| Tony Flanders |
| (Post Laureate) |
| 07/07/09 01:58 PM |
Re: lowest declination of a sky object at my latitude
|
Quote:
Can anyone offer a rule of thumb for the magnitude loss close to the horizon? (i.e. something more rough-and-ready than the extinction formula, which isn't much fun to work out in your head on a dark night).
There's an Excel spreadsheet for computing this in the online article Transparency and Atmospheric Extinction by me and Phil Creed.
According to the spreadsheet, at sea level in absolutely pristine air, extinction is 1.7 magnitudes 5 degrees above the horizon, 0.9 magnitudes at 10 degrees, and 0.6 magnitudes at 15 degrees -- all these at the peak wavelength for human night vision.
An aerosol optical depth of 0.10 -- average for the desert Southwest in the summer, and better than average in most parts of the U.S., raises the extinction to 2.9 at 5 degrees, 1.6 at 10 degrees, and 1.1 at 15 degrees.
Even at high altitudes in pristine air, there's an extremely heavy penalty for observing objects less than 15 degrees above the horizon.