Subject: Time: 18:47 OFFICE MEMO Tether brightness Date: 96/02/26 I may not have all the numbers right, but here's a first cut which others can modify/correct as needed. I'll use Bjorn Gimle's figure of 3 arcminutes for human visual resolution (though it seems to me a 9mm diameter, dark-adapted pupil should do much better than this), assume a range of 400 km, a tether surface reflectivity of 0.8, and inband solar illumination of 177 W/m^2. 3 arcminutes is 873 microradians, which at a range of 400 km translates to 350 meters. Thus, the 0.1" wide tether (2.54 mm) produces an equivalent visual response to an area of 0.89 square-meters (350m x 2.54 mm). .89 m^2 x 0.8 x 177 W/m^2 = 126 W. Assuming the tether is approximately Lambertian, this 126 W will be reflected into a hemisphere: Intensity = 126 W / (2 pi steradian) = 20 W/sr. At a range of 400 km, the resulting irradiance is 20 W/sr divided by the range squared, which gives 1.25 x 10^-10 W/m^2. Assuming the star Vega is magnitude 0, and Vega's power in the visual band is 3.34 x 10^-9 W/m^2, the tether is 26.7 times dimmer than Vega. This would give the tether a visual magnitude of about 3.6: 2.5 x LOG10(26.7) = 3.57. Of course, if you go back and change the visual resolution to 1 arcminute, the tether would be three times dimmer, or magnitude 4.8. Of course, this calculation doesn't take atmospheric absorption into consideration; nevertheless, the tether should be quite visible. My calculations show southern California will get a chance to see it the morning of March 6th, assuming it stays up that long. It's a low pass in the south, passing under Sagitarius...