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ANALYSIS OF FLIR 1 Video Dec 21, 2017 Graph of angular displacement relative to the center of the FOV. In the graph below the “dots” represent distances from the center of the field of view of the UFO as a function of frame number . Each frame represents a time span of 1/30 sec, so the graph shows the distance from the center as a function of time. On such a graph the slope of a line drawn between two points is the average velocity between the points: V = d/t where d is the distance between points and t is the time between the distance points. The actual distances are not known but the FLIR data do show angles so what is actually shown (raw data) are angles relative to the center of the screen of the FLIR. I assume the angular field of view is 0.75 deg so half the FOV (center to left edge) is 0.375 deg. On the vertical scale at the left 100 corresponds to 0.375 deg at the left side of the video image. The UFO image on the FLIR screen was initially “bouncing around” the center of the field of view (FOV) and then it moved for a short time, 9 frames or about 9/30 = 0.3 sec, at a constant velocity (constant slope) designated as V1. (Note: the solid black lines in the graph are extensions of the three main slopes of the data; they make it easy to estimate the values of the slopes or velocities.) Then the object “jumped” a sizeable angular distance during one frame time, a jump to the left. During that one frame time I assume the average velocity was V2. Then it abruptly slowed to a lower more or less constant velocity, V3, and maintained this velocity as it traveled to the left and reached the left edge of the screen. Of particular interest are the changes in velocity which indicate acceleration. The graph indicates a sudden acceleration from V1 to V2 which is approximated as (3.6 – 0.11)/(1/30) = 105 deg/sec^2. This sudden increase in velocity can be related to the actual acceleration if the angles are changed to radians and then multiplied by the known (or assumed) distance. Use 1 deg = 0.0174 radians so the acceleration would be 1.8 rad/sec^2 and, if the object was 5,000 ft away at that time then the leftward acceleration would have been about 9,000 ft/sec^2. This can be compared with gravitational acceleration, “g”, at the earth’s surface, about 32 ft/sec^2. It calculates to about 280 “g”. Then the object abruptly slowed to V3 = 0.34 deg/sec so the deceleration was (V2 – V3)/(1/30) = 98 deg/sec^2 or 1.7 rad/sec^2. If at 5,000 ft the deceleration was 8,500 ft/sec^2 or 265 g. A change in acceleration is called “jerk” and that is certainly what would have been felt by the pilot of the UFO!