How fast would the sweeps drift south in this scenario? Assuming a satellite elevation angle of 30 degrees from the ground (to simplify the math), and a range of 39,000 km, 2.3 deg/hour is the same as 0.038 deg/minute, and .038 deg at that range corresponds to a ground footprint of 52 km. So Robert Sheaffer in northern CA and Ron Lee in CO should see their peak flashes several minutes before I do in southern CA. This is testable. But even more importantly, if this description is accurate, then flashes will again be visible on the ascending node!! Superbird A is back at 1.7 degrees south latitude at around 5:44:30am PDT; fortunately, it is still dark in both CA locations at this time, though Colorado may be out of luck. Certainly worth a look-see. Furthermore, the flashes should be 3 times brighter due to the vastly improved geometry. Beta in the evening is 67 degrees for SoCal; beta in the morning is 151. Brightness is proportional to the square of SIN(Beta/2). (Beta being the angle from the sun to the earth to the satellite.) Of course, all this assumes the fast sweeps are roughly east-west oriented. If instead they are north-south, then it is the sun's relative motion and not the satellite's that dictates the duration of the flashes. But I think I can prove that isn't the case. The sun is moving roughly 15 degrees per hour relative to the earth-satellite line. That means the specular reflection moves a half-degree per minute. So, unless the specular image is considerably larger than a half-degree wide, flashes should could not possibly last 10 minutes. This is good news, because it means there's a good chance that flashes are actually visible on the ascending node... --Rob