Strange Flying Object in the Cave of Swallows

by Bruce Maccabee

DATE:  November 16, 1996
TIME:  about 11:30 AM local time
       in the State of San Luis Potosi, Mexico
       Latitude 20 deg, 30 min north
       Longitude 99 deg. 6 min west

LIGHTS: sun at about 140 deg azimuth, 45 deg elevation
CAMERA: Sony VX-1000 operated by professional cameraman Mark Lichtle 
       using a short shutter time, 1/2000 sec, to "stop motion" during a field or frame
ACTION:  a guy is about to jump into a cave that is hundreds of 
       feet deep while the cameraman films his descent

REACTION:  Huh? Did someone say jump into a cave hundreds of feet deep?
(Can You Swallow This?)

Yes, indeed.  A cheap way to get to get to use your parachute:
let yourself be swallowed up by the nearly vertical Cave of Swallows.
This has been called "base jumping."

The subject of interest to Mark was the base jumper/parachutist.  Mark pointed his camera at 
the man before the jump while both were at the mouth of the cave, the jumper being about 45 feet 
away.  Then Mark kept the jumper quite well centered in the field of view as he lept into the 
cave and commenced his downward fall, culminating in the opening of his parachute and a 
"gentle" landing at the bottom, about 500 ft below.

But as the jumper falls downward something else happens....
a strange image appears in the video, suggesting that a strange object traveled through the 
field of view of the camera.  It is that image which is the subject of this analysis.

The Images

Although the well-known frame rate for video is 30 per second, the picture can be created in much less than 1/30 sec. In this case the time to create one video "field," which shows the whole scene in front of the camera, was 1/2000 of a second. The video camera was operated with a short shutter time to "stop" the motion of the falling base jumper. That is, the shutter time was short enough so that the image of the base jumper was not smeared by his own motion (downward) during the time that a complete complete "field" was generated by the camera electronics. [Video Tutorial: in the far "olden days" of TV the picture was created by scanning a light- sensitive electronic device across a scene in such a way as to create a collection of horizontal lines, one above the other, with brightness varying along each line as determined by brightness variations of the scene being imaged by the camera. Typically there would be something like 460 horizontal lines from the top to the bottom of the picture. The 460 would be created in 1/30 of a second, but they weren't all created in serial order from top to bottom of the picture. In order to speed up the creation of representation of a complete scene the scanning mechanism created a top (first) line, then the third line down and then the fifth, seventh, 9, 11, etc. to line 459 at the bottom, all in 1/60 of a second. There were gaps between these scan lines, so the mechanism then returned to the top of the picture to fill in the gaps by creating the scan lines numbered 2, 4, 6, etc. down to 460, during the next 1/60 of a second. These even scan lines were in between the odd scan lines, thus making an "interlaced" scan. By the time the scanning mechanism had finished line 460 the time taken from the beginning of the first scan line to the end of the last was essentially 1/60 + 1/60 = 1/30 sec. Then the scanning mechanism returned to the top of the scene to begin the next field and frame. By convention the complete "frame" consists of the two "fields," the field of odd numbered scan lines plus the field of even scan lines. That makes a frame with vertical "resolution" of 460 lines. The main point to understand here, however, is that a field, obtained during a time of 1/60 sec (or less, e.g., 1/2000 sec), contains a complete picture of the scene but at 1/2 the vertical resolution of a frame. That is, the vertical resolution of the field would be 230 lines if the frame resolution is 460 (and more than 230 lines if the frame resolution is more than 460). With the new modern cameras the field can be created in less than 1/60 sec, say 1/2000 of a second. Then the camera waits 1/60 - 1/2000 = 0.01666666 - 0.0005 = 0.0161666 of a second before creating the next field in order to maintain the 60 fields or 30 frames per second standard frame rate. It is to be noted that a typical home video tape recorder - VTR - plays all fields when operating at normal speed. However, when operated at a frame by frame rate (one frame at a time) it skips every other field. In particular, it may show only the fields of odd lines, the "odd fields", rather than every field. Professional VTR are available which show the odd and even fields in their proper order.] In this particular video the anomalous image appears in 5 "frames" as played frame-by-frame on a home VTR, but it appears 11 fields. (Although I used a home VTR to capture the images shown here, I know what the successive fields show because Jose Escamillo provided me with pictures, made with a professional VTR, of the 11 fields which show the anomalous image.) The first field showing the image precedes the field (frame) shown at the beginning of this discussion (see above). Note in the upper right corner the white, elongated image with "things" sticking off at the sides. As time went on the jumper fell and the anomalous image moved from right to left and upward across the screen. Six fields (3 frames) later the image appears above the image of the diver.

Two fields (one frame) later a most fortuitous event happened: the front 3/4 of the image is dim with only the rear 1/4 still bright.

The anomalous image appears once more in the next field and then is gone. The picture below shows the scene one frame (two fields) after the previous picture. The strange image is gone.

The anomalous image maintains a rather consistent shape and brightness through the first nine fields of its appearance. Then the brightness of the front 3/4 diminishes considerably in the tenth frame and the whole image is dim in the eleventh. The consistency in shape appearance and the continuous motion across the field of view argues against any extremely unusual camera artifact. On the other hand, if one assumes there was a real object "out there" which caused the image one can explain the sudden darkening of the image in the tenth frame. The explanation is that the unusual object (UO) crossed the invisible "light/shadow boundary" and was no longer directly illuminated by the sun. If one further assumes that the light/shadow boundary was the same boundary that made the shadow line on the wall of the cave (see the sloping boundary between light and dark in the above images), namely the shadow of the mouth of the cave, then it is possible to estimate the distance to the UO. (NOTA BENE...note well: the following analysis assumes that the shadow which the UO entered was, in fact, the shadow made by the rim of the cave. However, if it were possible to demonstrate that some other artifact, rock, tree or whatever, could have created a shaded volume of air relatively close to the camera, within the field of view of the camera as it looked down into the cave, and far enough from the camera so that anything in that volume would be well focused. then this, too, could explain the entry of a moving object into a shaded area. In this case the object could be closer and hence smaller than calculated below. But it could not have been extremely close, because it seems to be in focus; see below. Since the cameraman was standing on the rim of the cave about 45 ft east of the jumper, it doesn't seem likely that there was any tree at his left side that could have cast a shadow in part of the volume of air in front of the camera. But perhaps a protruding rock could have. This can only be resolved by returning to the site and standing where the camera was and looking downward at the 57 degree angle calculated below to find out how close that sighting line was to the nearby rocks, etc. at the side of the cave rim. The UO seems to be quite well focused with edges that are not blurred out by motion or defocus. This raises the question, how close could the object be to the camera and still be in reasonably good focus? Although the camera was in broad daylight, the cave itself was quite dark. The camera f-stop opened to 2.8 to provide proper exposure looking down into the cave. This is probably why the image of the jumper is so bright as long he was in the direct sunlight and why the UO appears so bright when directly illuminated by the sun. It is possible to calculate the closest distance at which an object would be in reasonable focus based on the camera optics. The exact focal length is not known but if one assumes a reasonable value such as 100 mm, then the combination of the focal length, the f stop, the assumed circle of confusion of .1 mm [0.001 radians or about 1/30 of the angular length of the UO; see below], the diameter of the lens aperture [35 mm] and the focal distance, 93 ft [distance of the jumper; see below], indicate a near-focus distance of about 50 ft. Any object substantially closer than this would have rather fuzzy edges and be very blurred by defocus [as well as by any substantial motion during the shutter time]. Experiments should be done to determine just how close a small bright "rod- like" object could be to the camera and still be in focus.)

The Cave

The cave is a hole in the ground on the side of a mountain. The following pictures show the cave from different directions looking downward from far above. The black X marks the location of the cameraman and the yellow X marks the location of the jumper (before he jumped). Note that they were at the north side of the mouth of the cave.


The mountain rises toward the south so the higher part of the mouth of the cave is south of the cameraman and jumper. Of course the mouth of the cave is irregular so the shadow of the mouth is also irregular. However, an approximate model of the cave and its mouth can be created in order to estimate the distance from the camera to the light/shadow boundary. Jose Escamillo has provided me with the following dimensions for the cave mouth: about 205 ft (62.5 m) in the north-south direction by about 160 ft (48.8 m) in the east-west direction. The mouth of the cave is tilted with the south side higher than the north side by about 109 ft (33.2 m). The cave is therefore approximately an elliptical cylinder with the dimensions given. The mouth of the cave is approximately an elliptical surface, tilted to the horizontal by an angle of arctan(109 ft/205 ft) = 28 degrees. The shadow on the wall of the cave, as it appears in the video, is on the northwest wall. It is made by the southeastern edge of the cave mouth. Because of the complex geometry and the irregularities of the cave mouth it would be very difficult to determine the exact location of the light/shadow boundary or "shadow surface" in 3-D space within the cave. However, an approximation can be used to get a useful distance estimate. The method for estimating the minimum distance of the object from the camera when it crossed the light/shadow boundary are shown below. Even though the south rim of the cave is about 109 higher than the camera 205 ft away, the angular elevation of the sun is great enough to make the shadow of the rim be far below the camera level. In fact, the 45 degree elevation of the sun causes the shadow of the rim to occur about 205-109 = 96 ft below the camera. These numbers are not expected to be accurate since the sun was a bit east of due south and the cave rim is irregular (the 96 ft estimate could be off by 20 ft one way or the other). The light/shadow boundary surface is represented by a straight line sloping at 45 degrees from the south-eastern rim downward to the wall of the cave on the northwest side. The important fact to notice is that, because of the width of the cave mouth, the light/shadow boundary surface is at a considerable distance from the camera. The calculation presents an estimate of the minimum distance. The actual distance may have been several tens of feet more than estimated below.

The above calculation indicates that the point on the shadow surface that was closest to the camera may have been about 70 ft from the camera. However the direction from the camera to the apparent location of the UO was not toward the closest point and so the actual distance was may have been greater than 70 ft. This is important for estimating the length of the object based on the length of the image. To make this calculation it is necessary to determine the angular size calibration for the picture. Then the angular size (angular length) of the object can be multiplied by the estimated distance to get the estimated length. In order to determine the angle calibration it is necessary to appeal to gravity.

The Gravity of the Matter

In order to calibrate the angular sizes of images in the video it is necessary to know either the focal length of the camera and width of the film/video format or to know the actual size and distance of some object within the field of view. Because the cameraman was zooming and rotating the camera to follow the jumper, the exact focal length at the time of the appearance of the strange object cannot be determined from camera specifications. However, it is possible to estimate the angular size of an object in the film, namely, the jumper himself, because of gravity. Yes, a matter of "gravitas" (a political term often used in a derogatory manner a year before "911"). Since the jumper, well, jumped, he was falling under the effects of gravity and therefore experienced an acceleration of 32.2 ft/sec^2 (9.8 m/sec^2) (until his parachute opened several seconds after the strange object appeared in the video). Applying Newton's equations for free fall one finds that the distance fallen in a time t, seconds, is 1/2 g t^2, where g = 32.2 ft/sec^2. By counting frames and fields one finds that the time from the instant the jumper's feet left the rock at mouth of the cave to the time when the UO crossed the shadow line is about 2.2 seconds. During that time he fell about 78 ft. Unfortunately the exact dimensions of the jumper are not available, but assuming he is a man of typical stature (5.5 to 6.5 ft tall) one can estimate that from the top of his head to his crotch is about 3 ft (since his back and neck appear straight; note that his legs are curved by some amount so one cannot use his height from head to foot). The angular size of an object 3 ft long, measured perpendicular to the line of sight, and 78 ft away is 3/78 = 0.038 radians. However, the 3 ft dimension was not exactly perpendicular to the line of sight and the distance was greater than 78 ft (because the jumper was not directly below the camera). To get the correct angular size estimate some further calculations are needed. (The following may be skipped by those not familiar with trigonometry and projections onto the line of sight.) (FOR THE EXPERT: One may assume from the way the jumper oriented his body that his back was basically in a horizontal plane as he fell "face-first" into "the pit" (fortunately.... no pendulum!). But the camera was looking down at an angle. This look down angle can be estimated from the initial horizontal distance between the jumper and the camera, about 45 ft. The jumper did move horizontally somewhat away from the camera as he fell, so assume a horizontal distance of about 50 ft when the vertical distance was about 78 ft. The look down angle was therefore about arctan(78/50) = 57 degrees below horizontal. Imagine a triangle with the camera at the peak (on the rim of the cave)and drop a line straight down 78 ft. From the bottom of this line, 78 ft below the rim, draw a horizontal line (the baseline) over to the jumper. The third line of the triangle (hypotenuse) goes from the jumper to the camera, the line of sight. If the jumper's (horizontal) body were rotated so a line from his head to his crotch was perpendicular to the baseline, then the projected size would equal the actual size and the above angle calculation would be correct. On the other hand, if the crotch-head line were parallel to the baseline then the length of the jumper's body, as seen by the camera, would be the actual length projected onto a line perpendicular to the line of sight [and lying in the plane of the triangle]. Using the 57 degree angle, the projected length would be 3 sin 57 = 2.5 ft. The jumper's body is not oriented so that the crotch-head line is parallel to the baseline, nor is it perpendicular to the baseline. Hence the actual projected length lies between 3 and 2.5 ft. For purposes of this estimate assume that the projected length is about 2.8 ft.) The distance from the camera to the jumper (the hypotenuse) was 78/sin57 = 93 ft. Using 2.8 ft as an estimated projected head-crotch length, the angular size of the head-crotch length was 2.8/93 = 0.030 radians. The length of the UO image in when it crossed the light/shadow boundary is about 0.85 of the length of the crotch-head image. Hence the angular size (length) of the UO image, as projected onto the line of sight, is about 0.025 radians. Since the UO appears to be rotated at some angle in 3-D space such that it was not perpendicular to the line of sight, the angle if it were perpendicular to the line of sight would be greater than this. (The projected size is either equal to or less than the actual size of an object.) Radian measure can be projected outward a certain distance to obtain size at the assumed distance. For example, if the UO were 10 ft away its projected length would be 10 x .025 = 0.25 ft. If it were 100 ft away the projected size would be 10 times larger than this or 2.5 ft, and so on. Using the 93 ft hypotenuse distance calculated above yields a projected length of 2.3 ft. As pointed out above, this projected length is probably an underestimate of the actual length. (NOTA BENE: as pointed out before, this is based on the assumption that the shadow was made by the south rim of the cave; if the shadow was made by some artifact much closer to the camera then the length of the object could be much less. However, as pointed out in the previous NOTA BENE, the focal properties of the camera lens and the fact that the image seems to be in quite good focus, suggests that the UO was at least several tens of feet from the camera.)

The Need for Speed

The calculation above provides a "snapshot" of the UO having a particular length at a calculated distance based on the size of the image. What about it's speed? The field-by-field analysis shows that it traveled its own length between fields, i.e, its own length in 1/60 sec. If it were 2.3 ft long, then, it traveled at 2.3/(1/60) = 138 ft/sec = 94 mph. WOW! Some Bug.... if that's what it was. (NOTA BENE: if it were closer to the camera the speed would be lower.) (Dragonflies can approach 60 mph for short distances.) Incidently, at the same time, the jumper was falling at a speed equal to gt = 32.2 x 2.2 = 71 ft/sec. His speed would increase even further before he opened his parachute.)

Other Appearances of the UO

Jose Escamillo has written: "The most famous and perhaps one of the very best examples of rods on film was taken by Mark Lichtle who is a professional cameraman for US television. (also viewable on my site, He was making a film of parachutists jumping into a deep, vertical cave near San Luis Potosi, Mexico. As in most cases, the rods were only noticed when the film was reviewed later. On observing the slow-motion video, Lichtle saw numerous rods darting in and out of the frame and flying around the base jumpers as they fell through the air. In one shot, a rod avoids colliding with one of the jumpers by veering sharply away at the last second." Another appearance of this type of UO is shown below.

Two obvious differences between the latter two images of a UO and the former images are the size and brightness. Also, these images seem to be smeared a bit more. Neverless, the odd shape - a straight "rod" with some strange lateral appendage "wings" - is consistent. One presumes that the image is much larger in these frames because the UO was much closer to the camera. The UO moved from left to right and the image is larger in the second image (lower picture above) indicating that it moved even closer to the camera. Using the same method of calibrating the angular size as discussed above (based on the size of the falling jumper and the distance he has fallen), the angular size of this UO in the second of the two pictures above is about 0.063 radians. Unfortunately there is no way of determining the distance to the UO so an actual (projected) length cannot be calculated. But if it had been 40 ft from the camera its length would have been about 2.5 ft. (NOTA BENE: again, it seems to be in good focus indicating that it was several tens of feet from the camera.) The leading edge of this UO moved about 1.8 times its own length between fields. If it were 40 ft away and therefore 2.5 ft in length, then it moved about (1.8 x 2.5 ft)/(1/60 sec) = 270 ft/sec or 184 mph.


How much can you swallow? Sizeable objects moving at fantastic speeds, living only (perhaps) in the special environment of the Cave of Swallows? If there really is something it should be captureable. Where are the nets? More video should be obtained using two cameras running simultaneously and pointed in the same direction while separated by several feet to get stereo images of the Unusual Objects. Some sort of synchronization should be provided, such as a periodically flashing strobe light that appears in the fields of view of both cameras and an acoustic track that includes a description of what is happening. This could be done with or without a base jumper (although, maybe base jumping stirs up these "creatures" and makes them fly around.... who knows?). Question: is the possible discovery of a new species, whatever it's size, worthy of some real input (I mean $$$)? See for further descriptions and a possible explanation for the shapes of the images. I thank Jose Escamillo for providing the videos and technical information that were needed to perform the above analysis.