## THE DMSP SATELLITE PHOTOGRAPH

## PREFACE

It was March 3, 1986 when I got my first phone call from Alaska. The man on the phone wanted to tell me about a photo from a satellite which showed a UFO. He had gotten my name from Budd Hopkins and he wanted me to analyze the photo. He told me that Harley Rutledge had a studied it. (Harley Rutledge was chairman of South Eastern Missouri State University. More than 10 years earlier he had become known for his investigations of UFO sightings near Cape Girardeau, Mo. He had written a book in 1973, PROJECT IDENTIFICATION, which reported on the results of his investigations.) The man calling, Mr. James Bounds, told me a friend of his had swiped the picture from the weather office at Elmendorff Air Force Base in Anchorage, AK in October,1978. According to the anonymous friend, when this picture "came down" from the satellite NORAD went on full alert. The friend knew about this because of his job at the air force base. The photo sat in a drawer for years until this friend happened to tell Bounds about it. Then they decided to find out if it could be a UFO. They had spent three weeks with Rutledge and others investigating it. It seemed to be real. The National Enquirer was willing to pay $50,000 for the rights to publish the first satellite photo of a UFO if it could be proved to be a UFO. Could it have been a UFO? I asked him to describe the image. He said it was a rectangular vehicle with no wings and 4 contrails, flying at 45,000 ft and traveling at Mach 6 or more. The location of the photo had been determined by the land mass shown below. It was at Latitude 43 north, Longitude 131:30 East...about 40 km southwest of Vladivostok, just off the coast of the Soviet Union. Then came the punch line: would I fly immediately to Anchorage to help them analyze (and sell) this photo? (Based on phone notes of conversation with James Bounds, March 3, 1986 at 3:30 PM.) I was skeptical about this photo so I did nothing about it and I didn't hear from Bounds again until the summer of 1987!! But a little over a year (!) later, in late March 1987, I received from Philip Klass copies of some material that Bounds and his business partner, John T. Smith, had sent to Aviation Week Magazine. This material consisted of several newspaper stories that reported on a press conference that had been held in Cape Girardeau, MO, on Saturday, November 3, 1985 (4 months before he called me). At this press conference Mr. Bounds and Mr. Smith had announced the results of a three week investigation by Harley Rutledge of the satellite photo which had been obtained by none other than John T. Smith, who said he was on duty when the photo was taken. According to Smith, he kept the photo because he thought the strange image was "cute." It had lain in a desk drawer from the date of the photo, October 11, 1978, until 1985. According to Mr. Smith, the photo had been verified as coming from a defense weather satellite and he had determined that the strange object was at about 45,000 ft and traveling at 4,000 to 5,000 mph. Four months after receiving the newspaper stories about the November, 1985 press conference, I was called by Mr. Terry M. Slaughter of TMS Enterprises, the new business partner of James Bounds. He wondered if I would be willing to authenticate the photo. I said I would be willing to study it with no guarantee as to what I might find. A few days later I received the July 18, 1987 letter from Mr. Slaughter. With the letter he included a nearly 1:1 scale color copy (cropped slightly at the edges) and a blowup. Subsequently in August, 1987, Mr. Bounds supplied me with further technical information he obtained by visiting the weather office at Elmendorf Air Force Base. The most important information was the actual altitude of the orbit (460.9 nautical miles). He also gave me the name and number of the air force publication that describes the DMSP satellite. This enabled me to get a copy of that "Users Guide" for my own research purposes. Mr. Slaughter wrote in his July,18, letter "Following the inspection of the materials described above I would appreciate a written analysis of your findings. Please enclose the photos with your reply." It was late in 1987 when I finally finished my report on the satellite photo. I do not recall whether or not I sent a copy to Mr. Slaughter. (I did give the report to the Fund for UFO Research to sell.) I never heard from him or Mr. Bounds again. Incidently, I still have the prints that I was sent. And I never did go to Anchorage, Alaska!

## UNIDENTIFIED OBJECT RECORDED BY A DMSP SATELLITE

## by Bruce Maccabee (c) 1987, 2000 by B. Maccabee (NOTE: this article was written in 1987 and was only available from the Fund for UFO Research. Major revisions have been made for this web presentation.)

## INTRODUCTION

Two pictures made from a DMSP satellite negative (1) have been analyzed to determine the nature of an Anomalous Image (AI) which appears in the negative (see Figure 1 and the inset, Figure 2). Figure 2 is a blowup of the portion of the negative that surrounds and includes the AI.

Both pictures show the scan lines which formed the negative. Features in the photographs are distinguished by various "blue levels" (as opposed to grey levels). The cold tops of the clouds are white and the land features are dark blue, with other features being paler shades of blue. The full frame picture shows land features which have been tentatively identified as the shore line roughly 40 km southwest of Vladivostock, FSR (2). It seems likely that computer enhancement of the image in the original negative can provide further details of the land mass and thereby increase the accuracy with which the land mass is identified. According to information provided with the pictures, they show infra- red views of the earth. Information in the DMSP User's Guide (3) indicates that the original negative is probably a "Mode Infra-red" (MI) picture of the earth in the 8 - 13 micron infra-red band. Furthermore, since the very cloud tops are cold and the land mass is much warmer it is apparent that the MI output was inverted so that cold is white and warm is dark blue. (This is a typical mode of operation.) Because of the shape of its rather sharp outline and the dark rectangular area within the image, the AI looks somewhat like an automobile without wheels viewed obliquely from above. "Behind" (i.e., to the right of) this "space car" are curved white line images which look like contrails. Although the shape of the image resembles the shape of a recognizable object (a car), the image shape is not necessarily the same as the shape of a hypothetical Unidentified Object (UO) that could have caused the image. This is because of the way the DMSP satellite creates a picture. It does not photograph a whole area at once the way a camera does. Instead, it creates a picture line by line like a TV set, but much more slowly. Therefore if there is any relative motion between the satellite and the object it is photographing, the satellite picture of the object may be seriously distorted. This paper presents some examples of how distortions can occur as a result of the fact that the photograph was created by a line scanning mechanism. Knowledge of the way a scanning system can distort an image is needed in order for an analyst to make an "educated guess" as to what the true shape might have been. (Some of the public claims made in late 1986 that this photo shows a huge object at the level of the cloud tops result from the failure to take into account the distortions inherent in this sort of scanning system.) Although it is not possible to determine exactly what caused the AI, it is possible to determine what did not cause it. It is shown here that the AI was not created by random electronic or mechanical noise and therefore must be the image of something really out there, a UO. It is also shown that the UO was not a cloud nor an aircraft flying near the surface of the earth. It is also highly unlikely that it was some other satellite traveling in an orbit beneath the DMSP satellite. The general conclusion is that the exact shape of the UO and its exact nature cannot be determined.## TECHNICAL DETAILS OF THE DMSP SATELLITE SYSTEM

The satellite is in a nearly polar, sun-synchoronous orbit with an inclination to the equator at its ascending mode of 98.70 so it travels basically north during half of an orbit and south during the other half. Its altitude,A , is about (1,4) A = 461 nm (nm = nautical mile = 6077 ft = 1.854 km) and its orbital period is about l01.6 min = 6096 sec. The polar radius of the earth, Re is about 3432 nm (6356 km) so it travels about 2(pi)(3432+461) = 24460 nm in each orbit (pi = 3.1416). The orbital velocity, Vs, is therefore Vs = (24460 nm/101.6 min) = 240.7 nm/min or 4.012 nm/sec = 24,381 ft/sec (7.430 km/sec; Mach 21 if Mach 1 is taken as the speed of sound at sea level). As it travels (nearly due north or south) it scans perpendicular to the orbital track (i.e., nearly east-west) with an optical system that collects infra-red radiation and focusses the radiation on a detector. The output voltage of the detector at any instant is proportional to the average temperature of the area that is within the Instantaneous Field Of View (IFOV; the angular size of the area being viewed at any given time) of the optical system at that instant (see Figure 3).

The IFOV during MI operation is (3) 0.00533 radians (abbreviated "rad") or 0.305 degrees. Thus, at the surface of the earth directly below the satellite (the nadir), the instantaneous area is a circle roughly 0.00533 rad x 461 nm = 2.45 nm in diameter. This is the minimum width of the strip which is created by the scanning system. When the sensor scans east or west away from directly below the satellite the instantaneous area on the earth grows and becomes elliptical with a maximum width (measured in the north- south direction) of about 12 nm and a maximum (east-west) length that grows to about 30 nm as illustrated in Figures 3 and 4. The angular range of the scan, 55.6 degrees either side of center, corresponds to about 1600 nm (2960 km) along the surface of the earth (see Figure 4). However, the scan width was apparently less than this, as described below.

The time it takes to scan from east to west (or v.v.) is determined by a rotating mirror scanning system. The mirror rotates on a motor-driven shaft that is parallel to the orbit direction (tangent to the orbit). The rotation rate is 1.73 Hz, or 0.56 sec/revolution (3). This corresponds to 11.2 rad/sec (641 degrees/sec). Because the light rays bounce off the scanning mirror in such a way that the mirror is continually at 45 degrees to the centerline of the shaft and at 45 degrees to the incident and reflected rays, the scan rate of the IFOV is equal to that of the mirror rotation rate, i.e., the scan rate, SR, is 11.2 rad/sec. The total scan angle (as displayed on the photograph) is twice 55.6 degrees (0.97 rad) or 112 degrees (1.94 rad) so it takes 1.94 rad/(SR) = 0.173 sec for the sensor to scan from east to west (or west to east). (Actually the time was shorter than this if the scan width as smaller, as calculated below.) The rotating mirror also determines the time lag, TL, between successive scans (the time from the beginning of one scan to the beginning of the next scan). Since it takes O.56 sec for the mirror to rotate completely around, TL = 0.56 sec. Because the satellite is traveling along its orbit at a high speed each scan begins at a location further along the orbit than the previous scan, as illustrated in Figure 5. The distance that the satellite moves between equal scan angle points on successive scans is VsTL = 2.24 nm, where Vs = 4.012 nm/sec and TL = 0.56 sec. Because the optical system points toward the center of the earth at the center point of each scan (the nadir), the separation distance between scans at the earth's surface, S, is smaller than the spacing of the satellite positions by the ratio Re/(Re+A) = 0.88 (Re and A were given previously). Thus the spacing of the scan strips on the earth is S = VsTL(Re/[Re+A])= 2.24 x 0.88 = 1.97 or about 2.0 mm (3.6 km) and the speed of the scan along the earth is 2 nm/0.56 sec = 3.57 nm/sec. Since the instantaneous area of view on the earth is always larger than 2 nm, the scan strips overlap on the surface of the earth (see Figure 5).

As the optical system scans the earth the detector output voltage changes with temperature changes along the scan strip. At any instant the output voltage corresponds to the average temperature of all the objects viewed within the IFOV, i.e, all the objects within an area 2.45 mm wide or larger. The temperature changes that occur from one IFOV location to the next along a scan are processed electronically and ultimately are recorded as color and brightness changes along a scan line on the photographic negative that is created, line by line, by the DMSP electronic and photographic processing equipment. The temperature variations along a series of successive parallel scan strips on the earth become brightness variations along a series of parallel scan lines on the photograph. Thus the combination of the east-west scan in 0.17 sec (or less, see below) with the forward satellite motion, which causes 2.0 nm (at the earth's surface) separation of scan lines, makes a two dimensional picture of the temperature variations as seen by the infra-red sensor. Since the average temperatures of major earth features (land, ocean, clouds) are different the outlines of the features are clearly visible. This picture will be quite accurate if nothing changes temperature and nothing moves during the time of passage of the satellite over an area. (Because the earth rotates under the satellite there is a very small east-west smear along the direction of the scan. The distance the earth landforms would move due to rotation is about 0.25 cos(lat) in nm/sec, where lat is the latitude angle. So in 0.17 sec the landforms would move toward the east by about 0.04 cos(lat) or less than 300 ft, an entirely negligible amount.) The temperature sensor has a range from (3) 210 K (210 degrees Kelvin) to 310 K (-63 C to 37 C or -145 F to 99 F). This covers most of the range of temperatures found on the earth's surface and up to the altitude of the highest clouds. (Note that generally temperature decreases with increasing altitude.) Temperature differences as small as 1 C are sensed. However, the photographic display can be adjusted to show larger temperature differences, for example, 10 degree changes. The display can also be adjusted to show only a portion of the 100 degree maximum range, for example, 250 K to 300 K (a 50 degree range). The photographs of interest here show six or seven (it is difficult to determine whether it is six or seven) gradations in color between the land (dark blue) and the clouds tops (white). Assuming that the display was showing the whole temperature range (100 C) using 6 levels, then each temperature gradation was about 17 C.## DISCUSSION OF THE ANOMALOUS IMAGE (AI)

The clouds in the photograph have irregular shapes ("fractal" edges). The AI, on the other hand, has two nearly straight edges (see Figure 2). One edge lies along the scan direction and one edge is nearly at 45 degrees to the scan direction. Clouds do not have adjacent straight edges so this is a reason to reject the idea that the AI is a cloud. The outline of the AI is also more angular and more distinct than edges of cloud images and the shape is not striated or "blobby" the way other cloud images are. Hence the following analysis has been carried out under the assumption that the AI was not made by some exceedingly strangely shaped cloud. The curved "contrail" images at the upper right of the AI create the illusion of a speeding object that left vapor trails behind it. This might be true (but with "severe" consequences; see below), or it might be that these were simply elongated curved clouds that happened to appear next to the AI. One can find similar elongated or "striated" clouds with dark spaces (views of the earth's surface) between them in other portions of the picture (e.g., just below of center of the photo). The close proximity of the ends of these curved elongated clouds to one edge (the "back edge") of the AI combined with one's knowledge of what contrails behind a high-flying jet look like, make one think immediately of a multi-engine aircraft. However, if one covers up the AI itself leaving the "contrail" images one can see that there could be an association between the elongated clouds and the other cloud structure which is to the right of the AI. In other words, it seems reasonable to assume that the curved elongated images are of clouds that have no real association with the object that made the AI. Since there is a question as to the relevancy of the "contrail" clouds, for the purposes of the following analysis they will be ignored and the analytic effort will concentrate on the AI. The white image of the AI lies "on top of" the images of the warmer (i.e., bluer and hence lower altitude) clouds. The whiteness of the AI image seems equal to the whiteness of the larger areas of cloud and hence appears to be least as cold as the upper cloud tops. Thus it appears that the AI is above the clouds, or at least at the level of the cloud tops, i.e. at an altitude of several tens of km or more and so it cannot be some unusual feature on the surface of the earth. The extreme whiteness of the AI implies that whatever made it was at the "maximum cold" that could be registered, or colder, e.g., equal to or less than -60 C. The trapezoidal (nearly rectangular) "hole" in the AI image (see Figure 2), on the other hand, has the same dark blue color as the lower clouds and the earth's surface and so would appear to be at least as warm as the lower clouds and perhaps about as warm as the earth's surface beneath. At each end of the trapezoidal "hole" there is a region where the dark blue color appears to match the blue of the land mass. Between these darkest blue regions in the "hole" there is a less blue region. One gets the impression of perhaps looking through a hole in the object toward the earth and clouds below. A comparison with land features in the photograph suggests that some land may actually be visible within the trapezoidal area. (To prove or disprove this would require a very careful comparison with a map or a DMSP satellite photo of the same area.) At any rate, the temperature within the "hole" is comparable to the temperature of land and ocean features that are visible elsewhere in the photo.## THE LATERAL ANGULAR SIZE OF THE ANOMALOUS IMAGE

To find the angular size one must first "remove" from the AI the angle- dependent stretching in the scan direction that was introduced by the DMSP processing system as part of a standard "map rectification process" that compensates for the forshortening that occurs due to the oblique view of the earth at large scan angles. The map rectification results in a constant number of nm/inch along the photograph. The usual mode of operation (3) corresponds to 205 nm/inch (1:15,000,000 scale factor). Since the scan lines are spaced 2 nm on the surface of the earth this would correspond to (205 nm/inch)/(2 nm/line) = 102 lines/inch on the photograph. However, Figure 1 has 42.25 lines/inch. This corresponds to 42.25 lines/inch x 2 nm/line = 84.5 nm/inch. This is close to, but not as large as the other listed optional expansion (3), 102 nm/inch (1:7,500,000) which should produce a picture with 50 lines/inch. Why the number of lines/inch does not agree with either standard (102 or 51) is not known. However, there may have been a different scale factor used in the signal processor which created the photo from the electronic signals sent by the satellite. At any rate, I assume that the horizontal (east-west) and vertical (north- south) scale factors are the same (which they should be) and are, therefore, 84.5 nm/inch. (NOTE: this assumption applies only to the scan as projected to the surface of the earth. The vertical distance scale remains approximately constant with altitude up to the satellite but the horizontal scale shrinks with increasing altitude, as will be described. That is, a vertical dimension of one inch always represents 84.5 nm but a horizontal dimension represents 84.5 nm only along the earths's surface. At altitudes above the surface the horizontal scale factor is less.) With this assumption the picture does not show 1600 nm measured east to west across the surface of the earth, as illustrated in Figure 4, but rather about 680 nm. This variation in the data presentation does not affect the scan rate, 11.2 rad/sec (which is determined by the satellite and cannot be changed by ground operation control) but it does mean that the duration of a single scan was about 0.11 sec rather than the 0.17 sec calculated above. A "flat earth" approximation to the map rectification equations is presented in Figure 4. Let t be the angle away from vertical (nadir). (NOTE: angle t is represented as the Greek letter theta on Figure 4 and A, is represented as h) Then, x = distance on the earth's surface in nm = My, where M = 84.5 nm/inch and y is the distance in inches measured along a scan line on the photograph starting at the center of the scan line (the center of the photo represents the nadir, "straight down," or t = 0). By trigonometry, x = A tan (t). Now, imagine that a relatively small object on the earth's surface has a width (east-west dimension) given by Dx. The scanning process will create an image of angular width, Dt, that is related to Dx by the derivative of the equation, Dx/Dt = A/cos^2(t). Solving this for the angular size, Dt, corresponding to a length Dx, as viewed by the satellite, gives Dt = (Dx/A)cos^2(t) which shows that the angular size shrinks with increasing angle (cos(t) shrinks as the angle increases) even when Dx stays constant. The electronic processor "reverses" this equation to make the horizontal length scale constant. In other words, the processor multiplies the angular width of an image, Dt, by the factor [A/cos^2(t)] to give Dy, measured on the film, as Dy = Dx/M = A/[Mcos^2(t)] Dt. That is, a "stretch" was applied to each scan line, with the amount of stretch increasing with the angular distance from the nadir. To determine the actual angular width from a width measurement, Dy, on the photograph it is necessary to multiply Dy by the inverse of the "stretch factor," as indicated in the following equations. The angular width of an image along a scan line corresponding to an object of width Dx at the earth's surface is given by Dt = (Dx/A) cos^2(t). (1) To find the actual angular width of a small image (much smaller than the width of a scan line), first find the off-nadir angle, t, at the center of the image from t = arctan(My/A) (2) using M = 84.5 nm/inch and A = 461 nm: t = arctan(0.183y). The distance, y, is measured on the photo along any particular line that goes through the image, starting at the center of the scan line and measuring to a point along that scan line which is within the image. For example, measuring along a scan line that passes approximately through the widest part of the AI (which is close to the center of the image), from the center of the photo to a point at the center of the AI the distance is about y = 1.6 inches. Therefore at that point t = arctan(.183 x 1.6) = 16.3 degrees . (This is based on a print that is exactly 8 inches wide by 10 inches long. This print is apparently a direct 1:1 copy of the original negative created by the DMSP system. Figure 1 shows this print copied at 75 dpi and cropped about 0.6 inches at the left side, which shows only more cloud and a small amount of land, to save "bandwidth.") Let Dy be the width of any small image as measured along a scan line and let Dx = MDy. From the equation above (and see Figure 4, wherein h is used to denote altitude, represented here by A), the actual angular width of a small image, after removal of the "stretch" that was introduced by the DMSP electronics, is given by Dt = (Dx/A)cos^2(t) = (MDy/A)cos^2(t) = 0.183Dy cos^2(t). (3) Along the scan line that passes through the widest portion of the AI, Dy = 0.6 inches so Dt = 0.101 radians (0.101 rad) or about 5.8 degrees. This, then is the actual angular size of the AI as seen from the perspective of the DMSP. (Note: the stretch correction factor is only about 8% for the AI image, so one could proceed without including it.)## A REAL OBJECT "OUT THERE"

Finally, after all this preliminary calculation, the time to scan across the widest portion of the AI can be calculated: 0.101 rad/(11.2 rad/sec) = 0.0090 sec. For comparison, the time to scan across the third scan line up from the bottom of the AI, where y = 1.4 inches, Dy = 0.46 inches, t = 14.4 deg and Dt = 0.079 rad, was about 0.079 rad/(11.2 rad/sec) = 0.0070 sec (see the calculations at the bottom of Figure 4). Thus each scan across the AI required about 0.007/0.56 = l/80th of the time from one scan to the next. The total time required to create the AI is the time it took to create the 22 scan lines which make up the image (see Figure 6): 21 x 0.56 + 0.007 = 11.77 sec. (This is the sum of 21 "spaces" between the starting points for the 22 scan line plus the time it took for the 22nd scan line to cross the AI. This isn't quite correct because of the sloping boundary of the AI.)

During this nearly 12 sec period a substantially straight edge, at nearly 45 degrees to the scan direction, was created. This high degree of correlation means that the AI is not an artifact of random detector noise, mechanical vibration or electronic signal processor. Instead the AI was produced by a real thing "out there' an Unidentified Object (UO). Since the AI was not created by random system noise, nor was it created by a cloud it must have been made by a "real thing out there." The only conventional possibilities are that the UO was some man-made object at or above cloud level. Only aircraft and satellites travel above cloud level, so the remaining analysis is directed toward determining whether or not it is logical to conclude the UO was an aircraft or a satellite.

## COULD IT HAVE BEEN AN AIRCRAFT?

One "quick" conclusion is that the object could not have been a high- flying aircraft. To see that this is true in a simple way, consider that the IFOV is 0.00533 rad. Therefore the area covered by the IFOV at the altitude of the cloud tops, about 8 nm, would be (0.00533)(461 nm - 8 nm) = 2.4 nm = 14,670 ft. Any aircraft that could fly at 8 nm would be under 100 ft in size and therefore would only be a "dot" within a single IFOV. It would not register at all (unless it were a "ball of fire", in which case it might make a tiny blueish dot, one IFOV in size, on a picture such as Figure 1). Because of the width of the AI, another requirement for the hypothetical aircraft would be to make at least one scan line width. Consider the widest scan line in the AI, 0.101 rad. It required 0.009 sec to create. During that time a fast aircraft at 3,600 nm/hr = 1 nm/sec would move 0.009 nm = 54 ft, or about its own length. For the purposes of this calculation it could, therefore, be considered virtually stationary during the time it was scanned. How does this compare to the size at cloud-top level that would create a scan line .101 rad long? Projected to the cloud tops, at an angle of 16.3 degrees from vertical, this angular length corresponds to a distance (length) of X = 0.101 (461-8)/cos^2(16.3) = 49 nm. Clearly this single scan line by itself could not be caused by any stationary (zero velocity) man-made object at the altitude of the cloud tops!! But then one might suggest that the scan line was made by an aircraft which remained within the IFOV for a period of time by virtue of its speed (and, by extension, one might suggest that the 22 scan lines were created by 22 similar high speed, high altitude aircraft, each one of which happened to remain within an IFOV). If that were true, then the aircraft must have traveled 49 nm in the direction of the scan during the time of the scan, 0.009 sec (for the widest scan line) with the consequent velocity of 5520 nm/sec or about 20,000,000 nm/hr, not exactly a "moderate" speed. Another requirement for size of a slowly moving or stationary object at the cloud tops is based on the velocity of the DMSP satellite and the time duration represented by the 22 scan lies. That is, if stationary, it would have to be at least 11.77 sec x 3.6 nm/sec = 42 nm long, where 3.6 nm/sec is approximately the speed of the scan at 8 nm above the earth. If, on the other hand, one were to assume an object traveled along with the satellite and so appeared in a succession of scans the, besides having to be about 49 nm wide and 2 nm long (to fill the IFOV), it would have to travel at 3.6 nm/sec = 12,960 mph. The preceding analysis shows clearly that the UO could not have been any man-made object at the cloud tops for reasons related to size and speed. The only remaining man-made type of object is a satellite, and one that was at essentially the same altitude as the DMSP satellite, as will be shown.## COULD IT HAVE BEEN A SATELLITE ?

The Appendix provides a considerable amount of analysis that shows that if the object were an Unidentified Satellite (US), then it had to be moving nearly parallel to the DMSP. Any object in orbit would be traveling at a high speed and so satellites traveling in the opposite direction or perpendicular to the direction of motion of the DMSP would have to be much larger than any man-made satellite to create the AI. The only logical possibilty, then, is that a US passed, rather slowly, beneath the DMSP (actually about 16 degrees to the east of directly below). In order to determine the allowed range of satellite sizes and orbits it is necessary to consider first the following relationship between the orbital altitude and the velocity: Vp - Vs = -5.14E-4 Ro, where Vp = Vo cos(a) and Vo is the US velocity and 5.14E-4 = 5.14 x 10^-4 = 0.000514. For the purposes of this calculation it is sufficiently accurate to imagine that the US velocity vector lies in an imaginary horizontal plane that is at the altitude of the US and is perpendicular to the radius from the center of the earth to the US. (This is because any satellite in a (nearly) circular orbit has an instantaneous velocity vector that is essentially perpendicular to the radius from the center of the earth out to the satellite, i.e., tangent to the circular orbit, and any vector perpendicular to a radius vector lies in a plane defined as "locally horizontal.") The horizontal plane that contains the US velocity vector during the nearly 12 seconds of this "sighting" is slightly below the imaginary horizontal plane that contains the DMSP orbit velocity during the same time period. In the above equation, Vp is the component of the US velocity parallel to the DMSP orbit velocity, a is the acute angle between the US velocity vector and the DMSP velocity vector and Ro is the vertical distance between orbital altitudes, essentially the height difference between the imaginary horizontal planes described above. The above equation can be derived from the basic Newtonian force calculation for a circular orbit: F = ma where ma = mv^2/r and F = m(GM/r^2). The solution of these equations leads to v = k/r^0.5, where r is the radial distance from the center of the earth to the orbit altitude, r = Re+A = 3432 nm + 461 nm = 3893 nm. The derivative of this equation is dv/dr = - 0.5k/r^1.5. Because the variation in radial distance is small compared to the actual radius 0.5k/r^1.5 is a constant equal to 5.1E-4/sec. Hence dv = - 5.1E-4 dr. In this equation, dr = Ro where Ro is a negative number (positive is upward). For example, if the US were 1 nm below the DMSP, then Ro = -1 nm and Vp-Vs = (- 5.1E-4/sec) (-1 nm) = 5.1E-4 nm/sec = 3.12 ft/sec. In other words, Vp = Vs + 3.12 ft/sec so the US travels more rapidly than the DMSP. This is a small change to the orbital velocity, 24,381 ft/sec. (Note: from now on the (-) sign will be ignored as it is understood that distances Ro are measured downward from the DMSP.) Imagine a US traveling at some altitude Ro below the level of the DMSP satellite. If Ro = 1 nm, then Vp - Vs = 3.12 ft/sec. The US would approach the DMSP from behind below, catch up with the DMSP and then move ahead. As it approached it would eventually reach a point where the scan would pass over the leading edge of the US. As time went on the scan would move backward over the US as the US moved along, until eventually the lagging edge of the US would move ahead of the scan position. The image thus created would be a series of scan line segments with the width of each segment being a measure of the width of the US. But note that the US image would be reversed: the front (most forward in the DMSP orbit direction) edge of the image would correspond to the rear edge of the object (front-back mirror reversal). By the time of the last scan line the US would have moved, relative to the DMSP, a distance equal to its own length, Lo. Hence Lo = (time of scanning 22 lines) x ( the speed differential) or Lo = 11.77(Vp-Vs). The above relation between speed differential and altitude establishes a close relationship between the distance of the US below the DMSP and the length of the US: Lo = 11.77(-5.14E-4)Ro = 0.00605Ro. For R = -1 nm = -6,077 ft this yields a length (dimension along the orbit) of 36.8 ft (which is probably within the outer bound of man-made size). (Note that the length dimension is not affected by having the US not directly below the DMSP. However, the width is affected by the off-nadir angle, as shown below.) Consider the same US moving nearly parallel to the DMSP orbit but at a lower altitude. The angular width (as measured perpendicular to the orbit) is the angle corresponding to the width along a scan line. For the third line up from the bottom of the AI, for example, the angular size is 0.079 rad (see above). The lateral dimension of the US (a dimension lying within the imaginary horizontal plane as discussed above) is given by Wo = 0.079Ro/cos^2(t) (where the cosine factor corrects for the perspective effect of being off of vertical by t degrees). With t = 14.4 deg for the third line up and Ro = 1 nm = 6077 ft, Wo = 512 ft. (Decidedly NOT man-made size.) More generally, the equations for Lo and Wo can be combined: Ro = Lo/0.00605, so Wo = 0.079(Lo/0.00605)/cos^2(t) = [13/cos^2(t)]Lo. For t = 14.4 deg this becomes Wo = 13.8Lo or the width is about 14 times the length. Thus, although it is impossible to determine from the available information the exact value of either dimension, it is possible to determine their ratio. (This is a surprising result that is a consequence of the assumption that the US is in an orbit and hence constrained by orbital mechanics.) One can explore a range of values for Ro and calculate the resulting values of L and W: Ro = 0.1 nm = 607 ft, Lo = 3.68 ft, W = 50.8; Ro = 0.05 nm = 304 ft, Lo = 1.84 ft, W = 25.9 ft, etc. What one learns from this set of calculations is that to have both dimensions within the roughly 50 ft allowed maximum for man-made satellite structures in 1978, Ro can be no more than a few hundred feet! The above analysis leads to the conclusion that if the object was a US, then it was in an orbit very close to that of the DMSP. More specifically, the above analysis is based on the assumption (not stated until now) that the US was traveling exactly parallel to the DMSP for the 11.77 seconds of this encounter or "sighting." This, of course, could be possible but, if so, the orbital planes would have been very close to one another. Based on the above reasoning one might conclude that the image is that of some satellite with the shape of its outline equal to the outline indicated on the photo. Such a conclusion might be right or it might be wrong. If the US were a rotation-stabilized satellite the image shape might be a reasonable approximation of the actual shape. On the other hand, if it were a piece of space junk that was tumbling rapidly (more than 1 rotation per second, for example) as it traveled along, then there would be hardly any resemblance between the image shape and the object shape. Alternatively, if the object were a rotation stabilized satellite and traveling toward (or away from) the DMSP orbit the shape of the image could be distorted by the lateral motion. Note that the right edge is mostly a straight line at an angle to the direction of DMSP motion. In the Appendix it is shown that a rectangle could get distorted by motion and thereby create an image with a slanted edge as in the AI. More specifically, consider that the variation in end position of the successive scan lines at the right side is 0.022 inches per line for 13 of the 22 scan lines, a total of 0.286 inches of shift making a line at about 45 degrees. A reasonable assumption would be that this is a result of the US moving toward (or away from) the DMSP by some amount from one scan to the next. Using the method presented before for finding the angular size of an image an the photo (Equation 3), the shift in the image edge position corresponds to an angle shift Dt = (0.183) x Dy x cos^2(18) = (0.183 x 0.022 x 0.905) = 0.0036 rad, where the angle used, 18 deg, is at the (vertical) center of the right edge of the image. This amount of shift in angular position took place in 0.56 sec, so the angular transverse velocity that would produce the (nearly) 45 degree sloping right edge of the image is 0.0064 rad/sec. For an altitude difference of 300 ft, for example, this corresponds to a transverse velocity of 0.0064 x 300/cos^2(18) = 2.1 ft/sec. In other words, a US with a straight edge at its right (east) side and a transverse velocity component Vt = 2.1 ft/sec could make a slanted edge such as observed on the photo. This would make the UO orbit non-parallel to the DMSP, but not by much. One must not be confused by the "scale" of the photograph. What looks like a 45 degree angle is not that when projected onto real space because the distance moved by the UO satellite is essentially X = (0.56 sec)(Vp ft/sec) along the orbit and Y = (0.56 sec) (Vt ft/sec) transverse to the orbit. The acute angle, a, of the US velocity vector (and also the US orbit plane) relative to the DMSP velocity (and orbit plane) is therefore a = arctan(Vt/Vp), where Vt = 0.0064Ro/cos^2(t) and Vp = Vs + 5.14E-4Ro. In taking the ratio we can ignore the small contribution to Vs and simply write Vt/Vp = 0.0064Ro/[Vs Cos^2(t)] = 2.62E-7Ro/cos^2(t). For Ro = 300 ft the ratio is 8.7E-5 and the angle is 8.7E-5 radians or 0.005 degrees. Here is another way of looking at it: during the nearly 12 sec "sighting" both satellites would have travelled appoximately 12 sec x 4 nm/sec = 48 nm along their orbits, while the US would have moved only a comparatively miniscule 12 sec x 2.1 ft/sec = 25 ft toward (or away from) a point directly below the DMSP (i.e., a point lying in the DMSP orbit plane). For small angles such as these the arctan of an angle equals the value of the angle in radians. Hence one can write a = [2.62E-7Ro/cos^2(t)] which shows that the angle between the orbits increases with the vertical distance between the orbits. With 600 ft being the likely maximum value for Ro (see above; for satellites with maximum dimensions less than 50 ft) the maximum angle between the orbits would be about 1.74E-4 radians or about 0.01 deg. Since this would also be the acute angle between orbit planes, the maximum separation between these satellites, which would occur 1/4 of an orbit before or after this "sighting" would be only (Re+A)(1.74E-4) = 0.68 nm (where Re+A = 3893 nm). For smaller values of Ro the maximum separation decreases (.34 nm at Ro = 300 ft, etc.) This raises the question of whether or not the DMSP would detect the US at another location in its orbit. The period of a circular orbit is P = 2pi(Re+A)/v. In this case v = 4.012 nm/sec and 2 pi (Re+A) = 24,460 nm so P = 24,460/4.012 = 6096 sec = 101.6 min. With a speed differential of dV = 5.1E-4Ro, at 600 ft the US would travel .3084 ft/sec faster than the DMSP. One half an orbit later, when the DMSP and US would again be close together (where the orbits would cross on the other side of the earth), the US would be 3048 x .3084 = 940 ft ahead of the DMSP. Since the US travels faster the DMSP would never catch up with it, so it would not be detected. As the two satellites continued to orbit the earth the US would "lap" the DMSP after (24,460 nm x 6077 ft/nm)/(940 ft per 1/2 orbit) = 158,131 half orbits or about 79065 1/2 full orbits. The time for this many orbits would be about 134,000 hours or about 5,580 days (about 15 years). That would mean that within the lifetime of the DMSP (10 years or so) one might expect only one encounter with the UO satellite. This last conclusion explains why there was no similar image reported in any other photo from this DMSP satellite. (On the other hand, even if there had been such an image it might not have been noticed since many of the photos were simply thrown away. It was apparently a chance occurence that this particular picture was saved because of the "cute" image.) The above analysis was based on the seemingly reasonable assumption that the right hand edge of the AI was a straight line because the US was drifting slowly toward the orbit plane of the DMSP as the US passed nearly beneath the DMSP. The alternative possibility, mentioned above, is that the actual outline or edge of the US was a straight slanted line. If this were so then there was no motion of the US toward or away from the DMSP. (One would also imagine of mix of these explanations: the right edge of the US had a slant, but not as steep, and at the same time the US was moving very slowly toward the DMSP.) If the US were actually traveling eactly parallel to the DMSP orbit but at a slightly lower angle and slightly to the right, then because of orbital mechanics this would be the farthest apart, laterally, that their orbits would be. The orbits would cross 1/4 of an orbit later but, as with the situation above, the US would be ahead of the DMSP and so would never be imaged again (at least not for more than 15 years).## CONCLUSION

It is unfortunate that the DMSP photo was not made available for analysis until many years after it was taken because it should have been possible to determine which satellite, if any, could have been at the exact location indicated by the DMSP. On the other hand, it should still be possible to determine the dimensions of the largest satellites and pieces of space junk that were orbiting in 1978 and which of those satellites might have been in an orbit that would put it close to the DMSP. If there was no such satellite, then this could have truly been classified as an unidentifiable object. (Note: the meteor hypothesis is rejected because a meteor caught in a temporary arc about the earth would be traveling much faster than a satellite. If there were only on scan line image one might consider a meteor, but then such a small image would have been ignored and the question of identifying it would not have come up.)

## FOOTNOTES AND REFERENCES

1) Photographs made from the original negative were provided by Mr. James Bounds and Mr. Terry Slaughter of Anchorage, Alaska. According to Mr. Bounds the original negative was obtained in October, 1978, by a friend who worked at the weather center at Elmendorf Air Force Base where DMSP satellite data are received and processed. According to Mr. Bounds the original negative shows an infrared picture of the earth. Am effort was made, as part of this study, to verify the photographs as being derived from a DMSP satellite negative by locating in archival storage a magnetic tape containing the original electronic data. However, it was determined that no such tape exists. Therefore it is assumed in this analysis that the photographs are, in fact, derived from a DMSP negative. No information has been obtained which contradicts this assumption. 2)Letter entitled, "To Whom It May Concern, " by Dr. B. Ray Knox , Professor of Geology, Southeast Missouri State University dated Nov. 1, 1985 3) DMSP User's Guide, Air Weather Service (MAC), United Air Force Report # AWS-TR-74-250. Normal operation of the DMSP satellite is either "Mode infra- red" (MI) or High Resolution (HR) visibler. Because of information provided by Mr. Bounds it is assumed that the negative is a MI image of the earth. MI operates in the 8 - 13 micron band. Since the very cold cloud tops appear white it appears that the MI output was inverted to make colder regions appear whiter amd warmer regions appear darker. 4) The normal altitude according to Ref. 3 is 450 nm in a circular orbit. However, the altitude could range from 430 nm to 480 nm. The altitude used here, 460.9 or 461 nm was supplied by Mr. Bounds who was given the information on the actual altitude by a contact at Elmendorf Air Force Base near Anchorage, Alaska. The general conclusions stated here do not depend upon whether the satellite was at 450 or 461 nm altitude.

## APPENDIX

(Note: only the mathematically include aficionado should take the time to read this. However, do take a look at Figures 9, 10 and 11 below.) The size and actual shape of the Unidentified Object, "thing out there," the UO, can only be guessed at since the distance to it is unknown. (The DMSP satellite did not provide two simultaneous, independent views that could have allowed for a triangulation.) "Educated guessing" consists of creating model objects and then determining what such models would look like in DMSP photos. Any model which produces an image with the features of the AI is, potentially, an actual model of the object. Unfortunately there are many such models so one cannot be certain that any particular one is the actual shape of the object. Nevetheless this sort of analysis can help to determine with some level of confidence what the UO was not.## DISCUSSION OF A SINGLE LINE OF THE IMAGE

Because the satellite scans line by line at a low rate the picture that created is not like a "normal" photograph in which the whole picture is created "at once" (within the shutter time of the camera, for example). The slow scan would make no difference if the objects being scanned were at or near the earth's surface did not move or change in any way during the time that the picture was created. However, when one allows for the possibility that something, specifically, the UO, was (a) close to the DMSP and (b) did move relative to the DMSP, then it becomes necessary to proceed with caution in trying to understand the image. (Note: the DMSP scanning mechanism and electronics compensates for the motion of the MSP over the earth so that the views of the earth and slowly changing clouds are quite accurate.) Although the total picture of the UO took about 12 seconds to complete (with consequences that will be illustrated below), a given scan across the UO was rapid, taking less than 1/100 of a second. Let us, therefore try to understand the significance of this single line "snapshot" of the UFO. The maximum width of the AI as measured along a single scan line is about 0.6 inches and this corresponds to an angle of about 0.1 radians. The first question to be asked is, what size does this correspond to if the object were at the altitude of the cloud tops, say 8.2 nm (50,000 ft) above the earth? The satellite was about A = 461 nm above the earth so an object 8 nm above the earth would have been about A - H = 453 miles from the satellite. Imagine a scan line projected onto a plane that is parallel to the horizontal surface of the earth (flat earth approximation). The width of an object is measured within this imaginary plane, parallel to a scan line. If the line of sight to the object is not straight down but rather at angle t to the nadir (where t = 0), then the width must be projected onto an imaginary plane that is perpendicular to the the line of sight. This projection is given by x = X cos(t), where X is the actual width measured in the imaginary horizontal plane and x is the projected width in the imaginary plane. At the same time, the radial distance from the satellite to the object is R = (A-H)/cos(t). Therefore the projected length of the object is related to the angular size by x = R tan(a) = [(A-H)/cos(t)]tan(a) and the actual length is X = [(A-H)/cos^2(t)]tan(a). When an angle is given in radians, if the angle is less than 0.4 radians (about 23 degrees), tan(a) = a to better than 5% accuracy. In this case a = 0.1 rad, so at an altitude A - H = 452 miles an at an angle of abou 16 degrees from the nadir, the object dimension along the scan direction (the projected width on the photo) would be about X = (453 nm)(.1)/cos^2(16) 49 miles!! Hence the claim made by the promoters of this photo that the object was "climbing out of the atmosphere" only makes sense if the object was fantastically huge! It is no wonder that they never mentioned a size of the object (even though they did claim a huge speed of 4,000 - 5,000 mph). This size calculation immediately rules out an airplane. However, the angular width used in the calculation above is based on the assumption that there was no component of motion of the object along the scan line. What if there were motion of some small object, like an aircraft, along the scan direction? In that case it would spend too much time in the IFOV and create an image much longer than its actual length. (In the limit that the object moved at the scan speed it would create an "infinitely" long image.) In other words, one has to consider the possibility that the width of the AI image might be due, in part, to motion parallel to the scan direction. The following discussion is directed toward deciding whether or not the length of a scan line in the AI image was increased (or decreased) by motion of the object. All we know from the photo is that the scan line contacted one end of the UO at time To and maintained contact until it reached the other end at time T1. During this time the scan moved through the measured angle, Dt = 0.1 rad in 0.009 sec (for the longest scan line). If the object was moving with a component of its motion parallel to the scan line then the product of the total scan time, TS = T1 - To, multiplied by the scan angular velocity, SR, must equal the sum of the actual angular size plus the angular distance it moved during TS. Let the actual projected size along the scan direction (the projected width of the AI along a scan line) be Lp, let its projected velocity component along the scan line be Vp and its distance from the DMSP satellite be d = (A-H)/cos(t). Then its angular velocity along the scan line is Vp/d and its angular size along the scan line is Lp/d (for small angles less than 0.4 rad this is accurate to better than 5%). Then the above verbal equation can be written as (TS)(SR) = (Lp/d) +(Vp/d)TS, where SR > |Vp/d|. (4) In this equation we know SR (= 11.2 rad/sec), TS (= 0.009 sec for the longest scan line on the image) and d is assumed to be 452 nm. The equation becomes 0.1 rad = Lp/452 + 0.009(Vp/452) or 0.1 = 0.0022Lp + 0.00002 Vp = 2.2E-3Lp + 2E-5Vp, (5) where Lp is in nm and Vp is nm/sec. Since there is only one equation the actual length and velocity of the object cannot be determined without further information. However, one can easily see that if an image is constructed over a period of time rather than instantly, then the size of the image depends upon the relative velocity between the "camera" and the object. The image size could be greater or less than it should be depending upon whether the scan is in the same or opposite direction to the motion of the object. If the object velocity is very small and/or the "shutter time" (scan time) is very small the projected size is accurately given by ignoring the second term on the right side of the above equation. Using this equation one can test the "high flying aircraft" hypothesis by assuming a length, say 50 ft = 0.0082 nm, for the object, and then calculating how fast it would have to move (Vp) to create the 0.6 inch wide scan image, i.e., to satisfy equation (5). The answer is Vp = [0.1 - 2.2E- 3(0.0082)]/2E-5 = [0.1 - 1.8E-5]/2E-5 = 5E3 nm/sec which corresponds to a super duper meteoric speed of 1.8E7 nm/hr...that is 18 million nautical miles per hour! (Light speed is 1.6E5 nm/sec = 5.8E8 nm/hr). Scratch high flying aircraft!! The only remaining conventional explanation is that the DMSP happened to scan a passing satellite. But this is not as "simple" as it might seem. In order to understand why, it is necessary to delve more deeply into the consequences of the method by which DMSP pictures are made. This understanding can be gained by analyzing the effects of the scan system on creating images of "known" simple objects, i.e., model UOs.## ANALYSIS OF SIMPLE MODELS OF THE UNIDENTIFIED OBJECT

In the following discussion the terms "width" and "length" have specific meanings. The length of an object is a distance measured along the orbit of the DMSP satellite. The width is measured transverse (perpendicular) to the orbit and parallel to the east-west scan direction. First, a very simple model UO will be used to illustrate how the image shape and model size are related by the optical and scanning properties of the satellite. Assume that a flat rectangular (cold) object is directly below the satellite at altitude H above the surface of the earth (see Figure 7).

The width of the object is W, measured transverse to the orbit. It is shown at three possible heights above the earth. The angular size of the object as viewed by the satellite is the angle formed at the satellite by the lines of sight to the "left and right" edges of the object (see Figure 7). The angular size (i.e., the angle) increases as the altitude increases since the actual size is assumed to stay constant. The important fact to note is that for any assumed width there is only one height, H, which would give the same angular size as the angular size of the AI on the photograph.## A SINGLE LINE IMAGE MODEL

The third scan line up from the bottom of the AI will be used as a measure of the angular width of the AI. This width of this scan segment is about 0.46 inches on the photograph. Previously this width was calculated to be 0.079 rad (see the bottom of Figure 4). To create (using thought) the image that would result from this simple model UO, assume that the center of the flat object is directly below the satellite (the center is at some distance along the nadir) and observe that the ratio of the width, W, to the distance from the satellite, (A-H) is the angular size in radians as "seen" by the satellite (W << A-H is assumed to be less than 0.4 rad). For example, assume W = 1 nm (not a small size!). If H1 = 1 nm (see Figure 7), then A - H = 460 nm and the ratio is W/(A-H1) = 1/460 = 0.00217 rad. This is much smaller than the angular size of the AI along this scan line (0.079 rad). To increase the angular size move the object up to H2 = 100 nm altitude (not a small altitude!). Now the angle is 1/(461-100) = 11/361 = 0.0077 rad. This angle is bigger, but still not big enough. Let H3 = 361 nm (the altitude of a "low-flying" satellite). The ratio is now 1/(461-361) = 1/100 = 0.01. This is still not big enough. To find the correct value of H we can reverse the procedure and write H = A - (W/B), (6) where B is the angle in radians. With B = 0.079 rad, W = 1 nm (assumed) and A = 461 nm this equation gives H = 448 nm. In other words, if the UO were "only" 1 nm wide and directly below the DMSP satellite it would have to be at an altitude about 13 nm lower than the satellite itself. At this altitude and with this size it could only be some truly unconventional object. The mathematical procedure can be modified once again to provide the object width as a function of the assumed height keeping the angular width constant at 0.079 rad. The equation for this is W = B x (A-H)/cos^2(t), (7) where the cosine factor is introduced to correct for the oblique view (at angle t). (The model UO is supposed to be at the same angle away from the nadir as the real UO that made the AI.) This calculation also assumes that the model UO is oriented horizontally as in Figure 7. The center of the AI along the third scan line from the bottom is at t = 14.4 degrees from the nadir (see the bottom of Figure 4). Therefore, if the flat object were horizontal (the plane of the flat object is perpendicular to vertical) and if the altitude were, say, 460.95 nm, i.e., if it were 0.05 nm or about 291 ft lower than the DMSP satellite, then its width would be W = 0.079 rad x 291 ft/cos^2(14.4)= 24.5 ft. (8) Similarly, if it were at an altitude 50 ft lower than the satellite then it would be about 4 ft wide. The calculations just completed gave the sizes (widths along a scan line) for different altitudes assuming that the UO was essentially a flat plate oriented horizontally. More generally, the UO would be some three dimensional object. The shape of the AI (if there were no motion involved - see below) would be a projection of the outline of the UO onto a plane perpendicular to the line of sight. The 25 ft size just calculated for a 300 ft distance is the size scale of some of the the largest man-made orbiting structures. Could this have been a man-made satellite or "space junk?" If so it would have been traveling at a speed almost identical to that of the DMSP satellite, i.e., Vs = 4.012 nm/sec = 24,381 ft/sec in order to be picked up during the time it took to create 22 scans across the object. At this speed, during the time, 0.007 sec, required for the satellite optics to scan across the object (to make third scan line up from the bottom of the AI) the DMSP satellite would have traveled 0.007 sec x 24,381 ft/sec = l7l ft. This distance traveled along the orbit direction restricts the allowed size and speed of the object. Assume that a model UO was a flat-topped satellite of length Lo in the direction of its motion and a width, W, to be determined. Assume the flat top was horizontal (perpendicular to the radius from the center of the earth) and that it was at nearly the same altitude as the DMSP satellite but traveling in a direction perpendicular to the DMSP satellite orbit (i.e., traveling in an east-west direction). When the leading edge of the object was "met" by the leading edge of the scan (the leading edge of the IFOV) the scan line would brighten and the image on that scan line would appear. The brightness would continue until the trailing edge of the UO "met" the trailing edge of the IFOV, after which the line would darken and the image would "go away." This would create the length of the scan line on the image. In the case of the third scan line up from the bottom this is length is equivalent to 0.079 rad. To be more specific, assume, as an example, that the object was 300 ft away from the DMSP satellite at an angle of 14.4 degrees from the nadir and an altitude of 300 x cos(14.4) = 291 ft lower. During the time (0.007 sec) that IFOV was swept across the UO (by the scanning mechanism at 11.2 rad/sec), the IFOV area (the area viewed at any instant by the satellite sensor system), at that height difference, 291 ft, would have traveled a horizontal distance of about [291 ft/cos^2(14.4)] x 11.2 rad/sec x 0.007 sec = 24.5 ft along the scan direction, thereby making an image width of 0.079 radians. (At other ranges the distance scanned would be different. For example, if the object had been only 50 ft below the distance would have been .007 x 11.2 x 50/cos^2(14.4) = 4 ft.) The assumption that the flat-topped UO satellite was in orbit at essentially the same altitude as the DMSP satellite means that the UO would have been traveling at essentially the same speed, Vo = 24,381 ft/sec (specific calculations showing this are presented below). Hence, during the time that the scan was made, the leading and trailing edges of the object would travel 171 ft. In order to maintain continual "contact" with the IFOV (to make a continuous segment of the scan line) the UO would have to be long enough to cover this distance plus or minus the distance covered by the scan area, 24 ft. Specifically, if the UO were approaching the DMSP orbit and traveling against the scan direction (e.g., scanning toward the east while the UO travels toward the west), then the UO would have to be 171 + 25 = 196 ft long, whereas if the UO were traveling with the scan direction it would have to be 171-25 = 146 ft long. This is an application of equation 4 with Vo being a positive or negative number: (TS)(SR) = (0.007)(11.2) = |[Lo cos^2(14.4)]/291] + (0.007)[(+/-)Vo cos^2(14.4)]/291|, where the cosine factors project the horizontal UO length, Lo, and horizontal velocity, Vo, onto a plane tilted by 14.4 degrees to the horizontal in order to be perpendicular to the line of sight from the DMSP satellite. With Vo = (+/-)24,381 ft/sec the equation can be solved for Lo: Lo = |[(291)(11.2)(.007)/cos^2(14.4)] + (0.007) + (+/-)24381)| (9) = |25 + (+/-171)| = 146 or 196 ft The preceding calculation shows that for a UO satellite in an orbit perpendicular to but within a few hundred feet of the altitude of the DMSP orbit, the length of the UO in the direction of its motion is large. If the UO were moving opposite to the scan, then the minimum length would be about 171 ft if it were immediately below the DMSP satellite (say 3 ft below) and its size would increase with distance below the DMSP. On the other hand, if the UO moved with the scan direction then it would be possible to have the UO far enough away from the DMSP so that its angular rate of motion matched that of the scan, 11.2 rad/sec. At this distance the UO would only have to be large enough in the scan direction to substantially fill the IFOV. Equation 9 can be used to find that distance by setting Lo = 0 and replacing 291 ft with an unknown height difference distance, A-H = r. Then r = 171 cos^2(14.4)/(11.2)(.007) = 2,041 ft. (At this distance the speed of the UO in its orbit would be about 1 ft/sec faster than the DMSP orbit speed. ) To fill the IFOV at this distance the UO would have to have a size of appoximately 0.00533 rad x 2041 = 10.9 ft. At greater distances below the required UO size would increase with distance. In the above discussion the UO velocity component parallel to the motion of the DMSP satellite was assumed to be zero. This has a severe consequence for the required width of the UO (the dimension perpendicular to its own orbit). The consequence is that it must be at least 171 ft wide, and this dimension is roughly independent of the distance of the UO from the DMSP. This is because during the time that the DMSP scans the UO, the DMSP moves 171 ft along in its orbit. Therefore, in order to maintain continual contact with the flat, horizontal surface of the UO, the UO surface must be at least 171 ft wide (it could be wider). Thus, even for the "optimally small" dimension requirement of a UO moving with the scan and 2108 ft below the DMSP, for which the length could be as short as 11 ft (to fill the IFOV in the scan direction), the width would still have to be about 171 ft. The preceding analysis illustrates requirements for UO size based on the assumption of motion parallel or anti-parallel to the scan direction only. Let us now consider the requirements on UO size when the motion of the UO is only parallel or anti-parallel to the DMSP orbital motion. Again this analysis will be based on a single line of the image. If it moved anti-parallel, i.e., toward the DMSP satellite, at the same speed as the DMSP (because it is essentially at the same altitude), then the relative "closing" velocity between the two would be the sum of the velocities, 8 nm/sec or 48,762,ft/sec. Assuming it was at an altitude 291 ft lower and 300 ft away from the satellite (as before), then, in order to have an angular width (dimension measured along the scan direction) of 0.079 rad, it would have to be about 24 ft wide, which is acceptable for "man-made size." However, to also be visible within the IFOV along a scan line for 0.007 sec it would also have to be about 0.007 x 48,762 = 341 ft long! This is not acceptable for "man-made" size. (The tracing made by the scan area on the flat surface of the UO would be a diagonal line from the initial point of contact with the UO to the final point. The diagonal would be at an angle to the long edge of the large rectangular UO with the angle being arctan(24.5/341) = 4 degrees.) The object sizes calculated so far, based on the assumption that the UO was a satellite in an orbit just below that of the DMSP satellite, are much too large in at least one dimension (length or width) to be consistent with man-made satellites. However, this size problem can be "solved" by assuming that the UO satellite had a velocity component parallel to that of the DMSP satellite as well as a component perpendicular. In this case the relative velocity between the two could be low (but not zero). Since the UO was below the DMSP one might assume that the UO was traveling along in its orbit and passed nearly underneath the DMSP satellite. The assumption that the UO was in an orbit at a lower altitude has a consequence: to be at a lower altitude it would have to be traveling faster than the DMSP satellite. This is because orbital speed decreases with increasing altitude according to the equation V = K/R^0.5, where R is the altitude above the center of the earth, R = Re + A = 3,445 + 461 = 3,906 nm = 2.374E7 ft or 7,234 km. The value of K depends upon the units chosen for R. With r in meters, K = 2E7 m^1.5/sec in mks units; with r in feet, K = 1.18E8 ft^1.5/sec. One can show that for small altitude changes, dR, the derivative of the above equation yields dV = [(- 1/2)K/R^(1.5)]dR. For altitudes in feet starting at A = 461 nm altitude or 2.374E7 ft, [(-1/2) K/R^1.5] = [-5.1E-4]/sec so Vat R - Vdmsp = dV = - 5.1E-4 (Rdmsp - R), R

Suppose that the angular size of the sphere is 0.00533 rad (e.g., 1.6 ft in diameter at 300 ft distance) and that it travels at an angle to the DMSP orbit such that it continually fills the moving IFOV along 0.46 inches (0.079 rad) of a single scan line at an angular distance 14.4 degrees from the nadir. As pointed previously, the moving IFOV travels perpendicular to the DMSP orbit plane at an angular rate of 11.2 rad/sec. Therefore at 14.4 degrees from the nadir and 300 ft away the velocity of the scan area, as projected onto a horizontal flat surface 300 cos(14.4) = 291 ft lower than the DMSP, is about (11.2)[291/cos^2(14.4)] = 3469 ft/sec = 0.571 nm/sec. In order to remain within the IFOV the sphere must have a transverse velocity component equal to this: Vt = 0.57 nm/sec. Further assume that Vo = Vs = 4.012 nm/sec so that the UO sphere travels in its orbit at the same speed as the DMSP satellite (if in an orbit 291 ft lower, the sphere would travel only 0.15 ft/sec faster than the DMSP satellite, a negligible difference compared to the 24,381 ft/sec orbit speed). Then Vp = Vo cos(d) = Vs. The tangent of angle d is the is the ratio of the components: tan(d) = (Vt/Vp) which is the same as tan (d) = [Vo sin(d)]/[Vo cos(d)]. In this case we have Vo sin(d) = 0.571 nm/sec and Vo cos(d) = 4.012 nm/sec which leads to the equation tan (d) = 0.571/4.012, which has the solution, d = 8.098 degrees. With this angle we can find the required Vo: Vo = 4.012/cos(8.098) = 4.052 nm/sec or 24,626 ft/sec. The fact that a small object must travel at a higher orbital speed than the DMSP in order to satisfy the requirements is not surprising. However, the amount of increased speed is surprising and it means that the distance below the DMSP would have to be very large. Specifically, to have a speed difference of 24,626 - 24,381 = 245 ft/sec, using Equation 10 above, relating orbital speed difference to altitude difference, 245 ft/sec = 5.1E- 4 dR, where dR is the altitude difference, dR would have to be about 480,400 ft = 79 nm. Aside from the fact that this distance below would require the UO, in order to fill the IFOV, to be 0.00533 x 480,400 = 2,560 ft in size, which is way beyond reason for man-made objects, there is an inconsistency in the angular rate of the scan. It was required that the transverse component of velocity of the UO match the angular scan rate. This should happen regardless of the altitude of the UO if the UO is assumed to remain within the IFOV as it moves relative to the DMSP (crossing the track of the DMSP). The angular scan rate of an object traveling 24,626 ft/sec at a distance of 2,560 ft is 2560/24626 = 0.104 rad/sec which is two orders of magnitude less than the 11.2 rad/sec rate of the scan. Consider this situation another way. Imagine a horizontal surface at altitude H, which is distance Ro = A-H lower than the DMSP at altitude A. Another satellite's orbit would, for practcal purposes, be a straight line within this imaginary flat surface (the curvature of the orbit would be small over the lateral distances used here and, in any case, would project onto the plane as a straight line). The IFOV scans perpendicular to the DMSP orbit at a rate of 11.2 rad/sec. This corresponds to a horizontal, transverse component of scan velocity, Vts, projected onto that surface which is given by Vts = 11.2 r/cos(t) ft/sec, where r = Ro/cos(t), and t is the nadir angle, as before. The parallel component of the scan velocity is Vps = 4.012 nm/sec = 24,381 ft/sec. Thus on the imaginary flat surface at altitude (461 - H) in nm one can construct scan vector components Vts and Vps which make a net scan vector with a magnitude Vo = (Vts^2 + Vps^2)^0.5 and an angle relative to the DMSP orbit direction given by tan(ds) = Vts/Vps. For the previously assumed case of t = 14.4 deg and r = 300 ft, Ro = 291 ft and Vts = 3469 ft/sec the scan angle would be ds = arctan(3469/24381) = 8.098 deg as calculated before. Any object moving along this direction relative to the DMSP orbit and having velocity Vo, once it appears in a scan line, should stay within the single scan IFOV, at least for a short distance. More generally, ds = arctan[11.2Ro/24381cos^2(t)]. Note that the angle isn't constant; it increases as t increases. At Ro = 291 ft and t = 0 degrees, at the nadir, the angle is arctan(0.1336) = 7.6 deg, and at the maximum scan angle, ds = arctan(0.406) = 22.1 deg. Hence the projection of the scan line on the imaginary plane is a curve starting with a "slope" of 7.6 deg at the nadir, a slope 8.098 deg at t = 14.4 deg, 15 deg at t = 45 deg and finally reaching 22.1 deg at the end of the scan. Because of the curvature no small "point" satellite could stay within the scan for more than a short angular distance. The required Vo is given by Vo = (Vts^2 + 24,381^2)^0.5. When t = 14.4, Vts = 3469 ft/sec and Vo = 24,626 ft/sec, as calculated before. This speed is so large it requires that the imaginary horizontal surface, in which the UO satellite makes a straight line track, to be about 480,400 ft below the DMSP, and that, in turn, makes the UO satellite large (2,500 ft) in order to fill the IFOV. The bottom line is that a "point" satellite could not make a short line segment image such as the 0.46 inch scan line considered here. If it passed through the scan it would make only a "dot." A satellite of some size is necessary to make a sizeable image.## DISCUSSION OF MODEL OBJECTS WHICH CAN CREATE IMAGES IN SEVERAL CONSECUTIVE SCAN LINES

The image of interest appears not in a single scan line but in 22 consecutive lines. This means it "hung around" for a while. The following discussion illustrates the ranges of size, shape and velocity for an object to make an image that could appear in that number of scan lines. An object of the minimum size to make a single scan line, i.e., an object of 0.00533 rad angular size (e.g., 1.6 ft at 300 ft or 3.2 ft at 600 ft or 0.5 ft at 100 ft, etc.) cannot make two or more scan lines. This is because in order to make a single scan line by remaining within the IFOV long enough to make a reasonably long image on one scan lime it must be traveling at an angle to the orbit and therefore must have a non-zero transverse velocity. It could appear in one scan line, but by the time the scanning svstem went through its complete revolution, requiring 0.56 sec, the object would be so far from the spacecraft that it would not be within the field of view of the optical system. To see that this is true, consider the 0.53 ft sized object 100 ft away as an example. To be at an angle of 14.4 deg from the nadir it would have be at an altitude 100 cos 14.4 = 96.9 or about 97 ft lower than the satellite and 100 sin 14.4 = 24.9 or about 25 ft away from the orbit plane measured horizontally. (For the 1.6 ft sized object at 300 ft and 14.4 deg, considered previously, the distance down would be 291 ft and the distance from the orbit plane would be 300 sin(14.4) = 74.6 ft.) To remain at the same altitude and also within the IFOV long enough to make an image 0.079 rad wide along on a scan line it would have to travel at a transverse angular velocity of (at least) 11.2 rad/sec, corresponding to a transverse velocity of Vt = 100 x ll.2/cos 14.4 ft/sec = 1156 ft/sec. (For the 1.6 ft object this was 3469 ft/sec.) At this speed, by the time the satellite scanner completed a revolution the object would have moved 0.56 sec x ll56 ft/sec = 648 ft perpendicular to the orbit plane. (For the 1.6 ft object this would be 1940 ft.) If it were a man-made object in another orbit it would remain at an altitude 97 ft (291 ft) lower over this short time interval. However, the horizontal distance would now be 25 ft + 648 ft = 673 ft, if the object moved to the right, and the next scan line image would appear at an angle of arctan([648+25]/97) = 82 deg to the right of the center. If the object were imagined to be traveling to the left, the next scan line image would appear at arctan[(648-25)/97]= 81 deg to the left of center. (For the 1.6 ft object these angles are arctan([1940 - 79]/291) = 81 deg and 82 deg). The maximum scan angle left or right of the DMSP satellite is only about 56 deg (see Figure 4) so the object would be lost on the next scan. Since the smallest possible size which will make an image on a single scan line will not make an image on several scan lines, it is necessary to consider larger objects. The next larger size to consider is a rectangular object 0.079 rad in angular width by 0.00533 rad in angular length (e.g., at 300 ft distance, 24 ft by 1.6 ft). If an object such as this were to travel at exactly the same speed as the satellite it would appear in consecutive scans. If Vp = Vs and Vt = 0 the object would make a series of scan line images such as are illustrated in Figure 9, except the series would be continuous from the top of the photograph to the bottom and not finite in length as illustrated (i.e., the image would consist of many more than 22 scan lines). If the UO were traveling in an orbit parallel to but lower than, the DMSP satellite orbit it would be going slightly faster and could "catch up" with the DMSP and pass underneath it. While it was passing underneath, the DMSP would scan it a number of times. In this case a finite series of scan line images would be produced. The variables in this case are the object length, L (measured along the orbit direction), and the parallel component of velocity, Vp. To understand what is happening one should imagine the satellite and the object moving in the same direction with the object moving slightly faster (and at a lower attitude). The UO satellite would slowly catch up with DMSP satellite and eventually its leading edge, 0.079 rad wide and 14.4 degrees from directly below the DMSP, would be scanned. The first scan line would "paint" the object and create an image 0.46 inches wide. During the 0.56 sec needed for the scanner to revolve and begin another scan line across the object, the UO satellite would move forward relative to the DMSP satellite by an amount 0.56 sec(Vp - Vs). Thus the next scan line would be this distance from the first scan line, measured toward the rear end of the UO satellite. With each revolution of the scanner (i.e., with each scan line on the object) the scan would again move this distance along the object. Finally, after a number of scans the rear end of the UO would move ahead of the DMSP and the UO would not be detected again. To get N scans on the UO would require L = (N-1)(0.56)(Vp - Vs), where N-1 is the number of "spaces" between the scan lines. (It is the "spaces" which add together to make the length of the UO.) L would be at least this long but no longer than N(0.56)(Vp - Vs). In this equation neither L nor Vp are known. However, we know from the previous calculations that Vp is related to the distance below the DMSP satellite: the greater the distance down, the greater then speed. As an example, assume Vp - Vs = 1 ft/sec, corresponding to the UO satellite being at Ro = (1 ft/sec)/(5.1E-4/sec) = 1960 ft below the DMSP, and use N = 22. The L = (22-1) (0.56) (1 ft/sec) = 11.8 ft. Thus an 11.8 ft long UO satellite traveling parallel to the DMSP at a speed 1 ft/sec faster would create an image in 22 scan lines. However, because it would be 1960 ft below the DMSP, the width would have to be considerable: W = 0.079 rad x 1960 = 155 ft. If Vp - Vs = 0.1 ft sec, L = 1.18 ft, the distance down would be 196 ft and the width would be 15.5 ft. This would, of course, require a remarkable coincidence between satellite locations and hence between satellite orbits. For even lower speed differentials the calculated length of the UO would be unacceptably small (a fraction of a foot). A speed differential of 0.2 ft/sec provides a good compromise: L = 2.3 ft, Ro = 392 ft, W = 31 ft. If, in addition to having Vp nearly equal to Vs so that 22 scans appear on the object, it is also true that Vt = 0, then the object will appear at the same distance from the center of the photo in successive scans and the outline of the image will be rectangular, as illustrated in Figure 9. Note that to make this image the object was assumed to be rectangular with an anqular width of 0.079 radians an a length of (N-1)(0.56)(Vp - Vs) = L ft where Vp - Vs could be any number that qives a reasonable lenqth. It would be impossible to determine the actual size of a rectangular object from its image without knowing either the distance (down) or the velocity component, Vp, (or the speed differential) to within a fraction of a ft/sec. Because the length cannot be determined the actual object shape cannot be determined. However, the ratio between the length and width is known because, surprisingly, the speed vs altitude equation establishes a connection between the speed differential and the distance down. Specifically, we have Vp-Vs = (5.1E-4)Ro, L = 11.8(Vp-Vs) and W = 0.079(Ro), where (22-1)(0.56) = 11.8. Therefore L = 11.8(5.1E-4)Ro = 0.006(Ro) and so L/W = 0.006/0.079 = 0.076 or the width is about 13 times the length. The previous discussion considered a rectangular UO satellite making a rectangular AI. However, the actual AI is not rectangular. It has a "sloping" right hand edge. This, of course, could be an actual feature of the outline of the object. However, it could also be an artifact of the scanning system, as the following analysis shows. Assume that a rectangular object with an angular width of 0.079 rad moved at an angle to the DMSP orbit while keeping its "long" sides parallel and the leading edge perpendicular to the orbit. Its transverse component would cause the location of the image to shift "sideways" from scan to scan. Suppose that as the UO satellite "catches up" with the DMSP it is slightly to the right (and below) as illustrated in Figure 10.

The DMSP satellite scans the leading edge making an image 0.079 rad in angular width (scan #1- in Figure 10A). By the time of the next scan the satellite has moved 0.56 x (Vp - Vs) feet beyond the leading edge and at the same time the left and right edges have moved a distance 0.56 x Vt to the left and thus the second scan line image appears shifted sideways from the first (closer to the center of the photograph in Figure 10B, for example). The amount of shift is an angle in radians equal to the shift distance divided by the distance to the object. The same angle appears on the photograph as the shift (in radians) of the left and right edges of the image in one scan line relative to the next scan line. In the AI the right hand edge of the image shifts 0.022 inches from one scan line to the next (corresponding to 0.286 inches total shift over 13 scan lines). Using the method presented before for finding the angular size of an image an the photo, the shift in the image edge position corresponds to an angle shift D = (0.183) x Dy x cos^2(18) = (0.183 x 0.022 x 0.905) = 0.0036 rad, where the angle used, 18 deg, is at the (vertical) center of the right edge of the image. This amount of shift in angular position took place in 0.56 sec, so the angular transverse velocity that would produce the (nearly) 45 degree sloping right edge of the image is 0.0064 rad/sec. At a distance of 300 ft, for example, this corresponds to a transverse velocity of 0.0064 x 300/cos(18) = 2 ft/sec. If the rectangular object just discussed moved at a velocity such that Vp exactly equalled Vs and Vt = 2 ft/sec it would make a series of displaced images on the scan lines but the images would "never end." That is, the object would make images on scan lines from the upper right edge of the photo to the lower left because its parallel component is the same as that of the satellite. In order to make the image a finite size one must, as once before, assume that Vp is slightly greater than Vs and that the UO satellite catches up with the DMSP and passes under it in 11.77 sec. If Vp - Vs = 1 ft/sec the object would be 11.8 ft long as well as 23.7 ft wide (at a distance of 300 ft). Consider the photographic image that would be made by a model rectangular UO under these conditions (see Figure 10B). The impression one would gain from seeing such an image in a DMSP photograph is that the object itself was diamond shaped. However, using the above analysis one can show that a rectangular object 24 ft wide and 12 ft long, oriented with its "long" sides parallel to the DMSP orbit, moving with a parallel velocity component that is abou 1 ft/sec less than the satellite velocity and a transverse velocity component of 2 ft/sec and traveling 291 ft below the DMSP orbit at a radial distance of 300 ft from the satellite can make an image such as shown in Figure 11. The severe distortion present in the image of the rectangular object is an artifact of the scanning combined with the object motion relative to the satellite. The lesson to be learned is that the image in a photograph made with the slowly scanning system is not necessarily an accurate representation of the shape of the object. Incidently, if the object were rotating, which is the likely condition if the UO were space junk, then all bets are off as to the actual shape (see below).