Mystery High Altitude Cloud, May 13, 1999

Steven Dutch, Natural and Applied Sciences, University of Wisconsin - Green Bay
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The cloud pictured here (photos taken from Green Bay, Wisconsin looking northwest) was still sunlit long after cirrus clouds (5-6 km altitude) had gone dark. The cloud was about 5 degrees above the horizon. Although the last frames are quite dark, remnants of the cloud were visible and brighter than the sky even then. Thus the cloud must have been at a very high altitude.

The sun set at about 8:15 CDT in Green Bay that evening and was about 3.5 degrees below the horizon by the time the cloud vanished around 8:50 P.M. Call the location of the cloud C, the center of the Earth X and let H be the location of the horizon as seen from C (the farthest point on the ground just visible from the cloud). If the sun was just grazing the horizon as seen from the cloud, then angle CHX = 90 degrees and HCX = 90-3.5 = 86.5 degrees. Thus angle HXC (the angle at the center of the Earth) is 3.5 degrees. Put another way, an observer 3.5 degrees (almost 400 km) away would just see the cloud on the horizon. Distance XH = 6400 km, the radius of the Earth. Distance XC = 6400 + h (height of the cloud) = 6400/cos(3.5). Solving for h, we find that the cloud was at an altitude of at least 12 km. If we allow another degree for atmospheric refraction, the elevation was almost 20 km.

Possibilities include:

8:33 P.M. CDT

8:33 P.M. CDT

8:34 P.M. CDT

8:37 P.M. CDT

8:38 P.M. CDT

8:41 P.M. CDT

8:43 P.M. CDT

8:44 P.M. CDT

8:45 P.M. CDT

8:47 P.M. CDT

8:48 P.M. CDT

8:49 P.M. CDT

8:50 P.M. CDT

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Created 19 May 1998, Last Update 19 May 1998

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