track, the component parallel is 22 cos(30) = 19 and the component perpendicular is 22 sin 30 = 11; the target is 19miles ahead and 11 miles to the right.

17:15:59 FO: I have one at 9 and in front of it, approximately at 9 o’clock as well, are two luminous points. This refers to the “dim twins” and the Bright Light at the left, described in section II.F.

RO: At what position, Tellez?FO: At 9 o’clockRO: At 9? Hold on, let me check...UV: How far away is the one at 9 o’clock?PC: The one at 1 o’clock… how many miles?RO: 19.7 mi (17.1 nm), heading 95, speed 60 (52 kt)

The distance is now 17.1 nm = 19.7 mi. The radar target is traveling along 95o, almost due east, while the plane is traveling along a direction of 81o, so the tracks of the plane and the radar target are diverging by 14o. The speed was still 52 kt = 60 mph. The plane is going

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approximately 230 - 60 = 170 mph = 2.83 mi/min faster than the radar object. The conversation turned back to the lights at the left at 9:00:

17:16:35 FO: It's going along at our altitude. It can’t be possible.UV: Tellez, what’s the position of the one you have got right now?FO: Which one? The one below the cloud? That one?UV: Yes.UV: That one you have there, what's it's position?17:16:44 FO: I have it at 9.17:16:48 UV: Gentlemen, we are not alone. [Laughter]17:16:52 PC: Yes, we are not alone, right. It's true. How strange this is. They are surrounding us.

Clearly the crew was willing to inject some levity into the potential gravity of the situation.

17:16:55 FO: Hey, the one that you have at 1 o’clock?RO: At 1. I have it at 17 mi (14.8nm) at 1 o’clock, heading 99 with a speed of 60 mph (52kt).

This radar target at 1 o’clock and speed 52 kt = 60 mph is 14.8 nm = 17 mi away. The heading is now 9o south of due east, while the plane is still heading about 9o north of due east. Since the last message that provides a specific time and distance to this target, the message at 17:15:49, the plane traveled for 1 minute and 6 seconds (1.1 minutes) at 3.8 mi/min corresponding to 4.2 mi. If the object maintained its speed at 60 mph = 1 mi/min for 1.1 minutes, then it traveled 1.1 mi in nearly the same direction as the airplane. The distance to the target would then decrease to 21.9 – 4.2 + 1.1 = 18.8 mi. However, the measured separation distance was 17 mi, not 18.8. If the radar target had not moved at all, the separation distance would have been 21.9 – 4.2 = 17.7 mi. To get the distance as low as 17 mi one needs to assume that the target moved toward the plane a distance of 0.7 mi in 1.1 min or 0.64 miles/min or 38 mph. This seems to contradict both the heading (99) and the speed (60 mph)..

FO: You are checking one and I’m checking another one. I mean other ones.UV: Isn’t it a plane?UV: Hey, you have to record this.FO: I am recording it.17:17:31 PC: Watch your language gentlemen, but what is this?UV: Couldn't be a (-) plane. It's stopped there!UV: Ehh?FO: Do you have it in sight?UV: It seems so. Let’s wait for it to get closer.17:17:39 UV: Nah! It's stopped there!UV: It's stopped? On the ground?FO: No but the one that we have at 9...UV: It's in the clouds.

17:17:49 PC: The one that we have at 9 is following us. It is going parallel to us. It is in formation with us. Now we’ve got those clouds. It’s above us.FO: The one that you have at 1 where do you have it?RO: Hold on. I have it at... at 1.

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At this time light at the left was seen between propeller blades. The light was momentarily blocked by a cloud top estimated to be about 14 mi away (see Figure 55 and Appendix 3).

17:18:15 UV: There is one that is getting near to us.RO: At 1 of our position at 11.5 mi (10 nm) distance and speed of 60 mph (52 kt).PC: OK. Don’t lose it, radar. Look for the one at 1 o’clock. It’s coming closer to us.

At this time the distance to the target was 11.5 miles. The last mention of the target at 1 o’clock and speed 60 mph was at 17:16:55 when it was 17 mi away. Since then the plane has traveled toward the radar target for 1 minute 20 seconds = 1.33 minutes and covered a distance of 1.33 x 3.8= 5 miles. The radar target has traveled away from the plane a distance of 1.33 x 1 = 1.33 mi. Thus the separation should be 17 – 5 + 1.33 = 13.33 mi. Since the separation is actually 11.5 mi, it seems that the target moved toward the plane by 1.8 miles in 1.33 minutes at a speed of 1.35 miles/min = 81 mph.

17:18:53 RO: Now I get it at 2 o’clock at 10.35 mi (9 nm), heading 120, speed 60 mph (52 kt).

Evidently the radar target is turning southward from its initial eastward track.

PC: Quality 1?RO: No. Now I get it with quality 9, but I got it with quality 1 before.

(Note: “quality” is a measure of the strength of the radar reflection as compared to background noise. Higher quality numbers correlate with weaker signals.)

At this time the flir operator ended his search for a light at the right and happened to pick up the moon, as described previously. Initially there was some question as to what it was.

17:19:30: PC: Ahhh, that isn’t the one. That’s the moon. Is the moon in front?UV: There it is.PC: OK. It is the moon. We are not chasing the moon, are we?[Laughter]

(NOTE: THERE ARE NO TIMES GIVEN FOR THE FOLLOWING STATEMENTS)FO: One of them, was at 12 and the other one was at 9.RO: Now I get it almost at 3 o’clock, 7.9 mi (6.9 nm), heading 155, speed 60 mph (52 kt).

By this time the plane was flying past the target at the right which was now at its closest. It was heading southeast, almost perpendicular to the airplane’s flight path, at the same “ground speed.”

PC: OK, that isn’t the one. It’s a cloud. Yes, it is a cloud.FO: Where is the one you had at 1 o’clock now?RO: It’s now at 3 o’clock, 8.2 mi (7.1 nm), direction 182.PC: OK, it is pulling away now. Who knows what it was!

At this time the radar target was heading almost exactly south and it and the plane were separating rapidly. What might it have been? The low speed is a clue that it could have been a vehicle on the ground. Figure 1 shows a road running just south of eastward not far from the

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airplane track and several other roads that head directly southward. As investigator Brad Sparks was the first to point out, the radar reflector might have been a truck that was heading eastward and then turned south. The rear end of a pickup truck or a larger empty truck could be an effective corner reflector for the 3.2 cm wavelength radar radiation. This doesn’t prove it was a truck (or other vehicle) but it is a valid speculation.

II. J THE LIGHT BEHIND THE PLANE

After searching for a light on the right side of the plane the operator then turned the flir back to the left side to search for the previous light but couldn’t find it. He searched the clouds, occasionally, concentrating on small bright cloud areas, until 17:21:12 at which time he turned the recording off. He resumed at 17:22:53, taping what was probably a small cloud area until 17:23:42. He again turned off the recording and resumed at 17:26:30. He then recorded a bright light behind the plane until the end of the tape. This light was at about 175o azimuth and 1 – 2o

FIGURE 59 THE LAST BRIGHT LIGHT

elevation continually. A tracking gate (box) appears around some of the images, such as in Figure 59. The video shows this light continually getting smaller and dimmer until the end of the tape at 17:28:06. The nature of this light is unknown. It is so nearly directly behind the plane that there is no azimuth increase noted as time goes on, or any such increase is “covered

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FIGURE 60 THE LAST BRIGHT LIGHT

up” by the random fluctuations of the azimuth readings. Since it seems to shrink in size and diminish in brightness as time goes on, as illustrated in Figure 61, it is possible that it was stationary and the plane was flying away from it. The elevation angle of 1 - 2o suggests that the light was above the plane. However, as pointed out previously, the plane flies with an upward tilt so the elevation angles, when pointing rearward, may read as much as 2o too high. Hence the light might be on the ground. It was not possible to estimate the distance by parallax because it was almost exactly behind the plane.

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FIGURE 61 FINAL VIEW OF THE LAST BRIGHT LIGHT

During the appearance of this last light the air crew had a few more comments.

7:26:34 PC: (Reporting to ground control) The radar has them. The FLIR has them. We are seeing on the FLIR a luminous object. It has an opposite course to us but it doesn't leave us. It's two miles behind us.

(Note: what was the source of the information that it was 2 miles behind? Was this a guess?)

UV: It does not move apart from us, no, it doesn't move apart from us.

17:26:53 UV: Correct, who knows what it is.PC: Is it still with us, Marin?RO: Yes I still have it here.PC: The radar is catching now what it didn’t catch in its entire life.[Laughter]

17:27:11 RO: Now it lost a bit of quality, I have a quality of 5.UV: Again look it looks...RO: The one that we had at 1 o’clock is lost now.UV: Okay.RO: It just left the Radar.

17:27:58 PC: How weird is all this.…………………………………………………….

Weird, indeed!!

IV. CONCLUSION

A firm conclusion awaits a test flight to prove or disprove the ground light/gas flame theories that have been proposed. However, a conclusion can be made regarding the initial radar target and a tentative conclusion can be made regarding the twins, triplets and friends. The first radar target was odd in that it was close, yet not visible visually, as would be an airplane, nor was it visible in the flir, meaning that it didn’t generate heat. (One light seen just before the right turn might possibly have been associated with the radar target, but, if so, the question then is, why wasn’t it seen much earlier? Any man-made aircraft capable of traveling several hundred miles per hour generates and emits heat.) The first radar target remains unexplained. The single flir light seen after the right turn, the “twins and friends” and the “triplets and friends” may have been above the horizon. If so, they were probably moving and are unexplained. However, if the flir elevation calibration when looking to the left was off by several degrees (despite calibration using the moon elevation), then these groups of lights might have been distant (ca. 100 mi) ground lights (e.g., gas flames; see Appendix 4). The other flir lights and the second radar target are unidentified but could have been on the ground.

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ACKNOWLEDGEMENTSThis research could not have been accomplished without the help of the Department of

Defense, the air crew (pilot Maj. Magdaleno Castanon Munoz, radar operator Lt. German Martin Ramirez, flir operator Mario Adrian Vazquez Tellez), Major Sergio Soto, Jaime Maussan and translators Santiago Ytturia, John Velez , Patricio Elicer and Alexandro and Kurt Franz, who compiled the most complete translation. I also thank Thomas Strunch (Denmark), James smith (USA), Laurent Leger (France), Kento Mori (Brazil) and Brad Sparks (USA) for helpful discussions and analysis. Technical details of the radar system were provided by James Blumling of Telephonics, Inc. and information about the flir system was provided by John Miller of FLIR Systems, Inc.

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APPENDIX 1THE FLIR VIDEO DATA

TABULATION AND TRANSCRIPTBy Bruce Maccabee

Note: I sent early versions of this transcript to other researchers who used it as a basis for their

own analyses and conclusions,some of which have already been published on the web.

FLIR VIDEO DATA AND ANALYSIS REVISIONS:

Initial version May 20, 2004NOTE: Calculations presented below should be considered tentativeand data values may be changed or re-interpreted as the investigation continues.

Rev. 1 , May 24 distance calculations follow the basic data tableRev. 2, May 25, corrected measurements, added material (some basic data and speed calculation)Rev. 2B May 28 slight addition and word changeRev. 3 May 30 added info related to sun glint and moon locationRev. 4 June 1 added the available translation (and map of airplane)Rev 5 June 2 added answers to questions posed to the DOD (removed map of airline track)Rev. 6 June 4 analyzed the target at 1 o’clock at 52 ktRev. 7 June 24 …. further analysis of calibration and initial target speedRev. 8 July 8 corrections, elaborations, more on calibrations, more translationRev. 9 Aug 4,5,6 incorporated improved translation, corrections, removed calibration data (in Appendix 1);

removed distance calculations (placed in Appendix 3)Rev. 10 Aug 11 correctionsRev. 11 Aug 28 corrections, additions, etc. Still don’t have complete translation with times of all crew

commentsRev. 12 Oct 6, 2004 correctionsRev.Oct 28, 2004 incorporated more of the translation by Kurt Franz

In the following data table times are presented to the nearest second or tenth of a second, not to the nearest frame (30 frames/sec) or field (60 fields/sec with two fields per frame).

Measurements of image measurements given below are in mm. They were measured on a monitor which presented the full field width as 330 mm, unless otherwise noted. Therefore the scale factors are:

For WIDE FOV - 0.076o /mm; MED FOV – 0.0103o /mm; NRW FOV – 0.00242o /mm; NRWX2 – 0.0012o /mmThe azimuth and elevation angles listed in the table below, unless otherwise noted, are for the center of the field of view. The flir light(s) is (are) most often NOT exactly at the center of the field of view and hence, unless otherwise noted, the azimuth and elevation values given are not the actual values for the light(s). However, when the flir was set for the maximum magnification (maximum “zoom”) the width of the FOV was only 0.4o so in this case the azimuth of the light was never far from the azimuth of the center of the FOV.

Mileages and speeds are given by the crew in nautical miles (nm; 1 nm = 6,077 ft) and knots (kt; 1 knot = 1 nm/hour), respectively. These are converted in the commentary notes into statute miles (mi; 1 mi = 5,280 ft) and miles per hour (mph). Note that a nautical mile is almost exactly 15% greater than a statute mile.

The translations of the crew statements presented below are the combination of translations by Santiago Ytturia, John Velez, Patricio Elicer and Kurt Franz (who provided the most complete version of the translation). The statements have been inserted into the flir video timeline as closely as possible to the correct times as the translators can determine. Some of the translators didn’t indicate the exact times of groups of statements (Franz did provide a

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time for each statement)). Sometimes they disagreed on the exact time (to within a few seconds). Sometimes there were differences in the exact translation. There has been an attempt to identify the speakers who are heard on the tape, but this has not been entirely successful. However, together the translators have provided enough information for the proper analysis of the radar targets and flir lights. The notation (-) in the translation indicates “expletive deleted.” (Some translated statements have been left out as not necessary for analysis. For a complete transcript see www.alcione.org/CAMPECHE.)

Information in this “data sheet” is believed to be accurate. However, “nothing” is “caste in concrete”. Changes could happen in the future to information other than the flir readout numbers and their associated times,.

---------------------------------------------------------------------------------------------------------------------------------

CONTENTS OF THE FLIR DATA TAPE

TIME LATITUDE LONGITUDE AZIMUTH ELEVATION FIELD OF COMMENTShr:min:sec deg min. deg. min deg deg VIEW (FOV)

12:55:10 16o 23.66' 91o 9.91' -2.8 -25 TV format Non-IR video shows ground

scenery: fields, roads, trees, buildings

-------------------------Continuous to about 12:57:45. Then jump in time to 13:00:34.

13:00:35 16 27.16 91 15.85 -179 -5 TV format Showing ground scenes, clouds between plane and ground etc.

There are numerous jumps in time.This TV imagery continues until 13:29:11 which is about 5 min, 13 sec. into the video

Then it switches to black and white IR imagery obtained 3 3/4 hours later.

**************************************************************************UFO SEGMENT BEGINS (FLIR imagery and Radar Contact)

TIME Lat Long Az El FOV COMMENTS

16:42:20 18o 2.99' 90 56.61' -50.3 -11 WIDE Ground scenes and river, city, bright spots on the ground, buildings, roads, an airfield runway(?), etc. 16:42:30 3.44 56.90 -24.3 - 8 WIDE Plane is heading 327.8o at about 225 (statute) mph

16:42:20 - Radar Operator (RO) : It is now at 11 o'clock, 4 miles, 75 speed, direction 2-9-7. (Comment: A radar target has suddenly appeared and is now about 4 nm = 4.6 mi traveling at 75 kt = 86 mph at the 11:00 position, i.e., 30+/- 15o to the left of straight ahead, and heading in the direction 297o which is 27o north of due west.)

16:42:26 -- RO: At 3.6 and a speed of 75(Comment: The distance is now 3.6 nm = 4.1 mi, a decrease of 0.5 mi in 6 seconds, if the distances are considered to be exact. However, the distance readings are probably only accurate to a tenth of a mile or so. )

16:42:31 – RO: Heading 29716:42:34 -- Plane Commander (PC) : OK, let's see if it's going to land in Kuwi (sp?). 16:42:39 – PC: But we don't see it. Go on telling me the position.16:42:45 4.10 57.35

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16:42:42 -- RO: OK, it's almost at 12 o'clock, 2.9 distance, 75 speed.(Comment: The radar object is 2.9 nm = 3.3 mi ahead of the plane traveling at a speed relative to the ground of 75 kt = 86 mph = 1.43 mi/min. Compared to the distance about 22 seconds earlier, about 4.6 mi, the distance to the radar target has decreased by 1.3 mi during these (approximately) 22 seconds. The speed difference between the plane and the radar target was about 230 mph – 86 mph = 154 mph or 2.6 mi/minute. Hence one would expect the distance to decrease by about 1 mi in 22 seconds. The 0.3 mi discrepancy (1.3 mi vs 1 mi) demonstrates some level of inconsistency between the instantaneously measured speed and the average over many seconds. However, this inconsistency may be a result of imperfect distance readings from the radar scope.)

16:42:53-- Unknown Voice (UV): Turn to the left to avoid the sunlightUV: Pan more to the left because of the sun.FLIR OPERATOR (FO): Yes.UV: Ground personnel must have the info by now

16:43:00 4.88 57.80 -3.8 -10 “ Plane starts to turn left

16:43:06 - RO: Now it's 65 speed, 2.1 miles away(The radar target is now 2.1 nm = 2.4 mi away and traveling more slowly at 65 kt = 75 mph = 1.25 mi/min. The speed differential is now about 230- 75 = 165 mph = 2.75 mi/min. If the radar target speed had stayed constant at 1.25 mi/min during the 24 sec = 0.4 min from 16:42:42 to 16:43:06, one would expect the distance between the target and the plane to decrease by 2.75 x (0.4) = 1.1 mi. It actually decreased by 3.3 – 2.4 = 0.9 mi, which again demonstrates a slight inconsistency between the immediate speed measurement and the speed averaged over many seconds. At 16:42:20 the distance to the radar target was given as 4.6 mi, so during the 46 seconds between that time and 16:43:06 the distance decreased by 4.6 – 2.4 = 2.2 mi. This corresponds to a speed differential of 2.2 mi in 0.767 min or 172 mph. Since the plane speed was about 230 and both were traveling in the same direction the average speed of the radar target over this time interval was only 58 mph.. This average is even lower than the lowest measured speed.)

UV: There’s a runway about there.16:43:14 UV: I have here a runway called Candelaria16:43:17 FO: OK I see it. The runway is at 1 o’clock

16:43:15 5.38 58.3 Plane completes a nearly 30o turn to 301.6o at about 227 mph16:43:25 UV: Where is it?16:43:27 FO: twelve o’clock16:43:30 5.88 59.04 38.8 -25 “ Plane headed toward radar target (?) until about 16:51:3016:43:31 UV: the runway is to the right.16:43:34 RO: Now it’s the other way around, It’s marked at two fourteen,16:43:37 UV: Two fourteen what?16:43:39 RO: Two ninety seven of…correction, speed of 9716:43:41 UV: Speed of 97?16:43:42 RO: Yes(The distance is not given here. However the speed is greater indicating that the radar target was continually increasing its speed.)

16:43:47 UV: What chart is it, Juarez?16:43:57 PC: Go on giving me data.

16:43:55 – (Telephone tone begins. This covers up some statements.)16:43:57 - PC: Continue to give me readings.16:44:00 6.78 910 0.48 47.9 -22

16:44:00 - RO: Our twelve o'clock (target now) at 2.1 miles,(The radar distance is now 2.4 mi.)16:44:04 – RO: The direction is 2-9-8 with a speed of 113.

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(The radar system indicates the same heading (298) and with a speed of 130 mph. The object is still flying more slowly than the C-26, but the distance is not shrinking. The radar engineer said that it was unlikely to detect anything closer than 2 nm because the radar pulse is about that long and the receiver input is blocked during the transmit pulse. Note that the reading of speed is jumping around. Since the distance stays constant it must be that the object accelerated to about the speed of the plane over the 30 seconds from 16:43:34 to 16:44:04.)

16:44:05 - PC: 113? Speed?16:44:06 - RO: Yes.16:44:10 - RO: Right now the radar has it between 285 and 213…and 113(At 16:43:34 the target was 2.4 mi in front, at 16:44:04 it was 2.3 mi in front of the C-26 aircraft. Yet the radar system shows the target speed as increasing from 75 mph to 111 mph and finally to a variable 130-285 mph during this same time interval. These data raise the following question: how did the object manage to maintain a nearly constant distance from the C-26 during the portion of this 30 second interval when it was (according to the radar) at least part of the time traveling more slowly than the speed of the C-26, 230 mph? Or are the speed changes an artifact of the radar not working properly on a target that is at the minimum detectable distance?)

16:44:19 – RO: It is currently at speed 124. It was two fifteen. 16:44:25 -- PC: Speed?16:44:27 – RO: Yes, now its 194.UV: It’s a sudden change.Note: the object over the preceding 10 seconds of the radar sweep averaged 194 nm/hr = 223 mph, about the speed of the plane. This is consistent with the object maintaining a nearly constant distance in front of the plane for several minutes.)

16:44:30 – RO: It is at our twelve o'clock position at 2.1 miles16:44:35 7.77 91 2.21 96.4 -22 WIDE Ground scene but bottom of R.H. propeller at top right of screen

16:44:36 - 16:44:44Telephone noise/someone asking to speak to someone and other voices.) PC: Colonal Winchit(sp?) please. (Asking ground control to speak to the captain.)

16:44:52 –RO: It's marking a velocity of 177 at 2 point miles with a heading of 3-0-0.(Note: This velocity corresponds to 203 mph, which is slower than the plane. One wonders how the object can maintain its 2 nm distance from the plane while traveling more slowly. Perhaps the speed was greater than that of the plane during some of the time that the radar operator gives no speed information. If so, this would be an indication of erratic speed changes. The heading is about the same as before, however.)

16:44:59 - UV: The same direction as ours.16:45:01 - RO: Right, and the reading is 2.1 miles

16:45:00 8.51 3.36 Grounds scenes mostly. Some bright lights on ground.

16:45:29 - UV: What’s the frequency of Carmen ground control?UV: I'll tell you right away16:45:35 - UV: Give me the location.UV: A position.. 7- 4(Note: this is a question about the communication frequency of the tower at Carmen

16:45:10 8.88 3.86 -5.7 -10 “ First appearance of “small” cloud ahead and light to left which is silhouetted against the ground near a coastline(?) :14 8.92 4.11 -5.2 -8 TV There is a brief TV image of sky, hazy horizon, ground below.16:45:30 9.38 4.81 -4.7 -8 WIDE The scene shows the sky, distant clouds below horizon. It is dark above horizon.

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16:45:38 RO: At 12 of our position, I still have it thereUV: Can you get me the headquarters officer, Captain Danzon to let him know that we are reading a target.16:45:45 - UV: Point Carmen 18- 4(Note: this refers to the Carmen tower frequency of 118.4 MHz)

16:45:48 - RO: I keep having it at 12 o’clock, speed of 206 and heading of 301(NOTE: This target was traveling about 237 mph toward the west-northwest. The airplane was traveling at about 230 (statute) mph in the direction 301.6o, almost directly toward Cd. del Carmen. Hence this target was still directly ahead of the plane and flying in the same direction at about the same speed.)

16:46:00 10.24 6.21 -7.8 -11 “ The scene shows the sky and ground below. There is a “dot” of light to left of cloud. It is on the ground?16:46:28 11.15 7.69 -6.4 -8 WIDE Center of FOV pointed toward light at left of a small cloud :29 MED FOV This shows a large bright light, cloud at the right side of frame16:46:30 11.15 7.69 -6.1 -7 MED This shows a bright light and cloud at the right side, focused now. It looks as if silhouetted against the ground. :35 WIDE Shows light and cloud

16:45:54 – UV: Distance?

16:45:56: - Ground Operator (GO): Captain TellezPC: Colonel (name not certain; Hulci?) pleaseGO: He’s not here right now. May I help you?PC: Just to inform you that over Tenosique (sp?), we had a radar contactwith an aircraft that we've been unable to locate. It's still on the radar.We are trying to get eye contact with it, but we can't. Now we are at......where is the airplane?... 16:46:39 – PC: Radar, give me the information about the plane.16:46:42 - RO: From our position, 2.1 miles, uh, with 166 speed. Correction, 207 speed, direction 3-0-216:46:48 - PC:UV: It's heading directly to Carmen

:45 -18.3 -1 “ Camera searching; cloud and light off screen to the right

16:46:54 -1.8 4 “ Shows bottom of fuselage, straight ahead. This should be Az = 0.00 but it reads -1.8. This is the time and video frame for the first airplane bottom calibration.

16:46:59.5 12.01 9.09 -48.3 -15 WIDE There is a sun glint on the ground and glare from above. The glint is about 1o left of the center of the FOV, so the glint is at about -49.3o relative to the plane. This is the time and frame for the sun calibration.

16:46:59 - PC: Did you copy me, Tellez?16:47:04 - GO: No, negative16:47:06 PC: Check out Ciudad del Carmen. Any craft landing at present time?

16:47:06 12.18 9.34 -50.8 -21 WIDE This is another glint, about 2o above the center of the FOV and about 3o to the right of center. Hence the glint was about -21 +2 = -19o in elevation and -50.8 +3 = -47.8o azimuth relative to the plane. These numbers are closer to the angles predicted by the solar location: 302 – 257 = 45o azimuth and 19o elevation.16:47:15 WIDE Camera moving around, search mostly ground at left side

16:47:18 - GO: Was it heading to Ciudad del Carmen?

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16:47:20 - PC: That was apparently its direction, but we detected it doing low speedmaneuvers. We couldn't see it with the naked eye, but is still on radar,heading to Ciudad del Carmen, seemingly.16:47:39 - GO: OK, we're going to check out if Ciudad del Carmen is waiting forsomebody.16:47:52 - PC(?): This is going to delay my arrival at Chetumal.16:48:01 - GO: Let me check with Cd. del Carmen and also inform the duty officer. PC: Do we have enough fuel to get to Chetumal?UV: We are on the limit.PC: Safety limits?16:48:15 UV: That’s right.

16:47:30 12.88 10.53 0.3 -10 WIDE Shows sky with the light and cloud mentioned before; the light has not moved. There is also a pair of bright lights side by side at the lower left of the FOV at -9.1 az and -14.6 elevation16:47:43 audio interference :45 “ Camera searching ground at right side16:48:00 13.83 11.97 41.5 -20 " This shows the ground scenery.16:48:15 “ Sky and the same cloud and light as before left or straight ahead

16:48:19 PC: OK, where is the target? How many miles ahead?16:48:25 RO: - 2.1, speed 192(COMMUNICATION UNREADABLE DUE TO AN OVERRIDING TELEPHONE TONE)

16:48:30 14.69 13.38 -4.9 -9 " Shows cloud ahead and a bright dot to left of cloud; compared to background of the light (ground, coastline?) it does not seem to have moved

16:48:37 – RO: 2.1 miles16:48:44 PC: Give me the Carmen frequency. (This is for radio communication.)

16:48:48 15.27 14.36 -0.6 -16 " Shows cloud ahead and bright dot to upper left of cloud

16:48:49 UV: We have already asked Carmen (control) and it has no traffic reported. 16:48:51 PC: Who the hell is this guy, then? 16:48:57 UV: But we are heading direct toward Carmen. It is over there in front of us.UV: It is going (in that direction), too.(Note: the target is heading toward Carmen.)

16:49:00 15.56 14.82 -2.5 -11 " Still shows cloud but dot is a lot fainter.

16:49:03 - UV: Wouldn’t that be a helicopter of a King Air or something? 16:49:06 - UV: Who knows?16:49:07 - UV: Is the P.G.R. looking for something?16:49:10 - UV: Juan Solano. 16:49:12 - UV: Where is it? Can you still see it? Is it still in front of us?

16:49:18 The flir image still shows a cloud, but the dot mentioned above is barely visible and hasn’t moved relative to the ground.

16:49:19 - RO: Yes sir, I still have it in front.16:49:25 - UV: What a (-) lazy radar. Now it wants to work, no? Let’s proceed to Chetumal.

16:49:24 16.30 16.01 -0.6 -7 MED This shows a thin peninsula that lies on the other side of a body of water that is continuous to the right but ends a bit to the left of center. Projecting at 7o down from the plane’s altitude of 10,500 ft = 2.0 miles gives a horizontal distance of 2/tan(7) = 16.3 miles which is not far enough

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to show the water ahead of the plane. However, if the plane was flying with the nose tilted upward by 2 degrees – see Appendix 2 - then the depression angle was only about 5o. In this case the distance was 2/tan(5) = 23 miles. The distance from the plane location to south shore of Laguna de Terminos was about 23 miles

16:49:30 16.46 16.26 1.2 0 MED Shows coastline

***-----------THERE IS A TIME “JUMP” from 16:49:30 to 16:51:24. This is at 12 min 23 sec into the video. This is also the first(?) IR appearance(?) of the possible UFO (PUFO) as a small dot to the left of a cloud but not silhouetted against the coastline. By this time the plane has traveled about 25 miles from where it turned toward the radar target.

16:51:24 19.80 21.74 11.3 -6 WIDE During 26 frames there is a small, dim dot just above the center of screen to left of a cloud.

16:51:25 - UV: Perfect. Let’s proceed to…

16:51:25.2 ----------- 11.4 -6 The camera is switched to MED FOV. There is a large bright, slightly triangular image just to the right of center which very quickly (in a few frames) shrinks to 1.0 cm diameter on a screen 330 wide. There is a bright cloud at the lower right. The slightly triangular, large bright image has a smaller bright “dot” at its left side, contacting the larger one. The center of the dot is about 8 mm left of the center of the larger image, which has a quite irregular shape. By this time the plane has traveled about 31.6 miles from where it turned toward the radar target.

16:51:26 - UV: (-)16:51:27 - UV: (-)16:51:28 - UV: OK, then let’s go.16:51:30 - UV: Good. See you later.

16:51:30 19.96 21.99 11.7 -6 MED The bright image is not perfectly round, 10 mm wide by 8 mm high. The center of the “dot” at the left is about 8 mm from the center of the larger image. A few frames later camera switches to NRW. Note that the depression angle is not as great as -6o because the plane tilts upward about 2o (see Moon Calibration in Appendix 1). The light is about 0.2o below the center, so that actual elevation or depression angle is about -6-0.2 + 2 = -4.2o. The plane is also looking down at clouds and ground. Therefore if this object/light were close it could be at the altitude of clouds. However, if it were on the ground, with the plane altitude 2 mi, it would be about (2 mi)/tan(4.2) = 27 mi away. If it was on the ground at that distance it would have been an exceedingly bright source which raises the question, why wasn’t it detected before in the WIDE FOV? (And if in the air, why wasn’t it detected before? Did it suddenly appear?)

16:51:30 11.8 -6 NRW The large image is about 25 mm wide by about 30 high and surrounded by a striated halo 110 mm wide by 100 mm high. There is a bright dot at the left which now is separated from the large image by about 30 mm center to center.

* Question: why wasn’t this bright infrared light spotted before…. Or was it seen before but not recognized as an unknown? Did it suddenly “appear?” This is the only as-yet unidentified flir-detected light ahead of the plane. The azimuth (relative to true north) to this light is the azimuth of the airplane plus 11.8o or 301.6+11.8 = 313.4o . (This is almost exactly the same as the azimuth to the light after the turn to the right. See below.)

16:51:32 20.13 22.23

***----------------THERE IS A TIME JUMP starting at 20 fields into 16:51:32 and lasting to 46 fields into 16:52:33 for a total of 56 fields = 0.93 sec. plus 1 minute or 60.93 sec. During this time the plane traveled about 3.8 miles in an arc and turned onto an eastward heading.

BEGINNING OF THE FIRST CONTINUOUS SECTION OF FLIR UFO IMAGERY

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16:52:33 - PC: It's behind the clouds. UV: What could that be?16:52:40 - FLIR OPERATOR (FO): Don't know. It went on the other side - exactly where I had it.

16:52:33 22.52 22.97 -112.6 -20 MED Shows some cloud tops; view moves around over next many seconds :34 “ “ -114.7 “ Pointing at a particular cloud structure :35 “ “ -116.8 MED Still pointing at a cloud structure; plane is turning :36 22.64 22.73 -119.3 “ “ :37 “ “ -123.1 MED

AT THIS TIME THE PLANE IS COMPLETING THE TURN TO THE EAST. (Note that the elevation angle reached a large negative number, -20 degrees, even though the center of the camera showed the cloud tops; this is a result of banking to turn right. Coming out of the turn the elevation numbers become less and less negative and the azimuth numbers became more and more negative.)

16:52:40 22.64 22.73 16:52:43 22.73 22.48 -132.6 -8 MED Searching clouds and cloud tops

16:52:43 - UV: Go up, go up.16:52:46 - UV: Goes….

16:52:46 22.73 22.48 -134.2 -3 “ Still pointing at the same cloud structure.16:52:47 22.73 22.23 “ Constant latitude means that the plane is heading east. 16:52:49.17 “ “ -133 -1 MED A small faint dot of light appears to the right of a cloud top, moving to right relative to the clouds. It is near a tiny dark cloud to the right of the darker one. Then the camera continues to the right relative to the clouds (the operator hasn’t seen it?). This light is following the single bright light which appears ahead of it in 16:52:49.9 and after.

This point is 12 minutes, 48 seconds into the video.

16:52:49.9 “ “ “ “ “ For this frame and the next, see the faint light and also, at the very right side, see a very bright light, the one that was initially detected(?). Then the camera view moves to the left. Then it reverses again and moves to the right starting at 16:52:52.00.16:52:52 22.73 21.99 -135 0 MED The camera moved to the right again and shows the “dot” which, by this time, has moved to the right relative to the edge of a cloud. Motion of the “dot” is apparent.

16:52:51 - UV: Stop, stop, stop, stop.16:52:55 - FO: Come, come. Up, Up, There it goes, There it goes!.

16:52:54 “ “ “ “ MED The “dot” is perfectly left-right centered in the FOV and a bit above center16:52:55 22.73 21.99 -134.5 0 MED Note the constant latitude. The plane is flying directly east.The small dot is close to the center of the FOV. The big bright one now appears at the far right and looks like a bright cloud top, but it moves to the right.16:52:56 to 58 “ “ “ The big bright light moves behind a cloud top and re-emerges. The distance between the small dot and the bright one is about 160 mm at the MED FOV. This corresponds to 160 x .0103 deg/mm = 1.65o. At a distance of 10 miles this would be 0.29 miles.

16:53:00 22.73 21.74 -136.3 2 MED The small white light is moving over cloud tops.

16:53:00.5 “ “ -136.3 2 MED The small light, is nearly centered, the bright light is at the right side.

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16:53:00.6 “ “ -136.3 CHANGE FOV16:53:01 " " " " NRW The light is now bigger, 8-9 mm on 330 mm screen

16:53:01 - UV: Why didn’t you get it until now?16:52:02 - (Laugh)16:53:03 - UV: What is it Juarez? Juarez what is it?

16:53:04 “ 21.45 -136.4 3 NRW It appears that the plane has stopped turning. The light is exactly at the center of the screen. The azimuth of this light is the heading of the plane “minus” (rotate to the left) the flir azimuth. The plane is heading at 90o so the object is at 90 – 136 = -46 relative to north or azimuth 360-46 = 314o. This is essentially the same as the azimuth relative to north before the turn (see 16:51:30 above). Hence it is likely to be the same light.

16:53:06 22.68 21.24 -136 3 NRW The light moves behind faint cloud wisp; dims and shrinks to 3 mm16:53:07-09 22.68 21.24 -136 3 NRW The light is behind a cloud. It is not detected by the flir.

16:53:05 - FO: It's a point. It’s a point. There it is.(The flir shows a “point” or a round “dot” of light, too small to resolve.)16:53:08 - UV: There it is, directly behind us.(This refers to the radar target; see below)16:53:09 - RO: That is correct. It's now at our 7 o'clock position at 10.5 miles distance, heading in the direction 2-8-3 and with speed 334.(NOTE: The radar target is in the 7:00 direction or -150o +/- 15o relative to the plane. The 7:30 o’clock “position” is -135o +/- 15-o relative to the plane. At this time the flir light was close to the 7:30 position. Unfortunately the exact radar direction in degrees was not recorded on a radar data tape. However, it may be that the radar operator did not read the radar screen exactly so it may be that the radar target and the flir light were in the same direction. The distance to the radar target is 10.5 nm = 12 (statute) miles =19.4 km. At this time the plane was headed due east so the direction to the radar target from the plane was (-150o +/- 15o )+90 = - 60o +/- 15o or an azimuth of 300o +/- 15o relative to true north. This is barely consistent with the direction to the flir light, 314o (see previous note). The plane was about 32.5 nm = 37 mi from Carmen and the target was almost along the line from the plane to Carmen and 10.5 nm = 12 mi from the plane or about 25 mi from Carmen. The radar measured the target object speed at 334 kt = 384 mph = 6.4 mi/min. At this speed it would be expected to reach Carmen in 3.9 minutes, i.e. at about 16:57:09.

16:53:10 " " " " “ The light emerges from behind a dark portion of cloud. It takes 3 fields to reach maximum brilliance. The image is 5 mm diameter when between cloud tops. When viewed field by field (1/60 sec per field) the image pulsates rapidly in size and brightness in a seemingly random manner. This might be an atmospheric transmission effect known as “twinkling” (like the stars) which occurs when light travels a long distance through the atmosphere. (Note that for images like this the size of the image is related to the brightness: the brighter the light the larger the image.)16:53:12 22.68 21.00 -136.1 3 NRW As it passes between somewhat bright clouds the “dot” brightness dims but the size does not shrink to zero as it disappears behind the cloud. This might mean there is some size to it. The plane is flying almost due east so the azimuth relative to north is about -136 + 90 = -46 or 314o true from the plane location.

16:53:15 “ Traveling through clear space between widely separated clouds it still pulsates rapidly. Could be atmospheric twinkling?

16:53:22- UV: We can not do anything with this fuel we have.UV: NO. We can follow it and land in…

16:53:27 22.64 20.21 -135 2 NRW After passing a distance through clear (?)air; about 5 mm dia16:53:27.57 “ “ -135 2 Last NRW view

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***-------------THERE IS A TIME JUMP as the plane is turning slightly to the left to about 82 azimuth heading.

16:55:16.03 22.73 14.94 -132.4 3 NRWX2 This uses electronic “zoom”. Note the decrease in the magnitude of the azimuth as the plane turned a few degrees to the left.16:55:16.4 22.72 14.6516:55.18 22.73 14.65 -132.5 3 NRWX2 The light comes out of a cloud and has a diameter of about 8 mm.

16:55:17 - PC: At least we finally got it on the FLIR and we've had radar contact all the time.

16:55:20 “ 14.65 -132.3 3 NRWX2 The light is exactly centered at -132.3 az relative to the plane. This is the beginning of a 2.466 minute period of steady flight with a slow increase in azimuth of the light… see 16:57.48 or so.

16:55:22- UV: What was its altitude - more or less, chief?PC: I think it's at the same altitude as us.UV: A bit above us.16:55:27 UV: We were stupid, looking for it below us.16:55:29 UV: Yeah.

16:55:30 22.77 14.12 THIS IS 13 minutes 40 seconds into the video.

16:55:31 - UV: That is the flir angle, right?16:55:35 - FO: The FLIR angle is pointing level.(Note: the man said that the flir elevation angle is zero even though the actual reading is above zero!)16:55:36 - UV: Exactly.16:55:37 - PC: It’s just above usUV: That's right.16:55:41 - UV: Ah, that's why you couldn't see it.16:55:45 - RO: I'm almost losing it on the radar, it's almost off the screen.PC: OK, that’s too bad Now that we have a radar contact and….16:55:54 - RO: 31.8 miles now and I’m losing it(The radar target is now 31.8 nm = 36.5 mi behind the plane and approaching Carmen.)16:55:57 - PC: OK16:55:59 - FO: It is supposed to arrive at Carmen. Let’s see if we can make contact with ground control.16:56:01 22.89 12.63 -131.9 2 NRWX2 The light travels through clear space and behind clouds

16:56:04 - UV: Try to make contact wth the tower (air traffic control).16:56:06 - UV: You haven’t talked to the tower?16:56:07 - UV: No.16:56:08 - RO: It is exactly a mile from Carmen now. (Note: at 16:53:12 the radar target was about 25 miles from Carmen and heading in a direction slightly south of Carmen. Three minutes later, at 16:56:08, the radar target was one mile Carmen so it arrived about 1 minute earlier than expected (see note at 16:53:12). Apparently it changed its direction slightly and traveled that 25 – 1 = 24 miles in 3 minutes, corresponding to 8 mi/min or 480 mph. This is much faster than the previously measured speed of 384 mph. At this time the plane was about 47 mi from Carmen so the radar target was about 46 miles from the plane.

16:56:15 UV: How many miles?

16:56:15 22.97 11.81 -132.4 3 NRWX2 The light travels along, with the azimuth nearly constant, which means it appears to be pacing the aircraft which is traveling a straight line. It could also be at a very great distance and give the illusion of pacing the plane. The image is about 8 mm diameter image against a dark background.

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NOTE: At the location of the plane the direction to Carmen was about -67o relative to true north. The plane was flying at about 82o relative to true north. Therefore, relative to straight ahead of the airplane the azimuth to Carmen was - 82 - 67 = -149o. The flir light direction was presented as -132.4. There is about a 17o difference between the radar target angle and the flir target angle.

16:56:17 - RO: One or two miles from Carmen.16:56:22 - RO: I don't have it on the screen but the Radar is still marking the distance at 37 miles.(Note: 37 nm = 43 miles. Is this a persistence effect (target not on screen so the radar reads the last measured distance)? 16:56:28 - RO: It's now over Carmen. Carmen should see it now.16:56:32 – UV: At a minimum Carmen should be able to see it.16:56:34 - PC: Wow! What a good job the radar is doing.(Note: At this time the plane was about 41.5 nm = 48 mi from Carmen so a target “over Carmen” should be about 46-47 mi from the plane, not 43 mi. At 16:58:08 he said 1 mile from Carmen. At 16:56:19 he said 1 or 2 miles from Carmen. At 16:56:28 he said over Carmen. Is this an erratic object, changing speed and direction “randomly?” This could mean a distance of 1 – 2 nm traveled in 20 seconds or 7 seconds, corresponding to possible speeds in the range of about 3 - 8 nm/min to or about 180 – 514 nm/hr or 207 - 590mph. Note: if the radar beam at this time was still tilted downward at 30o, then the radar target was reflecting greatly reduced intensity as compared to what it would reflect had the beam been tilted much less or nearly horizontal. This suggests that the target was a very strong radar reflector.

16:56:30 23.05 10.99 -130.3 2 “ The image is about the same, but changes size as travels (absorption effect?)16:56:45 23.18 10.20 -133.1 2 NRWX2 The image size is 7-8 mm. It has grown and shrunk as travels . NOTE: the shrinking and growing (random pulsation) is likely a result of varying transmission of theatmosphere (a lot of water vapor which varies from location to location)

16:56:47 – UV: It’s not going down, or is it?16:56:56 - RO: OK, I just lost it.16:57:00 - PC: There it is.16:57:00 23.26 9.38 -132.0 1 NRWX2 The image diameter is about 8 mm.16:57:01 - RO: It's passing over Carmen. PC: It is still there.16:57:04 – RO: There it goes.16:57:08 – UV: Write down the time

16:57:15 23.39 8.47 -133.1 1 NRWX2 The image about 6 mm.

16:57:17 - PC: What is it doing?RO: It seems to be turning.16:57:20 - PC: It’s turning? RO: Yes.16:57:34 - PC: Eh? … No. Radar? 16:57:37 – PC: Did you lose it?16:57:41 – UV: He doesn’t have anything,

16:57:30 23.47 7.65 -133.4 2 " The image is about 1 cm diameter.16:57:45 23.59 6.78 -133.8 1 " The image is about 8 mm (irregular shape because of noise). The background is a bit brighter as there are more clouds in the picture. The light is below the clouds.16:57:45 “ “ -133.7 0 NRWX2 The light is centered in azimuth and slightly above center. Comparing this with the azimuth at 16:55:20, -132.3, we see that the object has dropped back a little with respect to the plane: the az increased by 1.4 degrees. During this time the plane traveled about 9.4 miles at about 230 mph. A graph of the azimuth of the center of the field of view as a function of time shows random oscillations but a

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general slow azimuth increase. If the object was stationary then it was over 100 miles away. Or else it was at a much shorter distance and moving along with the plane.

***------------THERE IS A TIME JUMP at 16:57:47.7 to 16:59:34.5. This is about 15 min, 57 sec into the videotape.

16:59:34 24.46 91 0.36 -136.7 1 NRW The light image is a dim, small dot in the black above the cloud tops.

16:59:36 - FO: Just a light. It could not make out what it was.PC: We couldn’t determine whether it was a plane or something else.16:59:39 - UV: It is the same one that we saw awhile ago.UV: Yes.UV: It’s what saw us a while ago.

16:59:45 24.54 90 59.78 -137.3 2 NRW The light image is a 3 mm dot moving over cloud tops16:59:49.0 24.58 59.49 -137.4 2 NRW The light image is exactly centered and silhouetted against a dim cloud. Note that the azimuth is larger than at 16:5756 implying it has dropped back more relative to the plane, which is traveling along a straight line.16:59:50 NRW The light appears to pass in front of a distant faint cloud, in which case it would not be at a huge distance. However, the image is also consistent with a very distant light was not completely blocked by the thin cloud, i.e., some radiation penetrated through the thin cloud. It does not prove that the light was as close as the cloud.16:59:51 24.62 59.20 -137.4 2 NRW The light moves behind a closer, brighter cloud.

***----------THERE IS A TIME JUMP at 16:59:54 to17:03:34.

This occurs at 16 minutes, 16 sec into the videotape.

THIS IS THE BEGINNING OF THE SECOND CONTINUOUS UFO SEGMENTThis is the first appearance of the "twins"... bright lights side by side. To the left of the twins is a single dot and then further left is a cluster of several. These are be the lights which later on clearly follow the “twins.” Also, to the right of the twins at some distance is “the leader”, the first in this line of 17(?) lights.At 17:03:35:

o oOoO o o (O) (O) o | --19 -- | -9- | --28-- | ---------90 ---------------| - 10 - | -----49--------| distances in mm on 330 wide screen (0.0103o/mm)|-----------------------------------205------------------------------------------| The total length of line is 2.1o or 0.37 miles at 10 miles or 0.74 miles at 20 miles, 3.7 mi at 100 mi, etc. Twin spacing is 0.103o which corresponds to 0.00179 radians or 0.179 mi (950 ft) at 100 mi. 17:03:35.0 26.43 46.8 -133.1 2 MED This shows the “twins” over a bright cloud. The center to center distance is about 11 mm. The diameter of each light image is about 6 mm.

17:03:36 - FO: Let me check.17:03:39 - PC: At 7 o’clock at the same altitude.FO: There in some clouds there we should be able to see this (-)PC: That's it but, let's see, look for what is coming behind us.

17:03:44 There is an abrupt change in FOV. Large blobby "twin" images shrink to a final steady size

17:03:45 26.52 46.27 -133.6 3 NRW twins separated by 42 mm center to center and 17 mm diameter. Each twin has a lower light about 15 mm below and smaller. Now the angles are based on .00242o per

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mm, so the twin spacing is calculated at 0.102o. The spacing between each twin and the fainter light below is 0.036o which corresponds to 0.00063 rad or 0.063 mi = 333 ft at 100 mi. Each twin has a striated "halo" around it. NOTE: the clouds look "fuzzy." It appears that the focus is not good. Defocus can explain some of the image size.

17:03:52 - UV:Looks like they are in two phases.17:03:56 - FO: Yes. It is at our 7 or 8 o'clock position. The strange thing is, that it has a small point traveling in front. A little point, in front of it. It's not there?

17:03:58 26.64 45.55 -132.7 2 NRW right hand twin about to enter cloud (16 min, 40 sec into video). As the twins enter the cloud they shrink appreciably in size but not to zero diameter. This suggests that there might be some size to the lights, perhaps a couple of feet if 10 mi away. (For “point sources” which are too small or far away to be resolved, the brightness creates an image size that grows with brightness. Hence decreasing the brightness would shrink size of the image. The cloud acts like a variable attenuator as the line between the camera and the light. moves into the cloud.)

17:03:58.5 AZ = -132.7 This is the beginning of a distance calculation by parallax measurement while the plane traveled in a straight line using a portion of a cloud edge as a reference just before the first twin reaches the edge of the cloud.

17:04:05 - FO: There goes another one.

17:04:07 26.76 44.87 -134.5 2 NRW The twins are behind a large cloud. The camera moves back and forth looking for them. At 17:04:07 there appear at the left of the screen other lights which appear to be moving to the right!! The furthest along is very faint. Then the next is brighter with a "donut" shape, dark center, implying out of focus optics.

This is 16 min, 48 sec into the video.

Two lights are seen moving into the cloud, then three more following along, of varying brightness, all apparently out of focus. Not in any real arrangement. I counted 14 that followed the twins, as the camera moved back and forth. Most were dim but some were barely detectable (very faint). They tend to come in pairs as follows: the twins – space – a pair – space – 5 pairs close together – larger space – final pair.

17:04:08 - PC: OK, radar, what do we have?17:04:12 - RO: Nothing sir. Hold on, let me look.17:04:17 UV: Commander what's that?RO: At what distance did you have them, Tellez?UV: Ehh?17:04:23 - UV: Hey, beware, this is like...

17:04:15-17 26.80 44.58 -135.0 2 NRW a group of 8 lights in a zig-zag line close together move into the cloud edge and two faint ones are a distance behind

17:04:24 - FO: (Almost shouting) There are more coming. They are following one behind the other. There are ten, eleven objects.17:04:30 - UV: There are many objects.UV: Ehh?

17:04:26 Another side by side pair move into the cloud.17:04:33.4 AZ = -137.3 This is the end of a time interval used for a distance calculation by parallax (see Appendix 3). The center of the FOV shows approximately the same cloud feature as at 17:03:58.5, i.e., the nearly vertical cloud edge that the light went behind. The AZ increased in magnitude by 5 deg in 34.9 sec. The plane speed was about 3.83 mi/min so it traveled in a straight line about 2.2 miles. Using the AZ angles in a triangle with

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the base being the distance the plane traveled and using the law of sines I get about 17 mi to the cloud. The lights were farther, but I don’t know how much farther.

17:04:34 - FO: They must be at our 9 o'clock position now.UV: At 9 o’clock? Let me.....

17:04:36 27.01 43.18 -136.2 2 MED Field of view change 17:04:37 “ Below a dark, curved “claw” shaped cloud, and against a dim white background, one can see a light that appears to be moving. It seems to come out of a dim white cloud and a few seconds later another comes. The end of the “claw” cloud is used as a reference point for a distance calculation (see Appendix 3). NOTE: the appearance of emergence from a cloud is an optical “illusion”. The grey level area that seems to be a cloud is actually a “hole” or clear area under the dark,curved “claw” cloud.17:04:38 MED Detect widely separated small moving lights coming from behind a cloud , one after another. They seem to be silhouetted against somewhat bright background cloud, but this could be a cloud illusion. It might be distant light that passes through thin cloud. 17:04:41 “ See a line of lights o 0 0 o o 17:04:42 MED One can see 5 in a row, distant, somewhat bright. One is faint and quite a distance in front. Followed by a cluster of a faint one followed by 2 bright ones and then another faint one. The brightness and image size fluctuates. The lead one goes behind a closer cloud bank after being silhouetted against a farther cloud bank.

17:04:37 - PC: At 9, almost at the same altitude .17:04:39 - UV: Okay, hold on.17:04:40 - PC: Affirmative. There they go.17:04:44 - FO: There they go.

17:04:43.9 Now one can see a faint one 12 mm behind. Evidently we are seeing part of the earlier line of objects.The line now looks like this: o o 0 0 o o | 12 | 4 | 9 | 10 | 39 | distances in mm at 0.0103o/mm.

The lights seem to be as far as or farther away than the dark cloud “claw” which is calculated by parallax to be about 25 miles away. If that distance is accurate, then the total length of this line of lights is 25 miles x tan (74 mm x .0103) = 0.33 miles. Note that changes in the relative spacings and orientations of the lights may not be consistent with the lights being distant fixed sources of heat. There can be some change in relative position due to parallax (shift in position of the airplane) but perspective change may not be large enough to cause the observed changes in the light pattern.

17:04:44 Switch field of view17:04:45 27.05 42.93 -135.6 2 NRW The camera is again. searching for the lights :46 “ Very briefly one can see 5 at the very left of the FOV about to enter a cloud. The images are horseshoe or inverted U shaped indicating defocus.

17:04:47 MED again, somewhat defocused?17:04:51 “ One can see two more faint light dots moving out of the cloud beneath the dark “hook” and crossing the space before entering the next cloud, following the previous group illustrated above. These are quite far behind.17:04:57 “ The leader light emerges from a cloud at the right side of the FOV as the farthest back light enters a cloud that the others previously entered, under the “hook”. There is approximately one whole FOV of 3.4o between these. If this line of lights is 25 miles away, then its total length is now about 1.5 miles.

17:04:59 - PC: What do we have on radar?RO: Nothing, Sir.

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17:05:00 27.30 41.78 -132.5 0 MED The camera is searching for the lights. The operator apparently didn’t notice the single faint dot light emerging from the stratus cloud.

17:05:00 - PC: But there are a lot, about 10.UV: Go one..17:05:03 - RO: I’m checking it out, but no... I have some tiny dots but I have no readings, no speed indication nothing.17:05:09 FO: There they go with the others bringing up the rear, ...the ones that follow.

17:05:06 “ The first “twin” (?) bright light emerges from a cloud17:05:14 “ Very brief sighting of lights through holes in the clouds

17:05:15 27.39 41.24 -137.8 2 MED still searching for the lights (behind clouds?)

17:05:16 MED Twins detected at a distance. They came from behind a “narrow” stratus cloud. They are very bright, about 5 mm in diameter. They are followed by a very dim, small light quite a distance behind.17:05 17 “ “Follower 1” emerges from the cloud; the twins are ahead of it at the right and the leader light has passed behind another cloud edge which is calculated to be about 23 miles distant.

17:05:17 – UV: Here they go.17:05:23 – FO: There goes one.

17:05:21 NRW The FOV narrowed and camera is searching.17:05:22 NRW The camera picks up Follower 1 just after it emerged from behind a small cloud. The center is dim (defocused). The diameter is 8 mm. It moves to the right relative to the clouds and is farther than the clouds. There is also a “ghost image” or anther very faint light just below and to the left. (An artifact of the defocus?)

17:05:25 – FO: There go the other ones behind.

17:05:28 NRWX2 The flir operator “zoomed in” and the image is now about 16 mm diameter, but it is defocused; cloud edges are very diffuse.

17:05:25 - UV: My God, what's that?17:05:31 - UV: Try to focus.UV: They are on the run. (nervous laughter)

17:05:30 27.50 40.38 -136.6 1 NRWX2 There is a single "thin" light image, 20 mm dia; defocused.17:05:37 NRWX2 As the light travels over the clouds the image shrinks and gets brighter and clouds look less diffuse because the flir optics are being focused. The image is about 9 mm in diameter before it goes into a cloud. One can see below and to the left just after the light disappears another light almost blocked by clouds, also moving to the right.

17:05:30 UV: Okay, don't let them leave you behind.FO: Okay.UV: Go.

17:05:39 NRWX2 The light, Follower 1, goes behind a rather sharp cloud edge. The light image does not appear to shrink in diameter

17:05:45 27.63 39.51 -137.4 2 NRWX2 The lead light has disappeared behind cloud. Others follow, lower down and nearly blocked by the top of a cloud layer.

17:05:44 - UV: Follow that one Tellez, don't lose it.

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17:05:46 - FO: There they go, all of them. There are approximately ten, did you see them? Have you seen them?PC: Look for the position, at what position, at 10 o’clock?17:06:01 - FO: Approximately 8 or 9 o’clock.

17:06:00 27.75 38.69 -137.5 2 NRWX2 The lights are behind clouds. The camera is searching for lights.

17:06:25 - FO: There they go. There they go.

17:06:25 27.96 37.24 -138.8 2 NRWX2 The lights reappear between clouds. There appears to be slight defocus. Now one sees 3 lights with the spacing O O O (not a straight line). These are apparently the second “triplet” (see below) These three lights are spaced by 18 mm and the trailing light is 25 mm behind second. They immediately go into another cloud and are followed by 3 other faint ones that come out of one cloud and enter another.

17:06:28- UV: Two, Three, four, five. six... There are six.RO: At the same altitude17:06:30 28.0 36.96 -138.6 2 NRWX2 No lights are visible. Searching.

:35 - 37 NRWX2 One can see a total of 8 lights traveling through breaks in the clouds before the camera suddenly changes direction. These lights are grouped as a triplet of bright lights followed by another triplet.17:06:36 “ The third light of the first triplet has split into two lights

17:06:35 - FO: There they go. Look, one, two, three, four, five, six, seven... no, there are eight.UV: Below or above? Above? Where?FO: In the middle. There they go, at the same altitude.17:06:43 - UV: This is terrible...FO: They are going to be seen better - there.UV: And they're going at a speed that…..! (a very high speed)

17:06:43 “ The lights emerge from a dark cloud background, seemingly between the cloud and the camera. However, it is more likely that the radiation from the lights penetrated the thin cloud that was not directly illuminated by the sun. As they move to the right they are seen against a bright cloud background and then clear sky. One can clearly see 8 lights. There is a faint “leader” followed by two bright triplets. The third light of the first triplet is now a single light (it was split into two 7 seconds before).

17:06:45 28.12 36.13 -138.8 2 NRWX2 Lights move out of cloudy area into a clear area .*NOTE: This is one of the few times when the lights SEEM to pass in front of a cloud. As the lights leave the cloud “background” their brightness fluctuates somewhat. If they were in front of a cloud their brightness shouldn’t change as they pass into a cloud-free area. Their brightness seems to change somewhat but it is difficult to quantify because of the image saturation. Probably the cloud in this area was thin enough for radiation from distant sources to penetrate the cloud (i.e., light is coming through the cloud).

17:06:51 28.20 35.55 -138.9 3 NRWX2 Now the lead light is faint, followed by two "triplets" followed by another faint light 0 0 0 0 0 0 0 0 (not a straight line) | 69 mm | 25 | 19 | 80 | 22 | 18 | 68 mm | The angle scale is now 0.0012o per mm so 18 mm corresponds to 0.0216o (0.000376 rad) , 22 mm corresponds to 0.026o (0.00045 rad), 80 mm corresponds to 0.096o (0.00167 rad) and the total distance across the array is 301 mm or 0.36o (0.00626 rad). At 100 miles distance the 18 mm spacing is 0.000376 x 100 = 0.0376 mi = 198 ft and the 80 mm spacing corresponds to nearly 900 ft..

17:06:52 - FO: One, two, three, four, five, six, seven, I have eight on the screen.

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17:06:55 - PC: Follow them, don't lose them!17:07:00 - UV: Are they at the same altitude, Tellez?FO: Affirmative. Above.

17:07:00 28.25 35.27 -139.1 2 NRWX2 same grouping as above. Background seems darker.17:07:03 “ the third light of the first triplet is starting to elongate.17:07:05 NRWX2 The camera turns to the left and shows 2 more lights trailing the left. hand light that is indicated above.17:07:06 “ The 4th light from the right end in the above line (the third in the first triplet) is starting to split into two lights.

17:07:07- FO: Eight, nine, ten, eleven, counting all the ones in the back. Two, four, six, eight, ten, eleven.

17:07:10 NRWX2 The 4th light in the line above has split into two again and the space between the two lights increases.17:07:15 28.37 34.40 -138.5 2 NRWX2 Now the spacing is 8 mm between the split lights. One is still 23 mm from its immediate "predecessor." The other is 31 mm.

17:07:16 - PC (talking to the ground control) Listen, we don't know what are we seeing. They are some luminous objects that are flying at the same altitude as us. We do not know what how far away they are.

17:07:18 “ The lights, one after another, enter another cloud.The spacing between the “split” lights is now about 9 mm. THE IS THE LAST TIME THIS GROUP OF LIGHTS IS SEEN (END OF THE SECOND UFO SEGMENT)

17:07:25 - RO: At what position do you have it, Tellez ?FO: At 8:30 o’clock. 8 or 9 o’clock.17:07:30 - RO: At 9 o’clock? I don't think the radar can catch them. We have a blind angle there, Sir.FO: At 8:30?RO: Yes.

17:07:30 28.49 33.58 -139.1 2 NRWX2 The lights are behind clouds. The camera is searching.

17:07:37 - RO: We have 20 degrees of blind angle there.FO: OK, But you should see them now.17:07:42 - RO: Yes, but it's strange, now I have nothing. I continue searching.PC: Don’t lose them, flir. Now they are going to appear again. They should be visible. Let’s zoom in to see what we can see at a distance.

17:07:45 28.62 32.71 -137.9 2 NRWX2 The lights are behind the clouds. The camera is searching.17:08:00 28.70 31.93 -139.1 2 NRWX2 " " " " " :15 28.82 31.02 -140.0 2 " " " " " " "

17:07:58 - FO: There’s a bunch of clouds. They went in there.17:08:06 - PC: Yes, a bunch of clouds, but on exiting that cloud bank they should be visible

17:08:10 - FO: We lost them.RO: I didn’t catch the, on radar, sir.(Background noise)17:08:34 - FO: Maybe a bit later we’ll see them again.17:08:49 - UV: Is that all the zoom, Tellez?FO: Affirmative.

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PC: At what position (direction) did we lose them, Tellez?17:09:08 - FO: At the 8 o’clock position (direction).

17:08:30 28.95 30.16 -140.9 2 " " " " " " " :45 29.11 29.38 -141.5 3 SAME17:09:00 29.23 28.51 -145.4 2 SAME17:09:08 29.28 27.93 -147.2 2 SAME

***-------THERE IS A JUMP in time at 17:09:08 to 17:14:58 . Here 5 minutes of the events were not recorded(?). This is 21 min, 50 sec into the video.

THIS IS THE THIRD UFO SEGMENT

17:14:58 31.87 7.62 -89.7 -1 NRWX2 There are two lights, the new “twins”, not as bright as the original twins.. They are 12 mm apart and about 3 mm in diameter. They are seen against the dim, grainy featureless background.17:14:59 " The two lights move to the right out of the FOV and a large bright light moves in from the left. It has an irregular shape, almost elliptical or triangular, 35 mm long by 30 high.

17:14:59 PC: [nervous giggle] Why are they following us? Maintain that.UV: It’s at 12 o’clock.

17:15:01 -90.2 -1 NRWX2 The light azimuth is 90.1o. 17:15:06 31.95 7.04 -90.8 -2 NRWX2 The large image almost triangular 40 long by 20 high.

17:15:05 RO: No, almost at 1 o’clock. (refers to a new radar target)PC: Well that's one, but we have another at 9.17:05:08 RO: Ah, I have one here. Let me check the one at 9 o’clock.17:15:12 FO: There are two others ahead of that one.PC: And they're the same ones, eh?FO: We have them in front, in the middle and at 8 o’clock.17:15:19 UV: [Laughing] Ahh, (-).

17:15:15 32.04 6.72 -91 -1 NRWX2 There appears to be a very faint light image between and below the "leading pair" (faint twins?) and the big bright image.

17:15:22 UV: [Laughing] Fasten your seat belts!UV: Let's see...UV: [nervous laughter] Don't scare me.17:15:26 FO: Can anyone check through the window? We have one going along at approximately 1o’clock and the rest are still going along at our 9 o’clock

17:15:30 32.16 5.81 -91.3 -1 NRWX2 The camera swings back and forth between the lights. There are occasional barely visible images that move through the field of view suggesting that there is some background, but very faint. Could the background be dim because the sun is low and behind the plane while the camera is looking almost northward?

17:15:33 UV: Does that means we are surrounded?FO: Well... I don't know.UV: At what distance?17:15:40 UV:[in a loud voice, as if making an announcement] Boys, today is your lucky day!!17:15:43 [Laughter]UV: At what distance?UV: And that one is getting near to the other one.

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UV: Do you mark the distance?FO: I don’t get distance readings.

17:15:45 32.28 4.94 -93.9 -2 NRWX2 The big one is very big, irregular and 50 mm long by 35 high. It has a dim "halo" around it As the image moves around in the field of view one sees shape changes suggesting that at least part of the shape and size is a result of some degree of defocus or lens aberration, although the new twins seem well focused.

17:15:49 RO: Okay, the one that is at 1 o’clock, I have it at 19 miles going at 52 knots.(Note: 19 nm = 21.9 miles; 52 kt = 60 mph)17:15:59 FO: I have one at 9 and in front of it, approximately at 9 o’clock as well, are two luminous points.RO: At what position, Tellez?FO: At 9 o’clockRO: At 9? Hold on, let me check...UV: How far away is the one at 9 o’clock?PC: The one at 1 o’clock… how many mile?17:16:20 RO: 17.1 miles, heading 95, speed 52(The distance is now 17.1 nm = 19.7 mi. The heading is 95o which is close to the airplane’s heading of about 81o and a speed of 52 kt = 60 mph. The plane is going approximately 230 - 60 = 170 mph = 2.83 mi/min faster than the radar object. Hence one would expect the plane to catch up to this target in 19.7/2.83 = 7 minutes, at about 17:23.)

17:16:00 32.41 4.12 -94.4 -1 SAME AS ABOVE17:16:15 32.57 3.25 -95.8 -1 SAME, The exact shape is slightly variable. Lens aberration?17:16:18 NRW The cameraman decreased the magnification Image looks the same but smaller. The shape is still irregular.17:16:27 MED Decreased magnification again. The image shape is now roughly circular and about 8 mm in diameter.17:16:30 32.7 2.39 -96.4 -1 SAME The new twins cannot be seen now.

17:16:24 FO: It's going along at our altitude. It can’t be possible.

17:16:42 UV: Tellez, what’s the position of the one you have got right now?FO: Which one? The one below the cloud? That one?UV: Yes.UV: That one you have there, what's it's position?17:16:44 FO: I have it at 9.17:16:48 UV: Gentlemen, we are not alone. [Laughter]17:16:52 PC: Yes, we are not alone, right. It's true. How strange this is. They are surrounding us.

17:16:45 32.90 1.28 -97.6 -1 MED The image of the light is the same. However, the camera swings to left and picks up the bottom of the left hand propeller! The UFO light seems to be between upper and lower cloud layers.

17:16:55 FO: And the one that you have at 1 o’clock?RO: I have it at 14.8 miles at 1 o’clock, heading 99 with a speed of 52.(Note: This radar target at 1 o’clock and speed 52 kt = 60 mph is 14.8 nm = 17 mi away. The heading is 9o south of due east, while the plane heading was about 9o north of due east. Since the last message that provides a specific time and distance to this target, the message at 17:15:49, the plane traveled for 1 minute and 6 seconds (1.1 minutes) at 3.8 mi/min corresponding to 4.2 mi. If the object maintained its speed at 60 mph = 1 mi/min for 1.1 minutes, then it traveled 1.1 mi in essentially the same direction as the airplane. The distance to the target would then decrease to 21.9 – 4.2 + 1.1 = 18.8 mi. However, the measured separation distance was 17 mi, not 18.8. If the radar target had not moved at all, the separation distance would have been 21.9 – 4.2 = 17.7 mi. To get the distance as low as 17 mi one needs to assume that the target moved toward the plane a distance of 0.7 mi in 1.1 min or 0.64 miles/min or 38 mph. This seems to contradict both the heading (99) and the speed (60 mph).

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17:17:00 32.98 0.70 -98.6 -1 MED It is the same slightly non round image, 6 mm long by 5 mm high.

17:17:04 FO: You are checking one and I’m checking another one. I mean other ones.UV: Isn’t it a plane?

17:17:15 33.11 89 59.88 -103.2 -1 MED Camera too far to left; gets edge on propeller blades. 17:17:20 UV: Hey, you have to record this.FO: I am recording it.17:17:28 PC: Watch your language gentlemen, but what is this?

17:17:30 33.23 59.01 -101.6 0 MED The image seen momentarily between rotating blades is small, 4 mm x 3 mm.

UV: Couldn't be a (-) plane. It's stopped there!(Note: is this a reference to the target at 1 o’clock that was traveling at 52 kt or 60 mph? If so, stopping would decrease the average speed over a time interval. )UV: Ehh?FO: Do you have it in sight?UV: It seems so. Let’s wait for it to get closer.17:17:39 UV: Nah! It's stopped there!UV: It's stopped? On the ground?FO: No but the one that we have at 9...UV: It's in the clouds.

17:17:45 33.35 58.19 -101.9 -1 MED The light, seen between propeller blades, is even smaller, 3 mm by 2 mm. The image size seems to pulsate but not in synch with the propeller. It seems to pulsate more slowly.

17:17:49 PC: The one that we have at 9 is following us. It is going parallel to us. It is in formation with us.

17:17:49 33.4 57.9 -102.6 -1 MED The light is still seen between propeller blades. The light is momentarily blocked by a cloud top estimated to be over 10 mi away (see Appendix 3). The image blinks out in one field and stays gone for 51 fields and seems to come back to full size in 1 field, but partial blockage by the propeller blade sweeping by makes it difficult to be certain.

17:17:55 PC: Now we’ve got those clouds. It’s above us.FO: The one that you have at 1 where do you have it?RO: Hold on. I have it at... at 1 o’clock. It is 10 miles away, speed 52.FO: There is one that is getting near to us.17:18:00 33.49 57.36 -103 0 MED Same image as before as seen between propeller blades

17:18:15 33.64 56.54 -104.1 -1 MED Same

17:18:15 PC: OK. Don’t lose it, radar. Look for the one at 1 o’clock. It’s coming closer to us.Note: The distance of the target is now 10 nm = 11.5 miles. The last mention of a target at 1 o’clock and speed 52 kt was at 17:16:55. Since then the plane has traveled for 1 minute 20 seconds = 1.33 minutes and covered a distance of 1.33 x 3.8= 5 miles. Previously the target was 17 miles away so the distance to the target has decreased by 17 - 11.5 = 5.5 miles, which is 0.5 miles more than the plane traveled. This implies that the target moved toward the plane by 0.5 miles in 1.333 minutes at a speed of 0.37 miles/min = 22.5 mph. Note: if this was the target that stopped moving at 17:17:39, this period of no motion could account for the lower speed averaged over the 1.33 minute time interval. If the target did not move the plane would overtake this target in 3 minutes, at about 17:21:15.

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17:18:30 33.77 55.67 -104.6 -1 MED Same. Pulsation nearly doubles size temporarily. A little faster than one pulsation per second.17:18:36 -105.3 “ The light is centered at the azimuth listed.17:18:39 MED The camera moves to right, light moves off screen17:18:41 WIDE The propeller blade appears almost edge on (the flir optical “head” is a little in front of the propeller?)17:18:45 33.89 54.89 -49.8 -4 WIDE The camera swings toward the front of the plane17:18:47 1.3 -15 WIDE The camera swings across straight ahead, shows sky, distant cloud

17:18:53 RO: Now I get it at 2 o’clock at 9 miles, heading 120, speed 52.PC: Quality 1?RO: No. Now I get it with quality 9, but I got it with quality 1 before.(Note: “quality” is a measure of the strength of the radar reflection as compared to background noise. Higher quality numbers correlate with weaker signals.)

17:19:00 34.05 54.03 53.7 -3 WIDE The camera shows diffuse layering of sky brightness and cloud layers. The camera seems to be searching.17:19:15 34.18 53.16 56.5 -2 WIDE The camera is moving left and right, shows clouds, etc. Searching?17:19:17 WIDE The camera moves to the left, heading back toward left side.

17:19:22 34.26 52.67 -2.9 7 WIDE The camera crosses nose of plane, rotating left, and shows bottom of fuselage and the moon. This is the time and frame of the second airplane bottom calibration.

17:19:24 -5.3 1 WIDE Now rotating to the right, the camera picks up the moon below and to the left of fuselage (which appears in upper right). This is the time and frame of the third airplane bottom calibration.17:19:24 -5.3 1The moon in this frame is (46 mm) 3.5o down and (18 mm) 1.37o left of the center. This is the time and fame of the first moonrise calibration. 17:19:32 34.38 52.09 -7.4 -1 MED The moon image is centered in the FOV. The image is elliptical, 55 mm wide by 48 mm high. This is the time and frame of the second moonrise calibration.

PC: Ahhh, that isn’t the one. That’s the moon. Is the moon in front?UV: There it is.PC: OK. It is the moon. We are not chasing the moon, are we?[Laughter]

17:19:38 -6.9 -1 TV Image. The TV shows a large moon image and craters.17:19:41 -6.7 0 MED FLIR The moon image is again 55 x 48 mm.

17:19:46 (Laughter)17:19:45 34.47 51.55 -22.7 -4 MED The camera rotates to the left17:19:51 WIDE The camera changes to wide FOV.

(NOTE: THERE ARE NO EXACT TIMES GIVEN FOR THE FOLLOWING STATEMENTS)FO: One of them, was at 12 and the other one was at 9.RO: Now I get it almost at 3 o’clock, 6.9 miles, heading 155, speed 52.PC: OK, that isn’t the one. It’s a cloud. Yes, it is a cloud.FO: Where is the one you had at 1 o’clock now?RO: It’s now at 3 o’clock, 7.1 miles, direction 182.PC: OK, it is pulling away now. Who knows what it was!

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17:19:55 WIDE The camera shows the propeller17:19:57 MED The camera changes to MED FOV.17:20:00 34.59 50.73 -99.8 0 MED The camera shows lower cloud tops between the propeller blades.17:20:07 NRW The camera shows a bright light, irregular shape which doesn’t move relative to the clouds.17:20:15 34.71 49.91 -96.8 1 NRW The camera still shows that bright light that does not move relative to the clouds. It is probably a bright spot on the cloud tops. 17:20:30 34.84 49.08 -98.0 -2 NRW The camera is searching and looking through propeller blades. It has left the previous bright light (cloud top).17:20:45 34.96 48.26 -99 0 NRW The camera is still searching clouds for the lights.17:21:00 35.04 47.43 -95.6 2 NRW Same as above

***--------------17:21:12 THERE IS A JUMP IN TIME This occurs at 28 min, 7 sec into the video.

17:22:59 35.66 40.84 -170.2 2 NRW The camera shows a small cluster of 3 bright lights.17:23:13 35.74 40.02 -172.4 2 NRWX2 The camera shows the same lights, not as bright. They haven't moved with respect to the clouds.17:23:30 35.83 39.19 -169.6 2 NRWX2 There is only one light now. It hasn't moved. Probably a bright cloud top. It seems strange that it would be a tiny bright cloud top in a "field" of dark clouds.17:23:42 NRWX2 This light has faded to near zero and camera has turned to the right and shows a much brighter cloud top.

***---------THERE IS A TIME JUMP 17:23:42 to 17:26:30 This is at 28 min, 48 sec into the video

THIS IS THE FOURTH UFO SEGMENT

17:26:30 36.73 29.31 -174.9 1 NRW The camera shows a bright, irregular image similar to the previous large image at 17:15:00. This background of this image is very dark. There are no well defined cloud images. There is a computer generated “box” or “tracking gate” around the image.

17:26:34 PC: (Reporting to ground control) The radar has them. The FLIR has them. We are seeing on the FLIR a luminous object. It has an opposite course to us but it doesn't leave us, it's two miles behind us.UV: It does not move apart from us, no, it doesn't move apart from us.

17:26:45 36.77 28.48 -174.0 3 NRW Same. One can see no cloud images in background to compare with. However, there may be a lower horizon line, very indistinct.

17:26:53 UV: Correct, who knows what it is.PC: Is it still with us, Marin?RO: Yes I still have it here.PC: The radar is catching now what it didn’t catch in its entire life.[Laughter]

17:27:00 36.86 27.66 -174.6 1 NRW The image has moved downward into a brighter area. There still are no cloud structures to compare with. There is no clear evidence of motion against the background.17:27:11 NRWX2 Perhaps there is something below the object, a faint bright area at the lower left. Is it a horizon with lights? The image looks large and almost like a flat surface with a sphere imbedded. The image is 30 mm wide by 16 mm high

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17:27:11 RO: Now it lost a bit of quality, I have a quality of 5.UV: Again look it looks...RO: The one that we had at 1 is lost now.UV: Okay.RO: It just left the Radar.

17:27:30 37.06 25.80 -175.4 1 NRWX2 Seems to be made of several lights together. There is still no apparent evidence of motion relative to the dim light images lower down and to the left.17:27:45 37.1 25.23 -175.7 1 NRWX2 Seems to be fading, "side blobs" fading

17:27:58 PC: How weird is all this.

17:28:00 37.19 24.45 -174.4 2 NRWX2 As the background gets lighter the image gets smaller and the side light images are nearly gone17:28:06 37.23 23.91 -174.7 2 NRWX2 The main light image is now only about 4mm in diameter and it has faded.---------------------------------END------------------------------- 30 min 24 sec into the tape

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APPENDIX 2 TECHNICAL DATA AND CALIBRATIONS

ANALYSTS: note well the discussion below about the potential errors in the elevation and azimuth readings from the forward-looking-infra-red (flir) system.Rev. Aug 28

WEATHER REPORT: at 10,000 ft, the wind was about 6 mph from the southeast. Cloud cover was broken with Cumulous (Cu) clouds with bases ranging from 1,500 to 2,500 ft and tops up to 10,000 to 12,000 ft. A higher layer of Cirro-stratus (Cs) was above 20,000 ft. The temperature at 10,000 ft was 11-13o C (52 – 55o F).

HORIZONAt the altitude of the plane, reported to be 10,500 ft, (some references gave 11,500 ft) the geometric

horizon is at a distance of about 127 mi. This is calculated from (2ReH)1/2, where Re = radius of earth and H is the height of the observer . Including the effects of atmospheric refraction can increase the optical horizon distance to 135 mi. The depression angle (elevation below horizontal) from the plane to the geometric horizon is given by cos-

1(Re/[Re+H]) = 1.8o (or -1.8o elevation angle) Atmospheric refraction reduces this by a small amount to perhaps 1.7o.

CENTER OF FIELD OF VIEWAzimuth and elevation values are for the center of the flir field of view. One would expect these to be

correct relative to the aircraft itself: azimuth = 0 is parallel to the main axis (fuselage) and elevation = 0 is horizontal when the plane is on the runway. However, when the plane is flying along the actual earth-referenced az and el values may different from the flir “readout” values. This is because there could be errors or “slop” in the flir mechanism and flight conditions may cause the direction of the main axis of the plane to be different from what is expected.

CRABBING ANGLE OR YAWWhen the plane flies in a stationary atmosphere the main axis of the plane (fuselage) points in the direction

that the airplane is flying and the main axis tilts upward a few degrees to maintain an optimum lift for the wing. This upward tilt (upward pitch) might vary with altitude and load distribution in the plane. Typically it is a few degrees (nose up), so the elevation angles listed below may need to be corrected in the manner described below. The correction for this upward tilt is given in the section below on ELEVATION CORRECTION.

When the plane flies in a moving atmosphere (wind), unless the wind is exactly from behind (tailwind) or exactly from in front (headwind), it turns slightly into the wind to keep from being blown off course. Thus, under crosswind conditions the axis of the plane would not be parallel to the direction of travel, as illustrated below.

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The angular difference between the axis and the direction of travel, the yaw or “crabbing angle,” varies with the wind speed and direction as measured relative to the desired course of the plane. The weather report indicates that at 10,000 ft the wind was about 6 mph from the southeast (and about 22 mph at ground level). This slight wind will cause the plane to turn slightly into it (“crabbing” or yaw). The plane was traveling slightly north of due east (actually at an angle of about 82o clockwise from north) during the time that parallax measurements were made. Therefore, during this time the axis of the plane was rotated clockwise to a slightly larger azimuth angle. To determine the amount of rotation it is necessary to determine the wind velocity components parallel and perpendicular to the track of the aircraft. Assume that the wind was from 135o azimuth (exactly southeast). Then the angle between the airplane track and the wind direction is 135 – 82 = 53o. The wind component along the track (toward the west) was 6 cos(53) = 3.6 mph and the component perpendicular to the track (toward the north) was 6 sin(53) = 4.8 mph. Since the speed along the track was determined from the flir azimuth values (based on GPS data) to be about 230 mph, and because the plane was traveling roughly eastward into the southeast wind, the 3.6 mph component of the wind was a headwind. Therefore the actual air speed of the aircraft must have been about 230 + 3.6 or about 234 mph along its track. The aircraft also had to have a component of its velocity pointed southward to counteract the 4.8 mph wind component perpendicular to the track. To produce the counteracting speed the airplane turned slightly southward. The amount of turn is the crabbing angle, C, which can be determined by solving the equation 234 sin( C ) = 4.8 mph Solving this equation for C one finds C = arcsin[4.8/234] = 1.2o, a very small angle.

If the wind velocity (speed and direction) changed as the plane flew, then the crabbing angle also changed. However, the wind speed is so small that the crabbing angle wouldn’t change very much. For example, if one imagines a “worst case” situation in which the southeast wind varied from 0 to 12 mph, then the crabbing angle would vary from 0o to about 2.4o. This much variation could introduce a spurious azimuth change of a distant light causing an error in estimating its distance by parallax, as described in the main text of this document. If the wind were perfectly steady the crabbing angle wouldn’t change and therefore it would not contribute to the change in azimuth of a distant light as the plane flew along.

Fortunately it has been possible to estimate some of the corrections to the elevation and azimuth from comparison of the flir direction readings with known directions: the three AIRPLANE BOTTOM CALIBRATIONs, the SUN GLINT CALIBRATION and two MOONRISE CALIBRATIONS. (Azimuth angles to the left of the plane’s heading are given as negative numbers, as are elevations below nominal horizontal.)

*(FIRST) AIRPLANE BOTTOM CALIBRATIONAt 16:46:54 , when the flir gave the azimuth as Az = -1.8o and the elevation as EL = 4o, the video

momentarily shows the bottom of fuselage at the center of the FOV while rotating rapidly to the right. Assuming the flir optic “head’ is centered on the bottom of the aircraft this should be Az = 0.00 but it reads -1.8. This could mean that one should add 1.8o to azimuth angles (but see sun and moon calibrations below). Note: because the flir was rotating to the right, the error could also be a result, in part, of a lag in the azimuth readout, i.e., an inability of

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the azimuth readout mechanism to keep up with the actual pointing direction when the flir was turning. [The bottom of the airplane is 4.6o above the centerline of the FOV, which itself is at 4o elevation. Therefore the bottom is at EL = 8.6o above (presumed) horizontal (EL = 0) when the plane is on the ground.]

*(SECOND) AIRPLANE BOTTOM CALIBRATION At 17:19:22, when Az = -2.9o and El = 7o the camera crosses the nose of plane, rotating left, and shows

the bottom of fuselage and the moon. The previous time (see above) the camera showed the center of the bottom of the fuselage while rotating to the right, The center was at -1.8 Az. Now, rotating left, the center is at -2.9 Az. (This contradicts the idea that the readout mechanism could not keep up with the rotation rate. If there had been a lag in readout the Az readout should have been positive.) Angular hysteresis? The bottom of the plane is about 1.9o above the center at 7 + 1.9 = 8.9o, so the elevation reading was not affected by the azimuth rotation.

*(THIRD) AIRPLANE BOTTOM CALIBRATION At 17:19:24, when Az = -5.3 and El = 1 the flir was rotating to the right and then stopping its motion. The

camera shows the moon below and to the left of fuselage (which appears at the upper right). (A “stationary frame” occurs when the motion momentarily stops as the rotation reverses direction. The moon image in a stationary frame is about 8 mm diameter. This is discussed below.) The bottom of the plane is (66 mm) 5.0o to the right and (100 mm) 7.6o above the center of the FOV so the azimuth and elevation of the bottom of the plane are Az = -5.3 + 5 = -0.3o and El = 1 + 7.6 = 8.6o. The azimuth number is again different (was -1.8 rotating right, then -2.9 rotating left and now -0.3 stationary). This suggests that the readout mechanism does have some “problem” with presenting azimuth values that are changing rapidly. The elevation is about the same as before, however.

*SUN GLINT CALIBRATION:Astronomical data presented below for the sun and moon calibrations were obtained from

http://aa.usno.navy.mil.)

*FOR SUN GLINT CALIBRATION:The sun’s azimuth relative to true north, Az*, and elevation, El*, above the horizon for Latitude 18o 12’

and Longitude 91o 09’ at 16:45 hours was at Az* = 256.9o, El* = 19.6o. At 16:50 hours and afterward these numbers are Az* = 257.4, El* = 18.4; at 17:00 hrs, 258.3, 16.1; at 17:15 hrs, 259.7, 12.6; and at 17:30 hrs, 261o, 9.2o . Sunset was at 18:11 hours at ground level.

At 16:46:59.5 the plane was at latitude 18o 12.01’ and longitude 91o 9.09’. The center of the flir FOV was at angles Azf = - 48.3o (rotation to the left from straight ahead) and Elf = - 15o. The FOV was WIDE. The video shows a sun glint on the ground and glare from above. The glint is about 1o left of the center of the FOV, so the glint is at about Azg = - 49.3o (rotation to the left from the zero direction of the flir). If the zero direction of the flir were along the axis of the plane, then -49.3 would be the angle with respect to the main axis of the plane. The continually changing latitude and longitude show that plane track was traveling in a straight line along Azp = 301.6 relative to true north. The weather report indicates that the wind was from the southeast (see above), i.e., a tailwind, so there was negligible crabbing or yaw; i.e., the main axis of the plane was pointing “straight ahead,” along 301.6o. Therefore, if the zero direction of the flir were along the axis of the plane., then the azimuth of the glint was in the direction 301.6 - 49.3 = 252.3 relative to true north. According to the U.S. Navy (http://aa.usno.navy.mil) calculations, the azimuth of the sun at this time and location was 257.7o. Agreement is improved by assuming that the flir is reading too negative. If 257.7 – 252.3 = 5.4o were added to -49.3 the agreement would be “perfect,” suggesting that one should add 5.4o to all azimuth readings. There could be a further error in the mechanism and there could also be an error in assuming that the azimuth of the maximum brightness of the reflection is the same as the angle of the specular reflection of the sun. (The presence of a crabbing angle or yaw would also effect this flir azimuth calibration, but that apparently was negligible as described above.) Assuming that the brightest glint was a mirror-like specular reflection in a (sufficiently large) lake one would expect the azimuth and depression (negative elevation) angles to be the same as the sun’s azimuth and elevation angles. But this was a small lake, and waves make the lakes create somewhat diffuse rather than mirror-like reflections. Therefore, if the lake is not large enough to include the specular reflection point, the maximum reflection brightness would occur at azimuth and elevation angles slightly different from the mirror-reflection angle (it could occur at a fraction of a degree left or right, up or down).

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The elevation of the glint is about 5 degrees below the center of the FOV, i.e., at a (depression) angle of -20o. The Navy calculation gives the elevation of the sun at about 19.4o, so the elevation angles are quite close. This could mean that the elevation is reasonably accurate and there is very small upward tilt of the airplane axis (see discussion below) at this time.

*(FIRST) MOONRISE CALIBRATION*FOR MOONRISE CALIBRATION

At latitude 18o 34’ and longitude 89o 52’ the moon rose at 17:13 hours. At 17:20 the moon was 1.4o above the horizon at ground level. At any altitude it would be 1.4o above horizontal. (Horizontal at any altitude is defined the plane perpendicular to line to the earth’s center.) The azimuth is 75.1 deg. The elevation increases at a rate of about 0.2o/min (or 0.1o in 30 seconds) and the azimuth increases at a rate of about 0.08o/min. A video frame picked for study occurred when the moon image had been steady for a short time (about a second) to allow the pointing system to stabilize. This frame, shown below, occurred at 17:19:25. At this time, about 35 sec before 17:20, the lunar elevation was about 1.3o.

At 17:19:25 the airplane latitude was 18o 34.26’ and the longitude was 89o 52.67’. The center of the WIDE FOV was at Azfc = -5.0 and Elfc = 1. The moon in this frame is 3o left of the center and 1.5o down. Hence it is at Azf = -5 – 3.1 = -8.1o and Elf = 1 – 1.5 = -0.5o elevation according to the flir. As shown in Figure 49 and again from analysis the data in Figure 56, the plane at this time was traveling along the direction 81o true so the direction to the moon relative to the airplane should be 81 - 75 = 6o to the left of straight ahead, i.e., it should read (-6o ) on the flir output. This is a two degrees less negative than the flir reading of -8o. About 1o of this difference can be explained by the small “crabbing angle” into the southeast wind (clockwise rotation of the axis of the plane by a degree or so as calculated above). The remaining 1o error could be a result of imperfect alignment of the flir zero direction with the main axis of the airplane. The implication is that one should add +2o to the (negative) flir azimuth reading, 1o to account for crabbing angle and 1o to account for imperfect alignment. (I suppose the surprising thing is how closely the flir reading agrees with what is calculated from the lunar location and the southeast wind.) (Note: the quantization or digitization of the brightness range into 256 levels causes there to be small, “jumps” in scene brightness where the brightness variation with position changes slowly. In the above image there is this sort of brightness jump with vertical distance just below the level of the moon and another such jump

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that just happens to occur at the level of the horizon. These vertical brightness jumps create jagged horizontal lines in the image because the brightness doesn’t vary much along any horizontal line drawn in the picture, but does vary noticeably along any vertical line in the picture. Hence the jagged horizontal lines are not characteristic of the vertical variations in atmospheric brightness but rather are artifacts of the quantization of brightness levels.)

The lunar elevation was 1.3o but the flir reads -0.5. This suggests that one should add 1.3 – (-0.5) = 1.8o to forward elevation angles. For other azimuth angles see the ELEVATION CORRECTION section below.

*(SECOND) MOONRISE CALIBRATIONAt 17:19:32, for essentially the same latitude and longitude, the center of the MED FOV was at Azf = -

7.4o and Elf = -1o. In this frame the moon image is large and almost perfectly centered in the FOV (see below)

The image is elliptical, 55 mm wide by 48 mm high. The width is 55 mm x .0103 deg/mm = 0.56 deg. The height is 0.49 deg. The actual lunar angular diameter is about 0.55o , so the width is essentially correct but the height is too small (astigmatism?).

As calculated above, the moon’s position in azimuth should be -6o relative to the plane. The flir actually reads about 1.4o more angle to the left than there should be (previously it was 2o more than it should be). The crabbing angle calculated above, 1.2o, can account for most of this, leaving only 0.4o for error in the alignment of the flir with the axis of the airplane (assuming that 301.6o is exactly the heading of the airplane).

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The flir should read the elevation of the moon as 1.3o relative to horizontal, but instead it reads -1o. Hence, it appears that one should add 2.3o to forward elevation angles. This is 0.5o more than the previous correction angle, 1.8o. Probably adding 2o is a “happy medium.” It seems unlikely that this difference results from a “boresight” error between the WIDE and MED fields of view, but maybe the WIDE FOV measurement was not as accurate as the MED FOV.

*ELEVATION CORRECTIONThe need to add about 2o to the elevation angle in the forward direction is probably, at least partially, a

result of the aircraft flying with the long axis (fuselage) tilted upward (so the nose is higher than the tail) by several degrees. When the axis tilts upward the correction to apply to the flir elevation reading depends upon the azimuth. For example, if the aircraft upward tilt were To, then at Azf = 0o (the flir azimuth reading of 0o) the flir output reading, Elf = 0o , would correspond to an actual elevation of To; at Azf = +/- 90o (right or left) the flir reading Elf = 0o would correspond to an actual elevation of 0o (if there is no “slop” or error in the flir mechanism); and at Azf = +/- 180o (pointing backward) the flir reading Elf = 0o would correspond to an actual elevation of -To. At intermediate angles the calculated elevation angle, Elc, can be found by solving the equation: Elc = Elf + Tcos(Azf,), with the understanding that because of errors in the mechanism and mechanical “slop” the actual elevation, Ela, could be different from Elc by perhaps plus or minus a degree or so (the actual amount of slop is not known as of this writing): Ela = Elc +/- Elr, where Elr is the (assumed) random mechanical error. Note that many of the “lights” detected by the flir were at an azimuth angle of about Azf = -135o, or 45o forward (and to the left) of directly behind the airplane. Therefore Elc = Elf + T(-0.707). Assuming (from the moon calibration) that T = 2o, Elc = Elf – 1.4o and the actual elevation would be Ela = Elf – 1.4 +/- Er. Thus, whenever the flir elevation was given as Elf = 1 to 2 degrees above horizontal (the flir does not provide precision to tenths of a degree in elevation), the actual elevation, Ela, could be zero or anywhere in the range +/- Elr. Since the horizon was at Elh = -1.7o, the elevation error would have to be about 2o to be consistent with lights on the ground. The magnitude of Elr is not known but it could be determined by experiment.

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APPENDIX 3CALCULATIONS OF DISTANCE AT VARIOUS TIMES BY PARALLAX

NOTE: ALL CALCULATIONS ARE SUBJECT TO CHANGE As will be seen, the calculations of cloud distance listed below indicate that the objects/lights were generally more than 10 miles away. These calculations assume that the crabbing or yaw angle, described in Appendix 2, is the same at the beginning and end of the time (or distance) interval used to create the parallax. (The wind speed was low so the crabbing angle would be small in any case; see Appedix 2.)

The measurements listed in this section were made on a screen display 300 mm wide.

16:52:50.1 to 16:53:00.45 This calculation uses a particular cloud shape, a vertical cloud “pillar,” as the reference point for parallax shift calculation to estimate the distance to a collection of clouds that the light passed behind. This sets a minimum distance to the light.NOTE: the plane was just coming out of an abrupt right turn and may still have been turning slowly when this time period starts. If so, the calculated azimuth change would be greater than it should be and the resulting distance would be less than it should be..

The time duration is 16:53:00.45 – 16:52:50.1 = 10.35 sec. The cloud “pillar” begins at the far right in the MED FOV or 1.7o right of center. The az reading at the center is -133.5. Therefore at the right side it would be –(133.5-1.7) = -131.8 at 16:52:50.1. At the end of the time interval the cloud pillar is about 15 mm right of center or (15/300) x 3.4 =0.17 deg right of center. The az reading at the center at this time is -136.3 so the az of the cloud pillar is –(136.3-.17) = -136.1, approximately. IF the yaw or “crabbing angle” (a slight turn into a cross-wind) were the same at the beginning and end of this time interval, then the total shift in azimuth was 136.1 – 131.8 = 4.3 degrees. (Unfortunately there is no way of knowing whether or not the crabbing angle was the same, but for the purposes of calculation it is assumed to have been the same.) IF during this period of time the plane flew in a perfectly straight line (a straight path is assumed for purposes of calculation) at 230 mph or 3.8 mi/min or 0.064 miles per second, then it traveled the distance D = 10.35 sec x .064 mi/sec = 0.66 mi. The track or path of the plane combined with the sighting lines to the light at the beginning and end of the interval form a triangle as illustrated below. Knowing the final azimuth (136.1) and the change in azimuth (4.3) one can determine the angles needed, = 180 – (final azimuth) and = (change in azimuth), for application of the “law of sines” to calculate the range at the beginning of the interval. The complete calculation is illustrated below. The same method is used in the following calculations of distance, R.

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NOTE: as pointed out above, the calculated distance, 6 miles, may be an underestimate since the plane may have been still turning slightly at the beginning of this calculation. In other words, the initial azimuth value, -131.8 may have been too small (in magnitude). Perhaps it should have been, for example, -1340. In this case the azimuth change would have been about 136 – 134 = 2 degrees and R = 0.66 sin 43.9 /sin 2 = 13 miles. Clearly the calculated range is very sensitive to the accuracy of the shift in azimuth, i.e., the angle , above. This sensitivity applies to all the calculations below. All of the following calculations also assume that the main axis of the aircraft lies exactly along the straight flight path or at least that the yaw or “crabbing angle” mentioned above, is the same at the beginning and end of each time interval. ……………………….

16:53:09.67 to 16:53:17.5 This calculation uses a vertical, right side edge of a cloud that crosses the NRW FOV of 0.8 o . Since the light passed behind these clouds this again provides a minimum distance to the light. The edge of the cloud starts at the right side of the FOV and ends at the left side. The azimuth of the center of the FOV stays constant at -136.1. Hence the azimuth shift is 0.8 degrees. The time difference is 17.5-9.67 = 7.8 sec approximately. During that time the plane traveled in a straight line 7.8 sec x 0.064 miles/sec = 0.5 mi. The angle = 180 – 136 = 440 approximately. Assuming that the FOV is exactly 0.8 deg and that there is no angular “slop”in the flir optics, the range at the beginning of the observation time was R = 0.5 mi x sin 44 / sin 0.8 = 25 miles. Comparing with the previous calculation the implication of this greater distance is either (a) the plane was still turning at the beginning of the previous calculation resulting in a calculated distance that is lower than it should have been or (b) the FOV angle used here, 0.8o, should be larger (the distance decreases as increases; if = 1.6o , R = 12 miles) or a combination of these.…………………………………………….

16:59:37.0 to 16:59:53.5This uses a small cloud “lump” on top of a cloud mass as the reference point. This lump is shown in Figures 20 and 21. It has the brightness label 225 in Figure 20 and 226 in Figure 21. As the plane flew along this lump crossed the NRW FOV of 0.8o. Because the light passed above this cloud this calculation provides no information about the distance to the light at this particular instant, but it does provide an estimate of the distance to the cloud. Seconds later the light passed behind the vertical cloud edge at the right side of Figures 20, 21 and 22. Since the cloud mass that contains the “lump” seems to be at the same distance as the vertical edge, it does appear that this calculation sets a minimum distance to the light. At the beginning of the time interval the lump is at the right side when the azimuth of the center is -137.0. At the end of the interval the bump is at the left side of the FOV and the center azimuth is -137.4. Hence the total shift in azimuth is (137.4 – 137.0)+0.8 = 1.20. The time interval is 16.5 sec so the plane

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D = 0.66 mi

R136.1o

43.9o

= 4.3

R = (D sin sin = (0.66 mi sin 43.9)/sin 4.3 = 6.1 miles

PLANE TRACK

CLOUD SHAPE (pillar)


traveled D = 16.5 x .064 = 1.1 miles. The angle = 180 – 137 = 43o, approximately. Therefore R = 1.1 x sin 43/ sin 1.2 = 34 miles, approximately. ……………………………………………………….

17:03:35.00 to 17:03:47.00This calculation uses the right hand peak of three peaks of a cloud which the below the “twins” pass over. The “twins” passed over these peaks, so they don’t provide a minimum distance to the twins. The right hand peak starts at the right side of the screen at the MED FOV (3.4o) and ends up at the left side of the NRW FOV of 0.8o. The time difference is 12 seconds. The distance traveled is D = 12 x ,064 = 0.77 mi. The azimuth of the center of the FOV at the beginning is -133.1, so the az of the cloud peak at the left side of the FOV is -(133.1 – 3.4/2) = -(133.1 – 1.7) = 131.4o. The az of the center of the FOV at the endof the time interval is -133.6 so the az of the cloud peak, now at the right side of the FOV, is –(133.6 + 0.8/2) = -(133.6+0.4) = -(134). The shift in az is = 134 – 131.4 = 2.6o. The angle is approximately 180-134 = 46o. Hence R is approximately 0.77 x sin 46 / sin 2.6 = 12 miles. …………………………………………………

17:03:35.00 to 17:03:45.2This calculation is based on a single peak cloud in the same scene as the previous calculation. The twins passed over this peak so it does not provide information about the distance to the twins. Because the triple peaked cloud moves backward (to the left) faster than the single peaked cloud, one can conclude that the single peak cloud is farther away. At the beginning of the time interval it is 40 mm right of center or 40 mm x 0.01130/mm = 0.450 to the right of the center azimuth of -133.1. That places it at az = -(133.1 – 0.45) = -132.6, approximately. At the end of the interval the peak is 60 mm left of the center or 0.68o left of the center azimuth of -133.7. That places it at az = -134.4, approximately. The shift in azimuth is = 134.4 – 132.6 = 1.8o. The time interval is 10.2 sec so D = 10.2 x 0.064= 0.65 miles. The angle = (180 – 134) = 460, approximately. Hence R = 0.65 x sin 46/ sin 1.8 = 14.9 miles, approximately. Hence the distance calculation by parallax shift is consistent with the observation that the single peak cloud is farther away than the triple peak cloud.………………………………………………..

17:03:58.5 to 17:04:33.4This calculation is based on a nearly vertical cloud edge that the “twins” and other lights pass behind. Thus this calculation sets a minimum distance to the lights.At the beginning of the time interval the az = -132.8. The cloud edge changes shape a bit during the =time interval but this is acceptable (one has to keep track of its location as the scenery keeps changing somewhat as all the lights pass by the edge). At the end of the 34.9 second interval the az = -137.0. The azimuth change of the center is 4.20. To this is added the azimuth change by going from the right side to the left side of the FOV, 0.8o, for a total change of 4.2+.8 = 50. During the time interval the plane traveled 34.9 x 0.064 mi/sec = 2.2 miles, approximately. The angle = (180 – 137) = 43, approximately, so R = 2.2 x sin 43/sin 5 =17 mi, approximately. ……………………………………………………

17:04:39.8 to 17:05:12.3

This calculation is based on the location of the left hand pillar of two, moderately large cloud pillars in a lower layer of clouds. This cloud pillar is just below a dark “cloud hook” in the upper layer of clouds (see below).At the beginning of this 33.5 sec time interval the left hand pillar is at the center of the field of view and the azimuth at the center of the MED FOV is -135.4o. At the end of the interval the pillar I at the left side of the field of view and the center is at -138.3o. Hence at the end of the interval the azimuth of the pillar is –(138.3 + 3.4/2) = -140o. The distance D = .064 x 33.5 = 2.1 miles, = (180 – 140) = 40o and = 140-135.4 = 4.6 so R = 2.1 x sin 40 / sin 4.6 = 17 miles, approximately.

……………………………………………………….

17:04:37.0 to 17:05:16.5

This calculation uses the end of a curved, dark, cloud “claw” as the reference point. It moves more slowly than the cloud pillar which appears below it, so it is farther away. Also, the lights emerge from a cloud wall which seems to be associated with the cloud hook rather than with the closer cloud pillar. They may even be farther away than the

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“claw”. At the beginning of the 39.6 second time interval the center of the MED FOV is -136.2 and the end of the “claw” is 90 mm to the right, or 90 x .0112 = 1.0 degrees less than 136.2,i.e., at 135.2o. At the end of the interval the end of the claw is at the left side of the screen and the center is at -137.3,so the end of the claw is at –(137.3+3.4/2) = -139. The angle = 139 – 135.2 = 3.8, D = .064 x 39.6 = 2.5 miles and = (180-139) = 41o. Hence R = 2.5 sin 41/ sin 3.8 = 25 miles approximately.…………………………………………

17:04:59.1 to 17:05:53.5This calculation is based on a vertical cloud edge that is preceded (i.e., on the left side) by a small, dim “pillar” which can be found in both MED FOV and NRWX2 FOV scenes. Since all the lights passed behind this edge this calculation sets a minimum distance for the lights. At the beginning of this 54.4 second time interval the azimuth at the center of the MED FOV is -133.6o . The cloud edge is at the right side of the MED FOV, so for the cloud edge, az = -( 133.6 – 3.4/2) = -131.9o. At the end of this time interval, just after all the lights have passed behind the cloud, the edge is at the left side of the NRWX2 FOV so the azimuth is 0.2o than the center FOV at -137.6o, that is az = -137.8o. The total azimuth shift is 137.8 – 131.9 = 5.9o. The distance traveled is D = 54.4 x 0.064 = 3.5 miles, approximately, and = (180 – 137.8) = 42o, approximately, so the distance to the cloud is R = 3.5 sin 42/sin 5.9 = 23 miles, approximately.……………………………………………..

17:08:05.0 to 17:08:34.0This calculation is based on the right hand vertical edge of a dark cloud. There are other clouds farther away. The lights went behind these clouds. At the beginning of this 29 second time interval the edge of the cloud is at the right side of the NRWX2 FOV for which the center azimuth is -139.1. Hence the initial direction to the cloud edge is az = -(139.1 – 0.2) = -138.9o. At the end of the interval the cloud edge is at the left side of the NRWX2 FOV and the center is at -141.1 so the ending az = -(141.1+.2) = -141.3. Hence = 141.3-138.9 = 2.4o, = 180 – 138.9 = 41o, approximately, D = .064 x 29 = 1.9 miles, approximately, and R = 1.9 sin 41 / sin 2.4 = 30 miles, approximately.………………………………………………..

17:17:38.5 to 17:18:01.0This calculation is based on the motion of a dim cloud peak, seen between propeller blades (!), which crosses the MED FOV. At the beginning of this 22.5 second time interval the peak is at the right side and the center azimuth is 101.3, so the peak is at az = -(101.3 – 3.4/2) = 99.6o. At the end of the interval the peak is at the left edge and the center azimuth is -103.4 so the peak is at –(103.4 + 3.4/2) = -105.1o. Hence = 105.1 - 99.6 = 5.5o, = 180 – 105.1 = 75o ,approximately, D = .064 x 22.5 = 1.4 miles, approximately, and R = 1.4 x sin 75/sin 5.5 = 14 miles, approximately.

CALCULATION OF SPEED BASED ON THE TRANSITTIME BETWEEN POINTS AT CALCULATED DISTANCE

This calculation is based on the transit time of “Follower 1” (F1) to cross part of the MED FOV and then disappear behind a cloud edge. It is also based on the assumption that the line of clouds is 25 miles away (see the section on calculation of distances) and runs parallel to the track of the airplane. At 17:05:18.0 the azimuth of -135o points toward the center of the distance that will be traveled in the next 21 seconds. F1 is 143 mm to the left of the cloud edge on 330 mm wide screen. (The “twins” are between F1 and the cloud edge.) Thus the 143 mm corresponds to a distance of 25 tan (143 x .0103) = 0.64 miles. However, this is a “projected” distance as measured perpendicular to the line of sight at a distance of 25 miles. The line of sight is at an azimuth of 135 o relative to the airplane track. The actual distance that will be traveled by the light, as measured parallel to the airplane track, is 0.64/cos(180 – 135) = -.64/.71 = 0.9 miles. At 17:05:39 a light that is assumed to be F1 (the light that the camera narrowed in on at 17:05:22 and then followed in the NRW and then NRWX2 fields of view) reaches the cloud edge. Thus it traveled about 0.9 miles in about 21 seconds corresponding to a speed of (0.9/21) x 3600 sec/hr = 150 mph, approximately. This is about 65% the speed of the airplane. Clearly, if the distance were different from 25 miles or if the light were not traveling parallel to the airplane track the calculated speed would be different.

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One observes in the basic data sheet that from 17:03:35 to 17:07:00 the azimuth to the light(s) continually increases from about 133 to about 139 degrees, that is an increase of 6o over a 4 minute time period. Since the plane was flying in a straight line this implies a slow “drop back” of the lights, i.e., they were traveling more slowly than the airplane or, at least, their component of velocity parallel to the airplane track was lower than the airplane speed.

On the other hand, the single light observed from 16:52:52 to 16:57:56 has a nearly constant azimuth in the range 133 to 136 and back to 133 or 134o. This light may have paced the plane.

During the third UFO segment, from 17:15.01 to 17:18:36, a time period of 3 min, 35 sec = 3.6 minutes, the azimuth continually increased from -90.1 to -105.3 o or 15 degrees, approximately. The increase in azimuth corresponds to the light dropping back. If the light were at range R, then the distance corresponding to the change in azimuth is X = R tan 15 = 0.27R. The light went behind a cloud at a calculated distance of about 14 miles. If it were 15 miles away, X = 4 miles. This distance difference was acquired in 3.6 minutes, corresponding to a speed difference of 67 mph, assuming that the light was traveling along a track parallel to the airplane. (If the light was not traveling parallel to the airplane the speed differential could be larger or smaller than this.) If the distance were 25 miles, such as estimated above, the distance would have been about 7 miles and the speed difference would have been about 112 mph, again assuming parallel tracks. Conversely, if the speed difference were 230 mph, meaning that the light was stationary, then the plane would have traveled 3.6 min x 3.8 miles/min = 14 miles, approximately, relative to the light at its left side, and the range to a stationary light would have been approximately 14/tan15 = 52 miles. Assuming the -1o elevation measurement is correct, at 52 miles the light would be at an altitude about 1 mile below the plane, which was flying at approximately 2 miles altitude, so the light would not be on the ground. (At -1 elevation with the plane at a bit above 2 miles altitude the ground level is about 115 miles away.)

……………………………………………………………………………………………………..

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APPENDIX 4ANALYSIS OF THE GAS FLAME EXPLANATION

A4.1 The Oil Field Proposal

Within days of the public announcement of these sightings on May 9, 2004, various scientists and investigators proposed several Candidate Explanatory Hypotheses (CEH) to explain the flir lights and radar targets. The CEH ranged from secret US aircraft to balloons and ball lightning. Many proposals assumed that the objects detected by the flir system were above ground and traveling at some speed. However, one proposal involved infrared radiation sources at ground level, namely, fires burning off gas from oil wells in the Gulf of Mexico northeast of Ciudad del Carmen.

I first learned of oil wells near Cd. del Carmen from an email message on May 22. However, the writer merely pointed out the possibility of fires associated with gas burnoff at the oil wells. The first person to formally conclude that the Cantarell Oil Field was the source of the flir images was Capt. Alexandro Franz on May 26. (Others may have reached the same conclusion at about the same time.) According to Franz, the Cantarell Field or Cantarell Complex is the largest oil field in Mexico, located 80 kilometers offshore in the Bay of Campeche. There are five major components of the field known as Akal, Nohoch, Chac, Kutz and the recently discovered Sihil. There are over 200 oil wells associated with 9 platforms run by PEMEX (Mexican Oil Company) spread out along a nearly north-south line about 22 mi long. The approximate area of the oil field is illustrated in Figure 1.

A4.2 The Direction of the Line of Sight to the Twins

The possibility that the oil field could be the source of at least some of the flir lights is based on the following fact: the flir was pointed in the general direction of the oil field as the “twins” and “triplets” were being recorded. This fact can be verified by comparing the geographical coordinates of the plane and one of the major oil platforms, called “Nohoch A” , at 17:03:45 when the flir video is focused on the “Twins” (see Figure 27). The geographic coordinates for Nohoch A are 19o 22.1’ north latitude by 92o 0.23’ west longitude, as illustrated in the following aeronautical navigation map, Figure A4A . (The map without the sighting lines and notations was provided by Capt. Franz.) On that map I have drawn sighting lines from the various wells toward the airplane and a geometric construction to illustrate how the 9 oil platforms would appear on the flir video if they could all be seen at the same time (see the small rectangular box at the bottom).

At 17:03:45 the geographic coordinates of the plane (see Figure 27) were 18o 26.52’ lat. by 90o 46.27’ long. Comparing oil platform and the plane we find that the latitude difference is 55.6’ and the longitude difference is 1o 13.9’. Translated into distances (1’ arc latitude = 1.15 mi north-south and, at 18o latitude, 1’ longitude = 1.094 mi east-west) these arc distances correspond to 63.9 mi north by 80.8 mi west. The distance between the plane and the oil platform was, therefore, the hypotenuse of this triangle (the sighting line from the plane to the platform), (63.92 + 80.82)0.5 = 103 mi. The angle between due north and the sighting line from

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the plane to the oil platform was arctan(80.8/63.9) = 51.7o to the west of due north (-51.7o relative to north). Since the plane was flying along a track of about 81o (east-northeast), the total angle between the airplane track and the sighting line to the platform was -51.7 – 81 = -132.7o. If the heading of the plane had been exactly along its plotted track (no “crabbing angle” or yaw) and if the flir pointing direction were accurately set to read zero when pointing along the heading of the plane, then the flir azimuth (center of the field of view) would read -132.7o when pointed toward the oil platform. The actual reading was about 1o larger at -133.6o.

FIGURE A4AMAP OF THE OIL PLATFORMS AND AN INDICATIONOF HOW THEY MIGHT APPEAR IN THE FLIR VIDEO

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Itr is probable that there was a slight wind from the southeast so that the plane might have a degree or so crabbing angle clockwise (as seen from above), this crabbing angle would increase the flir azimuth reading. For example, a 1o crabbing angle, which is certainly a possibility, would bring the flir reading into perfect agreement with the direction to Nohoch A. Whether or not agreement between the flir azimuth and the direction from the plane to Nohoch A is perfect, this calculation does show that the flir was looking in the general direction of the oil platforms.

A4.3 The Angular Elevation of the Twins

Although the sighting direction is consistent with Nohoch A, the flir elevation data are not. The flir elevation reads 3o above local horizontal. The angular elevation of the platform can be calculated from the geometry of the airplane and the earth’s surface. Taking into account the curvature of the earth because of the great distance, 103 mi, and using 2 mi as the altitude of the plane (but ignoring atmospheric refraction) the calculated elevation (depression) angle is -1.86o. As pointed out in Appendix 2 there could be a 2 degree or so correction to the flir reading of the elevation angle which would reduce it to about 1o or maybe even 0o, but this is still above the actual elevation angle. To obtain agreement would about a 2o error in the flir elevation reading.

The oil platforms are spread over a distance of 25-27 miles in a nearly north-south direction. When projected onto a line or plane that is perpendicular to the line of sight to the airplane (see Figure A4A) the platforms cover a distance of about 21 mi. At a roughly average distance of about 110 mi the angular width of the oil fire display as seen from the plane would be about arctan(21/110) = 10.7 degrees. As pointed out in Appendix 1, at 17:03:35, where there is a crude illustration of the “twins and friends,” the length of the array of lights from the rightmost “leader” to the leftmost “follower” is an angle of 2.1o. Hence, if the images are of oil flames they must be from only a few of the oil platforms.

A4.4 Comparison between the Closest Oil Platforms and the Twins

Nohoch A was not the closest oil platform to the plane. The closest was Xochitl at 92o 0’ longitude and 19o 17’ latitude. The latitude difference is 50.5’ and the longitude difference is 1o 13.7’. Hence the distance from the plane is (582 + 80.62) = 99.3 mi and the sighting line direction is 54.30 to the left of due north. The azimuth relative to the airplane heading of 81o was 135.3o. Adding 1o as an estimate of the “crabbing angle” the result is 136.3o. This is more than 2.1o larger than the azimuth to Nohoch A (calculated at 133.6o with crabbing angle). But the total angular width of the array of lights is only 2.1o. Therefore if Xochitl was in view, then Nohoch A was not and v.v. In other words, the combination of Nohoch A and Xochitl, the closest platforms which would (one presumes) make the brightest gas flame images, cannot be the “twins” because they are too far apart. At the distance of Nohoch A, 103 mi, the spacing between the twins, about 0.1o or 0.00174 rad, would be about 0.174 mi or 920 ft which is much smaller than the nearly 6 mi separation between these two platforms.

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A4.5 The Height of a Flame based on Assumed Ocean Reflection

A gas flame at the top of a chimney pipe at some height above the ocean would produce a reflection in the water which, from a great distance (100 mi), would appear to be directly beneath the bright flame. The image of the reflection would also appear somewhat diffuse and much dimmer than the image of the flame. Each twin does have a weaker light image directly below it, so assume for the moment that the lower image corresponds to the reflection of the gas flame in the water. The question then is, how high above the water surface would the flame have to be to create a reflection similar to the lower light in the video? As shown in Appendix 1 at 17:03:45, the spacing between the center of a twin image and the dimmer light below is about 0.00063 rad. This corresponds to about 340 ft at 103 mi. This does not mean that the flame was 340 ft above the water because the way a mirror reflects light: the angle of incidence equals the angle of reflection. In general, the reflection of a light (e.g., a flame) in a flat (horizontal) surface appears to come from a point below the surface that is at the same distance below as the light (flame) is above the surface. Of course the ocean is not a perfectly flat surface because of waves, but if the height of the light source and the distance to the viewing position are both much larger than the wave height, then the ocean surface will seem flat “on the average.” In this case the effect of the waves is to blur the reflected image of the light. Because of this characteristic of light reflected in a (rough surface) mirror, if the apparent spacing between the light and its reflection is 340 ft, then the light was about half that or 170 ft above the surface. The ocean, being mostly water, acts like a dielectric reflector rather than as a metallic reflector. Hence it reflects only a fraction of the light (much of it penetrating the surface of the water). From the preceding discussion one may conclude that, if the “twins” are images of flames associated with Nohoch A they must be about 900 ft apart and about 170 ft above the water.

A4.6 Comparison of the Triplets with Three Oil Platforms

Another test of the oil flame Candidate Explanatory Hypothesis can be carried out for the “triplets.” If the flir lights are images of the oil flames, then there should be a pattern of oil fires which matches each triplet. Specifically, the image spacings should be such that the spacing between the right end and middle fires, as projected onto a line perpendicular to the line of sight, should be less than the spacing between the left end and middle fires. Also, the right end and middle fires should appear higher in the video than the left end (see Figure 40). A candidate for creating the triplet images is the combination of three platforms: Nohoch A, Akal C and Akal J (see Figure A4A). At this time (17:06:51) the plane was farther away than it was when the “twins” were appearing. At this time the distance to Nohoch A was about 111 mi. The coordinates of Akal J are 19o 25.7’ by 92o 4.5’ and those of Akal C are 19o 24’ by 92o 2.3’. Figure A4A shows that these three platforms almost form a line directed toward the plane. (If they did form a straight line toward the airplane, and if all the oil flames were at the same height above the water, then they would make three images, one above the other in a vertical line. However, they are not in a straight line toward the plane and the deviations from straight are important.) As illustrated in Figure A4B (NOT TO SCALE), they form a “bent” line with the platforms about 3.1 mi apart. The line from Nohoch A to Akal C forms an angle of about 47o to the left of north and the line from Akal C to Akal J forms and angle of 50o to the left of north. The line of sight to the plane forms an angle of 56o to the left of north. From the location of the

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airplane Akal J would appear at the right end of a triplet, with Akal C in the middle and Nohoch A at the left end.

FIGURE A4BCOMPARISON OF THE THREE OIL PLATFORMS

WITH THE “TRIPLETS” IMAGES

The important question to be answered is, what would be the relative locations of the images of these platforms in the flir video? The answer is found by projecting the platform locations onto a plane (line) perpendicular to the line of sight as illustrated. The line from Nohoch A to Akal C, 3.1 mi long, is projected onto the line perpendicular to the sighting line by multiplying 3.1 mi by the cosine of the angle between the perpendicular line and the line connecting the platforms: 3.1 cos (47 + 34) = 0.485 mi = 2560 ft. Similarly, the distance between Akal C and Akal J, when projected onto the perpendicular line, is 3.1 cos (50 + 34) = 0.324 mi = 1710 ft. These projected distances appear as angular differences at the plane. The angles (in radians) are found by dividing each projected distance by the distance from the platforms to the airplane, about 111 mi for Nohoch A and 114 I for Akal C. Thus, between Nohoch A and Akal C the angle would be 0.485 mi/111 mi = 0.0044 rad or about 0.25o (divide the number of radians by 0.0174 rad/degree to get degrees) and the angle between Akal C and Akal J would be 0.324/114 = 0.0028 rad or 0.16o. One notes that the spacing between the right end (Akal J) and the middle (Akal C), 0.16o, is less than the spacing between the middle and the left end (Nohoch A), 0.25o. This is consistent with the actual relative spacings.

A4.7 Comparison of the Calculated and Actual Triplet Spacings Although the relative spacings of the projected platforms are consistent with the triplets,

the magnitudes of the actual angles are much larger. Consider the “leading” (right hand side) triplet. Between the right end and the middle the angle (from the video) is about 0.021o which

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corresponds to about 214 ft at 111 mi and between the middle and the left end the angle is about 0.026o which corresponds to about 265 ft at 111 mi. Similarly the spacings for the lagging triplet are 0.021o and 0.031o (0.031o corresponds to about 315 ft at 111 mi.). Thus the actual spacing angle between the right end and the middle, 0.021o is about 1/8 of the spacing calculated above (0.16) and the two spacings between the middle and the left hand light, 0.026o and 0.031o, are about 1/10 and 1/8 of the calculated spacing of the platforms (.25o). This rules out the combination of these three platforms as the source of one triplet.

A4.8 Analysis of the Relative Heights of the Triplet Images

The triplet images do not lie along a horizontal line in the video. If it is assumed that they are images of distance flames (or other lights) then, because the flir camera was looking down at them, lights that are farther away would appear higher up in the video than lights that are closer, assuming that the lights are all at the same height above the earth. Since Nohoch A was the closest of the three platforms, flames on Nohoch A would appear lowest in the video and the flames of Akal C would appear higher up and the highest image would be that of Akal J. The actual elevation angle differences can be measured from the video. Consider the “leading” triplet. The right end and middle images are at the same height in the video image so they would be at the same distance (assuming the flames are the same height above the water). However, they are both higher up than the left end light which is 0.0088o lower and, hence closer (again assuming that all the flames are the same height above the water).

Compare this with the angular elevation difference obtained by direct geometric calculation, again taking into account the curvature of the earth but ignoring the small differential atmospheric absorption. The elevation is calculated from the law of cosines combined with the geometric construction illustrated below which uses the ratio of a distance measured along the earth’s surface as compared to the earth’s radius to find an angle

FIGURE A4CCALCULATION OF THE ELEVATION ANGLE OF A GAS FLAME

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that can be used with the cosine law and then the sine law to find the elevation (depression) angle.

Nohoch A, 111 mi away, is at an elevation of -1.8404o and Akal C, 114 mi away, is at an elevation of -1.8351o. (The seemingly excessive five significant figure accuracy is needed to calculate to two significant figures the differences in angles.) The difference is 0.0053o. This is smaller than the 0.0088o actually measured. Akal J, which would be the source of the right end image of the triplet, is about 117 mi away at an elevation of 1.8312o. There is a difference of 0.0039o between this platform and Akal C. This difference, almost half the vertical difference between the left end image and the middle image, should be apparent if the middle image is that of Akal C and the right end image is that of Akal J. But no such difference appears in the “leading” triplet. There is a very slight height difference between the right end and middle images in the “lagging” triplet, but that difference is only 0.0013o, 1/3 of what is required to match Akal C and Akal J to the triplet image (along with Nohoch A as the left end). The main point of this discussion of image heights is that, if the images are of the three platforms, then the right end image of a triplet should be higher than the middle and the middle higher than the left end and the differences in angular elevation between the left end and the middle images should be larger than the calculated value listed above (i.e., larger than 0.0053o).

The measured vertical distance between the left end and middle image of a triplet is about 0.0088o = 0.000153 rad. At a distance of 111 mi this corresponds to a difference in vertical height of about 90 ft. If the flames of Nohoch A and Akal C were at the same height the vertical angle difference would be 0.0053o = 0.000092 rad. At 111 mi this corresponds to 54 ft. Thus more than half of the difference in the vertical angle on the video, 0.0088o, results from the different distances of the flames. The remainer of the difference in angle, 0.0088-0.0053 = 0.0035o must therefore result from having different heights of the flames. Specifically the flames must differ in height by 90 – 54 = 36 ft, with the flames of Akal C higher than those of Nohoch A. ………………………………………….NOTE (added for web publication, Dec. 27, 2004) This appendix is the beginning of a test of the oil flame hypothesis. As with any Candidate Explanatory Hypothesis, all key predictions of the CEH should be compared with the actual data. Any disagreements should be explainable. From the calculations presented above it seems that at least some of the predictions of this hypothesis do not seem to agree completely with the flir image data. In principle it should be possible to establish a 1:1 relationship between gas flames and flir images. The fact that the calculations above do not seem to show agreement with the oil platforms that were chosen for this mathematical/geometric exercise could simply mean that I chose the wrong platforms to calculate. More accurate calculations cannot be done, however, without exact data on the locations and heights of the flames themselves. The best way to obtain data for comparison with the oil flame hypothesis would be to refly at least part of the flight during a day of similar weather or a perfectly clear day. (A perfectly clear day would, one presumes, show at least as many and perhaps more bright images than the March 5, 2003 flir data tape shows, if the images are gas flames.) A comparison videotape could be made and, if the flir images are, in fact, of the oil wells (or is some of them are), then it should be possible to overlay the March 5 flir video images with the test video images and find an exact, or nearly exact, match.

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